Q
stringlengths 26
3.6k
| A
stringlengths 1
9.94k
| Result
stringclasses 3
values |
|---|---|---|
If \( f \) is as in the previous exercise, show that \( \left| {f\left( z\right) }\right| < M \) for all interior points \( z \in R \) , unless \( f \) is constant. | Null | No |
Show that the quotient group \( {H}_{0} \) of all nonzero fractional ideals modulo \( {P}_{0} \) is a subgroup of the ideal class group \( H \) of \( K \) and \( \left\lbrack {H : {H}_{0}}\right\rbrack \leq 2 \). | Null | No |
Proposition 9.34 Define a domain \( \operatorname{Dom}\left( \Delta \right) \) as follows:\n\n\( \operatorname{Dom}\left( \Delta \right) = \left\{ {\psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \left| {\;{\left| \mathbf{k}\right| }^{2}\widehat{\psi }\left( \mathbf{k}\right) \in {L}^{2}\left( {\mathbb{R}}^{n}\right) }\right. }\right\} . \) | The proof of Proposition 9.34 is extremely similar to that of Proposition 9.32 and is omitted. | No |
Consider the element \[
\alpha = {\left( x + y\right) }^{\ell - 2}\left( {x + {\zeta y}}\right)
\] Show that: (a) the ideal \( \left( \alpha \right) \) is a perfect \( \ell \) th power. | Null | No |
Proposition 3.2. In a chart \( U \times \mathbf{E} \) for \( {TX} \), let \( f : U \times \mathbf{E} \rightarrow \mathbf{E} \times \mathbf{E} \) represent \( F \), with \( f = \left( {{f}_{1},{f}_{2}}\right) \) . Then \( f \) represents a spray if and only if, for all \( s \in \mathbf{R} \) we have | The proof follows at once from the definitions and the formula giving the chart representation of \( s{\left( {s}_{TX}\right) }_{ * } \) . | No |
If \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| = 1 \), a.e., then \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = {\phi }_{2}{\widetilde{\mathbf{H}}}^{2} \) if and only if there is a constant \( c \) of modulus 1 such that \( {\phi }_{1} = c{\phi }_{2} \) . | Clearly \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = c{\phi }_{1}{\widetilde{\mathbf{H}}}^{2} \) when \( \left| c\right| = 1 \) . Conversely, suppose that \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = {\phi }_{2}{\widetilde{\mathbf{H}}}^{2} \) with \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| = 1 \), a.e. Then there exist functions \( {f}_{1} \) and \( {f}_{2} \) in \( {\widetilde{\mathbf{H}}}^{2} \) such that\n\[
{\phi }_{1} = {\phi }_{2}{f}_{2}\;\text{ and }\;{\phi }_{2} = {\phi }_{1}{f}_{1}
\]\nSince \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = 1 = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| \) a.e., it follows that\n\[
{\phi }_{1}\overline{{\phi }_{2}} = {f}_{2}\;\text{ and }\;{\phi }_{2}\overline{{\phi }_{1}} = {f}_{1}
\]\nthat is, \( {f}_{1} = \overline{{f}_{2}} \) . But since \( {f}_{1} \) and \( {f}_{2} \) are in \( {\widetilde{\mathbf{H}}}^{2},{f}_{1} = \overline{{f}_{2}} \) implies that \( {f}_{1} \) has Fourier coefficients equal to 0 for all positive and for all negative indices. Since the only nonzero coefficient is in the zeroth place, \( {f}_{1} \) and \( {f}_{2} \) are constants, obviously having moduli equal to 1 .\nSince the unilateral shift is a restriction of the bilateral shift to an invariant subspace, invariant subspaces of the unilateral shift are determined by Theorem 2.2.7: they are the invariant subspaces of the bilateral shift that are contained in \( {\widetilde{\mathbf{H}}}^{2} \) . In this case, the functions generating the invariant subspaces are certain analytic functions whose structure is important. | No |
Theorem 3.1. (Eisenstein’s Criterion). Let \( A \) be a factorial ring. Let \( K \) be its quotient field. Let \( f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0} \) be a polynomial of degree \( n \geqq 1 \) in \( A\left\lbrack X\right\rbrack \) . Let \( p \) be a prime of \( A \), and assume:\n\n{a}_{n} ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) ,\;{a}_{i} \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \;\text{ for all }\;i < n,\n\n{a}_{0} ≢ 0\;\left( {\;\operatorname{mod}\;{p}^{2}}\right)\nThen \( f\left( X\right) \) is irreducible in \( K\left\lbrack X\right\rbrack \) . | Proof. Extracting a g.c.d. for the coefficients of \( f \), we may assume without loss of generality that the content of \( f \) is 1 . If there exists a factorization into factors of degree \( \geqq 1 \) in \( K\left\lbrack X\right\rbrack \), then by the corollary of Gauss’ lemma there exists a factorization in \( A\left\lbrack X\right\rbrack \), say \( f\left( X\right) = g\left( X\right) h\left( X\right) \) ,\n\n{g}\left( X\right) = {b}_{d}{X}^{d} + \cdots + {b}_{0}\n\n{h}\left( X\right) = {c}_{m}{X}^{m} + \cdots + {c}_{0}\nwith \( d, m \geqq 1 \) and \( {b}_{d}{c}_{m} \neq 0 \) . Since \( {b}_{0}{c}_{0} = {a}_{0} \) is divisible by \( p \) but not \( {p}^{2} \), it follows that one of \( {b}_{0},{c}_{0} \) is not divisible by \( p \), say \( {b}_{0} \) . Then \( p \mid {c}_{0} \) . Since \( {c}_{m}{b}_{d} = {a}_{n} \) is not divisible by \( p \), it follows that \( p \) does not divide \( {c}_{m} \) . Let \( {c}_{r} \) be the coefficient of \( h \) furthest to the right such that \( {c}_{r} ≢ 0\left( {\;\operatorname{mod}\;p}\right) \) . Then\n\n{a}_{r} = {b}_{0}{c}_{r} + {b}_{1}{c}_{r - 1} + \cdots .\nSince \( p \nmid {b}_{0}{c}_{r} \) but \( p \) divides every other term in this sum, we conclude that \( p \nmid {a}_{r} \), a contradiction which proves our theorem. | Yes |
Proposition 5.46. Suppose \( M \) is a smooth manifold without boundary and \( D \subseteq M \) is a regular domain. The topological interior and boundary of \( D \) are equal to its manifold interior and boundary, respectively. | Suppose \( p \in D \) is arbitrary. If \( p \) is in the manifold boundary of \( D \), Theorem 4.15 shows that there exist a smooth boundary chart \( \left( {U,\varphi }\right) \) for \( D \) centered at \( p \) and a smooth chart \( \left( {V,\psi }\right) \) for \( M \) centered at \( p \) in which \( F \) has the coordinate representation \( F\left( {{x}^{1},\ldots ,{x}^{n}}\right) = \left( {{x}^{1},\ldots ,{x}^{n}}\right) \), where \( n = \dim M = \dim D \) . Since \( D \) has the subspace topology, \( U = D \cap W \) for some open subset \( W \subseteq M \), so \( {V}_{0} = V \cap W \) is a neighborhood of \( p \) in \( M \) such that \( {V}_{0} \cap D \) consists of all the points in \( {V}_{0} \) whose \( {x}^{m} \) coordinate is nonnegative. Thus every neighborhood of \( p \) intersects both \( D \) and \( M \smallsetminus D \), so \( p \) is in the topological boundary of \( D \) .
On the other hand, suppose \( p \) is in the manifold interior of \( D \) . The manifold interior is a smooth embedded codimension- 0 submanifold without boundary in \( M \) , so it is an open subset by Proposition 5.1. Thus \( p \) is in the topological interior of \( D \) .
Conversely, if \( p \) is in the topological interior of \( D \), then it is not in the topological boundary, so the preceding argument shows that it is not in the manifold boundary and hence must be in the manifold interior. Similarly, if \( p \) is in the topological boundary, it is also in the manifold boundary. | Yes |
Let \( f\left( z\right) \in {H}^{1} \). Then the Fourier transform \(\widehat{f}\left( s\right) = {\int }_{-\infty }^{\infty }f\left( t\right) {e}^{-{2\pi ist}}{dt} = 0\) for all \( s \leq 0 \). | By the continuity of \( f \rightarrow \widehat{f} \), we may suppose \( \int \in {\mathfrak{A}}_{N} \). Then for \( s \leq 0, F\left( z\right) = f\left( z\right) {e}^{-{2\pi isz}} \) is also in \( {\mathfrak{A}}_{N} \). The result now follows from Cauchy’s theorem because \({\int }_{0}^{\pi }\left| {F\left( {R{e}^{i\theta }}\right) }\right| {Rd\theta } \rightarrow 0\;\left( {R \rightarrow \infty }\right) .\) | No |
Let \( u = {\delta }_{{x}_{0}} \) and \( f \in \mathcal{S} \) . Then \( f * {\delta }_{{x}_{0}} \) is the function \( x \mapsto f\left( {x - {x}_{0}}\right) \) | for when \( h \in \mathcal{S} \), we have
\[
\left\langle {f * {\delta }_{{x}_{0}}, h}\right\rangle = \left\langle {{\delta }_{{x}_{0}},\widetilde{f} * h}\right\rangle = \left( {\widetilde{f} * h}\right) \left( {x}_{0}\right) = {\int }_{{\mathbf{R}}^{n}}f\left( {x - {x}_{0}}\right) h\left( x\right) {dx}.
\] | No |
Show that every element of R can be written as a product of irreducible elements. | Suppose b is an element of R . We proceed by induction on the norm of b . If b is irreducible, then we have nothing to prove, so assume that b is an element of R which is not irreducible. Then we can write b = ac where neither a nor c is a unit. By condition (i), n(b) = n(ac) = n(a)n(c) and since a, c are not units, then by condition (ii), n(a) < n(b) and n(c) < n(b). If a, c are irreducible, then we are finished. If not, their norms are smaller than the norm of b, and so by induction we can write them as products of irreducibles, thus finding an irreducible decomposition of b . | Yes |
Theorem 11.5 For each \( h > 0 \) the difference equations (11.10)-(11.11) have a unique solution. | The tridiagonal matrix \( A \) is irreducible and weakly row-diagonally dominant. Hence, by Theorem 4.7, the matrix \( A \) is invertible, and the Jacobi iterations converge. | Yes |
Corollary 3.4.6. Let \( 0 < {p}_{0} < \infty \) . Then for any \( p \) with \( {p}_{0} \leq p < \infty \) and for all locally integrable functions \( f \) on \( {\mathbf{R}}^{n} \) with \( {M}_{d}\left( f\right) \in {L}^{{p}_{0}}\left( {\mathbf{R}}^{n}\right) \) we have | Since for every point in \( {\mathbf{R}}^{n} \) there is a sequence of dyadic cubes shrinking to it, the Lebesgue differentiation theorem yields that for almost every point \( x \) in \( {\mathbf{R}}^{n} \) the averages of the locally integrable function \( f \) over the dyadic cubes containing \( x \) converge to \( f\left( x\right) \) . Consequently,\n\[
\left| f\right| \leq {M}_{d}\left( f\right) \;\text{ a.e. }\n\]Using this fact, the proof of (3.4.11) is immediate, since\n\[
\parallel f{\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \leq {\begin{Vmatrix}{M}_{d}\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) },\n\]and by Theorem 3.4.5 the latter is controlled by \( {C}_{n}\left( p\right) {\begin{Vmatrix}{M}^{\# }\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \) . | Yes |
The set of lines spanned by the vectors of \( {D}_{n} \) is star-closed. | Lemma 12.2.2 The set of lines spanned by the vectors of \( {D}_{n} \) is star-closed. | No |
If \( \left( {{g}_{1},{S}_{1}}\right) \sim \left( {{g}_{2},{S}_{2}}\right) \) then for all \( i\left( {{g}_{1},{e}_{i}\left( {S}_{1}\right) }\right) \sim \left( {{g}_{2},{e}_{i}\left( {S}_{2}\right) }\right) \) . | There exist \( e \) and \( T \) such that \( \left( {e\left( {S}_{i}\right), T}\right) \) is a balanced pair representing \( {g}_{i} \) for \( i = 1,2 \) . We write \( \left( {{g}_{1},{S}_{1}}\right) \underset{k}{ \sim }\left( {{g}_{2},{S}_{2}}\right) \) if the length of this \( e \) is \( \leq k \) . The lemma is proved by induction on \( k \) using 9.3.3. | No |
Every module can be embedded into an injective module. | By 4.11 there is a monomorphism of \( R/{L}_{n} \) into an injective \( R \) -module \( {J}_{n} \). By the hypothesis, \( J = {\bigoplus }_{n > 0}{J}_{n} \) is injective. Construct a module homomorphism \( \varphi : L \rightarrow J \) as follows. Let \( {\varphi }_{n} \) be the homomorphism \( R \rightarrow R/{L}_{n} \rightarrow {J}_{n} \). If \( x \in L \), then \( x \in {L}_{n} \) for some \( n \), and then \( {\varphi }_{k}\left( x\right) = 0 \) for all \( k \geqq n \). Let \( \varphi \left( x\right) = {\left( {\varphi }_{k}\left( x\right) \right) }_{k > 0} \in J \). Since \( J \) is injective, \( \varphi \) extends to a module homomorphism \( \psi : {}_{R}R \rightarrow J \). Then \( \psi \left( 1\right) = {\left( {t}_{k}\right) }_{k > 0} \in J \) for some \( {t}_{k} \in {J}_{k},{t}_{k} = 0 \) for almost all \( k \). Hence \( \psi \left( 1\right) \in {\bigoplus }_{k < n}{J}_{k} \) for some \( n \). Then \( \varphi \left( x\right) = \psi \left( {x1}\right) = {x\psi }\left( 1\right) \in {\bigoplus }_{k < n}{J}_{k} \) for all \( x \in L \); in particular, \( {\varphi }_{n}\left( x\right) = 0 \) and \( x \in {L}_{n} \), since \( R/{L}_{n} \rightarrow {J}_{n} \) is injective. Thus \( L = {L}_{n} \), and then \( {L}_{k} = {L}_{n} \) for all \( k \geqq n \). | Yes |
If \( \alpha \) and \( \beta \) are distinct variables, \( \alpha \) does not occur bound in \( \varphi \) , and \( \beta \) does not occur in \( \varphi \) at all, then \( \vdash \forall {\alpha \varphi } \leftrightarrow \forall \beta {\operatorname{Subf}}_{\sigma }^{\alpha }\varphi \) . | Proof\n\[
\vdash \forall {\alpha \varphi } \rightarrow {\operatorname{Subf}}_{\beta }^{\alpha }\varphi
\]\n10.53\n\[
\vdash \forall \beta \forall {\alpha \varphi } \rightarrow \forall \beta {\operatorname{Subf}}_{\beta }^{\alpha }\varphi
\]\nusing \( {10.23}\left( 2\right) \)\n\[
\vdash \forall {\alpha \varphi } \rightarrow \forall \alpha \forall {\alpha \varphi }
\]\n10.23(3)\n\[
\vdash \forall {\alpha \varphi } \rightarrow \forall \beta {\operatorname{Subf}}_{\beta }^{\alpha }\varphi
\]\nNow we can apply the result just obtained to \( \beta ,\alpha \), Subf \( {}_{\beta }^{\alpha }\varphi \) instead of \( \alpha ,\beta ,\varphi \) and obtain \( \vdash \forall \beta {\operatorname{Subf}}_{\beta }^{\alpha }\varphi \rightarrow \forall \alpha {\operatorname{Subf}}_{\beta }^{\alpha }{\operatorname{Subf}}_{\alpha }^{\beta }\varphi \) . Since \( {\operatorname{Subf}}_{\alpha }^{\beta }{\operatorname{Subf}}_{\beta }^{\alpha }\varphi = \varphi \), the desired theorem follows. | Yes |
Theorem 2.6 (Homomorphism Theorem). If \( \varphi : A \rightarrow B \) is a homomorphism of left \( R \) -modules, then \( A/\operatorname{Ker}\varphi \cong \operatorname{Im}\varphi \) | In Theorem 2.6, there is by Theorem I.5.2 a unique isomorphism \( \theta \) of abelian groups such that \( \varphi = \iota \circ \theta \circ \pi \) . Then \( \theta \left( {a + \operatorname{Ker}\varphi }\right) = \varphi \left( a\right) \) for all \( a \in A \) . Therefore \( \theta \) is a module homomorphism. | Yes |
Let \( p \) be an odd prime. Suppose that \( {2}^{n} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) and \( {2}^{n} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . Show that \( {2}^{d} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) where \( d \) is the order of 2 \( \left( {\;\operatorname{mod}\;p}\right) \) . | Solution. Suppose \( b \) is an element of \( R \) . We proceed by induction on the norm of \( b \) . If \( b \) is irreducible, then we have nothing to prove, so assume that \( b \) is an element of \( R \) which is not irreducible. Then we can write \( b = {ac} \) where neither \( a \) nor \( c \) is a unit. By condition (i),\n\n\( n\left( b\right) = n\left( {ac}\right) = n\left( a\right) n\left( c\right) \) and since \( a, c \) are not units, then by condition (ii), \( n\left( a\right) < n\left( b\right) \) and \( n\left( c\right) < n\left( b\right) \). \n\nIf \( a, c \) are irreducible, then we are finished. If not, their norms are smaller than the norm of \( b \), and so by induction we can write them as products of irreducibles, thus finding an irreducible decomposition of \( b \) . | No |
Suppose \( U \) is a simply connected domain in \( {\mathbb{R}}^{n} \) and \( \mathbf{F} \) is a smooth, \( {\mathbb{R}}^{n} \) -valued function on \( U \). Then \( \mathbf{F} \) is conservative if and only if \( \mathbf{F} \) satisfies \( \frac{\partial {F}_{j}}{\partial {x}_{k}} - \frac{\partial {F}_{k}}{\partial {x}_{j}} = 0 \) at each point in \( U \). | If \( \mathbf{F} \) is conservative, then \( \frac{\partial {F}_{j}}{\partial {x}_{k}} = - \frac{{\partial }^{2}V}{\partial {x}_{k}\partial {x}_{j}} = - \frac{{\partial }^{2}V}{\partial {x}_{j}\partial {x}_{k}} = \frac{\partial {F}_{k}}{\partial {x}_{j}} \) at every point in \( U \). In the other direction, if \( \mathbf{F} \) satisfies (2.8), \( V \) can be obtained by integrating \( \mathbf{F} \) along paths and using the Stokes theorem to establish independence of choice of path. | Yes |
Theorem 3.2.2. There exist finite constants \( {C}_{n} \) and \( {C}_{n}^{\prime } \) such that the following statements are valid:
(a) Given \( b \in {BMO}\left( {\mathbf{R}}^{n}\right) \), the linear functional \( {L}_{b} \) lies in \( {\left( {H}^{1}\left( {\mathbf{R}}^{n}\right) \right) }^{ * } \) and has norm at most \( {C}_{n}\parallel b{\parallel }_{BMO} \) . Moreover, the mapping \( b \mapsto {L}_{b} \) from BMO to \( {\left( {H}^{1}\right) }^{ * } \) is injective. (b) For every bounded linear functional \( L \) on \( {H}^{1} \) there exists a BMO function b such that for all \( f \in {H}_{0}^{1} \) we have \( L\left( f\right) = {L}_{b}\left( f\right) \) and also
\[
\parallel b{\parallel }_{BMO} \leq {C}_{n}^{\prime }{\begin{Vmatrix}{L}_{b}\end{Vmatrix}}_{{H}^{1} \rightarrow \mathbf{C}}
\] | Proof. We have already proved that for all \( b \in {BMO}\left( {\mathbf{R}}^{n}\right) ,{L}_{b} \) lies in \( {\left( {H}^{1}\left( {\mathbf{R}}^{n}\right) \right) }^{ * } \) and has norm at most \( {C}_{n}\parallel b{\parallel }_{BMO} \) . The embedding \( b \mapsto {L}_{b} \) is injective as a consequence of Exercise 3.2.2. It remains to prove (b). Fix a bounded linear functional \( L \) on \( {H}^{1}\left( {\mathbf{R}}^{n}\right) \) and also fix a cube \( Q \) . Consider the space \( {L}^{2}\left( Q\right) \) of all square integrable functions supported in \( Q \) with norm
\[
\parallel g{\parallel }_{{L}^{2}\left( Q\right) } = {\left( {\int }_{Q}{\left| g\left( x\right) \right| }^{2}dx\right) }^{\frac{1}{2}}.
\]
We denote by \( {L}_{0}^{2}\left( Q\right) \) the closed subspace of \( {L}^{2}\left( Q\right) \) consisting of all functions in \( {L}^{2}\left( Q\right) \) with mean value zero. We show that every element in \( {L}_{0}^{2}\left( Q\right) \) is in \( {H}^{1}\left( {\mathbf{R}}^{n}\right) \) and we have the inequality
\[
\parallel g{\parallel }_{{H}^{1}} \leq {c}_{n}{\left| Q\right| }^{\frac{1}{2}}\parallel g{\parallel }_{{L}^{2}}
\]
(3.2.4)
To prove (3.2.4) we use the square function characterization of \( {H}^{1} \) . We fix a Schwartz function \( \Psi \) on \( {\mathbf{R}}^{n} \) whose Fourier transform is supported in the annulus \( \frac{1}{2} \leq \left| \xi \right| \leq 2 \) and that satisfies (1.3.6) for all \( \xi \neq 0 \) and we let \( {\Delta }_{j}\left( g\right) = {\Psi }_{{2}^{-j}} * g \) . To estimate the \( {L}^{1} \) norm of \( {\left( \mathop{\sum }\limits_{j}{\left| {\Delta }_{j}\left( g\right) \right| }^{2}\right) }^{1/2} \) over \( {\mathbf{R}}^{n} \), consider the part of the integral over \( 3\sqrt{n}Q \) and the integral over \( {\left( 3\sqrt{n}Q\right) }^{c} \) . First we use Hölder’s inequality and an \( {L}^{2} \) estimate to prove that
\[
{\int }_{3\sqrt{n}Q}{\left( \mathop{\sum }\limits_{j}{\left| {\Delta }_{j}\left( g\right) \left( x\right) \right| }^{2}\right) }^{\frac{1}{2}}{dx} \leq {c}_{n}{\left| Q\right| }^{\frac{1}{2}}\parallel g{\parallel }_{{L}^{2}}.
\]
Now for \( x \notin 3\sqrt{n}Q \) we use the mean value property of \( g \) to obtain
\[
\left| {{\Delta }_{j}\left( g\right) \left( x\right) }\right| \leq \frac{{c}_{n}\parallel g{\parallel }_{{L}^{2}}{2}^{{nj} + j}{\left| Q\right| }^{\frac{1}{n} + \frac{1}{2}}}{{\left( 1 + {2}^{j}\left| x - {c}_{Q}\right| \right) }^{n + 2}},
\]
(3.2.5)
where \( {c}_{Q} \) is the center of \( Q \) . Estimate (3.2.5) is obtained in a way similar to that we obtained the corresponding estimate for one atom; see Theorem 2.3.11 for details. Now (3.2.5) implies that
\[
{\int }_{{\left( 3\sqrt{n}Q\right) }^{c}}{\left( \mathop{\sum }\limits_{j}{\left| {\Delta }_{j}\left( g\right) \left( x\right) \right| }^{2}\right) }^{\frac{1}{2}}{dx} \leq {c}_{n}{\left| Q\right| }^{\frac{1}{2}}\parallel g{\parallel }_{{L}^{2}},
\]
which proves (3.2.4).
Since \( {L}_{0}^{2}\left( Q\right) \) is a subspace of \( {H}^{1} \), it follows from (3.2.4) that the linear functional \( L : {H}^{1} \rightarrow \mathbf{C} \) is also a bounded linear functional on \( {L}_{0}^{2}\left( Q\right) \) with norm
\[
\parallel L{\parallel }_{{L}_{0}^{2}\left( Q\right) \rightarrow \mathbf{C}} \leq {c}_{n}{\left| Q\right| }^{1/2}\parallel L{\parallel }_{{H}^{1} \rightarrow \mathbf{C}}
\]
(3.2.6)
By the Riesz representation theorem for the Hilbert space \( {L}_{0}^{2}\left( Q\right) \), there is an element \( {F}^{Q} \) in \( {\left( {L}_{0}^{2}\left( Q\right) \right) }^{ * } = {L}^{2}\left( Q\right) /\{ \) constants \( \} \) such that
\[
L\left( g\right) = {\int }_{Q}{F}^{Q}\left( x\right) g\left( x\right) {dx}
\]
(3.2.7)
for all \( g \in {L}_{0}^{2}\left( Q\right) \), and this \( {F}^{Q} \) satisfies
\[
{\begin{Vmatrix}{F}^{Q}\end{Vmatrix}}_{{L}^{2}\left( Q\right) } \leq \parallel L{\parallel }_{{L}_{0}^{2}\left( Q\right) \rightarrow \mathbf{C}}
\]
(3.2.8)
Thus for any cube \( Q \) in \( {\mathbf{R}}^{n} \), there is square integrable function \( {F}^{Q} \) supported in \( Q \) such that (3.2.7) is satisfied. We observe that if a cube \( Q \) is contained in another cube \( {Q}^{\prime } \), then \( {F}^{Q} \) differs from \( {F}^{{Q}^{\prime }} \) by a constant on \( Q \) . Indeed, for all \( g \in {L}_{0}^{2}\left( Q\right) \) we have
\[
{\int }_{Q}{F}^{{Q}^{\prime }}\left( x\right) g\left( x\right) {dx} = L\left( g\right) = {\int }_{Q}{F}^{Q}\left( x\right) g\left( x\right) {dx}
\]
and thus
\[
{\int }_{Q}\left( {{F}^{{Q}^{\prime }}\left( x\right) - {F}^{Q}\left( x\right) }\right) g\left( x\right) {dx} = 0.
\]
Consequently,
\[
g \rightarrow {\int }_{Q}\left( {{F}^{{Q}^{\prime }}\left( x\right) - {F}^{Q}\left( x\right) }\right) g\left( x\right) {dx}
\]
is the zero functional on \( {L}_{0}^{2}\left( Q\right) \) ; hence \( {F}^{{Q}^{\prime }} - {F}^{Q} \) must be the zero function in the space \( {\left( {L}_{0}^{2}\left( Q\right) \right) }^{ * } \), i.e., \( {F}^{{Q}^{\prime }} - {F}^{Q} \) is a constant on \( Q \) . Let
\[
{Q}_{m} = {\left\lbrack -m/2, m/2\right\rbrack }^{n}
\]
for \( m = 1,2,\ldots \) . Then \( \left| {Q}_{1}\right| = 1 \) . We define a locally integrable function \( b\left( x\right) \) on \( {\mathbf{R}}^{n} \) by setting
\[
b\left( x\right) = {F}^{{Q}_{m}}\left( x\right) - \frac{1}{\left| {Q}_{1}\right| }{\int }_{{Q}_{1}}{F}^{{Q}_{m}}\left( t\right) {dt}
\]
(3.2.9)
whenever \( x \in {Q}_{m} \) . We check that this definition is unambiguous. Let \( 1 \leq \ell < m \) . Then for \( x \in {Q}_{\ell }, b\left( x\right) \) is also defined as in (3.2.9) with \( \ell \) in the place of \( m \) . The difference of these two functions is
\[
{F}^{{Q}_{m}} - {F}^{{Q}_{\ell }} - \underset{{Q}_{1}}{\operatorname{Avg}\left( {{F}^{{Q}_{m}} - {F}^{{Q}_{\ell }}}\right) } = 0,
\]
since the function \( {F}^{{Q}_{m}} - {F}^{{Q}_{\ell }} \) is constant in the cube \( {Q}_{\ell } \) (which is contained in \( {Q}_{m} \) ), as indicated earlier.
Next we claim that for any cube \( Q \) there is a constant \( {C}_{Q} \) such that
\[
{F}^{Q} = b - {C}_{Q}\;\text{ on }Q.
\]
(3.2.10)
Indeed, given a cube \( Q \) pick the smallest \( m \) such that \( Q \) is contained in \( {Q}^{m} \) and observe that
\[
{F}^{Q} = \underset{\text{constant on }Q}{\underbrace{{F}^{Q} - {F}^{{Q}_{m}}}} + \underset{\begin{matrix} {Q}_{1} \\ b\left( x\right) \end{matrix}}{\underbrace{{F}^{{Q}_{m}} - \operatorname{Avg}{F}^{{Q}_{m}}}} + \underset{\begin{matrix} {Q}_{1} \\ \text{ constant on }Q \end{matrix}}{\underbrace{\operatorname{Avg}{F}^{{Q}_{m}}}}
\]
and let \( - {C}_{Q} \) be the sum of the two preceding constant expressions on \( Q \) .
We have now found a locally integrable function \( b \) such that for all cubes \( Q \) and all \( g \in {L}_{0}^{2}\left( Q\right) \) we have
\[
{\int }_{Q}b\left( x\right) g\left( x\right) {dx} = {\int }_{Q}\left( {{F}^{Q}\left( x\right) + {C}_{Q}}\right) g\left( x\right) {dx} = {\int }_{Q}{F}^{Q}\left( x\right) g\left( x\right) {dx} = L\left( g\right) ,
\]
(3.2.11)
as follows from (3.2.7) and (3.2.10). We conclude the proof by showing that \( b \) lies in \( \operatorname{BMO}\left( {\mathbf{R}}^{n}\right) \) . By (3.2.10),(3.2.8), and (3.2.6) we have
\[
\mathop{\sup }\limits_{Q}\frac{1}{\left| Q\right| }{\int }_{Q}\left| {b\left( x\right) - {C}_{Q}}\right| {dx} = \mathop{\sup }\limits_{Q}\frac{1}{\left| Q\right| }{\int }_{Q}\left| {{F}^{Q}\left( x\right) }\right| {dx}
\]
\[
\leq \mathop{\sup }\limits_{Q}{\left| Q\right| }^{-1}{\left| Q\right| }^{\frac{1}{2}}{\begin{Vmatrix}{F}^{Q}\end{Vmatrix}}_{{L}^{2}\left( Q\right) }
\]
\[
\leq \mathop{\sup }\limits_{Q}{\left| Q\right| }^{-\frac{1}{2}}\parallel L{\parallel }_{{L}_{0}^{2}\left( Q\right) \rightarrow \mathbf{C}}
\]
\[
\leq {c}_{n}\parallel L{\parallel }_{{H}^{1} \rightarrow \mathbf{C}} < \infty .
\]
Using Proposition 3.1.2 (3), we deduce that \( b \in {BMO} \) and \( \parallel b{\parallel }_{BMO} \leq 2{c}_{n}\parallel L{\parallel }_{{H}^{1} \rightarrow \mathbf{C}} \) . Finally, (3.2.11) implies that
\[
L\left( g\right) = {\int }_{{\mathbf{R}}^{n}}b\left( x\right) g\left( x\right) {dx} = {L}_{b}\left( g\right)
\]
for all \( g \in {H}_{0}^{1}\left( {\mathbf{R}}^{n}\right) \), proving that the linear functional \( L \) coincides with \( {L}_{b} \) on a dense subspace of \( {H}^{1} \) . Consequently, \( L = {L}_{b} \), and this concludes the proof of part (b). | Yes |
Suppose that \( X \) is a Banach space with an unconditional basis. If \( X \) is not reflexive, then either \( {c}_{0} \) is complemented in \( X \), or \( {\ell }_{1} \) is complemented in \( X \) (or both). In either case, \( {X}^{* * } \) is nonseparable. | The first statement of the theorem follows immediately from Theorem 3.2.19, Theorem 3.3.1, and Theorem 3.3.2. Now, for the latter statement, if \( {c}_{0} \) were complemented in \( X \), then \( {X}^{* * } \) would contain a (complemented) copy \( {\ell }_{\infty } \) . If \( {\ell }_{1} \) were complemented in \( X \), then \( {X}^{ * } \) would be nonseparable, since it would contain a (complemented) copy of \( {\ell }_{\infty } \) . In either case, \( {X}^{* * } \) is nonseparable. | Yes |
Proposition 19. Let \( R \) be a commutative ring with 1 . (1) Prime ideals are primary. (2) The ideal \( Q \) is primary if and only if every zero divisor in \( R/Q \) is nilpotent. (3) If \( Q \) is primary then rad \( Q \) is a prime ideal, and is the unique smallest prime ideal containing \( Q \) . (4) If \( Q \) is an ideal whose radical is a maximal ideal, then \( Q \) is a primary ideal. (5) Suppose \( M \) is a maximal ideal and \( Q \) is an ideal with \( {M}^{n} \subseteq Q \subseteq M \) for some \( n \geq 1 \) . Then \( Q \) is a primary ideal with \( \operatorname{rad}Q = M \) . | The first two statements are immediate from the definition of a primary ideal. For (3), suppose \( {ab} \in \operatorname{rad}Q \) . Then \( {a}^{m}{b}^{m} = {\left( ab\right) }^{m} \in Q \), and since \( Q \) is primary, either \( {a}^{m} \in Q \), in which case \( a \in \operatorname{rad}Q \), or \( {\left( {b}^{m}\right) }^{n} \in Q \) for some positive integer \( n \), in which case \( b \in \operatorname{rad}Q \) . This proves that \( \operatorname{rad}Q \) is a prime ideal, and it follows that \( \operatorname{rad}Q \) is the smallest prime ideal containing \( Q \) (Proposition 12). To prove (4) we pass to the quotient ring \( R/Q \) ; by (2), it suffices to show that every zero divisor in this quotient ring is nilpotent. We are reduced to the situation where \( Q = \left( 0\right) \) and \( M = \operatorname{rad}Q = \operatorname{rad}\left( 0\right) \), which is the nilradical, is a maximal ideal. Since the nilradical is contained in every prime ideal (Proposition 12), it follows that \( M \) is the unique prime ideal, so also the unique maximal ideal. If \( d \) were a zero divisor, then the ideal \( \left( d\right) \) would be a proper ideal, hence contained in a maximal ideal. This implies that \( d \in M \), hence every zero divisor is indeed nilpotent. Finally, suppose \( {M}^{n} \subseteq Q \subseteq M \) for some \( n \geq 1 \) where \( M \) is a maximal ideal. Then \( Q \subseteq M \) so rad \( Q \subseteq \operatorname{rad}M = M \) . Conversely, \( {M}^{n} \subseteq Q \) shows that \( M \subseteq \operatorname{rad}Q \), so \( \operatorname{rad}Q = M \) is a maximal ideal, and \( Q \) is primary by (4). | Yes |
If \( u \in {W}^{1,2}\left( \Omega \right) \) is a weak solution of \( {\Delta u} = f \) with \( f \in {C}^{k,\alpha }\left( \Omega \right), k \in \mathbb{N},0 < \alpha < 1 \), then \( u \in {C}^{k + 2,\alpha }\left( \Omega \right) \), and for \( {\Omega }_{0} \subset \subset \Omega \), | Since \( u \in {C}^{2,\alpha }\left( \Omega \right) \) by Theorem 13.1.2, we know that it weakly solves \[ \Delta \frac{\partial }{\partial {x}^{i}}u = \frac{\partial }{\partial {x}^{i}}f \] Theorem 13.1.2 then implies \[ \frac{\partial }{\partial {x}^{i}}u \in {C}^{2,\alpha }\left( \Omega \right) \;\left( {i \in \{ 1,\ldots, d\} }\right) , \] and thus \( u \in {C}^{3,\alpha }\left( \Omega \right) \) . The proof is concluded by induction. | No |
Theorem 5.9 (Stone [1934]). Every Boolean lattice is isomorphic to the lattice of closed and open subsets of its Stone space. | Proof. Let \( L \) be a Boolean lattice and let \( X \) be its Stone space. For every \( a \in L, V\left( a\right) \) is open in \( X \), and is closed in \( X \) since \( X \smallsetminus V\left( a\right) = V\left( {a}^{\prime }\right) \) is open. Conversely, if \( U \in \mathrm{L}\left( X\right) \) is a closed and open subset of \( X \), then \( U \) is a union of basic open sets \( V\left( a\right) \subseteq U \) ; since \( U \) is closed, \( U \) is compact, \( U \) is a finite union \( V\left( {a}_{1}\right) \cup \cdots \cup V\left( {a}_{n}\right) = V\left( {{a}_{1} \vee \cdots \vee {a}_{n}}\right) \), and \( U = V\left( a\right) \) for some \( a \in L \) . Thus \( V \) is a mapping of \( L \) onto \( \mathrm{L}\left( X\right) \) . By 4.9 and 5.7, \( V : L \rightarrow \mathrm{L}\left( X\right) \) is a lattice isomorphism. \( ▱ \) | Yes |
Let \( M, B \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) be matrices, with \( M \) irreducible and \( \left| B\right| \leq M \) . Then \( \rho \left( B\right) \leq \rho \left( M\right) \) . | In order to establish the inequality, we proceed as above. If \( \lambda \) is an eigenvalue of \( B \), of modulus \( \rho \left( B\right) \), and if \( x \) is a normalized eigenvector, then \( \rho \left( B\right) \left| x\right| \leq \left| B\right| \cdot \left| x\right| \leq M\left| x\right| \), so that \( {C}_{\rho \left( B\right) } \) is nonempty. Hence \( \rho \left( B\right) \leq R = \rho \left( M\right) \) . | No |
Proposition 4.51. If \( S \) is compact, under assumptions (P1),(P2), one has\n\[
{\partial }_{F}m\left( \bar{x}\right) = \overline{\operatorname{co}}\left\{ {D{f}_{s}\left( \bar{x}\right) : s \in M\left( \bar{x}\right) }\right\}
\]\[
{\partial }_{F}p\left( \bar{x}\right) = \mathop{\bigcap }\limits_{{s \in P\left( \bar{x}\right) }}\left\{ {D{f}_{s}\left( \bar{x}\right) }\right\}
\] | The proof is left as an exercise that the reader can tackle while reading Sect. 4.7.1 | No |
Proposition 6.4.14. Assume that neither \( b/a, c/a \), nor \( c/b \) is the cube of \( a \) rational number. If the elliptic curve \( E \) with affine equation \( {Y}^{2} = {X}^{3} + {\left( 4abc\right) }^{2} \) has zero rank then the equation \( a{x}^{3} + b{y}^{3} + c{z}^{3} \) has no nontrivial rational solutions. | This is essentially a restatement of Corollary 8.1.15, and also immediately follows from the above proposition after rescaling. Note that Proposition 8.4.3 tells us that the elliptic curve \( {Y}^{2} = {X}^{3} - {432}{\left( abc\right) }^{2} \) is 3 -isogenous with the elliptic curve \( {Y}^{2} = {X}^{3} + {\left( 4abc\right) }^{2} \) . | No |
Let \( G \) be a \( k \) -connected graph, let \( x \) be a vertex of \( G \), and let \( Y \subseteq V \smallsetminus \{ x\} \) be a set of at least \( k \) vertices of \( G \) . Then there exists a \( k \) -fan in \( G \) from \( x \) to \( Y \) . | We now give the promised application of the Fan Lemma. By Theorem 5.1, in a 2-connected graph any two vertices are connected by two internally disjoint paths; equivalently, any two vertices in a 2-connected graph lie on a common cycle. Dirac (1952b) generalized this latter statement to \( k \) -connected graphs.\n\nTheorem 9.6 Let \( S \) be a set of \( k \) vertices in a \( k \) -connected graph \( G \), where \( k \geq 2 \) . Then there is a cycle in \( G \) which includes all the vertices of \( S \). \n\nProof By induction on \( k \) . We have already observed that the assertion holds for \( k = 2 \), so assume that \( k \geq 3 \) . Let \( x \in S \), and set \( T \mathrel{\text{:=}} S \smallsetminus x \) . Because \( G \) is \( k \) -connected, it is \( \left( {k - 1}\right) \) -connected. Therefore, by the induction hypothesis, there is a cycle \( C \) in \( G \) which includes \( T \) . Set \( Y \mathrel{\text{:=}} V\left( C\right) \) . If \( x \in Y \), then \( C \) includes all the vertices of \( S \) . Thus we may assume that \( x \notin Y \) . If \( \left| Y\right| \geq k \), the Fan Lemma (Proposition 9.5) ensures the existence of a \( k \) -fan in \( G \) from \( x \) to \( Y \) . Because \( \left| T\right| = k - 1 \), the set \( T \) divides \( C \) into \( k - 1 \) edge-disjoint segments. By the Pigeonhole Principle, some two paths of the fan, \( P \) and \( Q \), end in the same segment. The subgraph \( C \cup P \cup Q \) contains three cycles, one of which includes \( S = T \cup \{ x\} \) (see Figure 9.5). If \( \left| Y\right| = k - 1 \), the Fan Lemma yields a \( \left( {k - 1}\right) \) -fan from \( x \) to \( Y \) in which each vertex of \( Y \) is the terminus of one path, and we conclude as before. | Yes |
Theorem 1. \( {\lambda }_{m}\left( x\right) \) is irreducible in the rational field. | Proof. We observe first that \( {\lambda }_{m}\left( x\right) \) has integer coefficients. For, assuming this holds for every \( {\lambda }_{d}\left( x\right), d < m \), and setting \( p\left( x\right) = \mathop{\prod }\limits_{\substack{{d \mid m} \\ {1 \leq d < m} }}{\lambda }_{d}\left( x\right) \), we obtain by the usual division algorithm
that \( {x}^{m} - 1 = p\left( x\right) q\left( x\right) + r\left( x\right) \) where \( q\left( x\right) \) and \( r\left( x\right) \) e \( I\left\lbrack x\right\rbrack \) and \( \deg r\left( x\right) < \deg p\left( x\right) \) . On the other hand, we have \( {x}^{m} - 1 = \) \( p\left( x\right) {\lambda }_{m}\left( x\right) \), so by the uniqueness of the quotient and remainder, \( {\lambda }_{m}\left( x\right) = q\left( x\right) \) has integer coefficients. Now suppose that \( {\lambda }_{m}\left( x\right) = \) \( h\left( x\right) k\left( x\right) \) where \( h\left( x\right) \) is irreducible in \( {R}_{0}\left\lbrack x\right\rbrack \) and \( \deg h\left( x\right) \geq 1 \) . By Gauss’ lemma (Vol. I, p. 125) we may assume that \( h\left( x\right) \) and \( k\left( x\right) \) have integer coefficients and leading coefficients 1. Let \( p \) be a prime integer such that \( p \nmid m \) and let \( \zeta \) be a root of \( h\left( x\right) \) . We shall show that \( {\zeta }^{p} \) is a root of \( h\left( x\right) \) . Since \( \left( {p, m}\right) = 1,{\zeta }^{p} \) is a primitive \( m \) -th root of 1 and, if \( {\zeta }^{p} \) is not a root of \( h\left( x\right) ,{\zeta }^{p} \) is a root of \( k\left( x\right) \) ; consequently \( \zeta \) is a root of \( k\left( {x}^{p}\right) \) . Since \( h\left( x\right) \) is irreducible in \( {R}_{0}\left\lbrack x\right\rbrack \) and has \( \zeta \) as a root, \( h\left( x\right) \mid k\left( {x}^{p}\right) \) . It follows (as above) that \( k\left( {x}^{p}\right) = h\left( x\right) l\left( x\right) \), where \( l\left( x\right) \) has integer coefficients and leading coefficient 1. Also we have \( {x}^{m} - 1 = {\lambda }_{m}\left( x\right) p\left( x\right) = \) \( h\left( x\right) k\left( x\right) p\left( x\right) \) and all of these polynomials have integer coefficients and leading coefficients 1 . We now pass to congruences modulo \( p \) or, what is the same thing, to relations in the polynomial ring \( {I}_{p}\left\lbrack x\right\rbrack \) . Then we obtain
(5)
\[
{x}^{m} - \overrightarrow{1} = \overrightarrow{h}\left( x\right) \overrightarrow{k}\left( x\right) \overrightarrow{p}\left( x\right)
\]
where in general, if \( f\left( x\right) = {a}_{0}{x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n}{\varepsilon I}\left\lbrack x\right\rbrack \), then \( \bar{f}\left( x\right) = {\bar{a}}_{0}{x}^{n} + {\bar{a}}_{1}{x}^{n - 1} + \cdots + {\bar{a}}_{n},{\bar{a}}_{i} = {a}_{i} + \left( p\right) \) in \( {I}_{p} \) . Similarly, we have \( \bar{k}\left( {x}^{p}\right) = \bar{h}\left( x\right) \bar{l}\left( x\right) \) . On the other hand, using \( {\bar{a}}^{p} = \bar{a} \) for every integer \( a \), we see that
\[
\bar{f}{\left( x\right) }^{p} = {\left( {\bar{a}}_{0}{x}^{n} + \cdots + {\bar{a}}_{n}\right) }^{p} = {\bar{a}}_{0}{}^{p}{x}^{pn} + \cdots + {\bar{a}}_{n}{}^{p}
\]
\[
= {\bar{a}}_{0}{x}^{pn} + \cdots + {\bar{a}}_{n} = \bar{f}\left( {x}^{p}\right)
\]
for any polynomial \( f\left( x\right) \) . Hence \( \bar{k}{\left( x\right) }^{p} = \bar{k}\left( {x}^{p}\right) = \bar{h}\left( x\right) \bar{l}\left( x\right) \) which implies that \( \left( {h\left( x\right), k\left( x\right) }\right) \neq 1 \) . Then (5) shows that \( {x}^{m} - 1 \) has multiple roots in its splitting field over \( {I}_{p} \) . Since \( p \times m \) this is impossible and so we have proved that \( {\zeta }^{p} \) is a root of \( h\left( x\right) \) for every prime \( p \) satisfying \( p \times m \) . A repetition of this process shows that \( {\zeta }^{r} \) is a root of \( h\left( x\right) \) for every integer \( r \) prime to \( m \) . Since any primitive \( m \) -th root of 1 has the form \( {\zeta }^{r},\left( {r, m}\right) = 1 \) we see that every primitive \( m \) -th root of 1 is a root of \( h\left( x\right) \) . Hence \( h\left( x\right) = {\lambda }_{m}\left( x\right) \) and \( {\lambda }_{m}\left( x\right) \) is irreducible in \( {R}_{0}\left\lbrack x\right\rbrack \) . | Yes |
Theorem 1. Let \( J \) be a polygon in \( {\mathbf{R}}^{2} \). Then \( {\mathbf{R}}^{2} - J \) has exactly two components. | Proof. Let \( N \) be a "strip neighborhood" of \( J \), formed by small convex polyhedral neighborhoods of the edges and vertices of \( J \). Below and hereafter, pictures of polyhedra will not necessarily look like polyhedra. Only a sample of \( N \) is indicated in Figure 2.1.
Figure 2.1
Lemma 1. \( {\mathbf{R}}^{2} - J \) has at most two components.
Proof. Starting at any point \( P \) of \( N - J \), we can work our way around the polygon, along a path in \( N - J \), until we get to either \( {P}_{1} \) or \( {P}_{2} \). (See Figure 2.2.) From this the lemma follows, because every point \( Q \) of \( {\mathbf{R}}^{2} - J \) can be joined to some point \( P \) of \( N - J \) by a linear segment in \( {\mathbf{R}}^{2} - J \).
Figure 2.2
It is possible a priori that \( N - J \) has only one component. If so, \( N \) would be a Möbius band. (See Section 21 below.) But this is ruled out by the next lemma.
Lemma 2. \( {\mathbf{R}}^{2} - J \) has at least two components.
Proof. We choose the axes in general position, in the sense that no horizontal line contains more than one of the vertices of \( J \). (This can be done, because there are only a finite number of directions that we need to avoid. Hereafter, the phrase "in general position" will be defined in a variety of ways, in a variety of cases. In each case, the intuitive meaning will be the same: general position is a situation which occurs with probability 1 when certain choices are made at random.)
For each point \( P \) of \( {\mathbf{R}}^{2} \), let \( {L}_{P} \) be the horizontal line through \( P \). The index Ind \( P \) of a point \( P \) of \( {\mathbf{R}}^{2} - J \) is defined as follows. (1) If \( {L}_{P} \) contains no vertex of \( J \), then Ind \( P \) is the number of points of \( {L}_{P} \cap J \) that lie to the left of \( P \), reduced modulo 2. Thus Ind \( P \) is 0 or 1. (2) If \( {L}_{P} \) contains a vertex of \( J \), then Ind \( P \) is the number of points of \( {L}^{\prime } \cap J \), lying to the left of \( P \), reduced modulo 2, where \( {L}^{\prime } \) is a horizontal line lying "slightly above" or "slightly below" \( {L}_{P} \). Here the phrases in quotation marks mean that no vertex of \( J \) lies on \( {L}^{\prime } \), or between \( {L}_{P} \) and \( {L}^{\prime } \). It makes no difference whether \( {L}^{\prime } \) lies above or below. The three possibilities for \( J \), relative to \( L \), are shown in Figure 2.3. In each case, the two possible positions for \( {L}^{\prime } \) give the same index for \( P \).
Evidently the function
\[
f : {\mathbf{R}}^{2} - J \rightarrow \{ 0,1\}
\]
\[
f : P \mapsto \text{Ind}P
\]
is a mapping; if Ind \( P = i \), then Ind \( {P}^{\prime } = i \) when \( {P}^{\prime } \) is sufficiently close to \( P \). The set \( {f}^{-1}\left( 0\right) \) is nonempty; every point above all of \( J \) belongs to \( {f}^{-1}\left( 0\right) \). To show that \( {f}^{-1}\left( 1\right) \neq \varnothing \), let \( Q \) be a point of \( J \), such that \( {L}_{O} \) contains no vertex of \( J \). Let \( {P}_{1} \) be the leftmost point of \( J \) on \( {L}_{Q} \). Let \( P \) be a point of \( {L}_{Q} \), slightly to the right of \( {P}_{1} \), in the sense that \( P \notin J \), and no point between \( {P}_{1} \) and \( P \) belongs to \( J \). Then Ind \( P = 1 \).
Therefore \( {\mathbf{R}}^{2} - J \) is not connected; it is the union of the disjoint nonempty open sets \( {f}^{-1}\left( 0\right) \) and \( {f}^{-1}\left( 1\right) \).
The bounded component \( I \) of \( {\mathbf{R}}^{2} - J \) is called the interior of \( J \), and the unbounded component \( E \) is called the exterior. | Yes |
Proposition 8.6. \( {f}_{2k} \circ h = {\left( -1\right) }^{k}\mathop{\sum }\limits_{{l = 0}}^{{2k}}{\left( -1\right) }^{l}{f}_{l}^{i}{f}_{{2k} - l}^{i} \) , \( \operatorname{Pf} \circ h = {\left( -1\right) }^{\left\lbrack n/2\right\rbrack }{f}_{n}^{i} \) . | Proof. By Lemma 8.1,\[
{\left| \det \left( x{I}_{n} - M\right) \right| }^{2} = \det \left( {h\left( {x{I}_{n} - M}\right) }\right) = \det \left( {x{I}_{2n} - h\left( M\right) }\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{x}^{2\left( {n - k}\right) }{f}_{2k} \circ h\left( M\right) .
\]
On the other hand,\[
{\left| \det \left( x{I}_{n} - M\right) \right| }^{2} = {\left| \det \left( x{I}_{n} - i\left( -iM\right) \right) \right| }^{2} = {\left| \mathop{\sum }\limits_{{k = 0}}^{n}{\left( -1\right) }^{k}{\left( -i\right) }^{k}{f}_{k}^{i}\left( M\right) {x}^{n - k}\right| }^{2}
\]
\[
= {\left| {x}^{n} + i{x}^{n - 1}{f}_{1}^{i}\left( M\right) - {x}^{n - 2}{f}_{2}^{i}\left( M\right) - i{x}^{n - 3}{f}_{3}^{i}\left( M\right) + \cdots \right| }^{2}
\]
\[
= \mid \left( {{x}^{n} - {x}^{n - 2}{f}_{2}^{i}\left( M\right) + {x}^{n - 4}{f}_{4}^{i}\left( M\right) \cdots }\right)
\]
\[
+ i\left( {{x}^{n - 1}{f}_{1}^{i}\left( M\right) - {x}^{n - 3}{f}_{3}^{i}\left( M\right) + {x}^{n - 5}{f}_{5}^{i}\left( M\right) \cdots }\right) {|}^{2}
\]
\[
= {\left( {x}^{n} - {x}^{n - 2}{f}_{2}^{i}\left( M\right) + {x}^{n - 4}{f}_{4}^{i}\left( M\right) \cdots \right) }^{2}
\]
\[
+ {\left( {x}^{n - 1}{f}_{1}^{i}\left( M\right) - {x}^{n - 3}{f}_{3}^{i}\left( M\right) + {x}^{n - 5}{f}_{5}^{i}\left( M\right) \cdots \right) }^{2}.
\]
The coefficient of \( {x}^{{2n} - {2k}} \) in the last equality is
\[
\mathop{\sum }\limits_{{k - j\text{ even }}}{\left( -1\right) }^{\left( {k - j}\right) /2}{f}_{k - j}^{i}\left( M\right) {\left( -1\right) }^{\left( {k + j}\right) /2}{f}_{k + j}^{i}\left( M\right)
\]
\[
+ \mathop{\sum }\limits_{{k - j\text{ odd }}}{\left( -1\right) }^{\left( {k - j + 1}\right) /2}{f}_{k - j}^{i}\left( M\right) {\left( -1\right) }^{\left( {k + j + 1}\right) /2}{f}_{k + j}^{i}\left( M\right)
\]
\[
= {\left( -1\right) }^{k}\left\lbrack {\mathop{\sum }\limits_{{k - j\text{ even }}}{f}_{k - j}^{i}\left( M\right) {f}_{k + j}^{i}\left( M\right) - \mathop{\sum }\limits_{{k - j\text{ odd }}}{f}_{k - j}^{i}\left( M\right) {f}_{k + j}^{i}\left( M\right) }\right\rbrack
\]
\[
= {\left( -1\right) }^{k}\mathop{\sum }\limits_{{k - j}}{\left( -1\right) }^{k - j}{f}_{k - j}^{i}\left( M\right) {f}_{k + j}^{i}\left( M\right)
\]
\[
= {\left( -1\right) }^{k}\mathop{\sum }\limits_{l}{\left( -1\right) }^{l}{f}_{l}^{i}\left( M\right) {f}_{{2k} - l}^{i}\left( M\right) .
\]
This establishes the first identity in the proposition. For the one involving the Pfaffian, it suffices to check the formula in the case when \( M = \left( {i{\lambda }_{1}\ldots i{\lambda }_{n}}\right) \) . Then\[
\operatorname{Pf}\left( {h\left( M\right) }\right) = \operatorname{Pf}\left( \begin{matrix} & & - {\lambda }_{1} & & \\ & & & \ddots & \\ & & & & - {\lambda }_{n} \\ {\lambda }_{1} & & & & \\ & \ddots & & & \\ & & {\lambda }_{n} & & \end{matrix}\right)
\]
\[
= {\epsilon }^{1\left( {n + 1}\right) 2\left( {n + 2}\right) \ldots n\left( {2n}\right) }{\left( -1\right) }^{n}{\lambda }_{1}\cdots {\lambda }_{n}
\]
\[
= {\left( -1\right) }^{\left\lbrack n/2\right\rbrack }{\left( -1\right) }^{n}{\lambda }_{1}\cdots {\lambda }_{n} = {\left( -1\right) }^{\left\lbrack n/2\right\rbrack }\det {iM} = {\left( -1\right) }^{\left\lbrack n/2\right\rbrack }{f}_{n}^{i}\left( M\right) .
\]" | Yes |
Whitehead product 472, 482 | Whitehead Theorem 181 | No |
Theorem 2.10 A graph \( G \) is even if and only if \( \left| {\partial \left( X\right) }\right| \) is even for every subset \( X \) of \( V \) . | Suppose that \( \left| {\partial \left( X\right) }\right| \) is even for every subset \( X \) of \( V \) . Then, in particular, \( \left| {\partial \left( v\right) }\right| \) is even for every vertex \( v \) . But, as noted above, \( \partial \left( v\right) \) is just the set of all links incident with \( v \) . Because loops contribute two to the degree, it follows that all degrees are even. Conversely, if \( G \) is even, then Theorem 2.9 implies that all edge cuts are of even cardinality. | Yes |
If \( \mathrm{F} \) is a Galois extension field of \( \mathrm{K} \) and \( \mathrm{E} \) is a stable intermediate field of the extension, then \( \mathrm{E} \) is Galois over \( \mathrm{K} \) . | If \( u : E - K \), then there exists \( \sigma \in {\operatorname{Aut}}_{K}F \) such that \( \sigma \left( u\right) \neq u \) since \( F \) is Galois over \( K \) . But \( \sigma \mid E \) e Aut \( {}_{K}E \) by stability. Therefore, \( E \) is Galois over \( K \) by the Remarks after Definition 2.4. | Yes |
Corollary 10.7.7. Set \( b = \log \left( {2\pi }\right) - 1 - \gamma /2 \) . Then for all \( s \in \mathbb{C} \) we have the convergent product\n\[
\zeta \left( s\right) = \frac{{e}^{bs}}{s\left( {s - 1}\right) \Gamma \left( {s/2}\right) }\mathop{\prod }\limits_{\rho }\left( {1 - \frac{s}{\rho }}\right) {e}^{s/\rho },
\] | Proof. We apply Hadamard's theorem to the function\n\[
f\left( s\right) = s\left( {1 - s}\right) {\pi }^{-s/2}\Gamma \left( {s/2}\right) \zeta \left( s\right) = 2\left( {1 - s}\right) {\pi }^{-s/2}\Gamma \left( {s/2 + 1}\right) \zeta \left( s\right) .
\]\nSince the zeros of \( \zeta \left( s\right) \) for \( s = - {2k}, k \in {\mathbb{Z}}_{ \geq 1} \), are killed by the poles of \( \Gamma \left( {s/2 + 1}\right) \) and the pole of \( \zeta \left( s\right) \) is killed by \( 1 - s \), it follows that the zeros of \( f\left( s\right) \) are the nontrivial zeros of \( \zeta \left( s\right) \) . Thus for suitable constants \( {a}_{0} \) and \( {a}_{1} \)\nwe have\n\[
f\left( s\right) = {a}_{0}{e}^{{a}_{1}s}\mathop{\prod }\limits_{\rho }\left( {1 - \frac{s}{\rho }}\right) {e}^{s/\rho },
\]\nso that\n\[
\zeta \left( s\right) = \frac{{a}_{0}{e}^{bs}}{2\left( {1 - s}\right) \Gamma \left( {s/2 + 1}\right) }\mathop{\prod }\limits_{\rho }\left( {1 - \frac{s}{\rho }}\right) {e}^{s/\rho }
\]\nfor \( b = {a}_{1} + \log \left( \pi \right) /2 \) . We deduce that \( {a}_{0} = {2\zeta }\left( 0\right) = - 1 \), and by logarithmic differentiation that\n\[
\frac{{\zeta }^{\prime }\left( s\right) }{\zeta \left( s\right) } = b - \frac{1}{s - 1} - \frac{{\Gamma }^{\prime }\left( {s/2 + 1}\right) }{{2\Gamma }\left( {s/2 + 1}\right) } + \mathop{\sum }\limits_{\rho }\left( {\frac{1}{s - \rho } + \frac{1}{\rho }}\right) ,
\]\nso that\n\[
\frac{{\zeta }^{\prime }\left( 0\right) }{\zeta \left( 0\right) } = b + 1 - \frac{{\Gamma }^{\prime }\left( 1\right) }{\Gamma \left( 1\right) }
\]\nUsing \( {\zeta }^{\prime }\left( 0\right) = - \log \left( {2\pi }\right) /2 \) and \( {\Gamma }^{\prime }\left( 1\right) = - \gamma \) we obtain \( b = \log \left( {2\pi }\right) - 1 - \gamma /2 \) . | Yes |
Theorem 2. Let \( \chi \) be a nontrivial Dirichlet character modulo m. Then \( L\left( {1,\chi }\right) \neq 0 \) . | Having already proved that \( L\left( {1,\chi }\right) \neq 0 \) if \( \chi \) is complex we assume \( \chi \) is real.
Assume \( L\left( {1,\chi }\right) = 0 \) and consider the function
\[
\psi \left( s\right) = \frac{L\left( {s,\chi }\right) L\left( {s,{\chi }_{0}}\right) }{L\left( {{2s},{\chi }_{0}}\right) }
\]
The zero of \( L\left( {s,\chi }\right) \) at \( s = 1 \) cancels the simple pole of \( L\left( {s,{\chi }_{0}}\right) \) so the numerator is analytic on \( \sigma > 0 \) . The denominator is nonzero and analytic for \( \sigma > \frac{1}{2} \) . Thus \( \psi \left( s\right) \) is analytic on \( \sigma > \frac{1}{2} \) . Moreover, since \( L\left( {{2s},{\chi }_{0}}\right) \) has a pole at \( s = \frac{1}{2} \) we have \( \psi \left( s\right) \rightarrow 0 \) as \( s \rightarrow \frac{1}{2} \) .
We assume temporarily that \( s \) is real and \( s > 1 \) . Then \( \psi \left( s\right) \) has an infinite product expansion
\[
\psi \left( s\right) = \mathop{\prod }\limits_{p}{\left( 1 - \chi \left( p\right) {p}^{-s}\right) }^{-1}{\left( 1 - {\chi }_{0}\left( p\right) {p}^{-s}\right) }^{-1}\left( {1 - {\chi }_{0}\left( p\right) {p}^{-{2s}}}\right)
\]
\[
= \mathop{\prod }\limits_{{p \nmid m}}\frac{\left( 1 - {p}^{-{2s}}\right) }{\left( {1 - {p}^{-s}}\right) \left( {1 - \chi \left( p\right) {p}^{-s}}\right) }.
\]
If \( \chi \left( p\right) = - 1 \) the \( p \) -factor is equal to 1 . Thus
\[
\psi \left( s\right) = \mathop{\prod }\limits_{{\chi \left( p\right) = 1}}\frac{1 + {p}^{-s}}{1 - {p}^{-s}}
\]
where the product is over all \( p \) such that \( \chi \left( p\right) = 1 \) . Now,
\[
\frac{1 + {p}^{-s}}{1 - {p}^{-s}} = \left( {1 + {p}^{-s}}\right) \left( {\mathop{\sum }\limits_{{k = 0}}^{\infty }{p}^{-{ks}}}\right) = 1 + 2{p}^{-s} + 2{p}^{-{2s}} + \cdots + .
\]
Applying Lemma 3 we find that \( \psi \left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{n}^{-s} \) where \( {a}_{n} \geq 0 \) and the series converges for \( s > 1 \) . Note that \( {a}_{1} = 1 \) . (It is possible, but unnecessary to give an explicit formula for \( {a}_{n} \) ).
We once again consider \( \psi \left( s\right) \) as a function of a complex variable and expand it in a power series about \( s = 2,\psi \left( s\right) = \mathop{\sum }\limits_{{m = 0}}^{\infty }{b}_{m}{\left( s - 2\right) }^{m} \) . Since \( \psi \left( s\right) \) is analytic for \( \sigma > \frac{1}{2} \) the radius of convergence of this power series is at least \( \frac{3}{2} \) . To compute the \( {b}_{m} \) we use Taylor’s theorem, i.e., \( {b}_{m} = {\psi }^{\left( m\right) }\left( 2\right) /m \) ! where \( {\psi }^{\left( m\right) }\left( s\right) \) is the \( m \) th derivative of \( \psi \left( s\right) \) . Since \( \psi \left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{n}^{-s} \) we find \( {\psi }^{\left( m\right) }\left( 2\right) = \) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{\left( -\ln n\right) }^{m}{n}^{-2} = {\left( -1\right) }^{m}{c}_{m} \) with \( {c}_{m} \geq 0 \) . Thus \( \psi \left( s\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{m}{\left( 2 - s\right) }^{m} \) with \( {c}_{m} \) nonnegative and \( {c}_{0} = \psi \left( 2\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{n}^{-2} \geq {a}_{1} = \overline{1} \) . It follows that for real \( s \) in the interval \( \left( {\frac{1}{2},2}\right) \) we have \( \bar{\psi }\left( s\right) \geq 1 \) . This contradicts \( \psi \left( s\right) \rightarrow 0 \) as \( s \rightarrow \frac{1}{2} \), and so \( L\left( {1,\chi }\right) \neq 0 \) . | Yes |
Let \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \) be a set of generators for a finitely generated \( \mathbb{Z} \) -module \( M \), and let \( N \) be a submodule. \( \exists {\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m} \) in \( N \) with \( m \leq n \) such that \( N = \mathbb{Z}{\beta }_{1} + \mathbb{Z}{\beta }_{2} + \cdots + \mathbb{Z}{\beta }_{m} \) and \( {\beta }_{i} = \mathop{\sum }\limits_{{j \geq i}}{p}_{ij}{\alpha }_{j} \) with \( 1 \leq i \leq m \) and \( {p}_{ij} \in \mathbb{Z} \). | Proof. (a) We will proceed by induction on the number of generators of a \( \mathbb{Z} \) -module. This is trivial when \( n = 0 \). We can assume that we have proved the above statement to be true for all \( \mathbb{Z} \) -modules with \( n - 1 \) or fewer generators, and proceed to prove it for \( n \). We define \( {M}^{\prime } \) to be the submodule generated by \( {\alpha }_{2},{\alpha }_{3},\ldots ,{\alpha }_{n} \) over \( \mathbb{Z} \), and define \( {N}^{\prime } \) to be \( N \cap {M}^{\prime } \). Now, if \( n = 1 \), then \( {M}^{\prime } = 0 \) and there is nothing to prove. If \( N = {N}^{\prime } \), then the statement is true by our induction hypothesis. So we assume that \( N \neq {N}^{\prime } \) and consider \( A \), the set of all integers \( k \) such that \( \exists {k}_{2},{k}_{3},\ldots ,{k}_{n} \) with \( k{\alpha }_{1} + {k}_{2}{\alpha }_{2} + \cdots + {k}_{n}{\alpha }_{n} \in N \). Since \( N \) is a submodule, we deduce that \( A \) is a subgroup of \( \mathbb{Z} \). All additive subgroups of \( \mathbb{Z} \) are of the form \( m\mathbb{Z} \) for some integer \( m \), and so \( A = {k}_{11}\mathbb{Z} \) for some \( {k}_{11} \). Then let \( {\beta }_{1} = {k}_{11}{\alpha }_{1} + {k}_{12}{\alpha }_{2} + \cdots + {k}_{1n}{\alpha }_{n} \in N \). If we have some \( \alpha \in N \), then \( \alpha = \mathop{\sum }\limits_{{i = 1}}^{n}{h}_{i}{\alpha }_{i} \) with \( {h}_{i} \in \mathbb{Z} \) and \( {h}_{1} \in A \) so \( {h}_{1} = a{k}_{11} \). Therefore, \( \alpha - a{\beta }_{1} \in {N}^{\prime } \). By the induction hypothesis, there exist \( {\beta }_{i} = \mathop{\sum }\limits_{{j \geq i}}{k}_{ij}{\alpha }_{j} \) \( i = 2,3\ldots, m \), which generate \( {N}^{\prime } \) over \( \mathbb{Z} \) and which satisfy all the conditions above. It is clear that adding \( {\beta }_{1} \) to this list gives us a set of generators of \( N \). (b) Consider \( \alpha \), an arbitrary element of \( M \). Then \( \alpha = \sum {c}_{i}{\alpha }_{i} \). Recalling that \( {\beta }_{i} = \mathop{\sum }\limits_{{j \geq i}}{p}_{ij}{\alpha }_{j} \) we write \( {c}_{1} = {p}_{11}{q}_{1} + {r}_{1} \), with \( 0 \leq {r}_{1} < {p}_{11} \). Then \( \alpha - {q}_{1}{\beta }_{1} = \sum {c}_{i}^{\prime }{\alpha }_{i} \) where \( 0 \leq {c}_{1}^{\prime } < {p}_{11} \). Note that \( \alpha \equiv \alpha - {q}_{1}{\beta }_{1}\left( {\;\operatorname{mod}\;N}\right) \). Next we write \( {c}_{2}^{\prime } = {p}_{22}{q}_{2} + {r}_{2} \), where \( 0 \leq {r}_{2} < {p}_{22} \), and note that \( \alpha \equiv \alpha - {q}_{1}{\beta }_{1} - {q}_{2}{\beta }_{2}\;\left( {\;\operatorname{mod}\;N}\right) \). It is clear by induction that we can continue this process to arrive at an expression \( {\alpha }^{\prime } = \sum {k}_{i}{\alpha }_{i} \) with \( 0 \leq {k}_{i} < {p}_{ii} \) and \( \alpha \equiv {\alpha }^{\prime }\left( {\;\operatorname{mod}\;N}\right) \). It remains only to show that if we have \( \alpha = \sum {c}_{i}{\alpha }_{i} \) and \( \beta = \sum {d}_{i}{\alpha }_{i} \) where \( {c}_{i} \neq {d}_{i} \) for at least one \( i \) and \( 0 \leq {c}_{i},{d}_{i} < {p}_{ii} \), then \( \alpha \) and \( \beta \) are distinct mod \( N \). Suppose that this is not true, and that \( \sum {c}_{i}{\alpha }_{i} \equiv \sum {d}_{i}{\alpha }_{i}\;\left( {\;\operatorname{mod}\;N}\right) \) where \( {c}_{i} \neq {d}_{i} \) for at least one \( i \). Suppose \( {c}_{i} = {d}_{i} \) for \( i < r \) and \( {c}_{r} \neq {d}_{r} \). Then \( \sum \left( {{c}_{i} - {d}_{i}}\right) {\alpha }_{i} \in N \), so \( \mathop{\sum }\limits_{{i \geq r}}\left( {{c}_{i} - {d}_{i}}\right) {\alpha }_{i} = \mathop{\sum }\limits_{{i \geq r}}{k}_{i}{\beta }_{i} = \mathop{\sum }\limits_{{i \geq r}}{k}_{i}\left( {\mathop{\sum }\limits_{{j \geq i}}{p}_{ij}{\alpha }_{j}}\right) . \) Since \( {c}_{r},{d}_{r} \) are both less than \( {p}_{rr} \), we have \( {c}_{r} = {d}_{r} \), a contradiction. Thus, each coset in \( M/N \) has a unique representative \( \alpha = \sum {c}_{i}{\alpha }_{i} \) with \( 0 \leq {c}_{i} < {p}_{ii} \), and there are \( {p}_{11}{p}_{22}\cdots {p}_{nn} \) of them. So \( \left\lbrack {M : N}\right\rbrack = {p}_{11}{p}_{22}\cdots {p}_{nn} \) | Yes |
Let \( f : X \rightarrow Y \) be a map between two metric spaces. Suppose \( \omega : \lbrack 0,\infty ) \rightarrow \left\lbrack {0,\infty }\right\rbrack \) is a function such that \( d\left( {f\left( x\right), f\left( y\right) }\right) \leq \omega \left( {d\left( {x, y}\right) }\right) \) for every \( x, y \in X \), and \( \omega \left( s\right) \rightarrow 0 \) as \( s \rightarrow {0}^{ + } \) . Then \( f \) is uniformly continuous and \( \omega \geq {\omega }_{f} \). | We do (iii) and leave the other statements as an exercise. We need to show that for \( s > 0 \) there is \( {C}_{s} > 0 \) such that \( d\left( {f\left( x\right), f\left( y\right) }\right) \leq {C}_{s} \) whenever \( d\left( {x, y}\right) \leq s \) . From the definition of uniform continuity there exists \( {\delta }_{1} > 0 \) such that if \( 0 < \) \( d\left( {a, b}\right) < {\delta }_{1} \), then \( d\left( {f\left( a\right), f\left( b\right) }\right) < 1 \) . Let \( N = {N}_{s} \in \mathbb{N} \) be such that \( s/N < {\delta }_{1} \) . By the metric convexity of \( X \) one can find points \( x = {x}_{0},{x}_{1},\ldots ,{x}_{N} = y \) in \( X \) such that \( d\left( {{x}_{j},{x}_{j + 1}}\right) < d\left( {x, y}\right) /N < s/N < {\delta }_{1} \) for \( 0 \leq j \leq N - 1 \) . Therefore, by the triangle inequality,
\[
d\left( {f\left( x\right), f\left( y\right) }\right) \leq \mathop{\sum }\limits_{{j = 0}}^{{N - 1}}d\left( {f\left( {x}_{j}\right), f\left( {x}_{j + 1}\right) }\right) \leq N,
\]
and our claim holds with \( {C}_{s} = {N}_{s} \). | No |
Proposition 4.4. For all ideals \( \mathfrak{a},\mathfrak{b} \) of \( R \) and \( \mathfrak{A} \) of \( {S}^{-1}R \) : | The proofs make good exercises. | No |
Proposition 11.101. \( \mathfrak{B} \) is generated by \( {\mathfrak{B}}_{0} \) and any set of representatives for \( {W}^{\prime } \) in \( N \) . | Null | No |
If \( k \) is a finite field, then \( {k}^{ * } \) is cyclic. | Corollary 1.10. | No |
Proposition 1.1. Let \( \alpha : S \rightarrow T \) be a morphism of algebraic sets. If \( A \) is an irreducible subset of \( S \), then \( \alpha \left( A\right) \) is an irreducible subset of \( T \) . | This follows direcly from the definition of irreducibility; using only the continuity of \( \alpha \) . | No |
If \( \mu \) and \( \lambda \) are dominant, then \( \lambda \) belongs to \( \operatorname{Conv}\left( {W \cdot \mu }\right) \) if and only if \( \lambda \preccurlyeq \mu \) . | Since \( \operatorname{Conv}\left( {W \cdot \mu }\right) \) is convex and Weyl invariant, we see that if \( \lambda \) belongs to \( \operatorname{Conv}\left( {W \cdot \mu }\right) \), then every point in \( \operatorname{Conv}\left( {W \cdot \lambda }\right) \) also belongs to \( \operatorname{Conv}\left( {W \cdot \mu }\right) \) . Thus, Point 1 of the proposition may be restated as follows:\nIf \( \mu \) and \( \lambda \) are dominant, then \( \lambda \preccurlyeq \mu \) if and only if\n\[
\operatorname{Conv}\left( {W \cdot \lambda }\right) \subset \operatorname{Conv}\left( {W \cdot \mu }\right)
\] | Yes |
Let \( S \) be a closed subset of a Banach space \( X \) and let \( {W}_{0} \in \mathcal{S}\left( X\right) \). Then there exists \( W \in \mathcal{S}\left( X\right) \) containing \( {W}_{0} \) such that \( d\left( {x, S}\right) = d\left( {x, S \cap W}\right) \) for all \( x \in W \). | Starting with \( {W}_{0} \), we define inductively an increasing sequence \( {\left( {W}_{n}\right) }_{n \geq 1} \) of \( \mathcal{S}\left( X\right) \) such that \( d\left( {\cdot, S}\right) = d\left( {\cdot, S \cap {W}_{n}}\right) \) on \( {W}_{n - 1} \). Assuming that \( {W}_{1},\ldots ,{W}_{n} \) satisfying this property have been defined, in order to define \( {W}_{n + 1} \) we take a countable dense subset \( {D}_{n} \mathrel{\text{:=}} \left\{ {{w}_{p} : p \in \mathbb{N}}\right\} \) of \( {W}_{n} \); here, to avoid heavy notation we do not write the dependence on \( n \) of the elements of \( {D}_{n} \). For all \( p \in \mathbb{N} \) we pick sequences \( {\left( {x}_{k, p}\right) }_{k \geq 0} \) of \( S \) such that \( d\left( {{w}_{p},{x}_{k, p}}\right) \leq d\left( {{w}_{p}, S}\right) + {2}^{-k} \) for all \( k, p \in \mathbb{N} \). Let \( {W}_{n + 1} \) be the closure of the linear span of the set \( \left\{ {{x}_{k, p} : \left( {k, p}\right) \in {\mathbb{N}}^{2}}\right\} \cup {D}_{n} \). Then \( {W}_{n} \subset {W}_{n + 1},{W}_{n + 1} \in \mathcal{S}\left( X\right) \), and for all \( {w}_{p} \in {D}_{n} \) we have \( d\left( {{w}_{p}, S}\right) = \mathop{\inf }\limits_{k}d\left( {{w}_{p},{x}_{k, p}}\right) = d\left( {{w}_{p}, S \cap {W}_{n + 1}}\right) \). Since \( d\left( {\cdot, S}\right) \) and \( d\left( {\cdot, S \cap {W}_{n + 1}}\right) \) are continuous and since \( {D}_{n} \) is dense in \( {W}_{n} \), these two functions coincide on \( {W}_{n} \). Thus, our inductive construction is achieved.
Finally, we take for \( W \) the closure of the space spanned by the union of the family \( \left( {W}_{n}\right) \) and we use the inequalities \( d\left( {x, S}\right) = d\left( {x, S \cap {W}_{n + 1}}\right) \geq d\left( {x, S \cap W}\right) \geq d\left( {x, S}\right) \) for \( x \in {W}_{n} \), and a density argument as above extends the relation \( d\left( {\cdot, S}\right) = d\left( {\cdot, S \cap W}\right) \) from the union of the \( {W}_{n} \)'s to \( W \). | Yes |
Let \( F/K \) be a function field whose full constant field is \( K \). Suppose that \( {F}^{\prime } = {F}_{1}{F}_{2} \) is the compositum of two finite separable extensions \( {F}_{1}/F \) and \( {F}_{2}/F \). Assume that there exists a place \( P \in {\mathbb{P}}_{F} \) of degree one which splits completely in \( {F}_{1}/F \) and in \( {F}_{2}/F \). Then \( P \) splits completely in \( {F}^{\prime }/F \), and \( K \) is the full constant field of \( {F}^{\prime } \). | We only have to show that \( K \) is the full constant field of \( {F}^{\prime } = {F}_{1}{F}_{2} \); the remaining assertions follow immediately from Proposition 3.9.6. We choose a place \( {P}^{\prime } \) of \( {F}^{\prime } \) lying above \( P \), then \( f\left( {{P}^{\prime } \mid P}\right) = 1 \) and therefore the residue class field \( {F}_{{P}^{\prime }}^{\prime } \) of \( {P}^{\prime } \) is equal to the residue class field \( {F}_{P} = K \) of \( P \). Since the full constant field \( {K}^{\prime } \) of \( {F}^{\prime } \) satisfies \( K \subseteq {K}^{\prime } \subseteq {F}_{{P}^{\prime }}^{\prime } \), we conclude that \( {K}^{\prime } = K \). | Yes |
Corollary 10.3.13. We have\\n\\n{\\mathcal{H}}_{2}\\left( \\tau \\right) = \\frac{{5\\theta }\\left( \\tau \\right) {\\theta }^{4}\\left( {\\tau + 1/2}\\right) - {\\theta }^{5}\\left( \\tau \\right) }{480},\\n\\n{\\mathcal{H}}_{3}\\left( \\tau \\right) = - \\frac{7{\\theta }^{3}\\left( \\tau \\right) {\\theta }^{4}\\left( {\\tau + 1/2}\\right) + {\\theta }^{7}\\left( \\tau \\right) }{2016},\\n\\n{\\mathcal{H}}_{4}\\left( \\tau \\right) = \\frac{\\theta \\left( \\tau \\right) {\\theta }^{8}\\left( {\\tau + 1/2}\\right) + {14}{\\theta }^{5}\\left( \\tau \\right) {\\theta }^{4}\\left( {\\tau + 1/2}\\right) + {\\theta }^{9}\\left( \\tau \\right) }{3840}. | Null | Yes |
Theorem 5.4.4 The classification space \( \operatorname{irr}\left( n\right) / \sim \) is standard Borel. | Fix any irreducible \( A \) . Then the \( \sim \) -equivalence class \( \left\lbrack A\right\rbrack \) containing \( A \) equals
\[
{\pi }_{1}\left\{ {\left( {B, U}\right) \in \operatorname{irr}\left( n\right) \times U\left( n\right) : A = {UB}{U}^{ * }}\right\} ,
\]
where \( {\pi }_{1} : \operatorname{irr}\left( n\right) \times U\left( n\right) \rightarrow \operatorname{irr}\left( n\right) \) is the projection map to the first coordinate space. (Recall that \( U\left( n\right) \) denotes the set of all \( n \times n \) unitary matrices.) As the set
\[
\left\{ {\left( {B, U}\right) \in \operatorname{irr}\left( n\right) \times U\left( n\right) : A = {UB}{U}^{ * }}\right\}
\]
is closed and \( U\left( n\right) \) compact, \( \left\lbrack A\right\rbrack \) is closed by 2.3.24.
Now let \( \mathcal{O} \) be any open set in \( \operatorname{irr}\left( n\right) \) . Its saturation is
\[
\mathop{\bigcup }\limits_{{U \in U\left( n\right) }}\left\{ {A \in \operatorname{irr}\left( n\right) : {UA}{U}^{ * } \in \mathcal{O}}\right\}
\]
which is open. Thus \( \sim \) is a lower-semicontinuous partition of \( \operatorname{irr}\left( n\right) \) into closed sets. By 5.4.3, let \( C \) be a Borel cross section of \( \sim \) . Then \( q \mid C \) is a one-to-one Borel map from \( C \) onto \( \operatorname{irr}\left( n\right) / \sim \), where \( q : \operatorname{irr}\left( n\right) \rightarrow \operatorname{irr}\left( n\right) / \sim \) is the canonical quotient map. By the Borel isomorphism theorem (3.3.13), \( q \) is a Borel isomorphism, and our result is proved. | Yes |
Let \( {z}_{0} = {y}_{0}\mathrm{i} \) and \( {z}_{1} = {y}_{1}\mathrm{i} \) with \( 0 < {y}_{0} < {y}_{1} \) . Then | It is readily checked that the path \( \phi \) defined in the lemma has constant speed equal to \( \log {y}_{1} - \log {y}_{0} = \mathrm{L}\left( \phi \right) \) as claimed. It follows that
\[
\mathrm{d}\left( {{z}_{0},{z}_{1}}\right) \leq \log {y}_{1} - \log {y}_{0}
\]
Suppose now that \( \eta : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{H} \) is another path joining \( {z}_{0} \) to \( {z}_{1} \), and write \( \eta \left( t\right) = {\eta }_{x}\left( t\right) + \mathrm{i}{\eta }_{y}\left( t\right) \) with \( {\eta }_{x}\left( t\right) ,{\eta }_{y}\left( t\right) \in \mathbb{R} \) . Then
\[
\mathrm{L}\left( \eta \right) = {\int }_{0}^{1}\frac{{\begin{Vmatrix}{\eta }^{\prime }\left( t\right) \end{Vmatrix}}_{2}}{{\eta }_{y}\left( t\right) }\mathrm{d}t \geq {\int }_{0}^{1}\frac{\left| {\eta }_{y}^{\prime }\left( t\right) \right| }{{\eta }_{y}\left( t\right) }\mathrm{d}t \geq {\int }_{0}^{1}\frac{{\eta }_{y}^{\prime }\left( t\right) }{{\eta }_{y}\left( t\right) }\mathrm{d}t = \log {y}_{1} - \log {y}_{0}.
\]
Equality holds in the first inequality if and only if \( {\eta }_{x}^{\prime }\left( t\right) = {\eta }_{x}\left( t\right) = 0 \) for all \( t \in \left\lbrack {0,1}\right\rbrack \), and in the second if and only if \( {\eta }_{y}^{\prime }\left( t\right) \geq 0 \) for all \( t \in \left\lbrack {0,1}\right\rbrack \) . This implies the remaining statement of the lemma. | Yes |
Suppose \( X \) is \( F \) -smooth. If \( f \) is locally Lipschitzian and \( q \) is a Lyapunov function for \( S \), then \( S \) is stable. | In order to show the stability of \( S \) in both cases, it suffices to prove that for all \( z \in X \), the function \( t \mapsto {e}^{ct}q\left( {{x}_{z}\left( t\right) }\right) + {\int }_{0}^{t}{e}^{cs}p\left( {{x}_{z}\left( s\right) }\right) \mathrm{d}s \) is nonincreasing on \( {\mathbb{R}}_{ + } \) . Since \( {x}_{z} \) and \( p \) are continuous, the last term is differentiable. To study the first one, let us use a special case of the Leibniz rule for \( {\partial }_{F} \) : if \( g, h : W \rightarrow \mathbb{R} \) are two lower semicontinuous functions on an open subset \( W \) of a normed space, \( h \) being positive and differentiable, then
\[
\forall x \in W,\;\partial \left( {gh}\right) \left( x\right) = h\left( x\right) \partial g\left( x\right) + g\left( x\right) {h}^{\prime }\left( x\right) .
\]
Taking \( W \mathrel{\text{:=}} \left( {0, + \infty }\right), g\left( t\right) \mathrel{\text{:=}} q\left( {{x}_{z}\left( t\right) }\right), h\left( t\right) \mathrel{\text{:=}} {e}^{ct} \) and using Corollary 4.98, we are led to check that (L1) implies that for all \( t \in {\mathbb{R}}_{ + },{t}^{ * } \in \partial g\left( t\right) \) one has
\[
{e}^{ct}{t}^{ * } + c{e}^{ct}q\left( {{x}_{z}\left( t\right) }\right) + {e}^{ct}p\left( {{x}_{z}\left( t\right) }\right) \leq 0,
\]
(4.78)
\( \mathbb{R} \) being identified with its dual. Since \( {t}^{ * } \in {\partial }_{F}\left( {q \circ {x}_{z}}\right) \left( t\right) \) and since \( q \) is inf-compact on the image under \( {x}_{z} \) of a compact interval, Theorem 4.70 yields sequences \( \left( {t}_{n}\right) \rightarrow t,\left( {t}_{n}^{ * }\right) \rightarrow {t}^{ * },\left( {y}_{n}\right) { \rightarrow }_{q}{x}_{z}\left( t\right) ,\left( {y}_{n}^{ * }\right) ,\left( {v}_{n}^{ * }\right) \) such that \( \left( \begin{Vmatrix}{{y}_{n}^{ * } - {v}_{n}^{ * }}\end{Vmatrix}\right) \rightarrow 0 \) , \( \left( {\begin{Vmatrix}{v}_{n}^{ * }\end{Vmatrix} \cdot \begin{Vmatrix}{{x}_{z}\left( {t}_{n}\right) - {y}_{n}}\end{Vmatrix}}\right) \rightarrow 0 \), with \( {y}_{n}^{ * } \in \partial q\left( {y}_{n}\right) ,{t}_{n}^{ * } \in {D}_{F}^{ * }{x}_{z}\left( {t}_{n}\right) \left( {v}_{n}^{ * }\right) \) for all \( n \in \mathbb{N} \) . The last relation means that \( {t}_{n}^{ * } = \left\langle {{v}_{n}^{ * },{x}_{z}^{\prime }\left( {t}_{n}\right) }\right\rangle = \left\langle {{v}_{n}^{ * }, f\left( {{x}_{z}\left( {t}_{n}\right) }\right) }\right\rangle \) . Since \( f \) is locally Lipschitzian and \( \left( {{x}_{z}\left( {t}_{n}\right) }\right) \rightarrow {x}_{z}\left( t\right) ,\left( {\begin{Vmatrix}{v}_{n}^{ * }\end{Vmatrix} \cdot \begin{Vmatrix}{{x}_{z}\left( {t}_{n}\right) - {y}_{n}}\end{Vmatrix}}\right) \rightarrow 0 \), from (L1) we get
\[
{t}^{ * } = \mathop{\lim }\limits_{n}\left\langle {{v}_{n}^{ * }, f\left( {y}_{n}\right) }\right\rangle = \mathop{\lim }\limits_{n}\left\langle {{y}_{n}^{ * }, f\left( {y}_{n}\right) }\right\rangle \leq \mathop{\limsup }\limits_{n}\left( {-p\left( {y}_{n}\right) - {cq}\left( {y}_{n}\right) }\right) .
\]
Since \( p \) and \( q \) are lower semicontinuous, we obtain \( {t}^{ * } \leq - p\left( {{x}_{z}\left( t\right) }\right) - {cq}\left( {{x}_{z}\left( t\right) }\right) \) and (4.78), and hence the stability of \( S \) . | Yes |
Proposition 7.23. Let \[
\begin{matrix} \mathrm{X} = \left( {X,\mathcal{B},\mu, T}\right) \\ \downarrow \\ \mathrm{Y} = \left( {Y,\mathcal{A},\nu, S}\right) \end{matrix}
\] be a compact extension of invertible measure-preserving systems on Borel probability spaces. If Y is SZ, then so is X. | Proof of Proposition 7.23 using van der Waerden. By Lemma 7.24, we may assume that f = χB is AP and that there exists some set A ∈ A of positive measure with μy(A)(B) > 1/2μ(B) for y ∈ A. We will use SZ for A (for arithmetic progressions of quite large length K) to show SZ for B (for arithmetic progressions of length k). Given ε = μ(B)/6(k + 1) > 0 we may find (using the AP property of f) functions gs such that ... | Yes |
If \( F/\mathbb{Q} \) is an extension in which no finite prime ramifies, then \( F = \mathbb{Q} \). | A theorem of Minkowski (see Exercise 2.5) states that every ideal class of \( F \) contains an integral ideal of norm less than or equal to
\[
\frac{n!}{{n}^{n}}{\left( \frac{4}{\pi }\right) }^{{r}_{2}}\sqrt{{d}_{F}}
\]
where \( n = \left\lbrack {F : \mathbb{Q}}\right\rbrack ,{d}_{F} \) is the absolute value of the discriminant, and \( {r}_{2} \leq n/2 \) is the number of complex places. In particular, this quantity must be at least 1, so
\[
\sqrt{{d}_{F}} \geq \frac{{n}^{n}}{n!}{\left( \frac{\pi }{4}\right) }^{{r}_{2}} \geq \frac{{n}^{n}}{n!}{\left( \frac{n}{4}\right) }^{n/2}\overset{\text{ def }}{ = }{b}_{n}.
\]
Since \( {b}_{2} > 1 \) and
\[
\frac{{b}_{n + 1}}{{b}_{n}} = {\left( 1 + \frac{1}{n}\right) }^{n}\sqrt{\frac{\pi }{4}} \geq 2\sqrt{\frac{\pi }{4}} > 1
\]
we must have, if \( n \geq 2 \),
\[
{d}_{F} > 1\text{.}
\]
Consequently there exists a prime \( p \) dividing \( {d}_{F} \), which means \( p \) ramifies. This proves Lemma 14.3. | No |
Theorem 3.1.1. \( {W}_{G} \) is a finite group and the representation of \( {W}_{G} \) on \( {\mathfrak{h}}^{ * } \) is faithful. | Proof. Let \( s \in {\operatorname{Norm}}_{G}\left( H\right) \) . Suppose \( s \cdot \theta = \theta \) for all \( \theta \in \mathfrak{X}\left( H\right) \) . Then \( {s}^{-1}{hs} = h \) for all \( h \in H \), and hence \( s \in H \) by Theorem 2.1.5. This proves that the representation of \( {W}_{G} \) on \( {\mathfrak{h}}^{ * } \) is faithful.
To prove the finiteness of \( {W}_{G} \), we shall assume that \( G \subset \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) is in the matrix form of Section 2.4.1, so that \( H \) is the group of diagonal matrices in \( G \) . In the proof of Theorem 2.1.5 we noted that \( h \in H \) acts on the standard basis for \( {\mathbb{C}}^{n} \) by
\[
h{e}_{i} = {\theta }_{i}\left( h\right) {e}_{i}\text{ for }i = 1,\ldots, n,
\]
where \( {\theta }_{i} \in \mathcal{X}\left( H\right) \) and \( {\theta }_{i} \neq {\theta }_{j} \) for \( i \neq j \) . Let \( s \in {\operatorname{Norm}}_{G}\left( H\right) \) . Then
\[
{hs}{e}_{i} = s\left( {{s}^{-1}{hs}}\right) {e}_{i} = s{\theta }_{i}\left( {{s}^{-1}{hs}}\right) {e}_{i} = \left( {s \cdot {\theta }_{i}}\right) \left( h\right) s{e}_{i}.
\]
(3.1)
Hence \( s{e}_{i} \) is an eigenvector for \( h \) with eigenvalue \( \left( {s \cdot {\theta }_{i}}\right) \left( h\right) \) . Since the characters \( s \cdot {\theta }_{1},\ldots, s \cdot {\theta }_{n} \) are all distinct, this implies that there is a permutation \( \sigma \in {\mathfrak{S}}_{n} \) and there are scalars \( {\lambda }_{i} \in {\mathbb{C}}^{ \times } \) such that
\[
s{e}_{i} = {\lambda }_{i}{e}_{\sigma \left( i\right) }\;\text{ for }i = 1,\ldots, n.
\]
(3.2)
| Yes |
Lemma 5.1.3 With \( \Gamma \) as described in Theorem 5.1.2 satisfying conditions 1 and 2, condition 3 is equivalent to the following requirement: \( {3}^{\prime } \) . All embeddings \( \sigma \), apart from the identity and \( \mathbf{c} \), complex conjugation, are real and \( {A\Gamma } \) is ramified at all real places. | If condition \( {\mathcal{3}}^{\prime } \) holds and \( \sigma : {k\Gamma } \rightarrow \mathbb{R} \), then there exists \( \tau : {A\Gamma } \rightarrow \) \( \mathcal{H} \), Hamilton’s quaternions, such that \( \sigma \left( {\operatorname{tr}f}\right) = \operatorname{tr}\left( {\tau \left( f\right) }\right) \) for each \( f \in {\Gamma }^{\left( 2\right) } \) . Since \( \det \left( f\right) = 1,\tau \left( f\right) \in {\mathcal{H}}^{1} \), so that \( \operatorname{tr}\left( {\tau \left( f\right) }\right) \in \left\lbrack {-2,2}\right\rbrack \) .\n\nConversely, suppose condition 3 holds and \( \sigma : {k\Gamma } \rightarrow \mathbb{C} \) . Let \( f \in {\Gamma }^{\left( 2\right) } \) have eigenvalues \( \lambda \) and \( {\lambda }^{-1} \) and \( \mu \) be an extension of \( \sigma \) to \( {k\Gamma }\left( \lambda \right) \) . Then \( \sigma \left( {\operatorname{tr}{f}^{n}}\right) = \mu {\left( \lambda \right) }^{n} + \mu {\left( \lambda \right) }^{-n} \) . Thus\n\[
\left| {\sigma \left( {\operatorname{tr}{f}^{n}}\right) }\right| \geq \left| \right| \mu \left( \lambda \right) \left| {{}^{n} - }\right| \mu \left( \lambda \right) \left| {}^{-n}\right| .
\]\nSo, if \( \sigma \left( {\operatorname{tr}{f}^{n}}\right) \) is bounded, then \( \left| {\mu \left( \lambda \right) }\right| = 1 \) so that \( \sigma \left( {\operatorname{tr}f}\right) = \mu \left( \lambda \right) + \mu {\left( \lambda \right) }^{-1} \) is a real number in the interval \( \left\lbrack {-2,2}\right\rbrack \) . Now choose an irreducible subgroup \( \left\langle {{g}_{1},{g}_{2}}\right\rangle \) of \( {\Gamma }^{\left( 2\right) } \) such that \( {g}_{1} \) is not parabolic. Then\n\[
{A\Gamma } \cong \left( \frac{{\operatorname{tr}}^{2}{g}_{1}\left( {\operatorname{tr}{}^{2}{g}_{1} - 4}\right) ,\operatorname{tr}\left\lbrack {{g}_{1},{g}_{2}}\right\rbrack - 2}{k\Gamma }\right)
\]\nby (3.38). Since \( \sigma \left( {\operatorname{tr}f}\right) \in \left\lbrack {-2,2}\right\rbrack \) for all \( f \), it follows that \( {A\Gamma } \) is ramified at all real places (see Theorem 2.5.1). | Yes |
An almost greedy basis that is not greedy. | Aside from being quasi-greedy, the basis \( \mathcal{B} = {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) in Example 10.2.9 is democratic. Indeed, if \( \left| A\right| = m \), then
\[
{\left( \mathop{\sum }\limits_{{n \in A}}1\right) }^{1/2} = {m}^{1/2}
\]
while
\[
\mathop{\sum }\limits_{{n \in A}}\frac{1}{\sqrt{n}} \leq \mathop{\sum }\limits_{{n = 1}}^{m}\frac{1}{\sqrt{n}} \leq {\int }_{0}^{m}\frac{dx}{\sqrt{x}} = 2{m}^{1/2}.
\]
Hence,
\[
{m}^{1/2} \leq \begin{Vmatrix}{\mathop{\sum }\limits_{{n \in A}}{e}_{n}}\end{Vmatrix} \leq 2{m}^{1/2}
\]
Thus \( \mathcal{B} \) is almost greedy. It cannot be greedy, because it is not unconditional. | No |
Theorem 10.71 (Substitutivity of equivalence). Let \( \varphi ,\psi ,\chi \) be formulas and \( \alpha \in {}^{m}\operatorname{Rng}v \) . Suppose that if \( \beta \) occurs free in \( \varphi \) or in \( \psi \) but bound in \( \chi \) then \( \beta \in \left\{ {{\alpha }_{i} : i < m}\right\} \) . Let \( \theta \) be obtained from \( \chi \) by replacing zero or more occurrences of \( \varphi \) in \( \chi \) by \( \psi \) . Then
\[
\vdash \forall {\alpha }_{0}\cdots \forall {\alpha }_{m - 1}\left( {\varphi \leftrightarrow \psi }\right) \rightarrow \left( {\chi \leftrightarrow \theta }\right) .
\] | Proof. We proceed by induction on \( \chi \) . We may assume that \( \theta \neq \chi \) . If \( \chi \) is atomic, then \( \chi = \varphi \) and \( \psi = \theta \) ; this case is trivial. Suppose \( \chi \) is \( \neg {\chi }^{\prime } \) . Then \( \theta \) is of the form \( \neg {\theta }^{\prime } \), and the induction hypothesis easily gives the desired result. The induction steps involving \( \mathbf{v} \) and \( \land \) are similar. Now suppose \( \chi = \forall \beta {\chi }^{\prime } \) . Then by the induction hypothesis,
\[
\vdash \forall {\alpha }_{0}\cdots \forall {\alpha }_{m - 1}\left( {\varphi \leftrightarrow \psi }\right) \rightarrow \left( {{\chi }^{\prime } \leftrightarrow {\theta }^{\prime }}\right) .
\]
Note that \( \beta \) does not occur free in \( \forall {\alpha }_{0}\cdots \forall {\alpha }_{m - 1}\left( {\varphi \leftrightarrow \psi }\right) \) . Hence, using 10.61, we easily obtain
\[
\vdash \forall {\alpha }_{0}\cdots \forall {\alpha }_{m - 1}\left( {\varphi \leftrightarrow \psi }\right) \rightarrow \left( {\forall \beta {\chi }^{\prime } \leftrightarrow \forall \beta {\theta }^{\prime }}\right) ,
\]
as desired.
Again note that implicit in 10.71 is the assertion that the expression \( \theta \) formed from \( \chi \) is again a formula; this is easily established. | Yes |
Proposition 2.92. Let \( X \) and \( Z \) be Banach spaces, \( Z \) being finite-dimensional, let \( W \) be an open subset of \( X \), and let \( g : W \rightarrow Z \) be Hadamard differentiable at \( a \in W \) , with \( \operatorname{Dg}\left( a\right) \left( X\right) = Z \) . Then there exist open neighborhoods \( U \) of a in \( W, V \) of \( g\left( a\right) \) in \( Z \) and a map \( h : V \rightarrow U \) that is differentiable at \( g\left( a\right) \) and such that \( h\left( {g\left( a\right) }\right) = a \) , \( g \circ h = {I}_{V} \) . In particular, \( g \) is open at a. | Null | No |
one about a system with \( n - 1 \) variables. Namely, we determine necessary and sufficient conditions for which, given a vector \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \in {\mathbb{R}}^{n - 1} \), there exists \( {\bar{x}}_{n} \in \mathbb{R} \) such that \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n}}\right) \) satisfies \( {Ax} \leq b \) . Let \( I \mathrel{\text{:=}} \{ 1,\ldots, m\} \) and define \({I}^{ + } \mathrel{\text{:=}} \left\{ {i \in I : {a}_{in} > 0}\right\} ,\;{I}^{ - } \mathrel{\text{:=}} \left\{ {i \in I : {a}_{in} < 0}\right\} ,\;{I}^{0} \mathrel{\text{:=}} \left\{ {i \in I : {a}_{in} = 0}\right\} .\) Dividing the \( i \) th row by \( \left| {a}_{in}\right| \) for each \( i \in {I}^{ + } \cup {I}^{ - } \), we obtain the following system, which is equivalent to \( {Ax} \leq b \) : \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{x}_{j}\; + {x}_{n}\; \leq {b}_{i}^{\prime },\;i \in {I}^{ + }\) \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{x}_{j}\; - {x}_{n}\; \leq {b}_{i}^{\prime },\;i \in {I}^{ - }\) (3.1) \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}{x}_{j}\; \leq {b}_{i},\;i \in {I}^{0}\) where \( {a}_{ij}^{\prime } = {a}_{ij}/\left| {a}_{in}\right| \) and \( {b}_{i}^{\prime } = {b}_{i}/\left| {a}_{in}\right| \) for \( i \in {I}^{ + } \cup {I}^{ - } \). For each pair \( i \in {I}^{ + } \) and \( k \in {I}^{ - } \), we sum the two inequalities indexed by \( i \) and \( k \), and we add the resulting inequality to the system (3.1). Furthermore, we remove the inequalities indexed by \( {I}^{ + } \) and \( {I}^{ - } \) . This way, we obtain the following system: \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}\left( {{a}_{ij}^{\prime } + {a}_{kj}^{\prime }}\right) {x}_{j} \leq {b}_{i}^{\prime } + {b}_{k}^{\prime },\;i \in {I}^{ + }, k \in {I}^{ - },\) (3.2) \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}{x}_{j} \leq {b}_{i},\;i \in {I}^{0}.\) If \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1},{\bar{x}}_{n}}\right) \) satisfies \( {Ax} \leq b \), then \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfies (3.2). The next theorem states that the converse also holds. Theorem 3.1. A vector \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfies the system (3.2) if and only if there exists \( {\bar{x}}_{n} \) such that \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1},{\bar{x}}_{n}}\right) \) satisfies \( {Ax} \leq b \). Proof. We already remarked the "if" statement. For the converse, assume there is a vector \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfying (3.2). Note that the first set of inequalities in (3.2) can be rewritten as \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{kj}^{\prime }{x}_{j} - {b}_{k}^{\prime } \leq {b}_{i}^{\prime } - \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{x}_{j},\;i \in {I}^{ + }, k \in {I}^{ - }.\) (3.3) Let \( l : = \mathop{\max }\limits_{{k \in {I}^{ - }}}\{ \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{kj}^{\prime }{\bar{x}}_{j} - {b}_{k}^{\prime }\} \) and \( u : = \mathop{\min }\limits_{{i \in {I}^{ + }}}\{ {b}_{i}^{\prime } - \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{\bar{x}}_{j}\} , \) where we define \( l \mathrel{\text{:=}} - \infty \) if \( {I}^{ - } = \varnothing \) and \( u \mathrel{\text{:=}} + \infty \) if \( {I}^{ + } = \varnothing \). Since \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfies (3.3), we have that \( l \leq u \). Therefore, for any \( {\bar{x}}_{n} \) such that \( l \leq {\bar{x}}_{n} \leq u \), | o one about a system with \( n - 1 \) variables. Namely, we determine necessary and sufficient conditions for which, given a vector \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \in {\mathbb{R}}^{n - 1} \), there exists \( {\bar{x}}_{n} \in \mathbb{R} \) such that \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n}}\right) \) satisfies \( {Ax} \leq b \) . Let \( I \mathrel{\text{:=}} \{ 1,\ldots, m\} \) and define \({I}^{ + } \mathrel{\text{:=}} \left\{ {i \in I : {a}_{in} > 0}\right\} ,\;{I}^{ - } \mathrel{\text{:=}} \left\{ {i \in I : {a}_{in} < 0}\right\} ,\;{I}^{0} \mathrel{\text{:=}} \left\{ {i \in I : {a}_{in} = 0}\right\} .\) Dividing the \( i \) th row by \( \left| {a}_{in}\right| \) for each \( i \in {I}^{ + } \cup {I}^{ - } \), we obtain the following system, which is equivalent to \( {Ax} \leq b \) : \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{x}_{j}\; + {x}_{n}\; \leq {b}_{i}^{\prime },\;i \in {I}^{ + }\) \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{x}_{j}\; - {x}_{n}\; \leq {b}_{i}^{\prime },\;i \in {I}^{ - }\) (3.1) \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}{x}_{j}\; \leq {b}_{i},\;i \in {I}^{0}\) where \( {a}_{ij}^{\prime } = {a}_{ij}/\left| {a}_{in}\right| \) and \( {b}_{i}^{\prime } = {b}_{i}/\left| {a}_{in}\right| \) for \( i \in {I}^{ + } \cup {I}^{ - } \). For each pair \( i \in {I}^{ + } \) and \( k \in {I}^{ - } \), we sum the two inequalities indexed by \( i \) and \( k \), and we add the resulting inequality to the system (3.1). Furthermore, we remove the inequalities indexed by \( {I}^{ + } \) and \( {I}^{ - } \) . This way, we obtain the following system: \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}\left( {{a}_{ij}^{\prime } + {a}_{kj}^{\prime }}\right) {x}_{j} \leq {b}_{i}^{\prime } + {b}_{k}^{\prime },\;i \in {I}^{ + }, k \in {I}^{ - },\) (3.2) \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}{x}_{j} \leq {b}_{i},\;i \in {I}^{0}.\) If \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1},{\bar{x}}_{n}}\right) \) satisfies \( {Ax} \leq b \), then \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfies (3.2). The next theorem states that the converse also holds. Theorem 3.1. A vector \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfies the system (3.2) if and only if there exists \( {\bar{x}}_{n} \) such that \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1},{\bar{x}}_{n}}\right) \) satisfies \( {Ax} \leq b \). Proof. We already remarked the "if" statement. For the converse, assume there is a vector \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfying (3.2). Note that the first set of inequalities in (3.2) can be rewritten as \(\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{kj}^{\prime }{x}_{j} - {b}_{k}^{\prime } \leq {b}_{i}^{\prime } - \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{x}_{j},\;i \in {I}^{ + }, k \in {I}^{ - }.\) (3.3) Let \( l : = \mathop{\max }\limits_{{k \in {I}^{ - }}}\{ \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{kj}^{\prime }{\bar{x}}_{j} - {b}_{k}^{\prime }\} \) and \( u : = \mathop{\min }\limits_{{i \in {I}^{ + }}}\{ {b}_{i}^{\prime } - \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{a}_{ij}^{\prime }{\bar{x}}_{j}\} , \) where we define \( l \mathrel{\text{:=}} - \infty \) if \( {I}^{ - } = \varnothing \) and \( u \mathrel{\text{:=}} + \infty \) if \( {I}^{ + } = \varnothing \). Since \( \left( {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n - 1}}\right) \) satisfies (3.3), we have that \( l \leq u \). Therefore, for any \( {\bar{x}}_{n} \) such that \( l \leq {\bar{x}}_{n} \leq u \), the | Yes |
Theorem 11.1.1. Let \( G \) be a linear algebraic group. For every \( g \in G \) the map \( A \mapsto {\left( {X}_{A}\right) }_{g} \) is a linear isomorphism from \( \operatorname{Lie}\left( G\right) \) onto \( T{\left( G\right) }_{g} \) . Hence \( G \) is a smooth algebraic set and \( \dim \operatorname{Lie}\left( G\right) = \dim G \) . | Proof. We first show that for fixed \( g \in G \), the map \( A \mapsto {\left( {X}_{A}\right) }_{g} \) is injective from \( \operatorname{Lie}\left( G\right) \) to \( T{\left( G\right) }_{g} \) . Suppose \( {\left( {X}_{A}\right) }_{g} = 0 \) . Then for \( x \in G \) and \( f \in \mathcal{O}\left\lbrack {\mathbf{{GL}}\left( V\right) }\right\rbrack \) we have\\ \[
\left( {{X}_{A}f}\right) \left( x\right) = \left( {{X}_{A}f}\right) \left( {x{g}^{-1}g}\right) = \left( {L\left( {g{x}^{-1}}\right) \left( {{X}_{A}f}\right) }\right) \left( g\right)
\]
\[
= \left( {{X}_{A}\left( {L\left( {g{x}^{-1}}\right) f}\right) }\right) \left( g\right) = 0.
\]
Here we have used the left invariance of the vector field \( {X}_{A} \) on the second line. This shows that \( {X}_{A}f \in {\mathcal{I}}_{G} \) for all \( f \in \mathcal{O}\left\lbrack {\mathbf{{GL}}\left( V\right) }\right\rbrack \) . In particular, since \( I \in G \), we must have \( \left( {{X}_{A}f}\right) \left( I\right) = 0 \) for all regular functions \( f \) on \( \mathbf{{GL}}\left( V\right) \) . Hence \( A = 0 \) by Lemma 1.4.7.\\ The dual space \( \mathcal{O}{\left\lbrack G\right\rbrack }^{ * } \) is naturally identified with the subspace of \( \mathcal{O}\left\lbrack {\mathbf{{GL}}\left( V\right) }\right\rbrack \) consisting of the linear functionals that vanish on \( {\mathcal{J}}_{G} \) . In particular, each tangent vector to \( G \) at \( g \) is also a tangent vector to \( \mathbf{{GL}}\left( V\right) \) at \( g \) . To show that the map from \( \operatorname{Lie}\left( G\right) \) to \( T{\left( G\right) }_{g} \) is surjective, it suffices by left invariance to take \( g = I \) . If \( v \in T{\left( G\right) }_{I} \) then by Lemma 1.4.7 there is a unique \( A \in \operatorname{End}\left( V\right) \) such that \( v = {v}_{A} \) . We claim that \( A \in \operatorname{Lie}\left( G\right) \) . Take \( f \in {\mathcal{J}}_{G} \) and \( g \in G \) . Then\\ \[
\left( {{X}_{A}f}\right) \left( g\right) = \left( {L\left( {g}^{-1}\right) {X}_{A}f}\right) \left( I\right) = {X}_{A}\left( {L\left( {g}^{-1}\right) f}\right) \left( I\right) = v\left( {L\left( {g}^{-1}\right) f}\right) .
\]
But \( L\left( {g}^{-1}\right) f \in {\mathcal{I}}_{G} \) and \( v \) vanishes on \( {\mathcal{J}}_{G} \) . Hence \( \left( {{X}_{A}f}\right) \left( g\right) = 0 \) . This shows that \( {X}_{A}{\mathcal{J}}_{G} \subset {\mathcal{J}}_{G} \), proving that \( A \in \operatorname{Lie}\left( G\right) \) . | Yes |
Suppose \( f\left( {e}^{i\theta }\right) = \mathop{\sum }\limits_{{-\infty }}^{\infty }{a}_{n}{e}^{in\theta } \) lies in \( W \) . If \( f \) does not vanish on \( T \) , then \( 1/f \) is also in \( W\), that is, there exist \( \left\{ {b}_{n}\right\} \) with \( \mathop{\sum }\limits_{{-\infty }}^{\infty }\left| {b}_{n}\right| < \infty \) and \(\frac{1}{f\left( {e}^{i\theta }\right) } = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{b}_{n}{e}^{in\theta }\). | The hypothesis on \( f \) says that \( {\varphi }_{\lambda }\left( f\right) = f\left( \lambda \right) \) does not vanish as \( \lambda \) ranges over \( T \) . Since we have shown that the functionals \( {\varphi }_{\lambda } \) exhaust \( {\mathcal{M}}_{W} \), we may apply Corollary 5.29 to conclude that \( f \) is invertible in \( W \), which is the desired conclusion. | Yes |
If the function \( f \) has continuous first partial derivatives in a neighborhood of \( c \) that satisfy the \( \mathrm{{CR}} \) equations at \( c \), then \( f \) is (complex) differentiable at \( c \) . | The theorem is an immediate consequence of (2.12), since in this case \( {f}_{\bar{z}}\left( c\right) = 0 \) and hence \( {f}^{\prime }\left( c\right) = {f}_{z}\left( c\right) \) . | Yes |
Theorem 2.7. Let \( \mathrm{R} \) be a ring and \( \mathrm{I} \) an ideal of \( \mathrm{R} \) . Then the additive quotient group \( \mathrm{R}/\mathrm{I} \) is a ring with multiplication given by | SKETCH OF PROOF OF 2.7. Once we have shown that multiplication in \( R/I \) is well defined, the proof that \( R/I \) is a ring is routine. (For example, if \( R \) has identity \( {1}_{R} \), then \( {1}_{R} + I \) is the identity in \( R/I \) .) Suppose \( a + I = {a}^{\prime } + I \) and \( b + I = {b}^{\prime } + I \) . We must show that \( {ab} + I = {a}^{\prime }{b}^{\prime } + I \) . Since \( {a}^{\prime }\varepsilon {a}^{\prime } + I = a + I \) , \( {a}^{\prime } = a + i \) for some \( i \in I \) . Similarly \( {b}^{\prime } = b + j \) with \( j \in I \) . Consequently \( {a}^{\prime }{b}^{\prime } = \left( {a + i}\right) \left( {b + j}\right) = {ab} + {ib} + {aj} + {ij} \) . Since \( I \) is an ideal,
\[
{a}^{\prime }{b}^{\prime } - {ab} = {ib} + {aj} + {ij\varepsilon I}.
\]
Therefore \( {a}^{\prime }{b}^{\prime } + I = {ab} + I \) by Corollary I.4.3, whence multiplication in \( R/I \) is well defined. | No |
If \( V \) is an irreducible finite-dimensional representation of a Lie group \( G \), show that \( {V}^{ * } \) is also irreducible. | Null | No |
Proposition 6.45. Let \( \phi : {\mathbb{P}}^{1} \rightarrow {\mathbb{P}}^{1} \) be a Lattès map that fits into a commutative diagram (6.22). Then | The key to the proof of this proposition is the fact that the map \( \psi : E \rightarrow E \) is unramified, i.e., it has no critical points, see Remark 6.20. (In the language of modern algebraic geometry, the map \( \psi \) is étale.) More precisely, the map \( \psi \) is the composition of an endomorphism of \( E \) and a translation (Remark 6.19), both of which are unramified.
For any \( n \geq 1 \) we compute
\[
{\operatorname{CritVal}}_{\pi } = {\operatorname{CritVal}}_{\pi {\psi }^{n}}
\]
because \( \psi \) is unramified,
\[
= {\operatorname{CritVal}}_{{\phi }^{n}\pi }
\]
from the commutativity of (6.22),
\( = {\operatorname{CritVal}}_{{\phi }^{n}} \cup {\phi }^{n}\left( {\operatorname{CritVal}}_{\pi }\right) \; \) from the definition of critical value,
\[
\supseteq {\operatorname{CritVal}}_{{\phi }^{n}}
\]
This holds for all \( n \geq 1 \), which gives the inclusion
\[
{\operatorname{CritVal}}_{\pi } \supseteq \mathop{\bigcup }\limits_{{n = 0}}^{\infty }{\phi }^{n}\left( {\operatorname{CritVal}}_{\phi }\right) = {\operatorname{PostCrit}}_{\phi }.
\]
In order to prove the opposite inclusion, suppose that there exists a point \( {P}_{0} \in E \) satisfying
\[
{P}_{0} \in {\operatorname{CritPt}}_{\pi }\;\text{ and }\;\pi \left( {P}_{0}\right) \notin {\operatorname{PostCrit}}_{\phi }.
\]
(6.23)
Consider any point \( Q \in {\psi }^{-1}\left( {P}_{0}\right) \) . Then \( Q \) is a critical point of \( {\pi \psi } \), since \( \psi \) is unramified and \( \pi \) is ramified at \( \psi \left( Q\right) \) by assumption. But \( {\pi \psi } = {\phi \pi } \), so we see that \( Q \) is a critical point for \( {\phi \pi } \).
On the other hand,
\[
\phi \left( {\pi \left( Q\right) }\right) = \pi \left( {P}_{0}\right) \notin {\operatorname{CritVal}}_{\phi } = \phi \left( {\operatorname{CritPt}}_{\phi }\right) ,
\]
so \( \pi \left( Q\right) \) is not a critical point for \( \phi \) . It follows that \( Q \) is a critical point of \( \pi \) . Further, we claim that no iterate of \( \phi \) is ramified at \( \pi \left( Q\right) \) . To see this, we use the given fact that \( \pi \left( {P}_{0}\right) \) is not in the postcritical set of \( \phi \) to compute
\[
\pi \left( {P}_{0}\right) \notin {\mathsf{{PostCrit}}}_{\phi } \Rightarrow \pi \left( {P}_{0}\right) \notin {\phi }^{n}\left( {\mathsf{{CritPt}}}_{\phi }\right) \;\text{ for all }n \geq 1,
\]
\[
\Rightarrow \pi \left( {\psi \left( Q\right) }\right) \notin {\phi }^{n}\left( {\mathsf{{CritPt}}}_{\phi }\right) \;\text{ for all }n \geq 1,
\]
\[
\Rightarrow \phi \left( {\pi \left( Q\right) }\right) \notin {\phi }^{n}\left( {\operatorname{CritPt}}_{\phi }\right) \;\text{ for all }n \geq 1,
\]
\[
\Rightarrow \pi \left( Q\right) \notin {\phi }^{n}\left( {\mathsf{{CritPt}}}_{\phi }\right) \;\text{ for all }n \geq 0,
\]
\[
\Rightarrow \pi \left( Q\right) \notin {\operatorname{PostCrit}}_{\phi }.
\]
To recapitulate, we have now proven that every \( Q \in {\psi }^{-1}\left( {P}_{0}\right) \) satisfies
\[
Q \in {\operatorname{CritPt}}_{\pi }\;\text{ and }\;\pi \left( Q\right) \notin {\operatorname{PostCrit}}_{\psi }.
\]
In other words, every point \( Q \in {\psi }^{-1}\left( {P}_{0}\right) \) satisfies the same two conditions (6.23) that are satisfied by \( {P}_{0} \) . Hence by induction we find that if there is any point \( {P}_{0} \) satisfying (6.23), then the full backward orbit of \( \psi \) is contained in the set of critical points of \( \pi \), i.e.,
\[
{\operatorname{CritPt}}_{\pi } \supset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{\psi }^{-n}\left( {P}_{0}\right)
\]
But \( \psi \) is unramified and has degree at least 2 (note that \( \deg \psi = \deg \phi \) ), so
\[
\text{#}{\operatorname{CritPt}}_{\pi } \geq \# \left( {{\psi }^{-n}\left( {P}_{0}\right) }\right) = {\left( \deg \psi \right) }^{n}\xrightarrow[{n \rightarrow \infty }]{}\infty \text{.}
\]
This is a contradiction, since \( \pi \) has only finitely many critical points, so we conclude that there are no points \( {P}_{0} \) satisfying (6.23). Hence
\[
{P}_{0} \in {\operatorname{CritPt}}_{\pi } \Rightarrow \pi \left( {P}_{0}\right) \in {\operatorname{PostCrit}}_{\phi },
\]
which gives the other inclusion \( {\operatorname{CritVal}}_{\pi } \subseteq {\operatorname{PostCrit}}_{\phi } \) | Yes |
If \( G \) is an infinite connected solvable group of finite Morley rank with finite center, then \( G \) interprets an algebraically closed field. | We will prove this by induction on the rank of \( G \) . We first argue that we may, without loss of generality, assume that \( G \) is centerless. Suppose that \( Z\left( G\right) \) is finite. We claim that \( G/Z\left( G\right) \) is centerless. Let \( a \in G \) such that \( a/Z\left( G\right) \in Z\left( {G/Z\left( G\right) }\right) \) . For all \( g \in G,{a}^{-1}{g}^{-1}{ag} \in Z\left( G\right) \) . Thus, \( {a}^{-1}{a}^{G} \subseteq Z\left( G\right) \), and, hence, \( {a}^{G} \) is finite. Thus, \( \left\lbrack {G : C\left( a\right) }\right\rbrack \) is finite. Because \( G \) is connected, \( C\left( a\right) = G \) and \( a \in Z\left( G\right) \) . Thus, \( G/Z\left( G\right) \) is solvable and centerless. By Exercise 7.6.4, \( G/Z\left( G\right) \) is connected. Because \( G/Z\left( G\right) \) is interpretable in \( G \), if \( G/Z\left( G\right) \) interprets an algebraically closed field, so does \( G \) . | No |
If Dedekind's functional equation is satisfied for some A = (a b c d) in Γ with c > 0 and ε(A) given by (2), then it is also satisfied for ATm and for AS. | Replace τ by Tmτ in (5) to obtain η(ATmτ) = ε(A){-i(cTmτ + d)}1/2η(Tmτ) = ε(A){-i(cτ + mc + d)}1/2e^πim/12η(τ). Using Lemma 2 we obtain (6). Now replace τ by Sτ in (5) to get η(ASτ) = ε(A){-i(cSτ + d)}1/2η(Sτ). Using Theorem 3.1 we can write this as η(ASτ) = ε(A){-i(cSτ + d)}1/2{-iτ}1/2η(τ). If d > 0, we write cSτ + d = -c/τ + d = (dτ - c)/τ hence, -i(cSτ + d) = -i(dτ - c)/(-iτ)e^(-πi)/2, and therefore, { -i(cSτ + d)}1/2{-iτ}1/2 = e^(-πi)/4{-i(dτ - c)}1/2. Using this in (9) together with Lemma 3, we obtain (7). If d < 0, we write cSτ + d = -c/τ + d = -(dτ + c)/(-τ) so that in this case we have -i(cSτ + d) = -i(-dτ + c)/(-iτ)e^(πi)/2, and therefore, { -i(cSτ + d)}1/2{-iτ}1/2 = e^(πi)/4{-i(-dτ + c)}1/2. Using this in (9) together with Lemma 3, we obtain (8). | Yes |
Corollary 4.1.1. The discrete Dirichlet problem | This follows in the usual manner by applying the maximum principle to the difference of two solutions. | No |
Let \( g \) and \( h \) be hyperbolic with axes and translation lengths \( {A}_{g},{A}_{h},{T}_{g} \) and \( {T}_{h} \) respectively. Suppose that \( \langle g, h\rangle \) is discrete and nonelementary and that no images of \( {A}_{g} \) and \( {A}_{h} \) cross. Then | Proof of Theorem 11.6.9. Consider Figure 11.6.4. As \( g \) (or \( {g}^{-1} \) ) is \( {\sigma }_{3}{\sigma }_{1} \) and \( h\left( {\text{or}{h}^{-1}}\right) \) is \( {\sigma }_{2}{\sigma }_{3} \) we see that \( {\sigma }_{2}{\sigma }_{1} \) is in \( G \) . If \( G \) has no elliptic elements, then \( {L}_{1} \) and \( {L}_{2} \) cannot intersect (this case is not illustrated) and from Theorem 7.19.2 we obtain\n\n\( \sinh \left( {\frac{1}{2}{T}_{g}}\right) \sinh \left( {\frac{1}{2}{T}_{h}}\right) \cosh \rho \left( {{A}_{g},{A}_{h}}\right) = \cosh \left( {\frac{1}{2}{T}_{g}}\right) \cosh \left( {\frac{1}{2}{T}_{h}}\right) + \cosh \rho \left( {{L}_{1},{L}_{2}}\right) . \)\n\nThis yields the second inequality.\n\nIf \( {L}_{1} \) and \( {L}_{2} \) intersect, say at an angle \( \theta \), then \( \theta = {2\pi p}/q \) for some coprime integers \( p \) and \( q \) . If \( \theta > {2\pi }/q \) we can rotate \( {A}_{h} \) about the point of intersection to an image of itself which is closer to (but, by assumption, not intersecting) \( {A}_{g} \) . Thus if, in the argument above, we replace \( h \) by a conjugate \( {fh}{f}^{-1} \) of \( h \) with the property that its axis \( f\left( {A}_{h}\right) \) is as close as possible to (but distinct from) \( {A}_{g} \), we find that\n\n\( \rho \left( {{A}_{g},{A}_{h}}\right) \geq \rho \left( {{A}_{g}, f{A}_{h}}\right) \)\n\nand the corresponding \( \theta \) satisfies \( \theta \leq {2\pi }/q \leq {2\pi }/3 \) as obviously \( \theta < \pi \) . Thus from Theorem 7.18.1 we obtain the first inequality, namely\n\n\( \sinh \left( {\frac{1}{2}{T}_{g}}\right) \sinh \left( {\frac{1}{2}{T}_{h}}\right) \cosh \rho \left( {{A}_{g},{A}_{h}}\right) \geq \cosh \left( {\frac{1}{2}{T}_{g}}\right) \cosh \left( {\frac{1}{2}{T}_{h}}\right) + \cos \left( {{2\pi }/3}\right) . \)\n\n | Yes |
Show that every positive root is a nonnegative linear combination of simple roots. | Null | No |
Theorem 1 Given an edge \( {ab} \), denote by \( N\left( {s, a, b, t}\right) \) the number of spanning trees of \( G \) in which the (unique) path from \( s \) to \( t \) contains a and \( b \), in this order. Define \( N\left( {s, b, a, t}\right) \) analogously and write \( N \) for the total number of spanning trees. Finally, let \( {w}_{ab} = \{ N\left( {s, a, b, t}\right) - N\left( {s, b, a, t}\right) \} /N \) . Distribute currents in the edges of \( G \) by sending a current of size \( {w}_{ab} \) from a to \( b \) for every edge \( {ab} \) . Then there is a total current size 1 from \( s \) to \( t \) satisfying the Kirchhoff laws. | Proof. To simplify the situation, multiply all currents by \( N \) . Also, for every spanning tree \( T \) and edge \( {ab} \in E\left( G\right) \), let \( {w}^{\left( T\right) } \) be the current of size 1 along the unique \( s - t \) path in \( T \) : \[
{w}_{ab}^{\left( T\right) } = \left\{ \begin{array}{ll} 1 & \text{ if }T\text{ has a path }s\cdots {ab}\cdots t, \\ - 1 & \text{ if }T\text{ has a path }s\cdots {ba}\cdots t, \\ 0 & \text{ otherwise. } \end{array}\right.
\] Then \[
N\left( {s, a, b, t}\right) - N\left( {s, b, a, t}\right) = \mathop{\sum }\limits_{T}{w}_{ab}^{\left( T\right) },
\] where the summation is over all spanning trees \( T \) . Therefore, our task is to show that if we send a current of size \( \mathop{\sum }\limits_{T}{w}_{ab}^{\left( T\right) } \) from \( a \) to \( b \) for every edge \( {ab} \), then we obtain a total current of size \( N \) from \( s \) to \( t \) satisfying the Kirchhoff laws. Now, each \( {w}^{\left( T\right) } \) is a current of size 1 from \( s \) to \( t \) satisfying Kirchhoff’s current law, and so their sum is a current of size \( N \) from \( s \) to \( t \) satisfying Kirchhoff’s current law. All we have to show then is that the potential law is also satisfied. As all edges have the same resistance, the potential law claims that the total current in a cycle with some orientation is zero. To show this, we proceed as earlier, but first we reformulate slightly the definition of \( N\left( {s, a, b, t}\right) \) . Call a spanning forest \( F \) of \( G \) a thicket if it has exactly two components, say \( {F}_{s} \) and \( {F}_{t} \), such that \( s \) is in \( {F}_{s} \) and \( t \) is in \( {F}_{t} \) . Then \( N\left( {s, a, b, t}\right) \) is the number of thickets \( F = {F}_{s} \cup {F}_{t} \) for which \( a \in {F}_{s} \) and \( b \in {F}_{t} \), and \( N\left( {s, b, a, t}\right) \) is defined analogously. What is then the contribution of a thicket \( F = {F}_{s} \cup {F}_{t} \) to the total current in a cycle? It is the number of cycle edges from \( {F}_{s} \) to \( {F}_{t} \) minus the number of cycle edges from \( {F}_{t} \) to \( {F}_{s} \) ; so it is zero. Let us write out the second part of the proof more formally, to make it even more evident that we use the basic and powerful combinatorial principle of double counting, or reversing the order of summation. For a thicket \( F = {F}_{s} \cup {F}_{t} \) and an edge \( {ab} \in E\left( G\right) \), set \[
{w}_{ab}^{\left( F\right) } = \left\{ \begin{array}{ll} {w}_{ab}^{\left( F + ab\right) } & \text{ if }F + {ab}\text{ is a spanning tree,} \\ 0 & \text{ otherwise. } \end{array}\right.
\] Then \[
\mathop{\sum }\limits_{T}{w}_{ab}^{\left( T\right) } = \mathop{\sum }\limits_{F}{w}_{ab}^{\left( F\right) }
\] where the second summation is over all thickets \( F \) . Finally, the total current around a cycle \( {x}_{1}{x}_{2}\cdots {x}_{k} \) of \( G \), with \( {x}_{k + 1} = {x}_{1} \), is \[
\mathop{\sum }\limits_{{i = 1}}^{k}\mathop{\sum }\limits_{F}{w}_{{x}_{i}{x}_{i + 1}}^{\left( F\right) } = \mathop{\sum }\limits_{F}\mathop{\sum }\limits_{{i = 1}}^{k}{w}_{{x}_{i}{x}_{i + 1}}^{\left( F\right) } = 0
\] since \( \mathop{\sum }\limits_{{i = 1}}^{k}{w}_{{x}_{i}{x}_{i + 1}}^{\left( F\right) } = 0 \) for every thicket \( F \) . | Yes |
Proposition 1.2 Assume \( \varphi \in {\mathcal{D}}^{m} \), for some \( m \in \mathbb{N} \) . For every integer \( n \geq 1 \), the convolution \( \varphi * {\chi }_{n} \) belongs to \( \mathcal{D} \) and \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}\varphi * {\chi }_{n} = \varphi \;\text{ in }{\mathcal{D}}^{m} \) | Since the functions \( \varphi \) and \( {\chi }_{n} \) have compact support, so does \( \varphi * {\chi }_{n} \) . More precisely,
\[
\operatorname{Supp}\left( {\varphi * {\chi }_{n}}\right) \subset \operatorname{Supp}\varphi + \operatorname{Supp}{\chi }_{n} \subset \operatorname{Supp}\varphi + \bar{B}\left( {0,1/n}\right) \subset \operatorname{Supp}\varphi + \bar{B}\left( {0,1}\right) .
\]
At the same time, a classical theorem about differentiation under the integral sign easily implies, on the one hand, that \( \varphi * {\chi }_{n} \) is of class \( {C}^{\infty } \) and so \( \varphi * {\chi }_{n} \in \mathcal{D} \), and, on the other, that \( {D}^{p}\left( {\varphi * {\chi }_{n}}\right) = \left( {{D}^{p}\varphi }\right) * {\chi }_{n} \) for \( \left| p\right| \leq m \) . Now, since the support of \( {\chi }_{n} \) is contained in \( \bar{B}\left( {0,1/n}\right) \) and \( \int {\chi }_{n}\left( y\right) {dy} = 1 \) , we get
\[
\left( {{D}^{p}\varphi }\right) * {\chi }_{n}\left( x\right) - \left( {{D}^{p}\varphi }\right) \left( x\right) = {\int }_{\left| y\right| \leq 1/n}\left( {{D}^{p}\varphi \left( {x - y}\right) - {D}^{p}\varphi \left( x\right) }\right) {\chi }_{n}\left( y\right) {dy}
\]
and
\[
\mathop{\sup }\limits_{{x \in {\mathbb{R}}^{d}}}\left| {\left( {{D}^{p}\varphi }\right) * {\chi }_{n}\left( x\right) - \left( {{D}^{p}\varphi }\right) \left( x\right) }\right| \leq \mathop{\sup }\limits_{\substack{{x, z \in {\mathbb{R}}^{d}} \\ {\left| {z - x}\right| \leq 1/n} }}\left| {{D}^{p}\varphi \left( z\right) - {D}^{p}\varphi \left( x\right) }\right| .
\]
Since \( {D}^{p}\varphi \) is uniformly continuous (being continuous and having compact support), we deduce that the sequence \( {\left( {D}^{p}\left( \varphi * {\chi }_{n}\right) \right) }_{n \in \mathbb{N}} \) converges uniformly to \( {D}^{p}\varphi \) . | Yes |
If no free occurrence of \( \alpha \) in \( \varphi \) is within the scope of a quantifier on a variable occurring in \( \sigma \), then \( {\mathrm{{FSubf}}}_{\sigma }^{\alpha }\varphi \rightarrow \exists {\alpha \varphi } \) . | \[
\vdash \forall \alpha \sqsupset \varphi \rightarrow {\operatorname{Subf}}_{\sigma }^{\alpha } \sqsupset \varphi \;\text{ universal specification }
\]
\[
{ \vdash }^{ \Vdash }{\mathrm{{Subf}}}_{\sigma }^{\alpha }\varphi \rightarrow \exists {\alpha \varphi }
\]
suitable tautology | Yes |
Let \( \mathcal{X} \) be a set of 3-connected graphs. Let \( G \) be a graph with \( \kappa \left( G\right) \leq 2 \), and let \( {G}_{1},{G}_{2} \) be proper induced subgraphs of \( G \) such that \( G = {G}_{1} \cup {G}_{2} \) and \( \left| {{G}_{1} \cap {G}_{2}}\right| = \kappa \left( G\right) \) . If \( G \) is edge-maximal without a topological minor in \( \mathcal{X} \), then so are \( {G}_{1} \) and \( {G}_{2} \), and \( {G}_{1} \cap {G}_{2} = {K}^{2} \) . | Note first that every vertex \( v \in S \mathrel{\text{:=}} V\left( {{G}_{1} \cap {G}_{2}}\right) \) has a neighbour in every component of \( {G}_{i} - S, i = 1,2 \) : otherwise \( S \smallsetminus \{ v\} \) would separate \( G \), contradicting \( \left| S\right| = \kappa \left( G\right) \) . By the maximality of \( G \), every edge \( e \) added to \( G \) lies in a \( {TX} \subseteq G + e \) with \( X \in \mathcal{X} \) . For all the choices of \( e \) considered below, the 3-connectedness of \( X \) will imply that the branch vertices of this \( {TX} \) all lie in the same \( {G}_{i} \), say in \( {G}_{1} \) . (The position of \( e \) will always be symmetrical with respect to \( {G}_{1} \) and \( {G}_{2} \), so this assumption entails no loss of generality.) Then the \( {TX} \) meets \( {G}_{2} \) at most in a path \( P \) corresponding to an edge of \( X \) . | Yes |
The lightcone \( {\mathcal{L}}_{0} \) is a lightlike submanifold. | Suppose \( v \in {\mathcal{L}}_{0} \) ; then \( g\left( {v, v}\right) = 0 \) and \( v \neq 0 \) . Let \( \mathcal{U} \) be a neighborhood of \( v \) that does not contain the origin and define \( \widetilde{g} : \mathcal{U} \rightarrow \mathbb{R} \) by \( \widetilde{g}w = g\left( {w, w}\right) \forall w \in \mathcal{U} \) . Then \( {\mathcal{L}}_{0} \cap \mathcal{U} \) is defined by \( \widetilde{g} = 0 \) . Moreover, \( d\widetilde{g} \) is nowhere zero on \( \mathcal{U} \) because \( g \) is nondegenerate (Exercise 0.0.10). Thus \( {\mathcal{L}}_{0} \) is a submanifold by the implicit function theorem. To show \( {\mathcal{L}}_{0} \) is lightlike, suppose that \( w \in {V}_{v} \) for some \( v \in {\mathcal{L}}_{0} \), and let \( {\phi }_{v} : {V}_{v} \rightarrow V \) be the canonical isomorphism of Exercise 0.0.10 . Then \( w \) is in the tangent space \( {\left( {\mathcal{L}}_{0}\right) }_{v} \) iff \( w\widetilde{g} = 0 \), iff \( g\left( {{\phi }_{v}w, v}\right) = 0 \), and iff \( g\left( {w,{\phi }_{v}{}^{-1}v}\right) = 0 \) . Thus \( {\left( {\mathcal{L}}_{0}\right) }_{v} = {\left( {\phi }_{v}{}^{-1}v\right) }^{ \bot } \subset {V}_{v} \) . But \( {\phi }_{v}{}^{-1}v \in {V}_{v} \) is lightlike because \( g\left( {{\phi }_{v}{}^{-1}v,{\phi }_{v}{}^{-1}v}\right) = g\left( {v, v}\right) = 0 \), hence \( {\left( {\mathcal{L}}_{0}\right) }_{v} \) is lightlike by Proposition 1.1.3b. Since this holds for all \( v \in {\mathcal{L}}_{0},{\mathcal{L}}_{0} \) is lightlike. | Yes |
Show that \( \mathfrak{a} \) has an integral basis. | Let \( \mathfrak{a} \) be an ideal of \( {\mathcal{O}}_{K} \), and let \( {\omega }_{1},{\omega }_{2},\ldots ,{\omega }_{n} \) be an integral basis for \( {\mathcal{O}}_{K} \) . Note that for any \( {\omega }_{i} \) in \( {\mathcal{O}}_{K},{a}_{0}{\omega }_{i} = - \left( {{\alpha }^{r} + \cdots + {a}_{1}\alpha }\right) {\omega }_{i} \in \mathfrak{a} \) . Therefore \( \mathfrak{a} \) has finite index in \( {\mathcal{O}}_{K} \) and \( \mathfrak{a} \subseteq {\mathcal{O}}_{K} = \mathbb{Z}{\omega }_{1} + \mathbb{Z}{\omega }_{2} + \cdots + \mathbb{Z}{\omega }_{n} \) has maximal rank. Then since \( \mathfrak{a} \) is a submodule of \( {\mathcal{O}}_{K} \), by Theorem 4.2.2 there exists an integral basis for \( \mathfrak{a} \) . | Yes |
Prove that the group B = {(a, b) | a, b ∈ ℝ, a > 0} is amenable but not unimodular. | Null | No |
If \( K \) is a compact subset of \( \mathbb{C} \) then the algebra \( \mathcal{P}\left( K\right) \) coincides with the restriction to \( K \) of the subalgebra \( \mathcal{P}\left( \widehat{K}\right) \) of \( \mathcal{C}\left( \widehat{K}\right) \) . | A sequence \( \left\{ {p}_{n}\right\} \) of polynomials converging uniformly on \( K \) automatically converges uniformly on the outer boundary of \( K \) . But then, by the maximum modulus principle (Ex. 5M), \( \left\{ {p}_{n}\right\} \) converges uniformly on \( \widetilde{K} \), and the limit is therefore differentiable at each point of \( {\widehat{K}}^{ \circ } \) (see Problem 11U). | No |
Theorem 11.4.11. The action of \( {W}_{G} \) on \( \mathfrak{h} \) coincides with the action of \( W\left( {\mathfrak{g},\mathfrak{h}}\right) \) . Furthermore, every coset in \( {W}_{G} \) has a representative from \( U \ ). | Proof. For \( \alpha \in \Phi \) and \( {X}_{\alpha } \) as in (7.42), set
\[
{u}_{\alpha } = \frac{1}{2}\left( {{X}_{\alpha } - {X}_{-\alpha }}\right) \;\text{ and }\;{v}_{\alpha } = \frac{1}{2\mathrm{i}}\left( {{X}_{\alpha } + {X}_{-\alpha }}\right) .
\]
(11.13)
Then \( {X}_{\alpha } = {u}_{\alpha } + \mathrm{i}{v}_{\alpha } \), and \( {u}_{\alpha },{v}_{\alpha } \in \mathfrak{u} \) . We calculate the action of ad \( {u}_{\alpha } \) on \( h \in \mathfrak{h} \) as follows:
\[
\left\lbrack {{u}_{\alpha }, h}\right\rbrack = - \frac{1}{2}\langle \alpha, h\rangle \left( {{X}_{\alpha } + {X}_{-\alpha }}\right) = - \mathrm{i}\langle \alpha, h\rangle {v}_{\alpha },
\]
(11.14)
\[
\left\lbrack {{u}_{\alpha },{v}_{\alpha }}\right\rbrack = \frac{1}{4\mathrm{i}}\left\lbrack {{X}_{\alpha } - {X}_{-\alpha },{X}_{\alpha } + {X}_{-\alpha }}\right\rbrack = \frac{1}{2\mathrm{i}}\left\lbrack {{X}_{\alpha },{X}_{-\alpha }}\right\rbrack = \frac{1}{2\mathrm{i}}{H}_{\alpha }.
\]
(11.15)
From (11.14) we see that for all \( s \in \mathbb{C} \) ,
\[
\exp \left( {s\operatorname{ad}{u}_{\alpha }}\right) h = h\;\text{ if }\langle \alpha, h\rangle = 0.
\]
(11.16)
Taking \( h = {H}_{\alpha } \) in (11.14), we obtain \( \operatorname{ad}\left( {u}_{\alpha }\right) {H}_{\alpha } = \mathrm{i}\parallel \alpha {\parallel }^{2}{v}_{\alpha } \), and hence by (11.15) we have
\[
{\left( \operatorname{ad}{u}_{\alpha }\right) }^{2}{H}_{\alpha } = - {r}^{2}{H}_{\alpha }
\]
where \( r = \parallel \alpha \parallel /\sqrt{2} \) . Continuing in this way, we calculate that
\[
\exp \left( {s\operatorname{ad}{u}_{\alpha }}\right) {H}_{\alpha } = \cos \left( {rs}\right) {H}_{\alpha } - \left( {2\mathrm{i}/r}\right) \sin \left( {rs}\right) {v}_{\alpha }
\]
for all \( s \in \mathbb{C} \) . In particular, when \( s = \pi /r \) the second term vanishes and we have
\[
\exp \left( {\left( {\pi /r}\right) \operatorname{ad}{u}_{\alpha }}\right) {H}_{\alpha } = - {H}_{\alpha }.
\]
(11.17)
Thus by (11.16) and (11.17) we see that the element \( {g}_{\alpha } = \exp \left( {\left( {\pi /r}\right) {u}_{\alpha }}\right) \in U \) acts on \( \mathfrak{h} \) by the reflection \( {s}_{\alpha } \) . This proves that every element in the algebraic Weyl group \( W\left( {\mathfrak{g},\mathfrak{h}}\right) \) can be implemented by the adjoint action of an element \( k \in {\operatorname{Norm}}_{G}\left( H\right) \cap U \) | Yes |
Let \( W \) be the subgroup of the signed permutation group consisting of elements that change an even number of signs (Example 1.13). Then \( \mathcal{H} \) consists of the hyperplanes \( {x}_{i} - {x}_{j} = 0 \) and \( {x}_{i} + {x}_{j} = 0\left( {i \neq j}\right) \) . To figure out what the chambers look like, consider two coordinates, say \( {x}_{1} \) and \( {x}_{2} \) . From the fact that \( {x}_{1} \) is comparable to both \( {x}_{2} \) and \( - {x}_{2} \) on any given chamber \( C \), one can deduce that one of the coordinates is bigger than the other in absolute value and that this coordinate has a constant sign. In other words, we have an inequality of the form \( \epsilon {x}_{1} > \left| {x}_{2}\right| \) or \( \epsilon {x}_{2} > \left| {x}_{1}\right| \) on \( C \), where \( \epsilon = \pm 1 \ ) | It follows that there are \( {2}^{n - 1}n \) ! chambers, each defined by inequalities of the form
\[
{\epsilon }_{1}{x}_{\pi \left( 1\right) } > {\epsilon }_{2}{x}_{\pi \left( 2\right) } > \cdots > {\epsilon }_{n - 1}{x}_{\pi \left( {n - 1}\right) } > \left| {x}_{\pi \left( n\right) }\right|
\]
(1.21)
with \( {\epsilon }_{i} \in \{ \pm 1\} \) and \( \pi \in {S}_{n} \) . Note that the last inequality is equivalent to two linear inequalities, \( {\epsilon }_{n - 1}{x}_{\pi \left( {n - 1}\right) } > {x}_{\pi \left( n\right) } \) and \( {\epsilon }_{n - 1}{x}_{\pi \left( {n - 1}\right) } > - {x}_{\pi \left( n\right) } \), so we have \( n \) linear inequalities in all.
As fundamental chamber we take
\[
{x}_{1} > {x}_{2} > \cdots > {x}_{n - 1} > \left| {x}_{n}\right|
\]
with walls \( {x}_{1} = {x}_{2},{x}_{2} = {x}_{3},\ldots ,{x}_{n - 1} = {x}_{n} \), and \( {x}_{n - 1} = - {x}_{n} \) . Further analysis is left to the interested reader. | No |
Theorem 12.5.1 An internal set \( A \) is hyperfinite with internal cardinality \( N \) if and only if there is an internal bijection \( f : \{ 1,\ldots, N\} \rightarrow A \) . | Let \( A = \left\lbrack {A}_{n}\right\rbrack \) . If \( A \) is hyperfinite with internal cardinality \( N = \) \( \left\lbrack {N}_{n}\right\rbrack \), then we may suppose that for each \( n \in \mathbb{N},{A}_{n} \) is a finite set of cardinality \( {N}_{n} \) . Thus there is a bijection \( {f}_{n} : \left\{ {1,\ldots ,{N}_{n}}\right\} \rightarrow {A}_{n} \) . Let \( f = \) \( \left\lbrack {f}_{n}\right\rbrack \) . Then \( f \) is an internal function with domain \( \{ 1,\ldots, N\} \) that is injective \( \left( {{12.2}\left( 4\right) }\right) \) and has range \( A\left( {{12.2}\left( 1\right) }\right) \) .
Conversely, suppose that \( f = \left\lbrack {f}_{n}\right\rbrack \) is an internal bijection from \( \{ 1,\ldots, N\} \) onto \( A \) . Then
\[
\left\lbrack {\operatorname{dom}{f}_{n}}\right\rbrack = \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack = \{ 1,\ldots, N\} = \left\lbrack \left\{ {1,\ldots ,{N}_{n}}\right\} \right\rbrack ,
\]
so for \( \mathcal{F} \) -almost all \( n \) ,
\[
\operatorname{dom}{f}_{n} = \left\{ {1,\ldots ,{N}_{n}}\right\}
\]
(i)
Also, as \( A \) is the image of \( \{ 1,\ldots, N\} \) under \( \left\lbrack {f}_{n}\right\rbrack \), Exercise 12.2(1) implies that \( A = \left\lbrack {{f}_{n}\left( \left\{ {1,\ldots ,{N}_{n}}\right\} \right) }\right\rbrack \), so
\[
{f}_{n}\left( \left\{ {1,\ldots ,{N}_{n}}\right\} \right) = {A}_{n}
\]
(ii)
for \( \mathcal{F} \) -almost all \( n \) . Finally, by \( {12.2}\left( 4\right) \) ,
\[
{f}_{n}\text{is injective}
\]
(iii)
for \( \mathcal{F} \) -almost all \( n \) . Then the set \( J \) of those \( n \in \mathbb{N} \) satisfying (i)-(iii) must belong to \( \mathcal{F} \) . But for \( n \in J,{A}_{n} \) is finite of cardinality \( {N}_{n} \) . Hence \( A \) is hyperfinite of internal cardinality \( N \) . | No |
Consider the Lagrangian relaxation of the traveling salesman problem proposed in (8.7). As discussed in Sect. 8.1.1, the bound provided by (8.7) is equal to \[
\min \;\mathop{\sum }\limits_{{e \in E}}{c}_{e}{x}_{e}
\]
\[
\mathop{\sum }\limits_{{e \in \delta \left( i\right) }}{x}_{e} = 2\;i \in V \smallsetminus \{ 1\}
\]
\[
x \in \operatorname{conv}\left( Q\right)
\]
where \( Q \) is the set of incidence vectors of 1-trees. Let \( \mathcal{T} \) be the family of 1-trees of \( G \) and for any 1-tree \( T \in \mathcal{T} \) let \( c\left( T\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{e \in T}}{c}_{e} \) denote its cost. Held and Karp [197] give the following Dantzig-Wolfe relaxation of the traveling salesman problem. | ∎ | No |
Proposition 23.41 Let \( {\delta }_{P} \) be a fixed square root of \( {\mathcal{K}}_{P} \) . For any vector field \( X \) lying in \( P \), there is a unique linear operator \( {\nabla }_{X} \) mapping sections of \( {\delta }_{P} \) to sections of \( {\delta }_{P} \), such that | Proof. If \( V \) is a one-dimensional vector space, then the map \( \otimes : V \times V \rightarrow \) \( V \otimes V \) is commutative: \( u \otimes v = v \otimes u \) for all \( u, v \in V \) . Furthermore, if \( {u}_{0} \) is a nonzero element of \( V \), then the map \( u \mapsto u \otimes {u}_{0} \) is an invertible linear map of \( V \) to \( V \otimes V \) . Suppose \( {s}_{0} \) is a local nonvanishing section of \( {\delta }_{P} \) . Applying (23.28) with \( {s}_{1} = {s}_{2} = {s}_{0} \), we want \nabla_X(s_0) \otimes s_0 = \frac{1}{2} \nabla_X(s_0 \otimes s_0) . Since the operation of tensoring with \( s_0 \) is invertible, there is a unique section " \( \nabla_Xs_0 \) " of \( \delta_P \) for which this holds. Locally, any section \( s \) of \( \delta_P \) can be written as \( s = gs_0 \) for a unique function \( g \) . We then define \( \nabla_Xs \) by \( \nabla_Xs = X(g)s_0 + g\nabla_Xs_0 \) in which case,(23.27) is easily seen to hold. If \( s_1 = g_1s_0 \) and \( s_2 = g_2s_0 \) , then using (23.29) and the symmetry of the tensor product, it is easy to verify that (23.28) holds, with both sides of the equation equal to X(g_1g_2)\nabla_X(s_0 \otimes s_0) . Uniqueness of \( \nabla_X \) holds because both (23.29) and (23.30) are required by the definition of \( \nabla_X \) . The action of \( \nabla_X \) extends to sections of \( \delta_P^{\mathbb{C}} \), by writing such sections as complex-valued functions times \( s_0 \) . The analysis of the Lie derivative is similar and is omitted. | Yes |
Theorem 7. Let \( M \) and \( {M}^{\prime } \) be totally disconnected compact sets in \( {\mathbf{R}}^{2} \), and let \( f \) be a homeomorphism \( M \leftrightarrow {M}^{\prime } \). Then \( f \) has an extension \( F : {\mathbf{R}}^{2} \leftrightarrow {\mathbf{R}}^{2} \). | Theorem 7. Let \( M \) and \( {M}^{\prime } \) be totally disconnected compact sets in \( {\mathbf{R}}^{2} \), and let \( f \) be a homeomorphism \( M \leftrightarrow {M}^{\prime } \). Then \( f \) has an extension \( F : {\mathbf{R}}^{2} \leftrightarrow {\mathbf{R}}^{2} \). Proof. (1) Let \( A \) and \( {A}^{\prime } \) be 2-cells containing \( M \) and \( {M}^{\prime } \) respectively in their interiors. Let \( {N}_{1} \) be a frame of \( M \), lying in Int \( A \), and lying in a sufficiently small neighborhood of \( M \) so that every component of \( {N}_{1} \) has diameter \( < 1 \) . (Theorem 6.) Then the sets \( f\left( {M \cap C}\right) \), where \( C \) is a component of \( {N}_{1} \), are disjoint and compact. Let \( L \) be a frame of \( {M}^{\prime } \), lying in Int \( {A}^{\prime } \), with components \( {C}^{\prime } \) of sufficiently small diameter so that no \( {C}^{\prime } \) intersects two different sets \( f\left( {M \cap C}\right) \) . By repeated applications of Theorem 2 we get a frame \( {N}_{1}^{\prime } \) of \( {M}^{\prime } \), lying in \( \operatorname{Int}{A}^{\prime } \), such that each set \( f\left( {M \cap C}\right) \) is the intersection of \( {M}^{\prime } \) and a component of \( {N}_{1}^{\prime } \) . Now there is a homeomorphism \( {f}_{0} : {\mathbf{R}}^{2} - \operatorname{Int}A \leftrightarrow {\mathbf{R}}^{2} - \operatorname{Int}{A}^{\prime } \) . Let \( {E}_{1} = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{1},{E}_{1}^{\prime } = {\mathbf{R}}^{2} \) - Int \( {N}_{1}^{\prime } \) . By Theorem \( 1,{f}_{0} \) can be extended so as to give a homeomorphism \( {f}_{1} : {E}_{1} \leftrightarrow {E}_{1}^{\prime } \), such that if \( D \) and \( {D}^{\prime } \) are components of \( {N}_{1} \) and \( {N}_{1}^{\prime } \) , with \( {f}_{1}\left( {\operatorname{Bd}D}\right) = \operatorname{Bd}{D}^{\prime } \), then \( f\left( {M \cap D}\right) = {M}^{\prime } \cap {D}^{\prime } \).
(2) Suppose that we have given a frame \( {N}_{{2i} - 1} \) of \( M \), a frame \( {N}_{{2i} - 1}^{\prime } \) of \( {M}^{\prime } \), and a homeomorphism \( {f}_{{2i} - 1} : {E}_{{2i} - 1} \leftrightarrow {E}_{{2i} - 1}^{\prime } \), where
\[
{E}_{{2i} - 1} = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{{2i} - 1},\;{E}_{{2i} - 1}^{\prime } = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{{2i} - 1}^{\prime }.
\]
Suppose that the components of \( {N}_{{2i} - 1} \) have diameter less than \( 1/\left( {{2i} - 1}\right) \) . For each component \( A \) of \( {N}_{{2i} - 1} \), let \( {A}^{\prime } \) be the component of \( {N}_{{2i} - 1}^{\prime } \) bounded by \( {f}_{{2i} - 1}\left( {\operatorname{Bd}A}\right) \) . Suppose (as an induction hypothesis) that for each such \( A,{A}^{\prime } \) we have \( f\left( {M \cap A}\right) = {M}^{\prime } \cap {A}^{\prime } \) .
Let \( {N}_{2i}^{\prime } \) be a frame of \( {M}^{\prime } \), lying in Int \( {N}_{{2i} - 1}^{\prime } \), and lying in a sufficiently small neighborhood of \( {M}^{\prime } \) so that each component of \( {N}_{2i}^{\prime } \) has diameter less than \( 1/{2i} \) . (Theorem 6.) Then there is a frame \( {N}_{2i} \) of \( M \), lying in Int \( {N}_{{2i} - 1} \) , such that for each component \( {D}^{\prime } \) of \( {N}_{2i}^{\prime },{f}^{-1}\left( {{M}^{\prime } \cap {D}^{\prime }}\right) = M \cap D \), where \( D \) is a component of \( {N}_{2i} \) . (The construction of \( {N}_{2i} \) is like that of \( {N}_{1}^{\prime } \) . We work with the sets \( {M}^{\prime } \cap {A}^{\prime }\left( {A}^{\prime }\right. \) a component of \( \left. {N}_{{2i} - 1}^{\prime }\right) \) one at a time. For each such \( {A}^{\prime } \), let \( A \) be the component of \( {N}_{{2i} - 1} \) such that \( \operatorname{Bd}{A}^{\prime } = {f}_{{2i} - 1}\left( {\operatorname{Bd}A}\right) \) . In the construction of \( {N}_{1}^{\prime } \), described in (1), we use \( {f}^{-1},{M}^{\prime } \cap {A}^{\prime }, M \cap A,{A}^{\prime } \) , and \( A \) in place of \( f, M,{M}^{\prime }, A \) and \( {A}^{\prime } \) respectively.) Now extend \( {f}_{{2i} - 1} \) to get
\[
{f}_{2i} : {E}_{2i} \leftrightarrow {E}_{2i}^{\prime }
\]
where
\[
{E}_{2i} = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{2i},\;{E}_{2i}^{\prime } = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{2i}^{\prime },
\]
such that if \( D \) and \( {D}^{\prime } \) are components of \( {N}_{2i} \) and \( {N}_{2i}^{\prime } \), with \( {f}_{2i}\left( {\operatorname{Bd}D}\right) = \) Bd \( {D}^{\prime } \), then \( f\left( {M \cap D}\right) = {M}^{\prime } \cap {D}^{\prime } \) . (The construction of \( {f}_{2i} \) is like that of \( \left. {f}_{1}\right) \) . Thus, when we pass from \( {N}_{{2i} - 1},{N}_{{2i} - 1}^{\prime },{f}_{{2i} - 1} \) to \( {N}_{2i},{N}_{2i}^{\prime },{f}_{2i} \), the induction hypothesis is preserved.
(3) The recursive step from \( {N}_{2i},{N}_{2i}^{\prime },{f}_{2i} \) to \( {N}_{{2i} + 1},{N}_{{2i} + 1}^{\prime },{f}_{{2i} + 1} \) is entirely similar, and in fact the whole situation is logically symmetric. Thus we have sequences \( {N}_{1},{N}_{2},\ldots ,{N}_{1}^{\prime },{N}_{2}^{\prime },\ldots ,{f}_{1},{f}_{2},\ldots \) such that:
(a) \( {N}_{i} \) is a frame of \( M \), and \( {N}_{i}^{\prime } \) is a frame of \( {M}^{\prime } \) ;
(b) \( {N}_{i + 1} \subset \operatorname{Int}{N}_{i} \) and \( {N}_{i + 1}^{\prime } \subset \operatorname{Int}{N}_{i}^{\prime } \) ;
(c) Each component of \( {N}_{{2i} - 1} \) (or \( {N}_{2i}^{\prime } \) ) has diameter less than \( 1/\left( {{2i} - 1}\right) \) (or \( 1/{2i}) \) ;
(d) Each \( {f}_{i} \) is a homeomorphism \( {E}_{i} \leftrightarrow {E}_{i}^{\prime } \), where
\[
{E}_{i} = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{i},\;{E}_{i}^{\prime } = {\mathbf{R}}^{2} - \operatorname{Int}{N}_{i}^{\prime };
\]
(e) For each \( i,{f}_{i + 1} \) is an extension of \( {f}_{i} \).
Now let
\[
F = f \cup \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{f}_{i}
\]
By (e), \( F \) is a well-defined function. Since
\[
{\mathbf{R}}^{2} = M \cup \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{E}_{i} = {M}^{\prime } \cup \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{E}_{i}^{\prime },
\]
\( F \) is a bijection \( {\mathbf{R}}^{2} \leftrightarrow {\mathbf{R}}^{2} \) . It remains to show that \( F \) and \( {F}^{-1} \) are continuous. Given \( P \in {\mathbf{R}}^{2} - M, Q = F\left( P\right) \), we have \( Q \in {\mathbf{R}}^{2} - {M}^{\prime } \) . Given an open set \( U \) containing \( Q \), we may suppose that \( U \subset \operatorname{Int}{E}_{i}^{\prime } \) for some \( i \) . Since \( {f}_{i} \) is a homeomorphism, some neighborhood of \( P \) is mapped into \( U \) by \( F \) .
If \( P \in M \), then \( Q = F\left( P\right) = f\left( P\right) \in {M}^{\prime } \), and \( Q \) has arbitrarily small neighborhoods which are components \( {D}^{\prime } \) of sets \( {N}_{i}^{\prime } \) . It is now easy to check that \( {D}^{\prime } = F\left( D\right) \) for some component \( D \) of \( {N}_{i} \) . Thus \( F \) maps small neighborhoods of \( P \) onto small neighborhoods of \( Q \) . Therefore \( F \) is continuous. The continuity of \( {F}^{-1} \) can be shown similarly. (Again, the situation is logically symmetric.) | Yes |
If \( {h}_{\mathfrak{p}} \) is the natural homomorphism \( {h}_{\mathfrak{p}} : \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \rightarrow \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p}, \) then \( {h}_{\mathfrak{p}}{}^{-1} \) induces a natural lattice-embedding \( \mathfrak{a} \rightarrow {h}_{\mathfrak{p}}{}^{-1}\left( \mathfrak{a}\right) \) of \( \left( {\mathcal{I}\left( {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p}}\right) , \subset ,\cap , + }\right) \) into \( \left( {\mathcal{I}\left( {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack }\right) , \subset ,\cap , + }\right) \). | Let \( {\mathfrak{a}}_{1} \neq {\mathfrak{a}}_{2} \) be distinct ideals of \( \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p} \); say \( p \in {\mathfrak{a}}_{1} \) but \( p \notin {\mathfrak{a}}_{2} \). Then for any \( q \in \left\{ {{h}_{\mathfrak{p}}{}^{-1}\left( p\right) }\right\} \), we have that \( q \in {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{1}\right) \) and \( q \notin {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{2}\right) \)-that is, \( {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{1}\right) \neq {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{2}\right) \). Hence the mapping is 1:1 on the set of ideals.
Next, for any two ideals \( {\mathfrak{a}}_{1},{\mathfrak{a}}_{2} \) of \( \left. {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack }\right) /\mathfrak{p} \), we have
\[
{h}_{\mathfrak{p}}{}^{-1}\left( {{\mathfrak{a}}_{1} \cap {\mathfrak{a}}_{2}}\right) = {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{1}\right) \cap {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{2}\right)
\]
(11)
\[
{h}_{\mathfrak{p}}{}^{-1}\left( {{\mathfrak{a}}_{1} + {\mathfrak{a}}_{2}}\right) = {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{1}\right) + {h}_{\mathfrak{p}}{}^{-1}\left( {\mathfrak{a}}_{2}\right)
\]
(12)
these follow at once from the definitions of \( {h}_{\mathfrak{p}}{}^{-1}, \cap \), and + . Hence the embedding preserves the lattice structure. | Yes |
Singular homology satisfies Axiom 3. | Immediate from the definition of the boundary map on singular cubes and from the definition of the induced map on singular cubes as composition. | No |
Prove that \\[\\pi \\left( {x, z} \\right) = {xV}\\left( z\\right) + O\\left( {x\\left( \\log z\\right) }^{2}\\exp \\left( {-\\frac{\\log x}{\\log z}}\\right) }\\right) , | where \\[\\begin{aligned} V\\left( z\\right) &= \\mathop{\\prod }\\limits_{{p \\leq z}}\\left( {1 - \\frac{1}{p}}\\right) \\end{aligned} \\]and \\( z = z\\left( x\\right) \\rightarrow \\infty \\) as \\( x \\rightarrow \\infty \\) . | No |
Every convex function that is continuous on an open convex subset \( U \) of a normed space is locally Lipschitzian on \( U \). | Null | No |
Proposition 3.4 The image under \( T \) of the closed unit ball of \( C\left( Y\right) \) is a relatively compact subset of \( C\left( X\right) \) . | Proof. It is clear that \( T\left( {\bar{B}\left( {C\left( Y\right) }\right) }\right) \) is bounded by\n\[
M = \mu \left( Y\right) \mathop{\max }\limits_{{\left( {x, y}\right) \in X \times Y}}\left| {K\left( {x, y}\right) }\right| .
\]\nOn the other hand, \( K \) is uniformly continuous on \( X \times Y \) ; in particular, for all \( \varepsilon \), there exists \( \eta > 0 \) such that\n\[
\left| {K\left( {{x}_{1}, y}\right) - K\left( {{x}_{2}, y}\right) }\right| < \varepsilon \;\text{ for all }y \in Y\text{ and }{x}_{1},{x}_{2} \in X\text{ with }d\left( {{x}_{1},{x}_{2}}\right) < \eta .
\]\nThus, for all \( f \in \bar{B}\left( {C\left( Y\right) }\right) \), we have \( \left| {{Tf}\left( {x}_{1}\right) - {Tf}\left( {x}_{2}\right) }\right| \leq \mu \left( Y\right) \varepsilon \) . Therefore the subset \( T\left( {\bar{B}\left( {C\left( Y\right) }\right) }\right) \) of \( C\left( X\right) \) is equicontinuous, and we can apply Ascoli's Theorem. | Yes |
Let \( p \geq 3 \), let \( K = {\mathbb{F}}_{p} \), let \( E \) be a degenerate curve over \( {\mathbb{F}}_{p} \) as above, and let \( {c}_{6} = {c}_{6}\left( E\right) \) be the invariant defined in Section 7.1.2. Then \( E \) has a cusp (respectively a double point with tangents defined over \( {\mathbb{F}}_{p} \) , respectively a double point with tangents not defined over \( K \) ) if and only if \( \left( \frac{-{c}_{6}}{p}\right) = 0 \) (respectively \( \left( \frac{-{c}_{6}}{p}\right) = 1 \), respectively \( \left( \frac{-{c}_{6}}{p}\right) = - 1 \) ). | Since \( E \) is degenerate it has a singular point, and changing coordinates we may assume that it is at the origin, so that we can choose the equation of our curve to be \( {y}^{2} = {x}^{2}\left( {x + a}\right) \) for some \( a \in {\mathbb{F}}_{p} \) . One computes that \( {c}_{6}\left( E\right) = - {64}{a}^{3} \) . On the other hand, the general equation of a line through the origin is \( y = {tx} \), so such a line is tangent if and only if \( {t}^{2} = a \) . Thus, we have a cusp when \( a = 0 \) (so that \( \left( \frac{{c}_{6}}{p}\right) = 0 \) ), and the tangents are defined over \( {\mathbb{F}}_{p} \) if and only if \( a \) is a square; in other words, \( \left( \frac{-{c}_{6}}{p}\right) = 1 \) since \( - {c}_{6} = {\left( 8a\right) }^{2}a. \) | Yes |
Proposition 11.4. Let \( T = {\left( {S}^{1}\right) }^{k} \) and let \( t = \left( {{e}^{{2\pi }{\theta }_{1}},\ldots ,{e}^{{2\pi }{\theta }_{k}}}\right) \) be an element of \( T \) . Then \( t \) generates a dense subgroup of \( T \) if and only if the numbers 1, \( {\theta }_{1},\ldots ,{\theta }_{k} \) are linearly independent over the field \( \mathbb{Q} \) of rational numbers. | Proof of Proposition 11.4. In light of Lemma 11.5, we may reformulate the proposition as follows: The numbers \( 1,{\theta }_{1},\ldots ,{\theta }_{k} \) are linearly dependent over \( \mathbb{Q} \) if and only if there exists a nonconstant homomorphism \( \Phi : T \rightarrow {S}^{1} \) with \( \left( {{e}^{{2\pi i}{\theta }_{1}},\ldots ,{e}^{{2\pi i}{\theta }_{k}}}\right) \in \ker \left( \Phi \right) \) . Suppose first that there is a dependence relation among \( 1,{\theta }_{1},\ldots ,{\theta }_{k} \) over \( \mathbb{Q} \) . Then after clearing the denominators from this relation, we find that there exist integers \( {m}_{1},\ldots ,{m}_{k} \), not all zero, such that \({m}_{1}{\theta }_{1} + \cdots + {m}_{k}{\theta }_{k} \in \mathbb{Z}.\) Thus, we may define a nonconstant \( \Phi : T \rightarrow {S}^{1} \) by \(\Phi \left( {{u}_{1},\ldots ,{u}_{k}}\right) = {u}_{1}^{{m}_{1}}\cdots {u}_{k}^{{m}_{k}}\) (11.1) and the kernel of \( \Phi \) will contain \( \left( {{e}^{{2\pi i}{\theta }_{1}},\ldots ,{e}^{{2\pi i}{\theta }_{k}}}\right) \). In the other direction, Exercise 2 tells us that every continuous homomorphism \( \Phi : T \rightarrow {S}^{1} \) is of the form (11.1) for some set of integers \( {m}_{1},\ldots ,{m}_{k} \) . Furthermore, if \( \Phi \) is nonconstant, these integers cannot all be zero. Thus, if \({m}_{1}{\theta }_{1} + \cdots + {m}_{k}{\theta }_{k} = n\) for some integer \( n \), we must have \( 1,{\theta }_{1},\ldots ,{\theta }_{k} \) are linearly dependent over \( \mathbb{Q} \). | No |
Proposition 9.6.22 (Stirling’s formula). As \( n \rightarrow \infty \) we have\n\[
n! \sim {n}^{n}{e}^{-n}\sqrt{2\pi n}
\]\nor equivalently,\n\[
\log \left( {n!}\right) = \left( {n + \frac{1}{2}}\right) \log \left( n\right) - n + \frac{1}{2}\log \left( {2\pi }\right) + o\left( 1\right) .
\] | Proof. Once again there are several classical proofs. Certainly the most classical is as follows: if we set \( {u}_{n} = \log \left( {n!/\left( {{n}^{n}{e}^{-n}\sqrt{n}}\right) }\right) \) then\n\[
{u}_{n + 1} - {u}_{n} = 1 - \left( {n + \frac{1}{2}}\right) \log \left( {1 + \frac{1}{n}}\right) \sim - \frac{1}{{12}{n}^{2}};
\]\nhence this is the general term of an absolutely convergent series, so as \( n \rightarrow \infty \) , \( {u}_{n} \) tends to some limit \( \log \left( A\right) \), say (we could also apply the Euler-MacLaurin summation formula). To obtain \( A \) we can use Wallis’s formulas. We let \( {C}_{n} = \) \( {\int }_{0}^{\pi /2}{\cos }^{n}\left( t\right) {dt} \) . By integrating by parts, it is immediate that for \( n \geq 2 \) we have \( {C}_{n} = \left( {n - 1}\right) \left( {{C}_{n - 2} - {C}_{n}}\right) \), hence \( {C}_{n} = \left( {\left( {n - 1}\right) /n}\right) {C}_{n - 2} \) . Since \( {C}_{0} = \pi /2 \) and \( {C}_{1} = 1 \), we deduce that\n\[
{C}_{2k} = \frac{\left( {2k}\right) !}{{2}^{2k}{\left( k!\right) }^{2}}\frac{\pi }{2}\;\text{ and }\;{C}_{{2k} + 1} = \frac{{2}^{2k}{\left( k!\right) }^{2}}{\left( {{2k} + 1}\right) !}.
\]\nOn the other hand, the sequence \( {C}_{n} \) is clearly decreasing, so that in particular \( {C}_{{2k} + 1} \leq {C}_{2k} \leq {C}_{{2k} - 1} \) . If we replace \( {C}_{n} \) by its asymptotic value \( {n}^{n}{e}^{-n}{n}^{1/2}A \) (where \( A \) is the unknown nonzero constant above) a short computation shows that \( {A}^{2} = {2\pi } \), proving Stirling’s formula. Of course the \( o\left( 1\right) \) in the expression for \( \log \left( {n!}\right) \) can be given a complete asymptotic expansion by the Euler-MacLaurin summation formula; see Section 9.2.5. | Yes |
Theorem 5.23 (Jordan Curve Theorem \( {}^{2} \) ). If \( \gamma \) is a simple closed path in \( \mathbb{C} \), then | We shall not prove the above theorem. It is a deep result. In all of our applications, it will be obvious that our Jordan curves have the above properties. | No |
Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( B \) be a maximal linearly independent subset of \( V \) ; then \( B \) is a basis of \( V \) . | Proof. Let \( v \in V, v \notin B \) . Then \( B \cup \{ v\} \) is not linearly independent, by the maximality of \( B \) ; therefore, there exist \( {c}_{0},\ldots ,{c}_{t} \in k \) and (distinct) \( {b}_{1},\ldots ,{b}_{t} \in B \) such that
\[
{c}_{0}v + {c}_{1}{b}_{1} + \cdots + {c}_{t}{b}_{t} = 0,
\]
\( {}^{3} \) This terminology is also often used for free modules over any ring.
with not all \( {c}_{0},\ldots ,{c}_{t} \) equal to 0 . Now, \( {c}_{0} \neq 0 \) : otherwise we would get a linear dependence relation among elements of \( B \) . Since \( k \) is a field, \( {c}_{0} \) is a unit; but then
\[
v = \left( {-{c}_{0}^{-1}{c}_{1}}\right) {b}_{1} + \cdots + \left( {-{c}_{0}^{-1}{c}_{t}}\right) {b}_{t},
\]
proving that \( v \) is in the span of \( B \) . It follows that \( B \) generates \( V \), as needed. | Yes |
A scalar product satisfies the Cauchy-Schwarz inequality | The polynomial t ↦ q(tx + y) = q(x)t^2 + 2b(x, y)t + q(y) takes nonnegative values for t ∈ ℝ . Hence its discriminant 4(b(x, y)^2 - q(x)q(y)) is nonpositive. When the latter vanishes, the polynomial has a real root t0, which implies that t0x + y = 0 . The Cauchy-Schwarz inequality implies immediately q(x + y) ≤ (√q(x) + √q(y))^2, which means that the square root ∥⋅∥ := q^(1/2) satisfies the triangle inequality ∥x + y∥ ≤ ∥x∥ + ∥y∥ | Yes |
Proposition 7.5. Let \( X \) be a homogeneous space for \( G \) . If \( X \) is strictly unimodular, then there exists a left G-invariant volume form on \( X \), unique up to a constant multiple. | Proof. We want to define the invariant form on \( G/H \) by translating a given volume form \( {\omega }_{e} \) on \( {T}_{e}\left( {G/H}\right) \) . On \( G/H \), the left translation \( {L}_{h} \) is induced by conjugation \( {\mathbf{c}}_{h} \) on \( G \) . By Proposition 7.4 and the hypothesis, we have\\
\[
\text{det}T{L}_{h}\left( {e}_{G/H}\right) = \det T{\mathbf{c}}_{h}\left( {e}_{G/H}\right) = 1\text{.}
\]
Hence \( {L}_{h}{\omega }_{e}\left( {G/H}\right) = {\omega }_{e\left( {G/H}\right) } \), that is \( {\omega }_{e\left( {G/H}\right) } \) is invariant under translations by elements of \( H \) . Then for any \( g \in G \) we define\\
\[
{\omega }_{gH} = {L}_{g}{\omega }_{e}\left( {G/H}\right)
\]
The value on the right is independent of the coset representative \( g \), and it is then clear that translation yields the desired \( G \) -invariant volume form on \( G/H \) . The uniqueness up to a constant factor follows because the invariant forms are determined linearly from their values at the origin, and the forms at the origin constitute a 1-dimensional space. This concludes the proof. | Yes |
If \( \mathcal{M} \) is a countable model of \( {PA} \), then there is \( \mathcal{M} \prec \mathcal{N} \) such that \( \mathcal{N} \) is a proper end extension of \( \mathcal{M} \) . | Consider the language \( {\mathcal{L}}^{ * } \) where we have constant symbols for all elements of \( M \) and a new constant symbol \( c \) . Let \( T = {\operatorname{Diag}}_{\mathrm{{el}}}\left( \mathcal{M}\right) \cup \{ c > m \) : \( m \in M\} \), and for \( a \in M \smallsetminus \mathbb{N} \) let \( {p}_{a} \) be the type \( \{ v < a, v \neq m : m \in M\} \) . Any \( \mathcal{N} \vDash T \) is a proper elementary extension of \( \mathcal{M} \) . If \( \mathcal{N} \) omits each \( {p}_{a} \) , then \( \mathcal{N} \) is an end extension of \( \mathcal{M} \) . By Theorem 4.2.4, it suffices to show that each \( {p}_{a} \) is nonisolated. | Yes |
If \( q \) is nondegenerate and represents 0, then there exists a hyperbolic form \( h \) and a nondegenerate form \( {q}^{\prime } \) such that \( q \sim h \oplus {q}^{\prime } \) . Furthermore, \( q \) represents all elements of \( K \) . | This is Lemma 5.1.5 and Corollary 5.1.6. The fact that \( {q}^{\prime } \) is nondegenerate follows from Proposition 5.1.2. | No |
Using Dirichlet's hyperbola method, show that \[
\mathop{\sum }\limits_{{n \leq x}}\frac{f\left( n\right) }{\sqrt{n}} = {2L}\left( {1,\chi }\right) \sqrt{x} + O\left( 1\right)
\]
where \( f\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\chi \left( d\right) \) and \( \chi \neq {\chi }_{0} \) . | Null | No |
Let \( \left( {\Omega ,\mathcal{F},\mathrm{P}}\right) \) be a probability space, and let \( {\left( {X}_{n}\right) }_{n \in \mathbb{N}} \subseteq {\mathrm{L}}^{1}\left( {\Omega ,\mathcal{F},\mathrm{P}}\right) \) be a sequence of independent and identically distributed real random variables. Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n}\left( {{X}_{1} + \cdots + {X}_{n}}\right) = \mathrm{E}\left( {X}_{1}\right) \;\text{ P-almost surely. } \) | Since the \( {X}_{j} \) are identically distributed, \( v \mathrel{\text{:=}} {\mathrm{P}}_{{X}_{j}} \) (the distribution of \( {X}_{j} \) ) is a Borel probability measure on \( \mathbb{R} \), independent of \( j \), and \( \mathrm{E}\left( {X}_{1}\right) = {\int }_{\mathbb{R}}t\mathrm{\;d}v\left( t\right) \) is the common expectation. Define the product space \( \mathrm{X} \mathrel{\text{:=}} \left( {X,\sum ,\mu }\right) \mathrel{\text{:=}} \left( {{\mathbb{R}}^{\mathbb{N}},{\bigotimes }_{\mathbb{N}}\operatorname{Bo}\left( \mathbb{R}\right) ,{\bigotimes }_{\mathbb{N}}v}\right) \). As mentioned in Section 5.1.5, the left shift \( \tau \) is a measurable transformation of \( \left( {X,\sum }\right) \) and \( \mu \) is \( \tau \) -invariant. The measure-preserving system \( \left( {\mathrm{X};\tau }\right) \) is ergodic, and this can be shown in exactly the same way as it was done for the finite state space Bernoulli shift (Proposition 6.20). For \( n \in \mathbb{N} \) let \( {Y}_{n} : X \rightarrow \mathbb{R} \) be the \( {n}^{\text{th }} \) projection and write \( g \mathrel{\text{:=}} {Y}_{1} \) . Then \( {Y}_{j + 1} = g \circ {\tau }^{j} \) for every \( j \geq 0 \), hence Corollary 11.2 yields that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n}\left( {{Y}_{1} + \cdots + {Y}_{n}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n}\mathop{\sum }\limits_{{j = 0}}^{{n - 1}}\left( {g \circ {\tau }^{j}}\right) = {\int }_{X}g\mathrm{\;d}\mu \) pointwise \( \mu \) -almost everywhere. Note that \( {g}_{ * }\mu = v \) and hence \( {\int }_{X}g\mathrm{\;d}\mu = {\int }_{\mathbb{R}}t\mathrm{\;d}v\left( t\right) \). It remains to show that the \( {\left( {Y}_{n}\right) }_{n \in \mathbb{N}} \) are in a certain sense "the same" as the originally given \( {\left( {X}_{n}\right) }_{n \in \mathbb{N}} \) . This is done by devising an injective lattice homomorphism \( \Phi : {\mathrm{L}}^{0}\left( {\mathrm{X};\overline{\mathbb{R}}}\right) \rightarrow {\mathrm{L}}^{0}\left( {\Omega ,\mathcal{F},\mathrm{P};\overline{\mathbb{R}}}\right) \) which carries \( {Y}_{n} \) to \( \Phi \left( {Y}_{n}\right) = {X}_{n} \) for every \( n \in \mathbb{N} \) . Define \( \varphi : \Omega \rightarrow X,\;\varphi \left( \omega \right) \mathrel{\text{:=}} {\left( {X}_{n}\left( \omega \right) \right) }_{n \in \mathbb{N}}\;\left( {\omega \in \Omega }\right) \). Since \( {\left( {X}_{n}\right) }_{n \in \mathbb{N}} \) is an independent sequence, the push-forward measure satisfies \( {\varphi }_{ * }\mathrm{P} = \mu \) . Let \( \Phi = {T}_{\varphi } : f \mapsto f \circ \varphi \) be the Koopman operator induced by \( \varphi \) mapping \( {\mathrm{L}}^{0}\left( {\mathrm{X};\overline{\mathbb{R}}}\right) \) to \( {\mathrm{L}}^{0}\left( {\Omega ,\mathcal{F},\mathrm{P};\overline{\mathbb{R}}}\right) \) . The operator \( \Phi \) is well defined since \( \varphi \) is measure-preserving. By construction, \( \Phi {Y}_{n} = {Y}_{n} \circ \varphi = {X}_{n} \) for each \( n \in \mathbb{N} \) . Moreover, \( \Phi \) is clearly a homomorphism of lattices (see Chapter 7) satisfying \( \mathop{\sup }\limits_{{n \in \mathbb{N}}}\Phi \left( {f}_{n}\right) = \Phi \left( {\mathop{\sup }\limits_{{n \in \mathbb{N}}}{f}_{n}}\right) \;\text{ and }\;\mathop{\inf }\limits_{{n \in \mathbb{N}}}\Phi \left( {f}_{n}\right) = \Phi \left( {\mathop{\inf }\limits_{{n \in \mathbb{N}}}{f}_{n}}\right) \) for every sequence \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) in \( {\mathrm{L}}^{0}\left( {\mathrm{X};\overline{\mathbb{R}}}\right) \) . Since the almost everywhere convergence of a sequence can be described in purely lattice theoretic terms involving only countable suprema and infima (cf. also (11.1)), one has \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n} = f\;\mu \text{-a.e. } \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}\Phi \left( {f}_{n}\right) = \Phi \left( f\right) \;\text{ P-almost surely. } \) This, for \( {f}_{n} \mathrel{\text{:=}} \frac{1}{n}\left( {{Y}_{1} + \cdots + {Y}_{n}}\right) \) and \( f \mathrel{\text{:=}} \mathrm{E}\left( {X}_{1}\right) \mathbf{1} \), concludes the proof. | Yes |
As an illustration of Theorem 9.1.6 we consider a Goppa code \\[\\left. \\Gamma \\left( L, g\\left( z\\right) \\right) = {C}_{\\Omega }\\left( {D}_{L},{G}_{0} - {P}_{\\infty }\\right) \\right| \\limits_{{\\mathbb{F}}_{q}} \\] (notation as in Definition 2.3.10 and Proposition 2.3.11). Let \\( {g}_{1}\\left( z\\right) \\in {\\mathbb{F}}_{{q}^{m}}\\left\\lbrack z\\right\\rbrack \\) be the polynomial of maximal degree such that \\( {g}_{1}{\\left( z\\right) }^{q} \\) divides \\( g\\left( z\\right) \\) . We set \\( {G}_{1} \\mathrel{\\text{:=}} \\left( {g}_{1}\\left( z\\right) \\right) \\limits_{0} - {P}_{\\infty } \\) where \\( \\left( {g}_{1}\\left( z\\right) \\right) \\limits_{0} \\) is the zero divisor of \\( {g}_{1}\\left( z\\right) ,\\) and obtain from (9.12) the estimate \\[\\dim \\Gamma \\left( {L, g\\left( z\\right) }\\right) \\geq n - m\\left( {\\deg g\\left( z\\right) - \\deg {g}_{1}\\left( z\\right) }\\right) .\\] | Null | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.