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Let \( {\psi }_{1} \) denote the scaled Airy function in (15.26), let \( {\widetilde{\psi }}_{1} \) denote the same function with the Airy function replaced by the right-hand side of (15.33), and let \( {\psi }_{2} \) denote the oscillatory WKB function in (15.27). If \( x - a \) is positive and of order \( {\hslash }^{1/2} \), we have | Proof of Lemma 15.9. We consider only the estimates for the derivatives of the functions involved. The analysis of the functions themselves is similar (but easier) and is left as an exercise to the reader (Exercise 11).\n\nWe begin by considering \( {\psi }_{1}^{\prime } - {\widetilde{\psi }}_{1}^{\prime } \) . With a little algebra, we compute that\n\n\[
\frac{d{\psi }_{1}}{dx} - \frac{d{\widetilde{\psi }}_{1}^{\prime }}{dx} = - \sqrt{\pi }{\left( 2m{F}_{0}\right) }^{1/6}{\hslash }^{-5/6}\left( {{\operatorname{Ai}}^{\prime }\left( u\right) - {\widetilde{\operatorname{Ai}}}^{\prime }\left( u\right) }\right)
\]
(15.41)\n\nwhere \( u \) is as in (15.22) and where \( \widetilde{\mathrm{{Ai}}} \) is the function on the right-hand side of (15.33).\n\nNow, \( \operatorname{Ai}\left( u\right) \) has an asymptotic expansion for \( u \rightarrow - \infty \) given by\n\n\[
\operatorname{Ai}\left( u\right) = \widetilde{\operatorname{Ai}}\left( u\right) \left( {1 + C{u}^{-3/2} + \cdots }\right)
\]
and \( {\mathrm{{Ai}}}^{\prime }\left( u\right) \) has the asymptotic expansion obtained by formally differentiating this with respect to \( u \) . [See Eq. (7.64) in [30].] From this, we obtain\n\n\[
{\mathrm{{Ai}}}^{\prime }\left( u\right) - {\widetilde{\mathrm{{Ai}}}}^{\prime }\left( u\right) = {\widetilde{\mathrm{{Ai}}}}^{\prime }\left( u\right) O\left( {\left( -u\right) }^{-3/2}\right) + \widetilde{\mathrm{{Ai}}}\left( u\right) O\left( {\left( -u\right) }^{-5/2}\right) .
\]
(15.42)\n\nFrom the explicit formula for \( \widetilde{\mathrm{{Ai}}} \), we see that \( \widetilde{\mathrm{{Ai}}}\left( u\right) \) is of order \( {\left( -u\right) }^{-1/4} \) . Meanwhile, the formula \( {\widetilde{\mathrm{{Ai}}}}^{\prime }\left( u\right) \) will contain two terms, the larger of which will be of order \( {u}^{1/4} \) . Thus, the slower-decaying term on the right-hand side of (15.42) is the first one, which is of order \( {\left( -u\right) }^{-5/4} \) . Now, in the transition regions, \( u \) behaves like \( {\hslash }^{-2/3}{\hslash }^{1/2} = {\hslash }^{-1/6} \) . Thus,(15.42) goes like \( {\hslash }^{5/{24}} \) and so (15.41) goes like \( {\hslash }^{-5/6 + 5/{24}} = {\hslash }^{-5/8} \), as claimed.\n\nWe now consider \( {\widetilde{\psi }}_{1}^{\prime } - {\psi }_{2}^{\prime } \) . By direct calculation, the derivatives of \( {\widetilde{\psi }}_{1} \) and \( {\psi }_{2} \) each consist of two terms, a "dominant" obtained by differentiating the cosine factor and a "subdominant" term obtained by differentiating the coefficient of the cosine factor. In the case of \( {\widetilde{\psi }}_{1}^{\prime } \), the dominant term in the derivative may be simplified to\n\n\[
- \frac{1}{\hslash }{\left( \left( 2m{F}_{0}\right) \left( x - a\right) \right) }^{1/4}\sin \left( {\frac{2}{3}{\left( -u\right) }^{3/2} - \frac{\pi }{4}}\right) .
\]
(15.43)\n\nAccording to Exercise 12, we have, when \( x - a \) is of order \( {\hslash }^{1/2} \), the estimates\n\n\[
{\left( \left( 2m{F}_{0}\right) \left( a - x\right) \right) }^{1/4} = \sqrt{p} + \sqrt{p}O\left( {\hslash }^{1/2}\right)
\]
(15.44)\n\nand\n\n\[
\frac{2}{3}{\left( -u\right) }^{3/2} = \frac{1}{\hslash }{\int }_{a}^{x}p\left( y\right) {dy} + O\left( {\hslash }^{1/4}\right) .
\]
\( \left( {15.45}\right) \)\n\nSince the derivative of \( \sin \theta \) is bounded, a change of order \( {\hslash }^{1/4} \) in the argument of a sine function produces a change of order \( {\hslash }^{1/4} \) in the value of the sine. Thus, if we substitute (15.44) and (15.45) into (15.43), we find that the difference between the dominant term in \( {\widetilde{\psi }}_{1}^{\prime } \) and the dominant term in \( {\psi }_{1}^{\prime } \) is\n\n\[
\frac{1}{\hslash }\sqrt{p}O\left( {\hslash }^{1/4}\right) + \text{lower-order terms.}
\]
Since \( \sqrt{p} \) is of order \( {\left( x - a\right) }^{1/4} \) or \( {\hslash }^{1/8} \), we get an error of order \( {\hslash }^{-5/8} \), as claimed.\n\nFinally, the subdominant terms in the derivatives of \( {\widetilde{\psi }}_{1} \) and \( {\psi }_{2} \) are easily seen to be separately of order \( {\hslash }^{-5/8} \) . Thus, even without taking into account the cancellation between these terms, they do not change the order of the estimate. | No |
Theorem 8. Let \( 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \) be a short exact sequence of \( R \) -modules. Then there is a long exact sequence of abelian groups | Proof: Take a simultaneous projective resolution of the short exact sequence as in Proposition 7 and take homomorphisms into \( D \) . To obtain the cohomology groups \( {\operatorname{Ext}}_{R}^{n} \) from the resulting diagram, as noted in the discussion preceding Proposition 3 we replace the lowest nonzero row in the transformed diagram with a row of zeros to get the following commutative diagram:\nThe columns of (13) are cochain complexes, and the rows are split by Proposition 29(2) of Section 10.5 and the discussion following it. Thus (13) is a short exact sequence of cochain complexes. Theorem 2 then gives a long exact sequence of cohomology groups whose terms are, by definition, the groups \( {\operatorname{Ext}}_{R}^{n}\left( {\_, D}\right) \), for \( n \geq 0 \) . The \( {0}^{\text{th }} \) order terms are identified by Proposition 3, completing the proof. | Yes |
Singular homology satisfies Axiom 4. | We have defined \( {C}_{n}\left( {X, A}\right) = {C}_{n}\left( X\right) /{C}_{n}\left( A\right) \) . Thus for every \( n \), we have a short exact sequence\n\[
0 \rightarrow {C}_{n}\left( A\right) \rightarrow {C}_{n}\left( X\right) \rightarrow {C}_{n}\left( {X, A}\right) \rightarrow 0.
\]\nIn other words, we have a short exact sequence of chain complexes\n\[
0 \rightarrow {C}_{ * }\left( A\right) \rightarrow {C}_{ * }\left( X\right) \rightarrow {C}_{ * }\left( {X, A}\right) \rightarrow 0.
\]\nBut then we have a long exact sequence in homology by Theorem A.2.10. | Yes |
Every affine morphism of the multiplicative group \( {\mathbb{G}}_{m} \) has the form \( \psi \left( z\right) = a{z}^{d} \) for some nonzero \( a \) and some \( d \in \mathbb{Z} \) . More generally, for any commutative group \( G \), any \( a \in G \), and any \( d \in \mathbb{Z} \) there is an affine morphism \( \psi \left( z\right) = a{z}^{d} \) . | Notice that it is easy to compute the iterates of this map,\n\[
{\psi }^{n}\left( z\right) = {a}^{1 + d + \cdots + {d}^{n - 1}}{z}^{{d}^{n}}.
\] | No |
Theorem 3.4.23 (The isomorphism theorem for measure spaces) If \( \mu \) is a continuous probability on a standard Borel space \( X \), then there is a Borel isomorphism \( h : X \rightarrow I \) such that for every Borel subset \( B \) of \( I,\lambda \left( B\right) = \) \( \mu \left( {{h}^{-1}\left( B\right) }\right) \) . | Proof. By the Borel isomorphism theorem (3.3.13), we can assume that \( X = I \) . Let \( F : I \rightarrow I \) be the distribution function of \( \mu \) . So, \( F \) is a continuous, nondecreasing map with \( F\left( 0\right) = 0 \) and \( F\left( 1\right) = 1 \) . Let
\[
N = \left\{ {y \in I : {F}^{-1}\left( {\{ y\} }\right) }\right. \text{contains more than one point}\} \text{.}
\]
Since \( F \) is monotone, \( N \) is countable. If \( N \) is empty, take \( h = F \) . Otherwise, we take an uncountable Borel set \( M \subset I \smallsetminus N \) of Lebesgue measure 0, e.g., \( \mathcal{C} \smallsetminus N \) . So, \( \mu \left( {{F}^{-1}\left( M\right) }\right) = 0 \) . Put \( Q = M \cup N \) and \( P = {F}^{-1}\left( Q\right) \) . Both \( P \) and \( Q \) are uncountable Borel sets with \( \mu \left( P\right) = \lambda \left( Q\right) = 0 \) . Fix a Borel isomorphism \( g : P \rightarrow Q \) . Define
\[
h\left( x\right) = \left\{ \begin{array}{lll} g\left( x\right) & \text{ if } & x \in P, \\ F\left( x\right) & \text{ if } & x \in I \smallsetminus P. \end{array}\right.
\]
The map \( h \) has the desired properties. | Yes |
Theorem 2.4.1 For any \( y > 0 \), \(\mathop{\sum }\limits_{{n \leq x}}f\left( n\right) = \mathop{\sum }\limits_{{d \leq y}}g\left( d\right) H\left( \frac{x}{d}\right) + \mathop{\sum }\limits_{{d \leq \frac{x}{y}}}h\left( d\right) G\left( \frac{x}{d}\right) - G\left( y\right) H\left( \frac{x}{y}\right) \). | Proof. We have \(\mathop{\sum }\limits_{{n \leq x}}f\left( n\right) = \mathop{\sum }\limits_{{{de} \leq x}}g\left( d\right) h\left( e\right)\) \(\mathop{\sum }\limits_{{{de} \leq x}}g\left( d\right) h\left( e\right) + \mathop{\sum }\limits_{{{de} \leq x}}g\left( d\right) h\left( e\right)\) \(\mathop{\sum }\limits_{{d \leq y}}^{{d \leq y}}g\left( d\right) H\left( \frac{x}{d}\right) + \mathop{\sum }\limits_{{e \leq \frac{x}{y}}}^{{d > y}}h\left( e\right) \left\{ {G\left( \frac{x}{e}\right) - G\left( y\right) }\right\}\) \(\mathop{\sum }\limits_{{d \leq y}}g\left( d\right) H\left( \frac{x}{d}\right) + \mathop{\sum }\limits_{{e \leq \frac{x}{y}}}h\left( e\right) G\left( \frac{x}{e}\right) - G\left( y\right) H\left( \frac{x}{y}\right) \). | Yes |
Proposition 22.8 Take the symplectic potential \( \theta = {p}_{j}d{x}_{j} \) . Then the position, momentum, and holomorphic subspaces may be computed as follows. The position subspace consists of smooth functions \( \psi \) on \( {\mathbb{R}}^{2n} \) of the form | Proof. Since \( \theta \left( {\partial /\partial {p}_{j}}\right) = 0 \), we have \( {\nabla }_{\partial /\partial {p}_{j}} = \partial /\partial {p}_{j} \), so that functions that are covariantly constant in the \( \mathbf{p} \) -directions are actually constant in the \( \mathbf{p} \) -directions. Meanwhile, \( \theta \left( {\partial /\partial {x}_{j}}\right) = {p}_{j} \) and so \({\nabla }_{\partial /\partial {x}_{j}} = \frac{\partial }{\partial {x}_{j}} - \frac{i}{\hslash }{p}_{j}\) ... | No |
Lemma 5.9.1 An independent set \( C \) in a Moore graph of diameter two and valency seven contains at most 15 vertices. If \( \left| C\right| = {15} \), then every vertex not in \( C \) has exactly three neighbours in \( C \) . | Let \( X \) be a Moore graph of diameter two and valency seven. Suppose that \( C \) is an independent set in \( X \) with \( c \) vertices in it. Without loss of generality we may assume that the vertices are labelled so that the vertices \( \{ 1,\ldots ,{50} - c\} \) are the ones not in \( C \) . If \( i \) is a vertex not in \( C \), let \( {k}_{i} \) denote the number of its neighbours that lie in \( C \) . Since no two vertices in \( C \) are joined by an edge, we have
\[
{7c} = \mathop{\sum }\limits_{{i = 1}}^{{{50} - c}}{k}_{i}
\]
Now, consider the paths of length two joining two vertices in \( C \) . Since every pair of nonadjacent vertices in \( X \) has exactly one common neighbour, counting these in two ways yields
\[
\left( \begin{array}{l} c \\ 2 \end{array}\right) = \mathop{\sum }\limits_{{i = 1}}^{{{50} - c}}\left( \begin{matrix} {k}_{i} \\ 2 \end{matrix}\right)
\]
From these last two equations it follows that for any real number \( \mu \) ,
\[
\mathop{\sum }\limits_{{i = 1}}^{{{50} - c}}{\left( {k}_{i} - \mu \right) }^{2} = \left( {{50} - c}\right) {\mu }^{2} - {14c\mu } + {c}^{2} + {6c}.
\]
(5.1)
The right side here must be nonnegative for all values of \( \mu \), so regarding it as a quadratic in \( \mu \), we see that it must have at most one zero. Therefore, the discriminant
\[
{196}{c}^{2} - 4\left( {{50} - c}\right) \left( {{c}^{2} + {6c}}\right) = {4c}\left( {c - {15}}\right) \left( {c + {20}}\right)
\]
of the quadratic must be less than or equal to 0 . It follows that \( c \leq {15} \).
If \( c = {15} \), then the right side of (5.1) becomes
\[
{35}{\mu }^{2} - {210\mu } + {315} = {35}{\left( \mu - 3\right) }^{2}
\]
and so setting \( \mu \) equal to three in (5.1) yields that
\[
\mathop{\sum }\limits_{{i = 1}}^{{35}}{\left( {k}_{i} - 3\right) }^{2} = 0
\]
Therefore, \( {k}_{i} = 3 \) for all \( i \), as required. | Yes |
A connected s-arc transitive graph with girth \( {2s} - 2 \) is distance-transitive with diameter \( s - 1 \) . | Let \( X \) satisfy the hypotheses of the lemma and let \( \left( {u,{u}^{\prime }}\right) \) and \( \left( {v,{v}^{\prime }}\right) \) be pairs of vertices at distance \( i \) . Since \( X \) has diameter \( s - 1 \) by Lemma 4.1.4, we see that \( i \leq s - 1 \) . The two pairs of vertices are joined by paths of length \( i \), and since \( X \) is transitive on \( i \) -arcs, there is an automorphism mapping \( \left( {u,{u}^{\prime }}\right) \) to \( \left( {v,{v}^{\prime }}\right) \) . | No |
Proposition 3.2. In a ring extension \( E \) of \( R \), the elements of \( E \) that are integral over \( R \) constitute a subring of \( E \) . | Null | No |
If \( n \equiv 1\left( {\;\operatorname{mod}\;p}\right) \), prove that \( {n}^{{p}^{m}} \equiv 1\left( {\;\operatorname{mod}\;{p}^{m + 1}}\right) \) . | Null | No |
Proposition 8.5.2 Let \( v \in {\widetilde{S}}_{n}^{B} \) . Then, | Proof. This follows immediately from equations (8.68), (8.69), and (8.71) and the fact that \( v\left( 0\right) = 0 \) and \( v\left( {n + 2}\right) = N - v\left( {n - 1}\right) \) . \( ▱ \) | Yes |
Lemma 11.28. For \( t \in T \), let \( {\operatorname{Ad}}_{{t}^{-1}}^{\prime } \) denote the restriction of \( {\operatorname{Ad}}_{{t}^{-1}} \) to \( \mathfrak{f} \) . | The operator \( {\mathrm{{Ad}}}_{{t}^{-1}}^{\prime } - I \) is invertible provided that the restriction of \( {\mathrm{{Ad}}}_{{t}^{-1}} \) to \( f \) does not have an eigenvalue of 1 . Suppose, then, that \( {\operatorname{Ad}}_{{t}^{-1}}\left( X\right) = X \) for some \( X \in \mathfrak{f} \) . Then for every integer \( m \), we will have \( {\operatorname{Ad}}_{{t}^{m}}\left( X\right) = X \) . If \( t \) generates a dense subgroup of \( T \), then by taking limits, we conclude that \( {\operatorname{Ad}}_{s}\left( X\right) = X \) for all \( s \in T \) . But then for all \( H \in \mathfrak{t} \), we have
\[
\left\lbrack {H, X}\right\rbrack = {\left. \frac{d}{d\tau }{\operatorname{Ad}}_{{e}^{\tau H}}\left( X\right) \right| }_{\tau = 0} = 0.
\]
Since \( \mathfrak{t} \) is maximal commutative (Proposition 11.7), we conclude that \( X \in \mathfrak{f} \cap \mathfrak{t} = \) \( \{ 0\} \) . Thus, there is no nonzero \( X \in \mathfrak{f} \) for which \( {\operatorname{Ad}}_{{t}^{-1}}\left( X\right) = X \) .
For the second point, if \( w \in W \) is represented by \( x \in N\left( T\right) \), we have
\[
{\operatorname{Ad}}_{w \cdot {t}^{-1}}^{\prime } - I = {\operatorname{Ad}}_{x{t}^{-1}{x}^{-1}}^{\prime } - I
\]
\[
= {\operatorname{Ad}}_{x}\left( {{\operatorname{Ad}}_{{t}^{-1}}^{\prime } - I}\right) {\operatorname{Ad}}_{{x}^{-1}}.
\]
Thus, \( {\mathrm{{Ad}}}_{w \cdot {t}^{-1}}^{\prime } - I \) and \( {\mathrm{{Ad}}}_{{t}^{-1}}^{\prime } - I \) are similar and have the same determinant. | Yes |
Let \( A \) and \( B \) be Hopf algebras over a field. Then \({P}_{A \otimes B} = {P}_{A} + {P}_{B}\) | Let \( u = \sum a \otimes b \) be an element of \( {P}_{A \otimes B} \) . Then we have \(\sum \delta \left( a\right) \otimes \delta \left( b\right) = {s}_{23}\left( {\delta \left( u\right) }\right) = \sum a \otimes {1}_{A} \otimes b \otimes {1}_{B} + \sum {1}_{A} \otimes a \otimes {1}_{B} \otimes b\) where \( {s}_{23} \) is the switch of the 2nd and 3rd tensor factors. Now apply the linear map \( {\varepsilon }_{A} \otimes {i}_{A} \otimes {i}_{B} \otimes {\varepsilon }_{B} \) . This yields \(u = \sum \varepsilon \left( a\right) {1}_{A} \otimes b + \sum a \otimes \varepsilon \left( b\right) {1}_{B}.\) Since \( u \) is primitive, it is annihilated by the counit \( {\varepsilon }_{A} \otimes {\varepsilon }_{B} \) of \( A \otimes B \), whence we have \( \sum \varepsilon \left( a\right) \varepsilon \left( b\right) = 0 \) . Using this, we rewrite \( u \) by replacing \( b \) with \( b - \varepsilon \left( b\right) {1}_{B} \) in the first sum, and \( a \) with \( a - \varepsilon \left( a\right) {1}_{A} \) in the second sum. Then we have \(u = {1}_{A} \otimes s + r \otimes {1}_{B}\) with \( {\varepsilon }_{B}\left( s\right) = 0 = {\varepsilon }_{A}\left( r\right) \) . From the primitivity of \( u \), we get (using \( {s}_{23} \) as above) \({1}_{A} \otimes {1}_{A} \otimes \delta \left( s\right) + \delta \left( r\right) \otimes {1}_{B} \otimes {1}_{B} = {1}_{A} \otimes {1}_{A} \otimes \left( {s \otimes {1}_{B} + {1}_{B} \otimes s}\right) + \left( {r \otimes {1}_{A} + {1}_{A} \otimes r}\right) \otimes {1}_{B} \otimes {1}_{B}.\) Applying \( {\varepsilon }_{A} \otimes {\varepsilon }_{A} \otimes {i}_{B} \otimes {i}_{B} \), we find that \( \delta \left( s\right) = s \otimes {1}_{B} + {1}_{B} \otimes s \), so that \( s \) belongs to \( {P}_{B} \) . Similarly, \( r \) belongs to \( {P}_{A} \) . | Yes |
For all \( n,{P}_{n}^{\text{corr }} \) and \( {P}_{n + 1}^{\text{cut }} \) are linearly isomorphic. | Let \( f : {\mathbb{R}}^{n \times n} \rightarrow {\mathbb{R}}^{{E}_{n + 1}} \) be the linear function that maps each \( x \in {\mathbb{R}}^{n \times n} \) to the element \( y \in {\mathbb{R}}^{{E}_{n + 1}} \) defined by
\[
{y}_{ij} = \left\{ \begin{array}{ll} {x}_{ii} & \text{ if }1 \leq i \leq n, j = n + 1 \\ {x}_{ii} + {x}_{jj} - 2{x}_{ij} & \text{ if }1 \leq i < j \leq n. \end{array}\right.
\]
One can verify that \( f\left( {P}_{n}^{\text{corr }}\right) = {P}_{n + 1}^{\text{cut }} \) and, for every \( y \in {P}_{n + 1}^{\text{cut }} \), the point \( x \in {\mathbb{R}}^{n \times n} \) defined by
\[
{x}_{ij} = \left\{ {\begin{array}{ll} {y}_{i, n + 1} & \text{ if }i = j \\ \frac{1}{2}\left( {{y}_{i, n + 1} + {y}_{j, n + 1} - {y}_{ij}}\right) & \text{ if }i \neq j \end{array}\;i, j = 1,\ldots, n.}\right.
\]
is the only point in \( {P}_{n}^{\text{corr }} \) such that \( f\left( x\right) = y \) . | Yes |
Let \( X \) be a manifold with a spray or covariant derivative \( D \) . There exists a unique vector bundle morphism (over \( \pi \) )
\[
K : {TTX} \rightarrow {TX}
\]
such that for all vector fields \( \xi ,\zeta \) on \( X \), we have
(8)
\[
{D}_{\xi }\zeta = K \circ {T\zeta } \circ \xi ,\;\text{ in other words,}\;D = K \circ T
\]
as operators on vector fields, so the following diagram is commutative: | In a chart \( U \), we let the local representation
\[
{K}_{U,\left( {x, v}\right) } : \mathbf{E} \times \mathbf{E} \rightarrow \mathbf{E}
\]
be given by
\( \left( {8}_{U}\right) \)
\[
{K}_{U,\left( {x, v}\right) }\left( {z, w}\right) = w - {B}_{U}\left( {x;v, z}\right) ,
\]
so \( K = {S}_{2} \) satisfies the requirements of the lemma. | No |
Proposition 7.2. Let \( E \) be free, finite dimensional over \( R \) . Then we have an algebra-isomorphism \( T\left( {L\left( E\right) }\right) = T\left( {{\operatorname{End}}_{R}\left( E\right) }\right) \rightarrow {LT}\left( E\right) = {\bigoplus }_{r = 0}^{\infty }{\operatorname{End}}_{R}\left( {{T}^{r}\left( E\right) }\right) \) given by \( f \otimes g \mapsto T\left( {f, g}\right) \) | Proof. By Proposition 2.5, we have a linear isomorphism in each dimension, and it is clear that the map preserves multiplication. | No |
The spheres \( {S}_{r}\left( a\right) \left( {r > 0}\right) \) are both open and closed. | The spheres are closed in all metric spaces, since the distance function \( x \mapsto d\left( {x, a}\right) \) is continuous. A sphere of positive radius is open in an ultrametric space by part \( \left( c\right) \) of the previous lemma. | No |
Theorem 7.11. Let \( S \) be a subgroup of finite index in \( G \) . Let \( F \) be an \( S \) -module, and \( E \) a \( G \) -module (over the commutative ring \( R \) ). Then there is an isomorphism | The \( G \) -module \( {\operatorname{ind}}_{S}^{G}\left( F\right) \) contains \( F \) as a summand, because it is the direct sum \( \bigoplus {\lambda }_{i}F \) with left coset representatives \( {\lambda }_{i} \) as in Theorem 7.3. Hence we have a natural \( S \) -isomorphism\\
\[
f : {\operatorname{res}}_{S}\left( E\right) \otimes F\overset{ \approx }{ \rightarrow }E \otimes {\lambda }_{1}F \subset E \otimes {\operatorname{ind}}_{S}^{G}\left( F\right) .
\]
taking the representative \( {\lambda }_{1} \) to be 1 (the unit element of \( G \ )). By the universal property of induction, there is a \( G \) -homomorphism\\
\[
{\operatorname{ind}}_{S}^{G}\left( f\right) : {\operatorname{ind}}_{S}^{G}\left( {{\operatorname{res}}_{S}\left( E\right) \otimes F}\right) \rightarrow E \otimes {\operatorname{ind}}_{S}^{G}\left( F\right) ,
\]
which is immediately verified to be an isomorphism, as desired. (Note that here it only needed to verify the bijectivity in this last step, which comes from the structure of direct sum as \( R \) -modules.) | Yes |
Let \( \Omega \) be a bounded domain with \( {C}^{2} \) boundary and \( S \) its Szegő kernel. With \( \mathcal{P}\left( {z,\zeta }\right) \) as defined above, and with \( f \in C\left( \bar{\Omega }\right) \) holomorphic on \( \Omega \) , we have
\[
f\left( z\right) = {\int }_{\partial \Omega }\mathcal{P}\left( {z,\zeta }\right) f\left( \zeta \right) \mathrm{d}\sigma \left( \zeta \right)
\]
for all \( z \in \Omega \) . | See Proposition 1.2.9. | No |
If \( w \in W \) and \( s \in S \smallsetminus S\left( w\right) \) satisfy \( l\left( {sws}\right) < l\left( w\right) + 2 \), then \( s \) commutes with all elements of \( S\left( w\right) \) . | We have \( l\left( {sw}\right) = l\left( w\right) + 1 = l\left( {ws}\right) \) by Lemma 2.15. Therefore, in view of the folding condition (Section 2.3.1), the hypothesis \( l\left( {sws}\right) < l\left( w\right) + 2 \) is equivalent to the equation \( {sw} = {ws} \) . We now show by induction on \( l \mathrel{\text{:=}} l\left( w\right) \) that \( s \) commutes with all elements of \( S\left( w\right) \) . We may assume \( l \geq 1 \) . Choose a reduced decomposition \( w = {s}_{1}\cdots {s}_{l} \), and consider the equation
\[
s{s}_{1}\cdots {s}_{l} = {s}_{1}\cdots {s}_{l}s.
\]
By Theorem 2.33, we can perform M-operations of type (II) to the word on the left in order to convert it to the word on the right. One of these operations must involve the initial \( s \) . Prior to applying this operation, we have a word of the form \( s{t}_{1}\cdots {t}_{l} \) with \( w = {t}_{1}\cdots {t}_{l} \) and all \( {t}_{i} \in S \smallsetminus \{ s\} \) ; so the operation is possible only if \( m\left( {s,{t}_{1}}\right) = 2 \) . Thus \( s \) commutes with \( {t}_{1} \), hence also with \( {t}_{1}w = {t}_{2}\cdots {t}_{l} \), and an application of the induction hypothesis completes the proof. | Yes |
Theorem 12.24 (FRIEDRICHS). Let \( S \) be densely defined, symmetric and lower bounded in \( H \) . There exists a selfadjoint extension \( T \) with the same lower bound \( m\left( T\right) = m\left( S\right) \), and with \( D\left( T\right) \) contained in the completion of \( D\left( S\right) \) in the norm \( {\left( \left( Sv, v\right) + \left( 1 - m\left( S\right) \right) \parallel v{\parallel }^{2}\right) }^{\frac{1}{2}} \) . | Assume first that \( m\left( S\right) = c > 0 \) . The sesquilinear form
\[
{s}_{0}\left( {u, v}\right) = \left( {{Su}, v}\right)
\]
is then a scalar product on \( D\left( S\right) \) (cf. Theorem 12.12), and we denote the completion of \( D\left( S\right) \) with respect to this scalar product by \( V \) . Hereby \( {s}_{0}\left( {u, v}\right) \) is extended to a sesquilinear form \( s\left( {u, v}\right) \) with \( D\left( s\right) = V(s \) is the scalar product itself on \( V \) ). | No |
If \( \left( {X,\mu }\right) \) is a measure space, if \( f\left( x\right) \) is measurable, and if \( 0 < p < \infty \), then | We may assume \( f \) vanishes except on a set of \( \sigma \) -finite measure, because otherwise both sides of (4.1) are infinite. Then Fubini's theorem shows that both sides of (4.1) equal the product measure of the ordinate set \( \left\{ {\left( {x,\lambda }\right) : 0 < \lambda < {\left| f\left( x\right) \right| }^{p}}\right\} \) . That is,\[
\int {\left| f\right| }^{p}{d\mu } = {\iint }_{0}^{\left| f\right| }p{\lambda }^{p - 1}{d\lambda d\mu } = {\int }_{0}^{\infty }p{\lambda }^{p - 1}\mu \left( {\left| f\right| > \lambda }\right) {d\lambda }
\]
\[
= {\int }_{0}^{\infty }p{\lambda }^{p - 1}m\left( \lambda \right) {d\lambda }
\] | Yes |
Proposition 7.1. (1) \( {H}^{0}\left( {G, A}\right) \cong \{ a \in A \mid {xa} = a \) for all \( x \in G\} \) . | Null | No |
Let \( k \geq 2 \) be an even integer. We have \( {B}_{k} \equiv k + 1/2\left( {\;\operatorname{mod}\;{2}^{2 + {v}_{2}\left( k\right) }}\right) \), so in particular \( {B}_{k} \equiv k + 1/2 \) \( \left( {\;\operatorname{mod}\;4}\right) \) and \( {B}_{k} \equiv 1/2\left( {\;\operatorname{mod}\;2}\right) \) . | Null | No |
A (non-empty) closed, convex, locally compact, and line-free set \( A \) in \( X \) has an extreme point. | We may assume that \( A \) is not compact. Then \( {C}_{A} \) is a non-trivial closed cone in \( X \) (closure follows from equation (8.5)). Further \( {C}_{A} \) is itself locally compact since a translate of it lies in \( A \) . Let \( \phi \in {X}^{ * } \) be a strictly positive linear functional and let \( K \) be the half-space \( \{ x \in X : \phi \left( x\right) \leq 0\} \) . If we translate \( {C}_{A} \) and \( K \) to a point \( {x}_{o} \in A \) we see that the set \( A \cap \left( {{x}_{o} + K}\right) \equiv B \) must be compact, since otherwise, by its local compactness, it would contain a half-line outside the set \( {x}_{o} + {C}_{A} \), in contradiction to the definition of \( {C}_{A} \) . Finally, either \( B \) is contained in the hyperplane \( \left\{ {x \in \bar{X} : \phi \left( x\right) = \phi \left( {x}_{o}\right) }\right\} \) in which case \( \operatorname{ext}\left( B\right) \subset \operatorname{ext}\left( A\right) \left( \mathbf{{8A}}\right) \), or else \( B \) has an extreme point not in this hyperplane; but such a point must again be an extreme point of \( A \) . | Yes |
Theorem 10.33. Let \( S \) be defined as above with \( {b}_{j} = 0\left( {j = 1,2,\ldots, m}\right) \) , \( q \in {M}_{\rho ,\text{ loc }}\left( {\mathbb{R}}^{m}\right) \) and \( {q}_{ - } \in {M}_{\rho }\left( {\mathbb{R}}^{m}\right) \) for some \( \rho < 4 \) . Then \( S \) is bounded from below. If the lowest point of \( \sigma \left( S\right) \) is an eigenvalue, then it is simple. | By Theorem 10.29(a) the operators \( S \) and \( {S}_{0} - {q}_{ - } \) are bounded from below. The lower bound of \( {S}_{0} - {q}_{ - } \) is, at the same time, a lower bound of the operators \( {S}_{n} \) and \( S - {Q}_{n} \) used in steps 2 and 3 . These operators therefore have a common lower bound, so that Theorem 9.18(b) can be applied. | Yes |
Theorem 4.7.3 \( \left| \sum \right| < \infty \) . | Suppose that \( \left| \sum \right| = \infty \) . Since there are only finitely many elements in the root poset of any given depth, we conclude that there are small roots of arbitrarily large depth. For each small root \( \alpha \), we have (by the definition of \( \sum \) ) a saturated chain in the root poset, entirely contained in \( \sum \), from some simple root to \( \alpha \) . Hence, there are saturated chains (in the root poset) consisting entirely of small roots, of arbitrarily great length.
By Proposition 4.5.5, there are only finitely many pairs \( \left( {J, v}\right) \), with \( J \subseteq S \) and \( v \in {\mathbb{R}}^{J} \), such that there exists a \( \gamma \in \sum \) with \( \mathcal{N}\left( \gamma \right) = J \) and \( ((\gamma \mid \) \( \left. \left. {\alpha }_{s}\right) \right) {}_{s \in J} = v \) . Let \( C \) be a saturated chain in \( \sum \smallsetminus \Pi \) of length greater than the number of such pairs. Taking a suitable segment of \( C \), we conclude that there exists a saturated chain \( {\gamma }_{j} \vartriangleleft {\gamma }_{j + 1} \vartriangleleft \cdots \vartriangleleft {\gamma }_{k} \), such that \( \mathcal{N}\left( {\gamma }_{j}\right) = \mathcal{N}\left( {\gamma }_{k}\right) \) and
\[
\left( {{\gamma }_{j} \mid {\alpha }_{s}}\right) = \left( {{\gamma }_{k} \mid {\alpha }_{s}}\right)
\]
(4.42)
for all \( s \in \mathcal{N}\left( {\gamma }_{k}\right) \), with \( \operatorname{dp}\left( {\gamma }_{i}\right) = i \) for \( i = j,\ldots, k \) .
Let, for brevity, \( \mathcal{N}\overset{\text{ def }}{ = }\mathcal{N}\left( {\gamma }_{k}\right) \) . Let \( {s}_{i} \in S \) be such that \( {s}_{i}\left( {\gamma }_{i}\right) = {\gamma }_{i + 1} \) for \( i = j,\ldots, k - 1 \) . It follows from Lemma 4.7.2 that \( {s}_{j},{s}_{j + 1},\ldots ,{s}_{k - 1} \in \mathcal{N} \) . Let \( {\gamma }_{j - 1}\overset{\text{ def }}{ = }{s}_{k - 1}\left( {\gamma }_{j}\right) \) . By equation (4.42), we have that
\[
\left( {{\gamma }_{j - 1} \mid {\alpha }_{s}}\right) = \left( {{s}_{k - 1}\left( {\gamma }_{j}\right) \mid {\alpha }_{s}}\right)
\]
\[
= \left( {{\gamma }_{j} \mid {s}_{k - 1}\left( {\alpha }_{s}\right) }\right)
\]
\[
= \left( {{\gamma }_{j} \mid {\alpha }_{s} - 2\left( {{\alpha }_{{s}_{k - 1}} \mid {\alpha }_{s}}\right) {\alpha }_{{s}_{k - 1}}}\right)
\]
\[
= \left( {{\gamma }_{k} \mid {\alpha }_{s} - 2\left( {{\alpha }_{{s}_{k - 1}} \mid {\alpha }_{s}}\right) {\alpha }_{{s}_{k - 1}}}\right)
\]
\[
= \left( {{\gamma }_{k} \mid {s}_{k - 1}\left( {\alpha }_{s}\right) }\right)
\]
\[
= \left( {{s}_{k - 1}\left( {\gamma }_{k}\right) \mid {\alpha }_{s}}\right)
\]
\[
= \left( {{\gamma }_{k - 1} \mid {\alpha }_{s}}\right)
\]
(4.43)
for all \( s \in \mathcal{N} \) . In particular, since \( {\gamma }_{k - 1} < {\gamma }_{k} \), equation (4.43) implies by Lemma 4.6.2 that \( {\gamma }_{j - 1} \vartriangleleft {\gamma }_{j} \) .
Hence, we have obtained a saturated chain of roots \( {\gamma }_{j - 1} \vartriangleleft {\gamma }_{j} \vartriangleleft \cdots \vartriangleleft {\gamma }_{k - 1} \) such that
\[
\left( {{\gamma }_{j - 1} \mid {\alpha }_{s}}\right) = \left( {{\gamma }_{k - 1} \mid {\alpha }_{s}}\right)
\]
for all \( s \in \mathcal{N} \), and \( {s}_{k - 1}\left( {\gamma }_{j - 1}\right) = {\gamma }_{j},{s}_{i}\left( {\gamma }_{i}\right) = {\gamma }_{i + 1} \) for \( i = j,\ldots, k - 2 \) . Continuing in this way, we construct a saturated chain
\[
{\gamma }_{k} \vartriangleright {\gamma }_{k - 1} \vartriangleright \cdots \vartriangleright {\gamma }_{j} \vartriangleright {\gamma }_{j - 1} \vartriangleright \cdots \vartriangleright {\gamma }_{2} \vartriangleright {\gamma }_{1}
\]
in the root poset, with \( {\gamma }_{1} \) a simple root, and
\[
\left( {{\gamma }_{1} \mid {\alpha }_{s}}\right) = \left( {{\gamma }_{k - j + 1} \mid {\alpha }_{s}}\right)
\]
(4.44)
for all \( s \in \mathcal{N} \) .
Let \( r \in S \) be such that \( {\gamma }_{1} = {\alpha }_{r} \) . If \( r \in \mathcal{N} \), then from equation (4.44) we have that
\[
\left( {{\gamma }_{k - j + 1} \mid {\alpha }_{r}}\right) = \left( {{\gamma }_{1} \mid {\alpha }_{r}}\right) = 1
\]
whereas, on the other hand,
\[
\left| \left( {{\gamma }_{k - j + 1} \mid {\alpha }_{r}}\right) \right| < 1
\]
since \( \mathcal{N}\left( {\gamma }_{k - j + 1}\right) = \mathcal{N} \) . A contradiction!
Thus, assume that \( r \notin \mathcal{N} \) . Let \( j \leq i \leq k - 1 \) be such that \( {s}_{i}\left( {\gamma }_{k - j}\right) = \) \( {\gamma }_{k - j + 1} \) . Then, \( r \neq {s}_{i} \) and we conclude from equation (4.44) that
\[
0 \geq \left( {{\gamma }_{1} \mid {\alpha }_{{s}_{i}}}\right) = \left( {{\gamma }_{k - j + 1} \mid {\alpha }_{{s}_{i}}}\right) > 0
\]
thus again reaching a contradiction.
We close this section by giving a useful and interesting characterization of small roots. Let \( \beta ,\gamma \in {\Phi }^{ + } \) . We say that \( \beta \) dominates \( \gamma \), denoted \( \beta \) dom \( \gamma \) , if \( w\left( \beta \right) < 0 \) implies \( w\left( \gamma \right) < 0 \) for all \( w \in W \) . This relation is clearly transitive. Note that if \( \beta \) dom \( \gamma \), then \( \operatorname{dp}\left( \beta \right) \geq \operatorname{dp}\left( \gamma \right) \) . Also, if \( \beta \) dom \( \gamma \) , then \( w\left( \beta \right) \) dom \( w\left( \gamma \right) \) for all \( w \in W \) such that \( w\left( \gamma \right) \in {\Phi }^{ + } \) .
Lemma 4.7.4 Let \( \beta \in {\Phi }^{ + } \) and \( s \in S \) . Then, \( \beta \) dominates \( {\alpha }_{s} \) if and only if \( \left( {\beta \mid {\alpha }_{s}}\right) \geq 1 \) .
Proof. We may clearly assume that \( \beta \neq {\alpha }_{s} \) and, hence, that \( s\left( \beta \right) \in {\Phi }^{ + } \) . Suppose that \( \beta \) dominates \( {\alpha }_{s} \) and that \( \left( {\beta \mid {\alpha }_{s}}\right) < 1 \) . Since \( \beta \) dominates \( {\alpha }_{s} \) and \( {t}_{\beta }\left( \beta \right) = - \beta < 0 \), there follows that \( {t}_{\beta }\left( {\alpha }_{s}\right) = {\alpha }_{s} - 2\left( {{\alpha }_{s} \mid \beta }\right) \beta \in {\Phi }^{ - } \) and, therefore,
\[
\left( {{\alpha }_{s} \mid \beta }\right) > 0
\]
(4.45)
So we conclude that \( \left| \left( {{\alpha }_{s} \mid \beta }\right) \right| < 1 \) . By Proposition 4.5.4, this implies that the subgroup \( D \) generated by \( s \) and \( {t}_{\beta } \) is a finite dihedral group and that \( {t}_{\beta }s \) acts as a rotation of finite order on the subspace of \( V \) spanned by \( \beta \) and \( {\alpha }_{s} \) . However, it is not hard to see that this implies that there exists \( w \in D \) such that \( w\left( \beta \right) < 0 \) but \( w\left( {\alpha }_{s}\right) > 0 \), and this contradicts our hypothesis.
Conversely, suppose that \( \left( {\beta \mid {\alpha }_{s}}\right) \geq 1 \) . Then,
\[
\left( {s\left( \beta \right) \mid \beta }\right) = \left( {\beta - 2\left( {{\alpha }_{s} \mid \beta }\right) {\alpha }_{s} \mid \beta }\right) = 1 - 2{\left( {\alpha }_{s} \mid \beta \right) }^{2} \leq - 1.
\]
By Proposition 4.5.4, this implies that there are infinitely many positive roots of the form
\[
{\lambda \beta } + {\mu s}\left( \beta \right)
\]
(4.46)
with \( \lambda ,\mu > 0 \) . Now, let \( w \in W \) be such that \( w\left( \beta \right) < 0 \) . If \( w\left( {\alpha }_{s}\right) > 0 \), then we conclude that
\[
w\left( {s\left( \beta \right) }\right) = w\left( {\beta - 2\left( {{\alpha }_{s} \mid \beta }\right) {\alpha }_{s}}\right) = w\left( \beta \right) - 2\left( {{\alpha }_{s} \mid \beta }\right) w\left( {\alpha }_{s}\right) < 0,
\]
and, hence, that all the positive roots of the form (4.46) are also in \( N\left( w\right) \) , which contradicts Proposition 4.4.4. Hence, \( w\left( {\alpha }_{s}\right) < 0 \), showing that \( \beta \) dominates \( {\alpha }_{s} \), as desired. \( ▱ \)
A positive root \( \alpha \in {\Phi }^{ + } \) is said to be humble if \( \alpha \) dominates no positive root except itself. All simple roots are clearly humble. The preceding lemma has the following useful consequence.
Lemma 4.7.5 Let \( \beta ,\alpha \in {\Phi }^{ + } \) be such that \( \beta \vartriangleleft \alpha \) in the root poset. Then, we have that the following hold:
(i) If \( \beta \vartriangleleft \alpha \) is long, then \( \alpha \) is not humble.
(ii) If \( \beta \vartriangleleft \alpha \) is short, then \( \alpha \) is humble if and only if \( \beta \) is humble.
Proof. Let \( s \in S \) be such that \( \alpha = s\left( \beta \right) \) .
Assume that (i) holds. Then, \( \left( {\alpha \mid {\alpha }_{s}}\right) \geq 1 \) and, hence, \( \alpha \) dominates \( {\alpha }_{s} \) by Lemma 4.7.4. So \( \alpha \) is not humble.
Assume now that (ii) holds. Then, \( 0 > \left( {\beta \mid {\alpha }_{s}}\right) > - 1 \) . Let \( \gamma \in {\Phi }^{ + } \smallsetminus \{ \beta \} \) be such that \( \beta \) dom \( \gamma \) . Then, by Lemma 4.7.4, \( \gamma \neq {\alpha }_{s} \) . Hence, \( s\left( \beta \right) \) dom \( s\left( \gamma \right) \) and \( s\left( \gamma \right) \in {\Phi }^{ + } \smallsetminus \{ s\left( \beta \right) \} \) . So \( \beta \) is humble if \( \alpha \) is humble. For the converse statement a similar argument holds. \( ▱ \)
We can now prove the promised characterization of small roots.
Theorem 4.7.6 Let \( \alpha \in {\Phi }^{ + } \) . Then, \( \alpha \in \sum \) if and only if \( \alpha \) is humble.
Proof. Assume that \( \alpha \in \sum \) . Then, by definition, there is a saturated chain in the root poset, consisting entirely of short edges, from some simple root to \( \alpha \) . However, simple roots are humble, so \( \alpha \) is humble by part (ii) of Lemma 4.7.5.
Conversely, suppose that \( \alpha \) is humble and let \( {\alpha }_{1} \vartriangleleft {\alpha }_{2} \vartriangleleft \cdots \vartriangleleft {\alpha }_{p} = \alpha | Yes |
Theorem 15.2. Let \( {G}_{i}, i = 1,2 \) be two groups, and let \( G = {G}_{1} \times {G}_{2} \) be their direct product. Then the following sequence is exact:\[
{\bigoplus }_{p + q = n}{H}_{p}\left( {G}_{1}\right) \otimes {H}_{q}\left( {G}_{2}\right) \rightarrow {H}_{n}\left( G\right) \rightarrow {\bigoplus }_{p + q = n - 1}\operatorname{Tor}\left( {{H}_{p}\left( {G}_{1}\right) ,{H}_{q}\left( {G}_{2}\right) }\right) .
\] | Moreover the sequence splits by an unnatural splitting. | No |
Suppose \( M \) is a smooth \( n \) -manifold and \( D \subseteq {TM} \) is a distribution of rank \( k \). Then \( D \) is smooth if and only if each point \( p \in M \) has a neighborhood \( U \) on which there are smooth 1 -forms \( {\omega }^{1},\ldots ,{\omega }^{n - k} \) such that for each \( q \in U \), | First suppose that there exist such forms \( {\omega }^{1},\ldots ,{\omega }^{n - k} \) in a neighborhood of each point. The assumption (19.1) together with the fact that \( D \) has rank \( k \) implies that the forms \( {\omega }^{1},\ldots ,{\omega }^{n - k} \) are independent on \( U \) for dimensional reasons. By Proposition 10.15, we can complete them to a smooth coframe \( \left( {{\omega }^{1},\ldots ,{\omega }^{n}}\right) \) on a (possibly smaller) neighborhood of each point. If \( \left( {{E}_{1},\ldots ,{E}_{n}}\right) \) is the dual frame, it is easy to check that \( D \) is locally spanned by \( {E}_{n - k + 1},\ldots ,{E}_{n} \), so it is smooth by the local frame criterion.
Conversely, suppose \( D \) is smooth. In a neighborhood of any \( p \in M \), there are smooth vector fields \( {Y}_{1},\ldots ,{Y}_{k} \) spanning \( D \). By Proposition 10.15 again, we can complete these vector fields to a smooth local frame \( \left( {{Y}_{1},\ldots ,{Y}_{n}}\right) \) for \( M \) in a neighborhood of \( p \). With the dual coframe denoted by \( \left( {{\varepsilon }^{1},\ldots ,{\varepsilon }^{n}}\right) \), it follows easily that \( D \) is characterized locally by | Yes |
Theorem 18.6.2 For any nonzero \( a \in {}^{ * }R\left\lbrack x\right\rbrack \), the following are equivalent. | In general, \( \left| a\right| = {2}^{-o\left( a\right) } \) and \( o\left( a\right) \) is a nonnegative hyperinteger, so \( \left| a\right| \) will be appreciable iff \( o\left( a\right) \) is limited, or equivalently, \( \left| a\right| \) will be infinitesimal iff \( o\left( a\right) \) is unlimited. Thus (1) and (2) are equivalent.
Now, by transfer we have that for any nonzero \( a \in {}^{ * }R\left\lbrack x\right\rbrack \) ,
\[
\left( {\forall m \in {}^{ * }{\mathbb{Z}}^{ \geq }}\right) \left\lbrack {m < o\left( a\right) \leftrightarrow \left( {\forall n \in {}^{ * }{\mathbb{Z}}^{ \geq }}\right) \left( {n \leq m \rightarrow {a}_{n} = 0}\right) }\right\rbrack .
\]
From this,(2) implies (3) by putting \( N = o\left( a\right) \) . It is immediate that (3) implies (4). Finally, if (4) holds, then the above transferred sentence ensures that each standard \( m \) is smaller than \( o\left( a\right) \), so (2) follows. | Yes |
Suppose \( p \nmid m \) is prime. Show that \( p \mid {\phi }_{m}\left( a\right) \) for some \( a \in \mathbb{Z} \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;m}\right) \). Deduce from Exercise 1.2.5 that there are infinitely many primes congruent to 1 (mod \( m \) ). | Solution. If \( p \mid {\phi }_{m}\left( a\right) \), by the previous exercise the order of \( a\left( {\;\operatorname{mod}\;p}\right) \) is \( m \) so that \( m \mid p - 1 \).
Conversely, if \( p \equiv 1\left( {\;\operatorname{mod}\;m}\right) \), there is an element \( a \) of order \( m\left( {\;\operatorname{mod}\;p}\right) \) because \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ * } \) is cyclic. Again by the previous exercise \( p \mid {\phi }_{m}\left( a\right) \).
If there are only finitely many primes \( {p}_{1},\ldots ,{p}_{r} \) (say) that are congruent to \( 1\left( {\;\operatorname{mod}\;m}\right) \), then setting \( a = \left( {{p}_{1}\cdots {p}_{r}}\right) m \) we examine the prime divisors of \( {\phi }_{m}\left( a\right) \). Observe that the identity
\[
{x}^{m} - 1 = \mathop{\prod }\limits_{{d \mid m}}{\phi }_{d}\left( x\right)
\]
implies that \( {\phi }_{m}\left( 0\right) = \pm 1 \). Thus, the constant term of \( {\phi }_{m}\left( x\right) \) is \( \pm 1 \) so that \( {\phi }_{m}\left( a\right) \) is coprime to \( a \) and hence coprime to \( m \). (If \( {\phi }_{m}\left( a\right) = \pm 1 \), one can replace \( a \) by any suitable power of \( a \), so that \( \left| {{\phi }_{m}\left( a\right) }\right| > 1 \).)
By what we have proved, any prime divisor \( p \) of \( {\phi }_{m}\left( a\right) \) coprime to \( m \) must be congruent to \( 1\left( {\;\operatorname{mod}\;m}\right) \). The prime \( p \) is distinct from \( {p}_{1},\ldots ,{p}_{r} \). | Yes |
If \( p \) is a covering projection, so is \( {p}^{\prime } \) . | Let \( y \in Y \) and let \( U \) be a neighborhood of \( p\left( y\right) \) in \( B \) which is evenly covered by \( p \) . Then \( {p}^{-1}\left( U\right) = \bigcup \left\{ {{U}_{\alpha } \mid \alpha \in \mathcal{A}}\right\} \) where \( \mathcal{A} \) is an indexing set and \( \left\{ {{U}_{\alpha } \mid \alpha \in \mathcal{A}}\right\} \) consists of pairwise disjoint open subsets of \( E \) each mapped homeomorphically onto \( U \) by \( p \) . We claim that \( {g}^{-1}\left( U\right) \) is evenly covered by \( {p}^{\prime } \) . Indeed, \( {\left( {p}^{\prime }\right) }^{-1}{g}^{-1}\left( U\right) = \mathop{\bigcup }\limits_{\alpha }\left\{ {{\widetilde{g}}^{-1}\left( {U}_{\alpha }\right) \mid \alpha \in \mathcal{A}}\right\} \) . These sets \( {\widetilde{g}}^{-1}\left( {U}_{\alpha }\right) \) are pairwise disjoint open subsets of \( {g}^{ * }E \) . It is an exercise in the definitions to show that \( {p}^{\prime } \) maps \( {\widetilde{g}}^{-1}\left( {U}_{\alpha }\right) \) bijectively to \( {g}^{-1}\left( U\right) \) . Projections in a cartesian product map open sets to open sets, so \( {p}^{\prime } \) maps any open subset of \( {\widetilde{g}}^{-1}\left( {U}_{\alpha }\right) \) onto an open subset of \( {g}^{-1}\left( U\right) \), hence onto an open subset of \( Y \), since \( {g}^{-1}\left( U\right) \) is open in \( Y \) . Thus \( {p}^{\prime } \) maps \( {\widetilde{g}}^{-1}\left( {U}_{\alpha }\right) \) homeomorphically onto \( {g}^{-1}\left( U\right) \) as claimed. | No |
Prove that the map \( f\left( {x, y}\right) \mapsto f\left( {t,\tau }\right) \) defines an embedding \( k\left( {x, y}\right) \rightarrow k\left( \left( t\right) \right) \) . Thus, there is a discrete valuation on \( k\left( {x, y}\right) \) with residue field \( k \) . Show that this valuation is not obtained by the construction of Exercise 1.1. | Null | No |
Theorem 7.2.10. Let \( \mathcal{F} = \left( {{F}_{0},{F}_{1},{F}_{2},\ldots }\right) \) be a tower over \( {\mathbb{F}}_{q} \). | Proof. (a) Above each place \( P \in \operatorname{Split}\left( {\mathcal{F}/{F}_{0}}\right) \) there are exactly \( \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \) places of \( {F}_{n} \), and they are all rational. Hence \( N\left( {F}_{n}\right) \geq \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \cdot \left| {\operatorname{Split}\left( {\mathcal{F}/{F}_{0}}\right) }\right| \) , and (a) follows immediately.
(b) For simplity we set \( {g}_{n} \mathrel{\text{:=}} g\left( {F}_{n}\right) \) for all \( n \geq 0 \) . The Hurwitz Genus Formula for \( {F}_{n}/{F}_{0} \) gives
\[
2{g}_{n} - 2 = \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \left( {2{g}_{0} - 2}\right) + \mathop{\sum }\limits_{{P \in \operatorname{Ram}\left( {\mathcal{F}/{F}_{0}}\right) }}\mathop{\sum }\limits_{{Q \in {\mathbb{P}}_{{F}_{n}}, Q \mid P}}d\left( {Q \mid P}\right) \cdot \deg Q
\]
\[
\leq \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \left( {2{g}_{0} - 2}\right) + \mathop{\sum }\limits_{{P \in \operatorname{Ram}\left( {\mathcal{F}/{F}_{0}}\right) }}\mathop{\sum }\limits_{{Q \in {\mathbb{P}}_{{F}_{n}}, Q \mid P}}{a}_{P} \cdot e\left( {Q \mid P}\right) \cdot f\left( {Q \mid P}\right) \cdot \deg P
\]
\[
= \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \left( {2{g}_{0} - 2}\right) + \mathop{\sum }\limits_{{P \in \operatorname{Ram}\left( {\mathcal{F}/{F}_{0}}\right) }}{a}_{P} \cdot \deg P \cdot \mathop{\sum }\limits_{{Q \in {\mathbb{P}}_{{F}_{n}}, Q \mid P}}e\left( {Q \mid P}\right) \cdot f\left( {Q \mid P}\right)
\]
\[
= \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \left( {2{g}_{0} - 2 + \mathop{\sum }\limits_{{P \in \operatorname{Ram}\left( {\mathcal{F}/{F}_{0}}\right) }}{a}_{P} \cdot \deg P}\right) .
\]
Here we have used (7.7) and the Fundamental Equality
\[
\mathop{\sum }\limits_{{Q \mid P}}e\left( {Q \mid P}\right) \cdot f\left( {Q \mid P}\right) = \left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack
\]
see Theorem 3.1.11. Dividing the inequality above by \( 2\left\lbrack {{F}_{n} : {F}_{0}}\right\rbrack \) and letting \( n \rightarrow \infty \) we obtain the inequality
\[
\gamma \left( {\mathcal{F}/{F}_{0}}\right) \leq g\left( {F}_{0}\right) - 1 + \frac{1}{2}\mathop{\sum }\limits_{{P \in \operatorname{Ram}\left( {\mathcal{F}/{F}_{0}}\right) }}{a}_{P} \cdot \deg P.
\]
(c) follows immediately from (a) and (b), since \( \lambda \left( \mathcal{F}\right) = \nu \left( {\mathcal{F}/{F}_{0}}\right) /\gamma \left( {\mathcal{F}/{F}_{0}}\right) \) (see Equation (7.6)). | Yes |
Let \( {\left( {c}_{k}\right) }_{k \geq 0} \) be a sequence of elements of \( \mathbb{Z} \), set \( {a}_{k} = \mathop{\sum }\limits_{{0 \leq m \leq k}}{\left( -1\right) }^{k - m}\left( \begin{matrix} k \\ m \end{matrix}\right) {c}_{m} \), and assume that as \( k \rightarrow \infty \) we have \( {v}_{p}\left( {a}_{k}\right) \geq {\alpha k} + o\left( k\right) \) for some \( \alpha > 1/\left( {p - 1}\right) \) . Then there exists a function \( f \) having the following properties: | Clear from the above results. | No |
Theorem 3.1 (Divergence Theorem). \[
{\int }_{X}{\mathcal{L}}_{\xi }\Omega = {\int }_{\partial X}\Omega \circ \xi
\] | Suppose that \( \left( {X, g}\right) \) is a Riemannian manifold, assumed oriented for simplicity. We let \( \Omega \) or \( {\operatorname{vol}}_{g} \) be the volume form defined in Chapter XV, §1. Let \( \omega \) be the canonical Riemannian volume form on \( \partial X \) for the metric induced by \( g \) on the boundary. Let \( {\mathbf{n}}_{x} \) be a unit vector in the tangent space \( {T}_{x}\left( X\right) \) such that \( u \) is perpendicular to \( {T}_{x}\left( {\partial X}\right) \) . Such a unit vector is determined up to sign. Denote by \( {\mathbf{n}}_{x}^{ \vee } \) its dual functional, i.e. the component on the projection along \( {\mathbf{n}}_{x} \) . We select \( {\mathbf{n}}_{x} \) with the sign such that \[
{\mathbf{n}}_{x}^{ \vee } \land \omega \left( x\right) = \Omega \left( x\right)
\] We then shall call \( {\mathbf{n}}_{x} \) the unit outward normal vector to the boundary at \( x \) . In an oriented chart, it looks like this. Then by formula CON 3 of Chapter V, §5 we find \[
\Omega \circ \xi = \langle \mathbf{n},\;\xi \rangle \omega - {\mathbf{n}}^{ \vee }\; \land \;\left( {\omega \circ \xi }\right) ,
\] and the restriction of this form to \( \partial X \) is simply \( \langle \mathbf{n},\xi \rangle \omega \) . Thus we get: \[
{\int }_{X}{\mathcal{L}}_{\xi }\Omega = {\int }_{\partial X}\Omega \circ \xi
\] | Yes |
Let \( \Psi ,{\Delta }_{j}^{\Psi } \) be as above and \( \gamma > 0 \) . Then there is a constant \( C = \) \( C\left( {n,\gamma ,\Psi }\right) \) such that for all \( f \) in \( {\dot{\Lambda }}_{\gamma } \) we have the estimate | We begin with the proof of (1.4.7). We first consider the case \( 0 < \gamma < 1 \) , which is very simple. Since each \( {\Delta }_{j}^{\Psi } \) is given by convolution with a function with mean value zero, for a function \( f \in {\dot{\Lambda }}_{\gamma } \) and every \( x \in {\mathbf{R}}^{n} \) we write | No |
Corollary 18. If \( A \) is an abelian group then \( A \) is torsion free if and only if \( {\mathrm{{Tor}}}_{1}\left( {A, B}\right) = 0 \) for every abelian group \( B \) (in which case \( A \) is flat as a \( \mathbb{Z} \) -module). | By the proposition, if \( A \) has no elements of finite order then we have \( {\operatorname{Tor}}_{1}\left( {A, B}\right) = {\operatorname{Tor}}_{1}\left( {t\left( A\right), B}\right) = {\operatorname{Tor}}_{1}\left( {0, B}\right) = 0 \) for every abelian group \( B \) . Conversely, if \( {\operatorname{Tor}}_{1}\left( {A, B}\right) = 0 \) for all \( B \), then in particular \( {\operatorname{Tor}}_{1}\left( {A,\mathbb{Q}/\mathbb{Z}}\right) = 0 \), and this group is isomorphic to the torsion subgroup of \( A \) by the example above. | Yes |
The algebra \( B\left( {K, S}\right) \) contains the rational functions with poles in \( S \), and is closed under uniform limits in \( K \). | Null | No |
Corollary 25. Let \( R \) be a subring of the commutative ring \( S \) with \( 1 \in R \) . Then the integral closure of \( R \) in \( S \) is integrally closed in \( S \) . | Null | No |
Theorem 3.2. If \( N = p \) is prime \( \geqq 3 \), then for every admissible pair \( \left( {r, s}\right) \) the curve \( F\left( {r, s}\right) \) has genus \( \left( {p - 1}\right) /2 \), and \( K\left( {r, s}\right) = K\left( {1,{s}^{ * }}\right) \) for a uniquely determined integer \( {s}^{ * } \) such that the pair \( \left( {1,{s}^{ * }}\right) \) is admissible. | The genus can either be computed directly as we did for the Fermat curve, or one can use Theorem 3.1. The number of \( m \) such that \( \left( {\langle {mr}\rangle ,\langle {ms}\rangle }\right) \) is admissible is trivially computed to be \( \left( {p - 1}\right) /2 \), using the remark preceding the theorem. The statement that \( K\left( {r, s}\right) = K\left( {1,{s}^{ * }}\right) \) is clear. | Yes |
Corollary 15.2 Let \( A \) be an infinite set of positive integers with \( \gcd \left( A\right) = 1 \). Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} = \infty \) | For every sufficiently large integer \( k \) there exists a subset \( {F}_{k} \) of \( A \) of cardinality \( k \) such that \( \gcd \left( {F}_{k}\right) = 1 \) . By Theorem 15.2,\[
{p}_{A}\left( n\right) \geq {p}_{{F}_{k}}\left( n\right) = \frac{{n}^{k - 1}}{\left( {k - 1}\right) !\mathop{\prod }\limits_{{a \in {F}_{k}}}a} + O\left( {n}^{k - 2}\right) ,
\]
and so there exists a positive constant \( {c}_{k} \) such that
\[
{p}_{A}\left( n\right) \geq {c}_{k}{n}^{k - 1}
\]
for all sufficiently large integers \( n \) . Then
\[
\log {p}_{A}\left( n\right) \geq \log {p}_{{F}_{k}}\left( n\right) \geq \left( {k - 1}\right) \log n + \log {c}_{k}.
\]
Dividing by \( \log n \), we obtain
\[
\mathop{\liminf }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} \geq k - 1
\]
This is true for all sufficiently large \( k \), and so
\[
\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} = \infty
\] | Yes |
Theorem 9.5. Let the notation be as in Proposition 8.18 and Corollary 8.19. If there exists a prime \( l \equiv 1{\;\operatorname{mod}\;p} \) with \( l < {p}^{2} - p \) such that \({Q}_{i}^{k} ≢ 1{\;\operatorname{mod}\;l}\;\text{ for all }i \in \left\{ {{i}_{1},\ldots ,{i}_{s}}\right\} \), then the second case of Fermat’s Last Theorem has no solutions. | Proof. By Corollary \( {8.19}, p \nmid {h}^{ + }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \), so Assumption I is satisfied. Suppose that \({x}^{p} + {y}^{p} = {z}^{p},\;p \nmid {xy}, p \mid z, z \neq 0,\) where \( x, y, z \in \mathbb{Z} \) are relatively prime. Let \( l \) be as in the statement of the theorem. | No |
Corollary 23. The sequence \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) is exact if and only if \( \psi \) is injective, \( \varphi \) is surjective, and image \( \psi = \ker \varphi \), i.e., \( B \) is an extension of \( C \) by \( A \) . | Null | No |
Theorem 3.22 (Abel’s Limit Theorem). Assume that the power series \( \sum {a}_{n}{z}^{n} \) has finite radius of convergence \( \rho > 0 \) . If \( \sum {a}_{n}{z}_{0}^{n} \) converges for some \( {z}_{0} \) with \( \left| {z}_{0}\right| = \) \( \rho \), then \( f\left( z\right) = \sum {a}_{n}{z}^{n} \) is defined for \( \{ \left| z\right| < \rho \} \cup \left\{ {z}_{0}\right\} \) and we have
\[
\mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}f\left( z\right) = f\left( {z}_{0}\right)
\]
as long as \( z \) approaches \( {z}_{0} \) from inside the circle of convergence and
\[
\frac{\left| z - {z}_{0}\right| }{\rho - \left| z\right| }
\]
remains bounded. | Proof. By the change of variable \( w = \frac{z}{{z}_{0}} \) we may assume that \( \rho = 1 = {z}_{0} \) (replace \( \left. {{a}_{n}\text{by}{a}_{n}{z}_{0}^{n}}\right) \) . Thus \( \sum {a}_{n} \) converges to \( f\left( 1\right) \) . By changing \( {a}_{0} \) to \( {a}_{0} - f\left( 1\right) \), we may assume that \( f\left( 1\right) = \sum {a}_{n} = 0 \) . Thus we are assuming that \( \left| z\right| < 1 \) (with \( 1 - \left| z\right| \) small) and that \( \frac{\left| 1 - z\right| }{1 - \left| z\right| } \leq M \) for some fixed \( M > 0 \) . (Recall Exercise 2.4.) Let
\[
{s}_{n} = {a}_{0} + {a}_{1} + \cdots + {a}_{n}
\]
Then \( \mathop{\lim }\limits_{n}{s}_{n} = 0 \) ,
\[
{S}_{n}\left( z\right) = {a}_{0} + {a}_{1}z + \cdots + {a}_{n}{z}^{n}
\]
\[
= {s}_{0} + \left( {{s}_{1} - {s}_{0}}\right) z + \cdots + \left( {{s}_{n} - {s}_{n - 1}}\right) {z}^{n}
\]
\[
= {s}_{0}\left( {1 - z}\right) + {s}_{1}\left( {z - {z}^{2}}\right) + \cdots + {s}_{n - 1}\left( {{z}^{n - 1} - {z}^{n}}\right) + {s}_{n}{z}^{n}
\]
\[
= \left( {1 - z}\right) \left( {{s}_{0} + {s}_{1}z + \cdots + {s}_{n - 1}{z}^{n - 1}}\right) + {s}_{n}{z}^{n},
\]
and hence
\[
f\left( z\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n}\left( z\right) = \left( {1 - z}\right) \mathop{\sum }\limits_{{n = 0}}^{\infty }{s}_{n}{z}^{n}.
\]
Given \( \epsilon > 0 \), choose \( N \in {\mathbb{Z}}_{ > 0} \) such that \( \left| {s}_{n}\right| < \epsilon \) for \( n > N \) . Then
\[
\left| {f\left( z\right) }\right| \leq \left| {1 - z}\right| \left( {\left| {\mathop{\sum }\limits_{{n = 0}}^{N}{s}_{n}{z}^{n}}\right| + \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\left| {s}_{n}\right| {\left| z\right| }^{n}}\right)
\]
\[
\leq \left| {1 - z}\right| \left( {\left| {\mathop{\sum }\limits_{{n = 0}}^{N}{s}_{n}{z}^{n}}\right| + \epsilon \frac{{\left| z\right| }^{N + 1}}{1 - \left| z\right| }}\right)
\]
\[
\leq \left| {1 - z}\right| \left| {\mathop{\sum }\limits_{{n = 0}}^{N}{s}_{n}{z}^{n}}\right| + {\epsilon M}
\]
and thus we conclude that \( \mathop{\lim }\limits_{{z \rightarrow 1}}f\left( z\right) = 0 \). | Yes |
Let \( K \) be a function field over a perfect ground field \( k \). Then \( K \) is geometric. | Let \( {k}^{\prime } \) be a finite extension of \( k \). Then \( {k}^{\prime }/k \) is separable, so \( {k}^{\prime } = k\left( u\right) \) for some \( u \in {k}^{\prime } \) by (A.0.17). Moreover, \( u \) satisfies an irreducible separable polynomial \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) of degree \( n = \left| {{k}^{\prime } : k}\right| \). We claim that \( f \) remains irreducible over \( K \). Namely, any factor \( {f}_{0} \) has roots that are algebraic over \( k \), but since the coefficients are symmetric functions of the roots, the coefficients of \( {f}_{0} \) are also algebraic over \( k \). Since \( K/k \) is a function field, we see that \( {f}_{0}\left( X\right) \in k\left\lbrack X\right\rbrack \) and thus that \( {f}_{0} = f \).
Now identify \( {k}^{\prime } \) and \( K \) with their canonical images in \( {K}^{\prime } \mathrel{\text{:=}} {k}^{\prime } \otimes K \). Since \( u \) satisfies an irreducible polynomial of degree \( n \) over \( K, K\left\lbrack u\right\rbrack \) is a field and \( \mid K\left\lbrack u\right\rbrack \) : \( K \mid = n = {\dim }_{K}{K}^{\prime } \). We conclude that \( {K}^{\prime } = K\left( u\right) \) is a field, as required. | Yes |
Let \( f \) and \( g \) be two Krasner analytic functions on the set \( \mathcal{D} \) defined above. If \( f \) and \( g \) coincide on some nonempty open subset of \( \mathcal{D} \) then \( f = g \) . | Consider the map from \( \mathcal{D} \) to \( {\mathbb{C}}_{p} \) sending \( x \) to \( t = 1/\left( {x - 1}\right) \) . Since \( \left| {x - 1}\right| \geq 1 \) for \( x \in \mathcal{D} \), this map is a well-defined map from \( \mathcal{D} \) to \( {\mathcal{Z}}_{p} \), and since \( x = 1 + 1/t \) for \( t \neq 0 \), its image is equal to \( {\mathcal{Z}}_{p} \) minus the origin. A uniformly convergent sequence of rational functions of \( x \in \mathcal{D} \) gives a uniformly convergent sequence of rational functions \( {h}^{\left( n\right) } \) of \( t \in {\mathcal{Z}}_{p} \smallsetminus \{ 0\} \) . However, since all the poles of these \( {h}^{\left( n\right) } \) are outside \( {\mathcal{Z}}_{p} \), it follows that the sequence \( {h}^{\left( n\right) }\left( 0\right) \) is well defined and is a Cauchy sequence, hence converges, so that the sequence \( {h}^{\left( n\right) } \) is uniformly convergent on the whole of \( {\mathcal{Z}}_{p} \) . The map \( t = 1/\left( {x - 1}\right) \) thus gives a one-to-one correspondence between Krasner analytic functions on \( \mathcal{D} \) and on \( {\mathcal{Z}}_{p} \), so the lemma follows from Lemma 4.5.7. | Yes |
Proposition 4.11. The firm and the directional subdifferentials are homotone in the sense that for \( f \geq g \) with \( f\left( \bar{x}\right) = g\left( \bar{x}\right) \) finite one has | Proposition 4.11. If \( f \geq g \) with \( f\left( \bar{x}\right) = g\left( \bar{x}\right) \) finite one has \({\partial }_{F}g\left( \bar{x}\right) \subset {\partial }_{F}f\left( \bar{x}\right) ,\;{\partial }_{D}g\left( \bar{x}\right) \subset {\partial }_{D}f\left( \bar{x}\right) .\) | Yes |
Proposition 8.18. Let \( T \) be a Cesàro bounded operator on some Banach space \( E \) such that \( \frac{1}{n}{T}^{n}h \rightarrow 0 \) for each \( h \in E \) . Then for \( f, g \in E \) the following statements are equivalent: | The implication \( \left( \mathrm{v}\right) \Rightarrow \left( \mathrm{i}\right) \) follows from Theorem 8.5 while the implications (i) \( \Rightarrow \) (ii) \( \Rightarrow \) (iii) are trivial. If (iii) holds, then \( g \in \operatorname{fix}\left( T\right) \) by Lemma 8.17. Moreover,
\[
g \in {\operatorname{cl}}_{\sigma }\operatorname{conv}\left\{ {{T}^{n}f : n \in {\mathbb{N}}_{0}}\right\} = \overline{\operatorname{conv}}\left\{ {{T}^{n}f : n \in {\mathbb{N}}_{0}}\right\} ,
\]
where the last equality is due to Theorem C.7. Finally, suppose that (iv) holds and \( \mathop{\sum }\limits_{n}{t}_{n}{T}^{n}f \) is any convex combination of the vectors \( {T}^{n}f \) . Then
\[
f - \mathop{\sum }\limits_{n}{t}_{n}{T}^{n}f = \mathop{\sum }\limits_{n}{t}_{n}\left( {f - {T}^{n}f}\right) = \left( {\mathrm{I} - T}\right) \mathop{\sum }\limits_{n}n{t}_{n}{\mathrm{\;A}}_{n}f \in \operatorname{ran}\left( {\mathrm{I} - T}\right)
\]
by Lemma 8.2.c. It follows that \( f - g \in \overline{\operatorname{ran}}\left( {\mathrm{I} - T}\right) \), as was to be proved. | Yes |
Theorem 4.4.8 (Miljutin's Theorem). Suppose \( K \) is an uncountable compact metric space. Then \( \mathcal{C}\left( K\right) \) is isomorphic to \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \) . | The first step is to show that \( \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) is isomorphic to a complemented subspace of \( \mathcal{C}\left( \Delta \right) \) . By Lemma 4.4.7 there is a continuous surjection \( \psi : \Delta \rightarrow \left\lbrack {0,1}\right\rbrack \) , so that we can find a norm-one operator \( R : \mathcal{C}\left( \Delta \right) \rightarrow \mathcal{C}\left\lbrack {0,1}\right\rbrack \) with \( {Rf} \circ \psi = f \) for \( f \in \mathcal{C}\left\lbrack {0,1}\right\rbrack \) . Then \( R\left( {\chi }_{\Delta }\right) = {\chi }_{\left\lbrack 0,1\right\rbrack } \) . For fixed \( t \in \left\lbrack {0,1}\right\rbrack \) the linear functional \( f \rightarrow {Rf}\left( t\right) \) is given by a probability measure \( {\mu }_{t} \), so that
\[
{Rf}\left( t\right) = {\int }_{\Delta }{fd}{\mu }_{t}
\]
The map \( \widetilde{\psi } : {\Delta }^{\mathbb{N}} \rightarrow {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}} \) given by
\[
\widetilde{\psi }\left( {{s}_{1},\ldots ,{s}_{n},\ldots }\right) = \left( {\psi \left( {s}_{1}\right) ,\ldots ,\psi \left( {s}_{n}\right) ,\ldots }\right)
\]
is a continuous surjection. We will define \( \widetilde{R} : \mathcal{C}\left( {\Delta }^{\mathbb{N}}\right) \rightarrow \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) in such a way that \( \widetilde{R}f \circ \widetilde{\psi } = f \) for \( f \in \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) . Indeed, the subalgebra \( \mathcal{A} \) of \( \mathcal{C}\left( {\Delta }^{\mathbb{N}}\right) \) of all \( f \) that depend only on a finite number of coordinates is dense by the Stone-Weierstrass theorem. If \( f \in \mathcal{A} \) depends only on \( {s}_{1},\ldots ,{s}_{n} \), we define
\[
\widetilde{R}f\left( {{t}_{1},\ldots ,{t}_{n}}\right) = {\int }_{\Delta }\cdots {\int }_{\Delta }f\left( {{s}_{1},\ldots ,{s}_{n}}\right) d{\mu }_{{t}_{1}}\left( {s}_{1}\right) \cdots d{\mu }_{{t}_{n}}\left( {s}_{n}\right) .
\]
This map is clearly linear into \( {\ell }_{\infty }\left\lbrack {0,1}\right\rbrack \) and has norm one. It therefore extends to a norm-one operator \( \widetilde{R} : \mathcal{C}\left( {\Delta }^{\mathbb{N}}\right) \rightarrow {\ell }_{\infty }\left\lbrack {0,1}\right\rbrack \) . If \( f \in \mathcal{C}\left( {\Delta }^{\mathbb{N}}\right) \) is of the form \( {f}_{1}\left( {s}_{1}\right) \ldots {f}_{n}\left( {s}_{n}\right) \), then
\[
\widetilde{R}f\left( t\right) = R{f}_{1}\left( t\right) \ldots R{f}_{n}\left( t\right)
\]
so \( \widetilde{R}f \in \mathcal{C}\left\lbrack {0,1}\right\rbrack \) . The linear span of such functions is again dense by the Stone-Weierstrass theorem, so \( \widetilde{R} \) maps into \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \) .
If \( f \in \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) is of the form \( {f}_{1}\left( {t}_{1}\right) \ldots {f}_{n}\left( {t}_{n}\right) \), then it is clear that \( \widetilde{R}f \circ \widetilde{\psi } = f \) . It follows that this equation holds for all \( f \in \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) .
Thus \( \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) is isomorphic to a norm-one complemented subspace of \( \mathcal{C}\left( {\Delta }^{\mathbb{N}}\right) \) or \( \mathcal{C}\left( \Delta \right) \), since \( \Delta \) is homeomorphic to \( {\Delta }^{\mathbb{N}} \) .
Now, suppose \( K \) is an uncountable compact metric space. Then \( \mathcal{C}\left( K\right) \) is isomorphic to a complemented subspace of \( \mathcal{C}\left( {\left\lbrack 0,1\right\rbrack }^{\mathbb{N}}\right) \) by combining Proposition 4.4.3 and Theorem 4.4.4. Hence, by the preceding argument, \( \mathcal{C}\left( K\right) \) is isomorphic to a complemented subspace of \( \mathcal{C}\left( \Delta \right) \) . On the other hand, \( \mathcal{C}\left( \Delta \right) \) is isomorphic to a complemented subspace of \( \mathcal{C}\left( K\right) \), again by Proposition 4.4.3 and Theorem 4.4.4. We also have Proposition 4.4.5, which gives \( {c}_{0}\left( {\mathcal{C}\left( \Delta \right) }\right) \approx \mathcal{C}\left( \Delta \right) \) . We can apply Theorem 2.2.3 to deduce that \( \mathcal{C}\left( K\right) \approx \mathcal{C}\left( \Delta \right) \) . Of course, the same reasoning gives \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \approx \mathcal{C}\left( \Delta \right) \) | Yes |
Consider a sequence of fields \( {F}_{0} \subseteq {F}_{1} \subseteq {F}_{2} \subseteq \ldots \) where \( {F}_{0} \) is a function field with the exact constant field \( {\mathbb{F}}_{q} \) and \( \left\lbrack {{F}_{n + 1} : {F}_{n}}\right\rbrack \) \( < \infty \) for all \( n \geq 0 \) . Suppose that for all \( n \) there exist places \( {P}_{n} \in {\mathbb{P}}_{{F}_{n}} \) and \( {Q}_{n} \in {\mathbb{P}}_{{F}_{n + 1}} \) with \( {Q}_{n} \mid {P}_{n} \) and ramification index \( e\left( {{Q}_{n} \mid {P}_{n}}\right) > 1 \) . Then it follows that \( {F}_{n} \subsetneqq {F}_{n + 1} \) . | By the Fundamental Equality we have \( \left\lbrack {{F}_{n + 1} : {F}_{n}}\right\rbrack \geq e\left( {{Q}_{n} \mid {P}_{n}}\right) \) and therefore \( {F}_{n} \subsetneqq {F}_{n + 1} \) . If we assume the equality \( e\left( {{Q}_{n} \mid {P}_{n}}\right) = \left\lbrack {{F}_{n + 1} : {F}_{n}}\right\rbrack \), then \( {F}_{n} \) and \( {F}_{n + 1} \) have the same constant field (since constant field extensions are unramified by Theorem 3.6.3). | Yes |
Theorem 1. Let \( \rho : \mathrm{G} \rightarrow \mathrm{{GL}}\left( \mathrm{V}\right) \) be a linear representation of \( \mathrm{G} \) in \( \mathrm{V} \) and let \( \mathrm{W} \) be a vector subspace of \( \mathrm{V} \) stable under \( \mathrm{G} \) . Then there exists a complement \( {\mathrm{W}}^{0} \) of \( \mathrm{W} \) in \( \mathrm{V} \) which is stable under \( \mathrm{G} \) . | Let \( {\mathrm{W}}^{\prime } \) be an arbitrary complement of \( \mathrm{W} \) in \( \mathrm{V} \), and let \( p \) be the corresponding projection of \( \mathrm{V} \) onto \( \mathrm{W} \) . Form the average \( {p}^{0} \) of the conjugates of \( p \) by the elements of \( \mathrm{G} \):
\[
{p}^{0} = \frac{1}{g}\mathop{\sum }\limits_{{t \in \mathrm{G}}}{\rho }_{t} \cdot p \cdot {\rho }_{t}^{-1}\;\left( {g\text{ being the order of }\mathrm{G}}\right) .
\]
Since \( p \) maps \( \mathrm{V} \) into \( \mathrm{W} \) and \( {\rho }_{t} \) preserves \( \mathrm{W} \) we see that \( {p}^{0} \) maps \( \mathrm{V} \) into \( \mathrm{W} \) ; we have \( {\rho }_{t}^{-1}x \in \mathrm{W} \) for \( x \in \mathrm{W} \), whence
\[
p \cdot {\rho }_{t}^{-1}x = {\rho }_{t}^{-1}x,\;{\rho }_{t} \cdot p \cdot {\rho }_{t}^{-1}x = x,\;\text{ and }\;{p}^{0}x = x.
\]
Thus \( {p}^{0} \) is a projection of \( \mathrm{V} \) onto \( \mathrm{W} \), corresponding to some complement \( {\mathrm{W}}^{0} \) of \( \mathrm{W} \) . We have moreover
\[
{\rho }_{s} \cdot {p}^{0} = {p}^{0} \cdot {\rho }_{s}\;\text{ for all }s \in \mathrm{G}.
\]
Indeed, computing \( {\rho }_{s} \cdot {p}^{0} \cdot {\rho }_{s}^{-1} \), we find:
\[
{\rho }_{s} \cdot {p}^{0} \cdot {\rho }_{s}^{-1} = \frac{1}{g}\mathop{\sum }\limits_{{t \in \mathrm{G}}}{\rho }_{s} \cdot {\rho }_{t} \cdot p \cdot {\rho }_{t}^{-1} \cdot {\rho }_{s}^{-1} = \frac{1}{g}\mathop{\sum }\limits_{{t \in \mathrm{G}}}{\rho }_{st} \cdot p \cdot {\rho }_{st}^{-1} = {p}^{0}.
\]
If now \( x \in {\mathrm{W}}^{0} \) and \( s \in \mathrm{G} \) we have \( {p}^{0}x = 0 \), hence \( {p}^{0} \cdot {\rho }_{s}x = {\rho }_{s} \cdot {p}^{0}x \) \( = 0 \), that is, \( {\rho }_{s}x \in {\mathrm{W}}^{0} \), which shows that \( {\mathrm{W}}^{0} \) is stable under \( \mathrm{G} \), and completes the proof. | Yes |
a) Show that \( X \subseteq {M}^{n} \) is definable in \( \mathcal{M} \) if and only if it is definable in \( {\mathcal{M}}^{ * } \) if and only if it is definable in \( {\mathcal{M}}_{0} \) . | Null | No |
Theorem 6.3 Let \( A \) and \( B \) be \( n \times n \) Hermitian matrices. Let \( 1 \leq i, j, k \leq n \) be indices. If \( i + j = k + 1 \), we have \({\lambda }_{k}\left( {A + B}\right) \geq {\lambda }_{i}\left( A\right) + {\lambda }_{j}\left( B\right)\). If \( i + j = k + n \), we have \({\lambda }_{k}\left( {A + B}\right) \leq {\lambda }_{i}\left( A\right) + {\lambda }_{j}\left( B\right)\). | Proof. Once again, any inequality can be deduced from the other ones by means of \( \left( {A, B}\right) \leftrightarrow \left( {-A, - B}\right) \) . Thus it is sufficient to treat the case where \( i + j = k + n \) . From (6.4), we know that there exists an \( \left( {n - k + 1}\right) \) -dimensional subspace \( H \) such that \({\lambda }_{k}\left( {A + B}\right) = \mathop{\min }\limits_{{x \in H\smallsetminus \{ 0\} }}\frac{{x}^{ * }\left( {A + B}\right) x}{\parallel x{\parallel }_{2}^{2}}.\) From (6.3), there also exist \( i \) - and \( j \) -dimensional subspaces \( F, G \) such that \({\lambda }_{i}\left( A\right) = \mathop{\max }\limits_{{x \in F\smallsetminus \{ 0\} }}\frac{{x}^{ * }{Ax}}{\parallel x{\parallel }_{2}^{2}},\;{\lambda }_{j}\left( B\right) = \mathop{\max }\limits_{{x \in G\smallsetminus \{ 0\} }}\frac{{x}^{ * }{Bx}}{\parallel x{\parallel }_{2}^{2}}.\) We now use Corollary 1.1 twice: we have \(\dim F \cap G \cap H \geq \dim F + \dim G \cap H - n \geq \dim F + \dim G + \dim H - {2n}\) \(\quad = i + j + 1 - n - k = 1\). We deduce that there exists a unit vector \( z \) in \( F \cap G \cap H \) . From above, it satisfies \({\lambda }_{k}\left( {A + B}\right) \leq {z}^{ * }\left( {A + B}\right) z,\;{\lambda }_{i}\left( A\right) \geq {z}^{ * }{Az},\;{\lambda }_{j}\left( B\right) \geq {z}^{ * }{Bz},\) whence the conclusion. | Yes |
Theorem 3.7. The continued fraction map \( T\left( x\right) = \left\{ \frac{1}{x}\right\} \) on \( \left( {0,1}\right) \) is ergodic with respect to the Gauss measure \( \mu \) . | Proof of Theorem 3.7. The description of the continued fraction map as a shift on the space \( {\mathbb{N}}^{\mathbb{N}} \) described above suggests the method of proof: the measure \( \mu \) corresponds to a rather complicated measure on the shift space, but if we can control the measure of cylinder sets (and their intersections) well enough then we may prove ergodicity along the lines of the proof of ergodicity for Bernoulli shifts in Proposition 2.15. | No |
Let \( \psi : {\Omega }_{1} \rightarrow {\Omega }_{2} \) be a \( {C}^{j} \) diffeomorphism that satisfies (2.1.11.1), (2.1.11.2), and (2.1.11.3). Then there is a number \( J = J\left( j\right) \) such that whenever \( g \in {W}_{0}^{j + J}\left( {\Omega }_{2}\right) \), then \( g \circ \psi \in {W}_{0}^{j}\left( {\Omega }_{1}\right) \) . | The proof is given in the text. It involves showing that the composition of a function in \( {W}_{0}^{j + J}\left( {\Omega }_{2}\right) \) with a \( {C}^{j} \) diffeomorphism is in \( {W}_{0}^{j}\left( {\Omega }_{1}\right) \) by using the chain rule and Leibniz's rule, and then applying the Sobolev embedding theorem. | Yes |
Corollary 1.18. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is algebraic if and only if both \( k \subseteq E \) and \( E \subseteq F \) are algebraic. | If \( k \subseteq F \) is algebraic, then every element of \( F \) is algebraic over \( k \), hence over \( E \), and every element of \( E \) is algebraic over \( k \) ; thus \( E \subseteq F \) and \( k \subseteq E \) are algebraic.
Conversely, assume \( k \subseteq E \) and \( E \subseteq F \) are both algebraic, and let \( \alpha \in F \) . Since \( \alpha \) is algebraic over \( E \), there exists a polynomial
\[
f\left( x\right) = {x}^{n} + {e}_{n - 1}{x}^{n - 1} + \cdots + {e}_{0} \in E\left\lbrack x\right\rbrack
\]
such that \( f\left( \alpha \right) = 0 \) . This implies that in fact \( \alpha \) is ’already’ algebraic over the subfield \( k\left( {{e}_{0},\ldots ,{e}_{n - 1}}\right) \subseteq E \) ; therefore,
\[
k\left( {{e}_{0},\ldots ,{e}_{n - 1}}\right) \subseteq k\left( {{e}_{0},\ldots ,{e}_{n - 1},\alpha }\right)
\]
is a finite extension. On the other hand,
\[
k \subseteq k\left( {{e}_{0},\ldots ,{e}_{n - 1}}\right)
\]
is a finite extension by Proposition 1.15, since each \( {e}_{i} \) is in \( E \) and therefore algebraic over \( k \) . By Proposition 1.10
\[
k \subseteq k\left( {{e}_{0},\ldots ,{e}_{n - 1},\alpha }\right)
\]
is finite. This implies that \( \alpha \) is algebraic over \( k \), as needed, by Lemma 1.9 | Yes |
A closed subgroup of a Lie group is a Lie group in its own right with respect to the relative topology. | It is well known (see [8]) that if a smooth map \( \varphi \) has constant rank, then \( {\varphi }^{-1}\{ e\} \) is a closed regular submanifold of \( G \) of dimension \( \dim G - \operatorname{rk}\varphi \) . Since \( \ker \varphi \) is a subgroup, it suffices to show that \( \varphi \) has constant rank. Write \( {l}_{g} \) for left translation by \( g \) . Because \( \varphi \) is a homomorphism, \( \varphi \circ {l}_{g} = {l}_{\varphi \left( g\right) } \circ \varphi \), and since \( {l}_{g} \) is a diffeomorphism, the rank result follows by taking differentials. | Yes |
Theorem 6.3.3. The space \( {L}_{1} \) cannot be embedded in a Banach space with unconditional basis. | Let \( X \) be a Banach space with \( K \) -unconditional basis \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) and suppose that \( T : {L}_{1} \rightarrow X \) is an embedding. We can assume that for some constant \( M \geq 1 \) ,
\[
\parallel f{\parallel }_{1} \leq \parallel {Tf}\parallel \leq M\parallel f{\parallel }_{1},\;f \in {L}_{1}.
\]
By exploiting the unconditionality of \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) we are going to build an unconditional basic sequence in \( {L}_{1} \) using a gliding-hump-type argument.
Take \( {\left( {\delta }_{k}\right) }_{k = 1}^{\infty } \) a sequence of positive real numbers with \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{\delta }_{k} < 1 \) . Let \( {f}_{0} = \) \( 1 = {\chi }_{\left\lbrack 0,1\right\rbrack },{n}_{1} = 1,{s}_{0} = 0 \) and pick \( {s}_{1} \in \mathbb{N} \) such that
\[
\begin{Vmatrix}{\mathop{\sum }\limits_{{j = {s}_{1} + 1}}^{\infty }{e}_{j}^{ * }\left( {T\left( {{f}_{0}{r}_{{n}_{1}}}\right) }\right) {e}_{j}}\end{Vmatrix} < \frac{1}{2}{\delta }_{1}.
\]
Put
\[
{x}_{1} = \mathop{\sum }\limits_{{j = {s}_{0} + 1}}^{{s}_{1}}{e}_{j}^{ * }\left( {T\left( {{f}_{0}{r}_{{n}_{1}}}\right) }\right) {e}_{j}
\]
... (rest of the proof is omitted) | No |
Proposition 11.25. Let \( F : M \rightarrow N \) be a smooth map between smooth manifolds with or without boundary. Suppose \( u \) is a continuous real-valued function on \( N \) , and \( \omega \) is a covector field on \( N \) . Then | To prove (11.14) we compute
\[
{\left( {F}^{ * }\left( u\omega \right) \right) }_{p} = d{F}_{p}^{ * }\left( {\left( u\omega \right) }_{F\left( p\right) }\right) \;\left( {\text{by }\left( {11.13}\right) }\right)
\]
\[
= d{F}_{p}^{ * }\left( {u\left( {F\left( p\right) }\right) {\omega }_{F\left( p\right) }}\right) \;\text{(by (11.8))}
\]
\[
= u\left( {F\left( p\right) }\right) d{F}_{p}^{ * }\left( {\omega }_{F\left( p\right) }\right) \;\text{(by linearity of}d{F}_{p}^{ * }\text{)}
\]
\[
= u\left( {F\left( p\right) }\right) {\left( {F}^{ * }\omega \right) }_{p}\;\text{(by (11.13))}
\]
\[
= {\left( \left( u \circ F\right) {F}^{ * }\omega \right) }_{p}\;\text{ (by (11.8)). }
\]
For (11.15), we let \( v \in {T}_{p}M \) be arbitrary, and compute
\[
{\left( {F}^{ * }du\right) }_{p}\left( v\right) = \left( {d{F}_{p}^{ * }\left( {d{u}_{F\left( p\right) }}\right) }\right) \left( v\right) \;\text{(by (11.13))}
\]
\[
= d{u}_{F\left( p\right) }\left( {d{F}_{p}\left( v\right) }\right) \;\text{(by definition of}\left. {d{F}_{p}^{ * }}\right)
\]
\[
= d{F}_{p}\left( v\right) u\;\text{ (by definition of }{du}\text{ ) }
\]
\[
= v\left( {u \circ F}\right) \;\text{(by definition of}d{F}_{p}\text{)}
\]
\[
= d{\left( u \circ F\right) }_{p}\left( v\right) \;\text{(by definition of}d\left( {u \circ F}\right) \text{).} | Yes |
Show that \( {\Delta }_{p, q} \) is a \( G \) -map from \( {V}_{p, q}\left( {\mathbb{C}}^{n}\right) \) onto \( {V}_{p - 1, q - 1}\left( {\mathbb{C}}^{n}\right) \). | Null | No |
Let \( G \) be the additive group of residue classes mod \( k \) . Show that a sequence of natural numbers \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) is equidistributed in \( G \) if and only if \( \mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi ia}{x}_{n}/k} = o\left( N\right) \) for \( a = 1,2,\ldots, k - 1 \) . | Let \( n = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{k}^{{\alpha }_{k}} \) be the unique factorization of \( n \) as a product of powers of primes. Let \( N = {p}_{1}\cdots {p}_{k} \) . Then \( \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \mathop{\sum }\limits_{{d \mid N}}\mu \left( d\right) \) since the Möbius function vanishes on numbers that are not square-free. Any divisor of \( N \) corresponds to a subset of \( \left\{ {{p}_{1},\ldots ,{p}_{k}}\right\} \) . Thus, for \( n > 1 \) , \( \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \mathop{\sum }\limits_{{r = 0}}^{k}\left( \begin{array}{l} k \\ r \end{array}\right) {\left( -1\right) }^{r} = {\left( 1 - 1\right) }^{k} = 0. \) The result is clear if \( n = 1 \) . | No |
If \( \mathfrak{a} \) is an ideal of a commutative ring \( R \) [with identity], then \( R/\mathfrak{a} \) is a domain if and only if \( \mathfrak{a} \) is a prime ideal. | Null | No |
Show that \( \mathbb{Z}\left\lbrack \sqrt{-2}\right\rbrack \) is Euclidean. | Null | No |
Show that the sequence of non-zero rational numbers in \( \left\lbrack {0,1}\right\rbrack \) is u.d. mod 1. | Null | No |
Let \( X,\mathcal{B}, N, W \) be as in Chapter 2, and \( \sigma : X \rightarrow X \) an endomorphism such that \( \# {\sigma }^{-1}\left( {\{ x\} }\right) = N, x \in X \), and assume in addition that \( X \) is a compact Hausdorff space. Suppose branches of \( {\sigma }^{-1} \) may be chosen such that, for some measure \( {v}_{0} \) on \( X \), \(\mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{\left( {\omega }_{1},\ldots ,{\omega }_{n}\right) }\left( {{W}^{\left( n\right) } \cdot g}\right) \left( {{\tau }_{{\omega }_{n}}\cdots {\tau }_{{\omega }_{1}}x}\right) = {v}_{0}\left( g\right) = {\int }_{X}{gd}{v}_{0}\) for all \( g \in C\left( \Omega \right) \). The function \( {W}^{\left( n\right) } \) is defined by \( {W}^{\left( n\right) }\left( y\right) = W\left( y\right) W\left( {\sigma \left( y\right) }\right) \cdots W\left( {{\sigma }^{n - 1}\left( y\right) }\right) \). Let \( h \in C\left( \Omega \right) \) be a solution to (6.1.1) such that \( h \geq 0 \). Then \( {h}_{\min }\left( x\right) {v}_{0}\left( h\right) \leq h\left( x\right) ,\;x \in X\). | Proof of Theorem 6.1.1. Let \( k \in {\mathbb{N}}_{0} \), and consider the \( N \) -adic representation \( k = \) \( {i}_{1} + {i}_{2}N + \cdots + {i}_{n}{N}^{n - 1} \) . Note that \(\omega \left( k\right) = \left( {{i}_{1},\ldots ,{i}_{n},\underset{\infty \text{ string of zeroes }}{\underbrace{0,0,0,\ldots }}}\right) ,\) and that \(h\left( x\right) = {R}_{W}^{n + p}h\left( x\right)\) (6.1.6) \(\begin{matrix} = \mathop{\sum }\limits_{\left( {\omega }_{1},\ldots ,{\omega }_{n + p}\right) }W\left( {{\tau }_{{\omega }_{1}}x}\right) \cdots W\left( {{\tau }_{{\omega }_{n}}\cdots {\tau }_{{\omega }_{1}}x}\right) \cdots \\ \cdots W\left( {{\tau }_{{\omega }_{n + p}}\cdots {\tau }_{{\omega }_{1}}x}\right) h\left( {{\tau }_{{\omega }_{n + p}}\cdots {\tau }_{{\omega }_{1}}x}\right) . \end{matrix}\) For \( n \) fixed, let \( p \rightarrow \infty \) . Then \(\mathop{\lim }\limits_{{p \rightarrow \infty }}\mathop{\sum }\limits_{\left( {\omega }_{n + 1},\ldots ,{\omega }_{n + p}\right) }W\left( {{\tau }_{{\omega }_{1}}x}\right) \cdots W\left( {{\tau }_{{\omega }_{n}}\cdots {\tau }_{{\omega }_{1}}x}\right) \cdots\) (6.1.7) \(\cdots W\left( {{\tau }_{{\omega }_{n + p}}\cdots {\tau }_{{\omega }_{1}}x}\right) h\left( {{\tau }_{{\omega }_{n + p}}\cdots {\tau }_{{\omega }_{1}}x}\right) = {P}_{x}\left( {\{ \omega \left( k\right) \} }\right) {v}_{0}\left( h\right)\) \(\text{for}k = {\omega }_{1} + {\omega }_{2}N + \cdots + {\omega }_{n}{N}^{n - 1}\text{.}\) (6.1.8) We now exchange limits, and use Fatou's lemma in the form \(\mathop{\sum }\limits_{k}\left( {\mathop{\lim }\limits_{p}\cdots }\right) \leq \mathop{\lim }\limits_{p}\mathop{\sum }\limits_{k}\cdots\) Combining (6.1.6) and (6.1.7), we get \(\mathop{\sum }\limits_{{k \in {\mathbb{N}}_{0}}}{P}_{x}\left( {\{ \omega \left( k\right) \} }\right) {v}_{0}\left( h\right) \leq h\left( x\right)\) (6.1.9) But the factor on the left-hand side in (6.1.9) is \( {P}_{x}\left( {\mathbb{N}}_{0}\right) = {h}_{\min }\left( x\right) \), so the desired estimate (6.1.5) follows. | Yes |
Proposition 16. Let \( V \) and \( W \) be finite dimensional vector spaces over the field \( F \) with bases \( {v}_{1},\ldots ,{v}_{n} \) and \( {w}_{1},\ldots ,{w}_{m} \) respectively. Then \( V{ \otimes }_{F}W \) is a vector space over \( F \) of dimension \( {nm} \) with basis \( {v}_{i} \otimes {w}_{j},1 \leq i \leq n \) and \( 1 \leq j \leq m \) . | Remark: If \( v \) and \( w \) are nonzero elements of \( V \) and \( W \), respectively, then it follows from the proposition that \( v \otimes w \) is a nonzero element of \( V{ \otimes }_{F}W \), because we may always build bases of \( V \) and \( W \) whose first basis vectors are \( v, w \), respectively. In a tensor product \( M{ \otimes }_{R}N \) of two \( R \) -modules where \( R \) is not a field it is in general substantially more difficult to determine when the tensor product \( m \otimes n \) of two nonzero elements is zero. | No |
A compact Lie group \( G \) possesses a faithful representation, i.e., there exists a (finite-dimensional representation) \( \left( {\pi, V}\right) \) of \( G \) for which \( \pi \) is injective. | By the proof of the Peter-Weyl Theorem, for \( {g}_{1} \in {G}^{0},{g}_{1} \neq e \), there exists a finite-dimensional representation \( \left( {{\pi }_{1},{V}_{1}}\right) \) of \( G \), so that \( {\pi }_{1}\left( {g}_{1}\right) \) is not the identity operator. Thus \( \ker {\pi }_{1} \) is a closed proper Lie subgroup of \( G \), and so a compact Lie group in its own right. Since \( \ker {\pi }_{1} \) is a regular submanifold that does not contain a neighborhood of \( e \), it follows that \( \dim \ker {\pi }_{1} < \dim G \) . If \( \dim \ker {\pi }_{1} > 0 \), choose \( {g}_{2} \in {\left( \ker {\pi }_{1}\right) }^{0},{g}_{2} \neq e \), and let \( \left( {{\pi }_{2},{V}_{2}}\right) \) be a representation of \( G \), so that \( {\pi }_{2}\left( {g}_{2}\right) \) is not the identity. Then \( \ker \left( {{\pi }_{1} \oplus {\pi }_{2}}\right) \) is a compact Lie group with \( \ker \left( {{\pi }_{1} \oplus {\pi }_{2}}\right) < \) \( \dim \ker {\pi }_{1} \) . Continuing in this manner, there are representations \( \left( {{\pi }_{i},{V}_{i}}\right) ,1 \leq i \leq N \), of \( G \) , so that \( \dim \ker \left( {{\pi }_{1} \oplus \cdots \oplus {\pi }_{N}}\right) = 0 \) . Since \( G \) is compact, \( \ker \left( {{\pi }_{1} \oplus \cdots \oplus {\pi }_{N}}\right) = \) \( \left\{ {{h}_{1},{h}_{2},\ldots ,{h}_{M}}\right\} \) for \( {h}_{i} \in G \) . Choose representations \( \left( {{\pi }_{N + i},{V}_{N + i}}\right) ,1 \leq i \leq M \), of \( G \), so that \( {\pi }_{N + i}\left( {h}_{i}\right) \) is not the identity. The representation \( {\pi }_{1} \oplus \cdots \oplus {\pi }_{N + M} \) does the trick. | Yes |
Proposition 36. Let \( F \) be a field of characteristic not dividing \( n \) which contains the \( {n}^{\text{th }} \) roots of unity. Then the extension \( F\left( \sqrt[n]{a}\right) \) for \( a \in F \) is cyclic over \( F \) of degree dividing \( n \). | The extension \( K = F\left( \sqrt[n]{a}\right) \) is Galois over \( F \) if \( F \) contains the \( {n}^{\text{th }} \) roots of unity since it is the splitting field for \( {x}^{n} - a \) . For any \( \sigma \in \operatorname{Gal}\left( {K/F}\right) ,\sigma \left( \sqrt[n]{a}\right) \) is another root of this polynomial, hence \( \sigma \left( \sqrt[n]{a}\right) = {\zeta }_{\sigma }\sqrt[n]{a} \) for some \( {n}^{\text{th }} \) root of unity \( {\zeta }_{\sigma } \) . This gives a map\\ \[ \operatorname{Gal}\left( {K/F}\right) \rightarrow {\mu }_{n} \\ \sigma \mapsto {\zeta }_{\sigma } \\ \]where \( {\mu }_{n} \) denotes the group of \( {n}^{\text{th }} \) roots of unity. Since \( F \) contains \( {\mu }_{n} \), every \( {n}^{\text{th }} \) root of unity is fixed by every element of \( \operatorname{Gal}\left( {K/F}\right) \) . Hence\\ \[ {\sigma \tau }\left( \sqrt[n]{a}\right) = \sigma \left( {{\zeta }_{\tau }\sqrt[n]{a}}\right) \\ \[ = {\zeta }_{\tau }\sigma \left( \sqrt[n]{a}\right) \\ \[ = {\zeta }_{\tau }{\zeta }_{\sigma }\sqrt[n]{a} = {\zeta }_{\sigma }{\zeta }_{\tau }\sqrt[n]{a} \\ \]which shows that \( {\zeta }_{\sigma \tau } = {\zeta }_{\sigma }{\zeta }_{\tau } \), so the map above is a homomorphism. The kernel consists precisely of the automorphisms which fix \( \sqrt[n]{a} \), namely the identity. This gives an injection of \( \operatorname{Gal}\left( {K/F}\right) \) into the cyclic group \( {\mu }_{n} \) of order \( n \), which proves the proposition. | Yes |
Let \( A \) be a unital commutative Banach algebra and let \( \varphi \in \) \( \partial \left( A\right) \) . Then \( \ker \varphi \) consists of joint topological zero divisors. | It suffices to show that given \( {a}_{1},\ldots ,{a}_{q} \in A \) such that \( d\left( {{a}_{1},\ldots ,{a}_{q}}\right) > 0 \) , there is no maximal ideal of \( A \) containing all of \( {a}_{1},\ldots ,{a}_{q} \) and corresponding to some point in \( \partial \left( A\right) \) . Of course, we can assume \( d\left( {{a}_{1},\ldots ,{a}_{q}}\right) \geq 1 \) ; that is, \( \mathop{\sum }\limits_{{j = 1}}^{q}\begin{Vmatrix}{{a}_{j}y}\end{Vmatrix} \geq \parallel y\parallel \) for all \( y \in A \) . We claim that
\[
\mathop{\sum }\limits_{{j = 1}}^{q}r\left( {{a}_{j}y}\right) \geq r\left( y\right)
\]
for all \( y \in A \) . To verify this turns out to be quite intricate.
Let \( B \) denote the algebra of all formal power series
\[
\widetilde{x}\left( {{t}_{1},\ldots ,{t}_{q}}\right) = \sum {x}_{{n}_{1},\ldots ,{n}_{q}}{t}_{1}^{{n}_{1}} \cdot \ldots \cdot {t}_{q}^{{n}_{q}}
\]
in \( q \) variables \( {t}_{1},\ldots ,{t}_{q} \), where summation extends over all \( \left( {{n}_{1},\ldots ,{n}_{q}}\right) \in {\mathbb{N}}_{0}^{q} \) , \( {x}_{{n}_{1},\ldots ,{n}_{q}} \in A \) and \( \sum \begin{Vmatrix}{x}_{{n}_{1},\ldots ,{n}_{q}}\end{Vmatrix} < \infty \) . Recall that multiplication in \( B \) is given by
\[
\sum {x}_{{n}_{1},\ldots ,{n}_{q}}{t}_{1}^{{n}_{1}} \cdot \ldots \cdot {t}_{q}^{{n}_{q}} \cdot \sum {y}_{{m}_{1},\ldots ,{m}_{q}}{t}_{1}^{{m}_{1}} \cdot \ldots \cdot {t}_{q}^{{m}_{q}} = \sum {z}_{{p}_{1},\ldots ,{p}_{q}}{t}_{1}^{{p}_{1}} \cdot \ldots \cdot {t}_{q}^{{p}_{q}},
\]
where, for \( \left( {{p}_{1},\ldots ,{p}_{q}}\right) \in {\mathbb{N}}_{0}^{q} \) ,
\[
{z}_{{p}_{1},\ldots ,{p}_{q}} = \mathop{\sum }\limits_{\substack{{{n}_{j} + {m}_{j} = {p}_{j}} \\ {1 \leq j \leq q} }}{x}_{{n}_{1},\ldots ,{n}_{q}}{y}_{{m}_{1},\ldots ,{m}_{q}}.
\]
It is not difficult to check (compare the proof of Theorem 3.4.7 in the case \( q = 1) \) that \( B \), equipped with the norm
\[
\parallel \widetilde{x}\parallel = \sum \begin{Vmatrix}{x}_{{n}_{1},\ldots ,{n}_{q}}\end{Vmatrix}
\]
becomes a commutative Banach algebra. The map
\[
\phi : x \rightarrow \widetilde{x}\left( {{t}_{1},\ldots ,{t}_{q}}\right) = x
\]
is an isometric isomorphism from \( A \) into \( B \) . Using \( \phi \), we identify \( A \) with \( \phi \left( A\right) \) . Let
\[
z = \mathop{\sum }\limits_{{j = 1}}^{q}{a}_{j}{t}_{j} \in B
\]
We prove by induction that \( \begin{Vmatrix}{{z}^{k}y}\end{Vmatrix} \geq \parallel y\parallel \) for all \( y \in A \) and \( k \in \mathbb{N} \) . Clearly,
\[
\parallel {zy}\parallel = \begin{Vmatrix}{\mathop{\sum }\limits_{{j = 1}}^{q}\left( {{a}_{j}y}\right) {t}_{j}}\end{Vmatrix} = \mathop{\sum }\limits_{{j = 1}}^{q}\begin{Vmatrix}{{a}_{j}y}\end{Vmatrix} \geq \parallel y\parallel
\]
by assumption. For the inductive step, suppose that \( \begin{Vmatrix}{{z}^{k - 1}y}\end{Vmatrix} \geq \parallel y\parallel \) for all \( y \) and note that
3 Functional Calculus, Shilov Boundary, and Applications
\[
{z}^{k - 1} = \mathop{\sum }\limits_{{{n}_{1} + \ldots + {n}_{q} = k - 1}}{c}_{{n}_{1},\ldots ,{n}_{q}}{a}_{1}^{{n}_{1}} \cdot \ldots \cdot {a}_{q}^{{n}_{q}}{t}_{1}^{{n}_{1}} \cdot \ldots \cdot {t}_{q}^{{n}_{q}},
\]
where \( {c}_{{n}_{1},\ldots ,{n}_{q}} > 0 \) . For any \( y \in A \),
\[
\begin{Vmatrix}{{z}^{k - 1}y}\end{Vmatrix} = \mathop{\sum }\limits_{{{n}_{1} + \ldots + {n}_{q} = k - 1}}{c}_{{n}_{1},\ldots ,{n}_{q}}\begin{Vmatrix}{{a}_{1}^{{n}_{1}} \cdot \ldots \cdot {a}_{q}^{{n}_{q}}y}\end{Vmatrix}.
\]
This implies
\[
\begin{Vmatrix}{{z}^{k}y}\end{Vmatrix} = \begin{Vmatrix}{z \cdot \mathop{\sum }\limits_{{{n}_{1} + \ldots + {n}_{q} = k - 1}}{c}_{{n}_{1},\ldots ,{n}_{q}}{a}_{1}^{{n}_{1}} \cdot \ldots \cdot {a}_{q}^{{n}_{q}}{t}_{1}^{{n}_{1}} \cdot \ldots \cdot {t}_{q}^{{n}_{q}}}\end{Vmatrix}
\]
\[
= \begin{Vmatrix}{\mathop{\sum }\limits_{{j = 1}}^{q}\mathop{\sum }\limits_{{{n}_{1} + \ldots + {n}_{q} = k - 1}}{c}_{{n}_{1},\ldots ,{n}_{q}}{a}_{1}^{{n}_{1}} \cdot \ldots \cdot {a}_{j}^{{n}_{j} + 1} \cdot \ldots \cdot {a}_{q}^{{n}_{q}}y{t}_{1}^{{n}_{1}} \cdot \ldots \cdot {t}_{q}^{{n}_{q}}}\end{Vmatrix}
\]
\[
= \mathop{\sum }\limits_{{{n}_{1} + \ldots + {n}_{q} = k - 1}}{c}_{{n}_{1},\ldots ,{n}_{q}}\left( {\mathop{\sum }\limits_{{j = 1}}^{q}\begin{Vmatrix}{{a}_{j}{a}_{1}^{{n}_{1}} \cdot \ldots \cdot {a}_{q}^{{n}_{q}}y}\end{Vmatrix}}\right)
\]
\[
\geq \mathop{\sum }\limits_{{{n}_{1} + \ldots + {n}_{q} = k - 1}}{c}_{{n}_{1},\ldots ,{n}_{q}}\begin{Vmatrix}{{a}_{1}^{{n}_{1}} \cdot \ldots \cdot {a}_{q}^{{n}_{q}}y}\end{Vmatrix}
\]
\[
= \begin{Vmatrix}{{z}^{k - 1}y}\end{Vmatrix}\text{.}
\]
Thus, the inductive hypothesis shows that \( \begin{Vmatrix}{{z}^{k}y}\end{Vmatrix} \geq \parallel y\parallel \) . Replacing \( y \) with \( {y}^{k} \) ,
we get
\[
{\begin{Vmatrix}{\left( zy\right) }^{k}\end{Vmatrix}}^{1/k} \geq {\begin{Vmatrix}{y}^{k}\end{Vmatrix}}^{1/k}
\]
for all \( y \in A \) and \( k \in \mathbb{N} \) and hence \( r\left( {zy}\right) \geq r\left( y\right) \) for all \( y \in A \) . Using that the spectral radius is subadditive and submultiplicative and that \( r\left( {t}_{j}\right) = 1 \) , \( 1 \leq j \leq q \), we obtain
\[
r\left( y\right) \leq r\left( {zy}\right) = r\left( {\mathop{\sum }\limits_{{j = 1}}^{q}{a}_{j}y{t}_{j}}\right)
\]
\[
\leq \mathop{\sum }\limits_{{j = 1}}^{q}r\left( {{a}_{j}y{t}_{j}}\right) \leq \mathop{\sum }\limits_{{j = 1}}^{q}r\left( {{a}_{j}y}\right) r\left( {t}_{j}\right)
\]
\[
= \mathop{\sum }\limits_{{j = 1}}^{q}r\left( {{a}_{j}y}\right)
\]
for all \( y \in A \) . This establishes the above claim.
With \( \Gamma : A \rightarrow C\left( {\Delta \left( A\right) }\right), y \rightarrow \widehat{y} \) denoting the Gelfand homomorphism, we can reformulate what we have shown so far by
\[
\mathop{\sum }\limits_{{j = 1}}^{q}{\begin{Vmatrix}{\widehat{a}}_{j}\widehat{y}\end{Vmatrix}}_{\infty } = \mathop{\sum }\limits_{{j = 1}}^{q}r\left( {{a}_{j}y}\right) \geq r\left( y\right) = \parallel \widehat{y}{\parallel }_{\infty }
\]
for all \( y \in A \) ; that is, \( d\left( {{\widehat{a}}_{1},\ldots ,{\widehat{a}}_{q}}\right) \geq 1 \) .
Let \( C \) denote the closure of \( \Gamma \left( A\right) \) in \( C\left( {\Delta \left( A\right) }\right) \) . Then, by Theorem 3.4.10, the functions \( {\widehat{a}}_{1},\ldots ,{\widehat{a}}_{q} \) cannot simultaneously belong to \( \ker \psi \) for any \( \psi \in \) \( \partial \left( C\right) \) . Now, there is a continuous bijection \( \varphi \rightarrow \widetilde{\varphi } \) between \( \Delta \left( A\right) \) and \( \Delta \left( C\right) \) satisfying \( \widetilde{\varphi }\left( \widehat{a}\right) = \varphi \left( a\right) \) for all \( a \in A \) . In addition, since \( \Delta \left( A\right) \) is compact, this bijection is a homeomorphism and maps \( \partial \left( A\right) \) onto \( \partial \left( C\right) \) (Remark 3.3.7). It follows that none of the ideals \( \ker \varphi ,\varphi \in \partial \left( A\right) \), can contain all of \( {a}_{1},\ldots ,{a}_{q} \) .
This finishes the proof of the theorem. | Yes |
Corollary 7.56 Suppose \( \mathbf{A} \) is a balanced pre-Abelian category and \( {\mathbf{A}}^{\prime } \) is Abelian. Suppose \( F : \mathbf{A} \rightarrow {\mathbf{A}}^{\prime } \) is a functor. | a) If \( F \) is contravariant and left exact, and \( \mathbf{A} \) has enough projectives, then \( {\mathcal{L}}^{0}F \approx F. \) b) If \( F \) is covariant and right exact, and \( \mathbf{A} \) has enough projectives, then \( {\mathcal{L}}_{0}F \approx F \) . c) If \( F \) is contravariant and right exact, and \( \mathbf{A} \) has enough injectives, then \( {\mathcal{R}}^{0}F \approx F \) . d) If \( F \) is covariant and left exact, and \( \mathbf{A} \) has enough injectives, then \( {\mathcal{R}}_{0}F \approx F \) . Proof: (b), (c), and (d) follow from (a) by making substitutions of opposite categories for \( \mathbf{A} \) and/or \( {\mathbf{A}}^{\prime } \) . For (a), note that \( \mathbf{A} \) satisfies Ab-epic by Proposition 7.17. But this means that projective \( = \) quasiprojective, and epimorphism \( = \) cokernel \( = \) cokernel \( {}^{ \bullet } \) (Proposition 7.14(b)). Hence, \( {\mathcal{L}}^{0}F \approx Q{\mathcal{L}}^{0}F \approx F \) . | Yes |
Proposition 2.7.14. Let \( F : X \times I \rightarrow Y \) be a cellular homotopy from \( f \) to g. Define \( {D}_{n} : {C}_{n}\left( X\right) \rightarrow {C}_{n + 1}\left( Y\right) \) by \( {D}_{n}\left( {e}_{\alpha }^{n}\right) = {\left( -1\right) }^{n + 1}{F}_{\# }\left( {{e}_{\alpha }^{n} \times I}\right) \) . Then \( \left\{ {D}_{n}\right\} \) is a chain homotopy between \( {f}_{\# } \) and \( {g}_{\# } \) . | Proof. We use the proof of 2.7.10 and 2.7.9 to get:\n\[
\partial {D}_{n}\left( {e}_{\alpha }^{n}\right) = {\left( -1\right) }^{n + 1}\partial {F}_{\# }\left( {{e}_{\alpha }^{n} \times I}\right)
\]\n\[
= {\left( -1\right) }^{n + 1}{F}_{\# }\partial \left( {{e}_{\alpha }^{n} \times I}\right)
\]\n\[
= {F}_{\# }\left( {{e}_{\alpha }^{n}\times \{ 0\} - {e}_{\alpha }^{n}\times \{ 1\} }\right) + {\left( -1\right) }^{n + 1}{F}_{\# }\left( {\left( {\partial {e}_{\alpha }^{n}}\right) \times I}\right)
\]\n\[
= {F}_{\# }\left( {{e}_{\alpha }^{n}\times \{ 0\} - {e}_{\alpha }^{n}\times \{ 1\} }\right) - {D}_{n - 1}\left( {\partial {e}_{\alpha }^{n}}\right) .
\]\nSo \( \partial {D}_{n}\left( {e}_{\alpha }^{n}\right) + {D}_{n - 1}\partial \left( {e}_{\alpha }^{n}\right) = {f}_{\# }\left( {e}_{\alpha }^{n}\right) - {g}_{\# }\left( {e}_{\alpha }^{n}\right) \) . | Yes |
If \( M \) is an \( R \) -S-bimodule and \( A \) is an \( R \) -T-bimodule, then \( {\operatorname{Hom}}_{R}\left( {M, A}\right) \) is an \( S \) - \( T \) -bimodule, in which | In the above, \( {s\alpha } \) and \( {\alpha t} \) are homomorphisms of left \( R \) -modules, since \( M \) and \( A \) are bimodules. Moreover, \( s\left( {\alpha + \beta }\right) = {s\alpha } + {s\beta } \), and\\
\[
s\left( {{s}^{\prime }\alpha }\right) \left( x\right) = \left( {{s}^{\prime }\alpha }\right) \left( {xs}\right) = \alpha \left( {\left( {xs}\right) {s}^{\prime }}\right) = \alpha \left( {x\left( {s{s}^{\prime }}\right) }\right) = \left( {\left( {s{s}^{\prime }}\right) \alpha }\right) \left( x\right) ,
\]
so that \( s\left( {{s}^{\prime }\alpha }\right) = \left( {s{s}^{\prime }}\right) \alpha \) ; thus \( {\operatorname{Hom}}_{R}\left( {M, A}\right) \) is a left \( S \) -module. Similarly, \( {\operatorname{Hom}}_{R}\left( {M, A}\right) \) is a right \( T \) -module, and\\
\[
\left( {s\left( {\alpha t}\right) }\right) \left( x\right) = \left( {\alpha t}\right) \left( {xs}\right) = \alpha \left( {xs}\right) t = \left( {\left( {s\alpha }\right) \left( x\right) }\right) t = \left( {\left( {s\alpha }\right) t}\right) \left( x\right) ,
\]
so that \( s\left( {\alpha t}\right) = \left( {s\alpha }\right) t \) and \( {\operatorname{Hom}}_{R}\left( {M, A}\right) \) is a bimodule. | Yes |
Corollary 10.36 (The Normal Bundle to a Submanifold of \( {\mathbb{R}}^{n} \) ). If \( M \subseteq {\mathbb{R}}^{n} \) is an immersed m-dimensional submanifold with or without boundary, its normal bundle \( {NM} \) is a smooth rank- \( \left( {n - m}\right) \) subbundle of \( {\left. T{\mathbb{R}}^{n}\right| }_{M} \) . For each \( p \in M \), there exists a smooth orthonormal frame for \( {NM} \) on a neighborhood of \( p \) . | Apply Lemma 10.35 to the smooth subbundle \( {\left. TM \subseteq T{\mathbb{R}}^{n}\right| }_{M} \) . | Yes |
Proposition 13.19. Suppose \( \mathcal{U} \) is any open cover of \( X \) . Then the inclusion map \( {C}_{ * }^{\mathcal{U}}\left( X\right) \rightarrow {C}_{ * }\left( X\right) \) induces a homology isomorphism \( {H}_{p}^{\mathcal{U}}\left( X\right) \cong {H}_{p}\left( X\right) \) for all \( p \) . | The idea of the proof is simple, although the technical details are somewhat involved. If \( \sigma : {\Delta }_{p} \rightarrow X \) is any singular \( p \) -simplex, the plan is to show that there is a homologous \( p \) -chain obtained by "subdividing" \( \sigma \) into \( p \) -simplices with smaller images. If we subdivide sufficiently finely, we can ensure that each of the resulting simplices will be \( \mathcal{U} \) -small. The tricky part is to do this in a systematic way that allows us to keep track of the boundary operators. Before the formal proof, let us lay some groundwork. | No |
Theorem 9.21 Every chordal graph which is not complete has two nonadjacent simplicial vertices. | Let \( \left( {{V}_{1},{V}_{2},\ldots ,{V}_{k}}\right) \) be a simplicial decomposition of a chordal graph, and let \( x \in {V}_{k} \smallsetminus \left( {{ \cup }_{i = 1}^{k - 1}{V}_{i}}\right) \) . Then \( x \) is a simplicial vertex. Now consider a simplicial decomposition \( \left( {{V}_{\pi \left( 1\right) },{V}_{\pi \left( 2\right) },\ldots ,{V}_{\pi \left( k\right) }}\right) \), where \( \pi \) is a permutation of \( \{ 1,2,\ldots, k\} \) such that \( \pi \left( 1\right) = k \) . Let \( y \in {V}_{\pi \left( k\right) } \smallsetminus \left( {{ \cup }_{i = 1}^{k - 1}{V}_{\pi \left( i\right) }}\right) \) . Then \( y \) is a simplicial vertex nonadjacent to \( x \) . | Yes |
Theorem 2.9. Let \( A \) be an entire ring, integrally closed in its quotient field \( K \) . Let \( f\left( X\right) \in A\left\lbrack X\right\rbrack \) have leading coefficient 1 and be irreducible over \( K \) (or \( A \), it’s the same thing). Let \( \mathfrak{p} \) be a maximal ideal of \( A \) and let \( \bar{f} = f{\;\operatorname{mod}\;\mathfrak{p}} \) . Suppose that \( \bar{f} \) has no multiple roots in an algebraic closure of \( A/\mathfrak{p} \) . Let \( L \) be a splitting field for \( f \) over \( K \), and let \( B \) be the integral closure of \( A \) in L. Let \( \mathfrak{P} \) be any prime of \( B \) above \( \mathfrak{p} \) and let a bar denote reduction mod \( \mathfrak{p} \) . Then the map
\[
{G}_{\mathfrak{P}} \rightarrow {\bar{G}}_{\mathfrak{P}}
\]
is an isomorphism of \( {G}_{\mathfrak{P}} \) with the Galois group of \( \bar{f} \) over \( \bar{A} \) . | Proof. Let \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be the roots of \( f \) in \( B \) and let \( \left( {{\bar{\alpha }}_{1},\ldots ,{\bar{\alpha }}_{n}}\right) \) be their reductions mod \( \mathfrak{P} \) . Since
\[
f\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {X - {\alpha }_{i}}\right)
\]
it follows that
\[
\bar{f}\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {X - {\bar{\alpha }}_{i}}\right)
\]
Any element of \( G \) is determined by its effect as a permutation of the roots, and for \( \sigma \in {G}_{\mathfrak{P}} \), we have
\[
\bar{\sigma }{\bar{\alpha }}_{i} = \overline{\sigma {\alpha }_{i}}.
\]
Hence if \( \bar{\sigma } = \mathrm{{id}} \) then \( \sigma = \mathrm{{id}} \), so the map \( {G}_{\mathfrak{P}} \rightarrow {\bar{G}}_{\mathfrak{P}} \) is injective. It is surjective by Proposition 2.5, so the theorem is proved. | Yes |
If \( M \) is an orientable hyperbolic 3-manifold, then \( M \) is isometric to \( {\mathbf{H}}^{3}/\Gamma \), where \( \Gamma \) is a torsion-free Kleinian group. | Now let us suppose that the manifold \( M = {\mathbf{H}}^{3}/\Gamma \) has finite volume. This means that the fundamental domain for \( \Gamma \) has finite volume and so \( \Gamma \) has finite covolume. Thus \( \Gamma \) is finitely generated. Furthermore, if \( M \) is not compact, then the ends of \( M \) can be described following Theorem 1.2.12 and the remarks preceding it. | No |
n we carry over the max-flow min-cut theorem to this case? Yes, very easily, if we notice that a flow can be interpreted to flow in a vertex as well, namely from the part where all the currents enter it to the part where all the currents leave it. More precisely, we can turn each vertex of \( \overrightarrow{G} \) into an edge (without changing the nature of the directed graph) in such a way that any current entering (and leaving) the vertex will be forced through the edge. To do this, replace each vertex \( x \in V - \{ s, t\} \) by two vertices, say \( {x}_{ - } \) and \( {x}_{ + } \) ; send each incoming edge to \( {x}_{ - } \) and send each outgoing edge out of \( {x}_{ + } \) . Finally, for each \( x \), add an edge from \( {x}_{ - } \) to \( {x}_{ + } \) with capacity \( c\left( {{x}_{ - },{x}_{ + }}\right) = c\left( x\right) \) (see Fig. III | Can we carry over the max-flow min-cut theorem to this case? Yes, very easily, if we notice that a flow can be interpreted to flow in a vertex as well, namely from the part where all the currents enter it to the part where all the currents leave it. More precisely, we can turn each vertex of \( \overrightarrow{G} \) into an edge (without changing the nature of the directed graph) in such a way that any current entering (and leaving) the vertex will be forced through the edge. To do this, replace each vertex \( x \in V - \{ s, t\} \) by two vertices, say \( {x}_{ - } \) and \( {x}_{ + } \) ; send each incoming edge to \( {x}_{ - } \) and send each outgoing edge out of \( {x}_{ + } \) . Finally, for each \( x \), add an edge from \( {x}_{ - } \) to \( {x}_{ + } \) with capacity \( c\left( {{x}_{ - },{x}_{ + }}\right) = c\left( x\right) \) (see Fig. III.2). | Yes |
If \( u \in {H}_{0}^{k, p}\left( \Omega \right) \) for some \( p \) and all \( k \in \mathbb{N} \), then \( u \in {C}^{\infty }\left( \Omega \right) \) . | Null | No |
Let \( \left( {G, S}\right) \) be a Coxeter system and let \( d\left( { \geq 1}\right) \) be the largest number such that there is a d-element subset \( T \) of \( S \) with \( \langle T\rangle \) finite. Then every torsion free subgroup of finite index in \( G \) has geometric dimension \( \leq d \) and has type \( F \) . | The dimension of \( \left| K\right| \) is \( d - 1 \), so the dimension of \( \left| D\right| \) is \( d \) . The torsion free subgroup \( H \) acts freely on \( D \), and \( G \smallsetminus \left| D\right| \) is finite. | No |
Corollary 12.5.2. The Kuratowski set for any minor-closed graph property is finite. | Null | No |
Show that \( L\left( {s,\chi }\right) \) converges absolutely for \( \Re \left( s\right) > 1 \) and that \( L\left( {s,\chi }\right) = \mathop{\prod }\limits_{\wp }{\left( 1 - \frac{\chi \left( \wp \right) }{N{\left( \wp \right) }^{s}}\right) }^{-1} \), in this region. Deduce that \( L\left( {s,\chi }\right) \neq 0 \) for \( \Re \left( s\right) > 1 \) . | We have by multiplicativity of \( \chi \) , \( L\left( {s,\chi }\right) = \mathop{\prod }\limits_{\wp }{\left( 1 - \frac{\chi \left( \wp \right) }{N{\left( \wp \right) }^{s}}\right) }^{-1} \) and the product converges absolutely for \( \Re \left( s\right) > 1 \) if and only if \( \mathop{\sum }\limits_{\wp }\frac{1}{N{\left( \wp \right) }^{s}} \) converges in this region, which is certainly the case as there are only a bounded number of prime ideals above a given prime \( p \) . The non-vanishing is also clear. | Yes |
Theorem 4.3.1 Consider play sequences starting from some positive position \( p \in {\mathbb{R}}_{ + }^{S} \) . | The rule for changing the "position" \( p \) to \( {p}^{s} \) by firing node \( s \) coincides with the mapping \( {\sigma }_{s}^{ * } : p \mapsto s\left( p\right) \) considered in Section 4.1 (cf. equation (4.5)). Hence, the point denoted \( {p}^{w} \) here is the same as the point denoted by \( {w}^{-1}\left( p\right) \) there. This implies the "if" direction of part (i). The rest of the theorem follows from Corollary 4.2.6. | 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 |
Let \( N, H \) be normal subgroups of a group \( G \) . Then \(\left\lbrack {N, H}\right\rbrack \subseteq N \cap H\) | Proof. It suffices to verify this on generators; that is, it suffices to check that \(\left\lbrack {n, h}\right\rbrack = n\left( {h{n}^{-1}{h}^{-1}}\right) = \left( {{nh}{n}^{-1}}\right) {h}^{-1} \in N \cap H\) for all \( n \in N, h \in H \) . But the first expression and the normality of \( N \) show that \( \left\lbrack {n, h}\right\rbrack \in N \) ; the second expression and the normality of \( H \) show that \( \left\lbrack {n, h}\right\rbrack \in H \) . | Yes |
Theorem 11.3. Let \( \left( {A, X,\Omega, p}\right) \) be a maximum modulus algebra over \( \Omega \) . Fix \( F \in A \) . Then \( \lambda \mapsto \log {Z}_{F}\left( \lambda \right) \) is subharmonic on \( \Omega \). | Proof. In view of Exercise 11.3, it suffices to show that \( \log {Z}_{F} \) satisfies the inequality (17). We fix a disk \( \Delta = \left\{ {\left| {\lambda - {\lambda }_{0}}\right| \leq r}\right\} \) contained in \( \Omega \) and apply Theorem 11.2 to the function \( F \), a point \( {x}^{0} \in {p}^{-1}\left( {\lambda }_{0}\right) \), and the disk \( \Delta \) . This yields \( G \in {H}^{\infty }\left( {\operatorname{int}\Delta }\right) \) with \( G\left( {\lambda }_{0}\right) = F\left( {x}^{0}\right) \), and, by (16), \( \left| {G\left( \lambda \right) }\right| \leq \mathop{\max }\limits_{{y \in {p}^{-1}\left( \lambda \right) }}\left| {F\left( y\right) }\right| = {Z}_{F}\left( \lambda \right) \) for a.a. \( \lambda \in \partial \Delta \) . By Jensen’s inequality on int \( \Delta \), we have \[
\log \left| {G\left( {\lambda }_{0}\right) }\right| \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }\log \left| {G\left( {{\lambda }_{0} + r{e}^{i\theta }}\right) }\right| {d\theta } \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }\log {Z}_{F}\left( {{\lambda }_{0} + r{e}^{i\theta }}\right) {d\theta }.
\] The left-hand side \( = \log \left| {F\left( {x}^{0}\right) }\right| \), so inequality (17) holds, and we are done. | "No" |
We have \[ \Delta = \mathop{\sum }\limits_{{w \in W}}{\left( -1\right) }^{l\left( w\right) }{\mathrm{e}}^{w\left( \rho \right) }.\] | The irreducible representation \( \chi \left( 0\right) \) with highest weight vector 0 is obviously the trivial representation. Therefore, \( \chi \left( 0\right) = {\mathrm{e}}^{0} = 1 \) . The formula now follows from (22.4). | No |
Let \( I \subset S \) be a graded ideal. Then \({\alpha }_{ij}\left( {S/I}\right) = {\alpha }_{ij}\left( {S/{\operatorname{gin}}_{{ < }_{\text{rev }}}\left( I\right) }\right) \). | Let \( i < n \) . According to the definition of the generic annihilator numbers we have \( {\alpha }_{ij}\left( {S/I}\right) = {\dim }_{K}{A}_{i}\left( {{x}_{n},{x}_{n - 1},\ldots ,{x}_{1};S/{\operatorname{gin}}_{{ < }_{\text{rev }}}\left( I\right) }\right) \), and \({\alpha }_{ij}\left( {S/{\operatorname{gin}}_{{ < }_{\mathrm{{rev}}}}\left( I\right) }\right) = {\dim }_{K}{A}_{i}\left( {{x}_{n},{x}_{n - 1},\ldots ,{x}_{1};S/{\operatorname{gin}}_{{ < }_{\mathrm{{rev}}}}\left( {{\operatorname{gin}}_{{ < }_{\mathrm{{rev}}}}\left( I\right) }\right) }\right) = {\dim }_{K}{A}_{i}\left( {{x}_{n},{x}_{n - 1},\ldots ,{x}_{1};S/{\operatorname{gin}}_{{ < }_{\text{rev }}}\left( I\right) }\right) .\) The last equation follows from Corollary 4.2.7. Hence the conclusion. | Yes |
whether \( T\left( {{\psi \phi } - \phi }\right) = 0 \) for \( \phi \in \mathfrak{D} \) . We use the definition of the support of \( T \) to answer this. We must verify only that the support of \( \left( {1 - \psi }\right) \phi \) is contained in \( {\mathbb{R}}^{n} \smallsetminus K \) . This is true because \( 1 - \psi \) is zero on a neighborhood of \( K \) . The linearity of \( \bar{T} \) is trivial. For the continuity, suppose that \( {\phi }_{j} \rightarrow 0 \) in \( \mathcal{S} \) . Then for any \( \alpha ,{D}^{\alpha }{\phi }_{j} \) tends uniformly to 0, and \( {D}^{\alpha }\left( {{\phi }_{j}\psi }\right) \) tends uniformly to 0 by Leibniz’s Rule. Since there is one compact set containing the supports of all \( \psi {\phi }_{j} \), we can conclude that \( \psi {\phi }_{j} \rightarrow 0 \) in \( \mathbf{D} \) . By the continuity of \( T, T\left( {\psi {\phi }_{j}}\right) \rightarrow 0 \) and \( \bar{T}\left( {\phi }_{j}\right) \rightarrow 0 | s whether \( T\left( {{\psi \phi } - \phi }\right) = 0 \) for \( \phi \in \mathfrak{D} \) . We use the definition of the support of \( T \) to answer this. We must verify only that the support of \( \left( {1 - \psi }\right) \phi \) is contained in \( {\mathbb{R}}^{n} \smallsetminus K \) . This is true because \( 1 - \psi \) is zero on a neighborhood of \( K \) . The linearity of \( \bar{T} \) is trivial. For the continuity, suppose that \( {\phi }_{j} \rightarrow 0 \) in \( \mathcal{S} \) . Then for any \( \alpha ,{D}^{\alpha }{\phi }_{j} \) tends uniformly to 0, and \( {D}^{\alpha }\left( {{\phi }_{j}\psi }\right) \) tends uniformly to 0 by Leibniz’s Rule. Since there is one compact set containing the supports of all \( \psi {\phi }_{j} \), we can conclude that \( \psi {\phi }_{j} \rightarrow 0 \) in \( \mathbf{D} \) . By the continuity of \( T, T\left( {\psi {\phi }_{j}}\right) \rightarrow 0 \) and \( \bar{T}\left( {\phi }_{j}\right) \rightarrow 0 \) . | Yes |
Let \( {C}_{0} \) and \( {C}_{1} \) be disjoint coanalytic subsets of \( I = \left\lbrack {0,1}\right\rbrack \) that are not Borel separated; i.e., there is no Borel set containing \( {C}_{0} \) and disjoint from \( {C}_{1} \) . Let \( {A}_{0} = \left( {I\times \{ 0\} }\right) \bigcup \left( {{C}_{0} \times \left\lbrack {0,3/4}\right\rbrack }\right) \) and \( {A}_{1} = \left( {I\times \{ 1\} }\right) \bigcup \left( {{C}_{1} \times \left\lbrack {1/4,1}\right\rbrack }\right) \). Clearly, \( {A}_{0},{A}_{1} \) are disjoint coanalytic sets with sections closed. Suppose there is a Borel map \( u : I \times I \rightarrow I \) such that \( u \mid B \) is the characteristic function of \( {A}_{1} \), where \( B = {A}_{0}\bigcup {A}_{1} \). Then, the set \( E = \{ x \in I : u\left( {x,1/2}\right) = 0\} \) is Borel and separates \( {C}_{0} \) from \( {C}_{1} \). | Null | No |
Let \( {\left( {e}_{n}\right) }_{n = 1}^{\infty } \) be a basis for a Banach space \( X \) with biorthogonal functionals \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) . Then for every \( {x}^{ * } \in {X}^{ * } \) there is a unique sequence of scalars \( {\left( {a}_{n}\right) }_{n = 1}^{\infty } \) such that \( {x}^{ * } = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}_{n}^{ * } \) the convergence of the series being in the weak* topology of \( {X}^{ * } \) . More precisely, \( {x}^{ * } = {\text{weak}}^{ * } - \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{n = 1}}^{N}{x}^{ * }\left( {e}_{n}\right) {e}_{n}^{ * } \) . | For every \( x \in X \) , \[ \left| {\left( {{x}^{ * } - {S}_{N}^{ * }\left( {x}^{ * }\right) }\right) \left( x\right) }\right| = \left| \left( {{x}^{ * }\left( {x - {S}_{N}\left( x\right) }\right) \mid \leq \begin{Vmatrix}{x}^{ * }\end{Vmatrix}\begin{Vmatrix}{x - {S}_{N}\left( x\right) }\end{Vmatrix}\overset{N \rightarrow \infty }{ \rightarrow }0.}\right. \right| \]
From Proposition 3.2.1 (a) we infer that \( {\left( {e}_{n}^{ * }\right) }_{n = 1}^{\infty } \) will be a basis for \( {X}^{ * } \) (in the regular sense) if and only if \( \left\lbrack {e}_{n}^{ * }\right\rbrack = {X}^{ * } \) . Our next result provides a useful test for this, but first let us see a trivial case. | Yes |
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