Q
stringlengths 26
3.6k
| A
stringlengths 1
9.94k
| Result
stringclasses 3
values |
|---|---|---|
Theorem 2.5. \( \deg \left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) = \phi \left( n\right) \) and \( \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) \simeq {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \), with \( a{\;\operatorname{mod}\;n} \) corresponding to the map \( {\zeta }_{n} \mapsto {\zeta }_{n}^{a} \) . | Proof. Since \( \mathbb{Q}\left( {\zeta }_{m}\right) \) is normal over \( \mathbb{Q} \), Proposition 2.4 implies that if \( \left( {m, n}\right) = 1 \) then \( \deg \left( {\mathbb{Q}\left( {\zeta }_{mn}\right) /\mathbb{Q}}\right) = \deg \left( {\mathbb{Q}\left( {\zeta }_{m}\right) /\mathbb{Q}}\right) \cdot \deg \left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) \) . It therefore suffices to evaluate the degree for prime powers, which we have already done (Exercise 1.1). Since \( \phi \left( {p}^{n}\right) = \left( {p - 1}\right) {p}^{n - 1} \) and \( \phi \left( {mn}\right) = \phi \left( m\right) \phi \left( n\right) \) for \( \left( {m, n}\right) = 1 \), we obtain \( \deg \left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) = \phi \left( n\right) \) . It is a standard exercise in Galois theory to show that \( \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) \) is a subgroup of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \) . Since they are of the same order, they must be equal. This completes the proof. | No |
There exists a constant \( C = C\left( n\right) < \infty \) such that for all \( j \geq 1 \) and for all \( f \) in \( {L}^{1}\left( {\mathbf{R}}^{n}\right) \) we have | Let \( {K}^{\left( j\right) } = {\left( {\varphi }_{j}\right) }^{ \vee } * {d\sigma } = {\Phi }_{{2}^{-j}} * {d\sigma } \), where \( \Phi \) is a Schwartz function. Setting\n\[
{\left( {K}^{\left( j\right) }\right) }_{t}\left( x\right) = {t}^{-n}{K}^{\left( j\right) }\left( {{t}^{-1}x}\right)\n\]we have that\n\[
{\mathcal{M}}_{j}\left( f\right) = \mathop{\sup }\limits_{{t > 0}}\left| {{\left( {K}^{\left( j\right) }\right) }_{t} * f}\right|\n\](6.5.10)\nThe proof of the lemma is based on the estimate:\n\[
{\mathcal{M}}_{j}\left( f\right) \leq C{2}^{j}\mathcal{M}\left( f\right)\n\](6.5.11)\nand the weak type \( \left( {1,1}\right) \) boundedness of the Hardy-Littlewood maximal operator \( \mathcal{M} \) (Theorem 2.1.6). To establish (6.5.11), it suffices to show that for any \( M > n \) there is a constant \( {C}_{M} < \infty \) such that\n\[
\left| {{K}^{\left( j\right) }\left( x\right) }\right| = \left| {\left( {{\Phi }_{{2}^{-j}} * {d\sigma }}\right) \left( x\right) }\right| \leq \frac{{C}_{M}{2}^{j}}{{\left( 1 + \left| x\right| \right) }^{M}}.\n\](6.5.12)\nThen Theorem 2.1.10 yields (6.5.11) and hence the required conclusion. | Yes |
If \( \left( {p - 1}\right) \nmid i \), show that \( {\left| {B}_{i}/i\right| }_{p} \leq 1 \) . | Null | No |
Suppose \( X \) is a paracompact Hausdorff space. If \( \mathcal{U} = {\left( {U}_{\alpha }\right) }_{\alpha \in A} \) is an indexed open cover of \( X \), then \( \mathcal{U} \) admits a locally finite open refinement \( \mathcal{V} = {\left( {V}_{\alpha }\right) }_{\alpha \in A} \) indexed by the same set, such that \( {\bar{V}}_{\alpha } \subseteq {U}_{\alpha } \) for each \( \alpha \) . | Proof. By Lemma 4.80, each \( x \in X \) has a neighborhood \( {Y}_{x} \) such that \( {\bar{Y}}_{x} \subseteq {U}_{\alpha } \) for some \( \alpha \in A \) . The open cover \( \left\{ {{Y}_{x} : x \in X}\right\} \) has a locally finite open refinement. Let us index this refinement by some set \( B \), and denote it by \( Z = {\left( {Z}_{\beta }\right) }_{\beta \in B} \) . For each \( \beta \), there is some \( x \in X \) such that \( {Z}_{\beta } \subseteq {Y}_{x} \), and therefore there is some \( \alpha \in A \) such that \( {\bar{Z}}_{\beta } \subseteq {\bar{Y}}_{x} \subseteq {U}_{\alpha } \) . Define a function \( a : B \rightarrow A \) by choosing some such index \( \alpha \in A \) for each \( \beta \in B \), and setting \( a\left( \beta \right) = \alpha \) .
For each \( \alpha \in A \), define an open subset \( {V}_{\alpha } \subseteq X \) by
\[
{V}_{\alpha } = \mathop{\bigcup }\limits_{{\beta : a\left( \beta \right) = \alpha }}{Z}_{\beta }
\]
(If there are no indices \( \beta \) such that \( a\left( \beta \right) = \alpha \), then \( {V}_{\alpha } = \varnothing \) .) Because the family \( Z \) is locally finite, the closure of \( {V}_{\alpha } \) is equal to \( \mathop{\bigcup }\limits_{{\beta : a\left( \beta \right) = \alpha }}{\bar{Z}}_{\beta } \) (Lemma 4.75), which is contained in \( {U}_{\alpha } \) as required. | Yes |
We have \( {T}^{2}f = m{f}^{ - } \) . | Part (i) is proved as for the finite field case. For (ii), if \( y \) is not prime to \( m \), then \( {T\chi }\left( y\right) = 0 \) by Theorem 1.1. If \( y \) is prime to \( m \) then we can make the usual change of variables to get the right answer. Part (iii) is then proved as in the finite field case. | No |
Let \( {M}^{ \bullet } \) be an exact complex in \( \mathrm{C}\left( \mathrm{A}\right) \) . Then the complex \( \mathcal{F}\left( {M}^{ \bullet }\right) \), obtained by applying \( \mathcal{F} \) to the objects and morphisms of \( {M}^{ \bullet } \), is a zero-object in \( \mathrm{D} \) . | The zero-morphism: \( {M}^{ \bullet } \rightarrow \) \( {M}^{ \bullet } \) is a quasi-isomorphism; hence it is mapped to an invertible morphism by \( \mathcal{F} \) :
\[
\mathcal{F}\left( {M}^{ \bullet }\right) \underset{{\operatorname{id}}_{\mathcal{F}\left( {M}^{ \bullet }\right) }}{\underbrace{\xrightarrow[]{\mathcal{F}\left( 0\right) }\mathcal{F}\left( {M}^{ \bullet }\right) \xrightarrow[]{\mathcal{F}{\left( 0\right) }^{-1}}\mathcal{F}\left( {M}^{ \bullet }\right) }}
\]
Since \( \mathcal{F} \) is additive, \( \mathcal{F}\left( 0\right) = 0 \) . It follows that \( {\operatorname{id}}_{\mathcal{F}\left( {M}^{ \bullet }\right) } = 0 \) and hence that \( \mathcal{F}\left( {M}^{ \bullet }\right) \) is a zero-object of D, by Exercise 1.6. | No |
Theorem 10.10. Let \( D \) be a proper subdomain of \( \widehat{\mathbb{C}} \) . Let \( A \) be a subset of \( D \) that has no limit point in \( D \), and let \( v \) be a function mapping \( A \) to \( {\mathbb{Z}}_{ > 0} \) . Then there exists a function \( f \in \mathbf{H}\left( D\right) \) with \( {v}_{z}\left( f\right) = v\left( z\right) \) for all \( z \in A \), whose restriction to \( D - A \) has no zeros. | Proof. To begin, we make the following observations:
1. \( A \) is either finite or countable.
2. Without loss of generality, we may assume that \( \infty \in D - A \) and that \( A \) is nonempty.
3. If \( A \) is finite, let \( A = \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \) . Set \( {v}_{j} = v\left( {z}_{j}\right) \), for all \( 1 \leq j \leq n \), and choose \( {z}_{0} \in \mathbb{C} - D \) . In this case we set
\[
f\left( z\right) = \frac{{\left( z - {z}_{1}\right) }^{{v}_{1}}\cdots {\left( z - {z}_{n}\right) }^{{v}_{n}}}{{\left( z - {z}_{0}\right) }^{{v}_{1} + \cdots + {v}_{n}}},
\]
and note that \( f \) is holomorphic on \( D \), and does not vanish on \( \widehat{\mathbb{C}} - \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \) since \( f\left( \infty \right) = 1 \) .
Since \( D - A \subset \widehat{\mathbb{C}} - \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \), we have thus established the theorem for finite sets \( A \) .
To prove the theorem for infinite sets \( A \), let \( K = \widehat{\mathbb{C}} - D \) . Note that \( K \) is a nonempty compact subset of \( \mathbb{C} \) . Let \( {\left\{ {\alpha }_{n}\right\} }_{n \in \mathbb{N}} \) be a sequence whose terms consist of all \( \alpha \in A \), where each \( \alpha \) is repeated \( v\left( \alpha \right) \) times.
We first claim that, for each positive integer \( n \), we can choose a \( {\beta }_{n} \in K \) such that \( \left| {{\beta }_{n} - {\alpha }_{n}}\right| \leq \left| {\beta - {\alpha }_{n}}\right| \) for all \( \beta \in K \) . To see that this can be done, note that, for each positive integer \( n \), the function \( z \mapsto {l}_{n}\left( z\right) = \left| {z - {\alpha }_{n}}\right| \) is continuous on \( K \) and, therefore, achieves a minimum at some \( {\beta }_{n} \in K \) .
The function \( f \) we are seeking (whose existence must be established) is
\[
f\left( z\right) = \mathop{\prod }\limits_{{n = 1}}^{\infty }{E}_{n}\left( \frac{{\alpha }_{n} - {\beta }_{n}}{z - {\beta }_{n}}\right) .
\]
We show next that the product on the RHS converges in \( D \), by proving that \( \sum \left| {1 - {E}_{n}\left( \frac{{\alpha }_{n} - {\beta }_{n}}{z - {\beta }_{n}}\right) }\right| \) converges uniformly on compact subsets of \( D \) . For this we first prove that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| {{\beta }_{n} - {\alpha }_{n}}\right| = 0 \) . If we assume \( \left| {{\beta }_{n} - {\alpha }_{n}}\right| \geq \delta \) for some \( \delta > 0 \) and infinitely many \( n \), then for some subsequence \( \left\{ {\alpha }_{{n}_{j}}\right\} \) of \( \left\{ {\alpha }_{n}\right\} \) ,
\[
\left| {z - {\alpha }_{{n}_{j}}}\right| \geq \delta \text{ for all }z \in K.
\]
(10.2)
But a subsequence of this subsequence converges to some point \( \alpha \) in \( \widehat{\mathbb{C}} \) . From (10.2) we conclude that \( \alpha \notin K \) . Thus we arrive at the contradiction that \( \alpha \in D \) and is a limit point of \( A \) . Next, we put \( {r}_{n} = 2\left| {{\alpha }_{n} - {\beta }_{n}}\right| \) and observe that \( \left\{ {r}_{n}\right\} \) converges to zero. Let \( {K}_{0} \) be any nonempty compact subset of \( D \) ; since \( K \) and \( {K}_{0} \) are disjoint compact subsets of \( \widehat{\mathbb{C}} \), the distance between them must be positive. Therefore, the fact that \( {r}_{n} \rightarrow 0 \) implies there is an \( N \in {\mathbb{Z}}_{ > 0} \) such that \( \left| {z - {\beta }_{n}}\right| > {r}_{n} \) for all \( z \in {K}_{0} \) and all \( n > N \) . Thus
\[
\left| \frac{{\alpha }_{n} - {\beta }_{n}}{z - {\beta }_{n}}\right| \leq \frac{{r}_{n}}{2{r}_{n}} = \frac{1}{2}\;\text{ for all }n > N\text{ and all }z \in {K}_{0},
\]
and hence
\[
\left| {1 - {E}_{n}\left( \frac{{\alpha }_{n} - {\beta }_{n}}{z - {\beta }_{n}}\right) }\right| \leq {\left| \frac{{\alpha }_{n} - {\beta }_{n}}{z - {\beta }_{n}}\right| }^{n + 1} \leq {\left( \frac{1}{2}\right) }^{n + 1}
\]
for all \( n > N \) and all \( z \in {K}_{0} \), where the first inequality follows from Lemma 10.7. By Theorem 10.5, the infinite product defining \( f \) converges and \( f \in \mathbf{H}\left( D\right) \) . Finally, it follows from Lemma 10.7 that \( f\left( z\right) = 0 \) if and only if \( {E}_{n}\left( \frac{{\alpha }_{n} - {\beta }_{n}}{z - {\beta }_{n}}\right) = 0 \) for some \( n \in {\mathbb{Z}}_{ > 0} \) if and only if \( z = {\alpha }_{n} \) for some \( n \in {\mathbb{Z}}_{ > 0} \) | Yes |
If \( \left| \mathcal{K}\right| \) is a power of a prime for some nonidentity conjugacy class \( \mathcal{K} \) of \( G \) , then \( G \) is not a non-abelian simple group. | Suppose to the contrary that \( G \) is a non-abelian simple group and let \( \left| \mathcal{K}\right| = {p}^{c} \) . Let \( g \in \mathcal{K} \) . If \( c = 0 \) then \( g \in Z\left( G\right) \), contrary to a non-abelian simple group having a trivial center. As above, let \( {\chi }_{1},\ldots ,{\chi }_{r} \) be all the irreducible characters of \( G \) with \( {\chi }_{1} \) the principal character and let \( \rho \) be the regular character of \( G \) . By decomposing \( \rho \) into irreducibles we obtain
\[
0 = \rho \left( g\right) = 1 + \mathop{\sum }\limits_{{i = 2}}^{r}{\chi }_{i}\left( 1\right) {\chi }_{i}\left( g\right)
\]
(19.3)
If \( p \mid {\chi }_{j}\left( 1\right) \) for every \( j > 1 \) with \( {\chi }_{j}\left( g\right) \neq 0 \), then write \( {\chi }_{j}\left( 1\right) = p{d}_{j} \) . In this case (3) becomes
\[
0 = 1 + p\mathop{\sum }\limits_{j}{d}_{j}{\chi }_{j}\left( g\right)
\]
Thus \( \mathop{\sum }\limits_{j}{d}_{j}{\chi }_{j}\left( g\right) = - 1/p \) is an algebraic integer, a contradiction. This proves there is some \( j \) such that \( p \) does not divide \( {\chi }_{j}\left( 1\right) \) and \( {\chi }_{j}\left( g\right) \neq 0 \) . If \( \varphi \) is a representation whose character is \( {\chi }_{j} \), then \( \varphi \) is faithful (because \( G \) is assumed to be simple) and, by Lemma 6, \( \varphi \left( g\right) \) is a scalar matrix. Since \( \varphi \left( g\right) \) commutes with all matrices, \( \varphi \left( g\right) \in Z\left( {\varphi \left( G\right) }\right) \) . This forces \( g \in Z\left( G\right) \), contrary to \( G \) being a non-abelian simple group. The proof of the lemma is complete. | Yes |
Theorem 10.7.6. Let \( f\left( s\right) \) be an entire function of order at most equal to \( k \in {\mathbb{Z}}_{ \geq 0} \) . For all \( s \in \mathbb{C} \) we have the absolutely convergent product | f\left( s\right) = {s}^{r}{e}^{{P}_{k}\left( s\right) }\mathop{\prod }\limits_{\rho }\left( {1 - \frac{s}{\rho }}\right) \exp \left( {\mathop{\sum }\limits_{{1 \leq j \leq k}}\frac{{\left( s/\rho \right) }^{j}}{j}}\right) , | No |
Let \( K \) be a pointed convex cone that decomposes into the direct sum (13.7). If \( x \in {K}_{i} \) is a sum \( x = {x}_{1} + \cdots + {x}_{k} \) of elements \( {x}_{j} \in K \) , then each \( {x}_{j} \in {K}_{i} \) . | We have \( 0 = {\Pi }_{{\widehat{E}}_{i}}x = {\Pi }_{{\widehat{E}}_{i}}{x}_{1} + \cdots + {\Pi }_{{\widehat{E}}_{i}}{x}_{k} \) . Each term \( {\widehat{x}}_{j} \mathrel{\text{:=}} {\Pi }_{{\widehat{E}}_{i}}{x}_{j} \) belongs to \( {\widehat{K}}_{i} \subseteq K \), so that \( {\widehat{x}}_{j} \in K \) and \( - {\widehat{x}}_{j} = \mathop{\sum }\limits_{{l \neq j}}{\widehat{x}}_{l} \in K \) . Since \( K \) contains no lines, we have \( {\widehat{x}}_{j} = 0 \), that is, \( {x}_{j} = {\Pi }_{{E}_{i}}{x}_{j} \in {K}_{i}, j = 1,\ldots, k \) . | Yes |
Let \( \mathbf{f} \) be a class function on \( {S}_{k} \) . Write \( \mathbf{f} = \mathop{\sum }\limits_{\lambda }{c}_{\lambda }{\mathbf{s}}_{\lambda } \), where the sum is over the partitions of \( k \) . Then | The \( {\mathbf{s}}_{\lambda } \) are orthonormal by Schur orthogonality, so \( {\left| \mathbf{f}\right| }^{2} = \sum {\left| {c}_{\lambda }\right| }^{2} \) . By Theorem 36.2, \( {\mathrm{{Ch}}}^{\left( n\right) }\left( {\mathbf{s}}_{\lambda }\right) \) are distinct irreducible characters when \( \lambda \) runs through the partitions of \( k \) with length \( \leq n \), while, by Proposition 36.5, \( {\operatorname{Ch}}^{\left( n\right) }\left( {\mathbf{s}}_{\lambda }\right) = 0 \) if \( l\left( \lambda \right) > n \) . Therefore, we may write
\[
{\operatorname{Ch}}^{\left( n\right) }\left( \mathbf{f}\right) = \mathop{\sum }\limits_{{l\left( \lambda \right) \leq n}}{c}_{\lambda }{\operatorname{Ch}}^{\left( n\right) }\left( {\mathbf{s}}_{\lambda }\right)
\]
and the \( {\operatorname{Ch}}^{\left( n\right) }\left( {\mathbf{s}}_{\lambda }\right) \) in this decomposition are orthonormal by Schur orthogonality on \( \mathrm{U}\left( n\right) \) . Thus, \( {\left| {\mathrm{{Ch}}}^{\left( n\right) }\left( \mathbf{f}\right) \right| }^{2} = \mathop{\sum }\limits_{{l\left( \lambda \right) \leq n}}{\left| {c}_{\lambda }\right| }^{2} \) . | Yes |
Theorem 9.6.2 Let \( F \) be a maximal non-elementary Fuchsian subgroup of \( {\Gamma }_{d} \) . Then \( F \) is conjugate in \( \operatorname{PSL}\left( {2,\mathbb{C}}\right) \) to a Fuchsian group commensurable with \( {F}_{D} \) . | Theorem 9.6.2 completes the proof. | No |
Theorem 5.9. Let \( X \) be Stein, \( {\left( {U}_{\imath },{f}_{\imath }\right) }_{\imath \in I} \) a Cousin II distribution on \( X \) , \( h \in {Z}^{1}\left( {\mathfrak{U},{\mathcal{O}}^{ \star }}\right) \) the corresponding cocycle. Then \( {\left( {U}_{t},{f}_{t}\right) }_{t \in I} \) is solvable if and only if \( c\left( h\right) = 0 \) . | Theorem B and Theorem 4.9. | No |
Corollary 7.4 Let \( f \) be a flow and \( K \) a cut. If \( \operatorname{val}\left( f\right) = \operatorname{cap}\left( K\right) \), then \( f \) is a maximum flow and \( K \) is a minimum cut. | Let \( {f}^{ * } \) be a maximum flow and \( {K}^{ * } \) a minimum cut. By Theorem 7.3,\[
\operatorname{val}\left( f\right) \leq \operatorname{val}\left( {f}^{ * }\right) \leq \operatorname{cap}\left( {K}^{ * }\right) \leq \operatorname{cap}\left( K\right)
\]But, by hypothesis, \( \operatorname{val}\left( f\right) = \operatorname{cap}\left( K\right) \) . It follows that \( \operatorname{val}\left( f\right) = \operatorname{val}\left( {f}^{ * }\right) \) and \( \operatorname{cap}\left( {K}^{ * }\right) = \operatorname{cap}\left( K\right) \) . Thus \( f \) is a maximum flow and \( K \) is a minimum cut. | Yes |
Suppose \( B, C \in {}_{R}\mathbf{M} \) . Then \(\text{ SP-dim }\left( {B \oplus C}\right) = \max \{ \text{ SP-dim }B,\text{ SP-dim }C\} \). | SP-dim \( \left( {B \oplus C}\right) = \max \{ \) SP-dim \( \left( {B \oplus 0}\right) , \) SP-dim \( \left( {B \oplus 0, B \oplus C}\right) \) by Proposition 5.2. But any module between \( B \oplus 0 \) and \( B \oplus C \) corresponds to a submodule of \( C \approx B \oplus C/B \oplus 0 \) by the fundamental isomorphism theorems, so any module between \( B \oplus 0 \) and \( B \oplus C \) has the form \( B \oplus {C}^{\prime } \) for a submodule \( {C}^{\prime } \) of \( C \) . Also P-dim \( \left( {B \oplus {C}^{\prime }}\right) = \max \left\{ {\text{P-dim}B,\text{P-dim}{C}^{\prime }}\right\} \) by Exercise 11, Chapter 4. Taking suprema over \( {C}^{\prime } \), SP-dim \( (B \oplus 0, B \oplus C\} = \) \( \max \{ \mathrm{P} - \dim B,\mathrm{{SP}} - \dim C\} \), so that \(\text{ SP-dim }\left( {B \oplus C}\right) = \max \{ \text{ SP-dim }B,\text{ SP-dim }C\}\) (since SP-dim \( B \geq \) P-dim \( B \) by definition). | Yes |
Corollary 10.2. Under the hypotheses of the Seifert-Van Kampen theorem, the homomorphism \( \Phi \) descends to an isomorphism from the amalgamated free product \( {\pi }_{1}\left( {U, p}\right) { * }_{{\pi }_{1}\left( {U \cap V, p}\right) }{\pi }_{1}\left( {V, p}\right) \) to \( {\pi }_{1}\left( {X, p}\right) \) . | Null | No |
Theorem 16.11 (Stokes’s Theorem). Let \( M \) be an oriented smooth \( n \) -manifold with boundary, and let \( \omega \) be a compactly supported smooth \( \left( {n - 1}\right) \) -form on \( M \) . Then | We begin with a very special case: suppose \( M \) is the upper half-space \( {\mathbb{H}}^{n} \) itself. Then because \( \omega \) has compact support, there is a number \( R > 0 \) such that supp \( \omega \) is contained in the rectangle \( A = \left\lbrack {-R, R}\right\rbrack \times \cdots \times \left\lbrack {-R, R}\right\rbrack \times \left\lbrack {0, R}\right\rbrack \) (Fig. 16.9). We can write \( \omega \) in standard coordinates as
\[
\omega = \mathop{\sum }\limits_{{i = 1}}^{n}{\omega }_{i}d{x}^{1} \land \cdots \land \widehat{d{x}^{i}} \land \cdots \land d{x}^{n},
\]
where the hat means that \( d{x}^{i} \) is omitted. Therefore,
\[
{d\omega } = \mathop{\sum }\limits_{{i = 1}}^{n}d{\omega }_{i} \land d{x}^{1} \land \cdots \land \widehat{d{x}^{i}} \land \cdots \land d{x}^{n}
\]
\[
= \mathop{\sum }\limits_{{i, j = 1}}^{n}\frac{\partial {\omega }_{i}}{\partial {x}^{j}}d{x}^{j} \land d{x}^{1} \land \cdots \land \widehat{d{x}^{i}} \land \cdots \land d{x}^{n}
\]
\[
= \mathop{\sum }\limits_{{i = 1}}^{n}{\left( -1\right) }^{i - 1}\frac{\partial {\omega }_{i}}{\partial {x}^{i}}d{x}^{1} \land \cdots \land d{x}^{n}.
\]
Thus we compute
\[
{\int }_{{\mathbb{H}}^{n}}{d\omega } = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( -1\right) }^{i - 1}{\int }_{A}\frac{\partial {\omega }_{i}}{\partial {x}^{i}}d{x}^{1} \land \cdots \land d{x}^{n}
\]
\[
= \mathop{\sum }\limits_{{i = 1}}^{n}{\left( -1\right) }^{i - 1}{\int }_{0}^{R}{\int }_{-R}^{R}\cdots {\int }_{-R}^{R}\frac{\partial {\omega }_{i}}{\partial {x}^{i}}\left( x\right) d{x}^{1}\cdots d{x}^{n}.
\]
We can change the order of integration in each term so as to do the \( {x}^{i} \) integration first. By the fundamental theorem of calculus, the terms for which \( i \neq n \) reduce to
\[
\mathop{\sum }\limits_{{i = 1}}^{{n - 1}}{\left( -1\right) }^{i - 1}{\int }_{0}^{R}{\int }_{-R}^{R}\cdots {\int }_{-R}^{R}\frac{\partial {\omega }_{i}}{\partial {x}^{i}}\left( x\right) d{x}^{1}\cdots d{x}^{n}
\]
\[
= \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}{\left( -1\right) }^{i - 1}{\int }_{0}^{R}{\int }_{-R}^{R}\cdots {\int }_{-R}^{R}\frac{\partial {\omega }_{i}}{\partial {x}^{i}}\left( x\right) d{x}^{i}d{x}^{1}\cdots \widehat{d{x}^{i}}\cdots d{x}^{n}
\]
\[
= \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}{\left( -1\right) }^{i - 1}{\int }_{0}^{R}{\int }_{-R}^{R}\ldots {\int }_{-R}^{R}{\left\lbrack {\omega }_{i}\left( x\right) \right\rbrack }_{{x}^{i} = - R}^{{x}^{i} = R}d{x}^{1}\cdots \overset{⏜}{d{x}^{i}}\cdots d{x}^{n} = 0,
\]
because we have chosen \( R \) large enough that \( \omega = 0 \) when \( {x}^{i} = \pm R \) . The only term that might not be zero is the one for which \( i = n \) . For that term we have
\[
{\int }_{{\mathbb{H}}^{n}}{d\omega } = {\left( -1\right) }^{n - 1}{\int }_{-R}^{R}\cdots {\int }_{-R}^{R}{\int }_{0}^{R}\frac{\partial {\omega }_{n}}{\partial {x}^{n}}\left( x\right) d{x}^{n}d{x}^{1}\cdots d{x}^{n - 1}
\]
\[
= {\left( -1\right) }^{n - 1}{\int }_{-R}^{R}\cdots {\int }_{-R}^{R}{\left\lbrack {\omega }_{n}\left( x\right) \right\rbrack }_{{x}^{n} = 0}^{{x}^{n} = R}d{x}^{1}\cdots d{x}^{n - 1}
\]
\[
= {\left( -1\right) }^{n}{\int }_{-R}^{R}\ldots {\int }_{-R}^{R}{\omega }_{n}\left( {{x}^{1},\ldots ,{x}^{n - 1},0}\right) d{x}^{1}\cdots d{x}^{n - 1},
\]
(16.5)
because \( {\omega }_{n} = 0 \) when \( {x}^{n} = R \).
To compare this to the other side of (16.4), we compute as follows:
\[
{\int }_{\partial {\mathbb{H}}^{n}}\omega = \mathop{\sum }\limits_{i}{\int }_{A \cap \partial {\mathbb{H}}^{n}}{\omega }_{i}\left( {{x}^{1},\ldots ,{x}^{n - 1},0}\right) d{x}^{1} \land \cdots \land \widehat{d{x}^{i}} \land \cdots \land d{x}^{n}.
\]
Because \( {x}^{n} \) vanishes on \( \partial {\mathbb{H}}^{n} \), the pullback of \( d{x}^{n} \) to the boundary is identically zero (see Exercise 11.30). Thus, the only term above that is nonzero is the one for which \( i = n \), which becomes
\[
{\int }_{\partial {\mathbb{H}}^{n}}\omega = {\int }_{A \cap \partial {\mathbb{H}}^{n}}{\omega }_{n}\left( {{x}^{1},\ldots ,{x}^{n - 1},0}\right) d{x}^{1} \land \cdots \land d{x}^{n - 1}.
\]
Taking into account the fact that the coordinates \( \left( {{x}^{1},\ldots ,{x}^{n - 1}}\right) \) are positively oriented for \( \partial {\mathbb{H}}^{n} \) when \( n \) is even and negatively oriented when \( n \) is odd (Example 15.26), we find that this is equal to (16.5). | Yes |
Show that the set \({K}_{f}\left( X\right) = \{ L \in K\left( X\right) : L\text{ is finite }\}\) is an \( {F}_{\sigma } \) set. | Null | No |
Let the rod be the interval \( \left( {0,2}\right) \), and the end points are kept each at a constant temperature, but these are different at the two ends. To be specific, say that \( u\left( {0, t}\right) = 2 \) and \( u\left( {2, t}\right) = 5 \) . Let us take the initial temperature to be given by \( f\left( x\right) = 1 - {x}^{2} \) . The whole problem is | The unique solution is easily found to be \( \varphi \left( x\right) = \frac{3}{2}x + 2 \) . Substituting this into the initial condition of the original problem, we have
\[
1 - {x}^{2} = u\left( {x,0}\right) = v\left( {x,0}\right) + \varphi \left( x\right) = v\left( {x,0}\right) + \frac{3}{2}x + 2.
\]
We collect all the conditions to be satisfied by \( v \) :
\[
\left( {\mathrm{E}}^{\prime }\right) \;{v}_{xx} = {v}_{t},\;0 < x < 2,\;t > 0;
\]
\[
\left( {\mathrm{B}}^{\prime }\right) \;v\left( {0, t}\right) = 0,\;v\left( {2, t}\right) = 0,\;t > 0;
\]
(6.9)
\[
\left( {\mathrm{I}}^{\prime }\right) \;v\left( {x,0}\right) = - {x}^{2} - \frac{3}{2}x - 1,\;0 < x < 2.
\]
This problem is essentially of the sort considered and solved in Sec. 1.4 and 6.1. A slight difference is the fact that the length of the rod is 2 instead of \( \pi \), but the only consequence of this is that the sine functions in the solution will be adapted to this interval (as in Sec. 4.5). The reader is urged to perform all the steps that lead to the following formula for "general" solutions of \( \left( {\mathrm{E}}^{\prime }\right) + \left( {\mathrm{B}}^{\prime }\right) \) :
\[
v\left( {x, t}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}\exp \left( {-\frac{{n}^{2}{\pi }^{2}}{4}t}\right) \sin \frac{n\pi }{2}x.
\]
Next, the coefficients are adapted to \( \left( {\mathrm{I}}^{\prime }\right) \) :
\[
{a}_{n} = \frac{2}{2}{\int }_{0}^{2}\left( {-{x}^{2} - \frac{3}{2}x - 1}\right) \sin \frac{n\pi }{2}{xdx} = \frac{{16}{\left( -1\right) }^{n} - 2}{n\pi } + \frac{{16}\left( {1 - {\left( -1\right) }^{n}}\right) }{{n}^{3}{\pi }^{3}}.
\]
Finally, we put together the answer to the original problem:
\[
u\left( {x, t}\right) = \frac{3}{2}x + 2 + \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {\frac{{16}{\left( -1\right) }^{n} - 2}{n\pi } + \frac{{16}\left( {1 - {\left( -1\right) }^{n}}\right) }{{n}^{3}{\pi }^{3}}}\right) {e}^{-{n}^{2}{\pi }^{2}t/4}\sin \frac{n\pi }{2}x.
\] | Yes |
If \( E \) is a finitepotent subspace of \( {\operatorname{End}}_{k}\left( V\right) \), then \( \operatorname{tr} : E \rightarrow k \) is \( k \) -linear. | Take \( y, x \in E \) and for any nonnegative integer \( n \), put
\[
{V}_{n} \mathrel{\text{:=}} \mathop{\sum }\limits_{w}w\left( V\right)
\]
where the sum is taken over all words \( w \) of length \( n \) in \( x \) and \( y \) . If \( {w}_{0} \) is any initial segment of \( w \), then \( w\left( V\right) \subseteq {w}_{0}\left( V\right) \), and in particular, \( {V}_{n} \subseteq {V}_{n - 1} \) . This implies that \( {V}_{n} \) is invariant under \( y \) and \( x \) . For sufficiently large \( n \), it follows that \( {V}_{n} \) is a core subspace for both \( x \) and \( y \), and linearity of \( {\operatorname{tr}}_{V} \) follows from linearity of \( {\operatorname{tr}}_{{V}_{n}} \) . | No |
Let \( X \) be a standard Borel space, \( Y \) Polish, and \( A \subseteq X \times Y \) analytic with \( {\pi }_{X}\left( A\right) \) uncountable. Suppose that for every \( x \in {\pi }_{X}\left( A\right) \) , the section \( {A}_{x} \) is perfect. Then there is a \( C \subseteq {\pi }_{X}\left( A\right) \) homeomorphic to the Cantor set and a one-to-one Borel map \( f : C \times {2}^{\mathbb{N}} \rightarrow A \) such that \( {\pi }_{X}\left( {f\left( {x,\alpha }\right) }\right) = x \) for every \( x \) and every \( \alpha \) . | Fix a compatible complete metric on \( Y \) and a countable base \( \left( {U}_{n}\right) \) for the topology of \( Y \) . For each \( s \in {2}^{ < \mathbb{N}} \), we define a map \( {n}_{s}\left( x\right) : {\pi }_{X}\left( A\right) \rightarrow \mathbb{N} \) satifying the following conditions.
(i) \( x \rightarrow {n}_{s}\left( x\right) \) is \( \sigma \left( {\mathbf{\sum }}_{1}^{1}\right) \) -measurable,
(ii) diameter \( \left( {U}_{{n}_{s}\left( x\right) }\right) < \frac{1}{{2}^{\left| s\right| }} \) ,
(iii) \( {U}_{{n}_{s}\left( x\right) } \cap {A}_{x} \neq \varnothing \) for all \( x \in {\pi }_{X}\left( A\right) \) ,
(iv) \( \operatorname{cl}\left( {U}_{{n}_{s{}^{ \frown }\epsilon }\left( x\right) }\right) \subseteq {U}_{{n}_{s}\left( x\right) },\epsilon = 0 \) or \( 1 \), and
(v)
\[
\operatorname{cl}\left( {U}_{{n}_{{s}^{\prime }0}\left( x\right) }\right) \cap \operatorname{cl}\left( {U}_{{n}_{{s}^{\prime }1}\left( x\right) }\right) = \varnothing .
\]
Such a system of functions is defined by induction on \( \left| s\right| \) . This is a fairly routine exercise, which we leave for the reader. Now fix a continuous probability measure \( P \) on \( X \) such that \( P\left( {{\pi }_{X}\left( A\right) }\right) = 1 \) . Since every set in \( \sigma \left( {\mathbf{\sum }}_{1}^{1}\right) \) is \( P \) -measurable and since \( {\pi }_{X}\left( A\right) \) is uncountable, there is a homeomorph \( C \) of the Cantor set contained in \( {\pi }_{X}\left( A\right) \) such that \( {n}_{s} \mid C \) is Borel measurable for all \( s \in {2}^{ < \mathbb{N}} \) . Take \( x \in C \) and \( \alpha \in {2}^{\mathbb{N}} \) . Note that \( \mathop{\bigcap }\limits_{k}{U}_{{n}_{\alpha \mid k\left( x\right) }} \) is a singleton, say \( \{ y\} \) . Put \( f\left( {x,\alpha }\right) = \left( {x, y}\right) \) . The map \( f \) has the desired properties. | No |
Let \( A, B \) be sets, and suppose that \( \operatorname{card}\left( A\right) \leqq \operatorname{card}\left( B\right) \), and \( \operatorname{card}\left( B\right) \leqq \operatorname{card}\left( A\right) \) . Then | Then trivially, \( h \) is injective. We must prove that \( h \) is surjective. Let \( b \in B \) . If, when we try to lift back \( b \) by a succession of maps \[ \cdots \circ {f}^{-1} \circ {g}^{-1} \circ {f}^{-1} \circ {g}^{-1} \circ {f}^{-1}\left( b\right) \] we can lift back indefinitely, or if we get stopped in \( B \), then \( g\left( b\right) \) belongs to \( {A}_{2} \) and consequently \( b = h\left( {g\left( b\right) }\right) \), so \( b \) lies in the image of \( h \). On the other hand, if we cannot lift back \( b \) indefinitely, and get stopped in \( A \), then \( {f}^{-1}\left( b\right) \) is defined (i.e., \( b \) is in the image of \( f \) ), and \( {f}^{-1}\left( b\right) \) lies in \( {A}_{1} \). In this case, \( b = h\left( {{f}^{-1}\left( b\right) }\right) \) is also in the image of \( h \), as was to be shown. | Yes |
Theorem 17.21 (Hurewicz Isomorphism Theorem). Let \( X \) be a simply connected path-connected CW complex. Then the first nontrivial homotopy and homology occur in the same dimension and are equal, i.e., given a positive integer \( n \geq 2 \), if \( {\pi }_{q}\left( X\right) = 0 \) for \( 1 \leq q < n \), then \( {H}_{q}\left( X\right) = 0 \) for \( 1 \leq q < n \) and \( {H}_{n}\left( X\right) = {\pi }_{n}\left( X\right) \) | Proof. To start the induction, consider the case \( n = 2 \) . The \( {E}^{2} \) term of the homology spectral sequence of the path fibration is
Thus
\[
{H}_{2}\left( X\right) = {H}_{1}\left( {\Omega X}\right) \;\text{because}{PX}\text{has no homology}
\]
\[
= {\pi }_{1}\left( {\Omega X}\right) \;\text{because}{\pi }_{1}\left( {\Omega X}\right) = {\pi }_{2}\left( X\right) \text{is Abelian}
\]
\[
= {\pi }_{2}\left( X\right)
\]
Now let \( n \) be any positive integer greater than 2. By the induction hypothesis applied to \( {\Omega X} \) ,
\[
{H}_{q}\left( {\Omega X}\right) = 0\;\text{ for }q < n - 1
\]
and
\[
{H}_{n - 1}\left( {\Omega X}\right) = {\pi }_{n - 1}\left( {\Omega X}\right) = {\pi }_{n}\left( X\right) .
\]
The \( {E}_{2} \) term of the homology spectral sequence of the path fibration is
Since \( {PX} \) has trivial homology,
\[
{H}_{q}\left( X\right) = {H}_{q - 1}\left( {\Omega X}\right) = 0\;\text{ for }1 \leq q < n
\]
and
\[
{H}_{n}\left( X\right) = {H}_{n - 1}\left( {\Omega X}\right) = {\pi }_{n}\left( X\right)
\] | Yes |
Theorem 8.7. In a field extension, all transcendence bases have the same number of elements. | Theorem 8.7 is similar to the statement that all bases of a vector space have the same number of elements, and is proved in much the same way. First we establish an exchange property.
Lemma 8.8. Let \( B \) and \( C \) be transcendence bases of a field extension \( E \) of \( K \) . For every \( \beta \in B \) there exists \( \gamma \in C \) such that \( \left( {B\smallsetminus \{ \beta \} }\right) \cup \{ \gamma \} \) is a transcendence base of \( E \) over \( K \), and either \( \gamma = \beta \) or \( \gamma \notin B \) .
Proof. If \( \beta \in C \), then \( \gamma = \beta \) serves. Now let \( \beta \notin C \) . If every \( \gamma \in C \) is algebraic over \( K\left( {B\smallsetminus \{ \beta \} }\right) \), then, by \( {3.3},{3.5}, K\left( C\right) \) is algebraic over \( K\left( {B\smallsetminus \{ \beta \} }\right) \), and \( E \), which is algebraic over \( K\left( C\right) \), is algebraic over \( K\left( {B\smallsetminus \{ \beta \} }\right) \), contradicting 8.4. Therefore some \( \gamma \in C \) is transcendental over \( K\left( {B\smallsetminus \{ \beta \} }\right) \) . Then \( \gamma \notin B \smallsetminus \{ \beta \} \) ; in fact, \( \gamma \notin B \) since \( \gamma \neq \beta \) . By 8.3, \( {B}^{\prime } = \left( {B\smallsetminus \{ \beta \} }\right) \cup \{ \gamma \} \) is algebraically independent over \( K \) .
Since \( B \) is a maximal algebraically independent subset, \( {B}^{\prime } \cup \{ \beta \} = B \cup \{ \gamma \} \) is not algebraically independent over \( K \), and \( \beta \) is algebraic over \( K\left( {B}^{\prime }\right) \) by 8.3 . By \( {3.3},{3.5}, K\left( B\right) \) is algebraic over \( K\left( {B}^{\prime }\right) \), and \( E \), which is algebraic over \( K\left( B\right) \), is algebraic over \( K\left( {B}^{\prime }\right) \) . \( ▱ \)
We now prove 8.7. Let \( B \) and \( C \) be transcendence bases of \( K \subseteq E \) .
Assume that \( C \) is finite, with \( n = \left| C\right| \) elements. If \( B = \left\{ {{\beta }_{1},\ldots ,{\beta }_{n},{\beta }_{n + 1}}\right. \) , \( \ldots \} \) has more than \( n \) elements, then repeated applications of 8.8 yield transcendence bases \( \left\{ {{\gamma }_{1},{\beta }_{2},\ldots ,{\beta }_{n},{\beta }_{n + 1},\ldots }\right\} ,\left\{ {{\gamma }_{1},{\gamma }_{2},{\beta }_{3},\ldots ,{\beta }_{n},{\beta }_{n + 1},\ldots }\right\} ,\ldots \) , \( \left\{ {{\gamma }_{1},\ldots ,{\gamma }_{n},{\beta }_{n + 1},\ldots }\right\} \) . But \( C \) is a maximal algebraically independent subset. Hence \( B \) has at most \( n \) elements. Exchanging \( B \) and \( C \) then yields \( \left| B\right| = \left| C\right| \).
Now assume that \( C \) is infinite. Then \( B \) is infinite. In this case we use a cardinality argument. Every \( \beta \in B \) is algebraic over \( K\left( C\right) \) . Hence \( \beta \) is algebraic over \( K\left( {C}_{\beta }\right) \) for some finite subset \( {C}_{\beta } \) of \( C \) : indeed, \( f\left( \beta \right) = 0 \) for some polynomial \( f \in K\left( C\right) \left\lbrack X\right\rbrack \), and \( {C}_{\beta } \) need only include all the elements of \( C \) that appear in the coefficients of \( f \) . Then every \( \beta \in B \) is algebraic over \( K\left( {C}^{\prime }\right) \) , where \( {C}^{\prime } = \mathop{\bigcup }\limits_{{\beta \in B}}{C}_{\beta } \subseteq C \) . By \( {3.3},{3.5}, K\left( B\right) \) is algebraic over \( K\left( {C}^{\prime }\right) \), and \( E \) is algebraic over \( K\left( {C}^{\prime }\right) \) . Since \( C \) is minimal with this property, it follows that \( C = {C}^{\prime } = \mathop{\bigcup }\limits_{{\beta \in B}}{C}_{\beta } \) . Thus \( C \) is the union of \( \left| B\right| \) finite sets and \( \left| C\right| \leqq \left| B\right| {\aleph }_{0} = \) \( \left| B\right| \), by A.5.9. Exchanging \( B \) and \( C \) yields \( \left| B\right| = \left| C\right| \) . \( ▱ \) | Yes |
Theorem 7. Let \( p \) and \( {p}^{\prime } \) be PL paths in \( \mathrm{{CP}}\left( {U,{P}_{0}}\right) \), where \( U \) is open in \( {\mathbf{R}}^{3} \) and \( {P}_{0} \in U \) . If \( p \cong {p}^{\prime } \), then there is a PL mapping \( f : {\left\lbrack 0,1\right\rbrack }^{2} \rightarrow U \), under which \( p \cong {p}^{\prime } \) in \( \pi \left( {U,{P}_{0}}\right) \) . | Null | No |
If \( {H}_{\phi } \) and \( {H}_{\psi } \) are Hankel operators and \( U \) is the unilateral shift, then \( {H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = \left( {P\breve{\phi }}\right) \otimes \left( {P\bar{\psi }}\right) \) | Note that \( {H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = {H}_{\phi }{H}_{\psi } - {H}_{\phi }U{U}^{ * }{H}_{\psi }\;\text{ (by Theorem 4.1.7) } \) \[= {H}_{\phi }\left( {I - U{U}^{ * }}\right) {H}_{\psi } \] Recall that \( I - U{U}^{ * } \) is the projection of \( {\widetilde{\mathbf{H}}}^{2} \) onto the constant functions; i.e., \( I - {U}^{ * }U = {e}_{0} \otimes {e}_{0} \) . Therefore \[ {H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = {H}_{\phi }\left( {{e}_{0} \otimes {e}_{0}}\right) {H}_{\psi } \] \[= \left( {{H}_{\phi }{e}_{0}}\right) \otimes \left( {{H}_{{\psi }^{ * }}{e}_{0}}\right) \] \[= \left( {PJ\phi }\right) \otimes \left( {{PJ}{\psi }^{ * }}\right) \] \[= \left( {P\breve{\phi }}\right) \otimes \left( {P\bar{\psi }}\right) \] | Yes |
Corollary 14. (Eisenstein’s Criterion for \( \mathbb{Z}\left\lbrack x\right\rbrack \) ) Let \( p \) be a prime in \( \mathbb{Z} \) and let \( f\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \in \mathbb{Z}\left\lbrack x\right\rbrack, n \geq 1 \) . Suppose \( p \) divides \( {a}_{i} \) for all \( i \in \{ 0,1,\ldots, n - 1\} \) but that \( {p}^{2} \) does not divide \( {a}_{0} \) . Then \( f\left( x\right) \) is irreducible in both \( \mathbb{Z}\left\lbrack x\right\rbrack \) and \( \mathbb{Q}\left\lbrack x\right\rbrack \) . | This is simply a restatement of Proposition 13 in the case of the prime ideal \( \left( p\right) \) in \( \mathbb{Z} \) together with Corollary 6. | Yes |
Let \( \left( {\widetilde{P},\widetilde{H}}\right) \) be a general Hartogs figure in \( {\mathbb{C}}^{n} \) , \( f \) holomorphic in \( \widetilde{H} \) . Then there is exactly one holomorphic function \( F \) on \( \widetilde{P} \) with \( F \mid \widetilde{H} = f \) . | Let \( \left( {\widetilde{P},\widetilde{H}}\right) = \left( {g\left( P\right), g\left( H\right) }\right), g : P \rightarrow {\mathbb{C}}^{n} \) be biholomorphic. Then \( f \circ g \) is holomorphic in \( H \) and by Theorem 5.5 of Chapter I there is exactly one holomorphic function \( {F}^{ \star } \) on \( P \) with \( {F}^{ \star } \mid H = f \circ g \) . Let \( F = {F}^{ \star } \circ {g}^{-1} \) . Then \( F \) is holomorphic in \( \widetilde{P}, F \mid \widetilde{H} = f \), and the uniqueness of the continuation follows from the uniqueness of \( {F}^{ \star } \) . | Yes |
Corollary 2.5.4. Let \( f, g : \left( {{B}^{n},{S}^{n - 1}}\right) \rightarrow \left( {{e}_{\alpha }^{n},{\overset{ \bullet }{e}}_{\alpha }^{n}}\right) \) be characteristic maps inducing \( {f}^{\prime },{g}^{\prime } : {B}^{n}/{S}^{n - 1} \rightarrow {e}_{\alpha }^{n}/{e}_{\alpha }^{n} \) . For \( t \in \left( {0,1}\right) \), let \( {A}_{t} \) be the annulus \( \left\{ {x \in {B}^{n}\left| {t < }\right| x \mid < 1}\right\} \) and let \( {S}_{t} = \left\{ {x \in {B}^{n}\left| \right| x \mid = t}\right\} \) . Let \( z \in {\overset{ \circ }{e}}_{\alpha }^{n} \) . If there exists \( t \) such that \( f\left( {A}_{t}\right) \cup g\left( {A}_{t}\right) \subset {\overset{ \circ }{e}}_{\alpha }^{n} - \{ z\} \), and if \( {f}^{\prime } \simeq {g}^{\prime } \), then \( f\left| { \simeq g}\right| : {S}_{t} \rightarrow {e}_{\alpha }^{n} - \{ z\} . \) | Proof. Since \( {B}^{n}/{S}^{n - 1} \) and \( {e}_{\alpha }^{n}/{\mathbf{e}}_{\alpha }^{n} \) are homeomorphic to \( {S}^{n} \), the result follows from 2.5.3. In fact, \( f\left| { \simeq g}\right| \) as maps from \( {S}_{t} \) to \( {\overset{ \circ }{e}}_{\alpha }^{n} - \{ z\} \). | Yes |
Show that \( \mathbb{Q}\left( {\zeta }_{m}\right) \) is normal over \( \mathbb{Q} \) . | Solution. \( {\zeta }_{m} \) is a root of the \( m \) th cyclotomic polynomial, which we have shown to be irreducible. Thus, the conjugate fields are \( \mathbb{Q}\left( {\zeta }_{m}^{j}\right) \) where \( \left( {j, m}\right) = 1 \) and these are identical with \( \mathbb{Q}\left( {\zeta }_{m}\right) \) . | No |
Show that if P has the integer decomposition property, then P is an integral polyhedron. | 1. If P has the integer decomposition property, then P is an integral polyhedron. | No |
Theorem 2. Let \( \mathrm{K} \) be a field that is complete under a discrete valuation and has perfect residue field \( \overline{\mathbf{K}} \) . Let \( \mathfrak{g} \) be the Galois group of the algebraic closure of \( \overline{\mathbf{K}} \) over \( \overline{\mathbf{K}} \), and let \( \mathbf{X}\left( \mathrm{g}\right) \) be its character group. Then there is a split exact sequence
\[
0 \rightarrow {\mathrm{B}}_{\overline{\mathrm{K}}} \rightarrow {\mathrm{B}}_{\mathrm{K}} \rightarrow \mathrm{X}\left( \mathrm{g}\right) \rightarrow 0.
\] | As we have seen, the splitting of this sequence comes from choosing a uniformizer of \( \mathrm{K} \) . | No |
Let \( \operatorname{Conj}\left( G\right) \) be the set of conjugacy classes in \( G \) . For each \( C \in \operatorname{Conj}\left( G\right) \) let \( {\varphi }_{C} \) be the characteristic function of \( C \) . Then the set \( {\left\{ {\varphi }_{C}\right\} }_{C \in \operatorname{Conj}\left( G\right) } \) is a basis for \( \mathcal{A}{\left\lbrack G\right\rbrack }^{G} \), and every function \( f \in \mathcal{A}{\left\lbrack G\right\rbrack }^{G} \) has the expansion | f = \mathop{\sum }\limits_{{C \in \operatorname{Conj}\left( G\right) }}f\left( C\right) {\varphi }_{C} | Yes |
Let \( {w}_{2},\ldots ,{w}_{n} \) be smooth boundary functions with \( \left| {w}_{j}\right| < \delta \) for all \( j \) and such that \( {w}_{2} \) is schlicht, i.e., its analytic extension is one-one in \( \left| \zeta \right| \leq 1 \) . Put \( A = {A}_{w} \) . Suppose \( {x}^{ * } \in {H}_{1},\left| {x}^{ * }\right| < \delta \) on \( \Gamma \) and \( A{x}^{ * } = {x}^{ * } \) . Then \( \exists \) analytic disk \( E \) with \( \partial E \) contained in \( U \) . | Since \( A{x}^{ * } = {x}^{ * },{x}^{ * } = - T\left\{ {h\left( {{X}^{ * }, w}\right) }\right\} \), and so \( {x}^{ * } + {ih}\left( {{x}^{ * }, w}\right) \) is a boundary function by (2). Let \( \psi \) be the analytic extension of \( {x}^{ * } + {ih}\left( {{x}^{ * }, w}\right) \) to \( \left| \zeta \right| < 1 \) . The set defined for \( \left| \zeta \right| \leq 1 \) by \( {z}_{1} = \psi \left( \zeta \right) ,{z}_{2} = {w}_{2}\left( \zeta \right) ,\ldots ,{z}_{n} = {w}_{n}\left( \zeta \right) \) is an analytic disk \( E \) in \( {\mathbb{C}}^{n} \) . \( \partial E \) is defined for \( \left| \zeta \right| = 1 \) by \( {z}_{1} = {x}^{ * }\left( \zeta \right) + {ih}\left( {{x}^{ * }\left( \zeta \right) }\right. \), \( \left. {w\left( \zeta \right) }\right) ,{z}_{2} = {w}_{2}\left( \zeta \right) ,\ldots ,{z}_{n} = {w}_{n}\left( \zeta \right) \) and so by (1) lies on \( \mathop{\sum }\limits^{{{2n} - 1}} \) . Since by hypothesis \( \left| {x}^{ * }\right| < \delta \) and \( \left| {w}_{j}\right| < \delta \) for all \( j,\partial E \subset U \) . | Yes |
Theorem 4.2.1 (Duality). Each multiplicity space \( {E}^{\lambda } \) is an irreducible \( {\mathcal{R}}^{G} \) module. Furthermore, if \( \lambda ,\mu \in \operatorname{Spec}\left( \rho \right) \) and \( {E}^{\lambda } \cong {E}^{\mu } \) as an \( {\mathcal{R}}^{G} \) module, then \( \lambda = \mu \) . | We first prove that the action of \( {\mathcal{R}}^{G} \) on \( {\operatorname{Hom}}_{G}\left( {{F}^{\lambda }, L}\right) \) is irreducible. Let \( T \in {\operatorname{Hom}}_{G}\left( {{F}^{\lambda }, L}\right) \) be nonzero. Given another nonzero element \( S \in {\operatorname{Hom}}_{G}\left( {{F}^{\lambda }, L}\right) \) we need to find \( r \in {\mathcal{R}}^{G} \) such that \( {rT} = S \) . Let \( X = T{F}^{\lambda } \) and \( Y = S{F}^{\lambda } \) . Then by Schur’s lemma \( X \) and \( Y \) are isomorphic \( G \) -modules of class \( \lambda \) . Thus Lemma 4.2.3 implies that there exists \( u \in {\mathcal{R}}^{G} \) such that \( {\left. u\right| }_{X} \) implements this isomorphism. Thus \( {uT} : {F}^{\lambda } \rightarrow S{F}^{\lambda } \) is a \( G \) -module isomorphism. Schur’s lemma implies that there exists \( c \in \mathbb{C} \) such that \( {cuT} = S \), so we may take \( r = {cu} \) .
We now show that if \( \lambda \neq \mu \) then \( {\operatorname{Hom}}_{G}\left( {{F}^{\lambda }, L}\right) \) and \( {\operatorname{Hom}}_{G}\left( {{F}^{\mu }, L}\right) \) are inequivalent modules for \( {\mathcal{R}}^{G} \) . Suppose
\[
\varphi : {\operatorname{Hom}}_{G}\left( {{F}^{\lambda }, L}\right) \rightarrow {\operatorname{Hom}}_{G}\left( {{F}^{\mu }, L}\right)
\]
is an intertwining operator for the action of \( {\mathcal{R}}^{G} \) . Let \( T \in {\operatorname{Hom}}_{G}\left( {{F}^{\lambda }, L}\right) \) be nonzero and set \( S = \varphi \left( T\right) \) . We want to show that \( S = 0 \) . Set \( U = T{F}^{\lambda } + S{F}^{\mu } \) . Then since we are assuming \( \lambda \neq \mu \), the sum is direct. Let \( p : U \rightarrow S{F}^{\mu } \) be the corresponding projection. Then Lemma 4.2.3 implies that there exists \( r \in {\mathcal{R}}^{G} \) such that \( {\left. r\right| }_{U} = p \) . Since \( {pT} = 0 \), we have \( {rT} = 0 \) . Hence
\[
0 = \varphi \left( {rT}\right) = {r\varphi }\left( T\right) = {rS} = {pS} = S,
\]
which proves that \( \varphi = 0 \) . | Yes |
Theorem 27.2. Let \( G \) be the complexification of the compact connected Lie group \( K \) . With \( B, N, I \) as above, \( \left( {B, N, I}\right) \) is a Tits’ system in \( G \) . | The proof of this is identical to Theorem 27.1. The analog of Lemma 27.2 is true, and the proof is the same except that we use Lemma 27.3 instead of Lemma 27.1. All other details are the same. | No |
Proposition 24. Fix a monomial ordering on \( R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and let \( I \) be a nonzero ideal in \( R \). (1) If \( {g}_{1},\ldots ,{g}_{m} \) are any elements of \( I \) such that \( {LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right) \), then \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) is a Gröbner basis for \( I \). (2) The ideal \( I \) has a Gröbner basis. | Suppose \( {g}_{1},\ldots ,{g}_{m} \in I \) with \( {LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right) \). We need to see that \( {g}_{1},\ldots ,{g}_{m} \) generate the ideal \( I \). If \( f \in I \), use general polynomial division to write \( f = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} + r \) where no nonzero term in the remainder \( r \) is divisible by any \( {LT}\left( {g}_{i}\right) \). Since \( f \in I \), also \( r \in I \), which means \( {LT}\left( r\right) \) is in \( {LT}\left( I\right) \). But then \( {LT}\left( r\right) \) would be divisible by one of \( {LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) \), which is a contradiction unless \( r = 0 \). Hence \( f = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} \) and \( {g}_{1},\ldots ,{g}_{m} \) generate \( I \), so are a Gröbner basis for \( I \), which proves (1).\nFor (2), note that the ideal \( {LT}\left( I\right) \) of leading terms of any ideal \( I \) is a monomial ideal generated by all the leading terms of the polynomials in \( I \). By Exercise 1 a finite number of those leading terms suffice to generate \( {LT}\left( I\right) \), say \( {LT}\left( I\right) = \left( {{LT}\left( {h}_{1}\right) ,\ldots ,{LT}\left( {h}_{k}\right) }\right) \) for some \( {h}_{1},\ldots ,{h}_{k} \in I \). By (1), the polynomials \( {h}_{1},\ldots ,{h}_{k} \) are a Gröbner basis of \( I \), completing the proof. | Yes |
The function \( f\left( z\right) = \frac{1}{1 - z} \) is analytic on \( \mathbb{D} \) but is not in \( {\mathbf{H}}^{2} \) . | Since \( \frac{1}{1 - z} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{z}^{n} \), the coefficients of \( f \) are not square-summable. | Yes |
Show that conditions (1) and (2) in Definition 1.4 may be replaced by the single condition that the map \( \left( {{g}_{1},{g}_{2}}\right) \rightarrow {g}_{1}{g}_{2}^{-1} \) is smooth. | Null | No |
Theorem 2. Let \( \Omega \) be a bounded open set in \( {\mathbb{R}}^{2} \) whose boundary \( \partial \Omega \) consists of a finite number of simple closed curves. Assume the existence of a number \( r > 0 \) such that at each point of \( \partial \Omega \) there are two circles of radius \( r \) tangent to \( \partial \Omega \) at that point, one circle in \( \bar{\Omega } \) and the other in \( {\mathbb{R}}^{2} \smallsetminus \Omega \) . Let \( g \) be a twice continuously differentiable function on \( \Omega \) . Then the Dirichlet problem (4) has a unique solution. | Null | No |
If all colength n ideals were radical, then the Hilbert scheme Hn would be easy to describe, as follows. Every unordered list of n distinct points in ℂ2 corresponds to a set of n! points in (ℂ2)n, or alternatively, to a single point in the nth symmetric product Snℂ2, defined as the quotient (ℂ2)n/Sn by the symmetric group Sn. Of course, not every point of Snℂ2 corresponds to an unordered list of distinct points; for that, one needs to remove the diagonal locus | Null | No |
Theorem 18.2. (a) The value of \( {\Phi }_{0}\left( {d, p}\right) \) equals | Proof. (a) By the definition (2) we have\n\[
{\Phi }_{0}\left( {d, p}\right) = \mathop{\sum }\limits_{{i = 0}}^{n}\left( \begin{matrix} p - d + i - 1 \\ i \end{matrix}\right) + \mathop{\sum }\limits_{{i = 0}}^{m}\left( \begin{matrix} p - d + i - 1 \\ i \end{matrix}\right) .
\]\nThe desired expression then follows easily using Appendix 2, (9). | No |
If \( \deg f < \deg {gh} \) and \( \gcd \left( {g, h}\right) = 1 \), then there exist unique polynomials \( a, b \) such that \( \deg a < \deg g,\deg b < \deg h \), and \( f/\left( {gh}\right) = \) \( \left( {a/g}\right) + \left( {b/h}\right) \) . If \( \gcd \left( {f,{gh}}\right) = 1 \), then \( \gcd \left( {a, g}\right) = \gcd \left( {b, h}\right) = 1 \) . | Proof. Since \( \gcd \left( {g, h}\right) = 1 \), there exist polynomials \( s, t \) such that \( {gs} + {ht} = f \) . Polynomial division yields \( t = {gp} + a, s = {hq} + b \), where \( \deg a < \deg g \) and \( \deg b < \deg h \) . Then \( f = {gh}\left( {p + q}\right) + {ah} + {bg} \), with \( \deg \left( {{ah} + {bg}}\right) < \deg {gh} \) , and \( p + q = 0 \) : otherwise, \( \deg f \geqq \deg {gh} \), contradicting the hypothesis. Hence \( f = {ah} + {bg} \), and \( f/\left( {gh}\right) = \left( {a/g}\right) + \left( {b/h}\right) \) . If \( \gcd \left( {f,{gh}}\right) = 1 \), then a polynomial that divides \( a \) and \( g \), or divides \( b \) and \( h \), also divides \( f = {ah} + {bg} \) and \( {gh} \) ; hence \( \gcd \left( {a, g}\right) = \gcd \left( {b, h}\right) = 1 \) . | Yes |
The hull complex of a lattice module is locally finite. | We claim that the vertex \( \mathbf{0} \in L \) is incident to only finitely many edges of hull \( \left( {M}_{L}\right) \). This claim implies the theorem because (i) the lattice \( L \) acts transitively on the vertices of hull \( \left( {M}_{L}\right) \), so it suffices to consider the vertex \( \mathbf{0} \), and (ii) every face of hull \( \left( {M}_{L}\right) \) containing \( \mathbf{0} \) is uniquely determined by the edges containing \( \mathbf{0} \), so \( \mathbf{0} \in L \) lies in only finitely many faces.
To prove the claim we introduce the following definition. A nonzero vector \( \mathbf{u} = {\mathbf{u}}_{ + } - {\mathbf{u}}_{ - } \) in our lattice \( L \) is called primitive if there is no other vector \( \mathbf{v} \in L \smallsetminus \{ \mathbf{u},\mathbf{0}\} \) such that \( {\mathbf{v}}_{ + } \leq {\mathbf{u}}_{ + } \) and \( {\mathbf{v}}_{ - } \leq {\mathbf{u}}_{ - } \). The primitive vectors in \( L \) can be computed as follows. Fix any sign pattern in \( \{ - 1, + 1{\} }^{n} \) and consider the pointed affine semigroup consisting of all vectors in \( L \) whose nonzero entries are consistent with the chosen sign pattern. A vector in \( L \) is primitive if and only if it lies in the Hilbert basis of the semigroup associated to its sign pattern. Each of these Hilbert bases is finite by Theorem 7.16, and by taking the union over all sign patterns, we conclude that the set of primitive vectors in \( L \) is finite.
We will now prove that for any edge \( \{ \mathbf{0},\mathbf{u}\} \) of the hull complex hull \( \left( {M}_{L}\right) \), the vector \( \mathbf{u} \) is primitive. As the set of primitive vectors is finite, this proves the claim and hence the theorem. Suppose that \( \mathbf{u} \in L \smallsetminus \{ 0\} \) is not primitive, and choose \( \mathbf{v} \in L \smallsetminus \{ \mathbf{u},\mathbf{0}\} \) such that \( {\mathbf{v}}_{ + } \leq {\mathbf{u}}_{ + } \) and \( {\mathbf{v}}_{ - } \leq {\mathbf{u}}_{ - } \). This implies \( {\lambda }^{{v}_{i}} + {\lambda }^{{u}_{i} - {v}_{i}} \leq 1 + {\lambda }^{{u}_{i}} \) for all \( i \in \{ 1,\ldots, n\} \) and \( \lambda \gg 0 \). In other words, for \( \lambda \gg 0 \), the vector \( {\lambda }^{\mathbf{v}} + {\lambda }^{\mathbf{u} - \mathbf{v}} \) is componentwise smaller than or equal to the vector \( {\lambda }^{\mathbf{0}} + {\lambda }^{\mathbf{u}} \). This contradicts the fact that \( \{ \mathbf{0},\mathbf{u}\} \) is an edge of the hull complex hull \( \left( {M}_{L}\right) \), because the latter implies that \( {\lambda }^{\mathbf{v}} + {\lambda }^{\mathbf{u} - \mathbf{v}} \) is componentwise strictly smaller than \( {\lambda }^{\mathbf{0}} + {\lambda }^{\mathbf{u}} \) for \( \lambda \gg 0 \). This contradiction proves that \( \mathbf{u} \) is primitive, and hence the claim and the theorem. | Yes |
Determine the minimizers and the minimum value of the function f(x1,...,xn) = 1/2∑j=1nxj^2 - ∑1≤i<j≤nln|x_i - x_j|. | The solution to this differential equation is the Hermite polynomial of order n, Hn(x) = n!∑0≤k≤[n/2] ((-1)^k(2x)^(n-2k))/(k!(n-2k)!). Therefore, the solutions xj are the roots of the Hermite polynomial Hn(x). The discriminant of Hn is given by ∏i<j(xi - xj)^2 = 2^(-(n(n-1)/2)) ∏j=1n j^j. The above formula for Hn gives ∑j=1nxj^2 = n(n-1)/2. Thus, the minimum value of f is (1/4)n(n-1)(1 + ln 2) - 1/2 ∑j=1n j ln j. | Yes |
Theorem 3.1.9 (Bertrand’s postulate) For \( n \) sufficiently large, there is a prime between \( n \) and \( {2n} \) . | Proof: (S. Ramanujan) Observe that if \({a}_{0} \geq {a}_{1} \geq {a}_{2} \geq \cdots\) is a decreasing sequence of real numbers tending to zero, then \({a}_{0} - {a}_{1} \leq \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}{a}_{n} \leq {a}_{0} - {a}_{1} + {a}_{2}\) This is the starting point of Ramanujan's proof. We can write \(T\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\log n = \mathop{\sum }\limits_{{{de} \leq x}}\Lambda \left( d\right) = \mathop{\sum }\limits_{{e \leq x}}\psi \left( \frac{x}{e}\right) .\) We know that \(T\left( x\right) = x\log x - x + O\left( {\log x}\right) \) by Exercise 2.1.2. On the other hand, \(T\left( x\right) - {2T}\left( \frac{x}{2}\right) = \mathop{\sum }\limits_{{n \leq x}}{\left( -1\right) }^{n - 1}\psi \left( \frac{x}{n}\right) \leq \psi \left( x\right) - \psi \left( \frac{x}{2}\right) + \psi \left( \frac{x}{3}\right)\) by the observation above. Hence \(\psi \left( x\right) - \psi \left( \frac{x}{2}\right) + \psi \left( \frac{x}{3}\right) \geq \left( {\log 2}\right) x + O\left( {\log x}\right) .\) On the other hand, \(\psi \left( x\right) - \psi \left( \frac{x}{2}\right) \leq \left( {\log 2}\right) x + O\left( {\log x}\right)\) from which we deduce inductively \(\psi \left( x\right) \leq 2\left( {\log 2}\right) x + O\left( {{\log }^{2}x}\right) .\) Thus, \(\psi \left( x\right) - \psi \left( \frac{x}{2}\right) \geq \frac{1}{3}\left( {\log 2}\right) x + O\left( {{\log }^{2}x}\right) \) . Now, \(\psi \left( x\right) = \theta \left( x\right) + O\left( {\sqrt{x}{\log }^{2}x}\right) \) . Hence \(\theta \left( x\right) - \theta \left( \frac{x}{2}\right) \geq \frac{1}{3}\left( {\log 2}\right) x + O\left( {\sqrt{x}{\log }^{2}x}\right) .\) Therefore, for \( x \) sufficiently large, there is a prime between \( x/2 \) and \( x \) . | No |
Let \( L/K \) be a finite extension of algebraic number fields. Suppose that \( {\mathcal{O}}_{L} = {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \) for some \( \alpha \in L \) . If \( f\left( x\right) \) is the minimal polynomial of \( \alpha \) over \( {\mathcal{O}}_{K} \), show that \( {\mathcal{D}}_{L/K} = \left( {{f}^{\prime }\left( \alpha \right) }\right) \). | This result is identical to Exercises 5.6.6 and 5.6.7. More generally, one can show the following. For each \( \theta \in {\mathcal{O}}_{L} \) which generates \( L \) over \( K \), let \( f\left( x\right) \) be its minimal polynomial over \( {\mathcal{O}}_{K} \) . Define \( {\delta }_{L/K}\left( \theta \right) = {f}^{\prime }\left( \theta \right) \) . Then \( {\mathcalak D}_{L/K} \) is the ideal generated by the elements \( {\delta }_{L/K}\left( \theta \right) \) as \( \theta \) ranges over such elements. We refer the interested reader to \( \left\lbrack \mathrm{N}\right\rbrack \) . | No |
Corollary 3.111. The closed unit ball of the dual \( {X}^{ * } \) of a WCG space is sequentially compact for the weak* topology in the sense that every sequence of \( {B}_{{X}^{ * }} \) has a weak* convergent subsequence. | Given a bounded sequence \( \left( {x}_{n}^{ * }\right) \) of \( {X}^{ * } \), let \( F\left( n\right) \mathrel{\text{:=}} \left\{ {{x}_{p} : p \geq n}\right\} \) for \( n \in \mathbb{N} \) and let \( {x}^{ * } \) be a weak* cluster point of \( \left( {x}_{n}^{ * }\right) \), i.e., a point in \( {\operatorname{cl}}^{ * }\left( {F\left( n\right) }\right) \) for all \( n \in \mathbb{N} \) . Theorem 3.109 yields a sequence \( \left( {y}_{n}^{ * }\right) \overset{ * }{ \rightarrow }{x}^{ * } \) such that \( {y}_{n}^{ * } \in F\left( n\right) \) for all \( n \). It is the required subsequence of \( \left( {x}_{n}^{ * }\right) \) . | Yes |
Let \( R \) be a ring with derivation \( D \) and \( S = R\left\lbrack t\right\rbrack ,{tr} = {rt} + D\left( r\right) \) for all \( r \) in \( R \), the corresponding Ore extension. Several authors, among them [4] and [6], have noted the inequality \(\text{l.gl.dim}S \leq \text{l.gl.dim}R + 1\). | In this chapter we show that if \( R \) is left and right noetherian and of finite left global dimension, a necessary condition for equality to hold is the existence of a left \( S \) -module \( M \) that is finitely generated as an \( R \) -module and with \( {\operatorname{ldim}}_{R}M = \operatorname{l.gl.dim}R \) . Combining the result of this chapter with Corollary 1.7(b) of [9] shows that the given condition is sufficient as well as necessary. | No |
Theorem 4.3. Let \( K \) be an imaginary abelian extension of \( \mathbf{Q} \) . Then the norm map | We have to use class field theory, which gives the more general statement:\n\nLemma. Let \( K \) be an abelian extension of a number field \( F \) . Let \( H \) be the Hilbert class field of \( F \) (maximal abelian unramified extension of \( F \) ). If \( K \cap H = F \) then the norm map \( {N}_{K/F} : {C}_{K} \rightarrow {C}_{F} \) is surjective.\n\nProof. For any ideal class \( c \) in \( K \), the properties of the Artin symbol show that\n\n\[
\left( {c,{KH}/K}\right) \text{restricted to}H = \left( {{N}_{K/F}c, H/F}\right) \text{.}
\]\n\nWe have natural isomorphisms of Galois groups: \n\nHence the group \( \left( {{N}_{K/F}{C}_{K}, H/F}\right) \) is the whole Galois group \( \operatorname{Gal}\left( {H/F}\right) \), whence \( {N}_{K/F}{C}_{K} = {C}_{F} \) since the Artin symbol gives an isomorphism of the ideal class group with the Galois group. This proves the lemma.\n\nThe theorem follows at once, because \( K \) over \( {K}^{ + } \) is ramified at the archimedean primes, and hence cannot intersect the Hilbert class field of \( F \) except in \( F \). | Yes |
Show that two metrics \( d \) and \( \rho \) on a set \( X \) are equivalent if and only if for every sequence \( \left( {x}_{n}\right) \) in \( X \) and every \( x \in X \) , \( d\left( {{x}_{n}, x}\right) \rightarrow 0 \Leftrightarrow \rho \left( {{x}_{n}, x}\right) \rightarrow 0 \). | Null | No |
Theorem 19.1. Let \( \gamma \) be an oriented simple closed curve in \( {\mathbb{C}}^{2} \) with a finite number of self-intersections. Then a necessary and sufficient condition that there exists a bounded analytic variety \( \sum \) in \( {\mathbb{C}}^{2} \) with \( {b\sum } = \pm \gamma \) is that \( \gamma \) satisfies (9) (moment condition). | The complete proof of this theorem involves a considerable number of technical details, and we shall refer the reader to the paper of Harvey and Lawson [HarL2] for these. Here we shall present a sketch that we hope conveys the essential aspects of the construction. | No |
Proposition 17.9. Let \( \left( {X,\mathbf{S},\mu }\right) \) be a measure space, let \( \mathcal{E} \) be a normed space, and let \( p \) be a positive real number. Then the collection \( {\mathcal{L}}_{p}\left( {X;\mathcal{E}}\right) = {\mathcal{L}}_{p}\left( {X,\mathbf{S},\mu ;\mathcal{E}}\right) \) of all those measurable \( \mathcal{E} \) -valued mappings \( \Phi \) on \( X \) such that the function \( {N}_{\Phi } \) defined in (6) satisfies the condition | If \( \Phi \) and \( \Psi \) are arbitrary \( \mathcal{E} \) -valued mappings, then \( {N}_{\Phi + \Psi } \leq {N}_{\Phi } + {N}_{\Psi } \) by the triangle inequality in \( \mathcal{E} \), and since \( {\mathcal{L}}_{p}\left( X\right) \) is a linear space, it follows at once that \( {\mathcal{L}}_{p}\left( {X;\mathcal{E}}\right) \) is also a linear space. Furthermore, to show that \( \parallel {\parallel }_{p} \) is a pseudonorm for \( p \geq 1 \), it suffices to verify the triangle inequality, and this follows at once from the triangle inequalities in \( \mathcal{E} \) and \( {\mathcal{L}}_{p}\left( X\right) \) : | No |
Theorem 2.16. Let \( V \subset {\mathbb{C}}^{n} \) be any nonempty irreducible variety of dimension \( d \), and let \( {V}_{{d}_{1}} \supsetneq \ldots \supsetneq {V}_{{d}_{2}} \) be any strict chain of nonempty irreducible subvarieties of \( V \) . This chain can be extended (or refined) to a maximal chain of irreducible varieties | Proof of Theorem 2.16. It suffices to show that if \( {W}_{1} \subset {W}_{2} \) are irreducible nonempty subvarieties of \( V \) of dimension \( {d}_{1} \) and \( {d}_{2} \) respectively, then there is a strict chain of irreducible varieties from \( {W}_{2} \) to \( {W}_{1} \) of length \( {d}_{2} - {d}_{1} \) ; or what is the same, that there is a strict chain of prime ideals of length \( {d}_{2} - {d}_{1} \) in the coordinate ring \( {R}_{{W}_{2}} = \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), starting from (0) and ending in \( \mathfrak{p} \), where \( {R}_{{W}_{2}}/\mathfrak{p} = {R}_{{W}_{1}} = \mathbb{C}\left\lbrack {{y}_{1},\ldots ,{y}_{n}}\right\rbrack \). (This will ensure maximality, since for any irreducible variety \( V \), any strict chain of irreducible varieties of length \( d = \dim V \) starting with \( V \) and ending in a point must be maximal; otherwise, from Theorem 2.15 dim \( V \) would be greater than \( d \).) Now the transcendence degree of \( {R}_{{W}_{2}} \) is \( {d}_{2} \), and that of \( {R}_{{W}_{1}} \) is \( {d}_{1} \); we assume without loss of generality that \( {d}_{2} > {d}_{1} \), and that \( \left\{ {{x}_{1},\ldots ,{x}_{{d}_{2}}}\right\} \) and \( \left\{ {{y}_{1},\ldots ,{y}_{{d}_{1}}}\right\} \) are transcendence bases of \( {R}_{{W}_{2}} \) and \( {R}_{{W}_{1}} \) respectively. We may also assume that the elements \( {x}_{i} \) and \( {y}_{j} \) have been numbered so that the image ring of the homomorphism | Yes |
Theorem 2.2.15. If \( D \) is a fundamental discriminant, the Kronecker symbol \( \left( \frac{D}{n}\right) \) defines a real primitive character modulo \( m = \left| D\right| \) . Conversely, if \( \chi \) is a real primitive character modulo \( m \) then \( D = \chi \left( {-1}\right) m \) is a fundamental discriminant \( D \) and \( \chi \left( n\right) = \left( \frac{D}{n}\right) \) . | The definition of the Kronecker symbol and Theorem 2.2.9 show that \( \left( \frac{D}{n}\right) \) is a character modulo \( \left| D\right| \) . To show that it is primitive, it is sufficient to show that for any prime \( p \mid D \) it cannot be defined modulo \( D/p \) . Assume first that \( p \neq 2 \), and let \( a \) be a quadratic nonresidue modulo \( p \) . Since \( D \) is fundamental and \( p \) is odd we have \( \gcd \left( {p,4\left| D\right| /p}\right) = 1 \) ; hence by the Chinese remainder theorem there exists \( n > 0 \) such that \( n \equiv a\left( {\;\operatorname{mod}\;p}\right) \) and \( n \equiv 1 \) \( \left( {{\;\operatorname{mod}\;4}\left| D\right| /p}\right) \), and in particular \( n \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Thus by Theorem 2.2.9 and the quadratic reciprocity law for positive odd numbers we have
\[
\left( \frac{D}{n}\right) = \left( \frac{p}{n}\right) \left( \frac{D/p}{n}\right) = \left( \frac{p}{n}\right) \left( \frac{{4D}/p}{n}\right) = \left( \frac{p}{n}\right) = \left( \frac{n}{p}\right) = - 1,
\]
proving that \( \left( \frac{D}{n}\right) \) cannot be defined modulo \( D/p \) . Assume now that \( p = 2 \) , so that \( D \equiv 8 \) or 12 modulo 16, and choose \( n = 1 + \left| D\right| /2 \) . If \( D \equiv 8\left( {\;\operatorname{mod}\;{16}}\right) \) we have \( n \equiv 5\left( {\;\operatorname{mod}\;8}\right) \) and \( n \equiv 1\left( {{\;\operatorname{mod}\;\left| D\right| }/2}\right) \) and so
\[
\left( \frac{D}{n}\right) = \left( \frac{2}{n}\right) \left( \frac{D/2}{n}\right) = \left( \frac{2}{n}\right) = - 1
\]
since \( D/2 \equiv 0\left( {\;\operatorname{mod}\;4}\right) \) . If \( D \equiv {12}\left( {\;\operatorname{mod}\;{16}}\right) \) we have \( n \equiv 7\left( {\;\operatorname{mod}\;8}\right) \) and \( n \equiv 1\left( {{\;\operatorname{mod}\;D}/4}\right) \) hence
\[
\left( \frac{D}{n}\right) = \left( \frac{-4}{n}\right) \left( \frac{-D/4}{n}\right) = \left( \frac{-4}{n}\right) = - 1
\]
since \( - D/4 \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), proving in both cases that \( \left( \frac{D}{n}\right) \) cannot be defined modulo \( D/2 \) hence that it is a primitive character.
Conversely, let \( \chi \) be a real primitive character modulo \( m \) and let \( p \) be any odd prime such that \( p \nmid m \) . By Lemma 2.2.2 we have \( \chi \left( p\right) = \left( \frac{D}{p}\right) \) with \( D = \chi \left( {-1}\right) m \) . Since both sides are multiplicative in \( p \), we deduce that for any odd positive \( n \) we have \( \chi \left( n\right) = \left( \frac{D}{n}\right) \) . In addition, by definition of the Kronecker symbol we have \( \left( \frac{D}{-1}\right) = \operatorname{sign}\left( D\right) = \chi \left( {-1}\right) \) ; hence the equality \( \chi \left( n\right) = \left( \frac{D}{n}\right) \) is valid for any odd \( n \in \mathbb{Z} \) .
I now claim that \( D \equiv 0 \) or \( 1{\;\operatorname{mod}\;4} \) . Indeed, since \( \chi \) is periodic of period \( m = \left| D\right| \), by what we have just proved and the properties of the Kronecker symbol, we have
\[
1 = \chi \left( {1 + {2D}}\right) = \left( \frac{D}{1 + {2D}}\right) .
\]
Thus, if we had \( D \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) we would have
\[
1 = \left( \frac{-1}{1 + {2D}}\right) \left( \frac{-D}{1 + {2D}}\right) = {\left( -1\right) }^{D}\left( \frac{-D}{1}\right) = - 1,
\]
a contradiction, and if we had \( D \equiv 2\left( {\;\operatorname{mod}\;4}\right) \) we would have
\[
1 = \left( \frac{2}{1 + {2D}}\right) \left( \frac{2D}{1 + {2D}}\right) = \left( \frac{2}{1 + {2D}}\right) = \left( \frac{2}{5}\right) = - 1,
\]
also a contradiction.
We must now prove that \( \chi \left( 2\right) = \left( \frac{D}{2}\right) \) . We may of course assume \( D \) (or \( m) \) odd, otherwise both sides vanish. Thus \( \left( {D + 1}\right) /2 \) is odd; hence
\[
1 = \chi \left( {D + 1}\right) = \chi \left( 2\right) \chi \left( {\left( {D + 1}\right) /2}\right)
\]
\[
= \chi \left( 2\right) \left( \frac{D}{\left( {D + 1}\right) /2}\right) = \chi \left( 2\right) \left( \frac{D}{D + 1}\right) \left( \frac{D}{2}\right) = \chi \left( 2\right) \left( \frac{D}{2}\right) ,
\]
showing that \( \chi \left( 2\right) = \left( \frac{D}{2}\right) \) . By multiplicativity it follows that \( \chi \left( n\right) = \left( \frac{D}{n}\right) \) for all \( n \) .
Finally, since \( D \equiv 0 \) or 1 modulo 4, we can write (uniquely) \( D = {D}_{0}{f}^{2} \) , where \( {D}_{0} \) is a fundamental discriminant. It is clear that the character \( \left( \frac{{D}_{0}}{n}\right) \) takes the same values as the character \( \left( \frac{D}{n}\right) \) on integers \( n \) coprime to \( D \) ; hence \( \left( \frac{D}{n}\right) \) is primitive if and only if \( D = {D}_{0} \), i.e., \( D \) is a fundamental discriminant, finishing the proof of the theorem. | Yes |
The edges of \( {K}_{10} \) cannot be partitioned into three copies of the Petersen graph. | Let \( P \) and \( Q \) be two copies of Petersen’s graph on the same vertex set and with no edges in common. Let \( R \) be the subgraph of \( {K}_{10} \) formed by the edges not in \( P \) or \( Q \) . We show that \( R \) is bipartite.
Let \( {U}_{P} \) be the eigenspace of \( A\left( P\right) \) with eigenvalue 1, and let \( {U}_{Q} \) be the corresponding eigenspace for \( A\left( Q\right) \) . Then \( {U}_{P} \) and \( {U}_{Q} \) are 5-dimensional subspaces of \( {\mathbb{R}}^{10} \) . Since both subspaces lie in \( {\mathbf{1}}^{ \bot } \), they must have a nonzero vector \( u \) in common. Then
\[
A\left( R\right) u = \left( {J - I - A\left( P\right) - A\left( Q\right) }\right) u = \left( {J - I}\right) u - {2u} = - {3u},
\]
and so -3 is an eigenvalue of \( A\left( R\right) \) . Since \( R \) is cubic, it follows from Theorem 8.8.2 that it must be bipartite. | Yes |
Proposition 1. Aut \( \left( K\right) \) is a group under composition and \( \operatorname{Aut}\left( {K/F}\right) \) is a subgroup. | It is clear that \( \operatorname{Aut}\left( K\right) \) is a group. If \( \sigma \) and \( \tau \) are automorphisms of \( K \) which fix \( F \) then also \( {\sigma \tau } \) and \( {\sigma }^{-1} \) are the identity on \( F \), which shows that \( \operatorname{Aut}\left( {K/F}\right) \) is a subgroup. | Yes |
If \( 1 \leq p < \infty \) and if \( f\left( x\right) \in {L}^{p} \), then \( {\begin{Vmatrix}{P}_{y} * f - f\end{Vmatrix}}_{p} \rightarrow 0\;\left( {y \rightarrow 0}\right) \). | Let \( f \in {L}^{p},1 \leq p \leq \infty \) . When \( p = \infty \) we suppose in addition that \( f \) is uniformly continuous. Then
\[
{P}_{y} * f\left( x\right) - f\left( x\right) = \int {P}_{y}\left( t\right) \left( {f\left( {x - t}\right) - f\left( x\right) }\right) {dt}.
\]
Minkowski's inequality gives
\[
{\begin{Vmatrix}{P}_{y} * f - f\end{Vmatrix}}_{p} \leq \int {P}_{y}\left( t\right) {\begin{Vmatrix}{f}_{t} - f\end{Vmatrix}}_{p}{dt},
\]
when \( p < \infty \), because \( {P}_{y} \geq 0 \) . The same inequality is trivial when \( p = \infty \) . For \( \delta > 0 \), we now have
\[
{\begin{Vmatrix}{P}_{y} * f - f\end{Vmatrix}}_{p} \leq {\int }_{\left| t\right| \leq \delta }{P}_{y}\left( t\right) {\begin{Vmatrix}{f}_{t} - f\end{Vmatrix}}_{p}{dt} + {\int }_{\left| t\right| > \delta }{P}_{y}\left( t\right) {\begin{Vmatrix}{f}_{t} - f\end{Vmatrix}}_{p}{dt}.
\]
Since \( \int {P}_{y}\left( t\right) {dt} = 1 \), continuity of translations shows that \( {\int }_{\left| t\right| < \delta } \) is small provided \( \delta \) is small. With \( \delta \) fixed,
\[
{\int }_{\left| t\right| > \delta } \leq 2\parallel f{\parallel }_{p}{\int }_{\left| t\right| > \delta }{P}_{y}\left( t\right) {dt} \rightarrow 0\;\left( {y \rightarrow 0}\right)
\]
by property (vi) of the Poisson kernel. That proves (a) and (d). By Fubini's theorem, parts (b) and (c) follow from (a) and (d), respectively. | No |
Proposition 31.1. The group \( \widetilde{T} \) is connected and is a maximal torus of \( \widetilde{G} \) . | Let \( \Pi \subset \widetilde{G} \) be the kernel of \( p \) . The connected component \( {\widetilde{T}}^{ \circ } \) of the identity in \( \widetilde{T} \) is a torus of the same dimension as \( T \), so it is a maximal torus in \( \widetilde{G} \) . Its image in \( G \) is isomorphic to \( {\widetilde{T}}^{ \circ }/\left( {{\widetilde{T}}^{ \circ } \cap \Pi }\right) \cong {\widetilde{T}}^{ \circ }\Pi /\Pi \) . This is a torus of \( G \) contained in \( T \), and of the same dimension as \( T \), so it is all of \( T \) . Thus, the composition \( {\widetilde{T}}^{ \circ } \rightarrow \widetilde{T}\overset{p}{ \rightarrow }T \) is surjective. We see that \( \widetilde{T}/\Pi \cong T \cong {\widetilde{T}}^{ \circ }\Pi /\Pi \) canonically and therefore \( \widetilde{T} = {\widetilde{T}}^{ \circ }\Pi \).
We may identify \( \Pi \) with the fundamental group \( {\pi }_{1}\left( G\right) \) by Theorem 13.2. It is a discrete normal subgroup of \( \widetilde{G} \) and hence central in \( \widetilde{G} \) by Proposition 23.1.
Thus it is contained in every maximal torus by Proposition 18.14, particularly in \( {\widetilde{T}}^{ \circ } \) . Thus \( {\widetilde{T}}^{ \circ } = {\widetilde{T}}^{ \circ }\Pi = \widetilde{T} \) and so \( \widetilde{T} \) is connected and a maximal torus. | Yes |
Let \( \mathbf{H} = {L}^{2}\left( \left\lbrack {0,1}\right\rbrack \right) \) and let \( A \) be the operator on \( \mathbf{H} \) defined \( {by} \)
\[
\left( {A\psi }\right) \left( x\right) = {x\psi }\left( x\right)
\]
Then this operator is bounded and self-adjoint, and its spectrum is given by
\[
\sigma \left( A\right) = \left\lbrack {0,1}\right\rbrack
\] | Proof. It is apparent that \( \parallel {A\psi }\parallel \leq \parallel \psi \parallel \) and that \( \langle \phi ,{A\psi }\rangle = \langle {A\phi },\psi \rangle \) for all \( \phi ,\psi \in \mathbf{H} \), so that \( A \) is bounded and self-adjoint. Given \( \lambda \in \left( {0,1}\right) \), consider the functions \( {\psi }_{n} \mathrel{\text{:=}} {1}_{\left\lbrack \lambda \lambda + 1/n\right\rbrack } \), which satisfy \( {\begin{Vmatrix}{\psi }_{n}\end{Vmatrix}}^{2} = 1/n \) . On the other hand, since \( \left| {x - \lambda }\right| \leq 1/n \) on \( \left\lbrack {\lambda ,\lambda + 1/n}\right\rbrack \), we have
\[
{\begin{Vmatrix}\left( A - \lambda I\right) {\psi }_{n}\end{Vmatrix}}^{2} \leq 1/{n}^{3}
\]
Thus, by Proposition 7.7, \( \lambda \) belongs to the spectrum of \( A \) . Since this holds for all \( \lambda \in \left( {0,1}\right) \) and the spectrum of \( A \) is closed, \( \sigma \left( A\right) \supset \left\lbrack {0,1}\right\rbrack \).
Meanwhile, if \( \lambda \notin \left\lbrack {0,1}\right\rbrack \), then the function \( 1/\left( {x - \lambda }\right) \) is bounded on \( \left\lbrack {0,1}\right\rbrack \), and so \( A - {\lambda I} \) has a bounded inverse, consisting of multiplication by \( 1/\left( {x - \lambda }\right) \) . Thus, \( \sigma \left( A\right) = \left\lbrack {0,1}\right\rbrack \) . ∎ | Yes |
Corollary 2.28 A graph \( G \) is edge reconstructible if either \( m > \frac{1}{2}\left( \begin{array}{l} n \\ 2 \end{array}\right) \) or \( {2}^{m - 1} > n! \) | Null | No |
If the \( ON \) system \( \left\{ {\varphi }_{j}\right\} _{j = 1}^{\infty } \) is complete in \( V \), then \( \langle u, v\rangle = \mathop{\sum }\limits_{{j = 1}}^{\infty }\left\langle {u,{\varphi }_{j}}\right\rangle \overline{\left\langle v,{\varphi }_{j}\right\rangle } \) for all \( u, v \in V \) . | Let \( {P}_{n}\left( u\right) \) be the projection of \( u \) on to the subspace spanned by the \( n \) first \( \varphi \) ’s:\n\n\( {P}_{n}\left( u\right) = \mathop{\sum }\limits_{{j = 1}}^{n}\left\langle {u,{\varphi }_{j}}\right\rangle {\varphi }_{j} \)\n\nBy Theorem 5.2 we have\n\n\( \left\langle {{P}_{n}\left( u\right) ,{P}_{n}\left( v\right) }\right\rangle = \mathop{\sum }\limits_{{j = 1}}^{n}\left\langle {u,{\varphi }_{j}}\right\rangle \overline{\left\langle v,{\varphi }_{j}\right\rangle } \).\n\nUsing the triangle and Cauchy-Schwarz inequalities, we get\n\n\( \left| {\langle u, v\rangle - \left\langle {{P}_{n}\left( u\right) ,{P}_{n}\left( v\right) }\right\rangle }\right| \)\n\n\( = \left| {\langle u, v\rangle -\langle u,{P}_{n}\left( v\right) \rangle +\langle u,{P}_{n}\left( v\right) \rangle -\langle {P}_{n}\left( u\right) ,{P}_{n}\left( v\right) \rangle }\right| \)\n\n\( \leq \left| \left\langle {u, v - {P}_{n}\left( v\right) }\right\rangle \right| + \left| \left\langle {u - {P}_{n}\left( u\right) ,{P}_{n}\left( v\right) }\right\rangle \right| \)\n\n\( \leq \begin{Vmatrix}u\end{Vmatrix}\begin{Vmatrix}{v - {P}_{n}\left( v\right) }\end{Vmatrix} + \begin{Vmatrix}{u - {P}_{n}\left( u\right) }\end{Vmatrix}\begin{Vmatrix}{{P}_{n}\left( v\right) }\end{Vmatrix} \)\n\n\( \leq \parallel u\parallel \begin{Vmatrix}{v - {P}_{n}\left( v\right) }\end{Vmatrix} + \begin{Vmatrix}{u - {P}_{n}\left( u\right) }\end{Vmatrix}\parallel v\parallel . \)\n\nIn the last part we also used Bessel's inequality. Now we know, because of completeness, that \( \begin{Vmatrix}{v - {P}_{n}\left( v\right) }\end{Vmatrix} \rightarrow 0 \) as \( n \rightarrow \infty \), and similarly for \( u \), which implies that the final member of the estimate tends to zero. Then also the first member must tend to zero, and so\n\n\( \langle u, v\rangle = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left\langle {{P}_{n}\left( u\right) ,{P}_{n}\left( v\right) }\right\rangle = \mathop{\sum }\limits_{{j = 1}}^{\infty }\left\langle {u,{\varphi }_{j}}\right\rangle \overline{\left\langle v,{\varphi }_{j}\right\rangle } \),\n\nand the proof is complete. | Yes |
The coordinate frame \( \left( {\partial /\partial {x}^{i}}\right) \) is a global orthonormal frame for \( {\mathbb{R}}^{n} \) with the Euclidean metric. | Null | No |
Proposition 8.11 Let \( S \) be multigraded by a torsion-free abelian group \( A \) . All associated primes of multigraded S-modules are multigraded. | This is [Eis95, Exercise 3.5]. The proof, based on that of the corresponding \( \mathbb{Z} \) -graded statement in [Eis95, Section 3.5], is essentially presented in the aforementioned exercise from [Eis95]. It works because torsion-free grading groups \( A \cong {\mathbb{Z}}^{d} \) can be totally ordered, for instance lexicographically (so \( \mathbf{a} < \mathbf{b} \) when the earliest nonzero coordinate of \( \mathbf{a} - \mathbf{b} \) is negative). The proof fails for torsion groups because they admit no total orderings compatible with the group operation. However, Proposition 8.11 continues to hold for infinitely generated modules, because every associated prime is associated to a finitely generated submodule by definition. | No |
Proposition 4.39. Suppose \( X \) and \( Y \) are Hausdorff locally convex spaces, and suppose \( X \) is infrabarreled. Suppose \( T : X \rightarrow Y \) is a linear transformation for which \( f \circ T \in {X}^{ * } \) whenever \( f \in {Y}^{ * } \) Then \( T \) is continuous. | First of all, if \( f \in {Y}^{ * } \), then \( x \in \{ f{\} }_{ \circ } \Leftrightarrow \left| {f\left( x\right) }\right| \leq 1 \), so
\[
{T}^{-1}\left( {\{ f{\} }_{ \circ }}\right) = {T}^{-1}\left( {{f}^{-1}\left( {\{ z : \left| z\right| \leq 1\} }\right) }\right)
\]
\[
= {\left( f \circ T\right) }^{-1}\left( {\{ z : \left| z\right| \leq 1\} }\right)
\]
is a neighborhood of 0 in \( X \), and
\[
{T}^{-1}\left( {\left\{ {f}_{1},\ldots ,{f}_{n}\right\} }_{ \circ }\right) = \mathop{\bigcap }\limits_{{j = 1}}^{n}{\left( {f}_{j} \circ T\right) }^{-1}\left( {\{ z : \left| z\right| \leq 1\} }\right)
\]
is a neighborhood of 0 in \( X \) . Letting \( {Y}_{w} \) denote \( Y \) with the weak topology, \( T \in \) \( {\mathcal{L}}_{c}\left( {X,{Y}_{w}}\right) \) by Proposition 1.26(a).
Let \( B \) be a barrel neighborhood of 0 in \( Y \) ; then \( {T}^{-1}\left( B\right) \) is convex, balanced, and absorbent. [As before, \( {T}^{-1}\left( B\right) \) absorbs \( x \) because \( B \) absorbs \( T\left( x\right) \) .] But \( {T}^{-1}\left( B\right) \) is also closed, since \( B \) is weakly closed. Now suppose \( A \) is bounded in \( X \) . Then every \( f \circ T \) is bounded on \( A \) when \( f \in {Y}^{ * } \), so every \( f \in {Y}^{ * } \) is bounded on \( T\left( A\right) \) . Hence \( T\left( A\right) \) is bounded in \( Y \) by Corollary 3.31. Hence \( T\left( A\right) \subset {cB} \) for some \( c \), so \( A \subset {T}^{-1}\left( {T\left( A\right) }\right) \subset {T}^{-1}\left( {cB}\right) = c{T}^{-1}\left( B\right) \) . That is, \( {T}^{-1}\left( B\right) \) absorbs \( A \) . Since \( A \) was arbitrary, \( {T}^{-1}\left( B\right) \) is a neighborhood of 0 since \( X \) is infrabarreled. Since \( B \) was an arbitrary barrel neighborhood of 0 in \( Y, T \) is continuous by Proposition 1.26(a). | Yes |
Proposition 10.4. There are canonical isomorphisms \( {\Lambda }_{k}\left( {V}^{ * }\right) \cong {\Lambda }_{k}{\left( V\right) }^{ * } \cong {A}_{k}\left( V\right) \) . | The second isomorphism is the one induced from Exercise 26. For the first one, there is a unique bilinear map \( b : {\Lambda }_{k}\left( {V}^{ * }\right) \times {\Lambda }_{k}\left( V\right) \rightarrow \mathbb{R} \) which is given on decomposable elements by\\
\[
b\left( {{v}_{1}^{ * } \land \cdots \land {v}_{k}^{ * },{v}_{1} \land \cdots \land {v}_{k}}\right) = \det \left( {{v}_{i}^{ * }{v}_{j}}\right) .
\]
It determines a nonsingular pairing, and therefore an isomorphism \( {\Lambda }_{k}\left( {V}^{ * }\right) \cong {\Lambda }_{k}{\left( V\right) }^{ * } \) .\\
Observe that under the identification \( {\Lambda }_{k}\left( {V}^{ * }\right) \cong {A}_{k}\left( V\right) \) ,\\
\[
\left( {{v}_{1}^{ * } \land \cdots \land {v}_{k}^{ * }}\right) \left( {{v}_{1},\ldots ,{v}_{k}}\right) = \det \left( {{v}_{i}^{ * }{v}_{j}}\right) .
\] | No |
If \( f \) is in \( {BV}\left( {\mathbf{T}}^{1}\right) \), then \(\left| {\widehat{f}\left( m\right) }\right| \leq \frac{\operatorname{Var}\left( f\right) }{{2\pi }\left| m\right| }\) whenever \( m \neq 0 \) . | Integration by parts gives \(\widehat{f}\left( m\right) = {\int }_{{\mathbf{T}}^{1}}f\left( x\right) {e}^{-{2\pi imx}}{dx} = {\int }_{{\mathbf{T}}^{1}}\frac{{e}^{-{2\pi imx}}}{-{2\pi im}}{df},\) where the boundary terms vanish because of periodicity. The conclusion follows from the fact that the norm of the measure \( {df} \) is the total variation of \( f \). | Yes |
Show that the theory \( \mathcal{D} \) (Exercise 3.21) of dense linear order admits \( \Pi \) -reduction of quantifiers with \( \Pi = \left\{ {\left( {{x}_{i} = {x}_{j}}\right) ,\left( {{x}_{i} < {x}_{j}}\right) }\right. \) \( i, j \in \mathbf{N}\} \) . Hence show that \( \mathcal{D} \) is decidable and complete. | Null | No |
Suppose that the \( k \) -algebra \( \mathcal{O} \) is a complete discrete \( k \) -valuation ring with residue class map \( \eta : \mathcal{O} \rightarrow F \) . Assume further that \( F \) is a finite separable extension of \( k \) . Given any local parameter \( t \), there is a unique isometric isomorphism \( \widehat{\mu } : F\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \simeq \mathcal{O} \) such that \( \widehat{\mu }\left( X\right) = t \) . | Let \( \eta : \mathcal{O} \rightarrow F \) be the residue class map, and let \( \mu : F \rightarrow \mathcal{O} \) be the unique splitting given by (1.2.12). Define \( \widehat{\mu } : F\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \rightarrow \mathcal{O} \) via
\[
\mu \left( {\mathop{\sum }\limits_{i}{a}_{i}{X}^{i}}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{i}\mu \left( {a}_{i}\right) {t}^{i}
\]
This map is clearly well-defined and injective, and is uniquely determined by \( \mu \) and \( t \) . To show that it is surjective, put \( {F}^{\prime } \mathrel{\text{:=}} \operatorname{im}\left( \mu \right) \) . Then \( \mathcal{O} = {F}^{\prime } + P \), and \( {F}^{\prime } \cap P = 0 \) . Thus, for any \( x \in \mathcal{O} \) there exists a unique element \( {a}_{0} \in {F}^{\prime } \) with \( x \equiv {a}_{0} \) \( {\;\operatorname{mod}\;P} \) . Choose a local parameter \( t \in P \) . Then there exists a unique \( {r}_{1} \in \mathcal{O} \) such that \( x = {a}_{0} + {r}_{1}t \) . An easy induction now shows that for any integer \( n \) there exist uniquely determined elements \( {a}_{0},\ldots ,{a}_{n} \in {F}^{\prime } \) and a uniquely determined element \( {r}_{n + 1} \in \mathcal{O} \) such that
\[
x = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{t}^{i} + {r}_{n + 1}{t}^{n + 1}
\]
Put \( {x}_{n} \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{t}^{i} \) . Then \( \mathop{\lim }\limits_{n}{x}_{n} = x \) . It follows that \( x = \mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}{t}^{i} \in \operatorname{im}\left( \widehat{\mu }\right) \) . | Yes |
Theorem 5.17. 1. If \( G \) is a regular open subset of \( \mathbf{F} \) then \( {G}^{\Delta } \) is regular open. | Proof. 1. Let \( {G}_{1} = {\left( {G}^{\Delta }\right) }^{-0} \) . Then
\[
G = {G}^{\Delta * } \subseteq {\left( {G}^{\Delta }\right) }^{-0 * } = {G}_{1}^{ * }.
\]
If \( {G}_{2} \) is regular open and \( G \cap {G}_{2} = 0 \) then
\[
{\left( {G}^{\Delta } \cap {G}_{2}^{\Delta }\right) }^{ * } \subseteq {G}^{\Delta * } \cap {G}_{2}^{\Delta * } = G \cap {G}_{2} = 0.
\]
Therefore \( {G}^{\Delta } \cap {G}_{2}^{\Delta } = 0 \) and hence \( {G}_{1} \cap {G}_{2}^{\Delta } = 0 \) . Furthermore
\[
{\left( {G}_{1}^{ * } \cap {G}_{2}\right) }^{\Delta } \subseteq {G}_{1}^{*\Delta } \cap {G}_{2}^{\Delta } = {G}_{1} \cap {G}_{2}^{\Delta } = 0.
\]
Thus \( {G}_{1}^{ * } \cap {G}_{2} = 0 \) and hence, by Theorem 5.15, \( {G}_{1}^{ * } \subseteq {G}^{-0} = G \) . Consequently
\[
{G}^{\Delta } = {G}_{1}^{*\Delta } = {G}_{1}
\]
i.e., \( {G}^{\Delta } \) is regular open. | Yes |
Let \( f\left( x\right) \in k\left\lbrack x\right\rbrack, k \) a field. Suppose that \( \deg f\left( x\right) = n \) . Then \( f \) has at most \( n \) distinct roots. | The proof goes by induction on \( n \) . For \( n = 1 \) the assertion is trivial. Assume that the lemma is true for polynomials of degree \( n - 1 \) . If \( f\left( x\right) \) has no roots in \( k \), we are done. If \( \alpha \) is a root, \( f\left( x\right) = q\left( x\right) \left( {x - \alpha }\right) + r \), where \( r \) is a constant. Setting \( x = \alpha \) we see that \( r = 0 \) . Thus \( f\left( x\right) = q\left( x\right) \left( {x - \alpha }\right) \) and \( \deg q\left( x\right) = n - 1 \) . If \( \beta \neq \alpha \) is another root of \( f\left( x\right) \), then \( 0 = f\left( \beta \right) = \) \( \left( {\beta - \alpha }\right) q\left( \beta \right) \), which implies that \( q\left( \beta \right) = 0 \) . Since by induction \( q\left( x\right) \) has at most \( n - 1 \) distinct roots, \( f\left( x\right) \) has at most \( n \) distinct roots. | Yes |
Proposition 8.1.2. Let t be an element of F such that v_P(i)(t) = 1 for i = 1,…, n. Then the following hold: | Proof. (a) Since t is a prime element of P := P_i, the P-adic power series of η = dt/t with respect to t is
\[
\eta = \frac{1}{t}{dt}
\]
Hence v_P(η) = - 1 and res_P(η) = 1.
(b) Follows immediately from (a) and Proposition 2.2.10. | No |
For each \( r > 0 \), the image under \( q \) of the rectangle \( \left\lbrack {0,1}\right\rbrack \times \left\lbrack {-r, r}\right\rbrack \) is a Möbius band \( {M}_{r} \). Because \( q \) restricts to a smooth covering map from \( \mathbb{R} \times \left\lbrack {-r, r}\right\rbrack \) to \( {M}_{r} \), the same argument shows that a Möbius band is not orientable either. | Null | No |
Proposition 21.7 (Orbits of Proper Actions). Suppose \( \theta \) is a proper smooth action of a Lie group \( G \) on a smooth manifold \( M \) . For any point \( p \in M \), the orbit map \( {\theta }^{\left( p\right) } : G \rightarrow M \) is a proper map, and thus the orbit \( G \cdot p = {\theta }^{\left( p\right) }\left( G\right) \) is closed in \( M \) . If in addition \( {G}_{p} = \{ e\} \), then \( {\theta }^{\left( p\right) } \) is a smooth embedding, and the orbit is a properly embedded submanifold. | If \( K \subseteq M \) is compact, then \( {\left( {\theta }^{\left( p\right) }\right) }^{-1}\left( K\right) \) is closed in \( G \) by continuity, and since it is contained in \( {G}_{K\cup \{ p\} } \), it is compact by Proposition 21.5. Therefore, \( {\theta }^{\left( p\right) } \) is a proper map, which implies that \( G \cdot p = {\theta }^{\left( p\right) }\left( G\right) \) is closed by Theorem A.57. The final statement of the theorem then follows from Propositions 7.26 and 4.22. \( ▱ \) | Yes |
Corollary 25.5. Let \( \mathbf{K} \) be a class of \( {\mathcal{L}}^{\prime } \) -structures and let \( \mathcal{L} \) be a reduct of \( {\mathcal{L}}^{\prime } \) . If \( \mathbf{K} \) can be characterized by first-order sentences, then \( \mathbf{S}\left( {\mathbf{K} \upharpoonright \mathcal{L}}\right) \) is a universal class. | Proof. By 25.3 and 18.29, \( \mathbf{S}\left( {\mathbf{K} \upharpoonright \mathcal{L}}\right) \) is closed under \( \mathbf{S} \) and \( \mathbf{{Up}} \) . Hence 25.5 is immediate from 25.2. | Yes |
Let \( f \) be a function in \( {C}^{\infty }\left( \Omega \right) \). The multiplication by \( f \) defines a continuous operator \( {M}_{f} : \varphi \mapsto {f\varphi } \) on \( {C}_{0}^{\infty }\left( \Omega \right) \). Since | \[
{\int }_{\Omega }\left( {f\varphi }\right) {\psi dx} = {\int }_{\Omega }\varphi \left( {f\psi }\right) {dx},\varphi ,\psi \in {C}_{0}^{\infty }\left( \Omega \right) ,
\]
we define \( {M}_{f}u = {fu} \) by
\[
\left( {fu}\right) \left( \varphi \right) = u\left( {f\varphi }\right), u \in {\mathcal{D}}^{\prime }\left( \Omega \right) ,\varphi \in {C}_{0}^{\infty }\left( \Omega \right) ;
\]
\( {M}_{f} \) is a continuous operator on \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) | No |
If \( \pi \) is a prime of \( \mathbb{Z}\left\lbrack \rho \right\rbrack \), show that \( N\left( \pi \right) \) is a rational prime or the square of a rational prime. | Let \( N\left( \pi \right) = n > 1 \) . Then \( {\pi \pi } = n \) . Now \( n \) is a product of rational prime divisors. Since \( \pi \) is prime, \( \pi \mid p \) for some rational prime \( p \) . Write \( p = {\pi \gamma } \) . Then \( N\left( p\right) = N\left( \pi \right) N\left( \gamma \right) = {p}^{2} \) . Thus, either \( N\left( \pi \right) = p \) or \( N\left( \pi \right) = {p}^{2}. \) | Yes |
Proposition 8.2.1. For any \( P = \left( {x, y}\right) \in E \) set \(\phi \left( P\right) = \left( {\widehat{x},\widehat{y}}\right) = \left( {\frac{{y}^{2}}{{x}^{2}},\frac{y\left( {{x}^{2} - b}\right) }{{x}^{2}}}\right)\) for \( P \) not equal to \( T \) or \( \mathcal{O} \), and set \( \phi \left( T\right) = \phi \left( \mathcal{O}\right) = \widehat{\mathcal{O}} \). Then \( \phi \) is a group homomorphism from \( E \) to \( \widehat{E} \), whose kernel is equal to \( \{ \mathcal{O}, T\} \). | The proof consists in a series of explicit verifications, where in each case we must separate the points \( \mathcal{O} \) and \( T \) from the other points. It is done with utmost detail in [Sil-Tat], to which we refer. We will simply show that \( \phi \) maps \( E \) to \( \widehat{E} \), and that it maps three collinear points of \( E \) to three collinear points of \( \widehat{E} \). This is the essential part of the proof. Also, to simplify we will assume that all the points that occur are distinct and different from \( \mathcal{O}, T \), \( \widehat{\mathcal{O}} \), and \( \widehat{T} \). | No |
Let \( f \) and \( g \) be convex functions on a normed space \( X \). If \( f \) and \( g \) are finite at \( \bar{x} \) and if \( f \) is continuous at some point of \( \operatorname{dom}f \cap \operatorname{dom}g \), then | The inclusion \( \partial f\left( \bar{x}\right) + \partial g\left( \bar{x}\right) \subset \partial \left( {f + g}\right) \left( \bar{x}\right) \) is an immediate consequence of the definition of the subdifferential. Let us prove the reverse inclusion under the assumptions of the theorem. Let \( {\bar{x}}^{ * } \in \partial \left( {f + g}\right) \left( \bar{x}\right) \). Replacing \( f \) and \( g \) by the functions \( {f}_{0} \) and \( {g}_{0} \) given respectively by\n\n{f}_{0}\left( x\right) = f\left( {\bar{x} + x}\right) - f\left( \bar{x}\right) - \left\langle {{\bar{x}}^{ * }, x}\right\rangle\n\n{g}_{0}\left( x\right) = g\left( {\bar{x} + x}\right) - g\left( \bar{x}\right)\n\nwe may suppose \( \bar{x} = 0,{\bar{x}}^{ * } = 0, f\left( \bar{x}\right) = g\left( \bar{x}\right) = 0 \). Then we have \( f\left( x\right) + g\left( x\right) \geq 0 \) for every \( x \in X \) and \( f\left( 0\right) = 0 = g\left( 0\right) \). The interior \( C \) of the epigraph \( E \) of \( f \) is nonempty and contained in the strict epigraph of \( f \), hence is disjoint from the hypograph\n\nH \mathrel{\text{:=}} \{ \left( {x, s}\right) \in X \times \mathbb{R} : s \leq - g\left( x\right) \}\n\nof \( - g \). Let \( \left( {{u}^{ * }, c}\right) \in {\left( X \times \mathbb{R}\right) }^{ * } \smallsetminus \{ \left( {0,0}\right) \} \), which separates \( C \) and \( H \) :\n\n\left\langle {{u}^{ * }, w}\right\rangle + {cr} > 0 \geq \left\langle {{u}^{ * }, x}\right\rangle + {cs}\;\forall \left( {w, r}\right) \in C,\forall \left( {x, s}\right) \in H\n\n(we use the fact that \( 0 \in \operatorname{cl}\left( C\right) \cap H \) ). Let \( u \) be a point of dom \( g \) at which \( f \) is finite and continuous. Taking \( w = x = u \) and \( r \in \left( {f\left( u\right) , + \infty }\right) \) large enough, we see that \( c \geq 0 \). If we had \( c = 0 \), taking \( \left( {x, s}\right) = \left( {u, - g\left( u\right) }\right) \), we would have \( \left\langle {{u}^{ * }, w}\right\rangle > \left\langle {{u}^{ * }, u}\right\rangle \) for all \( w \) in a neighborhood of \( u \), an impossibility. Thus \( c > 0 \). Since \( E \subset \operatorname{cl}\left( C\right) \) we get\n\nr \geq \left\langle {-{c}^{-1}{u}^{ * }, w}\right\rangle \;\forall \left( {w, r}\right) \in E,\;g\left( x\right) \geq \left\langle {{c}^{-1}{u}^{ * }, x}\right\rangle \forall x \in \operatorname{dom}g,\n\nand since \( f\left( 0\right) = 0, g\left( 0\right) = 0 \), we get \( {x}^{ * } \mathrel{\text{:=}} - {c}^{-1}{u}^{ * } \in \partial f\left( 0\right) , - {x}^{ * } \in \partial g\left( 0\right) \). | Yes |
Proposition 1.6. For any tape description \( F \) and any \( e \in \mathbb{Z},\langle \left( {F,0, e}\right) \) , \( \left( {F,1, e + 1}\right) \rangle \) is a computation of \( {T}_{\text{left }} \) . | Thus \( {T}_{\text{left }} \) moves the tape one square to the left and then stops. | No |
For \( 1 \leq i \leq p - 2 \) we have \( \begin{Vmatrix}{f}_{i}\end{Vmatrix} = \left( {p - 1}\right) /2 \) . | For \( 1 \leq t \leq p - 1 \) and \( 1 \leq i \leq p - 1 \) we note that\\ \[ \lfloor {ti}/p\rfloor + \lfloor \left( {p - t}\right) i/p\rfloor = \lfloor {ti}/p\rfloor + i - \lceil {ti}/p\rceil = i - 1 \] \\ since \( p \nmid {ti} \) . It follows that\\ \[ \mathop{\sum }\limits_{{1 \leq t \leq p - 1}}\lfloor {ti}/p\rfloor = \mathop{\sum }\limits_{{1 \leq t \leq \left( {p - 1}\right) /2}}\left( {\lfloor {ti}/p\rfloor +\lfloor \left( {p - t}\right) i/p\rfloor }\right) = \left( {i - 1}\right) \left( {p - 1}\right) /2. \]\\ Since the coefficients of \( {f}_{i} \) are equal to 0 or 1, for \( 1 \leq i \leq p - 2 \) we have \( \begin{Vmatrix}{f}_{i}\end{Vmatrix} = i\left( {p - 1}\right) /2 - \left( {i - 1}\right) \left( {p - 1}\right) /2 = \left( {p - 1}\right) /2 \), proving (1) (note that this is false for \( i = p - 1 \) since in that case the above computation is not valid for \( i + 1 = p \), and in fact we know that \( \left. {\begin{Vmatrix}{f}_{p - 1}\end{Vmatrix} = \parallel s\left( G\right) \parallel = p - 1}\right) \) . | Yes |
Corollary 1. \( A,\mathfrak{m} \) and \( E \) being as in Theorem 7, suppose that \( {G}_{\mathfrak{m}}\left( E\right) \) is a finite \( {G}_{\mathfrak{m}}\left( A\right) \) -module. Then \( E \) is a finite \( A \) -module. | We apply Theorem 7 to the case \( F = E \). | No |
Let \( \left( {d, p}\right) = 1,{q}_{n} = {qd}{p}^{n} \), and \( {h}_{n}^{ - } = {h}^{ - }\left( {\mathbb{Q}\left( {\zeta }_{{q}_{n}}\right) }\right) \) . We assume \( d ≢ 2\left( {\;\operatorname{mod}\;4}\right) \) . Then | Let \( {q}_{n}^{\prime } = \operatorname{lcm}\left( {{q}_{n},2}\right) \) . Theorem 4.17 implies that
\[
{h}_{0}^{ - } = {q}_{0}^{\prime }Q\mathop{\prod }\limits_{\substack{{\theta \neq 1} \\ {{f}_{\theta } \mid {q}_{0}} \\ {\theta \text{ even }} }}\left( {-\frac{1}{2}{B}_{1,\theta {\omega }^{-1}}}\right)
\]
and
\[
{h}_{n}^{ - } = {q}_{n}^{\prime }Q\mathop{\prod }\limits_{\substack{{\chi \neq 1} \\ {{f}_{\chi } \mid {q}_{n}} \\ {\chi \text{ even }} }}\left( {-\frac{1}{2}{B}_{1,\chi {\omega }^{-1}}}\right) .
\]
The number \( Q \) equals 1 or 2, but is the same for all \( n \geq 0 \) by Corollary 4.13. Writing \( \chi = {\theta \psi } \), where \( \theta \) is of the first kind and \( \psi \) is of the second kind, we obtain
\[
\mathop{\prod }\limits_{{\chi \neq 1}}\left( {-\frac{1}{2}{B}_{1,\chi {\omega }^{-1}}}\right) = \mathop{\prod }\limits_{{\theta \neq 1}}\left( {-\frac{1}{2}{B}_{1,\theta {\omega }^{-1}}}\right) \mathop{\prod }\limits_{{\psi \neq 1}}\left( {-\frac{1}{2}{B}_{1,\psi {\omega }^{-1}}}\right) \mathop{\prod }\limits_{\substack{{\theta \neq 1} \\ {\psi \neq 1} }}\left( {-\frac{1}{2}{B}_{1,{\theta \psi }{\omega }^{-1}}}\right) .
\]
The first product is the same as that for \( {h}_{0}^{ - } \) . To treat the second product, note that
\[
- {B}_{1,\psi {\omega }^{-1}} = {L}_{p}\left( {0,\psi }\right) = \frac{g\left( {{\zeta }_{\psi } - 1,1}\right) }{h\left( {{\zeta }_{\psi } - 1,1}\right) }
\]
\( \left( {\psi {\omega }^{-1}\left( p\right) = 0}\right. \), so the Euler factor disappears. It is because of the Euler factors for the other characters that we must take the ratio \( {h}_{n}^{ - }/{h}_{0}^{ - } \) and require \( \zeta \neq 1 \) ; otherwise the formulas could reduce to \( 0 = 0 \) ). From Lemma 7.12, \( \frac{1}{2}g\left( {{\zeta }_{\psi } - 1,1}\right) \) is a unit; and
\[
h\left( {{\zeta }_{\psi } - 1,1}\right) = 1 - \frac{1 + q}{{\zeta }_{\psi }} \equiv 1 - {\zeta }_{\psi }^{-1}\left( {\;\operatorname{mod}\;q}\right) .
\]
Since \( {\zeta }_{\psi } \) equals \( \psi \) evaluated at a generator of \( {\Gamma }_{n},{\zeta }_{\psi } \) determines \( \psi \) . Since there are \( {p}^{n} \) elements of \( {\widehat{\Gamma }}_{n} \), it follows that as \( \psi \) runs through the characters of the second kind, \( {\zeta }_{\psi } \) runs through all \( {p}^{n} \) th roots of unity. Putting everything together, we find that
\[
{v}_{p}\left( {\mathop{\prod }\limits_{{\psi \neq 1}}\left( {-\frac{1}{2}{B}_{1,\psi {\omega }^{-1}}}\right) }\right) = {v}_{p}\left( {\mathop{\prod }\limits_{\substack{{{\zeta }^{p} = 1} \\ {\zeta \neq 1} }}{\left( 1 - {\zeta }^{-1}\right) }^{-1}}\right) = {v}_{p}\left( {p}^{-n}\right) = {v}_{p}\left( \frac{{q}_{0}^{\prime }}{{q}_{n}^{\prime }}\right) .
\]
For the third product, we proceed as above (again, since \( \psi \neq 1 \), the Euler factor disappears):
\[
- \frac{1}{2}{B}_{1,{\theta \psi }{\omega }^{-1}} = \frac{1}{2}f\left( {{\zeta }_{\psi } - 1,\theta }\right)
\]
so
\[
\mathop{\prod }\limits_{\substack{{\theta \neq 1} \\ {\psi \neq 1} }}\left( {-\frac{1}{2}{B}_{1,{\theta \psi }{\omega }^{-1}}}\right) = \mathop{\prod }\limits_{{\theta \neq 1}}\mathop{\prod }\limits_{\substack{{\zeta }^{p{n}_{ = 1}} \\ {\zeta \neq 1} }}\frac{1}{2}f\left( {\zeta - 1,\theta }\right) .
\]
Combining all the above, we obtain the theorem." | Yes |
Theorem 16.8.3 Let \( {Y}_{1} \) and \( {Y}_{2} \) be signed graphs that are related by a Whitney flip. Then their rank polynomials are equal. | The graphs \( {Y}_{1} \) and \( {Y}_{2} \) have the same edge set, and it is clear that a set \( S \subseteq E\left( {Y}_{1}\right) \) is independent in \( M\left( {Y}_{1}\right) \) if and only if it is independent in \( M\left( {Y}_{2}\right) \) . Therefore, the two graphs have the same cycle matroid. | Yes |
Let \( \Omega \) be a smoothly bounded, finite-type domain in \( {\mathbb{C}}^{2} \) . Equip Aut \( \left( \Omega \right) \) with the \( {C}^{k} \) topology, some integer \( k \geq 0 \) . Assume that \( \Omega \) has compact automorphism group in the \( {C}^{k} \) topology. Then there is an \( \epsilon > 0 \) so that if \( {\Omega }^{\prime } \) is a smoothly bounded, finite-type domain with \( {C}^{m} \) distance less than \( \epsilon \) from \( \Omega \) (with \( m \leq k) \), then \( \operatorname{Aut}\left( {\Omega }^{\prime }\right) \) can be realized as a subgroup of \( \operatorname{Aut}\left( \Omega \right) \). | The proof is just the same as that for the last theorem. The main point is to have a uniform bound for derivatives of automorphisms (Proposition 5.2.5), so that the smooth-to-the-boundary invariant metric can be constructed. | No |
The complete graph \( {K}_{n} \) is not the edge-disjoint union of \( n - 2 \) complete bipartite graphs. | Suppose that, contrary to the assertion, \( {K}_{n} \) is the edge-disjoint union of complete bipartite graphs \( {G}_{1},\ldots ,{G}_{n - 2} \) . For each \( i \), let \( {H}_{i} \) be obtained from \( {G}_{i} \) by adding to it isolated vertices so that \( V\left( {H}_{i}\right) = V\left( {K}_{n}\right) \) . Note that the Lagrangians of these graphs are such that \( {f}_{{K}_{n}} = \mathop{\sum }\limits_{{i = 1}}^{{n - 2}}{f}_{{H}_{i}} \) .
We know that each \( {f}_{{H}_{i}} \) is positive semi-definite on some subspace \( {U}_{i} \subset {C}_{0}\left( {K}_{n}\right) \) of dimension \( n - 1 \) . But then \( U = \mathop{\bigcap }\limits_{{i = 1}}^{{n - 2}}{U}_{i} \) is a subspace of dimension at least 2, on which each \( {f}_{{H}_{i}} \) is positive semi-definite. Hence \( {f}_{{K}_{n}} = \mathop{\sum }\limits_{{i = 1}}^{{n - 2}}{f}_{{H}_{i}} \) is positive semi-definite on \( U \), contradicting the fact that \( {f}_{{K}_{n}} \) is not positive semi-definite on any subspace of dimension 2. | Yes |
For \( n \geq 2 \) even we have \(\mathop{\sum }\limits_{{k \geq 1}}\frac{\cos \left( {2\pi kx}\right) }{{k}^{n}} = \frac{{\left( -1\right) }^{n/2 + 1}}{2}\frac{{\left( 2\pi \right) }^{n}{B}_{n}\left( {\{ x\} }\right) }{n!}\) and for \( n \geq 1 \) odd we have \(\mathop{\sum }\limits_{{k \geq 1}}\frac{\sin \left( {2\pi kx}\right) }{{k}^{n}} = \frac{{\left( -1\right) }^{\left( {n + 1}\right) /2}}{2}\frac{{\left( 2\pi \right) }^{n}{B}_{n}\left( {\{ x\} }\right) }{n!}\), except for \( n = 1 \) and \( x \in \mathbb{Z} \), in which case the left-hand side is evidently equal to 0 . | Proof. (1) and (2). Since \( {B}_{n}\left( 1\right) = {B}_{n}\left( 0\right) \) for \( n \neq 1 \), the function \( {B}_{n}\left( {\{ x\} }\right) \) is piecewise \( {C}^{\infty } \) and continuous for \( n \geq 2 \), with simple discontinuities at the integers if \( n = 1 \) . If \( n \geq 2 \) we thus have \({B}_{n}\left( {\{ x\} }\right) = \mathop{\sum }\limits_{{k \in \mathbb{Z}}}{c}_{n, k}{e}^{2i\pi kx},\) with \({c}_{n, k} = {\int }_{0}^{1}{B}_{n}\left( t\right) {e}^{-{2i\pi kt}}{dt}\) For \( n = 1 \), the same formula is valid for \( x \notin \mathbb{Z} \), and for \( x \in \mathbb{Z} \) we must replace \( {B}_{1}\left( {\{ x\} }\right) \) by \( \left( {{B}_{1}\left( {1}^{ - }\right) + {B}_{1}\left( {0}^{ + }\right) }\right) /2 = 0 \) . Using the definitions and the formulas \( {B}_{n}^{\prime }\left( x\right) = n{B}_{n - 1}\left( x\right) \) and \( {B}_{n}\left( 1\right) = \) \( {B}_{n}\left( 0\right) \) for \( n \neq 1 \), by integration by parts we obtain for \( k \neq 0 \) \({c}_{n, k} = \frac{n}{2i\pi k}{c}_{n - 1, k}\;\text{ and }\;{c}_{1, k} = - \frac{1}{2i\pi k},\) hence by induction \({c}_{n, k} = - \frac{n!}{{\left( 2i\pi k\right) }^{n}}.\) On the other hand, we clearly have \({c}_{n,0} = \frac{{B}_{n + 1}\left( 1\right) - {B}_{n + 1}\left( 0\right) }{n + 1} = 0\) as soon as \( n \geq 1 \) . Thus, with the above interpretation for \( x \in \mathbb{Z} \) when \( n = 1 \) , we obtain that for \( n \geq 1 \) we have \({B}_{n}\left( {\{ x\} }\right) = - \frac{n!}{{\left( 2i\pi \right) }^{n}}\mathop{\sum }\limits_{{k \neq 0}}\frac{{e}^{2i\pi kx}}{{k}^{n}}.\) Separating the cases \( n \) even and \( n \) odd, and grouping the terms \( k \) and \( - k \) proves (1) and (2). For (3) we proceed differently. We have \(\mathop{\sum }\limits_{{k \geq 1}}\frac{\cos \left( {2\pi kx}\right) }{k} = \Re \left( {\mathop{\sum }\limits_{{k \geq 1}}\frac{{e}^{2i\pi kx}}{k}}\right) = - \Re \left( {\log \left( {1 - {e}^{2i\pi x}}\right) }\right) = - \log \left( \left| {1 - {e}^{2i\pi x}}\right| \right) = - \log \left( {2\left| {\sin \left( {\pi x}\right) }\right| }\right) ,\) proving the theorem. | Yes |
Theorem 5.15 (Dynkin’s Formula). Let \( G \) be a Lie subgroup of \( {GL}\left( {n,\mathbb{C}}\right) \) . For \( X, Y \in \mathfrak{g} \) in a sufficiently small neighborhood of 0,\n\n{e}^{X}{e}^{Y} = {e}^{Z}\n\nwhere \( Z \) is given by the formula\n\nZ = \sum \frac{{\left( -1\right) }^{n + 1}}{n}\frac{1}{\left( {{i}_{1} + {j}_{1}}\right) + \cdots + \left( {{i}_{n} + {j}_{n}}\right) }\frac{\left\lbrack {X}^{\left( {i}_{1}\right) },{Y}^{\left( {j}_{1}\right) },\ldots ,{X}^{\left( {i}_{1}\right) },{Y}^{\left( {j}_{n}\right) }\right\rbrack }{{i}_{1}!{j}_{1}!\cdots {i}_{n}!{j}_{n}!},\n\nwhere the sum is taken over all 2n-tuples \( \left( {{i}_{1},\ldots ,{i}_{n},{j}_{1},\ldots ,{j}_{n}}\right) \in {\mathbb{N}}^{2n} \) satisfying \( {i}_{k} + {j}_{k} \geq 1 \) for positive \( n \in \mathbb{N} \) . | Proof. The approach of this proof follows [34]. Using Theorem 4.6, choose a neighborhood \( {U}_{0} \) of 0 in \( \mathfrak{g} \) on which exp is a local diffeomorphism and where \( \ln \) is well defined on \( \exp U \) . Let \( U \subseteq {U}_{0} \) be an open ball about of 0 in \( \mathfrak{g} \), so that \( {\left( \exp U\right) }^{2}{\left( \exp U\right) }^{-2} \subseteq \exp {U}_{0} \) (by continuity of the group structure as in Exercise 1.4). For \( X, Y \in U \), define \( \gamma \left( t\right) = {e}^{tX}{e}^{tY} \) mapping a neighborhood of \( \left\lbrack {0,1}\right\rbrack \) to \( \exp U \) . Therefore there is a unique smooth curve \( Z\left( t\right) \in {U}_{0} \), so that \( {e}^{Z\left( t\right) } = {e}^{tX}{e}^{tY} \) . Apply \( \frac{d}{dt} \) to this equation and use Theorem 5.14 to see that\n\n\left\lbrack {\left( \frac{{e}^{\operatorname{ad}Z\left( t\right) } - I}{\operatorname{ad}Z\left( t\right) }\right) \left( {{Z}^{\prime }\left( t\right) }\right) }\right\rbrack {e}^{Z\left( t\right) } = X{e}^{Z\left( t\right) } + {e}^{Z\left( t\right) }Y.\n\nSince \( Z\left( t\right) \in {U}_{0} \), exp is a local diffeomorphism near \( Z\left( t\right) \) . Thus the proof of Theorem 5.14 shows that \( \left( \frac{I - {e}^{-\operatorname{ad}Z\left( t\right) }}{\operatorname{ad}Z\left( t\right) }\right) \) is an invertible map on \( \mathfrak{g} \) . As \( {e}^{Z\left( t\right) } = {e}^{tX}{e}^{tY} \) , \( \operatorname{Ad}\left( {e}^{Z\left( t\right) }\right) = \operatorname{Ad}\left( {e}^{tX}\right) \operatorname{Ad}\left( {e}^{tY}\right) \), so that \( {e}^{\operatorname{ad}Z\left( t\right) } = {e}^{t\operatorname{ad}X}{e}^{t\operatorname{ad}Y} \) by Equation 4.11. Thus\n\n{Z}^{\prime }\left( t\right) = \left( \frac{\operatorname{ad}Z\left( t\right) }{{e}^{\operatorname{ad}Z\left( t\right) } - I}\right) \left( {X + \operatorname{Ad}\left( {e}^{Z\left( t\right) }\right) Y}\right) = \left( \frac{\operatorname{ad}Z\left( t\right) }{{e}^{\operatorname{ad}Z\left( t\right) } - I}\right) \left( {X + {e}^{\operatorname{ad}Z\left( t\right) }Y}\right)\n\n= \left( \frac{\operatorname{ad}Z\left( t\right) }{{e}^{\operatorname{ad}Z\left( t\right) } - I}\right) \left( {X + {e}^{t\operatorname{ad}X}{e}^{t\operatorname{ad}Y}Y}\right) = \left( \frac{\operatorname{ad}Z\left( t\right) }{{e}^{\operatorname{ad}Z\left( t\right) } - I}\right) \left( {X + {e}^{t\operatorname{ad}X}Y}\right) .\n\nUsing the relation \( A = \ln \left( {I + \left( {{e}^{A} - I}\right) }\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}{\left( {e}^{A} - I\right) }^{n} \) for \( A = \) ad \( Z\left( t\right) \) and \( {e}^{A} = {e}^{t\operatorname{ad}X}{e}^{t\operatorname{ad}Y} \), we get\n\n\frac{\operatorname{ad}Z\left( t\right) }{{e}^{\operatorname{ad}Z\left( t\right) } - I} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}{\left( {e}^{t\operatorname{ad}X}{e}^{t\operatorname{ad}Y} - I\right) }^{n - 1}.\n\nHence\n\n{Z}^{\prime }\left( t\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}{\left( {e}^{t\operatorname{ad}X}{e}^{t\operatorname{ad}Y} - I\right) }^{n - 1}\left( {X + {e}^{t\operatorname{ad}X}Y}\right)\n\n= \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}{\left\lbrack \mathop{\sum }\limits_{{i, j = 0,\left( {i, j}\right) \neq \left( {0,0}\right) }}^{\infty }\frac{{t}^{i + j}}{i!j!}{\left( \operatorname{ad}X\right) }^{i}{\left( \operatorname{ad}Y\right) }^{j}\right\rbrack }^{n - 1}\left( {X + \left( {\mathop{\sum }\limits_{{i = 0}}^{\infty }\frac{{t}^{i}}{i!}{\left( \operatorname{ad}X\right) }^{i}}\right) Y}\right)\n\n= \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}\left\lbrack {\sum \frac{{t}^{{i}_{1} + {j}_{1} + \cdots {i}_{n - 1} + {j}_{n - 1}}}{{i}_{1}!{j}_{1}!\cdots {i}_{n - 1}!{j}_{n - 1}!}\left\lbrack {{\left( X\right) }^{{i}_{1}},{\left( Y\right) }^{{j}_{1}},\ldots ,{\left( X\right) }^{{i}_{k - 1}},{\left( Y\right) }^{{j}_{k - 1}}, X}\right\rbrack }\right.\n\n\left. {+\sum \frac{{t}^{{i}_{1} + {j}_{1} + \cdots {i}_{n - 1} + {j}_{n - 1} + {i}_{n}}}{{i}_{1}!{j}_{1}!\cdots {i}_{n - 1}!{j}_{n - 1}!{i}_{n}!}\left\lbrack {{\left( X\right) }^{{i}_{1}},{\left( Y\right) }^{{j}_{1}},\ldots ,{\left( X\right) }^{{i}_{k - 1}},{\left( Y\right) }^{{j}_{k - 1}},{\left( X\right) }^{{i}_{n}}, Y}\right\rbrack }\right\rbrack\n\nwhere the second and third sum are taken over all \( {i}_{k},{j}_{k} \in \mathbb{N} \) with \( {i}_{k} + {j}_{k} \geq 1 \) . Since \( Z\left( 0\right) = 0, Z\left( 1\right) = {\int }_{0}^{1}\frac{d}{dt}Z\left( t\right) {dt} \) . Integrating the above displayed equation finishes the proof. | Yes |
When \( P \) is properly supported in \( \Omega \), there is a unique symbol \( p\left( {x,\xi }\right) \in {S}^{\infty }\left( \Omega \right) \) such that \( P = \operatorname{Op}\left( {p\left( {x,\xi }\right) }\right) \), namely, the one determined by (7.29). | Null | No |
Theorem 3.6.6. For all \( r \in \mathbb{Z} \) we have \[ \frac{\tau \left( {\omega }^{-r}\right) }{{\left( \zeta - 1\right) }^{s\left( r\right) }} \equiv - \frac{1}{t\left( r\right) }\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) . \] | Proof. By periodicity we may assume that \( 0 \leq r < q - 1 \) . We prove the theorem by induction on \( s\left( r\right) = {s}_{p}\left( r\right) \) . If \( s\left( r\right) = 0 \) we have \( r = 0 \), hence \( t\left( 0\right) = 1 \) and \( \tau \left( {\omega }^{0}\right) = \tau \left( \varepsilon \right) = - 1 \) by Lemma 2.5.8. The crucial case to be proved is the case \( s\left( r\right) = 1 \), in other words, \( r = {p}^{k} \) . In that case \( t\left( r\right) = 1 \), and since by Lemma 2.5.8 (3) we have \( \tau \left( {\omega }^{-{pa}}\right) = \tau \left( {\omega }^{-a}\right) \), it follows that we may assume that \( r = 1 \) . Since \( \omega \) is a nontrivial character we have \[ \tau \left( {\omega }^{-1}\right) = \mathop{\sum }\limits_{{x \in {\mathbb{F}}_{q}^{ * }}}{\omega }^{-1}\left( x\right) \left( {{\zeta }_{p}^{{\operatorname{Tr}}_{{\mathbb{F}}_{q}/{\mathbb{F}}_{p}}\left( x\right) } - 1}\right) . \] Since \( \mathfrak{p} = \left( {{\zeta }_{p} - 1}\right) {\mathbb{Z}}_{K} \) we have \[ \frac{{\zeta }_{p}^{m} - 1}{{\zeta }_{p} - 1} = \mathop{\sum }\limits_{{0 \leq j < m}}{\zeta }_{p}^{j} \equiv \mathop{\sum }\limits_{{0 \leq j < m}}1 \equiv m\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) , \] hence \[ \frac{\tau \left( {\omega }^{-1}\right) }{{\zeta }_{p} - 1} \equiv \mathop{\sum }\limits_{{x \in {\mathbb{F}}_{q}^{ * }}}{\omega }^{-1}\left( x\right) {\operatorname{Tr}}_{{\mathbb{F}}_{q}/{\mathbb{F}}_{p}}\left( x\right) \left( {\;\operatorname{mod}\;\mathfrak{p}}\right) . \] Now \( {\operatorname{Tr}}_{{\mathbb{F}}_{q}/{\mathbb{F}}_{p}}\left( x\right) = \mathop{\sum }\limits_{{0 \leq i < f}}{x}^{{p}^{i}} \in {\mathbb{F}}_{p} \), and on the other hand, by definition, \( {\omega }^{-1}\left( x\right) \equiv {x}^{-1}\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \) . It follows that \[ \frac{\tau \left( {\omega }^{-1}\right) }{{\zeta }_{p} - 1} \equiv \mathop{\sum }\limits_{{0 \leq i < f}}\mathop{\sum }\limits_{{x \in {\mathbb{F}}_{q}^{ * }}}{x}^{{p}^{i} - 1}\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) . \] Now again by Lemma 2.5.1 the inner sum vanishes if \( 1 \leq i < f \), and it is congruent to -1 modulo \( p \) if \( i = 0 \) . It follows that \( \tau \left( {\omega }^{-1}\right) /\left( {{\zeta }_{p} - 1}\right) \equiv - 1 \) (mod \( \mathfrak{P} \) ), proving the theorem when \( s\left( r\right) = 1 \) . Now let \( r \) be such that \( 0 \leq r < q - 1 \) with \( s\left( r\right) > 1 \), assume by induction that the theorem is true for all \( {r}^{\prime } < q - 1 \) such that \( s\left( {r}^{\prime }\right) < s\left( r\right) \), and let \( r = \mathop{\sum }\limits_{{0 \leq i < f}}{r}_{i}{p}^{i} \) with \( 0 \leq {r}_{i} \leq p - 1 \) . Thanks once again to Lemma 2.5.8 (3) we may assume that \( {r}_{0} \geq 1 \) . It follows in particular that \( s\left( {r - 1}\right) = s\left( r\right) - 1 \) and \( r - 1 \geq 1 \) . Since all the characters involved are nontrivial, by Corollary 2.5.17 we have \[ \tau \left( {\omega }^{-1}\right) \tau \left( {\omega }^{-\left( {r - 1}\right) }\right) = J\left( {{\omega }^{-1},{\omega }^{-\left( {r - 1}\right) }}\right) \tau \left( {\omega }^{-r}\right) . \] Using Proposition 3.6.4 we know that \[ J\left( {{\omega }^{-1},{\omega }^{-\left( {r - 1}\right) }}\right) \equiv - \left( \begin{array}{l} r \\ 1 \end{array}\right) \equiv - r \equiv - {r}_{0}\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) . \] Since \( 1 \leq {r}_{0} \leq p - 1,{r}_{0} \) is invertible modulo \( \mathfrak{P} \), so by induction and the case \( r = 1 \) we see that \[ \frac{\tau \left( {\omega }^{-r}\right) }{{\left( {\zeta }_{p} - 1\right) }^{s\left( r\right) }} \equiv - \frac{1}{{r}_{0}}\frac{\tau \left( {\omega }^{-1}\right) }{{\zeta }_{p} - 1}\frac{\tau \left( {\omega }^{-\left( {r - 1}\right) }\right) }{{\left( {\zeta }_{p} - 1\right) }^{s\left( {r - 1}\right) }} \equiv - \frac{1}{{r}_{0}}\frac{1}{t\left( {r - 1}\right) } \equiv - \frac{1}{t\left( r\right) }\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) , \] since \( t\left( r\right) = {r}_{0}t\left( {r - 1}\right) \) when \( {r}_{0} \neq 0 \), proving our induction hypothesis and hence the theorem. | Yes |
Let \( K \) be a geometric function field with \( \omega \in {\Omega }_{K} \) and \( P \in {\mathbb{P}}_{K} \) . Then, | Corollary 2.5.8. Let \( K \) be a geometric function field with \( \omega \in {\Omega }_{K} \) and \( P \in {\mathbb{P}}_{K} \) . Then,\n\n\[
{\nu }_{P}\left( {x\omega }\right) = {\nu }_{P}\left( x\right) + {\nu }_{P}\left( \omega \right)
\]\n\n\[
{\nu }_{P}\left( {\omega + {\omega }^{\prime }}\right) \geq \min \{ {\nu }_{P}\left( \omega \right) ,{\nu }_{P}\left( {\omega }^{\prime }\right) \} .
\] | Yes |
If \( 0 \leftarrow M \leftarrow {N}_{0} \leftarrow {N}_{1} \leftarrow \cdots \leftarrow {N}_{r} \leftarrow 0 \) is an exact sequence of finitely generated positively multigraded modules, then the Hilbert series of \( M \) equals the alternating sum of those for \( {N}_{0},\ldots ,{N}_{r} \) : | For each \( \mathbf{a} \in A \), the degree \( \mathbf{a} \) piece of the given exact sequence of modules is an exact sequence of finite-dimensional vector spaces over \( \mathbb{k} \) . The rank-nullity theorem from linear algebra says that the alternating sum of the dimensions of these vector spaces equals zero. | Yes |
Theorem 8.1.5. ( \( \mathbf{{Sp}}\left( n\right) \rightarrow \mathbf{{Sp}}\left( {n - 1}\right) \) Branching Law) The multiplicity \( m\left( {\lambda ,\mu }\right) \) is nonzero if and only if\n\[
{\lambda }_{j} \geq {\mu }_{j} \geq {\lambda }_{j + 2}\;\text{ for }j = 1,\ldots, n - 1\n\]
(8.4)\n(here \( {\lambda }_{n + 1} = 0 \) ). When these inequalities are satisfied let\n\[
{x}_{1} \geq {y}_{1} \geq {x}_{2} \geq {y}_{2} \geq \cdots \geq {x}_{n} \geq {y}_{n}\n\]be the nonincreasing rearrangement of \( \left\{ {{\lambda }_{1},\ldots ,{\lambda }_{n},{\mu }_{1},\ldots ,{\mu }_{n - 1},0}\right\} \) . Then\n\[
m\left( {\lambda ,\mu }\right) = \mathop{\prod }\limits_{{j = 1}}^{n}\left( {{x}_{j} - {y}_{j} + 1}\right) .\n\](8.5) | Null | No |
Let \( X \) be a path-connected space. The inclusion of basepoint preserving maps into the set of all maps induces a bijection \({\pi }_{q}\left( {X, x}\right) /{\pi }_{1}\left( {X, x}\right) \overset{ \sim }{ \rightarrow }\left\lbrack {{S}^{q}, X}\right\rbrack \) | Let \( h : {\pi }_{q}\left( {X, x}\right) \rightarrow \left\lbrack {{S}^{q}, X}\right\rbrack \) be induced by the inclusion of base point preserving maps into the set of all maps. If \( \left\lbrack \alpha \right\rbrack \in {\pi }_{q}\left( {X, x}\right) \) and \( \left\lbrack \gamma \right\rbrack \in {\pi }_{1}\left( {X, x}\right) \), it is laborious but not difficult to write down an explicit free homotopy between \( \alpha \) and \( {\gamma }_{ * }\alpha \) (see Figure 17.2 (b) for the cases \( q = 1 \) and \( q = 2 \) ). Hence \( h \) factors through the action of \( {\pi }_{1}\left( {X, x}\right) \) on \( {\pi }_{q}\left( {X, x}\right) \) and defines a map \[H : {\pi }_{q}\left( {X, x}\right) /{\pi }_{1}\left( {X, x}\right) \rightarrow \left\lbrack {{S}^{q}, X}\right\rbrack .\] Since \( X \) is path connected, any map in \( \left\lbrack {{S}^{q}, X}\right\rbrack \) can be deformed to a base-point preserving map. So \( H \) is surjective. To show injectivity, suppose \( \left\lbrack \alpha \right\rbrack \) in \( {\pi }_{q}\left( {X, x}\right) \) is null-homotopic in \( \left\lbrack {{S}^{q}, X}\right\rbrack \) . This means there is a map \( F : {I}^{q + 1} \rightarrow X \) such that \[{\left. F\right| }_{\text{top face }} = \alpha\] \[{\left. F\right| }_{\text{bottom face }}^{.} = \bar{x}\] and \( F \) is constant on the boundary of each horizontal slice (Figure 17.2 (c)). Let \( \gamma \) be the restriction of \( F \) to a vertical segment. Then \( \alpha = {\gamma }_{ * }\left( \bar{x}\right) \) . Therefore, \( H \) is injective. | Yes |
Proposition 8. Let \( \mathrm{A} \) be a ring that is Hausdorff and complete for the topology defined by a decreasing sequence \( {\mathfrak{a}}_{1} \supset {\mathfrak{a}}_{2} \supset \cdots \) of ideals such that \( {\mathfrak{a}}_{n} \cdot {\mathfrak{a}}_{m} \subset \) \( {\mathfrak{a}}_{n + m} \) . Assume that the residue ring \( \overline{\mathrm{K}} = \mathrm{A}/{\mathfrak{a}}_{1} \) is a perfect ring of characteristic p. Then: | Let \( \lambda \in \overline{\mathrm{K}} \) ; for all \( n \geq 0 \), denote by \( {\mathrm{L}}_{n} \) the inverse image of \( {\lambda }^{{p}^{-n}} \) in \( \mathrm{A} \), and by \( {\mathrm{U}}_{n} \) the set of all \( {x}^{{p}^{n}}, x \in {\mathrm{L}}_{n} \) ; the \( {\mathrm{U}}_{n} \) are contained in the residue class \( {\mathrm{L}}_{0} \) of \( \lambda \), and they form a decreasing sequence. We will show that they form a Cauchy filter base in A. Indeed, if \( a = {x}^{{p}^{n}} \) and \( b = {y}^{{p}^{n}} \), one shows by induction on \( n \) that \( a \equiv \mathrm{b}{\;\operatorname{mod}\;.}{\mathfrak{a}}_{n + 1} \), making use of the following lemma:
Lemma 1. If \( a \equiv b{\;\operatorname{mod}\;.}{\mathfrak{a}}_{n} \), then \( {a}^{p} \equiv {b}^{p}{\;\operatorname{mod}\;.}{\mathfrak{a}}_{n + 1} \) .
This lemma results from the binomial formula, taking into account that \( p \in {a}_{1} \), whence \( p{\mathfrak{a}}_{n} \subset {\mathfrak{a}}_{n + 1} \) .
Since the \( {\mathrm{U}}_{n} \) form a Cauchy filter base and \( \mathrm{A} \) is complete, one can set \( f\left( \lambda \right) = \lim {\mathrm{U}}_{n} \) . This defines a system of representatives. If \( \lambda = {\mu }^{p} \), the \( p \) th power operation in A maps \( {\mathrm{U}}_{n}\left( \mu \right) \) into \( {\mathrm{U}}_{n + 1}\left( \lambda \right) \), so passing to the limit shows that it maps \( f\left( \mu \right) \) on \( f\left( \lambda \right) \), and \( f \) does commute with the \( p \) th power. Conversely, if \( {f}^{\prime } \) is a system of representatives having this property, \( {f}^{\prime }\left( \lambda \right) \) is a \( {p}^{n} \) th power for all \( n \), hence \( {f}^{\prime }\left( \lambda \right) \in {\mathrm{U}}_{n}\left( \lambda \right) \) for all \( n \) ; as the \( {\mathrm{U}}_{n} \) form a Cauchy filter base, this implies the uniqueness of \( {f}^{\prime } \) as well as the fact that the intersection of the \( {\mathrm{U}}_{n} \) is non-empty and equal to \( f\left( \lambda \right) \) . This establishes (i) and (ii).
As for (iii), note that if \( x \) and \( y \) are \( {p}^{n} \) th powers for all \( n \), so is \( {xy} \) ; the same reasoning holds for (iv), taking into account that \( {\left( x + y\right) }^{{p}^{n}} = {x}^{{p}^{n}} + {y}^{{p}^{n}} \) if \( \mathrm{A} \) has characteristic \( p \) . | Yes |
Theorem 11.5.9. The map \( \Phi : U \times \mathfrak{u} \rightarrow G \) defined by \( \Phi \left( {u, X}\right) = u\exp \left( {\mathrm{i}X}\right) \), for \( u \in U \) and \( X \in \mathfrak{u} \), is a diffeomorphism onto \( G \) . In particular, \( U \) is connected. | Proof. By Lemma 11.5.8 we may assume that \( G \subset \mathbf{{GL}}\left( {n,\mathbb{C}}\right) \) and \( \tau \left( g\right) = {\left( {g}^{ * }\right) }^{-1} \) . If \( g \in G \) then \( {g}^{ * }g \) is positive definite. Since \( {\left( {g}^{ * }g\right) }^{m} \in G \) for all \( m \in \mathbb{Z} \), Lemma 11.5.6 implies that \( {\left( {g}^{ * }g\right) }^{s} \in G \) for all \( s \in \mathbb{R} \) . Also, \( s \mapsto {\left( {g}^{ * }g\right) }^{s} \) defines a one-parameter group of \( G \) as a real Lie group. Thus \( {\left( {g}^{ * }g\right) }^{s} = \exp \left( {sX}\right) \) for some \( X \in \mathfrak{g} \) by Theorem D.2.6. Clearly, \( \mathrm{d}\tau \left( X\right) = - X \) . Thus \( X \in \mathrm{{iu}} \) .
For \( g \in G \) define \( k\left( g\right) = g{\left( {g}^{ * }g\right) }^{-1/2} \) . Then
\[
k{\left( g\right) }^{ * } = {\left( {g}^{ * }g\right) }^{-1/2}{g}^{ * } = {\left( {g}^{ * }g\right) }^{-1/2}{g}^{ * }g{g}^{-1} = {\left( {g}^{ * }g\right) }^{1/2}{g}^{-1} = k{\left( g\right) }^{-1}.
\]
It is also evident that \( k\left( g\right) \in U \) . Thus the map \( \Phi \) in the theorem is surjective. If \( u\exp \left( {\mathrm{i}X}\right) = v\exp \left( {\mathrm{i}Y}\right) \) with \( u, v \in U \) and \( X, Y \in \mathfrak{u} \), then
\[
\exp \left( {2\mathrm{i}X}\right) = {\left( u\exp \left( \mathrm{i}X\right) \right) }^{ * }u\exp \left( {\mathrm{i}X}\right) = {\left( v\exp \left( \mathrm{i}Y\right) \right) }^{ * }v\exp \left( {\mathrm{i}Y}\right) = \exp \left( {2\mathrm{i}Y}\right) .
\]
Applying Lemma 11.5.6 yields \( \exp \left( {\mathrm{i}{tX}}\right) = \exp \left( {\mathrm{i}{tY}}\right) \) for all \( t \in \mathbb{R} \) . Thus \( X = Y \), and hence \( u = v \) . This proves that \( \Phi \) is injective.
If \( u \in U \) and \( X, Z, W \in \mathfrak{u} \), then
\[
\mathrm{d}{\Phi }_{\left( u, X\right) }\left( {Z, W}\right) = {uZ}\exp \left( {\mathrm{i}X}\right) + u\mathrm{\;d}{\Psi }_{\mathrm{i}X}\left( {\mathrm{i}W}\right)
\]
(notation as in Lemma 11.5.7). Now \( {Z}^{ * } = - Z \) and \( \mathrm{d}{\Psi }_{\mathrm{i}X}{\left( \mathrm{i}W\right) }^{ * } = \mathrm{d}{\Psi }_{\mathrm{i}X}\left( {\mathrm{i}W}\right) \) . Thus \( \mathrm{d}{\Phi }_{\left( u, X\right) } \) is injective (see the proof of Lemma 11.5.7). The theorem now follows from the inverse function theorem. | Yes |
Theorem 7. If \( M \) is orientable, then \(\chi \left( M\right) = 2 - {p}^{1}\left( M\right)\). If \( M \) is not orientable, then \(\chi \left( M\right) = 1 - {p}^{1}\left( M\right)\). | Proof. Let \( h \) be the number of handles in \( M \), and let \( m \) be the number of cross-caps, with \( 0 \leq m \leq 2 \). For \( m = 0 \), we have \(\chi \left( M\right) = 2 - {2h} = 2 - {p}^{1}\left( M\right)\). For \( m = 1,\chi \left( M\right) = 2 - \left( {{2h} + 1}\right),{p}^{1}\left( M\right) = {2h}\), and \(\chi \left( M\right) = 2 - \left\lbrack {{p}^{1}\left( M\right) + 1}\right\rbrack = 1 - {p}^{1}\left( M\right)\). For \( m = 2,\chi \left( M\right) = 2 - \left( {{2h} + 2}\right) = - {2h},{p}^{1}\left( M\right) = {2h} + 1\), and \(\chi \left( M\right) = 1 - {p}^{1}\left( M\right)\). | Yes |
Let \( Y \) be a regular CW complex structure on an n-manifold. Every cell of \( Y \) is a face of an \( n \) -cell of \( Y \) . Every \( \left( {n - 1}\right) \) -cell of \( Y \) is a face of at most two \( n \) -cells of \( Y \) . An \( \left( {n - 1}\right) \) -cell, \( e \), of \( Y \) is a face of exactly one \( n \) -cell of \( Y \) iff \( e \subset \partial Y \) . If \( e \) is a face of two \( n \) -cells of \( Y \), then \( \overset{ \circ }{e} \subset \overset{ \circ }{Y} \) . | We saw in Sect. 5.1 that every cell of \( Y \) has dimension \( \leq n \) . If some cell were not a face of an \( n \) -cell, there would be \( k < n \) and a \( k \) -cell \( \widetilde{e} \) of \( Y \) which is not a face of any higher-dimensional cell of \( Y \), implying \( \overset{ \circ }{e} \) open in \( Y \), contradicting 5.1.6(a). If the \( \left( {n - 1}\right) \) -cell \( e \) is a face of exactly one \( n \) -cell, then, since \( Y \) is regular, each \( x \in \overset{ \circ }{e} \) has a neighborhood in \( Y \) homeomorphic to \( {\mathbb{R}}_{ + }^{n} \) . Thus \( \overset{ \circ }{e} \subset \partial Y \), and, since \( \partial Y \) is closed in \( Y, e \subset \partial Y \) . On the other hand, if \( e \) is a face of two \( n \) -cells, then every \( x \in \overset{ \circ }{e} \) clearly has a neighborhood homeomorphic to \( {\mathbb{R}}^{n} \), so, by \( {5.1.6}\left( \mathrm{\;b}\right) ,\overset{ \circ }{e} \subset \overset{ \circ }{Y} \) . The proof that \( e \) is not a face of more than two \( n \) -cells is left as an exercise. | No |
Corollary 4.4.7. The product of two Hankel operators is 0 if and only if one of them is 0 . | If the product of two Hankel operators is the Toeplitz operator 0 , the previous corollary implies that at least one of the Hankel operators is zero. | No |
Let \( {a}^{ * } \in A \) and \( \widehat{A} \mathrel{\text{:=}} A \smallsetminus \left\{ {a}^{ * }\right\} \), and assume that \( G \) has no proper \( A \rightarrow B \) wave. Then \( {a}^{ * } \) is linkable for \( \left( {G,\widehat{A}, B}\right) \) . | Nu | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.