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We show now that \(\left\langle {{f}_{x}v;{sy}}\right\rangle = {\left( -1\right) }^{\deg x\deg y}\langle v;s\left\lbrack {x, y}\right\rbrack \rangle ,\;v \in V, y \in L.\) This formula exhibits \( {f}_{x} \) as the dual (up to sign) of ad \( x \) . In particular, \( {f}_{x} \) is locally conilpotent if and only if ad \( x \) is locally nilpotent.
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For the proof of (31.8) note that \( {f}_{x} = {d}_{1}{\eta }_{x} - {\left( -1\right) }^{\deg {\eta }_{x}}{\eta }_{x}{d}_{1} \), where \( {d}_{1} \) is the quadratic part of \( d \) . Then use \( §{21}\left( \mathrm{e}\right) \) to compute \(\left\langle {{f}_{x}v;{sy}}\right\rangle = - {\left( -1\right) }^{\deg {\eta }_{x}}\left\langle {{\eta }_{x}{d}_{1}v;{sy}}\right\rangle = {\left( -1\right) }^{\deg x}\left\langle {{d}_{1}v;{sy},{sx}}\right\rangle = {\left( -1\right) }^{\deg x\deg y}\langle v;s\left\lbrack {x, y}\right\rbrack \rangle .\)
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Yes
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Lemma 31.8 If some \( {E}_{i}\left( \sigma \right) \) is locally conilpotent then so is \( H\left( \sigma \right) \) .
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proof: If \( H\left( \sigma \right) \) is not locally conilpotent then there is an infinite sequence of non-zero classes \( {\alpha }_{k} \in H\left( M\right) \), such that \( H\left( \sigma \right) {\alpha }_{k + 1} = {\alpha }_{k}, k \geq 0 \) . Let \( p\left( k\right) \) be the greatest integer such that \( {\alpha }_{k} \) has a representing cocycle in \( {F}^{p\left( k\right) }M \) . Since \( p\left( k\right) \geq p\left( {k + 1}\right) \geq \cdots \) there is a \( p \) and a \( {k}_{0} \) such that \( p\left( k\right) = p, k \geq {k}_{0} \) .\n\nLet \( {Z}_{k} \) be the affine space of cocycles in \( {F}^{p} \) representing \( {\alpha }_{k}, k \geq {k}_{0} \) . Then \( {\sigma }^{n}\left( {Z}_{k + n}\right) \supset {\sigma }^{n + 1}\left( {Z}_{k + n + 1}\right) \supset \cdots \) is a decreasing sequence of finite dimensional affine subspaces and hence \( {A}_{k} = \mathop{\bigcap }\limits_{n}{\sigma }^{n}\left( {Z}_{k + n}\right) \) is non void subspace of \( {Z}_{k} \) . Exactly as in Lemma 31.7 it follows that \( \sigma \left( {A}_{k + 1}\right) = {A}_{k} \), and so there is a sequence of cocycles \( {z}_{k} \in {A}_{k} \) such that \( \sigma {z}_{k + 1} = {z}_{k} \) and \( {z}_{k} \) represents \( {\alpha }_{k} \) .\n\nSince \( p = p\left( k\right) \) and \( {z}_{k} \in {F}^{p},{z}_{k} \) represents a non-zero class \( \left\lbrack {z}_{k}\right\rbrack \in {E}_{i}^{p, * } \) . But clearly \( {E}_{i}\left( \sigma \right) \left\lbrack {z}_{k + 1}\right\rbrack = \left\lbrack {z}_{k}\right\rbrack, k \geq {k}_{0} \), and so if \( H\left( \sigma \right) \) is not locally conilpotent then \( {E}_{i}\left( \sigma \right) \) is not locally conilpotent either.
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Yes
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Lemma 31.9 With the notation and hypotheses above suppose ad \( x \) is locally nilpotent and \( {\mathrm{{hl}}}^{\prime }x \) is locally conilpotent. Then \( H\left( \theta \right) \) is locally conilpotent.
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proof: First observe that \( \theta \) preserves \( {\Lambda U} \) and also \( {\Lambda }^{ + }U \), because \( \left( {{\Lambda v} \otimes {\Lambda U}, D}\right) \) is a minimal Sullivan algebra. Next, let \( \eta \) be the derivation in \( {\Lambda v} \otimes {\Lambda U} \) defined by \( {\eta w} = \langle w;{sx}\rangle, w \in \mathbb{R}v \oplus U \) ; thus \( {\eta v} = 1 \) and \( \eta \left( U\right) = 0 \), and thus setting \( \xi = {D\eta } - {\left( -1\right) }^{\deg \eta }{\eta D} \) we have\n\n\[ \xi \left( {1 \otimes \Phi }\right) = 1 \otimes {\theta \Phi },\;\Phi \in {\Lambda U}. \]\n\nDefine \( f : U \rightarrow U \) by the condition \( \xi - f : U \rightarrow {\Lambda }^{ \geq 2}U \) . Then Example 2 states that \( f \) is dual (up to sign) to ad \( x \) . In particular \( f \) is locally conilpotent.\n\nNext, filter \( {\Lambda U} \otimes {\Lambda W} \) by the ideals \( {F}^{p} = {\Lambda }^{ \geq p}U \otimes {\Lambda W} \) . Since \( \theta \) preserves \( {\Lambda }^{ + }U \) it preserves this filtration and in the corresponding spectral sequence we have\n\n\[ {E}_{1}\left( \theta \right) = {\theta }_{f} \otimes {id} + {id} \otimes H\left( {\theta }_{v}\right) : {\Lambda U} \otimes H\left( {{\Lambda W},\bar{d}}\right) \rightarrow {\Lambda U} \otimes H\left( {{\Lambda W},\bar{d}}\right) ,\]\n\nwhere \( {\theta }_{f} \) is the derivation in \( {\Lambda U} \) extending \( f \) . Now \( H\left( {\theta }_{v}\right) \) is locally conilpotent by hypothesis, and we have just observed that so is \( f \), because ad \( x \) is locally nilpotent. Given a positive integer \( n \) choose \( N \) so that for \( k \geq 0 \) :\n\n\[ {U}^{ \leq n} \cap {f}^{k}\left( {U}^{ \geq N}\right) = 0\;\text{and}\;{H}^{ \leq n}\left( {{\Lambda W},\bar{d}}\right) \cap H{\left( {\theta }_{v}\right) }^{k}\left( {{H}^{ \geq N}\left( {{\Lambda W},\bar{d}}\right) }\right) = 0.\text{.}\]\n\nThen the formula above for \( {E}_{1}\left( \theta \right) \) implies that\n\n\[ {\left\lbrack {\Lambda }^{p}U \otimes H\left( \Lambda W,\bar{d}\right) \right\rbrack }^{n} \cap {E}_{1}{\left( \theta \right) }^{k}{\left\lbrack {\Lambda }^{p}U \otimes H\left( \Lambda W,\bar{d}\right) \right\rbrack }^{ \geq \left( {p + 1}\right) N} = 0,\;k \geq 0.\]\n\nHence \( {E}_{1}\left( \theta \right) \) is locally conilpotent and thus so is \( H\left( \theta \right) \), by Lemma 31.8.
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Yes
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Lemma 31.14 Suppose \( {\left( {x}_{n},{z}_{n}\right) }_{n \geq 0} \) is an infinite sequence of pairs of elements \( {x}_{n},{z}_{n} \in {\Lambda W} \) satisfying:\n\n\[ \bar{d}{z}_{n} = 0\;\text{ and }\;\theta {z}_{n + 1} = {z}_{n} + \bar{d}{x}_{n},\;n \geq 0. \]\n\nThen there is an infinite sequence of elements \( {y}_{n} \in {\Lambda W} \) such that\n\n\[ \bar{d}{y}_{n} = {z}_{n}\;\text{ and }\;\theta {y}_{n + 1} = {y}_{n} + {x}_{n},\;n \geq 0. \]
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proof: Observe first that each \( \left\lbrack {z}_{n}\right\rbrack \in \mathop{\bigcap }\limits_{k}\operatorname{Im}H{\left( \theta \right) }^{k} = 0 \), since \( \left\lbrack {z}_{n}\right\rbrack = H\left( \theta \right) \left\lbrack {z}_{n + 1}\right\rbrack = \) \( H{\left( \theta \right) }^{2}\left\lbrack {z}_{n + 2}\right\rbrack = \cdots \) . Thus the affine spaces \( {A}_{n} = {\bar{d}}^{-1}\left( {z}_{n}\right) \) are non-void and finite dimensional. Affine maps\n\n\[ \rightarrow {A}_{n + 1}\overset{{\alpha }_{n}}{ \rightarrow }{A}_{n} \rightarrow \cdots \overset{{\alpha }_{0}}{ \rightarrow }{A}_{0} \]\n\nare defined by \( {\alpha }_{n}\left( y\right) = {\theta y} - {x}_{n},\;y \in {A}_{n + 1} \) . For each \( n \) this gives the infinite decreasing sequence of affine spaces\n\n\[ {A}_{n} \supset \operatorname{Im}{\alpha }_{n} \supset \operatorname{Im}\left( {{\alpha }_{n} \circ {\alpha }_{n + 1}}\right) \supset \cdots \supset \operatorname{Im}\left( {{\alpha }_{n} \circ \cdots \circ {\alpha }_{n + k}}\right) \supset \cdots . \]\n\nFor dimension reasons this sequence must stabilize at some \( k\left( n\right) : \operatorname{Im}\left( {{\alpha }_{n} \circ \cdots \circ }\right. \) \( \left. {\alpha }_{n + p}\right) = \operatorname{Im}\left( {{\alpha }_{n} \circ \cdots {\alpha }_{n + k\left( n\right) }}\right) \) for \( p \geq k\left( n\right) \) . Put \( {E}_{n} = \operatorname{Im}\left( {{\alpha }_{n} \circ \cdots \circ {\alpha }_{n + k\left( n\right) }}\right) \) . Then \( {\alpha }_{n}\left( {E}_{n + 1}\right) = {E}_{n} \) since, for \( p \) sufficiently large, \( {\alpha }_{n}\left( {E}_{n + 1}\right) = {\alpha }_{n}\left( {\operatorname{Im}{\alpha }_{n + 1} \circ \cdots \circ }\right. \) \( \left. {\alpha }_{n + p}\right) = {E}_{n} \) . Thus we may choose an infinite sequence of elements \( {y}_{n} \in {E}_{n} \) such that \( {\alpha }_{n + 1}{y}_{n + 1} = {y}_{n}, n \geq 0 \) . Thus \( \theta \left( {y}_{n + 1}\right) = {y}_{n} + {x}_{n} \) . Since \( {E}_{n} \subset {A}_{n} = {\bar{d}}^{-1}\left( {z}_{n}\right) \) we also have \( \bar{d}{y}_{n} = {z}_{n}, n \geq 0 \) .
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Yes
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The fibration \( p : {S}^{{4m} + 3} \rightarrow \mathbb{H}{P}^{m} \).
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The unit sphere \( {S}^{3} \) of the quaternions \( \mathbb{H} \) acts freely by right multiplication on the unit sphere \( {S}^{{4m} + 3} \) of \( {\mathbb{H}}^{m + 1}\left( { \cong {\mathbb{R}}^{{4m} + 4}}\right) \) . This is the action of a classical principal \( {S}^{3} \) bundle ( \( §2\left( \mathrm{a}\right) \) ) \( p : {S}^{{4m} + 3} \rightarrow \mathbb{H}{P}^{m} \), with base space the quaternionic projective \( m \) -space.\n\nThe long exact rational homotopy sequence for this fibration reduces to\n\n\[ \n{\pi }_{{4m} + 3}\left( {S}^{{4m} + 3}\right) \otimes \mathbb{Q}\overset{ \cong }{ \rightarrow }{\pi }_{{4m} + 3}\left( {\mathbb{H}{P}^{m}}\right) \otimes \mathbb{Q}\;\text{ and }\;{\pi }_{4}\left( {\mathbb{H}{P}^{m}}\right) \otimes \mathbb{Q}\overset{ \cong }{ \rightarrow }{\pi }_{3}\left( {S}^{3}\right) \otimes \mathbb{Q}. \n\] \n\nThus \( \dim {\pi }_{ * }\left( {\mathbb{H}{P}^{m}}\right) \otimes \mathbb{Q} < \infty \) and \( \dim {\pi }_{\text{odd }}\left( {\mathbb{H}{P}^{m}}\right) \otimes \mathbb{Q} = 1 \) . Moreover, the holonomy representation is automatically locally nilpotent (but not trivial) since the fibre has finite rational homology.\n\nThus all the hypotheses but one of Theorem 31.10 are satisfied: \( {\partial }_{ * } \otimes \mathbb{Q} \neq 0 \) . Since \( {\operatorname{cat}}_{0}{S}^{{4m} + 3} < {\operatorname{cat}}_{0}{S}^{3} + \dim {\pi }_{\text{odd }}\left( {\mathbb{H}{P}^{m}}\right) \otimes \mathbb{Q} \) the conclusion of the theorem fails too, which shows that the hypothesis \( {\partial }_{ * } \otimes \mathbb{Q} = 0 \) is necessary. (Note, however that it intervenes only at the end, in Step 4, of the proof of Proposition 31.13).
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Yes
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The fibration associated with \( {S}^{3} \vee {S}^{3} \rightarrow {S}^{3} \) .
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Convert projection from \( {S}^{3} \vee {S}^{3} \) to the first factor to a fibration \( p : X \rightarrow {S}^{3} \) with \( X \simeq {S}^{3} \vee {S}^{3} \) . Since \( p \) is not a rational homology equivalence the fibre, \( F \), has non-trivial rational homology and so \( {\operatorname{cat}}_{0}F \geq 1 = {\operatorname{cat}}_{0}X \), and the conclusion of Theorem 31.10 fails. Here all the hypotheses hold, however, except for the hypothesis that the holonomy representation is locally nilpotent. This must therefore fail, and the example shows it is a necessary hypothesis.
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No
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Suppose \( \theta \) is a derivation of even (negative) degree in a commutative cochain algebra \( \left( {A,{d}_{A}}\right) \) such that \( {A}^{0} = \mathbb{R} \) and \( {H}^{1}\left( A\right) = 0 \) . Then we may construct the commutative cochain algebra \( \left( {{\Lambda v} \otimes A, d}\right) \), with \( \deg v + \deg \theta = 1 \) by setting \( {dv} = 0 \) and \( {da} = {d}_{A}a + v \otimes {\theta a}, a \in A \) .
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Now let \( \left( {{\Lambda W},\bar{d}}\right) \) be a minimal model of \( \left( {A, d}\right) \) . Then \( \left( {{\Lambda v} \otimes A, d}\right) \) has a Sullivan model of the form \( \left( {{\Lambda v} \otimes {\Lambda W}, d}\right) \) with \( {d\Phi } = 1 \otimes \bar{d}\Phi + v \otimes {\theta }^{\prime }\Phi \) . If \( \left( {{\Lambda v} \otimes {\Lambda W}, d}\right) \) is not a minimal Sullivan algebra then \( {\theta }^{\prime } \) restricts to a non-zero Gottlieb element \( {W}^{\deg v - 1} \rightarrow \mathbb{R} \) and Proposition 29.8(ii) asserts that \( \operatorname{cat}\left( {{\Lambda W},\bar{d}}\right) = \infty \) . On the other hand, the isomorphism \( H\left( {{\Lambda W},\bar{d}}\right) \cong H\left( {A,{d}_{A}}\right) \) identifies \( H\left( {\theta }^{\prime }\right) \) with \( H\left( \theta \right) \) . Thus if \( \bigcap \operatorname{Im}H{\left( \theta \right) }^{k} = 0 \) we can apply Proposition 31.13 to conclude that\n\n\[ \operatorname{cat}\left( {{\Lambda W},\bar{d}}\right) \text{ is }\left\{ \begin{array}{lll} \leq \operatorname{cat}\left( {{\Lambda v} \otimes {\Lambda W}, d}\right) - 1 & , & \text{ if }\left( {{\Lambda v} \otimes {\Lambda W}, d}\right) \text{ is minimal. } \\ \infty & , & \text{ otherwise. } \end{array}\right. \]
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Yes
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Example 4 A fibration \( X \rightarrow \mathbb{C}{P}^{m} \) with fibre \( {S}^{3} \) and \( {\operatorname{cat}}_{0}X = 2 \) .
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Let \( {S}^{1} \) act on \( {S}^{{2m} + 1} \) by complex multiplication: \( {S}^{1} \) is the unit circle of \( \mathbb{C} \) and \( {S}^{{2m} + 1} \) is the unit sphere in \( {\mathbb{C}}^{m + 1} \) . This is the action of a principal \( {S}^{1} \) -bundle, \( {S}^{{2m} + 1} \rightarrow \mathbb{C}{P}^{m}\left( {§2\left( \mathrm{\;d}\right) }\right) \) . For \( m = 1 \) we have the action of \( {S}^{1} \) on \( {S}^{3} \) and the associated bundle \( \left( {§2\left( \mathrm{e}\right) }\right) \)\n\n\[ p : X = {S}^{3}{ \times }_{{S}^{1}}{S}^{{2m} + 1} \rightarrow \mathbb{C}{P}^{m} \]\n\nis a Serre fibration with fibre \( {S}^{3} \) .\n\nSince \( {\pi }_{ * }\left( {\mathbb{C}{P}^{m}}\right) \otimes \mathbb{Q} \) is concentrated in degrees 2 and \( {2m} + 1 \) it follows for degree reasons that the rational connecting homomorphism is zero. Since the fibre has finite dimensional homology the holonomy representation is by locally nilpotent transformations. Thus Theorem 31.10 applies and gives \( {\operatorname{cat}}_{0}X \geq {\operatorname{cat}}_{0}{S}^{3} + 1 = 2 \) .\n\nBut we may also regard \( X \) as the total space of a Serre fibration \( X \rightarrow {S}^{2} \) with fibre \( {S}^{{2m} + 1} \), so that \( {\operatorname{cat}}_{0}X \leq 2 \) (Proposition 30.7). Altogether we have \( {\operatorname{cat}}_{0}X = 2 \) .
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Yes
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Example 1 A space with \( {e}_{0}X = 2 \) and \( {\operatorname{cat}}_{0}X = \infty \) .
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In Example 6, \( §{12}\left( \mathrm{\;d}\right) \), we constructed a minimal Sullivan model \( \left( {{\Lambda V}, d}\right) \) such that \( d : V \rightarrow {\Lambda }^{3}V \) and every cocycle in \( {\Lambda }^{ \geq 3}V \) is a coboundary. Moreover \( V = {V}^{ \geq 2} \) and has finite type and it is immediate from the construction that we may choose \( {V}^{\text{odd }} \) to be infinite dimensional.\n\nBecause every cocycle in \( {\Lambda }^{ \geq 3}V \) is a coboundary, \( e\left( {{\Lambda V}, d}\right) \leq 2 \) . Because there are no coboundaries in \( {\Lambda }^{2}V, e\left( {{\Lambda V}, d}\right) = 2 \) . Because \( d : {\Lambda }^{k}V \rightarrow {\Lambda }^{k + 3}V \) the Milnor-Moore spectral sequence collapses at the \( {E}_{3} \) -term. On the other hand, the homotopy Lie algebra is abelian, because the quadratic part of \( d \) is zero. Thus Theorem 31.18 shows that \( \operatorname{cat}\left( {{\Lambda V}, d}\right) = \infty \) .\n\nIn particular, if \( X \) is the geometric realization of \( \left( {{\Lambda V}, d}\right) \left( {§{17}}\right) \) then\n\n\[ \n{\operatorname{cat}}_{0}X = \operatorname{cat}\left( {{\Lambda V}, d}\right) = \infty \;\text{ and }\;{e}_{0}X = e\left( {{\Lambda V}, d}\right) = 2.\n\]\n\nFinally recall that the construction of \( \left( {{\Lambda V}, d}\right) \) begins with a graded subspace \( Z \subset V \) such that \( d\left( Z\right) = 0 \), and that \( V = Z \oplus W \) with \( d \) injective in \( W \) . Let \( I \) be the ideal \( {\Lambda }^{ \geq 2}V \oplus W \), and let \( \left( {A, d}\right) \) be the sub cochain algebra given by \( A = {\bigoplus }_{k}{\Lambda }^{2k}V \) . The inclusion \( \left( {A, d}\right) \rightarrow \left( {I \oplus \mathbb{R}, d}\right) \) is a quasi-isomorphism.\n\nMoreover, dividing by \( {\left( {A}^{ + }\right) }^{2} \) and by a complement of \( \ker d \) in \( {\Lambda }^{2}V \) defines a quasi-isomorphism \( \left( {A, d}\right) \overset{ \simeq }{ \rightarrow }H\left( A\right) \) and shows \( {H}^{ + }\left( A\right) \cdot {H}^{ + }\left( A\right) = 0 \) .\n\nThe surjection \( \left( {{\Lambda V}, d}\right) \rightarrow \left( {{\Lambda V}/I, d}\right) \) is a commutative representative for a continuous map \( q : X \rightarrow \bigvee {S}^{{n}_{\alpha }} \), where \( {S}^{{n}_{\alpha }} \) corresponds to a basis element of \( Z \) with degree \( {n}_{\alpha } \) . Moreover \( \left( {I \oplus \mathbb{R}}\right) \) is a commutative model for the cofibre of \( q \) as follows from a simple calculation using the Remark after Proposition 13.6. In view of the observations above \( I \oplus \mathbb{k} \) is the commutative model of a wedge of spheres (Example of \( §{12}\left( \mathrm{a}\right) \) ) and so we have constructed a cofibration\n\n\[ \n\mathop{\bigvee }\limits_{\alpha }{S}_{\mathbb{Q}}^{{n}_{\alpha }} \rightarrow X \rightarrow \mathop{\bigvee }\limits_{\beta }{S}_{\mathbb{Q}}^{{n}_{\beta }}\n\]
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Yes
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Proposition 32.1 If \( \left( {{\Lambda V}, d}\right) \) is a pure Sullivan algebra then \( H\left( {{\Lambda V}, d}\right) \) is finite dimensional if and only if \( {H}_{0}\left( {{\Lambda V}, d}\right) \) is finite dimensional.
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proof: Since \( {\Lambda V} \) is a finitely generated module over the (noetherian) polynomial algebra \( {\Lambda Q} \) any submodule is also finitely generated. Since \( d\left( {\Lambda Q}\right) = 0 \) , \( \ker d \) is a \( {\Lambda Q} \) -submodule of \( {\Lambda V} \) ; hence it is finitely generated. Thus \( H\left( {{\Lambda V}, d}\right) \) is finitely generated as a module over the subalgebra \( {H}_{0}\left( {{\Lambda V}, d}\right) \) . The Proposition follows.
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No
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Proposition 32.4 Let \( \left( {{\Lambda V}, d}\right) \) be a minimal Sullivan algebra in which \( V \) is finite dimensional and \( V = {V}^{ \geq 2} \) . Then the following conditions are equivalent:\n\n(i) \( \dim H\left( {{\Lambda V},{d}_{\sigma }}\right) < \infty \) .\n\n(ii) \( \dim H\left( {{\Lambda V}, d}\right) < \infty \) .\n\n(iii) \( \operatorname{cat}\left( {{\Lambda V}, d}\right) < \infty \) .
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proof: Since the odd spectral sequence converges from \( H\left( {{\Lambda V},{d}_{\sigma }}\right) \) to \( H\left( {{\Lambda V}, d}\right) \) the implication (i) \( \Rightarrow \) (ii) is immediate, while (ii) \( \Rightarrow \) (iii) follows from Corollary 1 to Proposition 29.3. To prove (iii) \( \Rightarrow \) (i) let \( {x}_{1},\ldots ,{x}_{n} \) be a basis of \( V \) such that \( \deg {x}_{1} \leq \deg {x}_{2} \leq \cdots \) and \( d{x}_{i} \in \Lambda \left( {{x}_{1},\ldots ,{x}_{i - 1}}\right) \) . We show first by induction on \( i \) that if \( \deg {x}_{i} \) is even then for some \( {N}_{i},{x}_{i}^{{N}_{i}} = {d}_{\sigma }{\Psi }_{i} \) .\n\nFor this, divide by the ideal \( {I}_{i} \) generated by \( {x}_{1},\ldots ,{x}_{i - 1} \) to obtain a quotient Sullivan algebra \( \left( {\Lambda \bar{V},\bar{d}}\right) \) . Then \( {x}_{i},\ldots ,{x}_{n} \) project to a basis \( \left( {\bar{x}}_{j}\right) \) of \( \bar{V} \), and \( \bar{d}{\bar{x}}_{i} = 0 \) . On the other hand, by the Mapping theorem \( {29.5}\left( {\Lambda \bar{V},\bar{d}}\right) \) has finite category. Thus there is an element \( \Phi \in \Lambda \left( {{x}_{i},\ldots ,{x}_{n}}\right) \) such that \( \bar{d}\bar{\Phi } = {\bar{x}}_{i}^{N} \) (some \( N) \), where \( \bar{\Phi } \) is the image of \( \Phi \) in \( \Lambda \bar{V} \) . Since \( \deg \Phi \) is odd we have \( \Phi = {\Phi }_{1} + {\Phi }_{3} + \cdots \) with \( {\Phi }_{j} \in {\Lambda }^{j}{V}^{\text{odd }} \otimes \Lambda {V}^{\text{even }} \) . It follows that \( {\left( \bar{d}\right) }_{\sigma }{\bar{\Phi }}_{1} = {\bar{x}}_{i}^{N} \), which implies that \( {d}_{\sigma }{\Phi }_{1} - {x}_{i}^{N} \in {I}_{i} \)\n\nDefine elements \( {y}_{j} \in V \) by setting \( {y}_{j} = {x}_{j} \) if \( \deg {x}_{j} \) is even and \( {y}_{j} = 0 \) if \( \deg {x}_{j} \) is odd. Since \( {\Phi }_{1} \in {V}^{\text{odd }} \otimes \Lambda {V}^{\text{even }} \) it follows that \( {d}_{\sigma }{\Phi }_{1} - {x}_{i}^{N} = \mathop{\sum }\limits_{{j < i}}{y}_{j}{\Omega }_{j} \) ,\n\nwith \( {\Omega }_{j} \in \Lambda {V}^{\text{even }} \) . The induction hypothesis gives \( {y}_{j}^{{N}_{j}} = {d}_{\sigma }{\Psi }_{j}, j < i \) . Since \( {d}_{\sigma }{\Omega }_{j} = 0 \) follows that some power of \( {x}_{i} \) is a \( {d}_{\sigma } \) -coboundary. This closes the induction.\n\nAs in \( §{32}\left( \mathrm{a}\right) \) put \( Q = {V}^{\text{even }} \) and \( P = {V}^{\text{odd }} \) . We have just shown that for a basis \( {y}_{j} \) of \( Q \) some power \( {y}_{j}^{{N}_{j}} \) is a \( {d}_{\sigma } \) -coboundary. It follows that \( {H}_{0}({\Lambda Q} \otimes \) \( \left. {{\Lambda P},{d}_{\sigma }}\right) = {\Lambda Q}/{\Lambda Q} \cdot {d}_{\sigma }\left( P\right) \) is finite dimensional. Hence \( H\left( {{\Lambda Q} \otimes {\Lambda P},{d}_{\sigma }}\right) \) is also finite dimensional (Proposition 32.1).
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Yes
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Example 1 \( \Lambda \left( {{a}_{2},{x}_{3},{u}_{3},{b}_{4},{v}_{5},{w}_{7};{da} = {dx} = 0,{du} = {a}^{2},{db} = {ax},{dv} = }\right. \) \( {ab} - {ux},{dw} = {b}^{2} - {vx}). \)
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Here subscripts denote degrees. The differential \( {d}_{\sigma } \) is given by \( {d}_{\sigma }a = {d}_{\sigma }b = \) \( {d}_{\sigma }x = 0,{d}_{\sigma }u = {a}^{2},{d}_{\sigma }v = {ab},{d}_{\sigma }w = {b}^{2} \) . Thus in \( H\left( {{\Lambda V},{d}_{\sigma }}\right) \) we have \( {\left\lbrack a\right\rbrack }^{2} = \) \( {\left\lbrack b\right\rbrack }^{2} = 0 \) and so \( \left( {{\Lambda V}, d}\right) \) is elliptic.
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Yes
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Example 3 Algebraic closure of \( \\mathbb{k} \) is necessary in Proposition 32.3.
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Indeed if \( \\mathbb{k} = \\mathbb{Q} \) the Sullivan algebra \( \\Lambda \\left( {{a}_{2},{b}_{2},{x}_{3};{dx} = {a}^{2} + {b}^{2}}\\right) \) admits no non-trivial morphism to \( \\mathbb{Q}\\left\\lbrack z\\right\\rbrack \), since we would have \( a \\mapsto {\\alpha z}, b \\mapsto {\\beta z} \) with \( \\alpha ,\\beta \\in \\mathbb{Q} \) satisfying \( {\\alpha }^{2} + {\\beta }^{2} = 0 \) .
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Yes
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A finite connected graph is n-colourable if each vertex can be assigned one of n distinct colours so that vertices connected by an edge have different colours.
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Indeed we lose no generality in assuming \\(\\mathbb{R}\\) is algebraically closed. If the graph is \\(n\\) -colourable identify the colours with the distinct \\({n}^{\\text{th }}\\) roots of unity \\({w}_{\\alpha }\\) and note by \\({w}_{\\alpha \\left( j\\right) }\\) the colour of the vertex \\({v}_{j}\\). If \\({w}_{\\alpha }\\) and \\({w}_{\\beta }\\) are distinct \\({n}^{\\text{th }}\\) roots of unity then \\(\\sum {w}_{\\alpha }^{s}{w}_{\\beta }^{n - s} = {w}_{\\alpha }^{n} - {w}_{\\beta }^{n}/{w}_{\\alpha } - {w}_{\\beta } = 0\\), and so a non-trivial morphism \\(\\left( {{\\Lambda Q} \\otimes {\\Lambda P}, d}\\right) \\rightarrow \\widetilde{\\mathbb{k}}\\left\\lbrack z\\right\\rbrack \\) is given by \\({y}_{j} \\mapsto {w}_{\\alpha \\left( j\\right) }z\\) and \\({x}_{i} \\mapsto 0\\). Conversely, given a non-trivial morphism \\(\\varphi : \\left( {{\\Lambda Q} \\otimes {\\Lambda P}, d}\\right) \\rightarrow \\mathbb{R}\\left\\lbrack z\\right\\rbrack \\) note that if \\({y}_{k}\\) and \\({y}_{\\ell }\\) correspond to the vertices of \\({e}_{i}\\) then \\(0 = \\varphi \\left( {d{x}_{i}}\\right) \\left( {\\varphi {y}_{k} - \\varphi {y}_{\\ell }}\\right) = \\) \\(\\sum {\\left( \\varphi {y}_{k}\\right) }^{s}{\\left( \\varphi {y}_{\\ell }\\right) }^{n - s}\\left( {\\varphi {y}_{k} - \\varphi {y}_{\\ell }}\\right) = {\\left( \\varphi {y}_{k}\\right) }^{n} - {\\left( \\varphi {y}_{\\ell }\\right) }^{n}\\). Since the graph is connected there is a single scalar \\(\\lambda \\) such that \\({\\left( \\varphi {y}_{j}\\right) }^{n} = \\lambda {z}^{n}\\), and since some \\(\\varphi {y}_{j} \\neq 0,\\lambda \\neq 0\\). Choose an \\({n}^{\\text{th }}\\) root of \\(\\lambda ,\\bar{\\lambda }\\), and define \\({w}_{\\alpha \\left( j\\right) }\\) by \\(\\varphi {y}_{j} = {w}_{\\alpha \\left( j\\right) }\\bar{\\lambda }z\\). This \\(n\\) -colours the graph.
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Yes
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Theorem 32.6 (Friedlander-Halperin [61] Suppose \( \left( {{\Lambda V}, d}\right) \) is an elliptic Sullivan algebra with formal dimension \( n \) and even and odd exponents \( {a}_{1},\ldots ,{a}_{q} \) and \( {b}_{1},\ldots ,{b}_{p} \) . Then\n\n(i) \( \mathop{\sum }\limits_{{i = 1}}^{p}\left( {2{b}_{i} - 1}\right) - \mathop{\sum }\limits_{{j = 1}}^{q}\left( {2{a}_{j} - 1}\right) = n \) .\n\n(ii) \( \mathop{\sum }\limits_{{j = 1}}^{q}2{a}_{j} \leq n \) .\n\n(iii) \( \mathop{\sum }\limits_{{i = 1}}^{p}\left( {2{b}_{i} - 1}\right) \leq {2n} - 1 \) .\n\n(iv) \( \dim {V}^{\text{even }} \leq \dim {V}^{\text{odd }} \leq \operatorname{cat}\left( {{\Lambda V}, d}\right) \) .
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proof: We have \( n = \mathop{\sum }\limits_{{i = 1}}^{p}\left( {2{b}_{i} - 1}\right) - \mathop{\sum }\limits_{{j = 1}}^{q}\left( {2{a}_{j} - 1}\right) \geq \mathop{\sum }\limits_{{i = 1}}^{p}\left( {2{b}_{i} - 1}\right) + q \geq p + q,\n\nwhere \( p = \dim {V}^{\text{odd }} \) and \( q = \dim {V}^{\text{even }} \) .
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Yes
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Corollary 2 If \( \left( {{\Lambda V}, d}\right) \) is an elliptic Sullivan algebra of formal dimension \( n \) then \( \dim V \leq n \) .
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proof: We have \( n = \mathop{\sum }\limits_{{i = 1}}^{p}\left( {2{b}_{i} - 1}\right) - \mathop{\sum }\limits_{{j = 1}}^{q}\left( {2{a}_{j} - 1}\right) \geq \mathop{\sum }\limits_{{i = 1}}^{p}\left( {2{b}_{i} - 1}\right) + q \geq p + q, \n\nwhere \( p = \dim {V}^{\text{odd }} \) and \( q = \dim {V}^{\text{even }} \) .
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Yes
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Lemma 32.7 \( \left( {{\Lambda V},{d}_{\sigma }}\right) \) and \( \left( {{\Lambda V}, d}\right) \) have the same formal dimension.
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proof: We argue by induction on \( \dim V \) . Write \( \left( {{\Lambda V}, d}\right) \) as a relative Sullivan algebra \( \left( {{\Lambda v} \otimes {\Lambda W}, d}\right) \) in which \( V = \mathbb{R}v \oplus W \) and \( v \) is an element in \( V \) of minimal degree. The Mapping theorem 29.5 asserts that the quotient Sullivan algebra \( \left( {{\Lambda W},\bar{d}}\right) \) has finite category; hence it too is elliptic (Proposition 32.4). Next observe that if \( m \) is the formal dimension of \( \left( {{\Lambda W},\bar{d}}\right) \) and \( n \) is the formal dimension of \( \left( {{\Lambda V}, d}\right) \) then \[ n = \left\{ \begin{array}{ll} m + \deg v & \text{ if }\deg v\text{ is odd } \\ m - \deg v + 1 & \text{ if }\deg v\text{ is even. } \end{array}\right. \] (32.8) Indeed if \( \deg v \) is odd (and therefore \( \geq 3 \) ) this follows from the long exact cohomology sequence associated with \[ 0 \rightarrow v \otimes \left( {{\Lambda W},\bar{d}}\right) \rightarrow \left( {{\Lambda v} \otimes {\Lambda W}, d}\right) \rightarrow \left( {{\Lambda W},\bar{d}}\right) \rightarrow 0. \] If \( \deg v \) is even extend \( \left( {{\Lambda V}, d}\right) \) to \( \left( {{\Lambda V} \otimes \Lambda \bar{v}, d}\right) \) by setting \( d\bar{v} = v \), and use the long exact cohomology sequence associated with \[ 0 \rightarrow \left( {{\Lambda V}, d}\right) \rightarrow \left( {{\Lambda V} \otimes \Lambda \bar{v}, d}\right) \rightarrow \left( {{\Lambda V}, d}\right) \otimes \bar{v} \rightarrow 0 \] to conclude that formal dimension \( \left( {{\Lambda V} \otimes \Lambda \bar{v}, d}\right) = n + \deg \bar{v} = n + \deg v - 1 \) . Then write \( \left( {{\Lambda V} \otimes \Lambda \bar{v}, d}\right) = \left( {{\Lambda v} \otimes \Lambda \bar{v} \otimes {\Lambda W}, d}\right) \) and observe that because \( H\left( {{\Lambda v} \otimes \Lambda \bar{v}, d}\right) = \) \( \mathbb{k} \) the surjection \( \left( {{\Lambda V} \otimes \Lambda \bar{v}, d}\right) \rightarrow \left( {{\Lambda W},\bar{d}}\right) \) is a quasi-isomorphism. This proves (32.8). The same argument applies to the relative Sullivan algebra \( \left( {{\Lambda v} \otimes {\Lambda W},{d}_{\sigma }}\right) \), and here the quotient Sullivan algebra \( \left( {{\Lambda W},{\bar{d}}_{\sigma }}\right) \) is just the pure Sullivan algebra associated with \( \left( {{\Lambda W},\bar{d}}\right) \) . By induction \( \left( {{\Lambda W},\bar{d}}\right) \) and \( \left( {{\Lambda W},{\bar{d}}_{\sigma }}\right) \) have the same formal dimension. Thus formula (32.8) for \( \left( {{\Lambda V}, d}\right) \) and for \( \left( {{\Lambda V},{d}_{\sigma }}\right) \) shows these have the same formal dimension too.
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Yes
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Lemma 32.11 If \( 2{a}_{1},\ldots ,2{a}_{q} \) are the degrees of a basis \( \left( {y}_{j}\right) \) of \( Q \) and if \( 2{b}_{1} - 1,\ldots ,2{b}_{p} - 1 \) are the degrees of a basis \( \left( {x}_{i}\right) \) of \( P \) then\n\n\[ \mathcal{U}\left( z\right) = \frac{\mathop{\prod }\limits_{{i = 1}}^{p}\left( {1 - {z}^{2{b}_{i}}}\right) }{\mathop{\prod }\limits_{{y = 1}}^{q}\left( {1 - {z}^{2{a}_{j}}}\right) } \]
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proof: Write \( {\Lambda Q} \otimes {\Lambda P} = {\Lambda V} \) . Clearly \( \mathcal{U} = {\mathcal{U}}_{\Lambda V} \) does not depend on the differential, and \( {\mathcal{U}}_{{\Lambda V} \otimes {\Lambda W}} = {\mathcal{U}}_{\Lambda V} \cdot {\mathcal{U}}_{\Lambda W} \) . Since \( {\Lambda V} = \Lambda {y}_{1} \otimes \cdots \otimes \Lambda {y}_{q} \otimes \Lambda {x}_{1} \otimes \cdots \otimes \Lambda {x}_{p} \) and since (trivially)\n\n\[ {\mathcal{U}}_{\Lambda {y}_{j}}\left( z\right) = \frac{1}{1 - {z}^{2{a}_{j}}}\;\text{ and }\;{\mathcal{U}}_{\Lambda {x}_{i}}\left( z\right) = 1 - {z}^{2{b}_{i}}, \]\nthe lemma follows.
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Yes
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Example 1 Simply connected finite \( H \) -spaces are rationally elliptic.
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If \( G \) is as in the title then its Sullivan model is an exterior algebra on a graded vector space \( {P}_{G} \) of finite dimension concentrated in odd degrees, and has zero differential (Example 3, \( §{12}\left( \mathrm{a}\right) \) ). The dimension of \( {P}_{G} \) is called the rank of \( G.▱ \)
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No
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Example 2 Simply connected compact homogeneous spaces \( G/K \) are rationally elliptic.
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Proposition 15.16 asserts that these spaces have a Sullivan model of the form \( \left( {\Lambda {V}_{{B}_{K}} \otimes \Lambda {P}_{G}, d}\right) \), where \( d = 0 \) in \( \Lambda {V}_{{B}_{K}},{V}_{{B}_{K}} \) is concentrated in even degrees, \( d\left( {P}_{G}\right) \subset \Lambda {V}_{{B}_{K}} \) and \( {P}_{G} \) and \( {V}_{{B}_{K}} \) are finite dimensional. Note that this is a pure Sullivan algebra.\n\nIn this example we have \( {\chi }_{\pi }\left( {G/K}\right) = \dim {V}_{{B}_{K}} - \dim {P}_{G} = \dim {P}_{K} - \dim {P}_{G} \) , (even though the Sullivan algebra may not be minimal). Thus\n\n\[ \n{\chi }_{\pi }\left( {G/K}\right) = \operatorname{rank}\left( K\right) - \operatorname{rank}\left( G\right) .\n\]
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Yes
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Suppose an \( r \) -torus \( T = {S}^{1} \times \cdots \times {S}^{1} \) ( \( r \) factors) acts smoothly and freely on a simply connected compact smooth manifold \( M \) . Then the projection \( M \rightarrow M/T \) onto the orbit space is a smooth principal bundle. Hence there is a classifying map \( M/T \rightarrow {BT} \) whose homotopy fibre is homotopy equivalent to \( M\left( {§2\left( \mathrm{e}\right) }\right) \) .
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Now assume \( M \) is rationally elliptic. Since \( {BT} = \mathbb{C}{P}^{\infty } \times \cdots \times \mathbb{C}{P}^{\infty } \) its homotopy groups are concentrated in degree 2, and since \( M/T \) is compact its homology is finite dimensional. Thus \( M/T \) is rationally elliptic. It is immediate from the long exact homotopy sequence that\n\n\[ \n{\chi }_{\pi }\left( {M/T}\right) = {\chi }_{\pi }\left( M\right) + {\chi }_{\pi }\left( {BT}\right) = {\chi }_{\pi }\left( M\right) + r.\n\]\n\nOn the other hand, Theorem 32.15 asserts that \( {\chi }_{\pi }\left( {M/T}\right) \leq 0 \) and so we conclude that\n\n\[ \nr \leq - {\chi }_{M}\n\]\n\ni.e., \( - {\chi }_{M} \) is an upper bound for the dimension of a free torus acting on \( M \) .
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Yes
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Proposition 33.8 Suppose \( {H}^{i}\left( {X;\mathbb{Q}}\right) = 0, i > {n}_{X} \) . Then the integers \( \dim {H}_{i}\left( {{\Omega X};\mathbb{Q}}\right) ,1 \leq i \leq 3\left( {{n}_{X} - 1}\right) \) determine whether \( X \) is rationally elliptic or rationally hyperbolic.
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proof: Theorem 33.3 asserts that \( X \) is rationally elliptic if and only if \( {\pi }_{j}\left( X\right) \otimes \) \( \mathbb{Q} = 0,2{n}_{X} \leq j < 3{n}_{X} - 1 \) . This only requires the calculation of \( {r}_{i},2{n}_{X} - 1 \leq \) \( i \leq 3{n}_{X} - 3 \) .
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No
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Proposition 33.10 The formal power series \( {P}_{\Omega X} \) and \( \sum {r}_{n}{z}^{n} \) have the same radius of convergence, \( R \) . Moreover\n\n(i) \( R = 1 \) if \( X \) is rationally elliptic and \( R < 1 \) if \( X \) is rationally hyperbolic.\n\n(ii) If \( X \) is rationally hyperbolic and if \( {H}^{i}\left( {X;\mathbb{Q}}\right) = 0, i > {n}_{X} \) then \( R < K < 1 \) for some constant \( K \) depending only on \( {n}_{X} \).
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proof: Write \( \sum {a}_{n}{z}^{n} \ll \sum {b}_{n}{z}^{n} \) if \( {a}_{n} \leq {b}_{n} \) for all \( n \) . Since\n\n\[ \sum {r}_{n}{z}^{n} \ll \frac{\mathop{\prod }\limits_{n}{\left( 1 + {z}^{{2n} + 1}\right) }^{{r}_{{2n} + 1}}}{\mathop{\prod }\limits_{n}{\left( 1 - {z}^{2n}\right) }^{{r}_{2n}}} \ll {e}^{\frac{\mathop{\sum }\limits_{{r}_{n}}{z}^{n}}{1 - z}} \]\n\nit follows by (33.7) that \( \sum {r}_{n}{z}^{n} \) is convergent if and only if \( {P}_{\Omega X} \) is convergent.\n\nIf \( X \) is rationally elliptic the products in (33.7) are finite and \( R = 1 \) . If \( X \) is rationally hyperbolic let \( {\operatorname{cat}}_{0}X = m \) . The proof of Theorem 33.2 shows that given \( N \) sufficiently large there is an infinite sequence of integers \( {N}_{i} \) such that \( N = {N}_{0},{N}_{i} \leq {N}_{i + 1} \leq \left( {m + 1}\right) {N}_{i} \) and\n\n\[ {\left( \mathop{\sum }\limits_{{j = {N}_{i}}}^{{2{N}_{i} - 2}}\dim {\pi }_{j}\left( X\right) \otimes \mathbb{Q}\right) }^{\frac{1}{{N}_{i}}} \geq {\left( \frac{\mathop{\sum }\limits_{{j = N}}^{{{2N} - 2}}\dim {\pi }_{j}\left( X\right) \otimes \mathbb{Q}}{{c}_{m}^{3}}\right) }^{\frac{1}{N}}, \]\n\nwhere \( {c}_{m} \) is a constant depending only on \( m \) . It is also shown that for \( N \) sufficiently large the right hand side is larger than one. It follows that \( \lim \sup {r}_{n}^{1/n} > \) 1. Moreover, Theorem 33.3 states that if \( {H}^{i}\left( {X;\mathbb{Q}}\right) = 0, i > {n}_{X} \), then when \( N = 2{n}_{X}{c}_{{n}_{X}}^{3} \) the right hand side is at least \( {2}^{1/N} \), from which assertion (ii) follows at once.
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Yes
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Example 2 \( X = {S}^{3} \vee {S}^{3} \) .
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As in Example 1, \( {P}_{\Omega X} = \frac{1}{1 - 2{z}^{2}} \) and hence \( {H}_{ * }\left( {{\Omega X};\mathbb{Q}}\right) \) is concentrated in even degrees. Thus formula (33.7) becomes\n\n\[ \frac{1}{1 - 2{z}^{2}} = \frac{1}{\mathop{\prod }\limits_{n}{\left( 1 - {z}^{2n}\right) }^{{r}_{2n}}} \]\n\nwhere \( {r}_{2n} = \dim {\pi }_{2n}\Omega \left( {{S}^{3} \vee {S}^{3}}\right) \otimes \mathbb{Q} \). Take logs of both sides to obtain\n\n\[ \mathop{\sum }\limits_{k}\frac{{2}^{k}{z}^{2k}}{k} = \mathop{\sum }\limits_{n}{r}_{2n}\mathop{\sum }\limits_{k}\frac{{z}^{2nk}}{k}. \]\n\nEquating the coefficients of \( {z}^{2N} \) gives \( \mathop{\sum }\limits_{{d \mid N}}{r}_{2d}d = {2}^{N} \). \n\nThe Möbius function \( \mu \left( n\right) \) is defined by: \( \mu \left( n\right) = 1,{\left( -1\right) }^{r} \) or 0 as \( n \) is 1, a product of \( r \) distinct primes or divisible by a prime squared. Elementary number theory gives\n\n\[ {r}_{2N} = \frac{1}{N}\mathop{\sum }\limits_{{d \mid N}}\mu \left( \frac{N}{d}\right) {2}^{d} \]\n\nas a precise formula for the dimension of \( {\pi }_{2N}\left( {\Omega X}\right) \otimes \mathbb{Q} \). As observed above, \( {\pi }_{\text{odd }}\left( {\Omega X}\right) \otimes \mathbb{Q} = 0. \)
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Yes
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Lemma 34.1 If \( M \) or \( N \) is \( {UL} \) -free then \( M \otimes N \) is \( {UL} \) -free.
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proof: Suppose \( N \) is \( {UL} \) -free on a basis \( {a}_{\alpha } \) . Since \( M \otimes N = {\bigoplus }_{\alpha }M \otimes \left( {{a}_{\alpha } \cdot {UL}}\right) \) it is sufficient to prove that \( M \otimes {UL} \) is \( {UL} \) -free. Let \( {F}_{p} \subset {UL} \) be the linear span of elements of the form \( {x}_{{i}_{1}}\cdots \cdots {x}_{{i}_{q}},{x}_{{i}_{\nu }} \in L, q \leq p \) . If \( m \in M \) and \( a \in {F}_{p} \) then \( \left( {m \otimes 1}\right) \cdot a - m \otimes a \in M \otimes {F}_{p - 1} \) . It follows that \( M \otimes {UL} \) is \( {UL} \) -free on a basis \( {m}_{\lambda } \otimes 1 \) where \( {m}_{\lambda } \) is any \( \mathbb{k} \) -basis of \( M \) .\n\nThe proof when \( M \) is \( {UL} \) -free is identical.
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Yes
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Example 1 \( {\operatorname{Tor}}_{1}^{UI}\left( {\mathbb{k},\mathbb{k}}\right) \cong s\left( {I/\left\lbrack {I, I}\right\rbrack }\right) . \)
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Again let \( I \subset L \) be an ideal. Denote by \( \left\lbrack {I, I}\right\rbrack \) the ideal in \( L \) which is the linear span of the elements \( \left\lbrack {y, z}\right\rbrack, y, z \in I \) . If \( x \in I \) denote by \( \left( x\right) \) the image of \( {sx} \) in the suspension \( s\left( {I/\left\lbrack {I, I}\right\rbrack }\right) \) . We shall establish an isomorphism\n\n\[ \n{\operatorname{Tor}}_{1}^{UI}\left( {\mathbb{k},\mathbb{k}}\right) \cong s\left( {I/\left\lbrack {I, I}\right\rbrack }\right) \n\] \n\nand show that the representation of \( L/I \) in \( {\operatorname{Tor}}_{1}^{UI}\left( {\mathbb{R},\mathbb{R}}\right) \) corresponds under this isomorphism to the representation \n\n\[ \n\left( x\right) \cdot y = {\left( -1\right) }^{\deg y}\left( \left\lbrack {x, y}\right\rbrack \right) . \n\] \n\nIn fact, in \( §{22}\left( \mathrm{\;b}\right) \) we constructed a \( {UI} \) -free resolution \( {P}_{ * } \) (Propositions 22.3 and 22.4) of \( \mathbb{k} \) explicitly, of the form \n\n\[ \n\mathbb{k} \leftarrow {UI}\overset{d}{ \leftarrow }{sI} \otimes {UI}\overset{d}{ \leftarrow }{\Lambda }^{2}{sI} \otimes {UI}\overset{d}{ \leftarrow }\cdots . \n\] \n\nApply \( - { \otimes }_{UI}\mathbb{k} \) and note from the definition that in this quotient \( d\left( {{sx} \land {sy}}\right) = \) \( {\left( -1\right) }^{\deg {sx}}s\left\lbrack {x, y}\right\rbrack \) . Thus \n\n\[ \n{\operatorname{Tor}}_{1}^{UI}\left( {\mathbb{R},\mathbb{R}}\right) = {sI}/d{\Lambda }^{2}{sI} = s\left( {I/\left\lbrack {I, I}\right\rbrack }\right) . \n\]
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Yes
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The representation of \( L/I \) in \( {\operatorname{Tor}}_{q}^{UI}\left( {\mathbb{R},\mathbb{R}}\right) \).
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As in Example 1 the inclusion of \( {P}_{ * } \) in \( {Q}_{ * } \) is a quasi-isomorphism of \( {UI} \) -free resolutions of \( \mathbb{R} \), and in particular applying \( - { \otimes }_{UI}\mathbb{R} \) gives a quasi-isomorphism \( \left( {{\Lambda sI},\bar{d}}\right) \overset{ \simeq }{ \rightarrow }{Q}_{ * }{ \otimes }_{UI}\mathbb{k} \) . Let \( a \in {\Lambda }^{q}{sI} \) be a cycle (representing a class \( \alpha \in \) \( {\operatorname{Tor}}_{q}^{UI}\left( {\mathbb{R},\mathbb{R}}\right) \) and let \( y \in L \) . Then the formulae of \( §{22}\left( \mathrm{\;b}\right) \) show that in \( {Q}_{ * }{ \otimes }_{UI}\mathbb{R} \) ,\n\n\[ \bar{d}\left( {a \land {sy} \otimes 1}\right) = a \cdot y \otimes 1 + {\left( -1\right) }^{\deg a}a \otimes y, \]\n\nwhere \( a \cdot y \) is defined by\n\n\[ \left( {s{x}_{1} \land \cdots \land s{x}_{q}}\right) \cdot y = {\left( -1\right) }^{\deg \left( {s{x}_{1} \land \cdots \land s{x}_{q}}\right) }\mathop{\sum }\limits_{{i = 1}}^{q}{\left( -1\right) }^{\deg y\deg \left( {s{x}_{i + 1} \land \cdots \land s{x}_{q}}\right) }s{x}_{1} \land \cdots \land \]\n\n\( s\left\lbrack {{x}_{i}, y}}\right\rbrack \land \cdots \land s{x}_{q}. \)\n\nIt follows that the representation of \( L/I \) in \( {\operatorname{Tor}}_{q}^{UI}\left( {\mathbb{R},\mathbb{R}}\right) \) is given by\n\n\[ \alpha \cdot y = {\left( -1\right) }^{\deg \alpha + 1}\left\lbrack {a \cdot y}\right\rbrack . \]
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Yes
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Lemma 34.4 If \( S \) is a right L-module then \( \varphi \) restricts to an isomorphism\n\n\[ \n{\operatorname{Hom}}_{UL}\left( {M \otimes S, N}\right) \overset{ \cong }{ \rightarrow }{\operatorname{Hom}}_{{UL}/I}\left( {M,{\operatorname{Hom}}_{UI}\left( {S, N}\right) }\right) .\n\]
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proof: It is immediate that \( \varphi \) is \( L \) -linear. Thus if \( f \in {\operatorname{Hom}}_{UL}\left( {M \otimes S, N}\right) \) we have \( f \cdot x = 0, x \in L \) and so \( \left( {\varphi f}\right) \cdot x = 0 \) . Thus \( {\varphi f} \) is \( L \) -linear. For \( x \in I \) and \( m \in M \) it follows that \( \left( {\varphi f}\right) \left( m\right) \cdot x = {\varphi f}\left( {m \cdot x}\right) = 0 \) . Thus \( {\varphi f}\left( m\right) \) is \( I - \) linear; i.e. \( \varphi : {\operatorname{Hom}}_{UL}\left( {M \otimes S, N}\right) \rightarrow {\operatorname{Hom}}_{{UL}/I}\left( {M,{\operatorname{Hom}}_{UI}\left( {S, N}\right) }\right) \) . Conversely if \( f \in \operatorname{Hom}\left( {M \otimes S, N}\right) \) and if \( {\varphi f} \in {\operatorname{Hom}}_{{UL}/I}\left( {M,{\operatorname{Hom}}_{UI}\left( {S, N}\right) }\right) \) then \( {\varphi f} \) is \( L \) -linear and hence so is \( f \) .
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Yes
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(i) \( W \otimes {\operatorname{Ext}}_{UI}^{q}\left( {\mathbb{k},{UI}}\right) \) is a free \( {UL}/I \) -module on \( 1 \otimes {\operatorname{Ext}}_{UI}^{q}\left( {\mathbb{k},{UI}}\right) \) .
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proof: Denote \( {\operatorname{Ext}}_{UI}^{q}\left( {\mathbb{k},{UI}}\right) \) by \( {E}^{q} \) . \n\n(i) An \( L/I \) -linear map \( \theta \) from the free \( L/I \) -module \( {E}^{q} \otimes {UL}/I \) to \( W \otimes {E}^{q} \) is given by \n\n\[ \n\theta \left( {\Phi \otimes a}\right) = \left( {1 \otimes \Phi }\right) \cdot a,\;\Phi \in {E}^{q}, a \in {UL}/I. \n\] \n\n(We identify \( {UL}/I = W \) .) It follows from formula (34.6) that \( \theta \left( {\Phi \otimes a}\right) - \) \( {\left( -1\right) }^{\deg \Phi \deg a}a \otimes \Phi \in {W}_{ < \deg a} \otimes {E}^{q} \) . This implies that \( \theta \) is an isomorphism.
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Yes
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Proposition 34.9 With the hypotheses and notation above assume the Lie algebra \( L/I \) is finitely generated and that \( \alpha \cdot {UL}/I \) is finite dimensional for each \( \alpha \in {\operatorname{Tor}}_{q}^{UI}\left( {N,\mathbf{k}}\right) \) . Then there is an isomorphism of \( {UL}/I \) -modules\n\n\[{\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UL\right) }^{\sharp }}\right) \cong {\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UI\right) }^{\sharp }}\right) \otimes {\left( UL/I\right) }^{\sharp }\]\n\nwhere \( L/I \) acts on the right via the action in \( {\left( UL/I\right) }^{\sharp } \) . In particular,\n\n\[{\mathrm{{Tor}}}_{p}^{{UL}/I}\left( {\mathbb{R},{\mathrm{{Tor}}}_{q}^{UI}\left( {N,{\left( UL\right) }^{\sharp }}\right) }\right) \cong {\mathrm{{Tor}}}_{p}^{{UL}/I}\left( {\mathbb{R},{\left( UL/I\right) }^{\sharp }}\right) \otimes {\mathrm{{Tor}}}_{q}^{UI}\left( {N,{\left( UI\right) }^{\sharp }}\right) \;.\]
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proof of Proposition 34.9: Dualize the inclusion \( {UI} \rightarrow {UL} \) to a surjection \( {\left( UL\right) }^{\sharp } \rightarrow {\left( UI\right) }^{\sharp } \) of \( {UI} \) -modules. This induces a linear map\n\n\[f : {\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UL\right) }^{\sharp }}\right) \rightarrow {\operatorname{Tor}}_{q}^{UL}\left( {N,{\left( UI\right) }^{\sharp }}\right) ,\]\n\nand (34.8) shows that \( f \) is surjective.\n\nRight multiplication in \( {UL}/I \) makes \( \operatorname{Hom}\left( {{UL}/I, - }\right) \) into a left (and thus right) \( {UL}/I \) -module. Define a morphism\n\n\[F : {\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UL\right) }^{\sharp }}\right) \rightarrow \operatorname{Hom}\left( {{UL}/I,{\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UI\right) }^{\sharp }}\right) }\right)\]\n\nof \( {UL}/I \) -modules by setting \( F\left( \alpha \right) \left( a\right) = f\left( {\alpha \cdot a}\right), a \in {\operatorname{Tor}}_{q}^{UI}\left( {\left( {N,{\left( UL\right) }^{\sharp }}\right), a \in }\right. \) \( {UL}/I \) . Recall that an inclusion\n\n\[\Phi : {\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UI\right) }^{\sharp }}\right) \otimes {\left( UL/I\right) }^{\sharp } \rightarrow \operatorname{Hom}\left( {{UL}/I,{\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UI\right) }^{\sharp }}\right) }\right)\]\n\nof \( {UL}/I \) -modules is given by \( \Phi \left( {\alpha \otimes g}\right) \left( a\right) = \langle g, a\rangle \alpha \) . the image of \( \Phi \) consists of the linear maps \( \varphi : {UL}/I \rightarrow {\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UI\right) }^{\sharp }}
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Yes
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Lemma 34.10 With the hypotheses of Proposition 34.9, \( \beta \cdot {UL}/I \) is finite dimensional for all \( \beta \in {\operatorname{Tor}}_{q}^{UI}\left( {N,{\left( UL\right) }^{\sharp }}\right) \) .
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proof: Write \( M = {\left( UL\right) }^{\sharp } \) . Then \( M = {M}_{ < 0} \) is the union of the submodules \( {M}_{ \geq - p} \) . It is thus sufficient to prove the lemma for \( \beta \in {\operatorname{Tor}}_{q}^{UI}\left( {N,{M}_{ \geq - p}}\right) \) . Consider the exact sequence\n\n\[ \n{\operatorname{Tor}}_{q}^{UI}\left( {N,{M}_{ > - p}}\right) \rightarrow {\operatorname{Tor}}_{q}^{UI}\left( {N,{M}_{ \geq - p}}\right) \rightarrow {\operatorname{Tor}}_{q}^{UI}\left( {N,{M}_{-p}}\right)\n\]\n\nwhere \( {UL} \) acts trivially in \( {M}_{-p} \) . By hypothesis the lemma holds for this trivial \( {UL} \) -module. Hence for \( \beta \in {\operatorname{Tor}}_{q}^{UI}\left( {N,{M}_{ \geq - p}}\right) \) and some \( n \geq 0,\beta \cdot {\left( UL/I\right) }_{ \geq n} \subset \) Image \( {\operatorname{Tor}}_{q}^{UI}\left( {N,{M}_{ > - p}}\right) \) . Because \( L/I \) is a finitely generated Lie algebra each \( {\left( UL/I\right) }_{ \geq n} \) is a finitely generated \( {UL}/I \) -module. Hence so is \( \beta \cdot {\left( UL/I\right) }_{ \geq n} \) . By induction on \( p \) it follows that \( \beta \cdot {\left( UL/I\right) }_{ \geq n} \) is finite dimensional. Hence so is \( \beta \cdot {UL}/I \) .
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Yes
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Lemma 35.1 Suppose \( {P}_{ * }\overset{ \simeq }{ \rightarrow }M \) is an \( A \) -projective resolution of an \( A \) -module \( M \) and suppose \( {Q}_{ * } = {\left\{ {Q}_{i}\right\} }_{0 \leq i \leq m} \) is a complex of free \( A \) -modules. Then\n\n\[ \n{H}_{i, * }\left( {{\operatorname{Hom}}_{A}\left( {{P}_{ * },{Q}_{ * }}\right) }\right) = 0,\;i > m - {\operatorname{projgrade}}_{A}M.\n\]
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proof: Set \( {Q}_{ * }^{\prime } = {\left\{ {Q}_{i}\right\} }_{0 \leq i \leq m - 1} \) . Because the \( {P}_{i} \) are \( A \) -projective the sequence\n\n\[ \n0 \rightarrow {\operatorname{Hom}}_{A}\left( {{P}_{ * },{Q}_{ * }^{\prime }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {{P}_{ * },{Q}_{ * }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {{P}_{ * },{Q}_{m}}\right) \rightarrow 0\n\]\n\nis exact. Since \( {Q}_{m} \) is \( A \) -free, \( {H}_{i, * }\left( {{\operatorname{Hom}}_{A}\left( {{P}_{ * },{Q}_{m}}\right) }\right) = {\operatorname{Ext}}_{A}^{m - i}\left( {M,{Q}_{m}}\right) = 0 \) , \( i > m - {\operatorname{projgrade}}_{A}M \) . By induction on \( m,{H}_{i, * }\left( {{\operatorname{Hom}}_{A}\left( {{\ddot{P}}_{ * },{Q}_{ * }^{\prime }}\right) }\right) = 0, i > \) \( m \) - proj grade \( {}_{A}M \) . The lemma follows.
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Yes
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Lemma 35.4 Let \( A \subset X \) be an inclusion of \( {\Omega Y} \) -spaces and give \( \left( {X, A}\right) \times {\Omega Y} \) the diagonal action, where \( {\Omega Y} \) acts by right multiplication on \( {\Omega Y} \) . If \( {H}_{ * }\left( {X, A}\right) \) is \( \mathbb{k} \) -free then \( {H}_{ * }\left( {\left( {X, A}\right) \times {\Omega Y}}\right) \) is \( {H}_{ * }\left( {\Omega Y}\right) \) -free.
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proof: If \( \gamma \in {\Omega Y} \) is a loop of length \( \ell \) let \( {\gamma }^{\prime } \) be the loop of length \( \ell \) given by \( {\gamma }^{\prime }\left( t\right) = \gamma \left( {\ell - t}\right) ,0 \leq t \leq \ell \) . Then \( \gamma \mapsto \gamma {\gamma }^{\prime } \) and \( \gamma \mapsto {\gamma }^{\prime }\gamma \) are homotopically constant maps. Thus the map\n\n\[ f : X \times {\Omega Y} \rightarrow X \times {\Omega Y},\;f\left( {x,\gamma }\right) = \left( {x \cdot \gamma ,\gamma }\right) \]\n\nis a homotopy equivalence with homotopy inverse \( \left( {x,\gamma }\right) \mapsto \left( {x \cdot {\gamma }^{\prime },\gamma }\right) \) .\n\nDenote by \( {\left( X \otimes \Omega Y\right) }_{\Delta } \) and \( {\left( X \times \Omega Y\right) }_{R} \) the \( {\Omega Y} \) -spaces in which \( {\Omega Y} \) acts respectively diagonally and by right multiplication on \( {\Omega Y} \) . The isomorphism (cf. \( §{35}\left( \mathrm{\;b}\right) ){H}_{ * }\left( {X, A}\right) \otimes {H}_{ * }\left( {\Omega Y}\right) \cong {H}_{ * }{\left( \left( X, A\right) \times \Omega Y\right) }_{R} \) identifies this latter as the free \( {H}_{ * }\left( {\Omega Y}\right) \) -module with basis a \( \mathbb{k} \) -basis of \( {H}_{ * }\left( {X, A}\right) \) . On the other hand, \( f : {\left( X \times \Omega Y\right) }_{R} \rightarrow {\left( X \times \Omega Y\right) }_{\Delta } \) is a map of \( {\Omega Y} \) -spaces and \( {H}_{ * }\left( f\right) \) is therefore an isomorphism of \( {H}_{ * }\left( {\Omega Y}\right) \) -modules.
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Yes
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Proposition 35.8 The sequence (35.7) is an \( {H}_{ * }\left( {\Omega Y}\right) \) -free resolution of \( \mathbf{k} \) ; i.e., the Milnor resolution is an Eilenberg-Moore resolution (§20(d)).
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proof: We need only show (35.7) is exact. Filter \( {C}_{ * }\left( {\left( \Omega Y\right) }^{*\infty }\right) \) by the submodules \( {C}_{ * }\left( {\left( \Omega Y\right) }^{*n}\right) \) . Then the quasi-isomorphism \( \varphi : V \otimes {C}_{ * }\left( {\Omega Y}\right) \rightarrow {C}_{ * }\left( {\left( \Omega Y\right) }^{*\infty }\right) \) preserves filtrations. In the construction of \( \varphi \) we showed that each \( H\left( {\bar{\varphi }\left( n\right) }\right) \) was an isomorphism. This means precisely that \( \varphi \) induces an isomorphism between the \( {E}^{1} \) -terms of the spectral sequences. It is thus sufficient to prove that\n\n\[ k \leftarrow {H}_{ * }\left( {\Omega Y}\right) \overset{{d}_{1}}{ \leftarrow }\cdots \overset{{d}_{1}}{ \leftarrow }{H}_{ * }\left( {{\left( \Omega Y\right) }^{*\left( {n + 1}\right) },{\left( \Omega Y\right) }^{*n}}\right) \overset{{d}_{1}}{ \leftarrow } \]\n\nis exact.\n\nSuppose by induction that\n\n\[ 0 \leftarrow \mathbb{k} \leftarrow {H}_{ * }\left( {\Omega Y}\right) \overset{{d}_{1}}{ \leftarrow }\cdots \overset{{d}_{1}}{ \leftarrow }{H}_{ * }\left( {{\left( \Omega Y\right) }^{*n},{\left( \Omega Y\right) }^{*\left( {n - 1}\right) }}\right) \]\n\nis exact. A simple spectral sequence argument then shows that any \( {d}_{1} \) -cycle, \( \alpha \) , in\n\n\( {H}_{ * }\left( {{\left( \Omega Y\right) }^{*n},{\left( \Omega Y\right) }^{*\left( {n - 1}\right) }}\right) \) is the image of some class \( \beta \in {H}_{ * }\left( {\left( \Omega Y\right) }^{*n}\right) \) . But \( {\left( \Omega Y\right) }^{*n} \) is contractible in \( {\left( \Omega Y\right) }^{*\left( {n + 1}\right) } \) . Thus a representing cycle, \( z \), for \( \beta \) has the form \( z = {dw} \) for some \( w \in {C}_{ * }\left( {\left( \Omega Y\right) }^{*\left( {n + 1}\right) }\right) \) . In particular, \( w \) represents a class \( \gamma \in {H}_{ * }\left( {{\left( \Omega Y\right) }^{*\left( {n + 1}\right) },{\left( \Omega Y\right) }^{*n}}\right) \) and \( {d}_{1}\gamma = \alpha \) . This closes the induction and completes the proof.
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Yes
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Theorem 35.10 If \( \left( {X,{x}_{0}}\right) \) is a normal path connected topological space and if each \( {H}_{i}\left( {\Omega X}\right) \) is \( \mathbb{k} \) -free on a finite basis then\n\n\[ \n\text{depth}{H}_{ * }\left( {\Omega X}\right) \leq \operatorname{cat}X\text{.\n\]\n\nIf equality holds then also \( \operatorname{cat}X = \operatorname{gl}\dim {H}_{ * }\left( {\Omega X}\right) \) .
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proof: Replace \( X \) by a well based space of the same homotopy type by adjoining an interval to \( X \) at the base point \( {x}_{0} \) . Then apply the Corollary above to Theorem 35.9 to \( f = i{d}_{X} \) .
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No
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Proposition 35.11 Let \( L = {\left\{ {L}_{i}\right\} }_{i > 1} \) be a graded Lie algebra with each \( {L}_{i} \) finite dimensional.\n\n(i) \( \operatorname{gl}\dim {UL} \) is the largest integer \( n \) (as \( \infty \) ) such that \( {\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{R},\mathbb{R}}\right) \neq 0 \) .\n\n(ii) \( \operatorname{depth}{UL} \leq \operatorname{gldim}{UL} \) .
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proof: (i) Clearly \( n \leq \operatorname{gl}\dim {UL} \), since \( {\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{k},\mathbb{k}}\right) \neq 0 \) . On the other hand, write \( {P}_{ * } = {C}_{ * }\left( L\right) \otimes {UL} \) . Then\n\n\[ \n{\operatorname{Ext}}_{UL}\left( {\mathbb{R},\mathbb{R}}\right) = H\left( {{\operatorname{Hom}}_{UL}\left( {{P}_{ * },\mathbb{R}}\right) }\right) = H\left( {{C}_{ * }{\left( L\right) }^{\sharp }}\right) = H{\left( {C}_{ * }\left( L\right) \right) }^{\sharp }, \n\]\n\nbecause \( \mathbb{k} \) is a field. It follows that \( H\left( {{C}_{ * }\left( L\right) }\right) \) is concentrated in homological degrees \( \leq n \) .\n\nNow choose a graded subspace \( E \subset {\Lambda }^{n}{sL} \) so that \( E \oplus d\left( {{\Lambda }^{n + 1}{sL}}\right) = {\Lambda }^{n}{sL} \) . Observe that \( \left( {{\Lambda }^{ < n}{sL} \oplus E}\right) \otimes {UL} \) is automatically a subcomplex of \( {P}_{ * } \), and that the inclusion defines a morphism of the spectral sequences derived from the filtrations \( {F}^{p}\left( -\right) = \left( -\right) \cdot {\left( UL\right) }_{ \geq p} \) . At the \( {E}^{0} \) -term the differentials are just \( d \otimes {id} \) in \( {C}_{ * }\left( L\right) \otimes {UL} \) and so this morphism is a quasi-isomorphism of \( {E}^{0} \) -terms. It follows that the inclusion of \( \left( {{\Lambda }^{ < n}{sL} \oplus E}\right) \otimes {UL} \) in \( {P}_{ * } \) is itself a quasi-isomorphism. This \( \left( {{\Lambda }^{ < n}{sL} \oplus E}\right) \otimes {UL} \) is a \( {UL} \) -free resolution of \( \mathbb{k} \) and so \( {\operatorname{Ext}}_{UL}^{ > n}\left( {\mathbb{k}, - }\right) = 0 \) ; i.e., \( \operatorname{gl}\dim {UL} \leq n \) .\n\n(ii) Let \( \operatorname{gl}\dim {UL} = n \) . Then \( {\operatorname{Ext}}_{UL}^{n + 1}\left( {\mathbb{R}, - }\right) = 0 \) . Apply \( {\operatorname{Hom}}_{UL}\left( {{P}_{ * }, - }\right) \)\nto the short exact sequence \( 0 \rightarrow {\left( UL\right) }_{ + } \rightarrow {UL} \rightarrow \mathbb{R} \rightarrow 0 \) and pass to homology to obtain an exact sequence\n\n\[ \n{\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{R},{UL}}\right) \rightarrow {\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{R},\mathbb{R}}\right) \rightarrow {\operatorname{Ext}}_{UL}^{n + 1}\left( {\mathbb{R},{\left( UL\right) }_{ + }}\right) .\n\]\n\nThus \( {\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{R},{UL}}\right) \) surjects onto \( {\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{R},\mathbb{R}}\right) \), which is non-zero by (i), and so \( \operatorname{depth}{UL} \leq n \) .
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Yes
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Proposition 35.12 \( \operatorname{cat}\left( {{\Lambda V}, d}\right) \leq \operatorname{gldim}{UL} \) .
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proof: Let \( n = \operatorname{gl}\dim {UL} \) ; according to Proposition 35.11 it is the largest integer such that \( {\operatorname{Ext}}_{UL}^{n}\left( {\mathbb{R},\mathbb{R}}\right) \neq 0 \) . Define an ideal \( I \subset {\Lambda V} \) by setting \( I = \) \( {\Lambda }^{ > n}V \oplus {I}^{n} \), where \( {I}^{n} \oplus {\left( \ker {d}_{1}\right) }^{n, * } = {\Lambda }^{n}V \) . Then \( H\left( {I,{d}_{1}}\right) = 0 \) . Filter \( I \) by wordlength and use the associated spectral sequence to deduce that \( H\left( {I, d}\right) = 0 \) . Conclude from Corollary 2 to Proposition 29.2 that \( \operatorname{cat}\left( {{\Lambda V}, d}\right) \leq \operatorname{nil}\left( {{\Lambda V}/I}\right) \leq n \) because \( \left( {{\Lambda V}, d}\right) \rightarrow \left( {{\Lambda V}/I, d}\right) \) is a quasi-isomorphism.
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Yes
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Proposition 36.2\n\n(i) If \( L \) is the direct sum of ideals \( I \) and \( J \) then\n\n\[ \text{depth}L = \operatorname{depth}I + \operatorname{depth}J.\]\n\n(ii) If \( L \) is the infinite direct sum of non-zero ideals then \( \operatorname{depth}L = \infty \).
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proof: \( \; \) (i) Because of (36.1) we may identify \( {\operatorname{Ext}}_{UL}^{p}\left( {\mathbb{R},{UL}}\right) \) with \( {\operatorname{Tor}}_{p}^{UL}{\left( \mathbb{R}, U{L}^{\sharp }\right) }^{\sharp } \), \( p \geq 0 \) (Lemma 34.3(iii)). Thus depth \( L \) is the least integer \( m \) such that \( {\operatorname{Tor}}_{m}^{UL}\left( {\mathbb{R}, U{L}^{\sharp }}\right) \) \( \neq 0 \). On the other hand, if \( {P}_{ * } \) and \( {Q}_{ * } \) are respectively \( {UI} - \) and \( {UJ} \) -free resolutions of \( \mathbb{k} \) then \( {P}_{ * } \otimes {Q}_{ * } \) is a \( {UI} \otimes {UJ} \) -free resolution of \( \mathbb{k} \). Since \( {UL} = {UI} \otimes {UJ} \) this gives\n\n\[ {\operatorname{Tor}}^{UL}\left( {\mathbb{R}, U{L}^{\sharp }}\right) = H\left( {\left( {{P}_{ * } \otimes {Q}_{ * }}\right) { \otimes }_{UL}\left( {U{I}^{\sharp } \otimes U{J}^{\sharp }}\right) }\right) \]\n\n\[ = \;H\left( {{P}_{ * }{ \otimes }_{UI}U{I}^{\sharp }}\right) \otimes H\left( {{Q}_{ * }{ \otimes }_{UJ}U{J}^{\sharp }}\right) \]\n\n\[ = {\operatorname{Tor}}^{UI}\left( {\mathbb{k}, U{I}^{\sharp }}\right) \otimes {\operatorname{Tor}}^{UJ}\left( {\mathbb{k}, U{J}^{\sharp }}\right) . \]\n\nAssertion (i) follows.\n\n(ii) Here there is no \( a \in {UL} \) such that \( {\left( UL\right) }_{ + } \cdot a = 0 \) and so \( {\operatorname{Ext}}_{UL}^{0}\left( {\mathbb{R},{UL}}\right) \) \( = {\operatorname{Hom}}_{UL}\left( {\mathbb{R},{UL}}\right) = 0 \). Thus \( \operatorname{depth}L \geq 1 \). But then for each \( n \geq 1 \) we may write \( L \) as the direct sum of \( n \) ideals \( L\left( i\right) \), each of which is itself an infinite direct sum. Thus by (i), depth \( L = \mathop{\sum }\limits_{{i = 1}}^{n}\operatorname{depth}L\left( i\right) \geq n \). Hence depth \( L = \infty \).
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Yes
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Proposition 36.3 Let \( I \subset L \) be an ideal.\n\n(i) \( \operatorname{depth}I \leq \operatorname{depth}L \) .
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proof: (i) The Hochschild-Serre spectral sequence (§34(b)) converges from \( {E}_{2}^{p, q} = {\operatorname{Ext}}_{{UL}/I}^{p}\left( {\mathbb{k},{\operatorname{Ext}}_{UI}^{q}\left( {\mathbb{k},{UL}}\right) }\right) \) to \( {\operatorname{Ext}}_{UL}^{p + q}\left( {\mathbb{k},{UL}}\right) \) . Since \( {UL} \) is \( {UI} \) -free it follows that \( {\operatorname{Ext}}_{UI}^{q}\left( {\mathbb{R},{UL}}\right) = 0, q < \operatorname{depth}I \) . Hence \( {\operatorname{Ext}}_{UL}^{r}\left( {\mathbb{R},{UL}}\right) = 0 \) , \( r < \operatorname{depth}I \) and \( \operatorname{depth}L \geq \operatorname{depth}I \) .
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Yes
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Theorem 36.4 The graded Lie algebra \( L \) is solvable and of finite depth if and only if \( L \) is finite dimensional. In this case \( {\operatorname{Ext}}_{UL}^{ * }\left( {\mathbb{k},{UL}}\right) \) is one dimensional, and \[ \text{depth}L = \dim {L}_{\text{even }}\text{.} \]
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proof: Suppose \( L \) is solvable and has finite depth. The ideal \( \left\lbrack {L, L}\right\rbrack \) has finite depth (Proposition 36.3(i)), and so by induction on solvlength, \( \left\lbrack {L, L}\right\rbrack \) is finite dimensional. In particular, for some \( k,{L}_{ \geq k} \) is an abelian ideal, also of finite depth. Write \( {L}_{ \geq k} = \bigoplus k{x}_{\alpha } \) and note that each \( k{x}_{\alpha } \) is an ideal in \( {L}_{ \geq k} \) . Since \( {L}_{ \geq k} \) has finite depth it cannot be an infinite direct sum. (Proposition 36.2(ii)) and hence \( {L}_{ \geq k} \) is finite dimensional. Thus so is \( L \) . Conversely, suppose \( L \) is finite dimensional. Since \( L = {L}_{ \geq 1}, L \) is trivially solvable. Moreover, a non-zero element \( x \) in \( L \) of maximal degree is central. Set \( {kx} = I \) and put \[ e = \left\{ \begin{array}{ll} 1 & \text{ if }\deg x\text{ is even } \\ 0 & \text{ if }\deg x\text{ is odd. } \end{array}\right. \] Since \( {UI} \) is either the exterior or polynomial algebra on \( {kx} \), a simple direct calculation gives \[ {\operatorname{Ext}}_{UI}^{ \neq e}\left( {\mathbb{R},{UI}}\right) = 0\;\text{ and }\;\dim {\operatorname{Ext}}_{UI}^{e}\left( {\mathbb{R},{UI}}\right) = 1. \] Thus Proposition 34.7 asserts that there are isomorphisms of \( {UL}/I \) -modules, \[ {\operatorname{Ext}}_{UI}^{q}\left( {\mathbb{I},{UL}}\right) \cong \left\{ \begin{array}{ll} {UL}/I & \text{ if }q = e \\ 0 & \text{ otherwise. } \end{array}\right. \] In particular the Hochschild-Serre spectral sequence collapses at \( {E}_{2} \), and \[ {\operatorname{Ext}}_{UL}^{r}\left( {\mathbb{R},{UL}}\right) \cong {\operatorname{Ext}}_{{UL}/I}^{r - e}\left( {\mathbb{R},{UL}/I}\right) . \] It follows by induction on \( \dim L \) that \( {\operatorname{Ext}}_{UL}^{ * }\left( {\mathbb{R},{UL}}\right) \) is one dimensional and that depth \( L = \dim {L}_{\text{even }} \) .
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Yes
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Theorem 36.5 [54] If L satisfying (36.1) has finite depth then its radical, \( R \) , is finite dimensional and \( \dim {R}_{\text{even }} \leq \operatorname{depth}L \) .
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proof: Every solvable ideal \( I \subset L \) satisfies \( \dim {I}_{\text{even }} \leq \operatorname{depth}L \), by Theorem 36.4. Choose \( I \) so that \( \dim {I}_{\text{even }} \) is maximized. For any solvable ideal \( J, I + J \) is solvable, and hence \( {J}_{\text{even }} \subset {I}_{\text{even }} \) . It follows that \( {R}_{\text{even }} = {I}_{\text{even }} \) and so, for some \( k,{R}_{ \geq k} \) is concentrated in odd degrees. Thus \( {R}_{ \geq k} \) is abelian and \( R \) itself is solvable.\n\nNow Theorem 36.4 asserts that \( {\operatorname{Ext}}_{UR}^{ * }\left( {\mathbb{R},{UR}}\right) \) is one-dimensional and concentrated in \( * = \dim {R}_{\text{even }} \) . It follows that the isomorphism \( {\operatorname{Ext}}_{UR}^{ * }\left( {\mathbb{R},{UL}}\right) \cong \) \( {\operatorname{Ext}}_{UR}^{ * }\left( {\mathbb{R},{UR}}\right) \otimes {UL}/R \) identifies the left hand side with \( {UL}/R \) as a \( L/R \) -module. Thus the \( {E}_{2} \) -term of the Hochschild-Serre spectral sequence converging from \( {\operatorname{Ext}}_{{UL}/R}^{p}\left( {\mathbb{I}k,{\operatorname{Ext}}_{UR}^{q}\left( {\mathbb{I}k,{UL}}\right) }\right) \) to \( {\operatorname{Ext}}_{UL}^{p + q}\left( {\mathbb{I}k,{UL}}\right) \) is given by\n\n\[ \n{E}_{2}^{p, q} = \left\{ \begin{matrix} {\operatorname{Ext}}_{{UL}/R}^{p}\left( {\mathbb{k},{UL}/R}\right) &, q = \dim R \\ 0 & ,\text{ otherwise. } \end{matrix}\right. \]\n\nIn particular, the spectral sequence collapses at \( {E}_{2} \), and depth \( L = \operatorname{depth}L/R + \) \( \dim {R}_{\text{even }} \) .
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Yes
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Theorem 36.8 A graded Lie algebra L satisfying (36.1) and of depth \( m \) contains at most \( m \) linearly independent Engel elements of even degree.
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proof: We show that if \( L = I \oplus \mathbb{R}x \), with \( I \) an ideal and \( x \) a non-zero Engel element of even degree, then \( \operatorname{depth}I < \operatorname{depth}L \) . (The theorem follows from this by an obvious argument.)\n\nTo establish this assertion note first that since ad \( x \) is locally nilpotent, \( x \) acts locally nilpotently in each \( {\operatorname{Tor}}_{q}^{UI}\left( {\mathbb{R},\mathbb{R}}\right) \), as follows directly from Example 2, §34(a). Thus Proposition 36.3(iii) asserts that\n\n\[ \text{depth}I + \operatorname{depth}{kx} = \operatorname{depth}L \]\n\nand the conclusion follows from depth \( {kx} = 1 \) (because \( x \) has even degree).
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Yes
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A graded Lie algebra \( L \) has depth 0 if and only if \( {\operatorname{Hom}}_{UL}\left( {\mathbb{R},{UL}}\right) \neq 0 \) ; i.e., if and only if \( a \cdot U{L}_{ + } = 0 \) for some non-zero \( a \in {UL} \).
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It follows at once from the Poincaré Birkoff Witt theorem 21.1 that this occurs if and only if \( L \) is finite dimensional and concentrated in odd degrees.
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No
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Example 2 Free products have depth 1.
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Let \( E \) and \( L \) be graded Lie algebras and recall the free product \( E \coprod L \) defined in \( §{21}\left( \mathrm{c}\right) \) . We shall show that \( \operatorname{depth}E \coprod L = 1 \) . Indeed, choose free resolutions of the form \[ \overset{d}{ \rightarrow }V\left( 2\right) \otimes {UE}\overset{d}{ \rightarrow }V\left( 1\right) \otimes {UE}\overset{d}{ \rightarrow }{UE} \rightarrow \mathbb{k}\text{, and} \] \[ \overset{d}{ \rightarrow }W\left( 2\right) \otimes {UE}\overset{d}{ \rightarrow }W\left( 1\right) \otimes {UL}\overset{d}{ \rightarrow }{UL} \rightarrow \mathbb{R}. \] Then as described in \( §{21}\left( \mathrm{c}\right) \) there is a \( U\left( {E \coprod L}\right) \) -free resolution of \( \mathbb{R} \) of the form \[ \left\lbrack {V\left( 2\right) \oplus W\left( 2\right) }\right\rbrack \otimes U\left( {E \coprod L}\right) \overset{d}{ \rightarrow }\left\lbrack {V\left( 1\right) \oplus W\left( 1\right) }\right\rbrack \otimes U\left( {E \coprod L}\right) \overset{d}{ \rightarrow }U\left( {E \coprod L}\right) \rightarrow \mathbb{k} \] in which the restriction of \( d \) to \( V\left( i\right) \) and \( W\left( i\right) \) is given by the resolutions above. Choose non-zero elements \( x \in E \) and \( y \in L \) and define a \( U\left( {E \coprod L}\right) \) -linear map \( f : \left\lbrack {V\left( 1\right) \oplus W\left( 1\right) }\right\rbrack \otimes U\left( {E \coprod L}\right) \rightarrow U\left( {E \coprod L}\right) \) by \( f\left( v\right) = \left( {dv}\right) \cdot {yx} \) and \( f\left( w\right) = \left( {dw}\right) \cdot {xy} \) . Trivial calculations show that \( f \circ d = 0 \) and that \( f \) is not a coboundary.
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Yes
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Example 3 \( X \vee Y \) .
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Let \( X \) and \( Y \) be simply connected spaces with rational homotopy of finite type. In Example 2 of \( §{24}\left( \mathrm{f}\right) \) we observed that \( {L}_{X \vee Y} = {L}_{X} \coprod {L}_{Y} \) . Thus depth \( {L}_{X \vee Y} = \) 1. On the other hand, \( {\operatorname{cat}}_{0}\left( {X \vee Y}\right) = \max \left( {{\operatorname{cat}}_{0}X,{\operatorname{cat}}_{0}Y}\right) \) as follows from the remarks at the start of \( \$ {27} \) . Thus the difference \( {\operatorname{cat}}_{0} \) -depth can be arbitrarily large.
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Yes
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If \( E \) and \( L \) are graded Lie algebras then\n\n\[ \operatorname{depth}\left( {E \oplus L}\right) = \operatorname{depth}E + \operatorname{depth}L \]
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as observed in Proposition 36.2. Since \( {L}_{X \times Y} = {L}_{X} \oplus {L}_{Y} \) we have that depth \( {L}_{X \times Y} \) \( = \operatorname{depth}{L}_{X} + \operatorname{depth}{L}_{Y} \) in analogy with \( {\operatorname{cat}}_{0}\left( {X \times Y}\right) = {\operatorname{cat}}_{0}X + {\operatorname{cat}}_{0}Y \) (Theorem 30.2)
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Yes
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Example 5 \( X = {S}_{a}^{3} \vee {S}_{b}^{3}{ \cup }_{{\left\lbrack a{\left\lbrack a, b\right\rbrack }_{W}\right\rbrack }_{W}}{D}^{8} \) .
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This CW complex was first discussed in Example 2, \( §{13}\left( \mathrm{\;d}\right) \) and subsequently in Example 4, \( §{24}\left( \mathrm{f}\right) \) and in Example 3, \( §{33}\left( \mathrm{c}\right) \) . In \( §{33}\left( \mathrm{c}\right) \) we showed that the homotopy fibre of the retraction \( X \rightarrow {S}_{a}^{3} \) was rationally \( {S}_{b}^{3} \vee {S}^{5} \), and that the fibre inclusion \( \varphi : {S}_{b}^{3} \vee {S}^{5} \rightarrow X \) restricted to \( {S}^{5} \) represented \( {\left\lbrack a, b\right\rbrack }_{W} \) . Recall that \( {L}_{X} \) is the rational homotopy Lie algebra of \( X \) . If \( \alpha ,\beta \in {\left( {L}_{X}\right) }_{2} \) correspond to \( a, b \in {\pi }_{3}\left( X\right) \) then it follows easily that \( {L}_{X} = \mathbb{L}\left( {\alpha ,\beta }\right) /\left\lbrack {\alpha ,\left\lbrack {\alpha ,\beta }\right\rbrack }\right\rbrack \) . As an immediate consequence we see that \( \alpha \) is an Engel element in \( {L}_{X} \) . Write \( {L}_{X} = \mathbb{k}\alpha \oplus I \), where \( I = \mathbb{k}\beta \oplus {\left( {L}_{X}\right) }_{ > 4} \) . The argument proving Theorem 36.8 shows that \( \operatorname{depth}I < \operatorname{depth}L \) . Since also \( \operatorname{depth}I > 0 \) (Example 1, above) we have depth \( {L}_{X} \geq 2 \) . On the other hand, \( X \) is a 2-cone and so cat \( X \leq 2 \) (Theorem 27.10). Since depth \( {L}_{X} \leq {\operatorname{cat}}_{0}X \) (Theorem 35.10) we conclude \[ \text{depth}{L}_{X} = {\operatorname{cat}}_{0}X = \operatorname{cat}X = 2\text{.} \]
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Yes
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Example 6 \( \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) .
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Because the inclusion \( i : \mathbb{C}{P}^{n} \rightarrow \mathbb{C}{P}^{\infty } \) induces the surjection \( {\Lambda x} \rightarrow {\Lambda x}/{x}^{n + 1} \) in cohomology \( \left( {\deg x = 2}\right) \), the cohomology algebra of \( \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) is just \( \mathbb{Q} \oplus \) \( {x}^{n + 1} \cdot {\Lambda x} \) . Moreover, \( \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) is \( {2n} + 1 \) connected and so \( {\pi }_{{2n} + 2}\left( {\mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n}}\right) = \) \( {H}_{{2n} + 2}\left( {\mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n}}\right) = \mathbb{Z} \) . This defines a continuous map \( f : \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \rightarrow \) \( K\left( {\mathbb{Z},{2n} + 2}\right) \) . Let \( g : F \rightarrow \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) be the homotopy fibre.\n\nNext, observe that a Sullivan representative for \( i \) is also a Sullivan representative for the surjection \( {\Lambda x} \rightarrow {\Lambda x}/{x}^{n + 1} \) ,(deg \( x = 2 \) ). Use Lemma 13.3 and 13.4 to deduce that \( A = \mathbb{R} \oplus {x}^{n + 1}{\Lambda x} \) is a commutative model for \( \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) .\n\nExtend the inclusion \( \Lambda {x}^{n + 1} \rightarrow A \) to a minimal Sullivan model \( \varphi : \left( {\Lambda {x}^{n + 1} \otimes }\right. \) \( {\Lambda V}, d)\overset{ \simeq }{ \rightarrow }\left( {A,0}\right) \) . This is a Sullivan model for \( \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) and the inclusion of \( \Lambda {x}^{n + 1} \) is a Sullivan representative for \( f \) . Thus the quotient Sullivan algebra \( \left( {{\Lambda V},\bar{d}}\right) \) is a minimal Sullivan model for \( F \) . Moreover since \( A \) is \( \Lambda {x}^{n + 1} \) -free on the basis \( 1,{x}^{n + 2},\ldots ,{x}^{{2n} + 1} \) it follows that \( \varphi \) induces a quasi-isomorphism\n\n\[ \bar{\varphi } : \left( {{\Lambda V},\bar{d}}\right) \overset{ \simeq }{ \rightarrow }\mathbb{k} \oplus {\bigoplus }_{i = n + 2}^{{2n} + 1}\mathbb{k}{x}^{i}. \]\n\nThus \( F \) is a formal space in which cup-products vanish and hence \( F \) has the rational homotopy type of \( \mathop{\bigvee }\limits_{{i = n + 2}}^{{{2n} + 1}}{S}^{2i} \) (Theorem 24.5).\n\nLet \( L \) be the rational homotopy Lie algebra of \( \mathbb{C}{P}^{\infty }/\mathbb{C}{P}^{n} \) . The discussion above establishes the short exact sequence\n\n\[ 0 \rightarrow \mathbb{L}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \rightarrow L \rightarrow \mathbb{Q}\beta \rightarrow 0 \]\n\nwith \( \deg \beta = {2n} + 1 \) and \( \deg {\alpha }_{i} = {2n} + {2i} + 1 \) . Now Proposition 36.3(iii) asserts\n\nthat\n\n\[ \operatorname{depth}L = \operatorname{depth}\mathbb{Q}\beta + \operatorname{depth}\mathbb{L}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = 1. \]
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Yes
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Proposition 37.2 ([S-T]) With the notation above,\n\n\[ \n{\pi }_{ * }\left( p\right) \gamma = {\left\lbrack {\pi }_{ * }\left( p\right) \beta ,\alpha \right\rbrack }_{W}\;\text{ and }\;\operatorname{hur}\gamma = \operatorname{hur}\beta \cdot \operatorname{hur}{\partial }_{ * }\alpha .\n\]
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proof: The first assertion is immediate from the definition of the Whitehead product \( \left( {§{13}\left( \mathrm{e}\right) }\right) \) . For the second, observe that \( c \) factors over the surjection \( \partial \left( {{D}^{m} \times {D}^{n + 1}}\right) \rightarrow \left( {{S}^{m} \times {S}^{n}}\right) \cup {D}^{m + 1} \) to define \( \widehat{c} : \left( {{S}^{m} \times {S}^{n}}\right) \cup {D}^{m + 1} \rightarrow E \) . Moreover, \( \operatorname{hur}\gamma = {H}_{ * }\left( c\right) \left\lbrack {\partial \left( {{D}^{m} \times {D}^{n + 1}}\right) }\right\rbrack = {H}_{ * }\left( \widehat{c}\right) \left( {\left\lbrack {S}^{m}\right\rbrack \otimes \left\lbrack {S}^{n}\right\rbrack }\right) = \operatorname{hur}\beta \cdot \operatorname{hur}\alpha \) , where \( \left\lbrack \;\right\rbrack \) denotes fundamental class.
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Yes
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Proposition 37.8 (Anick [6]) With the notation preceding Proposition 37.6,\n\n\[ U{L}_{Y}{\left( z\right) }^{-1} = \left( {1 + z}\right) {UL}{\left( z\right) }^{-1} - \left( {z - {H}_{ + }\left( Z\right) \left( z\right) + {H}_{ + }\left( X\right) \left( z\right) }\right) . \]
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proof of Proposition 37.8: If \( {C}_{0} \leftarrow {C}_{1} \leftarrow \cdots \leftarrow {C}_{n} \) is a finite dimensional chain complex then \( \sum {\left( -1\right) }^{p}\dim {H}_{p}\left( C\right) = \sum {\left( -1\right) }^{p}\dim {C}_{p} \), as follows by a trivial calculation. Hence if \( {C}_{0, * } \leftarrow {C}_{1, * } \leftarrow \cdots \leftarrow {C}_{n, * } \) is a finite complex of graded vector spaces of finite type with differential of bidegree \( \left( {-1,0}\right) \) then \( \sum {\left( -1\right) }^{p}\dim {C}_{p, q} = \sum {\left( -1\right) }^{p}\dim {H}_{p, q}, q \geq 0 \) . This implies that\n\n\[ \mathop{\sum }\limits_{p}{\left( -1\right) }^{p}{z}^{n - p}{C}_{p, * }\left( z\right) = \mathop{\sum }\limits_{p}{\left( -1\right) }^{p}{z}^{n - p}{H}_{p, * }\left( C\right) \left( z\right) . \]\n\nApply this (in the notation of the proof of Proposition 37.6) to the complex \( \left( {A \otimes {\left( UL\right) }^{\sharp }}\right) = {A}^{\sharp } \otimes {UL} \) to obtain\n\n\[ \left\lbrack {{A}^{2, * }\left( z\right) - z{A}^{1, * }\left( z\right) + {z}^{2}{A}^{0, * }\left( z\right) }\right\rbrack {UL}\left( z\right) = \left\lbrack {{sS}\left( z\right) + {z}^{2}}\right\rbrack , \]\n\nwhere \( I = {\mathbb{L}}_{S} \) . On the other hand since \( {C}^{ * }\left( {\mathbb{L}, d}\right) \overset{ \simeq }{ \rightarrow }\left( {A, d}\right) \) it follows that \( {H}^{ + }\left( A\right) \) is the dual of the homology of the complex\n\n\[ {\left( sV\right) }^{\sharp } \rightarrow {\left( sW\right) }^{\sharp } \]\n\nThus\n\n\[ {A}^{2, * }\left( z\right) - z{A}^{1, * }\left( z\right) + {z}^{2}{A}^{0, * }\left( z\right) \]\n\n\[ = {H}^{2, * }\left( A\right) \left( z\right) - z{H}^{1, * }\left( A\right) \left( z\right) + {z}^{2}{H}^{0, * }\left( A\right) \left( z\right) \]\n\n\[ = \left( {sW}\right) \left( z\right) - z\left( {sV}\right) \left( z\right) + {z}^{2} \]\n\n\[ = {H}_{ + }\left( Z\right) \left( z\right) - z{H}_{ + }\left( X\right) \left( z\right) + {z}^{2}. \]\n\nFinally, \( U{L}_{Y}\left( z\right) = {UI}\left( z\right) {UL}\left( z\right) \) and \( {UI}\left( z\right) = {TS}\left( z\right) = {\left( 1 - S\left( z\right) \right) }^{-1} \) . A short calculation completes the proof.
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Yes
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Proposition 37.9 With the notation above \( \left( {{s}^{2} = }\right. \) double suspension), (i) \( {sV} \cong {\operatorname{Tor}}_{1}^{UL}\left( {\mathbb{Q},\mathbb{Q}}\right) \) . (ii) \( {s}^{2}R \cong {\operatorname{Tor}}_{2}^{UL}\left( {\mathbb{Q},\mathbb{Q}}\right) \) .
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proof: (i) Indeed \( V \cong {\mathbb{L}}_{V}/\left\lbrack {{\mathbb{L}}_{V},{\mathbb{L}}_{V}}\right\rbrack = L/\left\lbrack {L, L}\right\rbrack \) and so \( {sV} \cong {\operatorname{Tor}}_{1}^{UL}\left( {\mathbb{Q},\mathbb{Q}}\right) \) by Example 1, \( §{34}\left( \mathrm{a}\right) \) . (ii) Consider the Hochschild-Serre spectral sequence converging from \( {\operatorname{Tor}}_{p}^{UL}\left( {\mathbb{Q},{\operatorname{Tor}}_{q}^{UJ}\left( {\mathbb{Q},\mathbb{Q}}\right) }\right) \) to \( {\operatorname{Tor}}_{p + q}^{U{\mathbb{L}}_{V}}\left( {\mathbb{Q},\mathbb{Q}}\right) \) . Again by Example 1 of \( §{34}\left( \mathrm{a}\right) \) we may identify \( {\operatorname{Tor}}_{1}^{UJ}\left( {\mathbb{Q},\mathbb{Q}}\right) = s\left( {J/\left\lbrack {J, J}\right\rbrack }\right) \) with the representation of \( {UL} \) induced by Lie bracket in \( J \) . It follows that \[ {E}_{0,1}^{2} = {\operatorname{Tor}}_{0}^{UL}\left( {\mathbb{Q},{\operatorname{Tor}}_{1}^{UJ}\left( {\mathbb{Q},\mathbb{Q}}\right) }\right) \cong s\left( {J/\left\lbrack {J, J}\right\rbrack { \otimes }_{UL}\mathbb{Q}}\right) \cong {sR}. \] On the other hand since \( {\mathbb{L}}_{V} \) is free we have \( {\operatorname{Tor}}_{U{\mathbb{L}}_{V}}^{1}\left( {\mathbb{Q},\mathbb{Q}}\right) = {sV} \) and \( {\operatorname{Tor}}_{U{\mathbb{L}}_{V}}^{ > 1}\left( {\mathbb{Q},\mathbb{Q}}\right) \) \( = 0 \) - cf. Proposition 21.4. Since \( {E}_{1,0}^{2} = {\operatorname{Tor}}_{1}^{\overline{UL}}\left( {\mathbb{Q},\mathbb{Q}}\right) = {sV} \) it follows that the differential \( {d}^{2} \) in the spectral sequence satisfies \[ {d}^{2} : {E}_{2,0}^{2} = {\operatorname{Tor}}_{2}^{UL}\left( {\mathbb{Q},\mathbb{Q}}\right) \overset{ \cong }{ \rightarrow }{E}_{0,1}^{2} \cong {sR}. \]
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Yes
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Proposition 38.3 If \( \left( {{\Lambda V}, d}\right) \) is an elliptic Sullivan algebra (introduction to §32) then \( H\left( {{\Lambda V}, d}\right) \) is a Poincaré duality algebra. Its formal dimension is \( n = \) \( \mathop{\sum }\limits_{i}\deg {x}_{i} - \mathop{\sum }\limits_{j}\left( {\deg {y}_{j} - 1}\right) \), where \( \left( {x}_{i}\right) \) is a basis of \( {V}^{\text{odd }} \) and \( \left( {y}_{j}\right) \) is a basis of \( {V}^{\text{even }} \) .
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proof: Recall the odd spectral sequence defined in \( §{32}\left( \mathrm{\;b}\right) \) . Its first term is \( \left( {{\Lambda V},{d}_{\sigma }}\right) \) with \( {d}_{\sigma }\left( {V}^{\text{even }}\right) = 0 \) and \( {d}_{\sigma }\left( {V}^{\text{odd }}\right) \subset {V}^{\text{even }} \) . Proposition 32.4 asserts that \( H\left( {{\Lambda V},{d}_{\sigma }}\right) \) is finite dimensional.\n\nChoose \( r \) sufficiently large that each \( {y}_{j}^{r} = {d}_{\sigma }{\Phi }_{j} \) and extend \( \left( {{\Lambda V},{d}_{\sigma }}\right) \) to \( ({\Lambda V} \otimes \) \( \left. {{\Lambda U},{d}_{\sigma }}\right) \) by assigning \( U \) the basis \( \left( {u}_{j}\right) \) and setting \( {d}_{\sigma }{u}_{j} = {y}_{j}^{r} \) . Then \( ({\Lambda V} \otimes \) \( \left. {{\Lambda U},{d}_{\sigma }}\right) \overset{ \simeq }{ \rightarrow }\left( {\left\lbrack {{\bigoplus }_{j}\Lambda {y}_{j}/{u}_{j}^{r}}\right\rbrack \otimes \Lambda {V}^{\text{odd }},{d}_{\sigma }}\right) \) and it follows from Lemma 38.2(ii) that \( H\left( {{\Lambda V} \otimes {\Lambda U},{d}_{\sigma }}\right) \) satisfies Poincaré duality. On the other hand, \( H({\Lambda V} \otimes \n\n\( \left. {{\Lambda U},{d}_{\sigma }}\right) = H\left( {{\Lambda V},{d}_{\sigma }}\right) \otimes \Lambda \left( {{u}_{1} - {\Phi }_{1},\ldots ,{u}_{q} - {\Phi }_{q}}\right) \) and so Lemma 38.1(ii) asserts that \( H\left( {{\Lambda V},{d}_{\sigma }}\right) \) satisfies Poincaré duality.\n\nFinally, Theorem 32.6 shows that \( n = \mathop{\sum }\limits_{i}\deg {x}_{i} - \mathop{\sum }\limits_{j}\left( {\deg {y}_{j} - 1}\right) \) is the top\n\ndegree in which both \( \left( {{\Lambda V},{d}_{\sigma }}\right) \) and \( \left( {{\Lambda V}, d}\right) \) have non-vanishing cohomology. Since \( H\left( {{\Lambda V},{d}_{\sigma }}\right) \) is a Poincaré duality algebra, \( {H}^{n}\left( {{\Lambda V},{d}_{\sigma }}\right) = \mathbb{R}\left\lbrack z\right\rbrack \) and so this class must survive through the spectral sequence. Thus if \( {E}_{p} \) is the \( p \) -th term of the spectral sequence then \( {\left( {E}_{p}^{n}\right) }^{\sharp } = \mathbb{k}{\omega }_{p}^{\sharp } \) and \( {d}^{\sharp }{\omega }_{p} = 0 \) . By Lemma 38.2(i), each \( {E}_{p} \) is a Poincaré duality algebra. Hence the bigraded algebra associated to \( H\left( {{\Lambda V}, d}\right) \) is a Poincaré duality algebra and thus so is \( H\left( {{\Lambda V}, d}\right) \) (Lemma 38.1(iii)).
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Yes
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Theorem 38.4 Let \( X \) be a simply connected topological space and let \( \left( {{\Lambda V}, d}\right) \) be a minimal simply connected Sullivan algebra such that \( H\left( X\right) \) and \( H\left( {{\Lambda V}, d}\right) \) are Poincaré duality algebras. Then \[ {\mathrm{e}}_{0}\left( X\right) = {\operatorname{cat}}_{0}X\;\text{ and }\;\mathrm{e}\left( {{\Lambda V}, d}\right) = \operatorname{cat}\left( {{\Lambda V}, d}\right) .
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proof of Theorem 38.4: Choose \( z \in {\left( \Lambda V\right) }^{\sharp } \) so that \( {d}^{\sharp }z = 0 \) and \( z \) represents a fundamental class of \( \left( {{\Lambda V}, d}\right) \) . Define \( \theta : \left( {{\Lambda V}, d}\right) \rightarrow \left( {{\left( \Lambda V\right) }^{\sharp },{d}^{\sharp }}\right) \) by setting \( {\theta \Phi }\left( \Psi \right) = z\left( {\Phi \land \Psi }\right) ,\Phi ,\Psi \in {\Lambda V} \) . Then \( \theta \) is a linear map of \( \left( {{\Lambda V}, d}\right) \) -modules and \( H\left( \theta \right) \) is an isomorphism because \( H\left( {{\Lambda V}, d}\right) \) satisfies Poincaré duality.\n\nOn the other hand in \( §{29}\left( \mathrm{\;h}\right) \) we introduced the invariants mcat and e for \( \left( {{\Lambda V}, d}\right) \) -modules \( \left( {M, d}\right) \) in terms of a semifree resolution of \( \left( {M, d}\right) \) . In particular, these invariants coincide for quasi-isomorphic modules. Thus according to Theorem 29.16, \[ \mathrm{e}\left( {{\Lambda V}, d}\right) = \mathrm{e}\left( {{\left( \Lambda V\right) }^{\sharp },{d}^{\sharp }}\right) = \operatorname{cat}\left( {{\Lambda V}, d}\right) . \] Finally, the assertion for \( X \) follows from this via the minimal Sullivan model for \( X \) .
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Yes
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Theorem 2.1.1 Suppose \( {\left\{ {a}_{n}\right\} }_{n = 1}^{\infty } \) is a sequence of complex numbers and \( f\left( t\right) \) is a continuously differentiable function on \( \left\lbrack {1, x}\right\rbrack \) . Set\n\n\[ \nA\left( t\right) = \mathop{\sum }\limits_{{n \leq t}}{a}_{n} \n\]\n\nThen\n\n\[ \n\mathop{\sum }\limits_{{n \leq x}}{a}_{n}f\left( n\right) = A\left( x\right) f\left( x\right) - {\int }_{1}^{x}A\left( t\right) {f}^{\prime }\left( t\right) {dt}. \n\]
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Proof. First, suppose \( x \) is a natural number. We write the left-hand side as\n\n\[ \n\mathop{\sum }\limits_{{n \leq x}}{a}_{n}f\left( n\right) = \mathop{\sum }\limits_{{n \leq x}}\{ A\left( n\right) - A\left( {n - 1}\right) \} f\left( n\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{n \leq x}}A\left( n\right) f\left( n\right) - \mathop{\sum }\limits_{{n \leq x - 1}}A\left( n\right) f\left( {n + 1}\right) \n\]\n\n\[ \n= A\left( x\right) f\left( x\right) - \mathop{\sum }\limits_{{n \leq x - 1}}A\left( n\right) {\int }_{n}^{n + 1}{f}^{\prime }\left( t\right) {dt} \n\]\n\n\[ \n= A\left( x\right) f\left( x\right) - \mathop{\sum }\limits_{{n \leq x - 1}}{\int }_{n}^{n + 1}A\left( t\right) {f}^{\prime }\left( t\right) {dt} \n\]\n\nsince \( A\left( t\right) \) is a step function. Also,\n\n\[ \n\mathop{\sum }\limits_{{n \leq x - 1}}{\int }_{n}^{n + 1}A\left( t\right) {f}^{\prime }\left( t\right) {dt} = {\int }_{1}^{x}A\left( t\right) {f}^{\prime }\left( t\right) {dt} \n\]\n\nand we have proved the result if \( x \) is an integer. If \( x \) is not an integer, write \( \left\lbrack x\right\rbrack \) for the greatest integer less than or equal to \( x \), and observe that\n\n\[ \nA\left( x\right) \{ f\left( x\right) - f\left( \left\lbrack x\right\rbrack \right) \} - {\int }_{\left\lbrack x\right\rbrack }^{x}A\left( t\right) {f}^{\prime }\left( t\right) {dt} = 0, \n\]\n\nwhich completes the proof.
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Yes
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Theorem 2.1.9 (Euler-Maclaurin summation formula) Let \( k \) be a nonnegative integer and \( f \) be \( \left( {k + 1}\right) \) times differentiable on \( \left\lbrack {a, b}\right\rbrack \) with \( a, b \in \mathbb{Z} \) . Then\n\n\[ \mathop{\sum }\limits_{{a < n \leq b}}f\left( n\right) = {\int }_{a}^{b}f\left( t\right) {dt} + \mathop{\sum }\limits_{{r = 0}}^{k}\frac{{\left( -1\right) }^{r + 1}}{\left( {r + 1}\right) !}\left( {{f}^{\left( r\right) }\left( b\right) - {f}^{\left( r\right) }\left( a\right) }\right) {B}_{r + 1} \]\n\n\[ + \frac{{\left( -1\right) }^{k}}{\left( {k + 1}\right) !}{\int }_{a}^{b}{B}_{k + 1}\left( t\right) {f}^{\left( k + 1}\right) }\left( t\right) {dt}. \]
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Example 2.1.10 For integers \( x \geq 1 \) , \n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n} = \log x + \gamma + \frac{1}{2x} + \frac{1}{{12}{x}^{2}} + O\left( \frac{1}{{x}^{3}}\right) . \]\n\nSolution. Put \( f\left( t\right) = 1/t \) in Theorem 2.1.9, \( a = 1 \) , \( b = x \), and \( k = 2 \) . Then\n\n\[ \mathop{\sum }\limits_{{2 \leq n \leq x}}\frac{1}{n} = \log x + \frac{1}{2}\left( {\frac{1}{x} - 1}\right) + \frac{1}{12}\left( {\frac{1}{{x}^{2}} - 1}\right) - {\int }_{1}^{x}\frac{{B}_{3}\left( t\right) }{{t}^{4}}{dt} \]\n\nso that\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n} = \log x + \frac{1}{2} - \frac{1}{12} - {\int }_{1}^{x}\frac{{B}_{3}\left( t\right) }{{t}^{4}}{dt} + \frac{1}{2x} - \frac{1}{{12}{x}^{2}}. \]\n\nSince\n\n\[ - \gamma = \mathop{\lim }\limits_{{x \rightarrow \infty }}\left( {\log x - \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n}}\right) \]\n\nwe must have\n\n\[ \gamma = \frac{1}{2} - \frac{1}{12} - {\int }_{1}^{\infty }\frac{{B}_{3}\left( t\right) {dt}}{{t}^{4}} \]\n\nAlso,\n\n\[ {\int }_{x}^{\infty }\frac{{B}_{3}\left( t\right) {dt}}{{t}^{4}} = O\left( \frac{1}{{x}^{3}}\right) \]\n\nso that the result is now immediate.
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Yes
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For integers \( x \geq 1 \) , \n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n} = \log x + \gamma + \frac{1}{2x} + \frac{1}{{12}{x}^{2}} + O\left( \frac{1}{{x}^{3}}\right) . \]
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Solution. Put \( f\left( t\right) = 1/t \) in Theorem 2.1.9, \( a = 1 \) , \( b = x \), and \( k = 2 \) . Then \n\n\[ \mathop{\sum }\limits_{{2 \leq n \leq x}}\frac{1}{n} = \log x + \frac{1}{2}\left( {\frac{1}{x} - 1}\right) + \frac{1}{12}\left( {\frac{1}{{x}^{2}} - 1}\right) - {\int }_{1}^{x}\frac{{B}_{3}\left( t\right) }{{t}^{4}}{dt} \] \n\nso that \n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n} = \log x + \frac{1}{2} - \frac{1}{12} - {\int }_{1}^{x}\frac{{B}_{3}\left( t\right) }{{t}^{4}}{dt} + \frac{1}{2x} - \frac{1}{{12}{x}^{2}}. \] \n\nSince \n\n\[ - \gamma = \mathop{\lim }\limits_{{x \rightarrow \infty }}\left( {\log x - \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n}}\right) \] \n\nwe must have \n\n\[ \gamma = \frac{1}{2} - \frac{1}{12} - {\int }_{1}^{\infty }\frac{{B}_{3}\left( t\right) {dt}}{{t}^{4}} \] \n\nAlso, \n\n\[ {\int }_{x}^{\infty }\frac{{B}_{3}\left( t\right) {dt}}{{t}^{4}} = O\left( \frac{1}{{x}^{3}}\right) \] \n\nso that the result is now immediate.
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Yes
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Theorem 2.4.1 For any \( y > 0 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}f\left( n\right) = \mathop{\sum }\limits_{{d \leq y}}g\left( d\right) H\left( \frac{x}{d}\right) + \mathop{\sum }\limits_{{d \leq \frac{x}{y}}}h\left( d\right) G\left( \frac{x}{d}\right) - G\left( y\right) H\left( \frac{x}{y}\right) . \]
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Proof. We have\n\n\[ \mathop{\sum }\limits_{{n \leq x}}f\left( n\right) = \mathop{\sum }\limits_{{{de} \leq x}}g\left( d\right) h\left( e\right) \]\n\n\[ = \mathop{\sum }\limits_{{{de} \leq x}}g\left( d\right) h\left( e\right) + \mathop{\sum }\limits_{{{de} \leq x}}g\left( d\right) h\left( e\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \leq y}}^{{d \leq y}}g\left( d\right) H\left( \frac{x}{d}\right) + \mathop{\sum }\limits_{{e \leq \frac{x}{y}}}^{{d > y}}h\left( e\right) \left\{ {G\left( \frac{x}{e}\right) - G\left( y\right) }\right\} \]\n\n\[ = \mathop{\sum }\limits_{{d \leq y}}g\left( d\right) H\left( \frac{x}{d}\right) + \mathop{\sum }\limits_{{e \leq \frac{x}{y}}}h\left( e\right) G\left( \frac{x}{e}\right) - G\left( y\right) H\left( \frac{x}{y}\right) . \]
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Yes
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Theorem 3.1.9 (Bertrand’s postulate) For \( n \) sufficiently large, there is a prime between \( n \) and \( {2n} \) .
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Proof: (S. Ramanujan) Observe that if\n\n\[ \n{a}_{0} \geq {a}_{1} \geq {a}_{2} \geq \cdots \n\] \n\nis a decreasing sequence of real numbers tending to zero, then\n\n\[ \n{a}_{0} - {a}_{1} \leq \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}{a}_{n} \leq {a}_{0} - {a}_{1} + {a}_{2} \n\] \n\nThis is the starting point of Ramanujan's proof. We can write\n\n\[ \nT\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\log n = \mathop{\sum }\limits_{{{de} \leq x}}\Lambda \left( d\right) = \mathop{\sum }\limits_{{e \leq x}}\psi \left( \frac{x}{e}\right) . \n\] \n\nWe know that \( T\left( x\right) = x\log x - x + O\left( {\log x}\right) \) by Exercise 2.1.2. On the other hand,\n\n\[ \nT\left( x\right) - {2T}\left( \frac{x}{2}\right) = \mathop{\sum }\limits_{{n \leq x}}{\left( -1\right) }^{n - 1}\psi \left( \frac{x}{n}\right) \leq \psi \left( x\right) - \psi \left( \frac{x}{2}\right) + \psi \left( \frac{x}{3}\right) \n\] \n\nby the observation above. Hence\n\n\[ \n\psi \left( x\right) - \psi \left( \frac{x}{2}\right) + \psi \left( \frac{x}{3}\right) \geq \left( {\log 2}\right) x + O\left( {\log x}\right) . \n\] \n\nOn the other hand,\n\n\[ \n\psi \left( x\right) - \psi \left( \frac{x}{2}\right) \leq \left( {\log 2}\right) x + O\left( {\log x}\right) \n\] \n\nfrom which we deduce inductively\n\n\[ \n\psi \left( x\right) \leq 2\left( {\log 2}\right) x + O\left( {{\log }^{2}x}\right) . \n\] \n\nThus, \( \psi \left( x\right) - \psi \left( \frac{x}{2}\right) \geq \frac{1}{3}\left( {\log 2}\right) x + O\left( {{\log }^{2}x}\right) \) . Now, \( \psi \left( x\right) = \theta \left( x\right) + \) \( O\left( {\sqrt{x}{\log }^{2}x}\right) \) . Hence\n\n\[ \n\theta \left( x\right) - \theta \left( \frac{x}{2}\right) \geq \frac{1}{3}\left( {\log 2}\right) x + O\left( {\sqrt{x}{\log }^{2}x}\right) . \n\] \n\nTherefore, for \( x \) sufficiently large, there is a prime between \( x/2 \) and \( x \) .
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No
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Theorem 4.1.4 Let \( \delta \left( x\right) \) be defined as above. Let\n\n\[ I\left( {x, R}\right) = \frac{1}{2\pi i}{\int }_{c - {iR}}^{c + {iR}}\frac{{x}^{s}}{s}{ds}. \]\n\nThen, for \( x > 0, c > 0, R > 0 \), we have\n\n\[ \left| {I\left( {x, R}\right) - \delta \left( x\right) }\right| < \left\{ \begin{array}{ll} {x}^{c}\min \left( {1,{R}^{-1}{\left| \log x\right| }^{-1}}\right) & \text{ if }x \neq 1, \\ \frac{c}{R} & \text{ if }x = 1. \end{array}\right. \]
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Proof. Suppose first \( 0 < x < 1 \) . Consider the rectangular contour \( {K}_{U} \) oriented counterclockwise with vertices \( c - {iR}, c + {iR}, U + {iR} \) , \( U - {iR}, U > 0 \) . By Cauchy’s theorem\n\n\[ \frac{1}{2\pi i}{\int }_{{K}_{U}}\frac{{x}^{s}}{s}{ds} = 0 = \delta \left( x\right) \]\n\nTo prove the theorem, we must estimate the three integrals\n\n\[ \frac{1}{2\pi i}{\int }_{c + {iR}}^{U + {iR}}\frac{{x}^{s}}{s}{ds},\;\frac{1}{2\pi i}{\int }_{c - {iR}}^{U - {iR}}\frac{{x}^{s}}{s}{ds},\;\frac{1}{2\pi i}{\int }_{U - {iR}}^{U + {iR}}\frac{{x}^{s}}{s}{ds}. \]\n\nNow,\n\n\[ \left| {{\int }_{c + {iR}}^{U + {iR}}\frac{{x}^{s}}{s}{ds}}\right| \leq \frac{1}{R}{\int }_{c}^{U}{x}^{\delta }{d\delta } \]\n\nAs \( U \rightarrow \infty \), this integral is bounded by\n\n\[ \frac{{x}^{c}}{R\left| {\log x}\right| } \]\n\nA similar estimate holds for the other integral. Now,\n\n\[ \left| {\frac{1}{2\pi i}{\int }_{U - {iR}}^{U + {iR}}\frac{{x}^{s}{ds}}{s}}\right| \leq \frac{{x}^{U}R}{U} \]\n\nwhich goes to zero as \( U \rightarrow \infty \), since \( 0 < x < 1 \) . This proves one of the two stated inequalities in the case \( 0 < x < 1 \) . For the other inequality, consider the circle of radius \( {\left( {c}^{2} + {R}^{2}\right) }^{1/2} \) centered at the origin. This circle passes through \( c - {iR} \) and \( c + {iR} \) . We can therefore replace the vertical line integral under consideration by a circular path on the right side of the line segment joining \( c - {iR} \) to \( c + {iR} \) . The integral is easily estimated:\n\n\[ \left| {I\left( {x, R}\right) }\right| \leq \frac{1}{2\pi }{\pi R} \cdot \frac{{x}^{c}}{R} < {x}^{c} \]\nsince \( \left| {x}^{s}\right| \leq {x}^{c} \) on the circular path.\n\nThe proof when \( x > 1 \) is similar but uses a rectangle or a circular arc to the left. The contour then includes the pole at \( s = 0 \), where the residue is \( 1 = \delta \left( x\right) \) . We leave the details as an exercise to the reader.\n\nFinally, the case \( x = 1 \) is handled directly as in Exercise 4.1.3. We\n\nhave\n\n\[ \frac{1}{2\pi i}{\int }_{c - {iR}}^{c + {iR}}\frac{ds}{s} = \frac{c}{\pi }{\int }_{0}^{R}\frac{dt}{{c}^{2} + {t}^{2}} \]\n\nwhich equals\n\n\[ \frac{1}{\pi }{\int }_{0}^{R/c}\frac{du}{1 + {u}^{2}} = \frac{1}{2} - \frac{1}{\pi }{\int }_{R/c}^{\infty }\frac{du}{1 + {u}^{2}}. \]\n\nThe last integral is less than \( c/R \), and this proves the theorem.
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Yes
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Theorem 4.2.7 Let \( s = \sigma + {it} \). There are positive constants \( {c}_{1} \) and \( {c}_{2} \) such that\n\n\[ 1 - \frac{{c}_{1}}{{\left( \log T\right) }^{9}} \leq \sigma \leq 2 \]\n\n\[ \left| {\zeta \left( s\right) }\right| > \frac{{c}_{2}}{{\left( \log T\right) }^{7}} \]\n\nwhere \( 1 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq T \).
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Proof. In Exercise 3.2.5, we proved\n\n\[ \left| {\zeta {\left( \sigma \right) }^{3}\zeta {\left( \sigma + it\right) }^{4}\zeta \left( {\sigma + {2it}}\right) }\right| \geq 1 \]\n\nfor \( \sigma > 1 \). Thus,\n\n\[ {\left| \zeta \left( \sigma + it\right) \right| }^{4} \geq {\left| \zeta \left( \sigma + 2it\right) \right| }^{-1}{\left| \zeta \left( \sigma \right) \right| }^{-3}. \]\n\nNow, \( \zeta \left( \sigma \right) \left( {\sigma - 1}\right) \) remains bounded as \( \sigma \rightarrow {1}^{ + } \) and being continuous for \( 1 \leq \sigma \leq 2 \) has an upper bound in that region. By Exercise 4.2.3, for some constant \( K \),\n\n\[ {\left| \zeta \left( \sigma + 2it\right) \right| }^{-1} \geq K{\left( \log T\right) }^{-1}. \]\n\nThus we get\n\n\[ {\left| \zeta \left( \sigma + it\right) \right| }^{4} \geq {K}_{1}{\left( \log T\right) }^{-1}{\left( \sigma - 1\right) }^{3}. \]\n\nIf\n\n\[ 1 + \frac{{c}_{1}}{{\left( \log T\right) }^{9}} \leq \sigma \leq 2 \]\n\nthen we obtain\n\n\[ \left| {\zeta \left( s\right) }\right| \gg \frac{1}{{\left( \log T\right) }^{7}} \]\n\nin this region. We can extend this result to the region\n\n\[ 1 - \frac{{c}_{1}}{{\left( \log T\right) }^{9}} \leq \sigma \leq 1 + \frac{{c}_{1}}{{\left( \log T\right) }^{9}} \]\n\nand \( 1 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq T \), by using the mean value theorem. Indeed, choose \( {s}^{\prime } \) such that \( {s}^{\prime } = {\sigma }^{\prime } + {it} \), with\n\n\[ {\sigma }^{\prime } = 1 + \frac{{c}_{1}}{{\left( \log T\right) }^{9}} \]\n\nThen\n\n\[ \zeta \left( {{\sigma }^{\prime } + {it}}\right) - \zeta \left( {\sigma + {it}}\right) = O\left( {\left( {\sigma - {\sigma }^{\prime }}\right) {\log }^{2}T}\right) \]\n\nby an application of the mean value theorem and Exercise 4.2.5. Thus, if \( {c}_{1} \) is chosen sufficiently small, we obtain\n\n\[ \left| {\zeta \left( s\right) }\right| \gg {\left( \log T\right) }^{-7} \]
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No
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Theorem 5.1.3 (Poisson summation formula) Let \( F \in {L}^{1}\left( \mathbb{R}\right) \) . Suppose that the series\n\n\[ \mathop{\sum }\limits_{{n \in \mathbb{Z}}}F\left( {n + v}\right) \]\n\nconverges absolutely and uniformly in \( v \), and that\n\n\[ \mathop{\sum }\limits_{{m \in \mathbb{Z}}}\left| {\widehat{F}\left( m\right) }\right| < \infty \]\n\nThen\n\n\[ \mathop{\sum }\limits_{{n \in \mathbb{Z}}}F\left( {n + v}\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}\widehat{F}\left( n\right) {e}^{2\pi inv}. \]
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Proof. The function\n\n\[ G\left( v\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}F\left( {n + v}\right) \]\n\nis a continuous function of \( v \) of period 1 . The Fourier coefficients of \( G \) are given by\n\n\[ {c}_{m} = {\int }_{0}^{1}G\left( v\right) {e}^{-{2\pi imv}}{dv} \]\n\n\[ = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}{\int }_{0}^{1}F\left( {n + v}\right) {e}^{-{2\pi imv}}{dv} \]\n\n\[ = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}{\int }_{n}^{n + 1}F\left( x\right) {e}^{-{2\pi imx}}{dx} \]\n\n\[ = {\int }_{-\infty }^{\infty }F\left( x\right) {e}^{-{2\pi imx}}{dx} = \widehat{F}\left( m\right) . \]\n\nSince \( \mathop{\sum }\limits_{{m \in \mathbb{Z}}}\left| {\widehat{F}\left( m\right) }\right| < \infty \), we can represent \( G \) by its Fourier series\n\n\[ \mathop{\sum }\limits_{{n \in \mathbb{Z}}}F\left( {n + v}\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}\widehat{F}\left( n\right) {e}^{2\pi inv} \]\n\nas desired.
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Yes
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Corollary 5.1.4 With \( F \) as above,\n\n\[ \mathop{\sum }\limits_{{n \in \mathbb{Z}}}F\left( n\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}\widehat{F}\left( n\right) \]
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Proof. Set \( v = 0 \) in the theorem.
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No
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Example 5.3.1 If \( \\left( {n, q}\\right) = 1 \\), then\n\n\\[ \n\\chi \\left( n\\right) \\tau \\left( \\bar{\\chi }\\right) = \\mathop{\\sum }\\limits_{{m = 1}}^{q}\\bar{\\chi }\\left( m\\right) e\\left( \\frac{mn}{q}\\right) .\n\\]
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Solution. We have\n\n\\[ \n\\chi \\left( n\\right) \\tau \\left( \\bar{\\chi }\\right) = \\mathop{\\sum }\\limits_{{m = 1}}^{q}\\bar{\\chi }\\left( m\\right) \\chi \\left( n\\right) e\\left( \\frac{m}{q}\\right)\n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{h = 1}}^{q}\\bar{\\chi }\\left( h\\right) e\\left( \\frac{nh}{q}\\right)\n\\]\n\non putting \( h \\equiv m{n}^{-1}\\left( {\\;\\operatorname{mod}\\;q}\\right) \\), which we can do, since \( \\left( {n, q}\\right) = 1 \) .
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Yes
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Theorem 5.3.3 If \( \chi \) is a primitive character \( \left( {\;\operatorname{mod}\;q}\right) \), then \( \left| {\tau \left( \chi \right) }\right| = {q}^{1/2} \) .
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Proof. By Exercise 5.3.2,\n\n\[ \chi \left( n\right) \tau \left( \bar{\chi }\right) = \mathop{\sum }\limits_{{m = 1}}^{q}\bar{\chi }\left( m\right) e\left( \frac{mn}{q}\right) . \]\n\nThus\n\n\[ {\left| \chi \left( n\right) \right| }^{2}{\left| \tau \left( \chi \right) \right| }^{2} = \mathop{\sum }\limits_{{{m}_{1} = 1}}^{q}\mathop{\sum }\limits_{{{m}_{2} = 1}}^{q}\bar{\chi }\left( {m}_{1}\right) \chi \left( {m}_{2}\right) e\left( \frac{n\left( {{m}_{1} - {m}_{2}}\right) }{q}\right) . \]\n\nSumming over \( n \) for \( 1 \leq n \leq q \) gives\n\n\[ \phi \left( q\right) {\left| \tau \left( \chi \right) \right| }^{2} = {q\phi }\left( q\right) \]\n\nso that \( {\left| \tau \left( \chi \right) \right| }^{2} = q \), as required.
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No
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Show that an entire function \( f\left( z\right) \) of finite order \( \beta \) without any zeros must be of the form \( f\left( z\right) = {e}^{g\left( z\right) } \), where \( g\left( z\right) \) is a polynomial and \( \beta = \deg g \) .
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## Solution.\n\nLet \( h\left( z\right) = \log f\left( z\right) - \log f\left( 0\right) \) . Then \( h\left( z\right) \) is entire, since \( f\left( z\right) \) has no zeros. Also, for any \( \epsilon > 0 \) ,\n\n\[ \n\operatorname{Re}h\left( z\right) = \log \left| {f\left( z\right) }\right| \ll {R}^{\beta + \epsilon }.\n\]\n\nWriting\n\n\[ \nh\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( {{a}_{n} + i{b}_{n}}\right) {z}^{n}\n\]\n\nwith \( {a}_{n},{b}_{n} \in \mathbb{R} \), we see that for \( z = R{e}^{i\theta } \),\n\n\[ \n\operatorname{Re}\left( {h\left( z\right) }\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{R}^{n}\cos {n\theta } - \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{R}^{n}\sin {n\theta }.\n\]\n\nBy Fourier analysis, we get\n\n\[ \n\left| {a}_{n}\right| {R}^{n} \ll {\int }_{0}^{2\pi }\left| {\operatorname{Re}\left( {h\left( {R{e}^{i\theta }}\right) }\right) }\right| {d\theta }.\n\]\n\nSince \( h\left( 0\right) = 0 \), we have \( {a}_{0} = 0 \), and therefore\n\n\[ \n{\int }_{0}^{2\pi }\operatorname{Re}\left( {h\left( {R{e}^{i\theta }}\right) }\right) {d\theta } = 0\n\]\n\nObserve that for \( x \in \mathbb{R} \), we have\n\n\[ \n\left| x\right| + x = \left\{ \begin{array}{lll} {2x} & \text{ if } & x \geq 0 \\ 0 & \text{ if } & x < 0 \end{array}\right.\n\]\n\nHence\n\n\[ \n\left| {a}_{n}\right| {R}^{n} \ll {\int }_{0}^{2\pi }\left\{ {\left| {\operatorname{Re}\left( {h\left( {R{e}^{i\theta }}\right) }\right) }\right| + \operatorname{Re}h\left( {R{e}^{i\theta }}\right) }\right\} {d\theta }\n\]\n\n\[ \n\ll {R}^{\beta + \epsilon }.\n\]\n\nLetting \( R \rightarrow \infty \) yields \( {a}_{n} = 0 \) if \( n > \beta \) .\n\nNotice that in this example the same result holds if the estimate\n\n\[ \n\left| {f\left( z\right) }\right| \ll {e}^{{R}_{i}^{\beta + \epsilon }}\n\]\n\nholds for \( \left| z\right| = {R}_{i} \) and \( {R}_{i} \) is a sequence tending to infinity.
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Yes
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Theorem 6.1.2 (Jensen’s theorem) Let \( f\left( z\right) \) be an entire function of order \( \beta \) such that \( f\left( 0\right) \neq 0 \) . If \( {z}_{1},{z}_{2},\ldots ,{z}_{n} \) are the zeros of \( f\left( z\right) \) in \( \left| z\right| < R \), counted with multiplicity, then
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Proof. We may assume, without loss of generality, that \( f\left( 0\right) = 1 \) . Also, it is clear that if the theorem is true for functions \( g \) and \( h \), that it is also true for the product \( {gh} \) . Thus, it suffices to prove it for functions with either no zero or one zero in \( \left| z\right| < R \) . Indeed, if \( f \) has no zeros in \( \left| z\right| < R \), the right-hand side is zero. The left-hand side is\n\n\[ \frac{1}{2\pi i}{\int }_{\left| z\right| = R}\left( {\log f\left( z\right) }\right) \frac{dz}{z} \]\n\nwhich by Cauchy's theorem is zero. Taking real parts gives the desired result.\n\nIf \( f \) has one zero \( z = {z}_{1} \) in \( \left| z\right| < R \), we consider the contour \( \left| z\right| = R \) taken in the counterclockwise direction and cut it from \( {z}_{1} \) to the boundary. We deform the contour so that we go around \( {z}_{1} \) in a clockwise direction along a circle of radius \( \epsilon \) (say). Then, by Cauchy’s theorem with \( g\left( z\right) = \log f\left( z\right) \),\n\n\[ 0 = \frac{1}{2\pi i}{\int }_{\mathcal{C}}g\left( z\right) \frac{dz}{z} \]\n\nwhere \( \mathcal{C} \) is the contour given above.\n\nSince the argument changes by \( - {2\pi i} \) when \( g\left( z\right) \) goes around the zero \( z = {z}_{1} \), we see that as \( \epsilon \rightarrow 0 \), we deduce\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi }\log \left| {f\left( {R{e}^{i\theta }}\right) }\right| {d\theta } = \log \frac{R}{\left| {z}_{1}\right| } \]\n\n as desired. This completes the proof.
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Yes
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Corollary 6.1.3 Let \( f \) be as in Theorem 6.1.2. Then\n\n\[ \log \left( \frac{{R}^{n}}{\left| {z}_{1}\right| \cdots \left| {z}_{n}\right| }\right) \leq \mathop{\max }\limits_{{\left| z\right| = R}}\log \left| {f\left( z\right) }\right| - \log \left| {f\left( 0\right) }\right| .\n\]
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Proof. This is clear from Jensen's theorem.
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No
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Theorem 6.5.6 There exists a constant \( c > 0 \) such that \( \zeta \left( s\right) \) has no zero in the region\n\n\[ \sigma \geq 1 - \frac{c}{\log \left| t\right| },\;\left| t\right| \geq 2 \]
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Proof. By Exercise 6.5.5,\n\n\[ - \operatorname{Re}\left( \frac{{\zeta }^{\prime }\left( {\sigma + {it}}\right) }{\zeta \left( {\sigma + {it}}\right) }\right) < {A}_{1}\log \left| t\right| - \frac{1}{\sigma - \beta }.\]\n\nWe also know, by Exercises 6.5.2 and 6.5.4, that\n\n\[ - \frac{{\zeta }^{\prime }\left( \sigma \right) }{\zeta \left( \sigma \right) } < \frac{1}{\sigma - 1} + {A}_{2} \]\n\nand\n\n\[ - \operatorname{Re}\left( \frac{{\zeta }^{\prime }\left( {\sigma + {2it}}\right) }{\zeta \left( {\sigma + {2it}}\right) }\right) < {A}_{3}\log \left| t\right| \]\n\nInserting these inequalities into\n\n\[ - 3\frac{{\zeta }^{\prime }\left( \sigma \right) }{\zeta \left( \sigma \right) } - 4\operatorname{Re}\left( \frac{{\zeta }^{\prime }\left( {\sigma + {it}}\right) }{\zeta \left( {\sigma + {it}}\right) }\right) - \operatorname{Re}\left( \frac{{\zeta }^{\prime }\left( {\sigma + {2it}}\right) }{\zeta \left( {\sigma + {2it}}\right) }\right) \geq 0 \]\n\n(Exercise 6.5.1), we obtain\n\n\[ \frac{4}{\sigma - \beta } < \frac{3}{\sigma - 1} + A\log \left| t\right| \]\n\nfor some constant \( A \) . Taking \( \sigma = 1 + \delta /\log \left| t\right| \) gives\n\n\[ \beta < 1 + \frac{\sigma }{\log \left| t\right| } - \frac{4\sigma }{\left( {3 + {A\delta }}\right) \log \left| t\right| } \]\n\nso that if \( \delta \) is sufficiently small, we get\n\n\[ \beta < 1 - \frac{c}{\log \left| t\right| } \]\n\nfor some suitable positive constant \( c \) .
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Yes
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Corollary 6.5.7 There exists a constant \( c > 0 \) such that \( \zeta \left( s\right) \) has no zero in the region\n\n\[ \sigma \geq 1 - \frac{c}{\log \left( {\left| t\right| + 2}\right) } \]
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Proof. The region \( \sigma \geq 1,\left| t\right| \leq 2 \) contains no zeros of \( \zeta \left( s\right) \) . Thus, there must be a constant \( {c}_{1} > 0 \) such that \( \zeta \left( s\right) \) has no zeros in \( \sigma \geq \) \( 1 - {c}_{1} \) and \( \left| t\right| \leq 2 \) . Combining such a region with the zero-free region provided by the theorem gives the result.
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No
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Theorem 6.5.12 There exists a positive absolute constant \( c \) such that if \( 0 < \delta < c \), then \( L\left( {s,\chi }\right) \) has no zeros in the region\n\n\[ \delta > 1 - \frac{c}{\log q\left( {\left| t\right| + 2}\right) } \]\n\nexcept possibly if \( \chi \) is real and nonprincipal, in which case there is at most one simple, real zero in the region.
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Proof. We need only consider the case where \( \chi \) is real and nonprin-cipal and \( \left| \gamma \right| < \delta /\log q \) . First suppose there are two complex zeros in the region. We have\n\n\[ - \frac{{L}^{\prime }\left( {\sigma ,\chi }\right) }{L\left( {\sigma ,\chi }\right) } < {c}_{1}\log q - \mathop{\sum }\limits_{\rho }\frac{1}{\sigma - \rho } \]\n\nthe sum over the zeros being real, since they occur in complex conjugate pairs. If \( \beta \pm {i\gamma } \) are zeros of \( L\left( {s,\chi }\right) \), with \( \gamma \neq 0 \), then\n\n\[ - \frac{{L}^{\prime }\left( {\sigma ,\chi }\right) }{L\left( {\sigma ,\chi }\right) } < {c}_{1}\log q - \frac{2\left( {\sigma - \beta }\right) }{{\left( \sigma - \beta \right) }^{2} + {\gamma }^{2}}. \]\n\nAlso,\n\n\[ - \frac{{L}^{\prime }\left( {\sigma ,\chi }\right) }{L\left( {\sigma ,\chi }\right) } = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\chi \left( n\right) \Lambda \left( n\right) }{{n}^{\sigma }} \geq - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\Lambda \left( n\right) }{{n}^{\sigma }} = \frac{{\zeta }^{\prime }\left( \sigma \right) }{\zeta \left( \sigma \right) } > - \frac{1}{\sigma - 1} - {c}_{0} \]\n\nfor some constant \( {c}_{0} \) . Thus\n\n\[ - \frac{1}{\sigma - 1} < {c}_{2}\log q - \frac{2\left( {\sigma - \beta }\right) }{{\left( \sigma - \beta \right) }^{2} + {\gamma }^{2}} \]\n\nand taking \( \sigma = 1 + {2\delta }/\log q \) gives\n\n\[ - \frac{1}{\sigma - 1} < {c}_{2}\log q - \frac{8}{5\left( {\sigma - \beta }\right) } \]\n\nbecause\n\n\[ \left| \gamma \right| < \frac{\delta }{\log q} = \frac{1}{2}\left( {\sigma - 1}\right) < \frac{1}{2}\left( {\sigma - \beta }\right) . \]\n\nTherefore,\n\n\[ \beta < 1 - \frac{\delta }{\log q} \]\n\nfor a sufficiently small \( \delta \) . The argument for two real zeros or a double real zero is the same. This completes the proof.
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Yes
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Theorem 7.1.7 Let \( N\left( T\right) \) be the number of zeros of \( \zeta \left( s\right) \) in the rectangle \( 0 < \sigma < 1,0 < t < T \) . Then\n\n\[ N\left( T\right) = \frac{T}{2\pi }\log \frac{T}{2\pi } - \frac{T}{2\pi } + \frac{7}{8} + S\left( T\right) + O\left( \frac{1}{T}\right) ,\] \n\nwhere \n\n\[ {\pi S}\left( T\right) = {\Delta }_{L}\arg \zeta \left( s\right) \] \n\nand \( L \) denotes the path of line segments joining 2 to \( 2 + {iT} \) and then to \( \frac{1}{2} + {iT} \) . We also have \n\n\[ S\left( T\right) = O\left( {\log T}\right) \]
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Proof. Let \( R \) be the rectangle with vertices \( 2,2 + {iT}, - 1 + {iT} \), and -1 , traversed in the counterclockwise direction. Then\n\n\[ {2\pi N}\left( T\right) = {\Delta }_{R}\arg \xi \left( s\right) \]\n\nThere is no change in the argument as \( s \) goes from -1 to 2 . Also, the change when \( s \) moves from \( \frac{1}{2} + {iT} \) to \( - 1 + {iT} \) and then to -1 is equal to the change as \( s \) moves from 2 to \( 2 + {iT} \) and then to \( \frac{1}{2} + {iT} \) , since\n\n\[ \xi \left( {\sigma + {it}}\right) = \xi \left( {1 - \sigma - {it}}\right) = \overline{\xi \left( {1 - \sigma + {it}}\right) }.\]\n\nHence \( {\pi N}\left( T\right) = {\Delta }_{L}\arg \xi \left( s\right) \), where \( L \) denotes the path of line segments joining 2 to \( 2 + {iT} \) and then to \( \frac{1}{2} + {iT} \) . By Exercises 7.1.1 and 7.1.3, we deduce\n\n\[ N\left( T\right) = \frac{T}{2\pi }\log \frac{T}{2\pi } - \frac{T}{2\pi } + \frac{7}{8} + S\left( T\right) + O\left( \frac{1}{T}\right) ,\]\n\nwhere\n\n\[ {\pi S}\left( T\right) = {\Delta }_{L}\arg \zeta \left( s\right) \]\n\nNow, the variation of \( \zeta \left( s\right) \) along \( \sigma = 2 \) is bounded, since \( \log \zeta \left( s\right) \) is bounded there. Thus,\n\n\[ {\pi S}\left( T\right) = O\left( 1\right) - {\int }_{\frac{1}{2} + {iT}}^{2 + {iT}}\operatorname{Im}\left( \frac{{\zeta }^{\prime }\left( s\right) }{\zeta \left( s\right) }\right) {ds}. \]\n\nWe now apply Exercise 7.1.6, which says that\n\n\[ \frac{{\zeta }^{\prime }\left( s\right) }{\zeta \left( s\right) } = \mathop{\sum }\limits_{\rho }^{\prime }\frac{1}{s - \rho } + O\left( {\log t}\right) \]\n\nwhere the dash on the summation means \( \left| {\operatorname{Im}\left( {s - \rho }\right) }\right| < 1 \) . Observing that\n\n\[ {\int }_{\frac{1}{2} + {iT}}^{2 + {iT}}\operatorname{Im}\left( \frac{1}{s - \rho }\right) {ds} = \Delta \arg \left( {s - \rho }\right) \]\n\nis at most \( \pi \) and noting that the number of terms in the sum above is \( O\left( {\log \left| t\right| }\right) \) by Exercise 7.1.5 gives us\n\n\[ S\left( T\right) = O\left( {\log T}\right) \]\n\nThis completes the proof.
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Yes
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Theorem 7.2.8 For some constant \( {c}_{1} > 0 \) ,\n\n\[ \psi \left( x\right) = x + O\left( {x\exp \left( {-{c}_{1}\sqrt{\log x}}\right) }\right) \]
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Proof. By the solution to Exercise 7.2.7, we know that\n\n\[ \psi \left( x\right) = x - \mathop{\sum }\limits_{{\left| \rho \right| < R}}\frac{{x}^{\rho }}{\rho } - \frac{{\zeta }^{\prime }\left( 0\right) }{\zeta \left( 0\right) } + \frac{1}{2}\log \left( {1 - {x}^{-2}}\right) + O\left( {\frac{x{\log }^{2}x}{R} + \frac{x{\log }^{2}R}{R\log x}}\right) . \]\n\nBy Theorem 6.5.6, we have \( \operatorname{Re}\left( \rho \right) = \beta < 1 - \frac{c}{\log R} \), so that the sum over the zeros is\n\n\[ x\exp \left( {-\frac{c\log x}{\log R}}\right) \mathop{\sum }\limits_{{\left| \gamma \right| < R}}\frac{1}{\left| \rho \right| }. \]\n\nBy partial summation and Theorem 7.1.7 we have\n\n\[ \mathop{\sum }\limits_{{\left| \gamma \right| < R}}\frac{1}{\left| \rho \right| } \ll {\int }_{1}^{R}\frac{\log t}{t}{dt} \ll {\log }^{2}R \]\n\nThe optimal choice for \( R \) satisfies\n\n\[ \log R = {c}_{2}{\left( \log x\right) }^{1/2} \]\n\nfor some appropriate constant \( {c}_{2} \) . It is now easily verified that this gives the desired result.
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Yes
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Theorem 7.3.2 (Weil’s explicit formula) Assume that \( h\left( s\right) \) satisfies the conditions of Lemma 7.3.1. In addition, assume that \( h\left( {it}\right) = {h}_{0}\left( {t/{2\pi }}\right) \) is a real-valued function for \( t \in \mathbb{R} \) whose Fourier transform\n\n\[ \n{\widehat{h}}_{0}\left( y\right) = {\int }_{-\infty }^{\infty }{h}_{0}\left( t\right) {e}^{-{2\pi ity}}{dt} \n\]\n\nsatisfies the bound\n\n\[ \n{\widehat{h}}_{0}\left( y\right) = O\left( {e}^{-\left( {\frac{1}{2} + \epsilon }\right) y}\right) \n\]\n\nfor fixed \( \epsilon > 0 \) as \( y \rightarrow \infty \) . Then we have\n\n\[ \n\mathop{\sum }\limits_{\gamma }h\left( {i\gamma }\right) + \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\Lambda \left( n\right) }{\sqrt{n}}{\widehat{h}}_{0}\left( {\log n}\right) \n\]\n\n\[ \n= h\left( \frac{1}{2}\right) - \frac{1}{2}\left( {\log \pi }\right) {\widehat{h}}_{0}\left( 0\right) + {\int }_{-\infty }^{\infty }\frac{{\Gamma }^{\prime }\left( {1/4 + {i\pi t}}\right) }{\Gamma \left( {1/4 + {i\pi t}}\right) }{h}_{0}\left( t\right) {dt} \n\]\n\nwhere the first sum is over all zeros \( 1/2 + {i\gamma } \) satisfying \( \operatorname{Im}\left( {i\gamma }\right) > 0 \), and \( \Lambda \left( n\right) \) is the von Mangoldt function, so that the second sum is over all prime powers.
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Proof. Recall that\n\n\[ \n\frac{{\xi }^{\prime }\left( {\frac{1}{2} + s}\right) }{\xi \left( {\frac{1}{2} + s}\right) } \n\]\n\n\[ \n= \frac{1}{s + 1/2} + \frac{1}{s - 1/2} - \frac{1}{2}\log \pi + \frac{{\Gamma }^{\prime }\left( {1/4 + s/2}\right) }{\Gamma \left( {1/4 + s/2}\right) } - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\Lambda \left( n\right) }{{n}^{s + 1/2}}, \n\]\nso that inserting this into Lemma 7.3.1 we see that\n\n\[ \n\frac{1}{2\pi i}{\int }_{\left( \frac{1}{2} + \epsilon \right) }\left\{ {\frac{1}{s + 1/2} + \frac{1}{s - 1/2} - \frac{1}{2}\log \pi }\right. \n\]\n\n\[ \n\left. {+\frac{{\Gamma }^{\prime }\left( {1/4 + s/2}\right) }{\Gamma \left( {1/4 + s/2}\right) } - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\Lambda \left( n\right) }{{n}^{s + 1/2}}}\right\} h\left( s\right) {ds} = \mathop{\sum }\limits_{\gamma }h\left( {i\gamma }\right) . \n\]\n\nObserve that by the growth condition on \( h \) ,\n\n\[ \n\frac{1}{2\pi i}{\int }_{\left( \frac{1}{2} + \epsilon \right) }h\left( s\right) {n}^{-s}{ds} = \frac{1}{2\pi i}{\int }_{\left( 0\right) }h\left( s\right) {n}^{-s}{ds} \n\]\n\nby moving the line of integration to the purely imaginary axis. Thus\n\n\[ \n\frac{1}{2\pi i}{\int }_{\left( \frac{1}{2} + \epsilon \right) }h\left( s\right) {n}^{-s}{ds} = \frac{1}{2\pi }{\int }_{-\infty }^{\infty }h\left( {it}\right) {e}^{-{it}\log n}{dt} \n\]\n\n\[ \n= \frac{1}{2\pi }{\int }_{-\infty }^{\infty }{h}_{0}\left( {t/{2\pi }}\right) {e}^{-{it}\log n}{dt} \n\]\n\n\[ \n= \frac{1}{2\pi }{\int }_{-\infty }^{\infty }{h}_{0}\left( t\right) {e}^{-{2\pi it}\log n}{dt} \n\]\n\n\[ \n= {\widehat{h}}_{0}\left( {\log n}\right) \text{.} \n\]\n\nSimilarly, we can also move the other integrals to \( \operatorname{Re}\left( s\right) = 0 \), which gives rise to the other terms of the formula. This completes the proof.
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Yes
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Theorem 8.1.3 (Phragmén - Lindelöf) Suppose that \( f\left( s\right) \) is entire in the region\n\n\[ \nS\left( {a, b}\right) = \{ s \in \mathbb{C} : a \leq \operatorname{Re}\left( s\right) \leq b\} \]\n\nand that as \( \left| t\right| \rightarrow \infty \),\n\n\[ \n\left| {f\left( s\right) }\right| = O\left( {e}^{{\left| t\right| }^{\alpha }}\right) \]\n\nfor some \( \alpha \geq 1 \) . If \( f\left( s\right) \) is bounded on the two vertical lines \( \operatorname{Re}\left( s\right) = a \) and \( \operatorname{Re}\left( s\right) = b \), then \( f\left( s\right) \) is bounded in \( S\left( {a, b}\right) \) .
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Proof. We first select an integer \( m > \alpha, m \equiv 2\left( {\;\operatorname{mod}\;4}\right) \) . Since arg \( s \rightarrow \) \( \pi /2 \) as \( t \rightarrow \infty \), we can choose \( {T}_{1} \) sufficiently large so that\n\n\[ \n\left| {\arg s - \pi /2}\right| < \pi /{4m} \]\n\nThen for \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{1} \), we find that \( \arg s = \pi /2 - \delta = \theta \) (say) satisfies\n\n\[ \n\cos {m\theta } = - \cos {m\delta } < - 1/\sqrt{2}. \]\n\nTherefore, if we consider\n\n\[ \n{g}_{\epsilon }\left( s\right) = {e}^{\epsilon {s}^{m}}f\left( s\right) \]\n\nthen\n\n\[ \n\left| {{g}_{\epsilon }\left( s\right) }\right| \leq K{e}^{{\left| t\right| }^{\alpha }}{e}^{-\epsilon {\left| s\right| }^{m}/\sqrt{2}}. \]\n\nThus, \( \left| {{g}_{\epsilon }\left( s\right) }\right| \rightarrow 0 \) as \( \left| t\right| \rightarrow \infty \) . Let \( B \) be the maximum of \( f\left( s\right) \) in the region\n\n\[ \na \leq \operatorname{Re}\left( s\right) \leq b,\;0 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq {T}_{1} \]\n\nLet \( {T}_{2} \) be chosen such that\n\n\[ \n\left| {{g}_{\epsilon }\left( s\right) }\right| \leq B \]\n\nfor \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{2} \) . Thus,\n\n\[ \n\left| {f\left( s\right) }\right| \leq B{e}^{-\epsilon {\left| s\right| }^{m}\cos \left( {m\arg s}\right) } \leq B{e}^{\epsilon {\left| s\right| }^{m}} \]\n\nfor \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{2} \) . Applying the maximum modulus principle to the region\n\n\[ \na \leq \operatorname{Re}\left( s\right) \leq b,\;0 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq {T}_{2}, \]\n\nwe find that \( \left| {f\left( s\right) }\right| \leq B{e}^{\epsilon {\left| s\right| }^{m}} \) . This estimate holds for all \( s \) in \( S\left( {a, b}\right) \) . Letting \( \epsilon \rightarrow 0 \) yields the result.
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Yes
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Corollary 8.1.4 Suppose that \( f\left( s\right) \) is entire in \( S\left( {a, b}\right) \) and that \( \left| {f\left( s\right) }\right| \) \( = O\left( {e}^{{\left| t\right| }^{\alpha }}\right) \) for some \( \alpha \geq 1 \) as \( \left| t\right| \rightarrow \infty \) . If \( f\left( s\right) \) is \( O\left( {\left| t\right| }^{A}\right) \) on the two vertical lines \( \operatorname{Re}\left( s\right) = a \) and \( \operatorname{Re}\left( s\right) = b \), then \( f\left( s\right) = O\left( {\left| t\right| }^{A}\right) \) in \( S\left( {a, b}\right) \) .
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Proof. We apply the theorem to the function \( g\left( s\right) = f\left( s\right) /{\left( s - u\right) }^{A} \) , where \( u > b \) . Then \( g \) is bounded on the two vertical strips, and the result follows.
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Yes
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Theorem 8.2.1 (Selberg) For any \( F \in \mathcal{S} \), let \( {N}_{F}\left( T\right) \) be the number of zeros \( \rho \) of \( F\left( s\right) \) satisfying \( 0 \leq \operatorname{Im}\left( \rho \right) \leq T \), counted with multiplicity. Then\n\n\[ \n{N}_{F}\left( T\right) \sim \left( {2\mathop{\sum }\limits_{{i = 1}}^{d}{\alpha }_{i}}\right) \frac{T\log T}{2\pi } \n\]\n\nas \( T \rightarrow \infty \) .
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Proof. This is easily derived by the method used to count zeros of \( \zeta \left( s\right) \) and \( L\left( {s,\chi }\right) \) as in Theorem 7.1.7 and Exercise 7.4.4.
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No
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Lemma 8.2.2 (Conrey and Ghosh) If \( F \in S \) and \( \deg F = 0 \), then \( F = 1 \) .
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Proof. We follow [CG]. A Dirichlet series can be viewed as a power series in the infinitely many variables \( {p}^{-s} \) as we range over primes \( p \) . Thus, if \( \deg F = 0 \), we can write our functional equation as\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{\left( \frac{{Q}^{2}}{n}\right) }^{s} = {wQ}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\overline{{a}_{n}}}{n}{n}^{s} \]\n\nwhere \( \left| w\right| = 1 \) .\n\nThus, if \( {a}_{n} \neq 0 \) for some \( n \), then \( {Q}^{2}/n \) is an integer. Hence \( {Q}^{2} \) is an integer. Moreover, \( {a}_{n} \neq 0 \) implies \( n \mid {Q}^{2} \), so that our Dirichlet series is really a Dirichlet polynomial. Therefore, if \( {Q}^{2} = 1 \), then \( F = 1 \), and we are done. So, let us suppose \( q \mathrel{\text{:=}} {Q}^{2} > 1 \) . Since \( {a}_{1} = \) 1, comparing the \( {Q}^{2s} \) term in the functional equation above gives \( \left| {a}_{q}\right| = Q \) . Since \( {a}_{n} \) is multiplicative, we must have for some prime power \( {p}^{r}\parallel q \) that \( \left| {a}_{{p}^{r}}\right| \geq {p}^{r/2} \) . Now consider the \( p \) -Euler factor\n\n\[ {F}_{p}\left( s\right) = \mathop{\sum }\limits_{{j = 0}}^{r}\frac{{a}_{{p}^{j}}}{{p}^{js}} \]\n\nwith logarithm\n\n\[ \log {F}_{p}\left( s\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }\frac{{b}_{{p}^{j}}}{{p}^{js}} \]\n\nViewing these as power series in \( x = {p}^{-s} \), we write\n\n\[ P\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{r}{A}_{j}{x}^{j} \]\n\n\[ \log P\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{B}_{j}{x}^{j} \]\n\nwhere \( {A}_{j} = {a}_{{p}^{j}},{B}_{j} = {b}_{{p}^{j}} \) . Since \( {a}_{1} = 1 \), we can factor\n\n\[ P\left( x\right) = \mathop{\prod }\limits_{{j = 1}}^{r}\left( {1 - {R}_{i}x}\right) \]\n\nso that\n\n\[ {B}_{j} = - \mathop{\sum }\limits_{{i = 1}}^{r}\frac{{R}_{i}^{j}}{j} \]\n\nWe also know that\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{r}\left| {R}_{i}\right| \geq {p}^{r/2} \]\n\nso that\n\n\[ \mathop{\max }\limits_{{1 \leq i \leq r}}\left| {R}_{i}\right| \geq {p}^{1/2} \]\n\nBut\n\n\[ {\left| {b}_{{p}^{j}}\right| }^{1/j} = {\left| {B}_{j}\right| }^{1/j} = {\left| \mathop{\sum }\limits_{{i = 1}}^{r}\frac{{R}_{i}^{j}}{j}\right| }^{1/j} \]\n\ntends to \( \mathop{\max }\limits_{{1 \leq i \leq r}}\left| {R}_{i}\right| \) as \( j \rightarrow \infty \), which is greater than or equal to \( {p}^{1/2} \) . This contradicts the condition that \( {b}_{n} = O\left( {n}^{\theta }\right) \) with \( \theta < 1/2 \) . Therefore, \( Q = 1 \) and hence \( F = 1 \) .
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Yes
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Theorem 8.2.3 (Selberg) If \( F \in \mathcal{S} \) and \( F \) is of positive degree, then \( \deg F \geq 1 \) .
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Proof. We follow [CG]. Consider the identity\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}^{-{nx}} = \frac{1}{2\pi i}{\int }_{\left( 2\right) }F\left( s\right) {x}^{-s}\Gamma \left( s\right) {ds}. \]\n\nBecause of the Phragmen - Lindelöf principle and the functional equation, we find that \( F\left( s\right) \) has polynomial growth in \( \left| {\operatorname{Im}\left( s\right) }\right| \) in any vertical strip. Thus, moving the line of integration to the left, and taking into account the possible pole at \( s = 1 \) of \( F\left( s\right) \) as well as the poles of \( \Gamma \left( s\right) \) at \( s = 0, - 1, - 2,\ldots \), we obtain\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}^{-{nx}} = \frac{P\left( {\log x}\right) }{x} + \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{F\left( {-n}\right) {\left( -1\right) }^{n}{x}^{n}}{n!}, \]\n\nwhere \( P \) is a polynomial. The functional equation relates \( F\left( {-n}\right) \) to \( F\left( {n + 1}\right) \) with a product of gamma functions. If \( 0 < \deg F < 1 \) , we find by Stirling's formula that the sum on the right-hand side converges for all \( x \) . Moreover, \( P\left( {\log x}\right) \) is analytic in \( \mathbb{C} \smallsetminus \{ x \leq 0 : x \in \mathbb{R}\} \) . Hence the left-hand side is analytic in \( \mathbb{C} \smallsetminus \{ x \leq 0 : x \in \mathbb{R}\} \) . But since the left-hand side is periodic with period \( {2\pi i} \), we find that\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}^{-{nz}} \]\n\nis entire. Thus, for any \( x \) ,\n\n\[ {a}_{n}{e}^{-{nx}} = {\int }_{0}^{2\pi }f\left( {x + {iy}}\right) {e}^{iny}{dy} \ll {n}^{-2} \]\n\nby integrating by parts. Choosing \( x = 1/n \) gives \( {a}_{n} = O\left( {1/{n}^{2}}\right) \) . Hence the Dirichlet series\n\n\[ F\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}} \]\n\nconverges absolutely for \( \operatorname{Re}s > - 1 \) . However, relating \( F\left( {-1/2 + {it}}\right) \) to \( F\left( {3/2 - {it}}\right) \) by the functional equation and using Stirling’s formula, we find that \( F\left( {-1/2 + {it}}\right) \) is not bounded. This contradiction forces \( \deg F \geq 1 \) .
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Yes
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Example 9.1.1 (Eratosthenes-Legendre) Let \( {P}_{z} \) be the product of the primes \( p \leq z \), and \( \pi \left( {x, z}\right) \) the number of \( n \leq x \) that are not divisible by any prime \( p \leq z \). Then\n\n\[ \pi \left( {x, z}\right) = \mathop{\sum }\limits_{{d \mid {P}_{z}}}\mu \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack . \]
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Solution. Clearly,\n\n\[ \pi \left( {x, z}\right) = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{d \mid \left( {n,{P}_{z}}\right) }}\mu \left( d\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \mid {P}_{z}}}\mu \left( d\right) \mathop{\sum }\limits_{\substack{{n \leq x} \\ {d \mid n} }}1 = \mathop{\sum }\limits_{{d \mid {P}_{z}}}\mu \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack \] \n\nas required.
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Yes
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Theorem 10.1.5 (Ostrowski) Every nontrivial norm \( \parallel \cdot \parallel \) on \( \mathbb{Q} \) is equivalent to \( {\left| \cdot \right| }_{p} \) for some prime \( {\left. p\text{or}\left| \cdot \right| \right| }_{\infty } \) .
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Proof. Case (i): Suppose there is a natural number \( n \) such that \( \left| \right| n\left| \right| > \) 1. Let \( {n}_{0} \) be the least such \( n \) . We know that \( {n}_{0} > 1 \), so we can write \( \begin{Vmatrix}{n}_{0}\end{Vmatrix} = {n}_{0}^{\alpha } \) for some positive \( \alpha \) . Write any natural number \( n \) in base \( {n}_{0} \) :\n\n\[ n = {a}_{0} + {a}_{1}{n}_{0} + \cdots + {a}_{s}{n}_{0}^{s},\;0 \leq {a}_{i} < {n}_{0}, \]\n\nand \( {a}_{s} \neq 0 \) . Then, by the triangle inequality,\n\n\[ \parallel n\parallel \leq \begin{Vmatrix}{a}_{0}\end{Vmatrix} + \begin{Vmatrix}{{a}_{1}{n}_{0}}\end{Vmatrix} + \cdots + \begin{Vmatrix}{{a}_{s}{n}_{0}^{s}}\end{Vmatrix} \]\n\n\[ \leq \begin{Vmatrix}{a}_{0}\end{Vmatrix} + \begin{Vmatrix}{a}_{1}\end{Vmatrix}{n}_{0}^{\alpha } + \cdots + \begin{Vmatrix}{a}_{s}\end{Vmatrix}{n}_{0}^{\alpha s}. \]\n\nSince all the \( {a}_{i} \) are less than \( {n}_{0} \), we have \( \begin{Vmatrix}{a}_{i}\end{Vmatrix} \leq 1 \) . Hence,\n\n\[ \parallel n\parallel \leq 1 + {n}_{0}^{\alpha } + \cdots + {n}_{0}^{\alpha s} \]\n\n\[ \leq {n}_{0}^{\alpha s}\left( {1 + \frac{1}{{n}_{0}^{\alpha }} + \cdots }\right) . \]\n\nSince \( n > {n}_{0}^{s} \), we deduce \( \parallel n\parallel \leq C{n}^{\alpha } \) for some constant \( C \) and for all natural numbers \( n \) . Thus, \( \left| \right| {n}^{N}\left| \right| \leq C{n}^{N\alpha } \), so that \( \parallel n\parallel \leq {C}^{1/N}{n}^{\alpha } \) . Letting \( N \rightarrow \infty \) gives \( \parallel n\parallel \leq {n}^{\alpha } \) for all natural numbers \( n \) . We can also get the reverse inequality as follows: since \( {n}_{0}^{s + 1} > n \geq {n}_{0}^{s} \), we have\n\n\[ \begin{Vmatrix}{n}_{0}^{s + 1}\end{Vmatrix} = \begin{Vmatrix}{n + {n}_{0}^{s + 1} - n}\end{Vmatrix} \]\n\n\[ \leq \parallel n\parallel + \begin{Vmatrix}{{n}_{0}^{s + 1} - n}\end{Vmatrix} \]\n\nso that\n\n\[ \parallel n\parallel \geq \begin{Vmatrix}{{n}_{0}^{s + 1}\parallel - }\end{Vmatrix}{n}_{0}^{s + 1} - n\parallel \]\n\n\[ \geq {n}_{0}^{\left( {s + 1}\right) \alpha } - {\left( {n}_{0}^{s + 1} - n\right) }^{\alpha }. \]\n\nThus,\n\n\[ \parallel n\parallel \geq {n}_{0}^{\left( {s + 1}\right) \alpha } - {\left( {n}_{0}^{s + 1} - {n}_{0}^{s}\right) }^{\alpha }, \]\nsince \( n \geq {n}_{0}^{s} \), so that\n\n\[ \parallel n\parallel \geq {n}_{0}^{\left( {s + 1}\right) \alpha }\left( {1 - {\left( 1 - \frac{1}{{n}_{0}}\right) }^{\alpha }}\right) \]\n\n\[ \geq \;{C}_{1}{n}^{\alpha } \]\n\nfor some constant \( {C}_{1} \) . Repeating the previous argument gives \( \parallel n\parallel \geq \) \( {n}^{\alpha } \) and therefore \( \parallel n\parallel = {n}^{\alpha } \) for all natural numbers \( n \) . Thus, \( \parallel \cdot \parallel \) is equivalent to \( {\left| \cdot \right| }_{\infty } \) .
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Yes
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Theorem 10.1.8 \( {\mathbb{Q}}_{p} \) is complete with respect to \( {\left| \cdot \right| }_{p} \) .
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Proof. Let \( {\left\{ {a}^{\left( j\right) }\right\} }_{j = 1}^{\infty } \) be a Cauchy sequence of equivalence classes in \( {\mathbb{Q}}_{p} \) . We must show that there is a Cauchy sequence to which it converges. We write \( {a}^{\left( j\right) } = {\left\{ {a}_{n}^{\left( j\right) }\right\} }_{n = 1}^{\infty } \) and set \( s = {\left\{ {a}_{j}^{\left( j\right) }\right\} }_{j = 1}^{\infty } \), the \
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No
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Theorem 10.1.11 Every equivalence class \( s \) in \( {\mathbb{Q}}_{p} \) for which \( {\left| s\right| }_{p} \leq 1 \) has exactly one representative Cauchy sequence \( {\left\{ {a}_{i}\right\} }_{i = 1}^{\infty } \) satisfying \( 0 \leq {a}_{i} < {p}^{i} \) and \( {a}_{i} \equiv {a}_{i + 1}\left( {\;\operatorname{mod}\;{p}^{i}}\right) \) for \( i = 1,2,3,\ldots \) .
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Proof. The uniqueness is clear, for if \( {\left\{ {a}_{i}^{\prime }\right\} }_{i = 1}^{\infty } \) is another such sequence, we have \( {a}_{i} \equiv {a}_{i}^{\prime }\left( {\;\operatorname{mod}\;{p}^{i}}\right) \), which forces \( {a}_{i} = {a}_{i}^{\prime } \) . Now let \( {\left\{ {c}_{i}\right\} }_{i = 1}^{\infty } \) be a Cauchy sequence of \( {\mathbb{Q}}_{p} \) in \( s \) . Then for each \( j \), there is an \( N\left( j\right) \) such that\n\n\[{\left| {c}_{i} - {c}_{k}\right| }_{p} \leq {p}^{-j}\]\n\nfor \( i, k \geq N\left( j\right) \) . Without loss of generality, we may take \( N\left( j\right) \geq j \) . Since \( {\left| s\right| }_{p} \leq 1 \), we have \( {\left| {c}_{i}\right| }_{p} \leq 1 \) for \( i \geq N\left( 1\right) \) because\n\n\[{\left| {c}_{i}\right| }_{p} \leq \max \left( {{\left| {c}_{k}\right| }_{p},{\left| {c}_{i} - {c}_{k}\right| }_{p}}\right)\]\n\n\[ \leq \max \left( {{\left| {c}_{k}\right| }_{p},1/p}\right)\]\n\nso that by choosing a sufficiently large \( k \) we are ensured that \( {\left| {c}_{k}\right| }_{p} \leq 1 \), since \( {\left| s\right| }_{p} = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\left| {c}_{k}\right| }_{p} \leq 1 \) . By Exercise 10.1.10, we can find a sequence of integers \( {a}_{j} \) such that\n\n\[{\left| {a}_{j} - {c}_{N\left( j\right) }\right| }_{p} \leq {p}^{-j}\]\n\nwith \( 0 \leq {a}_{j} < {p}^{j} \) . The claim is that \( {\left\{ {a}_{j}\right\} }_{j = 1}^{\infty } \) is the required sequence. First observe that by the triangle inequality,\n\n\[{\left| {a}_{j + 1} - {a}_{j}\right| }_{p} \leq \max \left( {{\left| {a}_{j + 1} - {c}_{N\left( {j + 1}\right) }\right| }_{p},{\left| {c}_{N\left( {j + 1}\right) } - {c}_{N\left( j\right) }\right| }_{p},}\right.\]\n\n\[\left. {\left| {c}_{N\left( j\right) } - {a}_{j}\right| }_{p}\right)\]\n\n\[\leq \max \left( {{p}^{-j - 1},{p}^{-j},{p}^{-j}}\right) = {p}^{-j}\]\n\nso that\n\n\[{a}_{j} \equiv {a}_{j + 1}\left( {\;\operatorname{mod}\;{p}^{j}}\right)\]\n\nfor \( i = 1,2,\ldots \) . Second, for any \( j \), and \( i \geq N\left( j\right) \), we have\n\n\[{\left| {a}_{i} - {c}_{i}\right| }_{p} \leq \max \left( {{\left| {a}_{i} - {a}_{j}\right| }_{p},{\left| {a}_{j} - {c}_{N\left( j\right) }\right| }_{p},{\left| {c}_{N\left( j\right) } - {c}_{j}\right| }_{p}}\right)\]\n\n\[\leq \max \left( {{p}^{-j},{p}^{-j},{p}^{-j}}\right) = {p}^{-j}\]\n\nso that \( \mathop{\lim }\limits_{{i \rightarrow \infty }}{\left| {a}_{i} - {c}_{i}\right| }_{p} = 0 \) .
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Yes
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Show that \( {x}^{2} = 6 \) has a solution in \( {\mathbb{Q}}_{5} \) .
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The equation \( {x}^{2} \equiv 6\left( {\;\operatorname{mod}\;5}\right) \) has a solution (namely \( x \equiv \) 1 (mod 5)). We will show inductively that \( {x}^{2} \equiv 6\left( {\;\operatorname{mod}\;{5}^{n}}\right) \) has a solution for every \( n \geq 1 \) . Suppose\n\n\[ \n{x}_{n}^{2} \equiv 6\left( {\;\operatorname{mod}\;{5}^{n}}\right) \n\]\n\nWe want to find \( {x}_{n + 1}^{2} \equiv 6\left( {\;\operatorname{mod}\;{5}^{n + 1}}\right) \) . Write \( {x}_{n + 1} = {5}^{n}t + {x}_{n} \) . So we must have\n\n\[ \n{\left( {5}^{n}t + {x}_{n}\right) }^{2} \equiv 6\left( {\;\operatorname{mod}\;{5}^{n + 1}}\right) \n\]\n\nwhich means that \( 2 \cdot {5}^{n}t{x}_{n} + {x}_{n}^{2} \equiv 6\left( {\;\operatorname{mod}\;{5}^{n + 1}}\right) \) . This reduces to\n\n\[ \n{2t}{x}_{n} + \frac{{x}_{n}^{2} - 6}{{5}^{n}} \equiv 0\left( {\;\operatorname{mod}\;5}\right) \n\]\n\nso that we can clearly solve for \( t \) . The method produces a sequence of integers \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) such that \( {x}_{n}^{2} \equiv 6\left( {\;\operatorname{mod}\;{5}^{n}}\right) \) and \( {x}_{n + 1} \equiv {x}_{n}\left( {\;\operatorname{mod}\;{5}^{n}}\right) \) . The sequence is therefore Cauchy and its limit \( x \) (which exists in \( {\mathbb{Q}}_{p} \) by completeness) satisfies \( {x}^{2} = 6 \) .
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Yes
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Theorem 10.2.8 \( {\left| \cdot \right| }_{p} \) is a nonarchimedean norm on \( K \) .
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Proof. It is clear that \( {\left| x\right| }_{p} = 0 \) if and only if \( x = 0 \) . It is also clear that \( {\left| xy\right| }_{p} = {\left| x\right| }_{p}{\left| y\right| }_{p} \), since the norm is multiplicative. To prove that\n\n\[ \n{\left| x + y\right| }_{p} \leq \max \left( {{\left| x\right| }_{p},{\left| y\right| }_{p}}\right) \n\] \n\nwe see (upon dividing by \( y \) ) that it suffices to prove for \( \alpha \in K \) ,\n\n\[ \n{\left| \alpha + 1\right| }_{p} \leq \max \left( {{\left| \alpha \right| }_{p},1}\right) \n\] \n\nIt is easily seen that this follows if we can show\n\n\[ \n{\left| \alpha \right| }_{p} \leq 1 \Rightarrow {\left| \alpha - 1\right| }_{p} \leq 1 \n\] \n\nThat is, we must show\n\n\[ \n{\left| {N}_{K/{\mathbb{Q}}_{p}}\left( \alpha \right) \right| }_{p} \leq 1 \Rightarrow {\left| {N}_{K/{\mathbb{Q}}_{p}}\left( \alpha - 1\right) \right| }_{p} \leq 1. \n\] \n\nThis reduces to showing\n\n\[ \n{N}_{K/{\mathbb{Q}}_{p}}\left( \alpha \right) \in {\mathbb{Z}}_{p} \Rightarrow {N}_{K/{\mathbb{Q}}_{p}}\left( {\alpha - 1}\right) \in {\mathbb{Z}}_{p}. \n\] \n\nIt is now necessary to use a little bit of commutative algebra. Clearly, \( {\mathbb{Q}}_{p}\left( \alpha \right) = {\mathbb{Q}}_{p}\left( {\alpha - 1}\right) \) . Now let\n\n\[ \nf\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \n\] \n\nbe the minimal polynomial for \( \alpha \) . The minimal polynomial for \( \alpha - 1 \) is clearly\n\n\[ \nf\left( {x + 1}\right) = {x}^{n} + \left( {{a}_{n - 1} + n}\right) {x}^{n - 1} + \cdots + \left( {1 + {a}_{n - 1} + \cdots + {a}_{1} + {a}_{0}}\right) . \n\] \n\nNow \( {N}_{K/{\mathbb{Q}}_{p}}\left( \alpha \right) = {\left( -1\right) }^{n}{a}_{0} \) and\n\n\[ \n{N}_{K/{\mathbb{Q}}_{p}}\left( {\alpha - 1}\right) = {\left( -1\right) }^{n}\left( {1 + {a}_{n - 1} + \cdots + {a}_{1} + {a}_{0}}\right) . \n\] \n\nWe now use the polynomial form of Hensel's lemma. If all the coefficients of \( f\left( x\right) \) are in \( {\mathbb{Z}}_{p} \), we are done. So, assume that\n\n\[ \nf\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \n\] \n\nis such that \( {a}_{0} \in {\mathbb{Z}}_{p} \) but some \( {a}_{i} \notin {\mathbb{Z}}_{p} \) . Choose \( m \) to be the smallest exponent such that \( {p}^{m}{a}_{i} \in {\mathbb{Z}}_{p} \) for all \( i \) and now \
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No
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Theorem 10.3.12 (Mahler,1961) Suppose \( f : {\mathbb{Z}}_{p} \rightarrow {\mathbb{Q}}_{p} \) is continuous. Let\n\n\[ \n{a}_{n}\left( f\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{\left( -1\right) }^{n - k}\left( \begin{array}{l} n \\ k \end{array}\right) f\left( k\right) .\n\]\n\nThen the series\n\n\[ \n\mathop{\sum }\limits_{{k = 0}}^{\infty }\left( \begin{array}{l} x \\ k \end{array}\right) {a}_{k}\left( f\right)\n\]\n\nconverges uniformly in \( {\mathbb{Z}}_{p} \) and\n\n\[ \nf\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\left( \begin{array}{l} x \\ k \end{array}\right) {a}_{k}\left( f\right)\n\]
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Proof. We know that given any positive integer \( s \), there exists a positive integer \( t \) such that for \( x, y \in {\mathbb{Z}}_{p} \), \n\n\[ \n{\left| x - y\right| }_{p} \leq {p}^{-t} \Rightarrow {\left| f\left( x\right) - f\left( y\right) \right| }_{p} \leq {p}^{-s}.\n\]\n\nIn particular,\n\n\[ \n{\left| f\left( k + {p}^{t}\right) - f\left( k\right) \right| }_{p} \leq {p}^{-s}\n\]\n\nfor \( k = 0,1,2,\ldots \) .\n\nSince \( f \) is continuous on \( {\mathbb{Z}}_{p} \), it is bounded there (recall that \( {\mathbb{Z}}_{p} \) is compact), and so we may suppose without loss of generality that \( {\left| f\left( x\right) \right| }_{p} \leq 1 \) for all \( x \in {\mathbb{Z}}_{p} \) . Hence,\n\n\[ \n{\left| {a}_{n}\left( f\right) \right| }_{p} \leq 1\;\text{ for }\;n = 0,1,2,\ldots\n\]\n\nNow by Exercise 10.3.10,\n\n\[ \n{a}_{n + {p}^{t}}\left( f\right) = - \mathop{\sum }\limits_{{j = 1}}^{{{p}^{t} - 1}}\left( \begin{matrix} {p}^{t} \\ j \end{matrix}\right) {a}_{n + j}\left( f\right) + \mathop{\sum }\limits_{{k = 0}}^{n}{\left( -1\right) }^{n - k}\left( \begin{array}{l} n \\ k \end{array}\right) \left\{ {f\left( {k + {p}^{t}}\right) - f\left( k\right) }\right\} .\n\]\n\nBy Exercise 10.3.6, \( p \mid \left( \begin{matrix} {p}^{t} \\ j \end{matrix}\right) \) for \( 1 \leq j \leq {p}^{t} - 1 \), so that\n\n\[ \n{\left| {a}_{n + {p}^{t}}\left( f\right) \right| }_{p} \leq \mathop{\max }\limits_{{1 \leq j < {p}^{t}}}\left\{ {{p}^{-1}{\left| {a}_{n + j}\left( f\right) \right| }_{p},{p}^{-s}}\right\} .\n\]\n\nSince \( {\left| {a}_{n}\left( f\right) \right| }_{p} \leq 1 \), we obtain\n\n\[ \n{\left| {a}_{n}\left( f\right) \right| }_{p} \leq {p}^{-1}\;\text{ for }\;n \geq {p}^{t}\n\]\n\nReplacing \( n \) by \( n + {p}^{t} \) in the penultimate inequality and using the above inequality, we obtain\n\n\[ \n{\left| {a}_{n}\left( f\right) \right| }_{p} \leq {p}^{-2}\;\text{ for }\;n \geq 2{p}^{t}\n\]\n\nRepeating the argument \( \left( {s - 1}\right) \) times gives\n\n\[ \n{\left| {a}_{n}\left( f\right) \right| }_{p} \leq {p}^{-s}\;\text{ for }\;n \geq s{p}^{t}\n\]\n\nThis proves \( {a}_{n}\left( f\right) \rightarrow 0 \) as \( n \rightarrow \infty \) . By Exercise 10.3.11, we have\n\n\[ \n{\left| \left( \begin{array}{l} x \\ n \end{array}\right) \right| }_{p} \leq 1\n\]\n\nfor \( x \in {\mathbb{Z}}_{p} \) . Therefore, the series\n\n\[ \n\mathop{\sum }\limits_{{k = 0}}^{\infty }\left( \begin{array}{l} x \\ k \end{array}\right) {a}_{k}\left( f\right)\n\]\n\nconverges uniformly on \( {\mathbb{Z}}_{p} \) and thus defines a continuous function. Since this function agrees with \( f\left( n\right) \) on the natural numbers and \( \mathbb{N} \) is dense in \( {\mathbb{Z}}_{p} \), we deduce the result.
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Yes
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Theorem 10.4.7 (Mazur, 1972)\n\n\\[ \n- \\left( {1 - {p}^{k - 1}}\\right) {B}_{k}/k = \\frac{1}{{\\alpha }^{-k} - 1}\\int _{{\\mathbb{Z}}_{p}^{ * }}{x}^{k - 1}d{\\mu }_{1,\\alpha }.\n\\]
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By Exercise 8.2.12, we can interpret the left hand side of the equation in Theorem 10.4.7 as\n\n\\[ \n\\left( {1 - {p}^{k - 1}}\\right) \\zeta \\left( {1 - k}\\right)\n\\]
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No
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Theorem 11.1.5 [Weyl,1916] A sequence \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) is u.d. if and only if\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}} = o\left( N\right) ,\;m = \pm 1, \pm 2,\ldots \]
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Proof. As observed earlier, the necessity is clear. For sufficiency, let \( \epsilon > 0 \) and \( f \) a continuous function \( f : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{C} \) . By the Weierstrass approximation theorem, there is a trigonometric polynomial \( \phi \left( x\right) \) such that \( \deg \phi \leq M \), with \( M \) depending on \( \epsilon \) such that\n\n\[ \mathop{\sup }\limits_{{0 \leq x \leq 1}}\left| {f\left( x\right) - \phi \left( x\right) }\right| \leq \epsilon \]\n\nThen,\n\n\[ \left| {{\int }_{0}^{1}f\left( x\right) {dx} - \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {x}_{n}\right) }\right| \]\n\n\[ \leq \left| {{\int }_{0}^{1}(f\left( x\right) - \phi \left( x\right) {dx}}\right| + \left| {{\int }_{0}^{1}\phi \left( x\right) {dx} - \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {x}_{n}\right) }\right| .\n\nBy (11.2), the first term is \( \leq \epsilon \) . The second term is\n\n\[ \leq \left| {{\int }_{0}^{1}\phi \left( x\right) {dx} - \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}\phi \left( {x}_{n}\right) }\right| + \left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}\left( {\phi \left( {x}_{n}\right) - f\left( {x}_{n}\right) }\right) }\right| .\n\nAgain by (11.2), the last term is \( \leq \epsilon \) . Writing\n\n\[ \phi \left( x\right) = \mathop{\sum }\limits_{{\left| m\right| \leq M}}{a}_{m}{e}^{2\pi imx} \]\n\nwe see that\n\n\[ {\int }_{0}^{1}\phi \left( x\right) {dx} = {a}_{0} \]\n\nand\n\n\[ \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}\phi \left( {x}_{n}\right) = {a}_{0} + \mathop{\sum }\limits_{{1 \leq \left| m\right| \leq M}}{a}_{m}\left( {\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}}\right) ,\n\nso that\n\n\[ \left| {{\int }_{0}^{1}\phi \left( x\right) {dx} - \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}\phi \left( {x}_{n}\right) }\right| \leq \mathop{\sum }\limits_{{1 \leq \left| m\right| \leq M}}\left| {a}_{m}\right| \left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}}\right| .\n\nLet \( T = \mathop{\sum }\limits_{{1 \leq \left| m\right| \leq M}}\left| {a}_{m}\right| \) . We may choose \( N \) (which depends on \( M \) )\nsufficiently large so that all of the inner terms above are \( \leq \epsilon /T \) by virtue of (11.1). Thus, this term is also \( \leq \epsilon \) . Hence,\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {x}_{n}\right) = {\int }_{0}^{1}f\left( x\right) {dx}. \]\n\nThis completes the proof.
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Yes
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Theorem 11.1.16 (van der Corput,1931) Let \( {y}_{1},\ldots ,{y}_{N} \) be complex numbers. Let \( H \) be an integer with \( 1 \leq H \leq N \). Then\n\n\[ \n{\left| \mathop{\sum }\limits_{{n = 1}}^{N}{y}_{n}\right| }^{2} \leq \n\]\n\n\[ \n\frac{N + H}{H + 1}\mathop{\sum }\limits_{{n = 1}}^{N}{\left| {y}_{n}\right| }^{2} + \frac{2\left( {N + H}\right) }{H + 1}\mathop{\sum }\limits_{{r = 1}}^{H}\left( {1 - \frac{r}{H + 1}}\right) \left| {\mathop{\sum }\limits_{{n = 1}}^{{N - r}}{y}_{n + r}{\bar{y}}_{n}}\right| .\n\]
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Proof. It is convenient to set \( {y}_{n} = 0 \) for \( n \leq 0 \) and \( n > N \). Clearly,\n\n\[ \n{\left( H + 1\right) }^{2}{\left| \mathop{\sum }\limits_{n}{y}_{n}\right| }^{2} = {\left| \mathop{\sum }\limits_{{h = 0}}^{H}\mathop{\sum }\limits_{n}{y}_{n + h}\right| }^{2} = {\left| \mathop{\sum }\limits_{n}\mathop{\sum }\limits_{{h = 0}}^{H}{y}_{n + h}\right| }^{2}.\n\]\n\nWe note that the inner sum is zero if \( n \geq N + 1 \) or \( n \leq - H \). Thus, in the outer sum, \( n \) is restricted to the interval \( \left\lbrack {-H + 1, N}\right\rbrack \). Applying the Cauchy-Schwarz inequality, we get that this is\n\n\[ \n\leq \left( {N + H}\right) \mathop{\sum }\limits_{n}{\left| \mathop{\sum }\limits_{{h = 0}}^{H}{y}_{n + h}\right| }^{2}\n\]\n\nExpanding the sum, we obtain\n\n\[ \n\mathop{\sum }\limits_{n}\mathop{\sum }\limits_{{h = 0}}^{H}\mathop{\sum }\limits_{{k = 0}}^{H}{y}_{n + h}{\bar{y}}_{n + k} = \left( {H + 1}\right) \mathop{\sum }\limits_{n}{\left| {y}_{n}\right| }^{2} + \mathop{\sum }\limits_{n}\mathop{\sum }\limits_{{h \neq k}}{y}_{n + h}{\bar{y}}_{n + k}.\n\]\n\nIn the second sum, we combine the terms corresponding to \( \left( {h, k}\right) \) and \( \left( {k, h}\right) \) to get that it is\n\n\[ \n2\operatorname{Re}\left( {\mathop{\sum }\limits_{n}\mathop{\sum }\limits_{{h = 0}}^{H}\mathop{\sum }\limits_{{k < h}}{y}_{n + h}{\bar{y}}_{n + k}}\right) .\n\]\n\nWe write \( m = n + k \) and re-write this as\n\n\[ \n2\operatorname{Re}\left( {\mathop{\sum }\limits_{m}\mathop{\sum }\limits_{{h = 0}}^{H}\mathop{\sum }\limits_{{k < h}}{y}_{m - k + h}{\bar{y}}_{m}}\right) = 2\operatorname{Re}\left( {\mathop{\sum }\limits_{m}\mathop{\sum }\limits_{{r = 1}}^{H}{y}_{m + r}{\bar{y}}_{m}\mathop{\sum }\limits_{{k < h;h - k = r}}1}\right) .\n\]\n\nThe innermost sum is easily seen to be \( H + 1 - r \). Therefore,\n\n\[ \n{\left| \mathop{\sum }\limits_{{n = 1}}^{N}{y}_{n}\right| }^{2} \leq \n\]\n\n\[ \n\frac{N + H}{H + 1}\mathop{\sum }\limits_{{n = 1}}^{N}{\left| {y}_{n}\right| }^{2} + \frac{2\left( {N + H}\right) }{H + 1}\mathop{\sum }\limits_{{r = 1}}^{H}\left( {1 - \frac{r}{H + 1}}\right) \left| {\mathop{\sum }\limits_{{n = 1}}^{{N - r}}{y}_{n + r}{\bar{y}}_{n}}\right| .\n\]\n\nThis completes the proof.
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Yes
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Corollary 11.1.17 (van der Corput,1931) If for each positive integer \( r \) , the sequence \( {x}_{n + r} - {x}_{n} \) is u.d. mod 1, then the sequence \( {x}_{n} \) is u.d. mod 1 .
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Proof. We apply Theorem 11.1.16 with \( {y}_{n} = {e}^{{2\pi im}{x}_{n}} \) to get\n\n\[ \n{\left| \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}\right| }^{2} \n\]\n\n\[ \n\leq \frac{1 + H/N}{H + 1} + \frac{2\left( {N + H}\right) }{{N}^{2}\left( {H + 1}\right) }\mathop{\sum }\limits_{{r = 1}}^{H}\left( {1 - \frac{r}{H + 1}}\right) \left| {\mathop{\sum }\limits_{{n = 1}}^{{N - r}}{e}^{{2\pi im}\left( {{x}_{n + r} - {x}_{n}}\right) }}\right| . \n\]\n\nTaking the limit as \( N \rightarrow \infty \) and using the fact that \( {x}_{n + r} - {x}_{n} \) is u.d. \( {\;\operatorname{mod}\;1} \) for every \( r \geq 1 \), we see that\n\n\[ \n\mathop{\lim }\limits_{{N \rightarrow \infty }}{\left| \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}\right| }^{2} \ll \frac{1}{H} \n\]\n\nfor any \( H \) . Choosing \( H \) arbitrarily large gives the result.
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Yes
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Theorem 11.2.2 The number \( x \) is normal to the base \( b \) if and only if the sequence \( \left( {x{b}^{n}}\right) \) is u.d. mod 1 .
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Proof. Let \( {B}_{k} = {b}_{1}{b}_{2}\cdots {b}_{k} \) be a block of \( k \) digits. The block\n\n\[ \n{a}_{m}{a}_{m + 1}\cdots {a}_{m + k - 1} \n\]\n\nin the \( b \) -adic expansion of \( x \) is identical with \( {B}_{k} \) if and only if\n\n\[ \n\frac{{B}_{k}}{{b}^{k}} \leq \left( {x{b}^{m - 1}}\right) < \frac{{B}_{k} + 1}{{b}^{k}}. \n\]\n\nLet \( {I}_{k} \) denote this interval of length \( 1/{b}^{k} \) . If the sequence \( {\left\{ x{b}^{n}\right\} }_{n = 1}^{\infty } \) is u.d. then\n\n\[ \n\# \left\{ {m \leq N - k + 1 : x{b}^{m - 1} \in {I}_{k}}\right\} \sim N/{b}^{k} \n\]\n\nas \( N \) tends to infinity. Thus, \( x \) is normal to the base \( b \) . Conversely, if \( x \) is normal to the base \( b \), then for any rational number of the form \( y = a/{b}^{k} \), we have\n\n\( \# \left\{ {n \leq N : \left( {x{b}^{n}}\right) < a/{b}^{k}}\right\} = \# \left\{ {m \leq N - k + 1 : \left( {x{b}^{m - 1}}\right) < a/{b}^{k}}\right\} + O\left( k\right) . \)\n\nThis is easily seen to be equal to\n\n\[ \n\mathop{\sum }\limits_{{{B}_{k} < a}}\# \left\{ {m \leq N - k + 1 : \left( {x{b}^{m - 1}}\right) \in {I}_{k}}\right\} + O\left( k\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{{B}_{k} < a}}\left( {N/{b}^{k} + o\left( N\right) }\right) + O\left( k\right) \n\]\n\nwhich is \( {aN}/{b}^{k} + o\left( N\right) \) since \( x \) is normal to the base \( b \) . Since the numbers of the form \( a/{b}^{k} \) are dense in \( \left\lbrack {0,1}\right\rbrack \), the asymptotic above extends to all \( y \) with \( 0 \leq y < 1 \) . This completes the proof.
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Yes
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Theorem 11.3.3 [Wiener - Schoenberg,1928] The sequence \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) has a continuous a.d.f. if and only if for every integer \( m \), the limit\n\n\[ \n{a}_{m} \mathrel{\text{:=}} \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}} \]\n\nexists and\n\n\[ \n\mathop{\sum }\limits_{{m = 1}}^{N}{\left| {a}_{m}\right| }^{2} = o\left( N\right) \]\n\n(11.3)
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Proof. Suppose the sequence has a continuous a.d.f. \( g\left( x\right) \) . The existence of the limits is clear. Now, by Exercise 11.3.2, we have\n\n\[ \n{a}_{m} = {\int }_{0}^{1}{e}^{2\pi imx}{dg}\left( x\right) \]\n\nThus,\n\n\[ \n\mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{N}\mathop{\sum }\limits_{{m = 1}}^{N}{\left| {a}_{m}\right| }^{2} = \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{N}\mathop{\sum }\limits_{{m = 1}}^{N}{\int }_{0}^{1}{\int }_{0}^{1}{e}^{{2\pi im}\left( {x - y}\right) }{dg}\left( x\right) {dg}\left( y\right) .\n\nThis is equal to\n\n\[ \n\mathop{\lim }\limits_{{N \rightarrow \infty }}{\int }_{0}^{1}{\int }_{0}^{1}\left( {\frac{1}{N}\mathop{\sum }\limits_{{m = 1}}^{N}{e}^{{2\pi im}\left( {x - y}\right) }}\right) {dg}\left( x\right) {dg}\left( y\right) .\n\nBy the Lebesgue dominated convergence theorem this is equal to\n\n\[ \n{\int }_{0}^{1}{\int }_{0}^{1}\left( {\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{N}\mathop{\sum }\limits_{{m = 1}}^{N}{e}^{{2\pi im}\left( {x - y}\right) }}\right) {dg}\left( x\right) {dg}\left( y\right) .\n\nThe integrand is zero unless \( x - y \in \mathbb{Z} \), in which case it is 1 . The set of such \( \left( {x, y}\right) \in {\left\lbrack 0,1\right\rbrack }^{2} \) is a set of measure zero. Therefore the limit is zero. Conversely, suppose that the limit is zero. By the Riesz representation theorem, there is a measurable function \( g\left( x\right) \) such that\n\n\[ \n{a}_{m} = {\int }_{0}^{1}{e}^{2\pi imx}{dg}\left( x\right) \]\n\nConsequently,\n\n\[ \n{\int }_{0}^{1}{\int }_{0}^{1}f\left( {x - y}\right) {dg}\left( x\right) {dg}\left( y\right) = 0 \]\n\nwhere \( f\left( {x - y}\right) = 0 \) unless \( x - y \in \mathbb{Z} \), in which case it is 1 . We want to show that this implies that \( g \) is continuous. Indeed, if \( g \) has a jump discontinuity at \( c \) (say), the double integral is at least \( \lbrack g\left( {c + }\right) - \) \( {\left. g\left( c - \right) \right\rbrack }^{2} > 0 \) . This completes the proof.
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Yes
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Theorem 11.4.7 Let \( M \) be a natural number. For any interval \( I = \\left\\lbrack {a, b}\\right\\rbrack \) with length \( b - a < 1 \), write\n\n\[ \n{\\Xi }_{I}\\left( x\\right) = \\mathop{\\sum }\\limits_{{n \\in \\mathbb{Z}}}{\\chi }_{I}\\left( {n + x}\\right) \n\]\n\nThen, there are trigonometric polynomials\n\n\[ \n{S}_{M}^{ \\pm }\\left( x\\right) = \\mathop{\\sum }\\limits_{{\\left| m\\right| \\leq M}}{\\widehat{S}}_{M}^{ \\pm }\\left( m\\right) e\\left( {mx}\\right) , \n\]\n\nsuch that for all \( x \)\n\n\[ \n{S}_{M}^{ - }\\left( x\\right) \\leq {\\Xi }_{I}\\left( x\\right) \\leq {S}_{M}^{ + }\\left( x\\right) \n\]\n\nand\n\n\[ \n{\\widehat{S}}_{M}^{ - }\\left( 0\\right) = b - a - \\frac{1}{M + 1},\\;{\\widehat{S}}_{M}^{ + }\\left( 0\\right) = b - a + \\frac{1}{M + 1}. \n\]
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Proof. Take \( \\delta = M + 1 \) in Exercise 11.4.5 and let \( {H}_{ \\pm } \) be the functions obtained by that exercise. Put\n\n\[ \n{V}_{ \\pm }\\left( x\\right) = \\mathop{\\sum }\\limits_{{n \\in \\mathbb{Z}}}{H}_{ \\pm }\\left( {n + x}\\right) \n\]\n\nBy Exercise 11.4.6, \( {V}_{ \\pm }\\left( x\\right) \\in {L}^{1}\\left( {0,1}\\right) \) and \( {\\widehat{V}}_{ \\pm }\\left( t\\right) = 0 \) for \( \\left| t\\right| \\geq M + 1 \) . Thus,\n\n\[ \n{V}_{ \\pm }\\left( x\\right) = \\mathop{\\sum }\\limits_{{\\left| m\\right| \\leq M}}{\\widehat{V}}_{ \\pm }\\left( m\\right) e\\left( {mx}\\right) \n\]\n\nalmost everywhere. Now set\n\n\[ \n{S}_{M}^{ \\pm } = \\mathop{\\sum }\\limits_{{\\left| m\\right| \\leq M}}{\\widehat{V}}_{ \\pm }\\left( m\\right) e\\left( {mx}\\right) . \n\]\n\nSince \( {\\chi }_{I}\\left( x\\right) \\geq {H}_{ - }\\left( x\\right) \), we get\n\n\[ \n\\Xi \\left( x\\right) = \\mathop{\\sum }\\limits_{{n \\in \\mathbb{Z}}}{\\chi }_{I}\\left( {n + x}\\right) \\geq \\mathop{\\sum }\\limits_{{n \\in \\mathbb{Z}}}{H}_{ - }\\left( {n + x}\\right) = {V}_{ - }\\left( x\\right) \n\]\n\nfor almost all \( x \) . By continuity, we deduce that \( {\\Xi }_{I}\\left( x\\right) \\geq {S}_{M}^{ - }\\left( x\\right) \) for all \( x \) . Similary, we deduce \( {\\Xi }_{I}\\left( x\\right) \\leq {S}_{M}^{ + }\\left( x\\right) \) for all \( x \) . In addition, we\n\nhave\n\[ \n{\\widehat{S}}_{M}^{ \\pm }\\left( 0\\right) = b - a \\pm \\frac{1}{M + 1}. \n\]
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No
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Theorem 11.4.8 (Erdös-Turán,1948) For any integer \( M \geq 1 \) ,\n\n\[ \n{D}_{N} \leq \frac{1}{M + 1} + 3\mathop{\sum }\limits_{{m = 1}}^{M}\frac{1}{Nm}\left| {\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}}\right| .\n\]
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Proof. Let \( {\chi }_{I} \) be the characteristic function of the interval \( I = \left\lbrack {a, b}\right\rbrack \) . Using Theorem 11.4.7, we have\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{N}{\Xi }_{I}\left( {x}_{n}\right) \leq \mathop{\sum }\limits_{{n = 1}}^{N}{S}_{M}^{ + }\left( {x}_{n}\right)\n\]\n\n\[ \n\leq N\left( {b - a}\right) + \frac{N}{M + 1} + \mathop{\sum }\limits_{{0 < \left| m\right| \leq M}}\left| {{\widehat{S}}_{M}^{ + }\left( m\right) }\right| \left| {\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}}\right| .\n\]\n\nTo estimate \( {\widehat{S}}_{M}^{ + }\left( m\right) \), we use\n\n\[ \n{\widehat{S}}_{M}^{ + }\left( m\right) = {\int }_{0}^{1}{\Xi }_{I}\left( t\right) e\left( {-{mt}}\right) {dt} + {\int }_{0}^{1}\left( {{S}_{M}^{ + }\left( t\right) - {\Xi }_{I}\left( t\right) }\right) e\left( {-{mt}}\right) {dt}\n\]\n\nto deduce\n\n\[ \n\left| {{\widehat{S}}_{M}^{ + }\left( m\right) }\right| \leq {\int }_{0}^{1}\left( {{S}_{M}^{ + }\left( x\right) - {\Xi }_{I}\left( x\right) }\right) {dx} + \left| {{\widehat{\Xi }}_{I}\left( m\right) }\right| .\n\]\n\nThe integral is \( 1/\left( {M + 1}\right) \) and\n\n\[ \n{\widehat{\Xi }}_{I}\left( m\right) = e\left( {-\frac{1}{2}m\left( {a + b}\right) }\right) \frac{\sin \pi \left( {b - a}\right) m}{\pi m},\n\]\n\nfrom which we get\n\n\[ \n\left| {{\widehat{S}}_{M}^{ + }\left( m\right) }\right| \leq \frac{1}{M + 1} + \left| \frac{\sin \pi \left( {b - a}\right) m}{\pi m}\right| \leq \frac{3}{2\left| m\right| }.\n\]\n\nThus,\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{N}{\Xi }_{I}\left( {x}_{n}\right) \leq N\left( {b - a}\right) + \frac{N}{M + 1} + 3\mathop{\sum }\limits_{{m = 1}}^{M}\frac{1}{m}\left| {\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}}\right| .\n\]\n\nSimilarly, we obtain\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{N}{\Xi }_{I}\left( {x}_{n}\right) \geq N\left( {b - a}\right) - \frac{N}{M + 1} - 3\mathop{\sum }\limits_{{m = 1}}^{M}\frac{1}{m}\left| {\mathop{\sum }\limits_{{n = 1}}^{N}{e}^{{2\pi im}{x}_{n}}}\right|\n\]\n\nfrom which the theorem follows.
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Yes
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Lemma 1.3.1 If \( x \) is a vertex of the graph \( X \) and \( g \) is an automorphism of \( X \), then the vertex \( y = {x}^{g} \) has the same valency as \( x \) .
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Proof. Let \( N\left( x\right) \) denote the subgraph of \( X \) induced by the neighbours of \( x \) in \( X \) . Then\n\n\[ N{\left( x\right) }^{g} = N\left( {x}^{g}\right) = N\left( y\right) \]\n\nand therefore \( N\left( x\right) \) and \( N\left( y\right) \) are isomorphic subgraphs of \( X \) . Consequently they have the same number of vertices, and so \( x \) and \( y \) have the same valency.\n\nThis shows that the automorphism group of a graph permutes the vertices of equal valency among themselves. A graph in which every vertex has equal valency \( k \) is called regular of valency \( k \) or \( k \) -regular. A 3-regular graph is called cubic, and a 4-regular graph is sometimes called quartic. In Chapter 3 and Chapter 4 we will be studying graphs with the very special property that for any two vertices \( x \) and \( y \), there is an automorphism \( g \) such that \( {x}^{g} = y \) ; such graphs are necessarily regular.
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Yes
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Lemma 1.4.1 The chromatic number of a graph \( X \) is the least integer \( r \) such that there is a homomorphism from \( X \) to \( {K}_{r} \) .
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Proof. Suppose \( f \) is a homomorphism from the graph \( X \) to the graph \( Y \) . If \( y \in V\left( Y\right) \), define \( {f}^{-1}\left( y\right) \) by\n\n\[ \n{f}^{-1}\left( y\right) \mathrel{\text{:=}} \{ x \in V\left( X\right) : f\left( x\right) = y\} .\n\] \n\nBecause \( y \) is not adjacent to itself, the set \( {f}^{-1}\left( y\right) \) is an independent set. Hence if there is a homomorphism from \( X \) to a graph with \( r \) vertices, the \( r \) sets \( {f}^{-1}\left( y\right) \) form the colour classes of a proper \( r \) -colouring of \( X \), and so \( \chi \left( X\right) \leq r \) . Conversely, suppose that \( X \) can be properly coloured with the \( r \) colours \( \{ 1,\ldots, r\} \) . Then the mapping that sends each vertex to its colour is a homomorphism from \( X \) to the complete graph \( {K}_{r} \) .
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Yes
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Lemma 1.6.1 If \( v \geq k \geq i \), then \( J\left( {v, k, i}\right) \cong J\left( {v, v - k, v - {2k} + i}\right) \) .
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Proof. The function that maps a \( k \) -set to its complement in \( \Omega \) is an isomorphism from \( J\left( {v, k, i}\right) \) to \( J\left( {v, v - k, v - {2k} + i}\right) \) ; you are invited to check the details.
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No
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Lemma 1.6.2 If \( v \geq k \geq i \), then \( \operatorname{Aut}\left( {J\left( {v, k, i}\right) }\right) \) contains a subgroup isomorphic to \( \operatorname{Sym}\left( v\right) \) .
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Note that \( \operatorname{Aut}\left( {J\left( {v, k, i}\right) }\right) \) is a permutation group acting on a set of size \( \left( \begin{array}{l} v \\ k \end{array}\right) \) , and so when \( k \neq 1 \) or \( v - 1 \), it is not actually equal to \( \operatorname{Sym}\left( v\right) \) . Nevertheless, it is true that \( \operatorname{Aut}\left( {J\left( {v, k, i}\right) }\right) \) is usually isomorphic to \( \operatorname{Sym}\left( v\right) \), although this is not always easy to prove.
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No
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Lemma 1.7.1 If \( X \) is regular with valency \( k \), then \( L\left( X\right) \) is regular with valency \( {2k} - 2 \) .
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Each vertex in \( X \) determines a clique in \( L\left( X\right) \) : If \( x \) is a vertex in \( X \) with valency \( k \), then the \( k \) edges containing \( x \) form a \( k \) -clique in \( L\left( X\right) \) . Thus if \( X \) has \( n \) vertices, there is a set of \( n \) cliques in \( L\left( X\right) \) with each vertex of \( L\left( X\right) \) contained in at most two of these cliques. Each edge of \( L\left( X\right) \) lies in exactly one of these cliques. The following result provides a useful converse:\n\nTheorem 1.7.2 A nonempty graph is a line graph if and only if its edge set can be partitioned into a set of cliques with the property that any vertex lies in at most two cliques.
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No
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Theorem 1.7.2 A nonempty graph is a line graph if and only if its edge set can be partitioned into a set of cliques with the property that any vertex lies in at most two cliques.
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Proof. Let \( C \) be a clique in \( L\left( X\right) \) containing exactly \( c \) vertices. If \( c > 3 \) , then the vertices of \( C \) correspond to a set of \( c \) edges in \( X \), meeting at a common vertex. Consequently, there is a bijection between the vertices of \( X \) and the maximal cliques of \( L\left( X\right) \) that takes adjacent vertices to pairs of cliques with a vertex in common. The remaining details are left as an exercise.
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No
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Lemma 1.7.3 Suppose that \( X \) and \( Y \) are graphs with minimum valency four. Then \( X \cong Y \) if and only if \( L\left( X\right) \cong L\left( Y\right) \).
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Proof. Let \( C \) be a clique in \( L\left( X\right) \) containing exactly \( c \) vertices. If \( c > 3 \) , then the vertices of \( C \) correspond to a set of \( c \) edges in \( X \), meeting at a common vertex. Consequently, there is a bijection between the vertices of \( X \) and the maximal cliques of \( L\left( X\right) \) that takes adjacent vertices to pairs of cliques with a vertex in common. The remaining details are left as an exercise.
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No
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