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Theorem 3.7.8 ([Scii 76]) Let \( P \) and \( Q \) be partial skew tableaux. Then\n\n\[ P \cong Q \Leftrightarrow P\overset{K}{ \cong }Q. \]\n | Proof. The only-if direction is Proposition 3.7.4. For the other implication, note that since \( P \cong Q \), their row words must have the same \( P \) -tableau (Theorem 3.4.3 again). So by the previous theorem, \( j\left( P\right) = j\left( Q\right) = {P}^{\prime } \), say. Thus we can take \( P \) into \( Q \) by p... | Yes |
Proposition 3.8.1 Let \( P \) and \( Q \) be standard with the same normal shape \( \lambda \vdash n \) . Then \( P\overset{{K}^{ * }}{ \cong }Q \) . | Proof. Induct on \( n \), the proposition being trivial for \( n \leq 2 \) . When \( n \geq 3 \) , let \( c \) and \( d \) be the inner corner cells containing \( n \) in \( P \) and \( Q \), respectively. There are two cases, depending on the relative positions of \( c \) and \( d \) .\n\nIf \( c = d \), then let \( {... | Yes |
Lemma 3.8.3 Let \( P \cong Q \) . If applying the same sequence of slides to both tableaux yields \( {P}^{\prime } \) and \( {Q}^{\prime } \), then \( {P}^{\prime } \cong {Q}^{\prime } \) . | ∎ | No |
Proposition 3.8.5 Let \( P \) and \( Q \) be distinct miniature tableaux of the same shape \( \lambda /\mu \) and content. Then\n\n\[ P\overset{{K}^{ * }}{ \cong }Q \Leftrightarrow P\overset{ * }{ \cong }Q. \] | Proof. Without loss of generality, let \( P \) and \( Q \) be standard.\n\n\ | Yes |
Lemma 3.8.7 ([Hai 92]) Let \( V, W, P \), and \( Q \) be standard skew tableaux with\n\n\[ \operatorname{sh}V = \mu /\nu ,\;\operatorname{sh}P = \operatorname{sh}Q = \lambda /\mu ,\;\operatorname{sh}W = \kappa /\lambda .\n\]\n\nThen\n\n\[ P\overset{ * }{ \cong }Q \Rightarrow V \cup P \cup W\overset{ * }{ \cong }V \cup ... | Proof. Consider what happens in performing a single forward slide on \( V \cup \) \( P \cup W \), say into cell \( c \) . Because of the relative order of the elements in the \( V \) , \( P \), and \( W \) portions of the tableau, the slide can be broken up into three parts. First of all, the slide travels through \( V... | Yes |
Theorem 3.8.8 ([Hai 92]) Let \( P \) and \( Q \) be standard tableaux of the same shape \( \lambda /\mu \) . Then\n\n\[ P\overset{{K}^{ * }}{ \cong }Q \Leftrightarrow P\overset{ * }{ \cong }Q. \]\n | Proof. \ | No |
Lemma 3.9.2 Let \( Q \) be any skew partial tableau; then\n\n\[ \n{j\Delta }\left( Q\right) = {\Delta j}\left( Q\right) \n\] | Proof. Let \( P \) be \( Q \) with its minimum element \( m \) erased from cell \( c \) . We write this as \( P = Q - \{ m\} \) . Then\n\n\[ \n{j\Delta }\left( Q\right) = j\left( P\right) \n\]\n\nby the definition of \( \Delta \) and the uniqueness of the \( j \) operator (Theorem 3.7.7).\n\nWe now show that any forwar... | Yes |
Proposition 3.9.3 ([Scii 63]) Suppose \( \pi = {x}_{1}{x}_{2}\ldots {x}_{n} \in {\mathcal{S}}_{n} \) and let\n\n\[ \n\bar{\pi } = \begin{matrix} 2 & 3 & \cdots & n \\ {x}_{2} & {x}_{3} & \cdots & {x}_{n} \end{matrix} \n\]\n\nThen\n\n\[ \nQ\left( \bar{\pi }\right) = {\Delta Q}\left( \pi \right) \n\] | Proof. Consider \( \sigma = {\pi }^{-1} \) and \( \bar{\sigma } = {\bar{\pi }}^{-1} \) . Note that the lower line of \( \bar{\sigma } \) is obtained from the lower line of \( \sigma \) by deleting the minimum element,1 .\n\nBy Theorem 3.6.6, it suffices to show that\n\n\[ \nP\left( \bar{\sigma }\right) = {\Delta P}\lef... | Yes |
Theorem 3.9.4 ([Sci 63]) If \( \pi \in {\mathcal{S}}_{n} \), then\n\n\[ Q\left( {\pi }^{r}\right) = \operatorname{ev}Q{\left( \pi \right) }^{t} \] | Proof. Let \( \pi ,\bar{\pi } \) be as in the previous proposition with\n\n\[ {\bar{\pi }}^{r} = \begin{matrix} 1 & 2 & \cdots & n - 1 \\ {x}_{n} & {x}_{n - 1} & \cdots & {x}_{2} \end{matrix} \]\n\nInduct on \( n \) . Now\n\n\[ Q\left( {\pi }^{r}\right) - \{ n\} = Q\left( {\bar{\pi }}^{r}\right) \;\left( {{\bar{\pi }}^... | Yes |
Theorem 3.10.3 ([NPS 97]) For fixed \( \lambda \), the map\n\n\[ T\overset{\mathrm{N} - \mathrm{P} - \mathrm{S}}{ \rightarrow }\left( {P, J}\right) \]\n\njust defined is a bijection between tableaux \( T \) and pairs \( \left( {P, J}\right) \) with \( P \) a standard tableau and \( J \) a hook tableau such that \( \ope... | Proof. As usual, we will create an inverse.\n\n\ | No |
Lemma 3.10.4 Suppose that all reverse paths in \( {\mathcal{R}}^{\prime } \) go through \( \left( {{i}_{0},{j}_{0}}\right) \) . Then \( {r}_{0}^{\prime } \) with initial cell \( \left( {{i}_{0}^{\prime },{j}_{0}^{\prime }}\right) \in {\mathcal{C}}^{\prime } \) is the largest reverse path in \( {\mathcal{R}}^{\prime } \... | Proof. Since both \( {r}^{\prime } \) and \( {r}_{0}^{\prime } \) end up at \( \left( {{i}_{0},{j}_{0}}\right) \), they must intersect somewhere and coincide after their intersection. For the forward implication, assume that neither \( \mathrm{R}1 \) nor \( \mathrm{R}2 \) hold. Then the only other possibilities force \... | Yes |
Lemma 3.10.5 If \( {r}_{0}^{\prime } \) goes through \( \left( {{i}_{0},{j}_{0}}\right) \) and is north of some cell of a reverse path \( {r}^{\prime \prime } \in {\mathcal{R}}^{\prime \prime } \), then \( {r}^{\prime \prime } \) goes through \( \left( {{i}_{0} + 1,{j}_{0}}\right) \) . | Proof. I claim that \( {r}_{0}^{\prime } \) is north of every cell of \( {r}^{\prime \prime } \) after the one given in the hypothesis of the lemma. This forces \( {r}^{\prime \prime } \) through \( \left( {{i}_{0} + 1,{j}_{0}}\right) \) as \( {r}^{\prime } \) goes through \( \left( {{i}_{0},{j}_{0}}\right) \) . If the... | Yes |
Proposition 3.10.6 For all \( k \), the hook tableau \( {J}_{k} \) is well defined and all reverse paths go through \( {c}_{k} = \left( {{i}_{0},{j}_{0}}\right) \) in \( {T}_{k} \) . | Proof. The proposition is true for \( {i}_{0} = 1 \) by definition of the algorithm. So by induction, we can assume that the statement is true for \( {J}^{\prime } \) and \( {T}^{\prime } \) and prove it for \( {J}^{\prime \prime } \) and \( {T}^{\prime \prime } \) . By Lemma 3.10.4 and (3.20), we see that if \( {r}^{\... | Yes |
Lemma 3.10.7 Suppose that \( {p}_{0}^{\prime } \) ends at \( \left( {{i}_{0}^{\prime },{j}_{0}^{\prime }}\right) \) and does not consist solely of \( E \) (east) steps. Then the initial cell \( \left( {{i}^{\prime \prime },{j}^{\prime \prime }}\right) \in {\mathcal{C}}^{\prime \prime } \) of any \( {r}^{\prime \prime }... | The proof of this result is by contradiction. It is also very similar to that of Lemma 3.10.5, and so is left to the reader. | No |
Proposition 4.1.4 If \( {f}_{i}\left( x\right) \in \mathbb{C}\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \) for \( i \geq 1 \) and \( \mathop{\lim }\limits_{{i \rightarrow \infty }}\deg \left( {{f}_{i}\left( x\right) - 1}\right) = \) \( \infty \), then \( \mathop{\prod }\limits_{{i \geq 1}}{f}_{i}\left( x\ri... | ∎ | No |
Theorem 4.1.5 For all \( n,{p}_{d}\left( n\right) = {p}_{o}\left( n\right) \) . | Proof. It suffices to show that \( {p}_{d}\left( n\right) \) and \( {p}_{o}\left( n\right) \) have the same generating function. But\n\n\[ \mathop{\prod }\limits_{{i \geq 1}}\left( {1 + {x}^{i}}\right) = \mathop{\prod }\limits_{{i \geq 1}}\left( {1 + {x}^{i}}\right) \mathop{\prod }\limits_{{i \geq 1}}\frac{1 - {x}^{i}}... | Yes |
Proposition 4.1.6 Let \( S \) and \( T \) be weighted sets.\n\n1. If \( S \cap T = \varnothing \), then\n\n\[ \n{f}_{S \uplus T}\left( x\right) = {f}_{S}\left( x\right) + {f}_{T}\left( x\right) \n\]\n\n2. Let \( S \) and \( T \) be arbitrary and weight \( S \times T \) by \( \operatorname{wt}\left( {s, t}\right) = \ope... | Proof. 1. If \( S \) and \( T \) do not intersect, then\n\n\[ \n{f}_{S \uplus T}\left( x\right) = \mathop{\sum }\limits_{{s \in S \uplus T}}\mathrm{{wt}}s \n\]\n\n\[ \n= \mathop{\sum }\limits_{{s \in S}}\operatorname{wt}s + \mathop{\sum }\limits_{{s \in T}}\operatorname{wt}s \n\]\n\n\[ \n= {f}_{S}\left( x\right) + {f}_... | Yes |
Theorem 4.2.2 Fix a partition \( \lambda \) . Then\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}{rp}{p}_{\lambda }\left( n\right) {x}^{n} = \mathop{\prod }\limits_{{\left( {i, j}\right) \in \lambda }}\frac{1}{1 - {x}^{{h}_{i, j}}}. \] | Proof. By the discussion after Proposition 4.1.4, the coefficient of \( {x}^{n} \) in this product counts partitions of \( n \), where each part is of the form \( {h}_{i, j} \) for some \( \left( {i, j}\right) \in \lambda \) . (Note that the part \( {h}_{i, j} \) is associated with the node \( \left( {i, j}\right) \in ... | Yes |
Lemma 4.2.3 In the decomposition of \( T \) into hooklengths, the hooklength \( {h}_{{i}^{\prime },{j}^{\prime }} \) was removed before \( {h}_{{i}^{\prime \prime },{j}^{\prime \prime }} \) if and only if\n\n\[ \n{i}^{\prime \prime } > {i}^{\prime },\;\text{ or }\;{i}^{\prime \prime } = {i}^{\prime }\text{ and }{j}^{\p... | Proof. Since (4.3) is a total order on the nodes of the shape, we need only prove the only-if direction. By transitivity, it suffices to consider the case where \( {h}_{{i}^{\prime \prime },{j}^{\prime \prime }} \) is removed directly after \( {h}_{{i}^{\prime },{j}^{\prime }} \) .\n\nLet \( {T}^{\prime } \) and \( {T}... | Yes |
Lemma 4.2.4 If \( {r}_{k} \) is the reverse path for \( {h}_{{i}_{k},{j}_{k}} \), then \( \left( {{i}_{k},{\lambda }_{{i}_{k}}}\right) \in {r}_{k} \) . | Proof. Use reverse induction on \( k \) . The result is obvious when \( k = f \) by the first alternative in step GH2.\n\nFor \( k < f \), let \( {r}^{\prime } = {r}_{k} \) and \( {r}^{\prime \prime } = {r}_{k + 1} \) . Similarly, define \( {T}^{\prime },{T}^{\prime \prime },{h}_{{i}^{\prime },{j}^{\prime }} \), and \(... | Yes |
Theorem 4.2.6 ([Stn 71]) Let \( A \) be a poset with \( \left| A\right| = n \) . Then the generating function for reverse \( A \) -partitions is\n\n\[ \frac{P\left( x\right) }{\left( {1 - x}\right) \left( {1 - {x}^{2}}\right) \cdots \left( {1 - {x}^{n}}\right) } \]\n\nwhere \( P\left( x\right) \) is a polynomial such t... | In the case where \( A = \lambda \), we can compare this result with Theorem 4.2.2 and obtain\n\n\[ \frac{P\left( x\right) }{\left( {1 - x}\right) \left( {1 - {x}^{2}}\right) \cdots \left( {1 - {x}^{n}}\right) } = \mathop{\prod }\limits_{{\left( {i, j}\right) \in \lambda }}\frac{1}{1 - {x}^{{h}_{i, j}}}. \]\n\n\nThus\n... | Yes |
Proposition 4.3.3 The space \( {\Lambda }^{n} \) has basis\n\n\[ \left\{ {{m}_{\lambda } : \lambda \vdash n}\right\} \]\n\nand so has dimension \( p\left( n\right) \), the number of partitions of \( n \) . | ∎ | No |
Proposition 4.3.5 We have the following generating functions\n\n\[ E\\left( t\\right) \\overset{\\text{ def }}{ = }\\mathop{\\sum }\\limits_{{n \\geq 0}}{e}_{n}\\left( \\mathbf{x}\\right) {t}^{n} = \\mathop{\\prod }\\limits_{{i \\geq 1}}\\left( {1 + {x}_{i}t}\\right) \]\n\n\[ H\\left( t\\right) \\overset{\\text{ def }}... | Proof. Work in the ring \\( \\mathbb{C}\\left\\lbrack \\left\\lbrack {\\mathbf{x}, t}\\right\\rbrack \\right\\rbrack \\) . For the elementary symmetric functions, consider the set \\( S = \\{ \\lambda : \\lambda \\) with distinct parts \\} \\) with weight\n\n\[ {\\mathrm{{wt}}}^{\\prime }\\lambda = {t}^{l\\left( \\lamb... | Yes |
Proposition 4.3.6 We have the following generating function:\n\n\[ \mathop{\sum }\limits_{{n \geq 1}}{p}_{n}\left( \mathbf{x}\right) \frac{{t}^{n}}{n} = \ln \mathop{\prod }\limits_{{i \geq 1}}\frac{1}{\left( 1 - {x}_{i}t\right) } \] | Proof. Using the Taylor expansion of \( \ln \frac{1}{1 - x} \), we obtain\n\n\[ \ln \mathop{\prod }\limits_{{i \geq 1}}\frac{1}{\left( 1 - {x}_{i}t\right) } = \mathop{\sum }\limits_{{i \geq 1}}\ln \frac{1}{\left( 1 - {x}_{i}t\right) } \]\n\n\[ = \mathop{\sum }\limits_{{i \geq 1}}\mathop{\sum }\limits_{{n \geq 1}}\frac{... | Yes |
Theorem 4.3.7 The following are bases for \( {\Lambda }^{n} \) . | 1. Let \( C = \left( {c}_{\lambda \mu }\right) \) be the matrix expressing the \( {p}_{\lambda } \) in terms of the basis \( {m}_{\mu } \) . If we can find an ordering of partitions such that \( C \) is triangular with nonzero entries down the diagonal, then \( {C}^{-1} \) exists, and the \( {p}_{\lambda } \) are also ... | Yes |
Proposition 4.4.2 The function \( {s}_{\lambda }\left( \mathbf{x}\right) \) is symmetric. | Proof 1. By definition of the Schur functions and Kostka numbers,\n\n\[ \n{s}_{\lambda } = \mathop{\sum }\limits_{\mu }{K}_{\lambda \mu }{\mathbf{x}}^{\mu }\n\]\n\n(4.11)\n\nwhere the sum is over all compositions \( \mu \) of \( n \) . Thus it is enough to show that\n\n\[ \n{K}_{\lambda \mu } = {K}_{\lambda \widetilde{... | Yes |
Proposition 4.4.3 We have\n\n\[ \n{s}_{\lambda } = \mathop{\sum }\limits_{{\mu \leq \lambda }}{K}_{\lambda \mu }{m}_{\mu }\n\]\n\nwhere the sum is over partitions \( \mu \) (rather than compositions) and \( {K}_{\lambda \lambda } = 1 \) . | Proof. By equation (4.11) and the symmetry of the Schur functions, we have\n\n\[ \n{s}_{\lambda } = \mathop{\sum }\limits_{\mu }{K}_{\lambda \mu }{m}_{\mu }\n\]\n\nwhere the sum is over all partitions \( \mu \) . We can prove that\n\n\[ \n{K}_{\lambda \mu } = \left\{ \begin{array}{ll} 0 & \text{ if }\lambda \ntriangler... | Yes |
Lemma 4.6.1 Let \( \mu = \left( {{\mu }_{1},{\mu }_{2},\ldots ,{\mu }_{l}}\right) \) be any composition. Consider the \( l \times l \) matrices \[ {A}_{\mu } = \left( {x}_{j}^{{\mu }_{i}}\right) ,{H}_{\mu } = \left( {h}_{{\mu }_{i} - l + j}\right) \;\text{ and }\;E = \left( {{\left( -1\right) }^{l - i}{e}_{l - i}^{\lef... | Proof. Consider the generating function for the \( {e}_{n}^{\left( j\right) } \) , \[ {E}^{\left( j\right) }\left( t\right) \overset{\text{ def }}{ = }\mathop{\sum }\limits_{{n = 0}}^{{l - 1}}{e}_{n}^{\left( j\right) }{t}^{n} = \mathop{\prod }\limits_{{i \neq j}}\left( {1 + {x}_{i}t}\right) . \] We can now mimic the pr... | Yes |
Corollary 4.6.2 Let \( \lambda \) have length \( l \) . Then\n\n\[ \n{s}_{\lambda } = \frac{{a}_{\lambda + \delta }}{{a}_{\delta }} \n\]\n\nwhere all functions are polynomials in \( {x}_{1},\ldots ,{x}_{l} \) . | Proof. Taking determinants in the lemma, we obtain\n\n\[ \n\left| {A}_{\mu }\right| = \left| {H}_{\mu }\right| \cdot \left| E\right| \n\]\n\n(4.19)\n\nwhere \( \left| {A}_{\mu }\right| = {a}_{\mu } \) . First of all, let \( \mu = \delta \) . In this case \( {H}_{\delta } = \left( {h}_{i - j}\right) \), which is upper u... | Yes |
Theorem 4.6.3 Let \( {\phi }_{\lambda }^{\mu } \) be the character of \( {M}^{\mu } \) evaluated on the class corresponding to \( \lambda \) . Then\n\n\[ \n{p}_{\lambda } = \mathop{\sum }\limits_{{\mu \geq \lambda }}{\phi }_{\lambda }^{\mu }{m}_{\mu }\n\] | Proof. Let \( \lambda = \left( {{\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{l}}\right) \) . Then we can write equation (4.6) as\n\n\[ \n\mathop{\prod }\limits_{i}\left( {{x}_{1}^{{\lambda }_{i}} + {x}_{2}^{{\lambda }_{i}} + \cdots }\right) = \mathop{\sum }\limits_{\mu }{c}_{\lambda \mu }{m}_{\mu }\n\]\n\nPick out... | Yes |
Theorem 4.6.4 If \( \lambda \vdash n \), then\n\n\[ \n{s}_{\lambda } = \frac{1}{n!}\mathop{\sum }\limits_{{\pi \in {\mathcal{S}}_{n}}}{\chi }^{\lambda }\left( \pi \right) {p}_{\pi }\n\]\n\n(4.22) | There are several other ways to write equation (4.22). Since \( {\chi }^{\lambda } \) is a class function, we can collect terms and obtain\n\n\[ \n{s}_{\lambda } = \frac{1}{n!}\mathop{\sum }\limits_{\mu }{k}_{\mu }{\chi }_{\mu }^{\lambda }{p}_{\mu }\n\]\n\nwhere \( {k}_{\mu } = \left| {K}_{\mu }\right| \) and \( {\chi ... | Yes |
Theorem 4.7.4 The map \( \operatorname{ch} : R \rightarrow \Lambda \) is an isomorphism of algebras. | Proof. By Proposition 4.7.2, it suffices to check that products are preserved. If \( \chi \) and \( \psi \) are characters in \( {\mathcal{S}}_{n} \) and \( {\mathcal{S}}_{m} \), respectively, then using part 2 of Theorem 1.11.2 yields\n\n\[ \operatorname{ch}\left( {\chi \cdot \psi }\right) \; = \;\langle \chi \cdot \p... | Yes |
There is a bijection between generalized permutations and pairs of semistandard tableaux of the same shape, such that \( \operatorname{cont}\check{\pi } = \operatorname{cont}T \) and \( \operatorname{cont}\widehat{\pi } = \operatorname{cont}U \) . | Proof. \ | No |
Theorem 4.8.5 ([Knu 70]) There is a bijection between \( \pi \in {\mathrm{{GP}}}^{\prime } \) and pairs \( \left( {T, U}\right) \) of tableaux of the same shape with \( T,{U}^{t} \) semistandard, such that \( \operatorname{cont}\check{\pi } = \operatorname{cont}T \) and \( \operatorname{cont}\widehat{\pi } = \operatorn... | Proof. \ | No |
Theorem 4.8.7 If \( M \in \operatorname{Mat} \) and \( M\overset{\mathrm{R} - \mathrm{S} - \mathrm{K}}{ \leftrightarrow }\left( {T, U}\right) \), then | \[ {M}^{t}\overset{\mathrm{R} - \mathrm{S} - \mathrm{K}}{ \leftrightarrow }\left( {U, T}\right) \text{. ∎} \] | Yes |
Theorem 4.8.9 ([Knu 70]) A pair of generalized permutations are Knuth equivalent if and only if they have the same \( T \) -tableau. | ∎ | No |
Theorem 4.8.11 ([Scii 76]) Let \( T \) and \( U \) be skew semistandard tableaux. Then \( T \) and \( U \) have Knuth equivalent row words if and only if they are connected by a sequence of slides. Furthermore, any such sequence bringing them to normal shape results in the first output tableau of the Robinson-Schensted... | ∎ | No |
Theorem 4.8.12 If \( T \) and \( U \) are semistandard of the same normal shape, then \( T \cong U \) . | ∎ | No |
Proposition 4.9.1 ([Mac 79]) Define \( {s}_{\lambda }\left( {\mathbf{x},\mathbf{y}}\right) = {s}_{\lambda }\left( {{x}_{1},{x}_{2},\ldots ,{y}_{1},{y}_{2},\ldots }\right) \) . Then \[ {s}_{\lambda }\left( {\mathbf{x},\mathbf{y}}\right) = \mathop{\sum }\limits_{{\mu \subseteq \lambda }}{s}_{\mu }\left( \mathbf{x}\right)... | Proof. The function \( {s}_{\lambda }\left( {\mathbf{x},\mathbf{y}}\right) \) enumerates semistandard fillings of the diagram \( \lambda \) with letters from the totally ordered alphabet \[ \left\{ {1 < 2 < 3 < \cdots < {1}^{\prime } < {2}^{\prime } < {3}^{\prime } < \cdots }\right\} . \] In any such tableau, the unpri... | Yes |
Theorem 4.9.2 If the \( {c}_{\mu \nu }^{\lambda } \) are Littlewood-Richardson coefficients, where \( \left| \mu \right| + \left| \nu \right| = \left| \lambda \right| \), then\n\n\[ \n{s}_{\lambda /\mu } = \mathop{\sum }\limits_{\nu }{c}_{\mu \nu }^{\lambda }{s}_{\nu }\n\] | Proof. Bring in yet a third set of variables \( \mathbf{z} = \left\{ {{z}_{1},{z}_{2},\ldots }\right\} \) . By using the previous proposition and Cauchy's formula (Theorem 4.8.4),\n\n\[ \n\mathop{\sum }\limits_{{\lambda ,\mu }}{s}_{\mu }\left( \mathbf{x}\right) {s}_{\lambda /\mu }\left( \mathbf{y}\right) {s}_{\lambda }... | Yes |
Lemma 4.9.5 If we can go from \( T \) to \( {T}^{\prime } \) by a sequence of slides, then \( {\pi }_{T} \) is a reverse lattice permutation if and only if \( {\pi }_{{T}^{\prime }} \) is. | Proof. It suffices to consider the case where \( T \) and \( {T}^{\prime } \) differ by a single move. If the move is horizontal, the row word doesn't change, so consider a vertical move. Suppose\n\n\n\nand\n\n![fe18... | Yes |
Theorem 4.10.2 (Murnaghan-Nakayama Rule [Mur 37, Nak 40]) If \( \lambda \) is a partition of \( n \) and \( \alpha = \left( {{\alpha }_{1},\ldots ,{\alpha }_{k}}\right) \) is a composition of \( n \), then we have\n\n\[ \n{\chi }_{\alpha }^{\lambda } = \mathop{\sum }\limits_{\xi }{\left( -1\right) }^{{ll}\left( \xi \ri... | Proof (of Theorem 4.10.2). Let \( m = {\alpha }_{1} \) . Consider \( {\pi \sigma } \in {\mathcal{S}}_{n - m} \times {\mathcal{S}}_{m} \subseteq {\mathcal{S}}_{n} \),\n\nwhere \( \pi \) has type \( \left( {{\alpha }_{2},\ldots ,{\alpha }_{k}}\right) \) and \( \sigma \) is an \( m \) -cycle. By part 2 of Theorem 1.11.3, ... | No |
Lemma 4.10.3 If \( \nu \vdash m \), then\n\n\[{\chi }_{\left( m\right) }^{\nu } = \left\{ \begin{array}{ll} {\left( -1\right) }^{m - r} & \text{ if }\nu = \left( {r,{1}^{m - r}}\right) , \\ 0 & \text{ otherwise. } \end{array}\right.\n\] | Proof (of Lemma 4.10.3). By equation (4.23), \( {\chi }_{\left( m\right) }^{\nu } \) is \( {z}_{\left( m\right) } = m \) times the coefficient of \( {p}_{m} \) in\n\n\[{s}_{\nu } = \mathop{\sum }\limits_{\mu }\frac{1}{{z}_{\mu }}{\chi }_{\mu }^{\nu }{p}_{\mu }\n\]\n\nUsing the complete homogeneous Jacobi-Trudi determin... | Yes |
Lemma 4.10.4 Let \( \lambda \vdash n,\mu \vdash n - m \), and \( \nu = \left( {r,{1}^{m - r}}\right) \). Then \( {c}_{\mu \nu }^{\lambda } = 0 \) unless each edgewise connected component of \( \lambda /\mu \) is a rim hook. In that case, if there are \( k \) component hooks spanning a total of \( c \) columns, then\n\n... | Proof. By the Littlewood-Richardson rule (Theorem 4.9.4), \( {c}_{\mu \nu }^{\lambda } \) is the number of semistandard tableaux \( T \) of shape \( \lambda /\mu \) containing \( r \) ones and a single copy each of \( 2,3,\ldots, m - r + 1 \) such that \( {\pi }_{T} \) is a reverse lattice permutation. Thus the numbers... | Yes |
Corollary 4.10.6 Let \( \\lambda \) be a partition of \( n \) and let \( \\alpha = \\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{k}}\\right) \) be any composition of \( n \) . Then\n\n\[ \n{\\chi }_{\\alpha }^{\\lambda } = \\mathop{\\sum }\\limits_{T}{\\left( -1\\right) }^{T}\n\]\n\nwhere the sum is over all rim hook ta... | ∎ | No |
Proposition 5.1.3 The poset \( Y \) satisfies the following two conditions.\n\n(1) If \( \lambda \in Y \) covers \( k \) elements for some \( k \), then it is covered by \( k + 1 \) elements.\n\n(2) If \( \lambda \neq \mu \) and \( \lambda \) and \( \mu \) both cover \( l \) elements for some \( l \), then they are bot... | Proof. (1) The elements covered by \( \lambda \) are just the partitions \( {\lambda }^{ - } \) obtained by removing an inner corner of \( \lambda \), while those covering \( \lambda \) are the \( {\lambda }^{ + } \) obtained by adding an outer corner. But along the rim of \( \lambda \), inner and outer corners alterna... | Yes |
Proposition 5.1.5 The operators \( D, U \) satisfy\n\n\[ \n{DU} - {UD} = I \n\]\n\nwhere \( I \) is the identity map. | Proof. By linearity, it suffices to prove that this equation holds when applied to a single \( \mathbf{\lambda } \in \mathbb{C}\mathbf{Y} \) . But by the previous proposition\n\n\[ \n{DU}\left( \mathbf{\lambda }\right) = \mathop{\sum }\limits_{\mathbf{\mu }}\mathbf{\mu } + \left( {k + 1}\right) \mathbf{\lambda } \n\]\n... | Yes |
Corollary 5.1.6 Let \( p\left( x\right) \in \mathbb{C}\left\lbrack x\right\rbrack \) be a polynomial. Then\n\n\[ \n{Dp}\left( U\right) = {p}^{\prime }\left( U\right) + p\left( U\right) D \n\]\n\nwhere \( {p}^{\prime }\left( x\right) \) is the derivative. | Proof. By linearity, it suffices to prove this identity for the powers \( {U}^{n}, n \geq 0 \) . The case \( n = 0 \) is trivial. So, applying induction yields\n\n\[ \nD{U}^{n + 1} = \left( {D{U}^{n}}\right) U \n\]\n\n\[ \n= \;\left( {n{U}^{n - 1} + {U}^{n}D}\right) U \n\]\n\n\[ \n= n{U}^{n} + {U}^{n}\left( {I + {UD}}\... | Yes |
Theorem 5.1.8 We have\n\n\[ \mathop{\sum }\limits_{{\lambda \vdash n}}{\left( {f}^{\lambda }\right) }^{2} = n! \] | Proof. In view of equation (5.5) we need only show that \( {D}^{n}{U}^{n}\left( \varnothing \right) = n!\varnothing \) . We will do this by induction, it being clear for \( n = 0 \) . But using Corollary 5.1.6 and the fact that \( \varnothing \) is the minimal element of \( Y \), we obtain\n\n\[ {D}^{n}{U}^{n}\left( \v... | Yes |
Lemma 5.1.11 If \( A \) is a poset satisfying DP1 and DP3, then \( l \leq 1 \) . | Proof. Suppose to the contrary that there are elements \( a, b \in A \) with \( l \geq 2 \) , and pick a pair of minimal rank. Then \( a, b \) both cover \( c, d \) one rank lower. But then \( c, d \) are both covered by \( a, b \) and so have an \( l \) value at least 2, a contradiction. - | No |
Lemma 5.1.12 If \( A \) is differential, then the nth rank, \( {A}_{n} \), is finite for all \( n \geq 0 \) . | Proof. Induct on \( n \), where the case \( n = 0 \) is taken care of by DP1. Assume that all ranks up through \( {A}_{n} \) are finite. Any \( a \in {A}_{n} \) can cover at most \( \left| {A}_{n - 1}\right| \) elements, and so can be covered by at most \( \left| {A}_{n - 1}\right| + 1 \) elements by DP2. It follows th... | Yes |
Proposition 5.1.14 Let \( A \) be a graded poset with \( {A}_{n} \) finite for all \( n \geq 0 \) . Then the following are equivalent. | 1. \( A \) is differential.\n\n2. \( {DU} - {UD} = I \) . ∎ | No |
Theorem 5.1.15 ([Stn 88]) In any differential poset \( A \) , \n\n\[ \mathop{\sum }\limits_{{a \in {A}_{n}}}{\left( {f}^{a}\right) }^{2} = n! \]\n\nwhere \( {f}^{a} \) is the number of saturated \( \varnothing - a \) chains. | ∎ | No |
Lemma 5.2.2 Rules LR1-3 construct a well-defined growth \( {g}_{\pi } : {C}_{n}^{2} \rightarrow Y \) in that \( \rho \) is a partition and satisfies \( \rho \succcurlyeq \mu ,\nu \) . | Proof. The only case where it is not immediate that \( \rho \) is a partition is LR2. But by the way \( \mu \) is obtained from \( \lambda \) we have \( {\mu }_{i} > {\mu }_{i + 1} \) . So adding 1 to \( {\mu }_{i + 1} \) will not disturb the fact that the parts are weakly decreasing.\n\nIt is clear that \( \rho \) cov... | Yes |
Lemma 5.2.3 The growth \( {g}_{\pi } \) has the following properties.\n\n1. \( {g}_{\pi }\left( {i, j - 1}\right) \prec {g}_{\pi }\left( {i, j}\right) \) if and only if there is an \( X \) in square \( \left( {k, j}\right) \) for some \( k \leq i \).\n\n2. \( {g}_{\pi }\left( {i - 1, j}\right) \prec {g}_{\pi }\left( {i... | Proof. We will prove that \( {P}_{\pi } = P\left( \pi \right) \) since the equation \( {Q}_{\pi } = Q\left( \pi \right) \) is derived using similar reasoning.\n\nFirst we will associate with each \( \left( {i, j}\right) \) in \( {C}_{n}^{2} \) a permutation \( {\pi }_{i, j} \) and a tableau \( {P}_{i, j} \). Note that ... | Yes |
Theorem 5.2.6 The map\n\n\[ \pi \rightarrow \left( {{\mathcal{C}}_{\pi },{\mathcal{D}}_{\pi }}\right) \]\n\nis a bijection between \( {\mathcal{S}}_{n} \) and pairs of saturated \( \widehat{0} - a \) chains as a varies over all elements of rank \( n \) in the differential poset \( A \) . | Proof. Since both \( \mathcal{C} \) and \( \mathcal{D} \) end at \( \left( {n, n}\right) \in {C}_{n}^{2} \), it is clear that the corresponding chains end at the same element which must be of rank \( n \) .\n\nTo show that this is a bijection, we construct an inverse. Given saturated chains\n\n\[ \mathcal{C} : \widehat... | No |
Theorem 5.3.3 Let \( V \) be a vector space and suppose we have two functions \( f, g : {B}_{n} \rightarrow V \) . Then the following are equivalent:\n\n\[ \text{(1)}f\left( S\right) = \mathop{\sum }\limits_{{T \subseteq S}}g\left( T\right) \text{for all}S \in {B}_{n}\text{,}\]\n\n\[ \text{(2)}g\left( S\right) = \matho... | Proof. We will prove that (1) implies (2), as the converse is similar. Now,\n\n\[ \mathop{\sum }\limits_{{T \subseteq S}}{\left( -1\right) }^{\left| S - T\right| }f\left( T\right) = \mathop{\sum }\limits_{{T \subseteq S}}{\left( -1\right) }^{\left| S - T\right| }\mathop{\sum }\limits_{{U \subseteq T}}g\left( U\right) \... | No |
Proposition 5.3.6 Let \( \lambda \vdash n, S = {\left\{ {n}_{1},{n}_{2},\ldots ,{n}_{k}\right\} }_{ < } \subseteq \{ 1,2,\ldots, n - 1\} \) , and \( \mu = \left( {{n}_{1},{n}_{2} - {n}_{1},\ldots, n - {n}_{k}}\right) \) . Then\n\n\[ \left| {\{ P : P\text{ a standard }\lambda \text{-tableau and }\operatorname{Des}P \sub... | Proof. We will give a bijection between the two sets involved, where \( {K}_{\lambda \mu } \) counts semistandard tableaux \( T \) of shape \( \lambda \) and content \( \mu \) . Given \( P \), replace the numbers \( 1,2,\ldots ,{n}_{1} \) with ones, then \( {n}_{1} + 1,{n}_{1} + 2,\ldots ,{n}_{2} \) with twos, and so o... | No |
For the action of \( {\mathcal{S}}_{n} \) on \( {B}_{n} \), decompose\n\n\[ \n{B}^{S} \cong \mathop{\sum }\limits_{\lambda }{b}^{S}\left( \lambda \right) {S}^{\lambda }\n\]\n\nThen the muliplicities are nonnegative integers, since\n\n\[ \n{b}^{S}\left( \lambda \right) = \left| {\{ P : P\text{ a standard }\lambda \text{... | Proof. Decompose \( {A}^{S} \) as\n\n\[ \n{A}^{S} \cong \mathop{\sum }\limits_{\lambda }{a}^{S}\left( \lambda \right) {S}^{\lambda }\n\]\n\nThen Corollary 5.3.4 implies\n\n\[ \n{a}^{S}\left( \lambda \right) = \mathop{\sum }\limits_{{T \subseteq S}}{b}^{T}\left( \lambda \right) \;\text{ for all }\;S \subseteq \{ 1,2,\ld... | Yes |
Proposition 5.4.2 For any \( n \), the sequence\n\n\[ \left( \begin{array}{l} n \\ 0 \end{array}\right) ,\left( \begin{array}{l} n \\ 1 \end{array}\right) ,\ldots ,\left( \begin{array}{l} n \\ n \end{array}\right) \]\n\nis symmetric and unimodal. It follows that the same is true of the Boolean algebra \( {B}_{n} \) . | Proof. For symmetry we have, from the factorial formula in Exercise 1b of Chapter 4,\n\n\[ \left( \begin{matrix} n \\ n - k \end{matrix}\right) = \frac{n!}{\left( {n - k}\right) !\left( {n - \left( {n - k}\right) }\right) !} = \frac{n!}{\left( {n - k}\right) !\left( k\right) !} = \left( \begin{array}{l} n \\ k \end{arr... | Yes |
Lemma 5.4.4 Let \( G \) be a group of automorphisms of a finite, graded poset A.\n\n1. If \( \operatorname{rk}X = \left| {A}_{k}\right| \), then \( \mathbb{C}{\mathbf{A}}_{k} \leq \mathbb{C}{\mathbf{A}}_{k + 1} \).\n\n2. If \( \operatorname{rk}X = \left| {A}_{k + 1}\right| \), then \( \mathbb{C}{\mathbf{A}}_{k} \geq \m... | Proof. We will prove the first statement, as the second is similar. We claim that \( {U}_{k} \) is actually a \( G \) -homomorphism. It suffices to show that \( {U}_{k}\left( {g\mathbf{a}}\right) = \) \( g{U}_{k}\left( \mathbf{a}\right) \) for \( g \in G \) and \( a \in {A}_{k} \) . In other words, it suffices to prove... | Yes |
Theorem 5.4.6 ([Stn 82]) Let \( A \) be a finite, graded poset of rank \( n \) . Let \( G \) be a group of automorphisms of \( A \) and \( V \) be an irreducible \( G \) -module. If \( A \) is unimodal and ample, then the following sequences are unimodal.\n\n(1) \( \mathbb{C}{\mathbf{A}}_{0},\mathbb{C}{\mathbf{A}}_{1},... | Proof. The fact that the first sequence is unimodal follows immediately from the definition of ample and Lemma 5.4.4. This implies that the second sequence is as well by definition of the partial order on \( G \) -modules. Finally, by Exercise \( 5\mathrm{\;b} \) in Chapter 2,(3) is the special case of (2) where one ta... | No |
Corollary 5.4.8 Let \( G \leq {\mathcal{S}}_{n} \) act on \( {B}_{n} \) and let \( V \) be an irreducible \( G \) - module. Then, keeping the notation in the statement and proof of the previous proposition, the following sequences are symmetric and unimodal.\n\n(1) \( {m}_{0}\left( V\right) ,{m}_{1}\left( V\right) ,{m}... | Proof. The fact that the sequences are unimodal follows from Theorem 5.4.6, and Propositions 5.4.2 and 5.4.7. For symmetry, note that the map \( f : {B}_{n, k} \rightarrow \) \( {B}_{n, n - k} \) sending \( S \) to its complement induces a \( G \) -module isomorphism. ∎ | Yes |
For fixed \( k, l \), the sequence\n\n\[ \n{p}_{k, l}\left( 0\right) ,{p}_{k, l}\left( 1\right) ,{p}_{k, l}\left( 2\right) ,\ldots ,{p}_{k, l}\left( {kl}\right) \n\]\n\nis symmetric and unimodal. So the poset \( {Y}_{k, l} \) is as well. | \( \\textbf{Proof. Identify the cells of }\\left( {l}^{k}\\right) \) with the elements of \( \\{ 1,2,\\ldots, k\\} \\times \\{ 1,2,\\ldots, l\\} \) in the usual way. Then \( {\\mathcal{S}}_{k}\\wr {\\mathcal{S}}_{l} \) has an induced action on subsets of \( \\left( {l}^{k}\\right) \), which are partially ordered as in ... | Yes |
Proposition 5.5.6 ([Stn 95]) Let \( \Gamma \) be any graph with vertex set \( V \). 1. \( {X}_{\Gamma } \) is homogeneous of degree \( d = \left| V\right| \). 2. \( {X}_{\Gamma } \) is a symmetric function. 3. If we set \( {x}_{1} = \cdots = {x}_{n} = 1 \) and \( {x}_{i} = 0 \) for \( i > n \), written \( \mathbf{x} = ... | Proof. 1. Every monomial in \( {X}_{\Gamma } \) has a factor for each vertex. 2. Any permutation of the colors of a proper coloring gives another proper coloring. This means that permuting the subscripts of \( {X}_{\Gamma } \) leaves the function invariant. And since it is homogeneous of degree \( d \), it is a finite ... | Yes |
Proposition 5.5.9 ([Stn 95]) The expansion of \( {X}_{\Gamma } \) in terms of monomial symmetric functions is\n\n\[ \n{X}_{\Gamma } = \mathop{\sum }\limits_{{\lambda \vdash d}}{i}_{\lambda }{y}_{\lambda }{m}_{\lambda }\n\] | Proof. In any proper coloring, the set of all vertices of a given color form an independent set. So given \( \kappa : V \rightarrow \mathbb{P} \) proper, the set of nonempty \( {\kappa }^{-1}\left( i\right) \) , \( i \in \mathbb{P} \), form an independent partition \( \beta \vdash V \) . Thus the coefficient of \( {\ma... | Yes |
Theorem 5.5.10 ([Stn 95]) We have\n\n\\[ \n{X}_{\\Gamma } = \\mathop{\\sum }\\limits_{{F \\subseteq E}}{\\left( -1\\right) }^{\\left| F\\right| }{p}_{\\lambda \\left( F\\right) }\n\\] | Proof. Let \( K\\left( F\\right) \) denote the set of all colorings of \( \\Gamma \) that are monochromatic on the components of \( F \) (which will usually not be proper). If \( \\beta \\left( F\\right) = \) \( {B}_{1}/\\ldots /{B}_{l} \), then we can compute the weight generating function for such colorings as\n\n\\[... | Yes |
Proposition 5.5.11 A tree \( T \) with \( \left| V\right| = d \) has chromatic polynomial\n\n\[ \n{P}_{T}\left( n\right) = n{\left( n - 1\right) }^{d - 1}.\n\] | Proof. Pick any \( v \in V \) that can be colored in \( n \) ways. Since \( T \) has no cycles, each of the neighbors \( v \) can now be colored in \( n - 1 \) ways. The same can be said of the uncolored neighbors of the neighbors of \( v \) . Since \( T \) is connected, we will eventually color every vertex this way, ... | Yes |
Example 5 Cubes and spheres.\n\nThe unit interval is a \( {CW} \) complex with three cells: \( \{ 0\} ,\{ 1\} \) and \( I \) . Thus the cube \( {I}^{n} \) inherits a (product) CW structure, and \( \partial {I}^{n} \) is a subcomplex. Note that the CW complex \( {I}^{n}/\partial {I}^{n} \) coincides, as a CW complex, wi... | There is also a homeomorphism\n\n\[ \n\partial {I}^{n + 1}\overset{ \cong }{ \rightarrow }{S}^{n},\;n \geq 1 \n\]\n\ndefined by translating \( {I}^{n + 1} \) in \( {\mathbb{R}}^{n + 1} \) to centre it at the origin, and then projecting the boundary onto \( {S}^{n} \) (explicitly, \( x \mapsto \left( {x - a}\right) /\pa... | Yes |
Any continuous map \( f : \left( {X, A}\right) \rightarrow \left( {Y, B}\right) \) between relative \( {CW} \) complexes is homotopic rel \( A \) to a cellular map. | proof: Step (i) Linear approximation. Suppose \( \varphi : Z \rightarrow {\mathbb{R}}^{k} \) is a continuous map from a finite \( n \) -dimensional CW complex. Write \( {Z}_{r} = {Z}_{r - 1}{ \cup }_{{f}_{r}}\left( {\mathop{\coprod }\limits_{\alpha }{D}_{\alpha }^{r}}\right) \) and let \( {o}_{r,\alpha } \) be the orig... | Yes |
Theorem 1.4 (Cellular models theorem) [160]\n\n(i) Every space \( Y \) has a cellular model \( f : X \rightarrow Y \) .\n\n(ii) If \( {f}^{\prime } : {X}^{\prime } \rightarrow Y \) is a second cellular model then there is a homotopy equivalence \( g : X\overset{ \simeq }{ \rightarrow }{X}^{\prime } \) such that \( {f}^... | To prove this theorem one needs the fundamental\n\nLemma 1.5 (Whitehead lifting lemma) Suppose given a (not necessarily commutative) diagram\n\n\n\ntogether with a with a homotopy \( H : A \times I \rightarrow Z \) fro... | Yes |
Lemma 1.5 (Whitehead lifting lemma) Suppose given a (not necessarily commutative) diagram\n\n\n\ntogether with a with a homotopy \( H : A \times I \rightarrow Z \) from \( {\psi i} \) to \( {f\varphi } \) . Assume \( \... | proof: As in the proof of Step (iii) in Theorem 1.2 it is enough, by induction on the cellular structure, to consider the case that \( A = {X}_{n} \) and \( X = {X}_{n + 1} \) . Then working one cell at a time reduces us to the case that \( A = {S}^{n}, X = {D}^{n + 1} \) and \( i \) is the standard inclusion. In this ... | Yes |
Corollary 1.6 If \( X \) is a \( {CW} \) complex and \( f : Y \rightarrow Z \) is a weak homotopy equivalence then composition with \( f \) induces a bijection \( {f}_{\# } : \left\lbrack {X, Y}\right\rbrack \rightarrow \left\lbrack {X, Z}\right\rbrack \) of homotopy classes of maps. Similarly, if \( \left( {X,{x}_{0}}... | proof: The lifting theorem, applied with the relative CW complex \( \left( {X,\phi }\right) \) , shows that \( {f}_{\# } \) is surjective. Regard \( I \) as a CW complex with \( {I}_{0} = \{ 0,1\} \) and \( {I}_{1} = I \) . Then the lifting theorem applied with the relative CW complex \( \left( {X \times I, X\times \{ ... | Yes |
Lemma 1.8 \( \left( {X, A}\right) \) is a cofibration if and only if \( X \times \{ 0\} \cup A \times I \) is a retract of \( X \times I \) . | proof: A continuous map \( f : X \rightarrow Y \) and a homotopy \( H : A \times I \) starting at \( {\left. f\right| }_{A} \) define a map \( \left( {f, H}\right) : X \times \{ 0\} \cup A \times I \rightarrow Y \) . If \( r : X \times I \rightarrow X \times \{ 0\} \cup A \times I \) is a retraction then \( \left( {f, ... | Yes |
Proposition 1.10 The following conditions are equivalent on a topological pair \( \left( {X, A}\right) \) :\n\n(i) \( A \) is closed in \( X \) and \( \left( {X, A}\right) \) is a cofibration.\n\n(ii) \( \left( {X, A}\right) \) is an NDR pair.\n\n(iii) \( \left( {X \times I, X\times \{ 0\} \cup A \times I}\right) \) is... | proof: \( \;\left( i\right) \Rightarrow \left( {ii}\right) \) : Let \( r = \left( {{r}_{X},{r}_{I}}\right) : X \times I \rightarrow X \times \{ 0\} \cup A \times I \) be a retraction (Lemma 1.8). Define \( h : X \rightarrow I \) by \( h\left( x\right) = \sup \left\{ {t - {r}_{I}\left( {x, t}\right) \mid t \in I}\right\... | Yes |
Proposition 1.11 Suppose \( \left( {X, A}\right) \) is an NDR pair. Then the inclusion \( i \) : \( A \rightarrow X \) is a homotopy equivalence if and only if \( \left( {X, A}\right) \) is a DR pair. In this case \( \left( {X \times I, X\times \{ 0,1\} \cup A \times I}\right) \) is a DR pair. | proof: If \( \left( {X, A}\right) \) is a DR pair then \( i \) is certainly a homotopy equivalence. Suppose that \( i \) is a homotopy equivalence. Choose \( \varrho : X \rightarrow A \) so that \( \varrho i \sim i{d}_{A} \) . Extend the homotopy to a homotopy \( X \times I \rightarrow A \) from \( \varrho \) to a map ... | Yes |
Lemma 1.12 If \( \\left( {X, A}\\right) \) is an NDR pair and \( {f}_{0},{f}_{1} : A \rightarrow Y \) are homotopic continuous maps then \( Y{ \cup }_{f_{0}}X \simeq Y{ \cup }_{f_{1}}X \) . | proof: Choose a homotopy \( H : A \times I \rightarrow Y \) from \( {f}_{0} \) to \( {f}_{1} \) . Denote \( X \times \\{ t\\} \cup A \times I \) by \( {B}_{t} \subset X \times I \) . Proposition 1.10 implies that each \( {B}_{t} \) is a strong deformation retract of \( X \times I \) . Hence \( Y{ \cup }_{H}{B}_{0} \) a... | Yes |
Lemma 1.14 If \( \left( {B, C}\right) \) is a DR pair and \( h : D \rightarrow W \) is a continuous map from a closed subspace \( D \subset C \) then \( \left( {W{ \cup }_{h}B, W{ \cup }_{h}C}\right) \) is a DR pair. In particular, the inclusion\n\n\[ W{ \cup }_{h}C \hookrightarrow W{ \cup }_{h}B \]\n\nis a homotopy eq... | proof of Theorem 1.13: Identify \( \left( {X, A}\right) \) with \( \left( {X, A}\right) \times \{ 1\} \subset X \times I \) . Then \( \left( {X \times I, X}\right) \) is a DR pair (trivially) and \( \left( {X \times I, X\times \{ 0\} \cup A \times I}\right) \) is a DR pair by Proposition 1.10. Moreover \( f \) is ident... | Yes |
Proposition 1.17 If \( \left( {X,{x}_{0}}\right) \) and \( \left( {Y,{y}_{0}}\right) \) are well based spaces then there is a homotopy equivalence, \( X * Y\overset{ \simeq }{ \rightarrow }\sum \left( {X \land Y}\right) \) . | proof: Embed \( I \) as \( I{x}_{0} \) and \( I{y}_{0} \) in \( {CX} \) and \( {CY} \) . Identify \( {CX}{ \cup }_{I}{CY} \) as the subspace of \( X * Y \) of points \( \left( {{tx},\left( {1 - t}\right) y}\right) \) such that either \( x = {x}_{0} \) or \( y = {y}_{0} \) . Now \( \left( {{CX}{ \cup }_{I}{CY}}\right) /... | Yes |
Proposition 2.1 (i) A fibration \( p : X \rightarrow Y \) has the lifting property with respect to any DR pair \( \left( {Z, A}\right) \) . In particular, if \( \left( {W, B}\right) \) is any NDR pair then \( p \) has the lifting property with respect to \( \left( {W \times I, W\times \{ 0\} \cup B \times I}\right) \) ... | proof: [157] In both cases we suppose given a commutative square\n\n\n\nand we have to construct \( k : Z \rightarrow X \) .\n\n(i) Let \( r : Z \rightarrow A \) be a retraction and let \( H : Z \times I \rightarrow Z ... | Yes |
Proposition 2.1 (i) A fibration \( p : X \rightarrow Y \) has the lifting property with respect to any DR pair \( \left( {Z, A}\right) \) . In particular, if \( \left( {W, B}\right) \) is any NDR pair then \( p \) has the lifting property with respect to \( \left( {W \times I, W\times \{ 0\} \cup B \times I}\right) \) ... | proof: [157] In both cases we suppose given a commutative square\n\n\n\nand we have to construct \( k : Z \rightarrow X \) .\n\n(i) Let \( r : Z \rightarrow A \) be a retraction and let \( H : Z \times I \rightarrow Z ... | Yes |
Proposition 2.1 (i) A fibration \( p : X \rightarrow Y \) has the lifting property with respect to any DR pair \( \left( {Z, A}\right) \) . In particular, if \( \left( {W, B}\right) \) is any NDR pair then \( p \) has the lifting property with respect to \( \left( {W \times I, W\times \{ 0\} \cup B \times I}\right) \) ... | proof: [157] In both cases we suppose given a commutative square\n\n\n\nand we have to construct \( k : Z \rightarrow X \) .\n\n(i) Let \( r : Z \rightarrow A \) be a retraction and let \( H : Z \times I \rightarrow Z ... | Yes |
Proposition 2.1 (i) A fibration \( p : X \rightarrow Y \) has the lifting property with respect to any DR pair \( \left( {Z, A}\right) \) . In particular, if \( \left( {W, B}\right) \) is any NDR pair then \( p \) has the lifting property with respect to \( \left( {W \times I, W\times \{ 0\} \cup B \times I}\right) \) ... | proof: [157] In both cases we suppose given a commutative square\n\n\n\nand we have to construct \( k : Z \rightarrow X \) .\n\n(i) Let \( r : Z \rightarrow A \) be a retraction and let \( H : Z \times I \rightarrow Z ... | Yes |
Proposition 2.2 [93] With the notation above the correspondence \( g \rightsquigarrow h \) defines natural set maps \( {\partial }_{n} : {\pi }_{n}\left( {Y,{y}_{0}}\right) \rightarrow {\pi }_{n - 1}\left( {F,{x}_{0}}\right) \) . When \( n \geq 2 \), these are group homomorphisms, and fit into a long exact sequence (th... | proof: Put \( W = {S}^{n} \times I \) and \( B = {S}^{n} \times \{ 0\} \cup \{ * \} \times I \) . Then the inclusion \( W \times \{ 0,1\} \cup B \times I \hookrightarrow W \times I \) is a weak homotopy equivalence. Hence it follows from Proposition 2.1(ii) that the based homotopy class of \( h \) is independent of the... | Yes |
Proposition 2.3 Suppose in the diagram above that \( p \) is a fibration.\n\n(i) If \( g \) is the inclusion of a DR pair (resp. an NDR pair), \( \left( {Y, A}\right) \) then \( \left( {X,{X}_{A}}\right) \) is also a DR pair (resp. on NDR pair).\n\n(ii) If \( g \) is a homotopy equivalence so is \( {g}_{X} \) . | proof: (i) Suppose \( \left( {Y, A}\right) \) is a DR pair. Let \( H : Y \times I \rightarrow Y \) be a homotopy rel \( A \) from \( i{d}_{Y} \) to a retraction \( r : Y \rightarrow A \) . Let \( h : Y \rightarrow I \) be a continuous function such that \( {h}^{-1}\left( 0\right) = A \) and define a homotopy \( {H}^{\p... | Yes |
Proposition 2.6 A fibre bundle is a Serre fibration. | proof: First note that if a continuous surjection \( p : X \rightarrow Y \) has the lifting property with respect to \( {\left( {I}^{n} \times I,{I}^{n}\times \{ 0\} \right) }_{n > 0} \), then it is a Serre fibration. Indeed, suppose given a relative CW complex \( \left( {W, A}\right) \) and a homotopy \( g : W \times ... | Yes |
Proposition 2.8 With the hypotheses above \( p : X \rightarrow Y \) is a principal \( G - \) bundle. | proof: It follows from Proposition 2.7 that \( p : X \rightarrow Y \) is a fibre bundle. Thus \( Y \) is covered by open sets \( {U}_{i} \) for which there are continuous maps \( {\tau }_{i} \) : \( {U}_{i} \rightarrow X \) such that \( p{\tau }_{i} = {id} \) . We have only to check that the continuous maps \( {h}_{i} ... | Yes |
Proposition 2.9 If \( p : X \rightarrow Y \) is a principal \( G \) -bundle then there is a weak homotopy equivalence \( {X}_{G}\overset{ \simeq }{ \rightarrow }Y \) . | proof: The associated fibre bundle with fibre \( {E}_{G} \) has the form\n\n\[ \n{q}^{\prime } : {X}_{G} = \left( {{E}_{G} \times X}\right) /G \rightarrow Y.\n\]\n\nSince \( {\pi }_{ * }\left( {E}_{G}\right) = 0 \) and since this is a Serre fibration, the long exact homotopy sequence shows that \( {q}^{\prime } \) is a... | Yes |
Proposition 2.10 \( \gamma \) and \( {\gamma }^{\prime } \) are weak homotopy equivalences, so that \( G \) and \( {\Omega B} \) are weakly equivalent topological monoids. | proof: Define an action of \( G{ \times }_{Z}{PZ} \) on \( {PZ} \) by setting \( u \cdot \left( {g, w}\right) = \) \( \left( {u \cdot g}\right) * w \) . Then \( \pi : {PZ} \rightarrow B,\pi \left( w\right) = {p}_{B}\left( {w\left( 0\right) }\right) \), is a \( G{ \times }_{Z}{PZ} \) -fibration. It fits in the diagram o... | Yes |
Proposition 2.11 The pullback fibre bundle \( p : Y{ \times }_{B}Z \rightarrow Y \) and the holonomy fibration of \( \varphi \) are connected by equivariant weak equivalences of fibrations: |  | No |
Lemma 3.2 A necessary and sufficient condition for \( H\left( f\right) \) to be an isomorphism (of degree \( i \) ) is that for each \( \left( {m, n}\right) \in M \times N \) satisfying \( {dm} = 0 \) and \( f\left( m\right) = \) dn there exist \( \left( {{m}^{\prime },{n}^{\prime }}\right) \in M \times N \) satisfying... | proof: Suppose the condition holds. If \( {dm} = 0 \) and \( H\left( f\right) \left\lbrack m\right\rbrack = 0 \) then \( f\left( m\right) = \) \( {dn} \) ; hence \( m = d{m}^{\prime } \) and \( \left\lbrack m\right\rbrack = 0 \) . If \( \left\lbrack n\right\rbrack \in H\left( N\right) \) then \( {dn} = 0 = f\left( 0\ri... | Yes |
A morphism \( \varphi : R \rightarrow S \) of graded algebras makes \( S \) into a left (and right) \( R \) -module | \[ x \cdot s = \varphi \left( x\right) s\;\text{ or }\;s \cdot x = {s\varphi }\left( x\right) ,\;x \in R, s \in S. \] If \( M \) is an \( R \) -module then \( S{ \otimes }_{R}M \) is an \( S \) -module via \( s \cdot \left( {{s}^{\prime }{ \otimes }_{R}m}\right) = s{s}^{\prime }{ \otimes }_{R}m \) . | Yes |
Let \( R \) be a graded algebra. An \( R \) -module \( M \) is free if \( M \cong R \otimes V \), with \( V \) a free graded module. A basis \( \left\{ {v}_{\alpha }\right\} \) for \( V \) is a basis for the free \( R \) -module \( M \) . If \( \varphi : R \rightarrow S \) is a morphism of graded algebras then | \[ S{ \otimes }_{R}M = S{ \otimes }_{R}\left( {R \otimes V}\right) = S \otimes V \] is a free \( S \) -module with the same basis. | Yes |
If \( R, S \) are graded algebras then \( R \otimes S \) denotes the graded algebra with multiplication | \[ \left( {x \otimes y}\right) \left( {{x}^{\prime } \otimes {y}^{\prime }}\right) = {\left( -1\right) }^{\deg y\deg {x}^{\prime }}x{x}^{\prime } \otimes y{y}^{\prime }. \] | Yes |
For any free graded module \( V \), the tensor algebra \( {TV} \) is defined by\n\n\[ \n{TV} = {\bigoplus }_{q = 0}^{\infty }{T}^{q}V\;{T}^{q}V = \underset{q}{\underbrace{V \otimes \cdots \otimes V}}. \n\]\n\nMultiplication is given by \( a \cdot b = a \otimes b \) . Note that \( q \) is not the degree: elements \( {v}... | Any linear map of degree zero from \( V \) to a graded algebra \( R \) extends to a unique morphism of graded algebras, \( {TV} \rightarrow R \) . Any degree \( k \) linear map \( V \rightarrow {TV} \) extends to a unique derivation of \( {TV} \) . | Yes |
A graded algebra \( A \) is commutative if\n\n\[ \n{xy} = {\left( -1\right) }^{\deg x\deg y}{yx},\;x, y \in A.\n\] | When \( \frac{1}{2} \in \mathbb{R} \) this condition implies that \( {x}^{2} = 0 \) if \( x \) has odd degree. If \( A \) is a commutative graded algebra, then a left \( A \) -module, \( M \), is automatically a right \( A \) -module, via\n\n\[ \n{mx} = {\left( -1\right) }^{\deg m\deg x}{xm}.\n\]\n\nIf \( N \) is a sec... | Yes |
Example 6 Free commutative graded algebras. | Suppose \( \mathbb{k} \) contains \( \frac{1}{2} \) . Let \( V \) be a free graded module. The elements \( v \otimes w - \) \( {\left( -1\right) }^{\deg v\deg w}w \otimes v\left( {v, w \in V}\right) \) generate an ideal \( I \subset {TV} \) . The quotient graded algebra\n\n\[ \n{\Lambda V} = {TV}/I \n\]\n\nis called th... | Yes |
If \( \left( {A, d}\right) \) and \( \left( {{A}^{\prime }, d}\right) \) are dga’s then the direct product \( \left( {A, d}\right) \times \left( {{A}^{\prime }, d}\right) \) is the dga \( \left( {A \times {A}^{\prime }, d}\right) \) given by \( \left( {a,{a}^{\prime }}\right) \cdot \left( {{a}_{1},{a}_{1}^{\prime }}\ri... | The direct product, \( \mathop{\prod }\limits_{\alpha }\left( {A\left( \alpha \right), d}\right) \), of a family of dga’s is defined in the same way. | No |
Proposition 3.3. If \( \mathbb{k} \) is a field these natural maps are isomorphisms:\n\n\[ H\left( M\right) \otimes H\left( N\right) = H\left( {M \otimes N}\right) \;\text{ and }\;H\left( {\operatorname{Hom}\left( {M, N}\right) }\right) = \operatorname{Hom}\left( {H\left( M\right), H\left( N\right) }\right) . | proof: This is a straightforward exercise using the fact that any complex \( M \) can be written \( M = \operatorname{Im}d \oplus H \oplus C \) with \( d : C\overset{ \cong }{ \rightarrow }\operatorname{Im}d \) and \( d = 0 \) in \( H \) . | No |
Proposition 4.10 AW and EZ are inverse chain equivalences. In fact, \( {AW} \circ \) \( {EZ} = {id} \) and \( {EZ} \circ {AW} \) is naturally homotopic to the identity. | proof: The first assertion is a simple computation, depending on the fact that we have divided by the degenerate simplices. For the second, we have to construct \( h : {C}_{n}\left( {X \times Y;\mathbb{R}}\right) \rightarrow {C}_{n + 1}\left( {X \times Y;\mathbb{R}}\right) \), natural in \( X \) and \( Y \), such that ... | Yes |
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