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Proposition 13.18 (Naturality of the Connecting Homomorphism). Suppose \n\n(13.12)\n\nis a commutative diagram of chain maps in which the horizontal rows are exact. Then the following diagram commutes for each \( p \... | Proof. Let \( \left\lbrack {e}_{p}\right\rbrack \in {H}_{p}\left( {E}_{ * }\right) \) be arbitrary. Then \( {\partial }_{ * }\left\lbrack {e}_{p}\right\rbrack = \left\lbrack {c}_{p - 1}\right\rbrack \), where \( F{c}_{p - 1} = \partial {d}_{p} \) for some \( {d}_{p} \) such that \( G{d}_{p} = {e}_{p} \) . Then by commu... | Yes |
Proposition 13.19. Suppose \( \mathcal{U} \) is any open cover of \( X \) . Then the inclusion map \( {C}_{ * }^{\mathcal{U}}\left( X\right) \rightarrow {C}_{ * }\left( X\right) \) induces a homology isomorphism \( {H}_{p}^{\mathcal{U}}\left( X\right) \cong {H}_{p}\left( X\right) \) for all \( p \) . | The idea of the proof is simple, although the technical details are somewhat involved. If \( \sigma : {\Delta }_{p} \rightarrow X \) is any singular \( p \) -simplex, the plan is to show that there is a homologous \( p \) -chain obtained by \ | No |
Lemma 13.20. If \( c \) is an affine chain, then\n\n\[ \partial \left( {w * c}\right) + w * \partial c = c. \] | Proof. For an affine simplex \( \alpha = A\left( {{v}_{0},\ldots ,{v}_{p}}\right) \), this is just a computation:\n\n\[ \partial \left( {w * \alpha }\right) = \partial A\left( {w,{v}_{0},\ldots ,{v}_{p}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{{p + 1}}{\left( -1\right) }^{i}A\left( {w,{v}_{0},\ldots ,{v}_{p}... | Yes |
Lemma 13.21. Suppose \( \alpha : {\Delta }_{p} \rightarrow {\mathbb{R}}^{n} \) is an affine simplex that is a homeomorphism onto a p-simplex \( \sigma \subseteq {\mathbb{R}}^{n} \) . Let \( \beta : {\Delta }_{p} \rightarrow {\mathbb{R}}^{n} \) be any one of the affine singular \( p \) - simplices that appear in the cha... | Proof. Part (a) follows immediately from the definition of the subdivision operator and an easy induction on \( p \) (see Fig. 13.9).\n\nTo prove (b), write \( \sigma = \alpha \left( {\Delta }_{p}\right) = \left\lbrack {{v}_{0},\ldots ,{v}_{p}}\right\rbrack \) and \( \tau = \beta \left( {\Delta }_{p}\right) = \left\lbr... | Yes |
Lemma 13.22. The singular subdivision operators \( s : {C}_{p}\left( X\right) \rightarrow {C}_{p}\left( X\right) \) have the following properties.\n\n(a) \( s \circ {f}_{\# } = {f}_{\# } \circ s \) for any continuous map \( f \) .\n\n(b) \( \partial \circ s = s \circ \partial \) .\n\n(c) Given an open cover \( \mathcal... | Proof. The first identity follows immediately from the definition of \( s \) :\n\n\[ s\left( {{f}_{\# }\sigma }\right) = s\left( {f \circ \sigma }\right) = {\left( f \circ \sigma \right) }_{\# }\left( {s{i}_{p}}\right) = {f}_{\# }{\sigma }_{\# }\left( {s{i}_{p}}\right) = {f}_{\# }\left( {s\sigma }\right) . \]\n\nThe se... | Yes |
Theorem 13.23 (Homology Groups of Spheres). For \( n \geq 1,{\mathbb{S}}^{n} \) has the following singular homology groups:\n\n\[ \n{H}_{p}\left( {\mathbb{S}}^{n}\right) \cong \left\{ \begin{array}{ll} \mathbb{Z} & \text{ if }p = 0, \\ 0 & \text{ if }0 < p < n, \\ \mathbb{Z} & \text{ if }p = n, \\ 0 & \text{ if }p > n.... | Proof. We use the Mayer-Vietoris sequence as follows. Let \( N \) and \( S \) denote the north and south poles, and let \( U = {\mathbb{S}}^{n} \smallsetminus \{ N\}, V = {\mathbb{S}}^{n} \smallsetminus \{ S\} \) . Part of the Mayer-Vietoris sequence reads\n\n\[ \n{H}_{p}\left( U\right) \oplus {H}_{p}\left( V\right) \r... | Yes |
Corollary 13.24 (Homology Groups of Punctured Euclidean Spaces). For \( n \geq 2 \) ,\n\n\( {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) has the following singular homology groups:\n\n\[ \n{H}_{p}\left( {{\mathbb{R}}^{n}\smallsetminus \{ 0\} }\right) \cong \left\{ \begin{array}{ll} \mathbb{Z} & \text{ if }p = 0, \\ 0 & \t... | Proof. Inclusion \( {\mathbb{S}}^{n - 1} \hookrightarrow {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) is a homotopy equivalence. | No |
Proposition 13.25. Suppose \( n \geq 1 \) and \( f, g : {\mathbb{S}}^{n} \rightarrow {\mathbb{S}}^{n} \) are continuous maps.\n\n(a) \( \deg \left( {g \circ f}\right) = \left( {\deg g}\right) \left( {\deg f}\right) \).\n\n(b) If \( f \simeq g \), then \( \deg f = \deg g \) . | Proof. Part (a) follows from the fact that \( {\left( g \circ f\right) }_{ * } = {g}_{ * } \circ {f}_{ * } \), and part (b) from the fact that homotopic maps induce the same homology homomorphism. | Yes |
Lemma 13.26. The homological degree and the homotopic degree of a continuous map \( f : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) are equal. | Proof. By (13.8), the following diagram commutes:\n\n\[ \n{\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \xrightarrow[]{{\left( \rho \circ f\right) }_{ * }}{\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \n\]\n\n\[ \n\gamma \downarrow \gamma \n\]\n\n\[ \n{H}_{1}\left( {\mathbb{S}}^{1}\right) \xrightarrow[{\left( \rho \circ ... | Yes |
Proposition 13.31. The antipodal map \( \alpha : {\mathbb{S}}^{n} \rightarrow {\mathbb{S}}^{n} \) is homotopic to the identity map if and only if \( n \) is odd. | Proof. If \( n = {2k} - 1 \) is odd, an explicit homotopy \( H : \mathrm{{Id}} \simeq \alpha \) is given by\n\n\[ H\left( {x, t}\right) = \left( {\left( {\cos {\pi t}}\right) {x}_{1} + \left( {\sin {\pi t}}\right) {x}_{2},\left( {\cos {\pi t}}\right) {x}_{2} - \left( {\sin {\pi t}}\right) {x}_{1},}\right.\n\n\[ \left. ... | Yes |
Theorem 13.32 (The Hairy Ball Theorem). There exists a nowhere vanishing vector field on \( {\mathbb{S}}^{n} \) if and only if \( n \) is odd. | Proof. Suppose there exists such a vector field \( V \) . By replacing \( V \) with \( V/\left| V\right| \), we can assume \( \left| {V\left( x\right) }\right| = 1 \) everywhere. We use \( V \) to construct a homotopy between the identity map and the antipodal map as follows:\n\n\[ H\left( {x, t}\right) = \left( {\cos ... | Yes |
Proposition 13.33 (Homology Effect of Attaching a Cell). Let \( X \) be any topological space, and let \( Y \) be obtained from \( X \) by attaching a closed cell \( D \) of dimension \( n \geq 2 \) along the attaching map \( \varphi : \partial D \rightarrow X \) . Let \( K \) and \( L \) denote the kernel and image, r... | Proof. First, assume that \( p \geq 2 \) . Let \( q : X \coprod D \rightarrow Y \) be a quotient map realizing \( Y \) as an adjunction space. Choose a point \( z \in \operatorname{Int}D \), and define open subsets \( U, V \subseteq Y \) by \( U = q\left( {\operatorname{Int}D}\right) \) and \( V = q\left( {X \coprod \l... | Yes |
Theorem 13.34 (Homology Properties of CW Complexes). Let \( X \) be a finite \( n \) - dimensional CW complex.\n\n(a) Inclusion \( {X}_{k} \hookrightarrow X \) induces isomorphisms \( {H}_{p}\left( {X}_{k}\right) \cong {H}_{p}\left( X\right) \) for \( p \leq k - 1 \) .\n\n(b) \( {H}_{p}\left( X\right) = 0 \) for \( p >... | Proof. Part (a) follows immediately from Theorem 13.33, because attaching an \( m \) - cell cannot change \( {H}_{p}\left( X\right) \) if \( p < m - 1 \) .\n\nTo prove (b), assume \( p > n \), and note that \( X \) is obtained from \( {X}_{0} \) by adding finitely many cells of dimensions less than or equal to \( n \),... | Yes |
Theorem 13.36. If \( X \) is a finite \( {CW} \) complex,\n\n\[ \chi \left( X\right) = \mathop{\sum }\limits_{p}{\left( -1\right) }^{p}\operatorname{rank}{H}_{p}\left( X\right) . \]\n\n(13.21)\n\nTherefore, the Euler characteristic is a homotopy invariant. | Proof. First let us assume that \( X \) is connected. We prove (13.21) by induction on the number of cells of dimension 2 or more. If \( X \) has no such cells, then it is a connected graph. Problem 10-20 shows that \( {\pi }_{1}\left( X\right) \) is a free group on \( 1 - \chi \left( X\right) \) generators, and then T... | Yes |
Lemma 13.40. Let \( \mathbb{F} \) be a field of characteristic zero.\n\n(a) For any abelian group \( G \), the set \( \operatorname{Hom}\left( {G,\mathbb{F}}\right) \) of group homomorphisms from G to \( \mathbb{F} \) is a vector space over \( \mathbb{F} \) with scalar multiplication defined pointwise: \( \left( {a\var... | Proof. The proofs of (a) and (b) are straightforward (and hold for any field, not just one of characteristic zero), and are left as an exercise. | No |
Corollary 13.44. If \( X \) is a topological space such that \( {H}_{p}\left( X\right) \) is finitely generated for all \( p \) and zero for \( p \) sufficiently large, then for any field \( \mathbb{F} \) of characteristic zero, | \[ \chi \left( X\right) = \mathop{\sum }\limits_{p}{\left( -1\right) }^{p}\dim {H}^{p}\left( {X;\mathbb{F}}\right) . \] | No |
Proposition 1.1.1 If \( \lambda = \left( {{1}^{{m}_{1}},{2}^{{m}_{2}},\ldots ,{n}^{{m}_{n}}}\right) \) and \( g \in {S}_{n} \) has type \( \lambda \), then \( \left| {Z}_{g}\right| \) depends only on \( \lambda \) and | \[ {z}_{\lambda }\overset{\text{ def }}{ = }\left| {Z}_{g}\right| = {1}^{{m}_{1}}{m}_{1}!{2}^{{m}_{2}}{m}_{2}!\cdots {n}^{{m}_{n}}{m}_{n}! \] Proof. Any \( h \in {Z}_{g} \) can either permute the cycles of length \( i \) among themselves or perform a cyclic rotation on each of the individual cycles (or both). Since the... | Yes |
All groups have the trivial representation, which is the one sending every \( g \in G \) to the matrix (1). | This is clearly a representation because \( X\left( \epsilon \right) = \left( 1\right) \) and\n\n\[ X\left( g\right) X\left( h\right) = \left( 1\right) \left( 1\right) = \left( 1\right) = X\left( {gh}\right) \]\n\nfor all \( g, h \in G \) . We often use \( {1}_{G} \) or just the number 1 itself to stand for the trivial... | Yes |
Let us find all one-dimensional representations of the cyclic group of order \( n,{C}_{n} \) . Let \( g \) be a generator of \( {C}_{n} \), i.e., \[ {C}_{n} = \left\{ {g,{g}^{2},{g}^{3},\ldots ,{g}^{n} = \epsilon }\right\} \] | If \( X\left( g\right) = \left( c\right), c \in \mathbb{C} \), then the matrix for every element of \( {C}_{n} \) is determined, since \( X\left( {g}^{k}\right) = \left( {c}^{k}\right) \) by property 2 in the preceding definition. But by property 1, \[ \left( {c}^{n}\right) = X\left( {g}^{n}\right) = X\left( \epsilon \... | Yes |
Consider the symmetric group \( {\mathcal{S}}_{n} \) with its usual action on \( S = \{ 1,2,\ldots, n\} \) . Now\n\n\[ \mathbb{{CS}} = \left\{ {{c}_{1}\mathbf{1} + {c}_{2}\mathbf{2} + \cdots + {c}_{n}\mathbf{n} : {c}_{i} \in \mathbb{C}\text{ for all }i}\right\} \]\n\nwith the action\n\n\[ \pi \left( {{c}_{1}\mathbf{1} ... | To make things more concrete, we can select a basis and determine the matrices \( X\left( \pi \right) \) for \( \pi \in {\mathcal{S}}_{n} \) in that basis. Let us consider \( {\mathcal{S}}_{3} \) and use the standard basis \( \{ \mathbf{1},\mathbf{2},\mathbf{3}\} \) . To find the matrix for \( \pi = \left( {1,2}\right)... | No |
We now describe one of the most important representations for any group, the (left) regular representation. Let \( G \) be an arbitrary group. Then \( G \) acts on itself by left multiplication: if \( g \in G \) and \( h \in S = G \), then the action of \( g \) on \( h,{gh} \), is defined as the usual product in the gr... | Thus if \( G = \left\{ {{g}_{1},{g}_{2},\ldots ,{g}_{n}}\right\} \), then we have the corresponding \( G \) -module\n\n\[ \mathbb{C}\left\lbrack \mathbf{G}\right\rbrack = \left\{ {{c}_{1}{\mathbf{g}}_{1} + {c}_{2}{\mathbf{g}}_{2} + \cdots + {c}_{n}{\mathbf{g}}_{n} : {c}_{i} \in \mathbb{C}\text{ for all }i}\right\} \]\n... | Yes |
Example 1.3.5 Let group \( G \) have subgroup \( H \), written \( H \leq G \) . A generalization of the regular representation is the (left) coset representation of \( G \) with respect to \( H \) . Let \( {g}_{1},{g}_{2},\ldots ,{g}_{k} \) be a transversal for \( H \) ; i.e., \( \mathcal{H} = \left\{ {{g}_{1}H,{g}_{2}... | Let us consider \( G = {\mathcal{S}}_{3} \) and \( H = \{ \epsilon ,\left( {2,3}\right) \} \) . We can take\n\n\[ \mathcal{H} = \{ H,\left( {1,2}\right) H,\left( {1,3}\right) H\} \]\n\nand\n\n\[ \mathbb{C}\mathcal{H} = \left\{ {{c}_{1}\mathbf{H} + {c}_{2}\left( {\mathbf{1},\mathbf{2}}\right) \mathbf{H} + {c}_{3}\left( ... | Yes |
For a nontrivial example of a submodule, consider \( G = {\mathcal{S}}_{n} \) , \( n \geq 2 \), and \( V = \mathbb{C}\{ \mathbf{1},\mathbf{2},\ldots ,\mathbf{n}\} \) (the defining representation). Now take\n\n\[ W = \mathbb{C}\{ \mathbf{1} + \mathbf{2} + \cdots + \mathbf{n}\} = \{ c\left( {\mathbf{1} + \mathbf{2} + \cd... | But\n\n\[ \pi \left( {\mathbf{1} + \mathbf{2} + \cdots + \mathbf{n}}\right) = \pi \left( \mathbf{1}\right) + \pi \left( \mathbf{2}\right) + \cdots + \pi \left( \mathbf{n}\right) \]\n\n\[ = \mathbf{1} + \mathbf{2} + \cdots + \mathbf{n} \in W \]\n\nbecause applying \( \pi \) to \( \{ 1,2,\ldots, n\} \) just gives back th... | Yes |
Next, let us look again at the regular representation. Suppose \( G = \left\{ {{g}_{1},{g}_{2},\ldots ,{g}_{n}}\right\} \) with group algebra \( V = \mathbb{C}\left\lbrack \mathbf{G}\right\rbrack \) . Using the same idea as in the previous example, let\n\n\[ W = \mathbb{C}\left\lbrack {{\mathbf{g}}_{1} + {\mathbf{g}}_{... | \[ g\left( {{\mathbf{g}}_{1} + {\mathbf{g}}_{2} + \cdots + {\mathbf{g}}_{n}}\right) = g{\mathbf{g}}_{1} + g{\mathbf{g}}_{2} + \cdots + g{\mathbf{g}}_{n} \]\n\n\[ = {\mathbf{g}}_{1} + {\mathbf{g}}_{2} + \cdots + {\mathbf{g}}_{n} \in W \]\n\nbecause multiplying by \( g \) merely permutes the elements of \( G \), leaving ... | Yes |
Proposition 1.5.2 Let \( V \) be a \( G \) -module, \( W \) a submodule, and \( \langle \cdot , \cdot \rangle \) an inner product invariant under the action of \( G \) . Then \( {W}^{ \bot } \) is also a \( G \) -submodule. | Proof. We must show that for all \( g \in G \) and \( \mathbf{u} \in {W}^{ \bot } \) we have \( g\mathbf{u} \in {W}^{ \bot } \) . Take any \( \mathbf{w} \in W \) ; then\n\n\[ \langle g\mathbf{u},\mathbf{w}\rangle = \left\langle {{g}^{-1}g\mathbf{u},{g}^{-1}\mathbf{w}}\right\rangle \;\text{ (since }\langle \cdot , \cdot... | Yes |
Theorem 1.5.3 (Maschke’s Theorem) Let \( G \) be a finite group and let \( V \) be a nonzero \( G \) -module. Then\n\n\[ V = {W}^{\left( 1\right) } \oplus {W}^{\left( 2\right) } \oplus \cdots \oplus {W}^{\left( k\right) } \]\n\nwhere each \( {W}^{\left( i\right) } \) is an irreducible \( G \) -submodule of \( V \) . | Proof. We will induct on \( d = \dim V \) . If \( d = 1 \), then \( V \) itself is irreducible and we are done \( \left( {k = 1}\right. \) and \( \left. {{W}^{\left( 1\right) } = V}\right) \) . Now suppose that \( d > 1 \) . If \( V \) is irreducible, then we are finished as before. If not, then \( V \) has a nontrivia... | No |
Corollary 1.5.4 Let \( G \) be a finite group and let \( X \) be a matrix representation of \( G \) of dimension \( d > 0 \) . Then there is a fixed matrix \( T \) such that every matrix \( X\left( g\right), g \in G \), has the form\n\n\[ \n{TX}\left( g\right) {T}^{-1} = \left( \begin{matrix} {X}^{\left( 1\right) }\lef... | Proof. Let \( V = {\mathbb{C}}^{d} \) with the action\n\n\[ \ng\mathbf{v} = X\left( g\right) \mathbf{v} \n\]\n\nfor all \( g \in G \) and \( \mathbf{v} \in V \) . By Maschke’s theorem,\n\n\[ \nV = {W}^{\left( 1\right) } \oplus {W}^{\left( 2\right) } \oplus \cdots \oplus {W}^{\left( k\right) } \n\]\n\neach \( {W}^{\left... | Yes |
Proposition 1.6.4 Let \( \theta : V \rightarrow W \) be a \( G \) -homomorphism. Then\n\n1. \( \ker \theta \) is a \( G \) -submodule of \( V \), and\n\n2. \( \operatorname{im}\theta \) is a \( G \) -submodule of \( W \) . | Proof. We prove only the first assertion, leaving the second one for the reader. It is known from the theory of vector spaces that \( \ker \theta \) is a subspace of \( V \) since \( \theta \) is linear. So we only need to show closure under the action of \( G \) . But if \( \mathbf{v} \in \ker \theta \), then for any ... | No |
Theorem 1.6.5 (Schur’s Lemma) Let \( V \) and \( W \) be two irreducible \( G \) - modules. If \( \theta : V \rightarrow W \) is a \( G \) -homomorphism, then either\n\n1. \( \theta \) is a \( G \) -isomorphism, or\n\n2. \( \theta \) is the zero map. | Proof. Since \( V \) is irreducible and \( \ker \theta \) is a submodule (by the previous proposition), we must have either \( \ker \theta = \{ \mathbf{0}\} \) or \( \ker \theta = V \) . Similarly, the irreducibility of \( W \) implies that \( \operatorname{im}\theta = \{ \mathbf{0}\} \) or \( W \) . If \( \ker \theta ... | Yes |
Corollary 1.6.6 Let \( X \) and \( Y \) be two irreducible matrix representations of \( G \) . If \( T \) is any matrix such that \( {TX}\left( g\right) = Y\left( g\right) T \) for all \( g \in G \), then either | 1. \( T \) is invertible, or\n2. \( T \) is the zero matrix. ∎ | Yes |
Corollary 1.6.7 Let \( V \) and \( W \) be two \( G \) -modules with \( V \) being irreducible. Then \( \dim \operatorname{Hom}\left( {V, W}\right) = 0 \) if and only if \( W \) contains no submodule isomorphic \( {to}V. \) | ∎ | No |
Corollary 1.6.8 Let \( X \) be an irreducible matrix representation of \( G \) over the complex numbers. Then the only matrices \( T \) that commute with \( X\left( g\right) \) for all \( g \in G \) are those of the form \( T = {cI} \) -i.e., scalar multiples of the identity matrix. | ∎ | No |
Suppose that \( X \) is a matrix representation such that\n\n\[ X = \left( \begin{matrix} {X}^{\left( 1\right) } & 0 \\ 0 & {X}^{\left( 2\right) } \end{matrix}\right) = {X}^{\left( 1\right) } \oplus {X}^{\left( 2\right) },\]\n\nwhere \( {X}^{\left( 1\right) },{X}^{\left( 2\right) } \) are inequivalent and irreducible o... | Suppose that\n\n\[ T = \left( \begin{array}{ll} {T}_{1,1} & {T}_{1,2} \\ {T}_{2,1} & {T}_{2,2} \end{array}\right)\]\n\nis a matrix partitioned in the same way as \( X \) . If \( {TX} = {XT} \), then we can multiply out each side to obtain\n\n\[ \left( \begin{array}{ll} {T}_{1,1}{X}^{\left( 1\right) } & {T}_{1,2}{X}^{\l... | Yes |
Example 1.7.3 Suppose that\n\n\[ X = \left( \begin{matrix} {X}^{\left( 1\right) } & 0 \\ 0 & {X}^{\left( 1\right) } \end{matrix}\right) = 2{X}^{\left( 1\right) } \]\n\nwhere \( {X}^{\left( 1\right) } \) is irreducible of degree \( d \) . Take \( T \) partitioned as before. Doing the multiplication in \( {TX} = {XT} \) ... | Corollaries 1.6.6 and 1.6.8 come into play again to reveal that, for all \( i \) and \( j \) ,\n\n\[ {T}_{i, j} = {c}_{i, j}{I}_{d} \]\n\nwhere \( {c}_{i, j} \in \mathbb{C} \) . Thus\n\n\[ \operatorname{Com}X = \left\{ {\left( \begin{array}{ll} {c}_{1,1}{I}_{d} & {c}_{1,2}{I}_{d} \\ {c}_{2,1}{I}_{d} & {c}_{2,2}{I}_{d} ... | Yes |
Proposition 1.7.6 The center of \( {\operatorname{Mat}}_{d} \) is\n\n\[ \n{Z}_{{\text{Mat }}_{d}} = \left\{ {c{I}_{d} : c \in \mathbb{C}}\right\} \n\] | Proof. Suppose that \( C \in {Z}_{{\text{Mat }}_{d}} \) . Then, in particular,\n\n\[ \nC{E}_{i, i} = {E}_{i, i}C \n\]\n\n\( \left( {1.15}\right) \)\n\nfor all \( i \) . But \( C{E}_{i, i} \) (respectively, \( {E}_{i, i}C \) ) is all zeros except for the \( i \) th column (respectively, row), which is the same as \( C \... | Yes |
Lemma 1.7.7 Suppose \( A, X \in {\operatorname{Mat}}_{d} \) and \( B, Y \in {\operatorname{Mat}}_{f} \) . Then\n\n1. \( \left( {A \oplus B}\right) \left( {X \oplus Y}\right) = {AX} \oplus {BY} \),\n\n2. \( \left( {A \otimes B}\right) \left( {X \otimes Y}\right) = {AX} \otimes {BY} \) . | Proof. Both assertions are easy to prove, so we will do only the second. Suppose \( A = \left( {a}_{i, j}\right) \) and \( X = \left( {x}_{i, j}\right) \) .\n\n\[ \left( {A \otimes B}\right) \left( {X \otimes Y}\right) = \left( {{a}_{i, j}B}\right) \left( {{x}_{i, j}Y}\right) \;\text{ (definition of } \otimes \text{ ) ... | No |
Theorem 1.7.8 Let \( X \) be a matrix representation of \( G \) such that\n\n\[ X = {m}_{1}{X}^{\left( 1\right) } \oplus {m}_{2}{X}^{\left( 2\right) } \oplus \cdots \oplus {m}_{k}{X}^{\left( k\right) },\]\n\nwhere the \( {X}^{\left( i\right) } \) are inequivalent, irreducible and \( \deg {X}^{\left( i\right) } = {d}_{i... | ∎ | No |
Theorem 1.7.9 Let \( V \) be a \( G \) -module such that\n\n\[ V \cong {m}_{1}{V}^{\left( 1\right) } \oplus {m}_{2}{V}^{\left( 2\right) } \oplus \cdots \oplus {m}_{k}{V}^{\left( k\right) },\]\n\nwhere the \( {V}^{\left( i\right) } \) are pairwise inequivalent irreducibles and \( \dim {V}^{\left( i\right) } = {d}_{i} \)... | ∎ | No |
Proposition 1.7.10 Let \( V \) and \( W \) be \( G \) -modules with \( V \) irreducible. Then \( \dim \operatorname{Hom}\left( {V, W}\right) \) is the multiplicity of \( V \) in \( W \) . | ∎ | No |
Suppose we consider the defining representation of \( {S}_{n} \) with its character \( {\chi }^{\text{def }} \) . If we take \( n = 3 \), then we can compute the character values directly by taking the traces of the matrices in Example 1.2.4. The results are\n\n\[ \n{\chi }^{\text{def }}\left( {\;\left( 1\right) \left(... | \[ \n{\chi }^{\text{def }}\left( \pi \right) = \text{the number of ones on the diagonal of}X\left( \pi \right) \n\]\n\n\[ \n= \text{the number of fixedpoints of}\pi \text{. ∎} \n\] | Yes |
Example 1.8.4 Let \( G = \left\{ {{g}_{1},{g}_{2},\ldots ,{g}_{n}}\right\} \) and consider the regular representation with module \( V = \mathbb{C}\left\lbrack \mathbf{G}\right\rbrack \) and character \( {\chi }^{\text{reg }} \) . Now \( X\left( \epsilon \right) = {I}_{n} \), so \( {\chi }^{\text{reg }}\left( \epsilon ... | To compute the character values for \( g \neq \epsilon \), we will use the matrices arising from the standard basis \( \mathcal{B} = \left\{ {{\mathbf{g}}_{1},{\mathbf{g}}_{2},\ldots ,{\mathbf{g}}_{n}}\right\} \) . Now \( X\left( g\right) \) is the permutation matrix for the action of \( g \) on \( \mathcal{B} \), so \... | Yes |
Proposition 1.8.5 Let \( X \) be a matrix representation of a group \( G \) of degree \( d \) with character \( \chi \) .\n\n1. \( \chi \left( \epsilon \right) = d \) .\n\n2. If \( K \) is a conjugacy class of \( G \), then\n\n\[ g, h \in K \Rightarrow \chi \left( g\right) = \chi \left( h\right) . \]\n\n3. If \( Y \) i... | Proof. 1. Since \( X\left( \epsilon \right) = {I}_{d} \) ,\n\n\[ \chi \left( \epsilon \right) = \operatorname{tr}{I}_{d} = d \]\n\n2. By hypothesis \( g = {kh}{k}^{-1} \), so\n\n\[ \chi \left( g\right) = \operatorname{tr}X\left( g\right) = \operatorname{tr}X\left( k\right) X\left( h\right) X{\left( k\right) }^{-1} = \o... | Yes |
If \( G = {C}_{n} \), the cyclic group with \( n \) elements, then each element of \( {C}_{n} \) is in a conjugacy class by itself (as is true for any abelian group). Since there are \( n \) conjugacy classes, there must be \( n \) inequivalent irreducible representations of \( {C}_{n} \) . But we found \( n \) degree ... | Since the representations are one-dimensional, they are equal to their corresponding characters. Thus the table we displayed on page 5 is indeed the complete character table for \( {C}_{4} \) . ∎ | No |
Example 1.8.9 Recall that a conjugacy class in \( G = {\mathcal{S}}_{n} \) consists of all permutations of a given cycle type. In particular, for \( {\mathcal{S}}_{3} \) we have three conjugacy classes,\n\n\[ \n{K}_{1} = \{ \epsilon \} ,\;{K}_{2} = \{ \left( {1,2}\right) ,\left( {1,3}\right) ,\left( {2,3}\right) \} ,\;... | We will be able to fill in the last line using character inner products. ∎ | No |
Proposition 1.9.2 Let \( \chi \) and \( \psi \) be characters; then\n\n\[ \langle \chi ,\psi \rangle = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\chi \left( g\right) \psi \left( {g}^{-1}\right) .\n\] | When the field is arbitrary, equation (1.19) is taken as the definition of the inner product. In fact, for any two functions \( \chi \) and \( \psi \) from \( G \) to a field, we can define\n\n\[ \langle \chi ,\psi {\rangle }^{\prime } = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\chi \left( g\right) \p... | No |
Corollary 1.9.4 Let \( X \) be a matrix representation of \( G \) with character \( \chi \) . Suppose\n\n\[X \cong {m}_{1}{X}^{\left( 1\right) } \oplus {m}_{2}{X}^{\left( 2\right) } \oplus \cdots \oplus {m}_{k}{X}^{\left( k\right) },\] \n\nwhere the \( {X}^{\left( i\right) } \) are pairwise inequivalent irreducibles wi... | Proof. 1. Using the fact that the trace of a direct sum is the sum of the traces, we see that\n\n\[ \chi = \operatorname{tr}X = \operatorname{tr}{\bigoplus }_{i = 1}^{k}{m}_{i}{X}^{\left( i\right) } = \mathop{\sum }\limits_{{i = 1}}^{k}{m}_{i}{\chi }^{\left( i\right) }.\]\n\n2. We have, by the previous theorem,\n\n\[ \... | Yes |
Example 1.9.5 Let \( G = {\mathcal{S}}_{3} \) and consider \( \chi = {\chi }^{\text{def }} \) . Let \( {\chi }^{\left( 1\right) },{\chi }^{\left( 2\right) },{\chi }^{\left( 3\right) } \) be the three irreducible characters of \( {\mathcal{S}}_{3} \), where the first two are the trivial and sign characters, respectively... | \[ {m}_{1} = \left\langle {\chi ,{\chi }^{\left( 1\right) }}\right\rangle = \frac{1}{3!}\mathop{\sum }\limits_{{\pi \in {\mathcal{S}}_{3}}}\chi \left( \pi \right) {\chi }^{\left( 1\right) }\left( \pi \right) = \frac{1}{6}\left( {3 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 + 0 \cdot 1}\right) = 1, \]\n\n\[... | Yes |
Proposition 1.10.2 The irreducible characters of a group \( G \) form an orthonormal basis for the space of class functions \( R\left( G\right) \) . | Proof. Since the irreducible characters are orthonormal with respect to the bilinear form \( \langle \cdot , \cdot \rangle \) on \( R\left( G\right) \) (Theorem 1.9.3), they are linearly independent. But part 3 of Proposition 1.10.1 and equation (1.18) show that we have \( \dim R\left( G\right) \) such characters. Thus... | Yes |
Theorem 1.10.3 (Character Relations of the Second Kind) Let \( K, L \) be conjugacy classes of \( G \) . Then\n\n\[ \mathop{\sum }\limits_{\chi }{\chi }_{K}\overline{{\chi }_{L}} = \frac{\left| G\right| }{\left| K\right| }{\delta }_{K, L} \]\n\nwhere the sum is over all irreducible characters of \( G \) . | Proof. If \( \chi \) and \( \psi \) are irreducible characters, then the character relations of the first kind yield\n\n\[ \langle \chi ,\psi \rangle = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{K}\left| K\right| {\chi }_{K}\overline{{\psi }_{K}} = {\delta }_{\chi ,\psi }, \]\n\nwhere the sum is over all conjugac... | Yes |
Theorem 1.11.2 Let \( X \) and \( Y \) be matrix representations for \( G \) and \( H \), respectively.\n\n1. Then \( X \otimes Y \) is a representation of \( G \times H \) .\n\n2. If \( X, Y \) and \( X \otimes Y \) have characters denoted by \( \chi ,\psi \), and \( \chi \otimes \psi \), respectively, then\n\n\[ \lef... | Proof. 1. We verify the two conditions defining a representation. First of all,\n\n\[ \left( {X \otimes Y}\right) \left( {\epsilon ,\epsilon }\right) = X\left( \epsilon \right) \otimes Y\left( \epsilon \right) = I \otimes I = I. \]\n\nSecondly, if \( \left( {g, h}\right) ,\left( {{g}^{\prime },{h}^{\prime }}\right) \in... | Yes |
Theorem 1.11.3 Let \( G \) and \( H \) be groups.\n\n1. If \( X \) and \( Y \) are irreducible representations of \( G \) and \( H \), respectively, then \( X \otimes Y \) is an irreducible representation of \( G \times H \) . | Proof. 1. If \( \phi \) is any character, then we know (Corollary 1.9.4, part 4) that the corresponding representation is irreducible if and only if \( \langle \phi ,\phi \rangle = 1 \) . Letting \( X \) and \( Y \) have characters \( \chi \) and \( \psi \), respectively, we have\n\n\[ \langle \chi \otimes \psi ,\chi \... | Yes |
Proposition 1.12.3 Let \( H \leq G \) have transversal \( \left\{ {{t}_{1},\ldots ,{t}_{l}}\right\} \) with cosets \( \mathcal{H} = \left\{ {{t}_{1}H,\ldots ,{t}_{l}H}\right\} \) . Then the matrices of \( 1{ \uparrow }_{H}^{G} \) are identical with those of \( G \) acting on the basis \( \mathcal{H} \) for the coset mo... | Proof. Let the matrices for \( 1{ \uparrow }^{G} \) and \( \mathbb{C}\mathcal{H} \) be \( X = \left( {x}_{i, j}\right) \) and \( Z = \left( {z}_{i, j}\right) \) , respectively. Both arrays contain only zeros and ones. Finally, for any \( g \in G \) ,\n\n\[ \n{x}_{i, j}\left( g\right) = 1 \Leftrightarrow {t}_{i}^{-1}g{t... | Yes |
Theorem 1.12.4 Suppose \( H \leq G \) has transversal \( \left\{ {{t}_{1},\ldots ,{t}_{l}}\right\} \) and let \( Y \) be a matrix representation of \( H \) . Then \( X = Y{ \uparrow }_{H}^{G} \) is a representation of \( G \) . | Proof. Analogous to the case where \( Y \) is the trivial representation, we prove that \( X\left( g\right) \) is always a block permutation matrix; i.e., every row and column contains exactly one nonzero block \( Y\left( {{t}_{i}^{-1}g{t}_{j}}\right) \) . Consider the first column (the other cases being similar). It s... | Yes |
Proposition 1.12.5 Consider \( H \leq G \) and a matrix representation \( Y \) of H. Let \( \left\{ {{t}_{1},\ldots ,{t}_{l}}\right\} \) and \( \left\{ {{s}_{1},\ldots ,{s}_{l}}\right\} \) be two transversals for \( H \) giving rise to representation matrices \( X \) and \( Z \), respectively, for \( Y{ \uparrow }^{G} ... | Proof. Let \( \chi ,\psi \), and \( \phi \) be the characters of \( X, Y \), and \( Z \), respectively. Then it suffices to show that \( \chi = \phi \) (Corollary 1.9.4, part 5). Now\n\n\[ \chi \left( g\right) = \mathop{\sum }\limits_{i}\operatorname{tr}Y\left( {{t}_{i}^{-1}g{t}_{i}}\right) = \mathop{\sum }\limits_{i}\... | Yes |
Theorem 1.12.6 (Frobenius Reciprocity) Let \( H \leq G \) and suppose that \( \psi \) and \( \chi \) are characters of \( H \) and \( G \), respectively. Then\n\n\[ \left\langle {\psi { \uparrow }^{G},\chi }\right\rangle = \left\langle {\psi ,\chi { \downarrow }_{H}}\right\rangle \]\n\nwhere the left inner product is c... | Proof. We have the following string of equalities:\n\n\[ \left\langle {\psi { \uparrow }^{G},\chi }\right\rangle = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\psi { \uparrow }^{G}\left( g\right) \chi \left( {g}^{-1}\right) \]\n\n\[ = \frac{1}{\left| G\right| \left| H\right| }\mathop{\sum }\limits_{{x \i... | Yes |
Now consider \( \lambda = \left( {1}^{n}\right) \) . Each equivalence class \( \{ t\} \) consists of a single tableau, and this tableau can be identified with a permutation in one-line notation (by taking the transpose, if you wish). Since the action of \( {\mathcal{S}}_{n} \) is preserved, | \[ {M}^{\left( {1}^{n}\right) } \cong \mathbb{C}{\mathcal{S}}_{n} \] and the regular representation presents itself. | Yes |
Consider \( \lambda \vdash n \) with Young subgroup \( {\mathcal{S}}_{\lambda } \) and tabloid \( \{ {t}^{\lambda }\} \) , as before. Then \( {V}^{\lambda } = \mathbb{C}{\mathcal{S}}_{n}{\mathcal{S}}_{\lambda } \) and \( {M}^{\lambda } = \mathbb{C}{\mathcal{S}}_{n}\left\{ {t}^{\lambda }\right\} \) are isomorphic as \( ... | Proof. Let \( {\pi }_{1},{\pi }_{2},\ldots ,{\pi }_{k} \) be a transversal for \( {\mathcal{S}}_{\lambda } \) . Define a map\n\n\[ \theta : {V}^{\lambda } \rightarrow {M}^{\lambda } \]\n\nby \( \theta \left( {{\pi }_{i}{\mathcal{S}}_{\lambda }}\right) = \left\{ {{\pi }_{i}{t}^{\lambda }}\right\} \) for \( i = 1,\ldots,... | Yes |
Theorem 2.1.12 Consider \( \lambda \vdash n \) with Young subgroup \( {\mathcal{S}}_{\lambda } \) and tabloid \( \{ {t}^{\lambda }\} \) , as before. Then \( {V}^{\lambda } = \mathbb{C}{\mathcal{S}}_{n}{\mathcal{S}}_{\lambda } \) and \( {M}^{\lambda } = \mathbb{C}{\mathcal{S}}_{n}\left\{ {t}^{\lambda }\right\} \) are is... | Proof. Let \( {\pi }_{1},{\pi }_{2},\ldots ,{\pi }_{k} \) be a transversal for \( {\mathcal{S}}_{\lambda } \) . Define a map\n\n\[ \theta : {V}^{\lambda } \rightarrow {M}^{\lambda } \]\n\nby \( \theta \left( {{\pi }_{i}{\mathcal{S}}_{\lambda }}\right) = \left\{ {{\pi }_{i}{t}^{\lambda }}\right\} \) for \( i = 1,\ldots,... | Yes |
Lemma 2.2.4 (Dominance Lemma for Partitions) Let \( {t}^{\lambda } \) and \( {s}^{\mu } \) be tableaux of shape \( \lambda \) and \( \mu \), respectively. If, for each index \( i \), the elements of row \( i \) of \( {s}^{\mu } \) are all in different columns in \( {t}^{\lambda } \), then \( \lambda \trianglerighteq \m... | Proof. By hypothesis, we can sort the entries in each column of \( {t}^{\lambda } \) so that the elements of rows \( 1,2,\ldots, i \) of \( {s}^{\mu } \) all occur in the first \( i \) rows of \( {t}^{\lambda } \) . Thus\n\n\( {\lambda }_{1} + {\lambda }_{2} + \cdots + {\lambda }_{i} = \) number of elements in the firs... | Yes |
Proposition 2.2.6 If \( \lambda ,\mu \vdash n \) with \( \lambda \trianglerighteq \mu \), then \( \lambda \geq \mu \) . | Proof. If \( \lambda \neq \mu \), then find the first index \( i \) where they differ. Thus \( \mathop{\sum }\limits_{{j = 1}}^{{i - 1}}{\lambda }_{j} = \) \( \mathop{\sum }\limits_{{j = 1}}^{{i - 1}}{\mu }_{j} \) and \( \mathop{\sum }\limits_{{j = 1}}^{i}{\lambda }_{j} > \mathop{\sum }\limits_{{j = 1}}^{i}{\mu }_{j} \... | Yes |
Lemma 2.3.3 Let \( t \) be a tableau and \( \pi \) be a permutation. Then\n\n1. \( {R}_{\pi t} = \pi {R}_{t}{\pi }^{-1} \) ,\n\n2. \( {C}_{\pi t} = \pi {C}_{t}{\pi }^{-1} \) ,\n\n3. \( {\kappa }_{\pi t} = \pi {\kappa }_{t}{\pi }^{-1} \) ,\n\n4. \( {\mathbf{e}}_{\pi t} = \pi {\mathbf{e}}_{t} \) . | Proof. 1. We have the following list of equivalent statements:\n\n\[ \sigma \in {R}_{\pi t}\; \leftrightarrow \;\sigma \{ {\pi t}\} = \{ {\pi t}\} \]\n\n\[ \leftrightarrow \;{\pi }^{-1}{\sigma \pi }\{ t\} = \{ t\} \]\n\n\[ \leftrightarrow \;{\pi }^{-1}{\sigma \pi } \in {R}_{t} \]\n\n\[ \leftrightarrow \sigma \in \pi {R... | Yes |
Proposition 2.3.5 The \( {S}^{\lambda } \) are cyclic modules generated by any given poly-tabloid. | ∎ | No |
Example 2.3.6 Suppose \( \lambda = \left( n\right) \) . Then \( {e}_{{12}\cdots n} = \underline{\overline{\mathbf{{12}}\cdots \mathbf{n}}} \) is the only polytabloid and \( {S}^{\left( n\right) } \) carries the trivial representation. This is, of course, the only possibility, since \( {S}^{\left( n\right) } \) is a sub... | ∎ | No |
Lemma 2.4.1 (Sign Lemma) Let \( H \leq {\mathcal{S}}_{n} \) be a subgroup.\n\n1. If \( \pi \in H \), then\n\n\[ \pi {H}^{ - } = {H}^{ - }\pi = \left( {\operatorname{sgn}\pi }\right) {H}^{ - }.\]\n\nOtherwise put: \( {\pi }^{ - }{H}^{ - } = {H}^{ - } \).\n\n2. For any \( \mathbf{u},\mathbf{v} \in {M}^{\lambda } \), \n\n... | Proof. 1. This is just like the proof that \( \pi {\mathbf{e}}_{t} = \left( {\operatorname{sgn}\pi }\right) {\mathbf{e}}_{t} \) in Example 2.3.7.\n\n2. Using the fact that our form is \( {\mathcal{S}}_{n} \) -invariant,\n\n\[ \left\langle {{H}^{ - }\mathbf{u},\mathbf{v}}\right\rangle = \mathop{\sum }\limits_{{\pi \in H... | Yes |
Corollary 2.4.2 Let \( t = {t}^{\lambda } \) be a \( \lambda \) -tableau and \( s = {s}^{\mu } \) be a \( \mu \) -tableau, where \( \lambda ,\mu \vdash n \) . If \( {\kappa }_{t}\{ s\} \neq \mathbf{0} \), then \( \lambda \geq \mu \) . And if \( \lambda = \mu \), then \( {\kappa }_{t}\{ s\} = \pm {\mathbf{e}}_{t} \) . | Proof. Suppose \( b \) and \( c \) are two elements in the same row of \( {s}^{\mu } \) . Then they cannot be in the same column of \( {t}^{\lambda } \), for if so, then \( {\kappa }_{t} = k\left( {\epsilon - \left( {b, c}\right) }\right) \) and \( {\kappa }_{t}\{ s\} = 0 \) by parts 3 and 4 in the preceding lemma. Thu... | Yes |
Corollary 2.4.3 If \( \mathbf{u} \in {M}^{\mu } \) and \( \operatorname{sh}t = \mu \), then \( {\kappa }_{t}\mathbf{u} \) is a multiple of \( {\mathbf{e}}_{t} \) . | Proof. We can write \( \mathbf{u} = \mathop{\sum }\limits_{i}{c}_{i}\left\{ {s}_{i}\right\} \), where the \( {s}_{i} \) are \( \mu \) -tableaux. By the previous corollary, \( {\kappa }_{t}\mathbf{u} = \mathop{\sum }\limits_{i} \pm {c}_{i}{\mathbf{e}}_{t} \) . ∎ | Yes |
Theorem 2.4.4 (Submodule Theorem [Jam 76]) Let \( U \) be a submodule of \( {M}^{\mu } \). Then\n\n\[ U \supseteq {S}^{\mu }\;\text{ or }\;U \subseteq {S}^{\mu \bot }.\]\n\nIn particular, when the field is \( \mathbb{C} \), the \( {S}^{\mu } \) are irreducible. | Proof. Consider \( \mathbf{u} \in U \) and a \( \mu \) -tableau \( t \). By the preceding corollary, we know that \( {\kappa }_{t}\mathbf{u} = f{\mathbf{e}}_{t} \) for some field element \( f \). There are two cases, depending on which multiples can arise.\n\nSuppose that there exits a \( \mathbf{u} \) and a \( t \) wi... | Yes |
Proposition 2.4.5 Suppose the field of scalars is \( \mathbb{C} \) and \( \theta \in \operatorname{Hom}\left( {{S}^{\lambda },{M}^{\mu }}\right) \) is nonzero. Thus \( \lambda \trianglerighteq \mu \), and if \( \lambda = \mu \), then \( \theta \) is multiplication by a scalar. | Proof. Since \( \theta \neq 0 \), there is some basis vector \( {\mathbf{e}}_{t} \) such that \( \theta \left( {\mathbf{e}}_{t}\right) \neq \mathbf{0} \) . Because \( \langle \cdot , \cdot \rangle \) is an inner product with complex scalars, \( {M}^{\lambda } = {S}^{\lambda } \oplus {S}^{\lambda \bot } \) . Thus we can... | Yes |
Theorem 2.4.6 The \( {S}^{\lambda } \) for \( \lambda \vdash n \) form a complete list of irreducible \( {\mathcal{S}}_{n} \) - modules over the complex field. | Proof. The \( {S}^{\lambda } \) are irreducible by the submodule theorem and the fact that \( {S}^{\lambda } \cap {S}^{\lambda \bot } = \mathbf{0} \) for the field \( \mathbb{C} \) .\n\nSince we have the right number of modules for a full set, it suffices to show that they are pairwise inequivalent. But if \( {S}^{\lam... | Yes |
Corollary 2.4.7 The permutation modules decompose as\n\n\[ \n{M}^{\mu } = {\bigoplus }_{\lambda \geq \mu }{m}_{\lambda \mu }{S}^{\lambda } \]\n\nwith the diagonal multiplicity \( {m}_{\mu \mu } = 1 \) . | Proof. This result follows from Proposition 2.4.5. If \( {S}^{\lambda } \) appears in \( {M}^{\mu } \) with nonzero coefficient, then \( \lambda \trianglerighteq \mu \) . If \( \lambda = \mu \), then we can also apply Proposition 1.7.10 to obtain\n\n\[ \n{m}_{\mu \mu } = \dim \operatorname{Hom}\left( {{S}^{\mu },{M}^{\... | Yes |
Lemma 2.5.5 (Dominance Lemma for Tabloids) If \( k < l \) and \( k \) appears in a lower row than \( l \) in \( \{ t\} \), then\n\n\[ \{ t\} \vartriangleleft \left( {k, l}\right) \{ t\} \] | Proof. Suppose that \( \{ t\} \) and \( \left( {k, l}\right) \{ t\} \) have composition sequences \( {\lambda }^{i} \) and \( {\mu }^{i} \) , respectively. Then for \( i < k \) or \( i \geq l \) we have \( {\lambda }^{i} = {\mu }^{i} \).\n\nNow consider the case where \( k \leq i < l \) . If \( r \) and \( q \) are the... | Yes |
Corollary 2.5.6 If \( t \) is standard and \( \{ s\} \) appears in \( {e}_{t} \), then \( \{ t\} \geq \{ s\} \) . | Proof. Let \( s = {\pi t} \), where \( \pi \in {C}_{t} \), so that \( \{ s\} \) appears in \( {\mathbf{e}}_{t} \) . We induct on the number of column inversions in \( s \), i.e., the number of pairs \( k < l \) in the same column of \( s \) such that \( k \) is in a lower row than \( l \) . Given any such pair,\n\n\[ \... | Yes |
Lemma 2.5.8 Let \( {\mathbf{v}}_{1},{\mathbf{v}}_{2},\ldots ,{\mathbf{v}}_{m} \) be elements of \( {M}^{\mu } \) . Suppose, for each \( {\mathbf{v}}_{i} \) , we can choose a tabloid \( \left\{ {\mathbf{t}}_{i}\right\} \) appearing in \( {\mathbf{v}}_{i} \) such that\n\n1. \( \left\{ {t}_{i}\right\} \) is maximum in \( ... | Proof. Choose the labels such that \( \left\{ {t}_{1}\right\} \) is maximal among the \( \left\{ {t}_{i}\right\} \) . Thus conditions 1 and 2 ensure that \( \left\{ {t}_{1}\right\} \) appears only in \( {\mathbf{v}}_{1} \) . (If \( \left\{ {t}_{1}\right\} \) occurs in \( {\mathbf{v}}_{i}, i > 1 \), then \( \left\{ {t}_... | Yes |
Proposition 2.5.9 The set\n\n\[ \n\\left\\{ {{e}_{t} : t\\text{ is a standard }\\lambda \\text{-tableau }}\\right\\} \n\]\n\nis independent. | Proof. By Corollary 2.5.6, \\( \\{ t\\} \\) is maximum in \\( {\\mathbf{e}}_{t} \\), and by hypothesis they are all distinct. Thus Lemma 2.5.8 applies. | No |
Proposition 2.6.3 Let \( t, A \), and \( B \), be as in the definition of a Garnir element. If \( \left| {A \cup B}\right| \) is greater than the number of elements in column \( j \) of \( t \) , then \( {g}_{A, B}{e}_{t} = \mathbf{0} \) . | Proof. First, we claim that\n\n\[ \n{\mathcal{S}}_{A \cup B}^{ - }{e}_{t} = 0 \n\]\n\n(2.4)\n\nConsider any \( \sigma \in {C}_{t} \) . By the hypothesis, there must be \( a, b \in A \cup B \) such that \( a \) and \( b \) are in the same row of \( {\sigma t} \) . But then \( \left( {a, b}\right) \in {\mathcal{S}}_{A \c... | Yes |
Theorem 2.6.4 The set\n\n\[ \n\\left\\{ {{\\mathbf{e}}_{t} : t\\text{ is a standard }\\lambda \\text{-tableau }}\\right\\} \n\]\n\nspans \( {S}^{\\lambda } \) . | First note that if \( {\\mathbf{e}}_{t} \) is in the span of the set (2.6), then so is \( {\\mathbf{e}}_{s} \) for any \( s \\in \\left\\lbrack t\\right\\rbrack \), by the remarks at the beginning of this section. Thus we may always take \( t \) to have increasing columns.\n\nThe poset of column tabloids has a maximum ... | Yes |
Theorem 2.6.5 For any partition \( \lambda \) :\n\n1. \( \left\{ {{\mathbf{e}}_{t} : t}\right. \) is a standard \( \lambda \) -tableau \( \} \) is a basis for \( {S}^{\lambda } \),\n\n2. \( \dim {S}^{\lambda } = {f}^{\lambda } \), and\n\n3. \( \mathop{\sum }\limits_{{\lambda \vdash n}}{\left( {f}^{\lambda }\right) }^{2... | Proof. The first two parts are immediate. The third follows from the fact (Proposition 1.10.1) that for any group \( G \) ,\n\n\[ \mathop{\sum }\limits_{V}{\left( \dim V\right) }^{2} = \left| G\right| \]\n\nwhere the sum is over all irreducible \( G \) -modules. ∎ | No |
Lemma 2.8.2 We have\n\n\[ \n{f}^{\lambda } = \mathop{\sum }\limits_{{\lambda }^{ - }}{f}^{{\lambda }^{ - }} \n\] | Proof. Every standard tableau of shape \( \lambda \vdash n \) consists of \( n \) in some inner corner together with a standard tableau of shape \( {\lambda }^{ - } \vdash n - 1 \) . The result follows. | Yes |
Proposition 2.9.4 If \( t \) is the fixed \( \lambda \) -tableau and \( T \in {\mathcal{T}}_{\lambda \mu } \), then \( {\kappa }_{t}\mathbf{T} = \mathbf{0} \) if and only if \( T \) has two equal elements in some column. | Proof. If \( {\kappa }_{t}\mathbf{T} = \mathbf{0} \), then\n\n\[\n\mathbf{T} + \mathop{\sum }\limits_{\substack{{\pi \in {C}_{t}} \\ {\pi \neq \epsilon } }}\left( {\operatorname{sgn}\pi }\right) \pi \mathbf{T} = \mathbf{0}\n\]\n\nSo we must have \( \mathbf{T} = \pi \mathbf{T} \) for some \( \pi \in {C}_{t} \) with \( \... | Yes |
Lemma 2.10.4 Let \( V \) and \( \mathcal{B} \) be as before and consider a set of vectors \( {\mathbf{v}}_{1},{\mathbf{v}}_{2},\ldots ,{\mathbf{v}}_{m} \in V \) . Suppose that, for all \( i \), there exists \( {\mathbf{b}}_{i} \in \mathcal{B} \) appearing in \( {\mathbf{v}}_{i} \) such that\n\n1. \( \left\lbrack {\math... | ∎ | No |
Lemma 2.10.5 Let \( V \) and \( W \) be vector spaces and let \( {\theta }_{1},{\theta }_{2},\ldots ,{\theta }_{m} \) be linear maps from \( V \) to \( W \) . If there exists \( a\mathbf{v} \in V \) such that \( {\theta }_{1}\left( \mathbf{v}\right) ,{\theta }_{2}\left( \mathbf{v}\right) ,\ldots ,{\theta }_{m}\left( \m... | Proof. Suppose not. Then there are constants \( {c}_{i} \), not all zero, such that \( \mathop{\sum }\limits_{i}{c}_{i}{\theta }_{i} \) is the zero map. But then \( \mathop{\sum }\limits_{i}{c}_{i}{\theta }_{i}\left( \mathbf{v}\right) = \mathbf{0} \) for all \( \mathbf{v} \in V \), a contradiction to the hypothesis of ... | Yes |
Proposition 2.10.6 The set\n\n\\[ \n\\left\\{ {{\\bar{\\theta }}_{T} : T \\in {\\mathcal{T}}_{\\lambda \\mu }^{0}}\\right\\} \n\\]\n\nis independent. | Proof. Let \\( {T}_{1},{T}_{2},\\ldots ,{T}_{m} \\) be the elements of \\( {\\mathcal{T}}_{\\lambda \\mu } \\) . By the previous lemma, it suffices to show that \\( {\\bar{\\theta }}_{{T}_{1}}{\\mathbf{e}}_{t},{\\bar{\\theta }}_{{T}_{2}}{\\mathbf{e}}_{t},\\ldots ,{\\bar{\\theta }}_{{T}_{m}}{\\mathbf{e}}_{t} \\) are ind... | Yes |
Lemma 2.10.7 Consider \( \theta \in \operatorname{Hom}\left( {{S}^{\lambda },{M}^{\mu }}\right) \) . Write\n\n\[ \theta {e}_{t} = \mathop{\sum }\limits_{T}{c}_{T}T \]\n\nwhere \( t \) is the fixed tableau of shape \( \lambda \) .\n\n1. If \( \pi \in {C}_{t} \) and \( {T}_{1} = \pi {T}_{2} \), then \( {c}_{{T}_{1}} = \l... | Proof. 1. Since \( \pi \in {C}_{t} \), we have\n\n\[ \pi \left( {\theta {\mathbf{e}}_{t}}\right) = \theta \left( {\pi {\kappa }_{t}\{ \mathbf{t}\} }\right) = \theta \left( {\left( {\operatorname{sgn}\pi }\right) {\kappa }_{t}\{ \mathbf{t}\} }\right) = \left( {\operatorname{sgn}\pi }\right) \left( {\theta {\mathbf{e}}_{... | Yes |
Proposition 2.10.8 The set\n\n\\[ \n\\left\\{ {{\\bar{\\theta }}_{T} : T \\in {\\mathcal{T}}_{\\lambda \\mu }^{0}}\\right\\} \n\\]\n\nspans \\( \\operatorname{Hom}\\left( {{S}^{\\lambda },{M}^{\\mu }}\\right) \\) . | Proof. Pick any \\( \\theta \\in \\operatorname{Hom}\\left( {{S}^{\\lambda },{M}^{\\mu }}\\right) \\) and write\n\n\\[ \n\\theta {\\mathbf{e}}_{t} = \\mathop{\\sum }\\limits_{T}{c}_{T}\\mathbf{T} \n\\]\n\n\\( \\left( {2.13}\\right) \\)\n\nConsider\n\n\\[ \n{L}_{\\theta } = \\left\\{ {S \\in {\\mathcal{T}}_{\\lambda \\m... | Yes |
Theorem 2.11.2 (Young’s Rule) The multiplicity of \( {S}^{\lambda } \) in \( {M}^{\mu } \) is equal to the number of semistandard tableaux of shape \( \lambda \) and content \( \mu \), i.e., | \[ {M}^{\mu } \cong {\bigoplus }_{\lambda }{K}_{\lambda \mu }{S}^{\lambda } \] | No |
Example 2.11.3 Suppose \( \mu = \left( {2,2,1}\right) \) . Then the possible \( \lambda \trianglerighteq \mu \) and the associated \( \lambda \) -tableaux of type \( \mu \) are as follows: | \[ {\lambda }^{1} = \left( {2,2,1}\right) \;{\lambda }^{2} = \left( {3,1,1}\right) \;{\lambda }^{3} = \left( {3,2}\right) \;{\lambda }^{4} = \left( {4,1}\right) \;{\lambda }^{5} = \left( 5\right) \] Thus \[ {M}^{\left( 2,2,1\right) } \cong {S}^{\left( 2,2,1\right) } \oplus {S}^{\left( 3,3,1\right) } \oplus 2{S}^{\left(... | Yes |
Example 2.11.4 For any \( \mu ,{K}_{\mu \mu } = 1 \) . | This is because the only \( \mu \) -tableau of content \( \mu \) is the one with all the 1’s in row 1, all the 2’s in row 2, etc. Of course, we saw this result in Corollary 2.4.7. | No |
Example 2.11.6 For any \( \lambda ,{K}_{\lambda \left( {1}^{n}\right) } = {f}^{\lambda } \) (the number of standard tableaux of shape \( \lambda \) ). This says that | \[ {M}^{\left( {1}^{n}\right) } \cong {\bigoplus }_{\lambda }{f}^{\lambda }{S}^{\lambda } \] But \( {M}^{\left( {1}^{n}\right) } \) is just the regular representation (Example 2.1.7) and \( {f}^{\lambda } = \dim {S}^{\lambda } \) (Theorem 2.5.2). Thus this is merely the special case of Proposition 1.10.1, part 1, where... | Yes |
Theorem 3.1.1 ([Rob 38, Sch 61]) The map\n\n\\[ \pi \\overset{\\mathrm{R} - \\mathrm{S}}{ \\rightarrow }\\left( {P, Q}\\right) \\]\n\nis a bijection between elements of \\( {\\mathcal{S}}_{n} \\) and pairs of standard tableaux of the same shape \\( \\lambda \\vdash n \\) . | Proof. To show that we have a bijection, it suffices to create an inverse.\n\n\ | No |
Theorem 3.2.3 ([Sch 61]) If \( P\left( \pi \right) = P \), then \( P\left( {\pi }^{r}\right) = {P}^{t} \), where \( t \) denotes transposition. | Proof. We have\n\n\[ P\left( {\pi }^{r}\right) = {r}_{{x}_{1}}\cdots {r}_{{x}_{n - 1}}{r}_{{x}_{n}}\left( \varnothing \right) \;\text{ (definition of }P\left( {\pi }^{r}\right) \text{ ) }\n\]\n\n\[ = {r}_{{x}_{1}}\cdots {r}_{{x}_{n - 1}}{c}_{{x}_{n}}\left( \varnothing \right) \;\text{(initial tableau is empty)}\n\]\n\n... | Yes |
Theorem 3.3.2 ([Sch 61]) Consider \( \pi \in {\mathcal{S}}_{n} \) . The length of the longest increasing subsequence of \( \pi \) is the length of the first row of \( P\left( \pi \right) \) . The length of the longest decreasing subsequence of \( \pi \) is the length of the first column of \( P\left( \pi \right) \) . | Proof. Since reversing a permutation turns decreasing sequences into increasing ones, the second half of the theorem follows from the first and Theorem 3.2.3. To prove the first half, we actually demonstrate a stronger result. In what follows, \( {P}_{k - 1} \) is the the tableau formed after \( k - 1 \) insertions of ... | Yes |
Lemma 3.3.3 If \( \pi = {x}_{1}{x}_{2}\ldots {x}_{n} \) and \( {x}_{k} \) enters \( {P}_{k - 1} \) in column \( j \), then the longest increasing subsequence of \( \pi \) ending in \( {x}_{k} \) has length \( j \) . | Proof. We induct on \( k \) . The result is trivial for \( k = 1 \), so suppose it holds for all values up to \( k - 1 \) .\n\nFirst we need to show the existence of an increasing subsequence of length \( j \) ending in \( {x}_{k} \) . Let \( y \) be the element of \( {P}_{k - 1} \) in cell \( \left( {1, j - 1}\right) ... | Yes |
Theorem 3.4.3 ([Knu 70]) If \( \pi ,\sigma \in {\mathcal{S}}_{n} \), then\n\n\[ \pi \overset{K}{ \cong }\sigma \Leftrightarrow \pi \overset{P}{ \cong }\sigma \] | Proof. \ | No |
Lemma 3.6.2 Let the shadow diagram of \( \pi = {x}_{1}{x}_{2}\ldots {x}_{n} \) be constructed as before. Suppose the vertical line \( x = k \) intersects \( i \) of the shadow lines. Let \( {y}_{j} \) be the \( y \) -coordinate of the lowest point of the intersection with \( {L}_{j} \) . Then the first row of the \( {P... | Proof. Induct on \( k \), the lemma being trivial for \( k = 0 \) . Assume that the result holds for the line \( x = k \) and consider \( x = k + 1 \) . There are two cases. If\n\n\[ \n{x}_{k + 1} > {y}_{i} \n\]\n\nthen the box \( \left( {k + 1,{x}_{k + 1}}\right) \) starts a new shadow line. So none of the values \( {... | Yes |
Corollary 3.6.3 ([Vie 76]) If the permutation \( \pi \) has Robinson-Schensted tableaux \( \left( {P, Q}\right) \) and shadow lines \( {L}_{j} \), then, for all \( j \), \n\n\[ \n{P}_{1, j} = {y}_{{L}_{j}}\;\text{ and }\;{Q}_{1, j} = {x}_{{L}_{j}}. \n\] | Proof. The statement for \( P \) is just the case \( k = n \) of Lemma 3.6.2.\n\nAs for \( Q \), the entry \( k \) is added to \( Q \) in cell \( \left( {1, j}\right) \) when \( {x}_{k} \) is greater than every element of the first row of \( {P}_{k - 1} \) . But the previous lemma’s proof shows that this happens precis... | Yes |
Theorem 3.6.5 ([Vie 76]) Suppose \( \pi \overset{\mathrm{R} - \mathrm{S}}{ \rightarrow }\left( {P, Q}\right) \) . Then \( {\pi }^{\left( i\right) } \) is a partial permutation such that\n\n\[ \n{\pi }^{\left( i\right) }\overset{\mathrm{R} - \mathrm{S}}{ \rightarrow }\left( {{P}^{\left( i\right) },{Q}^{\left( i\right) }... | ∎ | No |
Theorem 3.6.6 ([Sci 63]) If \( \pi \in {\mathcal{S}}_{n} \), then\n\n\[ P\left( {\pi }^{-1}\right) = Q\left( \pi \right) \;\text{ and }\;Q\left( {\pi }^{-1}\right) = P\left( \pi \right) . | Proof. Taking the inverse of a permutation corresponds to reflecting the shadow diagram in the line \( y = x \) . The theorem now follows from Theorem 3.6.5. ∎ | Yes |
Theorem 3.6.10 If \( \pi ,\sigma \in {\mathcal{S}}_{n} \), then\n\n\[ \pi \overset{{K}^{ * }}{ \cong }\sigma \Leftrightarrow \pi \overset{Q}{ \cong }\sigma . \]\n | Proof. We have the following string of equivalences:\n\n\[ \pi \overset{{K}^{ * }}{ \cong }\sigma \; \Leftrightarrow \;{\pi }^{-1}\overset{K}{ \cong }{\sigma }^{-1}\;\text{ (Lem }\n\n\[ \Leftrightarrow \;P\left( {\pi }^{-1}\right) = P\left( {\sigma }^{-1}\right) \;\text{(Theorem 3.4.3)}\n\n\[ \Leftrightarrow \;Q\left( ... | Yes |
Proposition 3.7.4 ([Scii 76]) If \( P, Q \) are standard skew tableaux, then\n\n\[ P \cong Q \Rightarrow P\overset{K}{ \cong }Q. \]\n | Proof. By induction, it suffices to prove the theorem when \( P \) and \( Q \) differ by a single slide. In fact, if we call the operation in steps Fb or Rb of the slide definition a move, then we need to demonstrate the result only when \( P \) and \( Q \) differ by a move. (The row word of a tableau with a hole in it... | Yes |
Theorem 3.7.7 ([Scii 76]) If \( P \) is a partial skew tableau that is brought to a normal tableau \( {P}^{\prime } \) by slides, then \( {P}^{\prime } \) is unique. In fact, \( {P}^{\prime } \) is the insertion tableau for \( {\pi }_{P} \) . | Proof. By the previous proposition, \( {\pi }_{P}\overset{K}{ \cong }{\pi }_{{P}^{\prime }} \) . Thus by Knuth’s theorem on \( P \) -equivalence (Theorem 3.4.3), \( {\pi }_{P} \) and \( {\pi }_{{P}^{\prime }} \) have the same insertion tableau. Finally, Lemma 3.4.5 tells us that the insertion tableau for \( {\pi }_{{P}... | Yes |
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