Lecture 18

Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 27 August 2013

On the other hand, since $z = \sum_{a \in 𝒜} χ_{V} (a) a^{*}$ it follows that $\begin{matrix} [z] & = & \sum_{a \in 𝒜} χ_{V} (a) [a^{*}] \\ = & \sum_{b \in ℬ} \sum_{a \in 𝒜} χ_{V} (a) ⟨ [a^{*}], b ⟩ b^{*} \end{matrix}$ In fact, these scalars $⟨ [a^{*}], b ⟩$ are equal to $⟨ a^{*}, [b] ⟩,$ for $\begin{matrix} ⟨ [a^{*}], b ⟩ & = & \sum_{h \in ℬ} ⟨ h a^{*} h^{*}, b ⟩ \\ = & \sum_{h \in ℬ} \vec{t} (h a^{*} h^{*} b) \\ = & \sum_{h \in ℬ} \vec{t} (a^{*} h^{*} b h) \\ = & \sum_{h \in ℬ} ⟨ a^{*}, h^{*} b h ⟩ \\ = & \sum_{h^{*} \in ℬ^{*}} ⟨ a^{*}, h^{*} b {(h^{*})}^{*} ⟩ \\ = & ⟨ a^{*}, [b] ⟩ \end{matrix}$ (This last equality follows by the basis-independence of the $[\cdot]$ operator.) This now gives us a second formula for $[z],$ namely $[z] = \sum_{b \in ℬ} \sum_{a \in 𝒜} χ_{V} (a) ⟨ a^{*}, [b] ⟩ b^{*}$ Comparing this with our previous formula for $[z],$ and equating coefficients of $b^{*},$ we obtain $χ_{V ↑_{A}^{B}} (b) = \sum_{a \in 𝒜} χ_{V} (a) ⟨ a^{*}, [b] ⟩ .$

Now let’s consider the special case of group algebras, with $H \subseteq G$ for $H$ and $G$ finite groups. We know that $ℂ H \subseteq ℂ G$ is a subalgebra. Let $V$ be a representation of $H,$ with $V : ℂ H \to M_{d} (ℂ) : h \mapsto V (h) .$ Let $𝔳 = {v_{1}, \dots, v_{d}}$ be a basis for the $ℂ H -module$ $V,$ such that $h v_{k} = \sum_{j = 1}^{d} v_{j} V_{j k} (h) .$ Let us denote $V ↑_{ℂ H}^{ℂ G}$ by $V ↑_{H}^{G} .$ Note that $V ↑_{H}^{G} = ℂ G \otimes_{ℂ H} V = span {g \otimes v | g \in G, v \in V} = span {g \otimes v_{k} | g \in G, v_{k} \in 𝔳} .$ Let $𝔤 = {g_{0} = 1, g_{1}, \dots, g_{m}}$ be a set of (right) coset representatives for $H$ in $G,$ so that ${H, g_{1} H, \dots, g_{m} H} = G / H .$ Now if $g \in G,$ then there is some $g_{j} \in 𝔤$ and some $h \in H$ with $g = g_{j} h;$ one has that for $v_{k} \in 𝔳$ $g \otimes v_{k} = g_{j} h \otimes v_{k} = g_{j} \otimes h v_{k} = \sum_{l = 1}^{d} g_{j} \otimes v_{l} V_{l k} (h) .$ Thus the elements $g_{j} \otimes v_{l}$ form a basis for $V ↑_{H}^{G} .$

Note that $G$ acts on $G / H$ on the left, by permuting the cosets. For $g \in G$ and $g_{i} \in 𝔤,$ we can produce $g_{j} \in 𝔤$ and $h \in H$ with $g g_{i} = g_{j} h .$ Let $π (g)$ be the $m + 1$ by $m + 1$ permutation matrix determined by the action of $g$ on $G / H .$ $π : G \to M_{m + 1} (ℂ)$ is the permutation representation of $G .$ Now $\begin{matrix} g (g_{i} \otimes v_{k}) & = & (g g_{i}) \otimes v_{k} \\ = & (g_{j} h) \otimes v_{k} \\ = & g_{j} \otimes (h v_{k}) \\ = & \sum_{l = 1}^{d} (g_{j} \otimes v_{l}) V_{l k} (h) \\ g (g_{i} \otimes v_{k}) & = & \sum_{l = 1}^{d} (g_{j} \otimes v_{l}) V_{l k} (g_{j}^{- 1} g g_{i}) \end{matrix}$ Taking traces, we find that $\begin{matrix} χ_{V ↑_{H}^{G}} (g) & = & \sum_{g_{i} \in 𝔤} \sum_{v_{k} \in 𝔳} g (g_{i} \otimes v_{k}) |_{g_{i} \otimes v_{k}} \\ = & \sum_{\underset{g_{i}^{- 1} g g_{i} \in H}{g_{i} \in 𝔤}} \sum_{v_{k} \in 𝔳} V_{k k} (g_{i}^{- 1} g g_{i}) \\ = & \sum_{\underset{g_{i}^{- 1} g g_{i} \in H}{g_{i} \in 𝔤}} χ_{V} (g_{i}^{- 1} g g_{i}) \\ = & \sum_{h \in H} \frac{1}{H} \sum_{\underset{g_{i}^{- 1} g g_{i} \in H}{g_{i} \in 𝔤}} χ_{V} (h^{- 1} g_{i}^{- 1} g g_{i} h) \\ = & \sum_{\underset{w^{- 1} g w \in H}{g_{i} h = w \in G}} \frac{1}{H} χ_{V} (w g w^{- 1}) \\ = & \sum_{\underset{g' \in 𝒞_{g} \cap H}{g' \in G}} \frac{1}{H} χ_{V} (g') \end{matrix}$ (where $𝒞_{g}$ is the conjugacy class of $g).$ Our previous (general) formula for $χ_{V ↑_{A}^{B}}$ would have us believe that $χ_{V ↑_{H}^{G}} = \sum_{a \in 𝒜 = H} χ_{V} (a) ⟨ a^{*}, [b] ⟩ .$ But for group algebras $⟨ a^{*}, [b] ⟩ = {\begin{matrix} 1 & if a \in 𝒞_{b} \\ 0 & otherwise \end{matrix}$ so our two formulae are in agreement.

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

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