13 250520

Degree of Irreducible Character Divides $|G|$

Proposition

Let $G$ be a finite group, and let $\chi$ be a character of $G$. Then $\chi (g)$ is an algebraic integer. In particular, if $\chi (g)\in \mathbb{Q}$, then $\chi (g)\in \mathbb{Z}$.

\begin{proof} See ^e7uw1z. \end{proof}

Theorem

The degree of a simple representation of $G$ over $\mathbb{C}$ divides $|G|$.

\begin{proof} See The degree of every irreducible character divides G. \end{proof}

Induced Representation

然后开始讲 8 Induced Representation，没有听。

Lemma

Let $A\to B$ be a ring homomorphism such that $1_A\mapsto 1_B$. Let $V$ be an $A$-module and let $W$ be a $B$-module, then

• ${\mathrm{Hom}}_B(B{\otimes }_{A}V,W)\simeq {\mathrm{Hom}}_A(V,W{\downarrow }_A)$
• ${\mathrm{Hom}}_A(W{\downarrow }_A,V)\simeq {\mathrm{Hom}}_B(W,{\mathrm{Hom}}_A(B,V))$
• if $B\to C$ is another ring homomorphism, then $C{\otimes }_B(B{\otimes }_{A}V)\simeq C{\otimes }_{A}V$.

\begin{proof} Note that

$${\mathrm{Hom}}_B({}_B{B}_A,W)\simeq W{\downarrow }_A\simeq {}_A{B}_B\otimes W$$

as $A$-modules, then by tensor-hom adjunction, we proved i) and ii).

iii) is proved by ^489zft. \end{proof}

Corollary

Let $H⩽K⩽G$. Let $V$ be an $RH$-module and let $W$ be an $RG$-module, where $R$ is a commutative ring. Then

• ${\mathrm{Hom}}_{RG}({\mathrm{Ind}}_H^{G}V,W)\simeq {\mathrm{Hom}}_{RH}(V,{\mathrm{Res}}_H^G(W))$, and ${\mathrm{Hom}}_{RG}(W,{\mathrm{Ind}}_H^{G}V)\simeq {\mathrm{Hom}}_{RH}({\mathrm{Res}}_H^{G}W,V)$;
• ${\mathrm{Ind}}_K^G({\mathrm{Ind}}_H^{K}V)\simeq {\mathrm{Ind}}_H^{G}V$ as $RG$-modules;
• ${\mathrm{Res}}_H^K({\mathrm{Res}}_H^{K}W)\simeq {\mathrm{Res}}_H^G(W)$ as $RH$-modules;
• ${\mathrm{Ind}}_H^G(V){\otimes }_{R}W\simeq {\mathrm{Ind}}_H^G(V{\otimes }_R{\mathrm{Res}}_H^G(W))$ as $RG$-modules. In particular, ${\mathrm{Ind}}_H^G(R){\otimes }_{R}W\simeq {\mathrm{Ind}}_H^G({\mathrm{Res}}_H^{G}W)$;
• ${\mathrm{Ind}}_H^G(V)\simeq {\mathrm{Hom}}_{RH}(RG,V)$ as $RG$-modules.

\begin{proof} Note that if v) holds, then by ^752a42 we can prove i)-ii). Also iii) is trivial. So it remains to show iv) and v).

iv) Define

$$\phi :(RG{\otimes }_{RH}V){\otimes }_{R}W\to RG{\otimes }_{RH}(V{\otimes }_{R}W),g\otimes v\otimes w\mapsto g\otimes v\otimes g^{-1}w,$$

and ${\phi }^{-1}(g\otimes v\otimes w)=g\otimes v\otimes gw$.

v) Define

$$RG{\otimes }_{RH}V\to {\mathrm{Hom}}_{RH}(RG,V),g\otimes v\mapsto {\varphi }_{g,v},$$

where ${\varphi }_{g,v}(x)=(xg)v$ if $xg\in H$ or ${\varphi }_{g,v}(x)=0$ if $xg\notin H$. This map is an isomorphism, as its inverse is $\theta \mapsto {\sum }_{g\in [G/H]}g\otimes \theta (g^{-1})$.

Now we finish the proof. \end{proof}

Remark. I have not verified the proof for (v) myself. However, Gemini suggested that the map ${\varphi }_{g,v}$ as defined might not be $RH$-linear.

Corollary

${\mathrm{Ind}}_H^G(\chi )\cdot \psi ={\mathrm{Ind}}_H^G(\chi \cdot {\mathrm{Res}}_H^G(\psi ))$, where $\psi$ is a character of $G$ and $\chi$ is a character of $H$.

Mackey Decomposition Formula

Let $H,K⩽G$ be subgroups, and let $V$ be a $kK$-module. Consider ${\mathrm{Res}}_H^G({\mathrm{Ind}}_K^G(V))$.

Definition

If $\Omega$ is a left $G$-set, write $G\backslash \Omega$ the set of $G$-orbits on $\Omega$ and $[G\backslash \Omega ]$ the representatives. Similarly for right $G$-set $\Omega$, write $\Omega /G$ and $[\Omega /G]$.

Recall Double Cosets, and we have lemmas as follows.

Lemma

The set of $(H,K)$-double cosets is in bijection with the orbits $H\backslash (G/K)$, and with the orbits $(H\backslash G)/K$, under the mapping

$$\begin{array}{rl}HgK& \mapsto H(gK)\in H\backslash (G/K)\\ HgK& \mapsto (Hg)K\in (H\backslash G)/K.\end{array}$$

The lemma below is same as ^vy5iem.

Lemma

$H,K⩽G$, $g\in G$. Then $HgK/K\simeq H/H\cap {}^{g}K$ as left $H$-sets, and $H\backslash HgK\simeq (H^g\cap K)\backslash K$ as right $K$-sets. So $[G:K]={\sum }_{g\in [H\backslash G/K]}[H:H\cap {}^{g}K]$, $[G:H]={\sum }_{g\in [H\backslash G/K]}[K:H^g\cap K]$.

\begin{proof} $HgK$ is single $H$-orbits on left $K$-cosets. ${\mathrm{Stab}}_H(gK)=\{h\in H:hgK=gL\}=\{h\in H:g^{-1}hg\in K\}=H\cap {}^{g}K$. \end{proof}

Define $H^g=g^{-1}Hg$ and ${}^{g}H=gH{g}^{-1}$. Let $H⩽G$ be a subgroup and let $V$ be a $kH$-module. Define ${}^{g}V$ be a $k{}^{g}H$-module as follows: set ${}^{g}V=V$, and for $v\in V,h\in H$, define ${}^{g}hv=hv$. If $\phi :H\to \mathrm{GL}(V)$ is a representation, then the representation of ${}^{g}V$ is given by

$${}^{g}H\to H\stackrel{\phi }{\to }\mathrm{GL}(V),{}^{g}h\mapsto h\mapsto \phi (h).$$

Recall that ${\mathrm{Ind}}_H^G(V)=kG{\otimes }_{kH}V={\oplus }_{g\in [G/H]}g\otimes V$, where $g\otimes V$ is a $k{}^{g}H$-module by

$${}^{g}h(g\otimes V)=gh{g}^{-1}g\otimes V=g\otimes hv.$$

When $g\otimes V$ is identified with $V$ via $g\otimes v\leftrightarrow v$, the action of ${}^{g}H$ on $V$ coincide with the action of ${}^{g}H$ on ${}^{g}V$.

Mackey decomposition formula

Let $H,K⩽G$ be subgroups, and let $V$ be a $kK$-module, then

$${\mathrm{Res}}_H^G({\mathrm{Ind}}_K^G(V))\simeq {\oplus }_{g\in [H\backslash G/K]}{\mathrm{Ind}}_{H\cap {}^{g}K}^H({}^g{\mathrm{Res}}_{H^g\cap K}^{K}V).$$

\begin{proof} Note that ${\mathrm{Ind}}_K^{G}V={\oplus }_{x\in [G/K]}x\otimes V$. For a double coset $HgK$, we know $M_g:={\oplus }_{x\in [G/K],x\in HgK}x\otimes V$ is an $H$-invariant subspace, and

$${\mathrm{Ind}}_K^G(V)={\oplus }_{g\in [H\backslash G/K]}({\oplus }_{x\in [G/K],x\in HgK}x\otimes V)={\oplus }_{g\in [H\backslash G/K]}{M}_g.$$

On the other hand, note that

$${\mathrm{Stab}}_H(g\otimes V)=\{h\in H:hg\otimes v=g\otimes v\}=\{h\in H:h^g(1\otimes v)=1\otimes v\}=H\cap {}^{g}K:=S_g.$$

Define $S_g=H\cap {}^{g}K$, and consider a $k{S}_g$-module $g\otimes V$. We claim that $g\otimes V\simeq {}^g{\mathrm{Res}}_{H^g\cap K}^{K}V$ as $k{S}_g$-modules. On the one hand, for any $s\in S_g$, $s(g\otimes v)=sg\otimes v$. Since $s\in H\cap {}^{g}K$, there exists $k_s\in K$ such that $s^g=k_s$ and so $sg\otimes v=g\otimes k_{s}v=g\otimes (g^{-1}sgv)$. On the other hand, remark that for any $K_0$-module $M$, ${}^{g}M$ is defined as a ${}^g{K}_0$-module (where $k^{\prime }=gk{g}^{-1}\in {}^g{K}_0$ acts on $v$ by $kv$ and ${}^{g}M=M$ as sets). Thus, in $k{S}_g$-module ${}^g{\mathrm{Res}}_{H^g\cap K}^{K}V$, for any $v\in V$ there is $s\cdot v=(g^{-1}sg)v$. Therefore, we know $g\otimes V\simeq {}^g{\mathrm{Res}}_{H^g\cap K}^{K}V$ as $k{S}_g$-modules.

It remains to show there is a isomorphism between $H$-modules $M_g$ and ${\mathrm{Ind}}_{S_g}^H(g\otimes V)$ for each $g\in [H\backslash G/K]$. Define

$${\mathrm{Ind}}_{S_g}^H(g\otimes V)\to M_g,h_i\otimes (g\otimes v_j)\mapsto h_{i}g\otimes v_j,$$

which is an isomorphism between $H$-modules. Now we finish the proof. \end{proof}