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Definition

Two matrix representations $(r,\rho )$ and $(s,\sigma )$ are isomorphic iff $r=s$ and there is an element $f\in {\mathrm{GL}}_r(k)$ such that $\sigma (g)=f\rho (g)f^{-1}$ for all $g\in G$.

Extension of scalars

Definition

Assume $k$ is a subfield of a field $k^{\prime }$. For a $k$-representation $(M,\rho )$ of G, its extension of scalars is a $k^{\prime }$-representation $(k^{\prime }{\otimes }_{k}M,k^{\prime }{\otimes }_k\rho )$ where $k^{\prime }{\otimes }_k\rho$ is the morphism

$$G\to \mathrm{GL}(k^{\prime }{\otimes }_{k}M),g\mapsto {\mathrm{id}}_{k^{\prime }}\otimes \rho (g).$$

Remark. A basis of $M$ over $k$ is identified with a basis of $k^{\prime }{\otimes }_{k}M$ over $k^{\prime }$. From the point of a matrix representation $\rho :G\to {\mathrm{GL}}_r(k)$, extending the scalars consists in viewing the matrix entries of $\rho (g)$ as elements of $k^{\prime }$.

Definition

A $k^{\prime }$-representation $(M^{\prime },{\rho }^{\prime })$ of $G$ is said to be rational over $k$, if there is $k$-representation $(M,\rho )$ such that $(M^{\prime },{\rho }^{\prime })\simeq (k^{\prime }{\otimes }_{k}M,k^{\prime }{\otimes }_k\rho )$.

A matrix representation $(r,\rho )$ of $G$ over $k^{\prime }$ is rational over $k$ if there exists $U\in {\mathrm{GL}}_r(k^{\prime })$ such that for all $g\in G$, the entries of $U{\rho }^{\prime }(g)U^{-1}$ lie in $k$. Similarly, see Frobenius-Schur Indicator.

For $k$-representations $(M,\rho )$ and $(N,\sigma )$, define

• $(M,\rho )\oplus (N,\sigma )=(M\oplus N,\rho \oplus \sigma )$, and
• $(M,\rho )\otimes (N,\sigma )=(M\otimes N,\rho \otimes \sigma )$.

Definition

We say $(M,\rho )$ is indecomposable if $M\ne 0$ and $(M,\rho )\simeq (M_1,{\rho }_1)\oplus (M_2,{\rho }_2)$ implies either $M_1=0$ or $M_2=0$.

Definition

If $M_1$ is a subspace of $M$, which is stable under the actions of all $\rho (g)$ with $g\in G$, we get a subrepresentation $(M_1,{\rho }_1)$ of $(M,\rho )$ where ${\rho }_1(g)$ is a restriction of $\rho (g)$ to $M_1$.

Definition

A representation $(M,\rho )$ is called simple (or irreducible) if $M\ne 0$ and the only $G$-stable subspaces of $M$ are $M$ and $0$.

Krull-Schmidt theorem

Let $(M,\rho )$ be a representation of finite degree.

• It is a direct sum of indecomposable representation.
• If $G$ is finite, and if $(M,\rho )\simeq {\oplus }_{i\in I}(M_i,{\rho }_i)\simeq {\oplus }_{j\in J}(M_j^{\prime },{\rho }_j^{\prime })$ where $(M_i,{\rho }_i)$ and $(M_j^{\prime },{\rho }_j^{\prime })$ are indecomposable, there is a bijection $\alpha :I\stackrel{\sim }{\to }J$ such that for all $i\in I$. $(M_i,{\rho }_i)\simeq (M_{\alpha (i)}^{\prime },{\rho }_{\alpha (i)}^{\prime })$.

Let $(\Omega ,\rho )$ be a set representation of $G$. (Remark that “Set” contains all sets and “sets” means it is a finite set.)

Define by $k\Omega$ be a $k$-vector space with basis $\Omega$. Then $\rho$ induces a group morphism $k\rho :G\to \mathrm{GL}(k\Omega )$, which is a ${\text{vect}}_k$-representation of $G$. It gives a function $\text{Rep}(G,\text{set})\to \text{Rep}(G,{\text{vect}}_k)$. This construction transforms:

• disjoint union of set-representations into direct sum of ${\text{vect}}_k$-representation
• product of set-representations into tensor product of ${\text{vect}}_k$-representation

Remark. For a transitive/irreducible $(\Omega ,\rho )\in \text{Rep}(G,\text{set})$ with $|\Omega |⩾2$, we have $(k\Omega ,k\rho )\in \text{Rep}(G,{\text{vect}}_k)$. Note that $S\Omega ={\sum }_{\omega \in \Omega }\in k\Omega$ is invariant under all $\rho (g)$, so $(k\Omega ,k\rho )$ is not irreducible.

Definition

The group $G$ acts on itself by left multiplication. The corresponding $k$-linear representation is called the regular representation of $G$.

Lemma

Assume that $k$ has characteristic $p$ dividing the order of $G$, then the line $k(SG)=\mathrm{span}\{{\sum }_{g\in G}g\}$ has no $G$-stable complement in the regular representation of $G$.

\begin{proof} Let $kG=k(SG)\oplus V^{\prime }$ as $k$-vector spaces. Then we have a projection $\pi :kG\to k(SG)$. In particular, $\pi (k(SG))=k(SG)$. If $V^{\prime }$ is $G$-stable, then $\pi (gx)=g\pi (x)$ for all $x\in kG$ and $g\in G$ and so $\pi (g\cdot 1)=g\pi (1)$, which implies that $\pi (SG)=(SG)\cdot \pi (1)$ with $\pi (1)=c\cdot SG$ for some $c\in k$. It deduces that $\pi (SG)=(c|G|)SG=0$, leading to a contradiction. \end{proof}

For any $\alpha \in {\mathrm{Hom}}_R(M,N)$, define $g\cdot \alpha =\rho (g)\alpha \rho (g{)}^{-1}$. Thus ${\mathrm{Hom}}_{\alpha }(M,N)$ becomes a representation of $G$. Note that $\alpha \in {\mathrm{Hom}}_R(M,N)$ is invariant under all $g\in G$ iff $\alpha$ is a morphism $(M,\rho )\to (N,\sigma )$. In particular, there is a representation associated with $M^{\ast }={\mathrm{Hom}}_k(M,k)$. For any $\alpha \in {\mathrm{Hom}}_k(M,k)$, $g\cdot \alpha =\alpha \cdot \rho (g{)}^{-1}$, which is called the contragredient representation of $(M,\rho )$.

For a $1$-dimensional $G\to {\mathrm{GL}}_1(k)=k^{\ast }$, its kernel contains $[G,G]$. It deduces that $\mathrm{Hom}(G,k^{\ast })\simeq \mathrm{Hom}(G/[G,G],k^{\ast })$.

Lemma

Let $(S,\rho )$ be an irreducible representation of $G$ and let $(k,\sigma )$ be a degree $1$ representation. Then the tensor representation $(S,{\sigma }_{\rho })$ is still irreducible.

the Group Algebra

Recall the definition of $kG$, which is a $k$-vector space with basis $G$. Let $\tau :kG\to k$ be the linear form of $kG$ defined by $\tau (g)=0$ if $g\ne 1$ and $\tau (1)=1$.

Lemma

• $\tau$ is a class function. For all $x,y\in kG$, $\tau (xy)=\tau (yx)$.
• The map $\hat{\tau }:kG\to (kG{)}^{\ast }={\mathrm{Hom}}_k(kG,k),x\mapsto (y\mapsto \tau (xy))$ is a $k$-linear morphism.

\begin{proof}

• Exercise \end{proof}

There is a $1$-$1$ corresponding

$$\{\text{functions }G\to k\}⟷{\mathrm{Hom}}_k(kG,k),$$

and denote by $F(G,k)$ the $k$-vector space of functions $G\to k$.

For $f\in F(G,k)$, the inverse image of $f$ under the isomorphism $\hat{\tau }$ is $f^0={\sum }_{g\in G}f(g^{-1})g$.

Let $f\in F$ be a class function, that is, $f(gh)=f(hg)$ for all $g,h\in G$. Identify $f$ with its correspondent in ${\mathrm{Hom}}_k(kG,k)$. Then $f(xy)=f(yx)$ for all $x,y\in kG$.

Denote by $CF(G,k)$ the space of class function on $G$ (or $kG$).

Remark. Let $a\in kG,f\in F(G,k)$. Define by $a\cdot f$ the element of $F(G,k)$ defined by $(a\cdot f)(x)=f(xa)$, for all $x\in kG$. Then $(a\cdot f{)}^0=a\cdot f^0$.