Group Representation

When we have a group $G$, and want to know how the elements in the group act.

In my opinion, key things are knowing how the map is act on vector space or modules.

Linear Represenation of $G$ in $V$

Linear Representation of $G$ in $V$ is a group homomorphism $\rho :G\to Au{t}_k(V)$.

In this case, $kG$-module $V$ is given. Then we can define the homomorphism ${\rho }_V(g)v:=g\cdot v$.

When $\rho$ is given, $RG$-module $V$ as a Representation space of $G$ is can be defined as

$$\left(\sum _i{r}_i{g}_i\right)\cdot v:=\sum _i{r}_i\rho (g_i)v.$$

Definitions

Irreducible Representation

Group representation $\rho :G\to G{L}_k(V)$ is irreducible if $\forall W⊲V$, $\forall g\in G$, $\rho (g)W\subset W$ implies that $W=0$ or $W=V$.

So it has a correspondence with Simple Module.

Subrepresentation

Let representation $\rho :G\to G{L}_k(V)$, if $\forall W⊲V$, $\forall g\in G$, $\rho (g)W\subset W$, when we restrict $V$ to $W$, such that ${\rho }^W:G\to G{L}_k(W)$ is a subrepresentation.

Isomorphism and Intertwine Operator

Let $\rho :G\to G{L}_k(V)$ and ${\rho }^{\prime }:G\to G{L}_k(V^{\prime })$. These two represenations are isomorphic(or similar) if there exists a linear isomorphism $\tau :V\to V^{\prime }$ such that

$$\tau \circ \rho (g)={\rho }^{\prime }(g)\circ \tau \text{ for all }g\in G.$$

We call $\tau$ as a intertwine operator.

Corollary

We can write, ${\rho }^{\prime }(g)=\tau \circ \rho (g)\circ {\tau }^{-1}$. So, characters are same because

$${\chi }^{\prime }(g)=Tr({\rho }^{\prime }(g))=Tr(\tau \circ \rho (g)\circ {\tau }^{-1})=Tr(\rho (g))=\chi (g).$$

Direct Sum

Let $\rho :G\to G{L}_k(V)$ and ${\rho }^{\prime }:G\to G{L}_k(W)$. The direct sum $\rho \oplus {\rho }^{\prime }$ to be the representation on the direct sum $V\oplus W$ such that

$${\rho }_{V\oplus W}:G\to G{L}_k(V\oplus W)$$ $${\rho }_{V\oplus W}=\left(\begin{array}{cc}{\rho }_V& 0\\ 0& {\rho }_W\end{array}\right).$$

Tensor Product

Let ${\rho }_V:G\to G{L}_k(V)$ and ${\rho }_W:G\to G{L}_k(W)$. The tensor product of the given representations is

$${\rho }_{V\otimes W}:G\to G{L}_k(V\otimes W)$$

such that $g\mapsto {\rho }_V(g)\otimes {\rho }_W(g)$.

Two Important Examples

Trivial Representation

Let $\rho :G\to G{L}_1(V)$ as $g\mapsto 1$ for all $g\in G$.

So in character language, ${\chi }_{\rho }\equiv 1$ for $\forall g\in G$.

Regular Representation

Let $G$ be a finite group order $n$ and $k$ be a field. With the group ring $kG$ and as a module. Regular Representation is the map $\rho :G\to G{L}_n(kG)$ which is injective.

In Serre’s book,

Let $V$ be a vector space of dimension $n$ with a basis $(e_t{)}_{t\in G}$ indexed by the elements $t$ of $G$. For $s\in G$, let ${\rho }_s$ be the linear map of $V$ into $V$ which sends $e_t$ to $e_{st}$. Note that the images under ${\rho }_s$‘s of $e_1$ form a basis of $V$.

Dual Representation

If $\rho :G\to Au{t}_k(E)$ is a representation of $G$ on $E$, then dual representation is ${\rho }^{\ast }:G\to Au{t}_k(E^{\ast })$, where $E^{\ast }=Ho{m}_k(E,k)$.[^1]

The group action $g\cdot V^{\ast }$ is $gf(g^{-1}v)$ for $f\in V^{\ast },v\in V$. SInce our case is $W=k$, so $g\cdot f(v)=f(g^{-1}\cdot v)$.

Also note that $f\in V^{\ast }$, so $f(v)=f^{T}v$ as a matrix multiplication. So $$ g \cdot f (v) = f^{T}A^{-1}(v) = (\rho(g^{-1})^{T}f)^{T}v= \rho(g^{-1})^{T}f(v).

Therefore, $\rho^{\ast} (g) = \rho(g^{-1})^{T}$. [^1]: [[Hom#dual-representation|Hom]]