Induced Representation #1

 

1. Abelian Subgroups

Definition. $G$ is abelian $($or commutative $)$ if $st = ts$ for all $s, t \in G$.

Abelian group implies that each conjugacy class of $G$ consists of a single element, also that each function on $G$ is a class function.

 

Theorem 9. 

$G$ is abelian $\iff$ All the irreducible representation of $G$ have degree $1$.

Let $ (n_{1}, \ldots, n_{h}) $ be the degrees of the distinct irreducible representations of $G$.
$h$ = #$($irreducible representations of $G$$)$ = #$($class of $G$$)$.
By Corollary 2 of Proposition 5, $g = n_{1}^{2} + \cdots + n_{h}^{2} \Rightarrow g = h \iff n_{i} = 1$ for all $i$. $_\Box$ 

 

Corollary. Let $A$ be an abelian subgroup of $G$, let $a=|A|, g=|G|$. Each irreducible representation of $G$ has degree $\le g/a$.

Let $\rho: G \rightarrow GL(V)$ be an irreducible representation of $G$. Through restriction to the subgroup $A$, $\rho_{A}:A\rightarrow GL(V)$ is a representation of $A$.
Let suppose $W \subset V$ be an irreducible subrepresentation of $\rho_{A}$, by Theorem 9, dim$ (W) = 1$.Let $V'$ be the vector subspace of $V$ as follows.$$ V' = \underset{s \in G}{\bigcup} \rho_{s} W \subset V$$$V'$ easily verify $G$-stable and $\rho$ is an irreducible representation $\Rightarrow$ $V'=V$.For $^\forall s \in G, ^\forall t \in A$,$$ \rho_{st} W = \rho_{s} \rho_{t} W = \rho_{s} W.$$$ \Rightarrow$ the number of distinct $\rho_{s}W$ is at most equal to $g/a$. $\Rightarrow$ dim$(V) \le g/a.\ \Box$

 

2. Product of Two Groups

Definition. Let $G_{1}$ and $G_{2}$ be two groups, and let $G_{1} \times G_{2}$ be their product given follows.

$$ (s_{1},\ t_{1}) \cdot (s_{2},\ t_{2}) = (s_{1}s_{2},\ t_{1}t_{2})\quad for\ every\ s_{i} \in G_{1}, t_{i} \in G_{2}. $$

Conversely, let $G$ be a group containing $G_{1}$ and $G_{2}$ as subgroups $s.t.$

$($i$)$ $^\forall s \in G$ can be written uniquely in the form $s = s_{1}t_{1}$ with $s_{1} \in G_{1}, s_{2} \in G_{2}.$

$($ii$)$ $^\forall s_{1} \in G_{1}, ^\forall t_{1} \in G_{2} \Rightarrow s_{1}t_{1}=t_{1}s_{1}.$

$ st = (s_{1}t_{1}) =(s_{1}s_{2})(t_{1}t_{2}) \Rightarrow G \cong G_{1} \times G_{2}. $

 

Let let $\rho^{1}:G\rightarrow GL(V_{1})$ and $\rho^{2}:G\rightarrow GL(V_{2})$ be linear representations of $G_{1}$ and $G_{2}$ respectively. We define a linear representation $\rho^{1} \otimes \rho^{2}:\ G_{1}\times G_{2} \rightarrow V_{1} \otimes V_{2}$ as 

$$ (\rho^{1} \otimes \rho^{2} ) (s_{1}, s_{2}) = \rho^{1}(s_{1}) \otimes \rho^{2}(s_{2}).$$

This representation is called the tensor product of the representations $\rho^{1}$ and $\rho^{2}$.

 

Theorem 10.

$($i$)$ $\rho^{1}$ and $\rho^{2}$ are irreducible $\Rightarrow$ $\rho^{1} \otimes \rho^{2}$ is an irreducible representation of $G_{1} \times G_{2}.$

$($ii$)$ Each irreducible representation of $G_{1} \times G_{2}$ is isomorphic to a representation $\rho^{1} \otimes \rho^{2}$, where $\rho^{i}$ is an irreducible representation of $G_{i}.$

$($i$)$ $\rho^{1}$ and $\rho^{2}$ are irreducible $\Rightarrow$ By Theorem 5, 
$$ {1 \over g_{1} } \underset{s_{1}}{\sum} | \chi_{1} (s_{1})|^{2} = 1, {1 \over g_{2} } \underset{s_{2}}{\sum} | \chi_{2} (s_{2})|^{2} = 1 $$
$ \Rightarrow {1 \over g } \underset{s_{1}, s_{2}}{\sum} | \chi (s_{1}, s_{2})|^{2} = {1 \over g_{1} } \underset{s_{1}}{\sum} | \chi_{1} (s_{1})|^{2} \times {1 \over g_{2} } \underset{s_{2}}{\sum} | \chi_{2} (s_{2})|^{2} = 1 \times 1 = 1.$
$\Rightarrow \rho^{1} \otimes \rho^{2}$ is irreducible from Theorem 5.
$($ii$)$ It suffices to show that each class function $f$ on $G_{1} \times G_{2}$, which is orthogonal to the character of the form $\chi(s_{1}, s_{2})$, is zero. Because it implies that any class function satisfies $\left(f|\chi \right) \neq 0.$
Let suppose:
$$ \underset{s_{1}, s_{2}}{\sum} f(s_{1}, s_{2}) \chi( s_{1}, s_{2} )^{*} = 0. $$
Let $g(s_{1}) = \sum_{s_{2}} f(s_{1}, s_{2}) \chi_{2}(s_{2})^{*},$ then 
$$ \underset{s_{1}}{\sum} q(s_{1}) \chi_{1}(s_{1})^{*} = 0\quad for\ all\ \chi_{1}. $$
By $q$ is a class function of $G_{1}$, and same argument is true for each $\chi_{2}.$
$\therefore f(s_{1}, s_{2}) = 0.\ _\Box$

 

As Theorem 10, we conclude that any representation of $G_{1} \times G_{2}$ completely reduces into representations of $G_{1}$ and $G_{2}.$