# Group Theory/Representations

Definition (representation):

Let ${\displaystyle G}$ be a group, ${\displaystyle {\mathcal {C}}}$ a category, ${\displaystyle A}$ an object of ${\displaystyle {\mathcal {C}}}$. Then a representation (also called action) of ${\displaystyle G}$ on ${\displaystyle A}$ by automorphisms of ${\displaystyle {\mathcal {C}}}$ is a group homomorphism

${\displaystyle G\to \operatorname {Aut} (A)}$.

Example (symmetric group acting on a product set):

Let ${\displaystyle S}$ be a set and let ${\displaystyle S^{n}}$ be the product of ${\displaystyle n}$ copies of ${\displaystyle S}$. The symmetric group acts on ${\displaystyle S^{n}}$ via

${\displaystyle \sigma (x_{1},\ldots ,x_{n}):=(x_{\sigma (1)},\ldots ,x_{\sigma (n)})}$.

Note that in this notation, we identified the element of ${\displaystyle G}$ with the automorphism of ${\displaystyle A}$ to which the element is sent via the homomorphism of the representation. This convention is followed throughout group theory and will be understood by every mathematician.

Definition (equivalence of representations):

Let ${\displaystyle G}$ be a group, ${\displaystyle {\mathcal {C}}}$ a category, ${\displaystyle A,B}$ objects of ${\displaystyle {\mathcal {C}}}$ and ${\displaystyle \rho :G\to A}$, ${\displaystyle \psi :G\to B}$ two representations of ${\displaystyle G}$ on ${\displaystyle A}$ resp. ${\displaystyle B}$. Then an equivalence of representations is an isomorphism ${\displaystyle f:A\to B}$ such that

${\displaystyle \forall g\in G:g\circ f=f\circ g}$.

Proposition (inverse of equivalence of representations is equivalence of representations):

Let ${\displaystyle G}$ be a group, ${\displaystyle {\mathcal {C}}}$ a category, ${\displaystyle A,B}$ objects of ${\displaystyle {\mathcal {C}}}$ and ${\displaystyle \rho :G\to A}$, ${\displaystyle \psi :G\to B}$ two representations of ${\displaystyle G}$ on ${\displaystyle A}$ resp. ${\displaystyle B}$. Let ${\displaystyle f:A\to B}$ be an equivalence of representations. Then ${\displaystyle f^{-1}:B\to A}$ is also an equivalence of representations.

Proof: We have

${\displaystyle g\circ f^{-1}=f^{-1}\circ g\Leftrightarrow f\circ g\circ f^{-1}=g\Leftrightarrow g=g}$,

since ${\displaystyle f}$ is an equivalence of representations. ${\displaystyle \Box }$