Representation of a group

Last updated January 27, 2022

Definition

Traditional definition

Definition. Let $G$ be a group and $\mathbb{F}$ a field. Let $V$ be a vector space over $\mathbb{F}$. Then a homomorphism $$ T:G\longrightarrow GL(V) $$ is called a representation of $G$. The dimension of $T$ is the dimension of the vector space $V$.

The image forms a vector space under composition.

Definition. A representation is said to be faithful if its kernel is trivial.

Morphisms

Definition. A morphism of representations $T:G\to GL(V)$ and $S:G\to GL(U)$ is a map $A\in [V,U]$ such that $$ AT(g)=S(g)A$$$$ for all $$\in $G$.

Definition. The representations $T:G\to GL(V)$ and $S:G\to GL(U)$ are isomorphic if there is a morphism between them that is bijective with respect to their vector spaces.

Irreducible representations

Definition. Assume $T:G\to GL(V)$ is a representation. A subspace $U\subset V$ is called invariant with respect to $T$ if $$ T(g)u\in U,\quad \forall g\in G,\ u\in U. $$ We can construct the representation $S\to GL(U)$ setting $S(g)$ to be the restriction of the automorphism $T(g)$ to the subspace $U$. $S$ is called a subrepresentation of the representation $T$.

Definition. Assume $T:G\to GL(V)$ is a representatioin. $T$ is called irreducible if it has no non-trivial invariant subspaces.

Definition. Assume $T:G\to GL(V)$ is a representation. $T$ is called completely reducible if for any invariant subspace $U$ there exists another invariant subspace $W$ such that $V=U\oplus W$.

It is an open question whether every representation is completely reducible, but the situation has been proven in the affirmative in the finite dimensional case.

Proposition. Let $A\in\text{Mor}(T,S)$ be a morphism between irreducible representations.

Categorical definition

Every group $G$ can be regarded as a category with a single object whose morphisms are precisely the elements of the group.

Definition. A linear representation of $G$ over $F$ is a functor $G\to\text{Vect}_F$. The functor category is called the category of linear representations of $G$ over $F$.

• At first it may not be clear how this lines up with the traditional definition, in which a representation is a map from $G$ to the automorphism group of $V$ (as opposed to just $V$). A functor between categories covariantly induces a group homomorphism between the automorphism groups (see Automorphisms of a category). So a functor contains exactly the information of the traditional definition.
• Morphisms of representations are natural transformations in the corresponding category of linear representations. By definition, if $S$ and $T$ are representations, then a morphism $\phi:S\to T$ is a commuting diagram of the form $$\begin{array} sG & \overset{S}{\rightarrow} & \text{Vect}_F \ \downarrow & & \downarrow \ G & \overset{T}{\rightarrow} & \text{Vect}_F \end{array}.$$ The map on the left is the identity. Also, every representation defined as a functor is in bijection with a group homomorphism from $G$ to the automorphism group of the vector space $V$ it is acting on, which are (invertible) maps $V\overset{g}{\to}V$. So diagrams of the form above are in bijection with commuting diagrams of the form $$\begin{array} sV & \overset{\phi}{\rightarrow} & W \ \llap{g}{\downarrow} & & \llap{g}{\downarrow} \ V & \overset{\phi}{\rightarrow} & W \end{array}$$ for all $g\in G$, which matches the traditional definition.

Definition. A faithful representation is a faithful functor in the category of representations.

• We should check that this matches with the notion of a faithful representation in the traditional definition. Let $T$ be a representation, in the sense that it is a functor. In particular $T$ can be regarded as a morphism of groups by our remarks above. Suppose the kernel of $T$ (regarded as a morphism of groups) is just the identity on $G$. Now suppose $T(g)=T(h)$. Then $T(g)T(g^{-1})=T(h)T(g^{-1})$ so $T(hg^{-1})=1$ so $h=g$. Conversely, suppose $T$ is a faithful functor. If $T(g)=1$, then by definition of a faithful functor $g$ must be the identity morphism on $G$, which is precisely the identity group element.

Irreducible representations

We claim that the traditional definition of irreducible representations coincides with that of simple objects in the category of representations.

One direction is clear. If a representation $T$ is reducible, then there is an epimorphism from it to one of its nontrivial subrepresentations. It remains to show that a simple object is an irreducible representation. Let $S$ be a simple object, say acting on a vector space $V$. Then its only quotient objects are itself and 0. This means that there are no nontrivial subspaces $U\subset V$ such that $S'=S\mid_U$ is a representation, for then $S\to S'$ would be an epimorphism. Hence $S$ is irreducible.

Dual

The dual object in the category is the internal hom from the object to the unit object. We claim that the traditional definition of the dual representation is equivalent to taking the dual object of a representation when viewed as an object in the category of representations.

By the definition of a dual object that we have just given, we need to check $[T,I]$ where $T$ is an arbitrary representation and $I$ is the unit (below we spell out that this is the trivial representation on the base field $K$). By the above discussion, the set of morphisms $[T,I]$ is in bijection with squares of the form $$ \begin{array} zV & \rightarrow & K \
\llap{{}^g}{\downarrow} & & \llap{{}^{1_K}}{\downarrow} \
V & \rightarrow & K \end{array} $$ for all $g\in G$. These are equivalent to $\text{Hom}(V\overset{g}{\to}V,K)$. By contravariance this is equivalently a map $\text{Hom}(V,K)\to\text{Hom}(V,K)$ defined by sending $f\mapsto g^{-1}f$. (The map defined by $g$ lives in the opposite category of $G$ hence we need to invert it).

So we can see that for $T^\ast(g)$ it must be the case that $f(gv)=f(g^{-1}(v))$. We can then write this as in the traditional definition: $T^\ast(g)=(T(g^{-1}))^\ast$.

Properties

• Monoidal category
  • The tensor product of linear representations makes the category of linear representations a monoidal category.
  • The unit is the trivial representation on the base field (see below)
  • Generally, this is true if the target category is monoidal, for example $\text{Vect}_F$ is monoidal.

Reducibility and Decomposability

A representation is decomposable if it can be written as the direct sum of nontrivial subrepresentations. This might seem like the same thing as a reducable representation, but it is not.

While every irreducible representation is indecomposable, not every indecomposable representation is irreducible. This is the intuition that a representation may only have a single non-trivial representation.

However, if $G$ is finite and $\text{char}(k)\nmid |G|$, then the two notions coincide. In particular, every indecomposable representation is irreducible. (Property 5)

If you notice, this is a condition under which Maschke’s Theorem is stated. It is because it is a direct corollary of (Property 5): just use induction. If the representation is infinite dimensional, you need the Axiom of Choice to say anything.

Operations

• Direct sum (of representations is a representation)
  • $(T\oplus S)(g)(v,u)\coloneqq (T(g)v,S(g)u),\quad\forall g\in G,\ v\in V,\ u\in U$
• Tensor product
  • If $S:G\to V$ and $T:G\to W$ are representations, then $S\otimes T$ is a representation defined by \begin{gather} T\otimes S: G\longrightarrow \text{GL}(V\otimes W) \ g(v\otimes w)=gv\otimes gw. \end{gather}
  • That this a representation follows from property 2 here.
  • The unit object is the trivial representation sending each $g\in G$ to the identity automorphism on the base field $K$. To see this, consider that in $T\otimes 1$, we have $g(v\otimes k)=gv\otimes gk=gv\otimes k$. Thus there is a natural isomorphism $T\otimes 1\to T$ given by the projection $p_V$ onto the $V$ factor: $$ \begin{array} zV\otimes K & \rightarrow & V \ \llap{{}^{g\otimes 1}}{\downarrow} & & \rlap{\ \ {}^{g}}{\downarrow} \ V\otimes K & \rightarrow & V \end{array}.$$ This is well defined since every element in $V\otimes W$ is a linear combinations of elementary tensors (these have the form $v\otimes w$) so the projection map will send an element to the sum of the $V$ factors in its expression as a linear combination of elementary tensors.
• Dual
  • Define $T^\ast:G\to\text{GL}(V^\ast)$ by $$ T^\ast(g)=(T(g^{-1}))^\ast. $$
  • $T^\ast$ is a representation over $V^\ast$.
  • $(T(g))^\ast = (T(-n))^t$
• Symmetric / antisymmetric ( definition)
  • $(S^2 T)(g)\coloneqq S^2(T(g))\coloneqq T(g)\mid_{S^2(V)}$
  • $(\Lambda^2 T)(g)\coloneqq\Lambda^2(T(g))\coloneqq T(g)\mid_{\Lambda^2(V)}$
• Equivalent representations: multiplying by an invertible matrix
  • Given a representation $\rho$ and an invertible matrix $T$, you can define a representation $\sigma(g) = T^{-1}\rho(g)T$.
  • If the degree of a representation $\rho$ is 1, or the representation is trivial, then the only equivalent representation to $\rho$ is $\rho$ itself.

Properties

• Given a morphism between representations, the kernel and image are invariant subspaces (wrt to the source and target representations)
• A morphism between irreducible representations is either trivial or an isomorphism.
• The image of a faithful representation is isomorphic to the group
• Schur’s Lemma All endomorphisms of irreducible complex representations are one-dimensional

5. If $G$ is finite and $\text{char}(k)\nmid |G|$, then every indecomposable representation is irreducible.
   1. If $U$ is a nontrivial invariant subrepresentation of $T$, then there exists a complementary invariant subresentation $V$ such that $T=U\oplus V$.

• Maschke’s Theorem The representation of a finite group whose order is not divisible by the characteristic of the field is completely reducible.