Representation Theory, Part I: Basics

October 8th, 2017

Definition: A representation of a finite group $G$ on a finite dimensional vector space $V$ (WLOG the vector space is assumed to be over the complex numbers).

This map gives $V$ the structure of a module over $G$, because for $g\in G$, we have:

$$ g(v+w) = gv+gw \\ g(hv) = (gh)v $$

Sometimes, $V$ is itself called the representation of the group; thus, we identify a representation of a group as a vector space on which $G$ acts linearly.

Definition: A map $\varphi$ between two representations $V,W$ of $G$ (also called a $G$-linear map) is a vector space map $\varphi: V \rightarrow W$ such that for any $g \in G$ and $v \in V$:

$$ g\varphi(v) = \varphi(gv) $$

Definition: A subrepresentation of a representation $V$ is a vector subpace of $W$ of $V$ which is invariant under $G$.

Definition: A representation $V$ is called irreducible if there is no proper nonzero invariant subspace $W$ of $V$.

Given two represetnations, the direct sum $V\oplus W$ and the tensor product $V \otimes W$ are also representations. The latter is given by:

$$ g(v\otimes w) = gv \otimes gw $$

Similarly, the $n$th tensor power can be constructed from a representation, and similarly the exterior powers and symmetric powers as subrepresentations.

Duals and Tensor Products of Representations; Representation of $Hom(V,W)$

The dual $V^$ of a vector space is a representation as well. We wish to respect the natural pairing between $V^$ and $V$, given by:

$$ \langle v^*, v\rangle = v^*(v) $$

So we need to define the dual representation such that:

$$ \langle \rho^*(g)v^*, \rho(g)v\rangle = \langle v^*, v\rangle $$

And this forces us to define the representation as follows. Note that by the definition of the transpose:

$$ \rho(g^{-1})^Tv^*(gv) = v^*(g^{-1}gv) = v^*(v) $$

So we define:

$$ \rho^*(g) = \rho(g^{-1})^T: V^* \rightarrow V^* $$

Now that we have defined the dual and the tensor product of representations, we can show that $Hom(V,W)$ is a representation. Note that there is a natural identification:

$$ V^{*} \otimes W \rightarrow Hom(V,W) \\ a^*\otimes b \rightarrow (v \mapsto a^*(v)b) $$

It is not hard to show that this identification is surjective and injective, and hence an isomorphism of vector spaces. Now, we take an arbitrary element $a^* \otimes b \in V^* \otimes W$. We identify this element naturally with $\varphi \in Hom(V,W)$:

$$ \varphi: v \mapsto a^*(v)b $$

Now we consider $g\varphi = g(a^* \otimes b)$. We have:

$$ g(a^*\otimes b) = ga^* \otimes gb = (g^{-1})^Ta^* \otimes gb $$

Where we have used the definition of the dual representation. By the natural identification again we have:

$$ g\varphi: v \mapsto (g^{-1})^Ta^*(v) gb = ga^*(g^{-1}v) b $$

But this is simply telling us that:

$$ (g\varphi)(v) = g\varphi(g^{-1}v) $$

And this gives us the representation of the space $Hom(V,W)$.

Proposition 1

The vector space of $G$-linear maps between two representations $V,W$ of $G$ is the subspace of $Hom(V,W)$ which is fixed by $G$, often denoted $Hom_G(V,W)$

Note that if we have a $G$-linear map $\varphi$, then by definition:

$$ g\varphi(v) = \varphi(gv) $$

Note that the representation of $Hom(V,W)$ however is given by:

$$ (g\varphi)(v) = g\varphi(g^{-1}v) = \varphi(gg^{-1}v) = \varphi(v) $$

So indeed $\varphi$ is fixed under the action of $G$. The converse holds evidently as well; if $\varphi$ is fixed by $G$, then it follows that $\varphi$ is $G$-linear.

Finally, if $X$ is any finite set and $G$ acts on $X$, then $G$ naturally is embedded into the permutation group $Aut(X)$ of $X$. So we can construct a vector space with basis $\{e_x\ :\ x\in X\}$ and the action of $G$ is then given by:

$$ g \sum a_x e_x = \sum a_x e_{gx} $$

Definition: The regular representation $R_G$ or $R$ corresponds to the action of $G$ on itself. We could alternatively define it as the space of complex-valued functions on $G$ where:

$$ (g\alpha)(h) = \alpha(g^{-1}h) $$

To prove that these are equivalent, we identify $e_x$ with the function $f_x$ which takes the value $1$ on $x$ and $0$ elsewhere. Then we have:

$$ (gf_x)(h) = f_x(g^{-1}h) $$

And evidently this function takes value $1$ where $g^{-1}h = x$ or equivalently $h = gx$. Thus we can write:

$$ (gf_x) = f_{gx} $$

Complete Reducibility; Schur’s Lemma

Proposition 2 (Maschke’s Theorem)

If $W$ is a subrepresentation of a representation $V$ of a finite group $G$, then there is a complementary invariant subspace $W’$ of $V$, so that $V= W\oplus W’$

We define the complement as follows. Chose an arbitary subspace $U$ which is complementary to $W$. Then we can write:

$$ V \cong W \oplus U $$

So for any $v \in V$, we can identify it with some pair $(w,u)$. Define the natural projection map $\pi_0: V \mapsto W$ as:

$$ \pi_0(w,u) = w $$

This map is $G$-linear. Then, we define a new map $\pi$:

$$ \pi(v) = \sum\limits_{g\in G} g\pi_0(g^{-1}v) $$

Since $\pi_0$ is $G$-linear, it follows that this map is $G$ linear. In fact on $W$, we have:

$$ \pi(w) = \sum\limits_{g\in G} g\pi_0(g^{-1}w) = \sum\limits_{g\in G} gg^{-1}\pi_0(w) = |G|w $$

So this map is nothing more than multiplication by $|G|$ on $W$. Therefore, its kernel is a subspace of $V$ which is invariant under $G$ and is complementary to $W$.

Corollary

Any representation is a direct sum of irreducible representations.

Now we move on to Schur’s Lemma, one of the more useful theorems in basic representation theory.

Proposition 3 (Schur’s Lemma)

If $V, W$ are irreducible representations of $G$ and $\varphi: V \rightarrow W$ is a $G$-module homomorphism, then:

The first claim follows from the fact that if $\varphi$ is a module homomorphism, then its kernel and image are subspaces of $V, W$ respectively. Furthermore, for $v \in \ker \varphi$:

$$ \varphi(gv) = g\varphi(v) = 0 $$

So that the kernel is invariant under $G$. Similarly, for $\varphi(v)$ in the image we have:

$$ g\varphi(v) = \varphi(gv) $$

And so $g\varphi(v)$ also lies in the image. Thus, we have shown the kernel and image of $\varphi$ are subrepresentations of $V$ and $W$ respectively. The only possibilities are that the kernel is trivial and the image is $W$ (yielding an isomorphism), or the kernel is $V$ and the image is trivial (i.e. $\varphi = 0$).

To prove the second claim, $\varphi$ must have an eigenvalue $\lambda$ so that $\varphi - \lambda I$ has nonzero kernel. But if the kernel is nonzero, then by the above argument, the kernel is the $V$. So identically we indeed have:

$$ \varphi - \lambda I = 0 $$

Proposition 4

For any representation $V$ of a finite group $G$, there is a decomposition:

$$ V = V_1^{\oplus a_1} \oplus \dots \oplus V_k^{\oplus a_k} $$

Where $V_i$ are distinct irreducible representations. The decomposition is furthermore unique.

This is a straightforward consequence of Schur’s Lemma. Occasionally this decomposition is written:

$$ V = a_1V_1 \oplus \dots \oplus a_kV_k = a_1V_1 + \dots + a_kV_k $$

Where the $a_i$ denote multiplicities.

Examples: Abelian Groups; $S_3$

In general,if $V$ is a repreentation of a finite group $G$, then each $g\in G$ gives a map $\rho(g): V \rightarrow V$. However, in general, this map is not a $G$-module homomorphism ($G$-linear), i.e. in general we do not have:

$$ g(h(v)) = h(g(v)) $$

Indeed, $\rho(g)$ is $G$-linear for every $\rho$ iff $g$ is in $Z(G)$. Then $g$ commutes with $h$ and the above holds. In particular if $G$ is abelian, the above holds. But if $V$ is an irreducible representation, by Schur’s Lemma each $g\in G$ acts on $V$ by a scalar multiple, so every subspace is invariant. Thus, $V$ is one dimensional.

Therefore, the irreducible representations of an abelian group $G$ correspond to homomorphisms:

$$ \rho: G \rightarrow \mathbb{C} $$

Next, we look at $S_3$. There are two one dimensional representations, given by the trivial representation ($U$) and the alternating representation $U’$ given by:

$$ gv = \text{sgn}(g)v $$

Naturally, we ask if there are any others. Since $G$ is a permutation group, it has a natural permutation representation, where it acts on $\mathbb{C}^3$ by permuting the basis vectors. The representation is not irreducible since it has the invariant subspace spanned by $(1,1,1)$. The complementary subspace is given by:

$$ V = \{(z_1, z_2, z_3)\ :\ z_1+z_2+z_3=0\} $$

And this is irreducible since it has no invariant subspaces. It is called the standard representation.

In general, we take a representaiton $W$ of $S_3$ and look at the action of the abelian subgroup $\mathbb{Z}/3$ on $W$. If $\tau$ is a generator of this subgroup (a 3-cycle), then the space $W$ is spanned by eigenvectors for the action of $\tau$. Furthermore, since $\tau^3 = 1$, the eigenvalues are all third roots of unity. We write $\tau(v) = \omega^i v$ where $\omega^i$ is one of the roots of unity.

Let $\sigma$ be a transposition in $S_3$. Then we have the relation:

$$ \sigma \tau \sigma = \tau^2 $$

So therefore we can write:

$$ \begin{aligned} \tau(\sigma(v)) &= \sigma(\tau^2(v)) \\ &= \sigma (\omega^{2i}v) \\ &= \omega^{2i}\sigma(v) \end{aligned} $$

So if $v$ is an eigenvector for $\tau$ with eigenvalue $\omega^i$, then $\sigma(v)$ is an eigenvector for $\tau$ with eigenvalue $\omega^{2i}$.

If $v$ is an eigenvector of $\tau$ with eigenvalue $\omega^i \neq 1$, then $\sigma(v)$ is an eigenvector with a different eigenvalue and hence independent. Thus, $v, \sigma(v)$ span a two dimensional subspace of $W$ which is invariant under $S_3$.

On the other hand, if $w^i = 1$, then $\sigma(v)$ may or may not be linearly independent to $v$. If it is not, then $v$ spans a one-dimensional subrepresentation, isomorphic to the trivial representation if $\sigma(v) = v$ and the alternating representation if $\sigma(v) = -v$. If $\sigma(v)$ and $v$ are linearly independent, then $v+\sigma(v)$ and $v-\sigma(v)$ span one dimensional represnetations of $W$ isomorphic to the trivial and alternating representations, respectively.

This is not the best approach to find the decomposition of any representation of $S_3$, but it is one way to do it.

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