## Tuesday, 23 February 2016

### Representation theory, for the impatient

$\def\C{\mathbf{C}}$

Let $G$ be a finite group. The group algebra $\C G$ is the complex algebra with basis $G$, with multiplication defined by linearly extending the group law of $G$. The description of $\C G$ as an algebra is called representation theory. Relatedly, we want to understand the structure of $\C G$-modules. We often call $\C G$-modules (complex) representations of $G$, and simple $\C G$-modules irreducible representations.

Lemma: (Schur's lemma) If $V$ and $V'$ are simple $\C G$-modules, then every nonzero homomorphism $V\to V'$ is an isomorphism. Moreover every homomorphism $V\to V$ is a scalar multiple of the identity.
Proof: Suppose that $f:V\to V'$ is a nonzero homomorphism. Then $\ker f$ is a submodule, so $\ker f = 0$, and similarly $f(V) = V'$, so $f$ is an isomorphism. Now suppose $V=V'$. Since every eigenspace of $f$ is a submodule, by simplicity every eigenspace is either $0$ or $V$. From the spectral theorem we deduce that $f$ is a scalar multiple of the identity.

It turns out that we can ask for a unitary structure in $\C G$-modules for free.

Lemma: (Weyl's averaging trick) Every $\C G$-module $V$ has a $G$-invariant inner product. Moreover if $V$ is simple then this inner product is unique up to scaling.
Proof: Let $V$ be a $\C G$-module and let $(,)$ be any inner product on $V$. Define a new inner product $\langle,\rangle$ by
$$\langle u, v\rangle = \frac{1}{|G|} \sum_{g\in G} (g\cdot u, g\cdot v).$$
Clearly $\langle,\rangle$ has the desired properties. Now suppose that $V$ is simple, and that $\langle,\rangle'$ is another $G$-invariant inner product. Let $f:V\to V$ be the adjoint of the formal identity
$$(V,\langle,\rangle)\to(V,\langle,\rangle').$$
In other words let $f$ be the unique function $V\to V$ satisfying
$$\langle u,v\rangle' = \langle u, f(v)\rangle$$
for all $u,v\in V$. Then $f$ is a homomorphism, so by Schur's lemma it must be a multiple of the identity, so $\langle,\rangle'$ must be a multiple of $\langle,\rangle$.

Lemma: Let $U$ be a finite-dimensional $\C G$-module with $G$-invariant inner product $\langle,\rangle$, and for each simple $\C G$-module $V$ let $U_V$ be the sum of all submodules of $U$ isomorphic to $V$ (the isotypic component of $U$ corresponding to $V$). Then $U_V\cong V^{\oplus m}$ for some $m$, and $U$ is the orthogonal direct sum of the submodules $U_V$.
Proof: Since the orthogonal complement of a submodule is a submodule, it follows by induction on dimension that $U$ is an orthogonal direct sum of simple submodules, so $U = \bigoplus U'_V$, where $V$ runs over simple $\C G$-modules and each $U'_V \cong V^{\oplus m}$ for some $m\geq 0$. Moreover by Schur's lemma any two nonisomorphic simple submodules must be orthogonal, as the orthogonal projection from one to the other is a homomorphism, so every submodule of $U$ isomorphic to $V$ must be contained in $U'_V$, so $U'_V=U_V$.

The above lemma is particularly interesting when applied to the $\C G$-module $\C G$ itself, the regular representation. We give $\C G$ a $G$-invariant inner product by declaring the basis $G$ to be orthonormal. From the above lemma we know then that $\C G = \bigoplus \C G_V$, where $V$ runs over simple $\C G$-modules, and $\C G_V\cong V^{\oplus m}$ for some $m\geq 0$. Note then by Schur's lemma that $\dim\text{Hom}(\C G,V)=m$. On the other hand every homomorphism $\C G\to V$ is determined uniquely by the destination of the unit $1$ in $V$, so $\dim\text{Hom}(\C G,V)=\dim V$. We deduce that as $\C G$-modules
$$\C G \cong \bigoplus V^{\oplus \dim V}.$$
In particular there are only finitely many simple $\C G$-modules, and their dimensions obey
$$|G| = \sum (\dim V)^2.$$

One can also prove $\C G_V\cong V^{\oplus\dim V}$ in a more informative way, as follows. Fix an invariant inner product $\langle,\rangle_V$ on $V$, and consider any homomorphism $f:V\to\C G$. The adjoint $f^*:\C G\to V$ is also a homomorphism, so
$$f(v) = \sum_{g\in G}\langle f(v),g\rangle_{\C G} g = \sum_{g\in G}\langle v, f^*(g)\rangle_V g = \sum_{g\in G}\langle v,g f^*(1)\rangle_V g.$$
Conversely for any $u\in V$ we may define a homomorphism $V\to\C G$ by
$$f_{V,u}(v) = \sum_{g\in G} \langle v,g\cdot u\rangle_V g.$$
We deduce therefore that $\C G_V$ is the subspace of $\C G$ spanned by the elements $f_{V,u}(v)$, where $u,v\in V$. Moreover using Schur's lemma one can show that the images of $f_{V,u}$ and $f_{V,u'}$ are orthogonal whenever $u$ and $u'$ are orthogonal in $V$, so by letting $u$ range over a basis of $V$ we thus see that $\C G_V$ is the orthogonal direct sum of $\dim V$ copies of $V$.

In any case we now understand the structure of $\C G$ as a $\C G$-module, and we are only a short step away from understanding its structure as an algebra. Consider the obvious map
$$\C G \longrightarrow \bigoplus \text{End}(V),$$
where as always the sum runs over a complete set of irreducible representations up to isomorphism. We claim this map is an isomorphism. Since we already know the dimensions agree, it suffices to prove injectivity. Thus suppose $x\in\C G$ maps to zero, i.e., that $x$ acts trivially on each simple $\C G$-module $V$. Then $x$ acts trivially on $\C G$, so $x = x1 = 0$. Thus we have proved the following theorem.

Theorem: As complex algebras, $\C G \cong \bigoplus \text{End}(V)$.
$\def\tr{\text{tr}}$
Finally, it is useful to understand how to project onto isotypic components. Given $g\in G$, we can compute the trace of $g$ as an operator on $\C G$ in two different ways. On the one hand, by looking at the basis $G$,
$$\tr_{\C G} (g) = \begin{cases} |G| & \text{if }g=1,\\ 0 &\text{if }g\neq 1.\end{cases}$$
On the other hand from the decomposition $\C G\cong\bigoplus V^{\oplus\dim V}$ we have
$$\tr_{\C G} (g) = \sum_V (\dim V) \tr_V(g).$$
As a consequence, for every $x\in \C G$ we have
$$x = \sum_V P_V x,$$
where $P_V : \C G \to \C G$ is the operator defined by
$$P_V x = \frac{\dim V}{|G|} \sum_{g\in G} \tr_V(xg^{-1}) g = \frac{\dim V}{|G|} \sum_{g\in G} \tr_V(g^{-1}) gx.$$
These identities are most easily verified first for $x\in G$, then extending to all of $\C G$ by linearity. Now if $x \in \C G_U$ for $U\neq V$ then $x$ acts as zero on $V$, so $\tr_V(x g^{-1}) = 0$, so $P_V x = 0$. On the other hand one can verify directly that $P_V$ is a homomorphism, so by Schur's lemma the image of $P_V$ must be contained in $\C G_V$. We deduce therefore from $x = \sum_V P_V x$ that $P_V$ is the orthogonal projection onto $\C G$.

The function $\chi_V(g) = \tr_V(g)$ is usually called the character of $V$. From the relations $P_V^2 = P_V$ and $P_U P_V = 0$ for $U\neq V$ one can deduce the well known orthogonality relations for characters. In fact the distinction between $P_V$ and $\chi_V$ is hardly more than notational. Often we identify functions $f:G\to \C$ with elements $\sum_{g\in G} f(g) g \in \C G$, in which case the operation of convolution corresponds to multiplication in the group algebra. The operator $P_V$ then is just convolution with $(\dim V)\chi_V$. So, in brief, to project onto the $V$-isotypic component you convolve with the character of $V$ and multiply by $\dim V$.

We have kept to almost the bare minimum in the above discussion: the complex numbers $\C$ and finite groups $G$. There are a number of directions we could try to move in. We could replace $\C$ with a different field, say one which is not algebraically closed, or one which has positive characteristic. Alternatively we could replace $G$ with an infinite group, say with a locally compact topology. We mention two such generalisations.

Theorem: (Artin--Wedderburn) Every semisimple ring is isomorphic to a product $\prod_{i=1}^k M_{n_i}(D_i)$ of matrix rings, where the $n_i$ are integers and the $D_i$ are division rings. In particular every semisimple $\C$-algebra is isomorphic to a product $\prod_{i=1}^k M_{n_i}(\C)$.
When defining unitary representations for compact groups $G$ we demand that the map $G\to U(V)$ be continuous, where $U(V)$ is given the strong operator topology.$\def\HS{\text{HS}}$

Theorem: (Peter--Weyl) Let $G$ be a compact group and $\mu$ its normalised Haar measure. Let $\widehat{G}$ be the set of all irreducible unitary representations of $G$ up to isomorphism. Then $\widehat{G}$ is countable, every $V\in\widehat{G}$ is finite-dimensional, and the algebra $L^2(G)$ of square-integrable functions with the operation of convolution decomposes as a Hilbert algebra as $$L^2(G) \cong \bigoplus_{V\in\widehat{G}} (\dim V) \cdot \HS(V),$$ where $\HS(V)$ is the space $\text{End}(V)$ together with the Hilbert-Schmidt inner product.