Suspicious Characters, Part 1

For those familiar with linear algebra and group theory...

Suppose we have a group that acts on a vector space.  This situation comes up fairly often; every time we have an object situated in euclidean space, the symmetries of the object are a subgroup of the symmetries of the space itself.  Even if that seems kind of obvious, this is a genuinely useful insight: the symmetries (a.k.a. automorphisms) of $n$-dimensional euclidean space can be represented by matrices, and there are a lot of powerful techniques out there (e.g. determinants, eigenvalues) that only work on matrices.  So studying the matrix representation ought to be a useful technique for understanding the action of the group.

This is more effective than you might realize at first: if both the group and the dimension of our vector space are finite, there is a nice way to decompose the vector space into orthogonal orbit-like-subspaces just by looking at the traces of the matrices.  Over the next few blog posts, we'll be investigating this business with decomposition using trace operators--known as character theory.

With some character theory, Rainbow Dash might have noticed

the different subrepresentations of Mare-Do-Well beforehoof.

First, the formalism.  For the sake of elegance, (and algebraic closure) all our vector spaces will be over $\mathbb{C}$.

Jargon:

A representation of a finite group $G$ over a vector space $V$ is a homomorphism $\rho : G \rightarrow GL(V)$, where $GL(V)$ is the group of all invertible linear maps $f: V \rightarrow V$

Since we're assuming that $V \cong \mathbb{C}^n$, $GL(V)$ is the same thing as the set of all invertible $n \times n$ matrices with complex-valued entries.

To simplify notation, whenever we are describing the action of each $g \in G$ on vectors (because we will be using them as functions!) we will write $\rho(g)$ as  $\rho_g$.

Jargon:

A subspace $W \subset V$ is invariant under $G$ or $G$-invariant if $\rho_g(W) = W \,  $ $\, \forall \, g \in G$

This means every $\rho_g$ acts as an automorphism of $W$.  But it get's better: there's some basis for which the orthogonal complement of $W$, call it $W^\perp$, is also $G$-invariant.

Lemma:  If the subspace $W \subset V$ is $G$-invariant, there is a basis of $V$ for which $W^\perp$ is $G$-invariant

Let $\phi$ be a projection of $V$ onto $W$, i.e.

$\phi(\vec{v}) \in W \hspace{1.5mm} \forall \, \vec{v} \in V, \hspace{1.5mm}$ and $\hspace{1.5mm} \phi(\vec{w}) = \vec{w} \hspace{1.5mm} \forall \, \vec{w} \in W$ 

We cleverly define $\phi' = \frac{1}{|G|}\displaystyle \sum_{g\in G}\rho_g\circ\phi\circ\rho_{g^{-1}}$
Using the fact that every $h \in G$ is a permutation on $G$, we get:
$\rho_h \circ \phi'\circ \rho_{h^{-1}} =
\frac{1}{|G|}\displaystyle \sum_{g\in G}\rho_{hg}\circ\phi\circ\rho_{g^{-1}h^{-1}} =\frac{1}{|G|}\displaystyle \sum_{g\in G}\rho_g\circ\phi\circ\rho_{g^{-1}} = \phi' $ 

Then for all $g\in G$, and $\vec{x} \in Ker(\phi')$,  we have $\phi'\left( \rho_g(\vec{x}) \right) = \rho_g \circ \phi' \circ \rho_{g^{-1}} \left( \rho_g(\vec{x}) \right) =
\rho_g \circ \phi'(x) = \rho_g (\vec{0}) = \vec{0}$
That is, we have, $\rho_g(\vec{x}) \in Ker(\phi')$.   

So $Ker(\phi')$ is invariant under $G$.  As as this is the kernel of a projection onto $W$, there is some basis for which this is the orthogonal complement of $W$.

$\Box$

So in these cases, $\rho(g)$ can be "split" into an automorphism of $W$ and an automorphism of $W^\perp$.  In a very natural way, the action of $G$ on $W$ (or $W^\perp$) is, by itself, a representation of $G$.  Since this arises out of a subspace of $V$, we call the corresponding map of $G$ into $GL(W)$ a subrepresentation of $\rho$.

One Last Bit of Jargon:

We call a representation of $G$ over $V$ irreducible if the only $G$-invariant subspaces of $V$ are itself and the zero subspace.

Essentially, a representation is irreducible when there are no subrepresentations for us to look at.
Since vector spaces of dimension $1$ have no proper nontrivial subspaces, representations over them must be irreducible.  And it should be intuitive that every finite-dimensional vector space--if it is not already irreducible--can be written as the direct sum of finitely many $G$-invariant subspaces.

So much math to learn!

Now before we wrap this installment up, we'll take a look at the nice structures these irreducible representations will afford us:


Proposition:

For any linear map $f: V \rightarrow \mathbb{C}$,
And any irreducible linear representation $\rho: G \rightarrow GL(V)$,
If  $\exists \, \vec{v} \in V$ such  that $\displaystyle \sum_{g \in G} f(\rho_g(\vec{v})) \neq 0$,

Then $\rho_g$ is the identity matrix $\forall \, g \in G$

Proof:

For such a $\vec{v} \in V$, define $\vec{v_0} = \displaystyle \sum_{g \in G} \rho_g(\vec{v})$.  By linearity, $f(\vec{v_0}) = \displaystyle \sum_{g \in G} f\left( \rho_g(\vec{v}) \right)$.
So  $f(\vec{v_0}) \neq 0$ and therefore $\vec{v_0} \neq \vec{0}$ 

Again using the fact that every $h \in G$ is a permutation on $G$, we get:
$\rho_h(\vec{v_0}) = \rho_h \left( \displaystyle \sum_{g \in G} \rho_g(\vec{v}) \right) = \displaystyle \sum_{hg \in G} \rho_{hg}(\vec{v})= \displaystyle \sum_{g \in G} \rho_g(\vec{v}) = \vec{v_0}$
And therefore $\vec{v_0}$ is fixed by the action of every $\rho_h$.

So therefore the subspace generated by $\vec{v_0}$, call it $Span(\vec{v_0})$, is both nontrivial (because $\vec{v_0} \neq \vec{0}$) and $G$-invariant (because $\rho_g(\vec{v_0})=\vec{v_0} \,  \forall \, g \in G$). 

Now $V$ is irreducible, so it contains no proper nontrivial $G$-invariant subspaces. Therefore $Span(\vec{v_0})$ cannot be proper, i.e. it equals $V$ itself.  

Thus $\rho_g$ fixes the generators of $V$, and so $\rho_g$ must be the identity matrix $\forall \, g \in G$

$\Box$

Now what does all this stuff have to do with the trace operator?  Find out in the next installment!