Suspicious Characters, Part 2

It's trace operator time!  For those who don't know already, the trace of a matrix is the sum of the entries on the main diagonal.

Why is it importart?  We won't be proving this fact today, but the trace of a matrix $\rho_g$ is also the sum of its eigenvalues!

Eigenvalues?  Wündebar!  Time to make de magics.

The trace operator also plays well with the composition of spaces!  If I have a linear transformation $\rho_g$ of a vector space $V$, and another linear transformation $\sigma_h$ of a vector space $W$, the corresponding linear transformation $\rho_g \oplus \sigma_h$ of the space $V \oplus W$ has this property:

$Tr(\rho_g \oplus \sigma_h) = Tr(\rho_g) + Tr(\sigma_h)$

Which is nice.  So if we had some sort of decomposition of a space $V$ into some sort of irreducible components $W_1, W_2,... W_m$ and a family of linear transformations $\rho_g^{W_1}, \rho_g^{W_2},... \rho_g^{W_m}$ on those spaces...

$V=\displaystyle \bigoplus_{i=1}^m W_i  \, \, \Rightarrow \, \, Tr\left( \bigoplus_{i=1}^m \rho_g^{W_i}\right) = \sum_{i=1}^m Tr(\rho_g^{W_i})$

Neat!  So if we have a representation $\rho: G \rightarrow V$, a decomposition of $V$ into $G$-invariant subspaces also induces a sort of decomposition on the values of $Tr(\rho_g)$.  But to study $Tr(\rho(g))$ as a function on $G$, we're going to face some clunky notation.  So let's give this function its own name, $\chi_\rho $.

This map we've created, $\chi_\rho : G \rightarrow \mathbb{C}$ such that $\chi_\rho(g) = Tr(\rho(g)) \, \, \forall \, g \in G$, is called the character of $\rho$.

Hence why this stuff is called "Character Theory."  And now we can make this come together around some very neat properties of $\chi_\rho$ when $\rho$ is an irreducible representation!  Remember that proposition we proved at the end of the last installment?  We're going to use that result.

Proposition:

If $\rho: G \rightarrow V$ is an irreducible representation, and we write each $\rho_g$ as  matrix in the basis $\{ \vec{e_1}, \vec{e_2}, ... \vec{e_n}\}$ of $V$, with $g_{ij}$ to denote the entry in the $i$th row and $j$th column,

$\frac{1}{|G|}\displaystyle \sum_{g\in G} g_{ij} = \left\{\displaystyle \begin{array}{lc} \delta_{ij} & \mbox{ if } \rho \mbox{ is the trivial representation} \\ 0 & \mbox{ otherwise} \end{array} \right. $  

Where $\delta_{ij}$ is the Kroenecker delta.

Proof:

If $\rho$ is the trivial representation (i.e. $\rho_g$ is the identity matrix $\, \forall \, g \in G$) then the above sum is just the entry in the $i$th row and $j$th column of the identity matrix. 

If $\rho$ is not trivial, define $u_k: V \rightarrow \mathbb{C}$ as $\hspace{2mm} u_i \left( \displaystyle \sum_{j=1}^n c_j \vec{e_j} \right) = c_i \hspace{2mm}$ and rewrite the sum,

$ \frac{1}{|G|} \displaystyle \sum_{g\in G}g_{ij}= $$\frac{1}{|G|} \displaystyle \sum_{g\in G} u_i \left( \rho_g (\vec{e_j}) \right)$

But by the proposition at the end of the last blogpost, the expression on the right can only be nonzero if $\rho$ is trivial.  Thus it is zero in these nontrivial cases.

$\Box$

Now let's use this result with the trace operator:

$\frac{1}{|G|}\displaystyle\sum_{g\in G}Tr(\rho_g) = $$\frac{1}{|G|}\displaystyle\sum_{i=1}^n\left(\sum_{g\in G} g_{ii}\right)$

When $\rho$ is irreducible and nontrivial, this sum we made is zero.  When $\rho$ is irreducible and trivial, this sum becomes $1$.  And... remember that additive bit about the trace operator earlier?  When $\rho$ is not irreducible, this sum is the 'multiplicity' of trivial subrepresentations in the decomposition of $V$ and $\rho$.

If only there were similarly powerful techniques

for keeping track of dressmaking supplies...

Just you wait, we'll get some orthogonality relations out of this...