The Unapologetic Mathematician

The Character of a Representation

Now we introduce a very useful tool in the study of group representations: the “character” of a representation. And it’s almost effortless to define: the character $\chi$ of a matrix representation $X$ of a group $G$ is a complex-valued function on $G$ defined by $\displaystyle\chi(g)=\mathrm{Tr}(X(g))$

That is, the character is “the trace of the representation”. But why this is interesting is almost completely opaque at this point. I’m still not entirely sure why this formula has so many fabulous properties.

First of all, we need to recall something about the trace: it satisfies the “cyclic property”. That is, given an $m\times n$ matrix $A$ and an $n\times m$ matrix $B$, we have $\displaystyle\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$

Indeed, if we write out the matrices in components we find \displaystyle\begin{aligned}(AB)_i^j&=\sum\limits_{k=1}^nA_i^kB_k^j\\(BA)_k^l&=\sum\limits_{i=1}^mB_k^iA_i^l\end{aligned}

Then since the trace is the sum of the diagonal elements we calculate \displaystyle\begin{aligned}\mathrm{Tr}(AB)&=\sum\limits_{i=1}^m\sum\limits_{k=1}^nA_i^kB_k^i\\\mathrm{Tr}(BA)&=\sum\limits_{k=1}^n\sum\limits_{i=1}^mB_k^iA_i^k\end{aligned}

but these are exactly the same!

We have to be careful, though, that we don’t take this to mean that we can arbitrarily reorder matrices inside the trace. If $A$, $B$, and $C$ are all $n\times n$ matrices, we can conclude that \displaystyle\begin{aligned}\mathrm{Tr}(ABC)&=\mathrm{Tr}(BCA)&=\mathrm{Tr}(CAB)\\\mathrm{Tr}(ACB)&=\mathrm{Tr}(CBA)&=\mathrm{Tr}(BAC)\end{aligned}

but we cannot conclude in general that any of the traces on the upper line are equal to any of the traces on the lower line. We can “cycle” matrices around inside the trace, but not rearrange them arbitrarily.

So, what good is this? Well, if $A$ is an invertible $n\times n$ matrix and $X$ is any matrix, then we find that $\mathrm{Tr}(AXA^{-1})=\mathrm{Tr}(XA^{-1}A)=\mathrm{Tr}(X)$. If $A$ is a change of basis matrix, then this tells us that the trace only depends on the linear transformation $X$ represents, and not on the particular matrix. In particular, if $X$ and $Y$ are two equivalent matrix representations then there is some intertwining matrix $A$ so that $AX(g)=Y(g)A$ for all $g\in G$. The characters of $X$ and $Y$ are therefore equal.

If $V$ is a $G$-module, then picking any basis for $V$ gives a matrix $X(g)$ representing each linear transformation $\rho(g)$. The previous paragraph shows that which particular matrix representation we pick doesn’t matter, since they’re all give us the same character $\chi(g)$. And so we can define the character of a $G$-module to be the character of any corresponding matrix representation.

October 14, 2010 -

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