# 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.

## 4 Comments	»

1. [...] Our first observation about characters takes our work from last time and spins it in a new [...]

Pingback by | October 15, 2010 | Reply

2. [...] we’re dealing with characters, there’s something we can do to rework our expression for the inner product on the space of [...]

Pingback by | October 18, 2010 | Reply

3. [...] let’s try to compute the character of this representation. If we write the representing homomorphisms and , then we get a [...]

Pingback by | November 4, 2010 | Reply

4. [...] Characters of Induced Representations We know how to restrict and induce representations. Now we want to see what this looks like on the level of characters. [...]

Pingback by | November 29, 2010 | Reply

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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