Lifting and Descending Representations

Let’s recall that a group representation is, among other things, a group homomorphism. This has a few consequences.

First of all, we can consider the kernel of a matrix representation $X$ . This is not the kernel we’ve talked about recently, which is the kernel of a $G$ -morphism. This is the kernel of a group homomorphism. In this context, it’s the collection of group elements $g\in G$ so that the image $X(g)$ is the identity transformation. We call this subgroup $N\subseteq G$ . If $N$ is the trivial subgroup, we say that $X$ is a “faithful” representation, since it doesn’t send any two group elements to the same matrix.

Now, basic group theory tells us that $N$ is a normal subgroup, and so we can form the quotient group $G/N$ . I say that the representation $X$ “descends” to a representation of this quotient group. That is, we can define a representation $Y$ by $Y(gN)=X(g)$ for all cosets $gN\in G/N$ . We have to check that this doesn’t depend on which representative $g$ of the coset we choose, but any other one looks like $g'=gn$ for $n\in N$ . Then $Y(g')=Y(g)Y(n)=Y(g)$ , since $Y(n)$ is the identity matrix by definition.

I say that $Y$ is a faithful representation. That is, the only coset that $Y$ sends to the identity matrix is the one containing the identity: $N$ itself. And indeed, if $Y(gN)=I$ , then $X(g)=I$ , and so $g\in N$ and $gN=N$ in the first place.

Next, $Y$ is irreducible if and only if $X$ is. We’ll prove this one by using the properties we proved a few days ago. In particular, we’ll calculate the inner product of the character of $Y$ with itself. Writing $\psi$ for the character of $Y$ and $\chi$ for that of $X$ , we find that

$\displaystyle\psi(gN)=\mathrm{Tr}(Y(gN))=\mathrm{Tr}(X(g))=\chi(g)$

and we use this to calculate:

$\displaystyle\begin{aligned}\langle\psi,\psi\rangle&=\frac{1}{\lvert G/N\rvert}\sum\limits_{gN\in G/N}\overline{\psi(gN)}\psi(gN)\\&=\frac{\lvert N\rvert}{\lvert G\rvert}\sum\limits_{gN\in G/N}\overline{\chi(g)}\chi(g)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{gN\in G/N}\sum\limits_{n\in N}\overline{\chi(gn)}\chi(gn)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\overline{\chi(g)}\chi(g)\\&=\langle\chi,\chi\rangle\end{aligned}$

Essentially, the idea is that each group element in the kernel $N$ contributes the same to the sum, and this is exactly compensated for by the difference between the sizes of $G$ and $G/N$ . Since the two inner products are the same, either both are $1$ or neither one is, and so either both representations are irreducible or neither one is.

We can also run this process in reverse: let $G$ be a finite group, let $N$ be any normal subgroup so we have the quotient group $G/N$ , and let $Y$ be any representation of $G/N$ . We will use this to define a representation of the original group $G$ by “lifting” the representation $Y$ .

So, the obvious choice is to define $X(g)=Y(gN)$ . This time there’s no question that $X$ is well-defined, since here we start with a group element and find its coset, rather than starting with a coset and picking a representative element. And indeed, this is easily verified to be a representation.

If $Y$ is faithful, then the kernel of $X$ is exactly $N$ . Indeed, if $X(g)=I$ , then $Y(gN)=I$ . If $Y$ is faithful, then we must have $gN=N$ , and thus $g\in N$ .

And finally, $X$ is irreducible if and only if $Y$ is. The proof runs exactly as it did before.

About this weblog

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”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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The Unapologetic Mathematician

Mathematics for the interested outsider