The Unapologetic Mathematician

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.

October 29, 2010 -