5
$\begingroup$

Let $G$ be a finite subgroup of the group $U_d(\mathbb{C})$ of unitary transformations of $\mathbb{C}^d$. Suppose that $G$ acts irreducibly but is imprimitive, meaning that there is a nontrivial direct sum decomposition $\mathbb{C}^d = \bigoplus_{i = 1}^r V_i$ such that each $g \in G$ permutes the $V_i$.

Then it seems that the $V_i$ are necessarily orthogonal: I wrote up a proof here. However I'm happy to admit that it took me quite some time to find this proof, and I still don't know of a reference. This must surely be well-known, and was probably known to Frobenius. Can anyone supply me with a reference?

$\endgroup$
1
  • 1
    $\begingroup$ Interestingly this is not necessarily true over finite fields, where it is possible for the $V_i$ to be totally singular. For example, ${\rm SU}(2n,q)$ has subgroups that are extensions of ${\rm SL}(n,q^2)$, which interchange two totally singular subspaces of dimension $n$. $\endgroup$
    – Derek Holt
    Jan 10, 2018 at 13:26

1 Answer 1

6
$\begingroup$

This is not an answer, but here is another proof in the same spirit as yours. Write $(-\vert -)$ for the canonical scalar product of ${\mathbb C}^d$. Since the $G$-representation ${\mathbb C}^d$ is irreducible, we have $$ {\mathbb C}^d = \bigoplus_{g\in G/G_1} g.V_1 $$ where $G_1$ is the stabilizer of $V_1$ in $G$. Write $(-\vert -)_1$ for the restriction of $(-\vert -)$ to $V_1\times V_1$; it is $G_1$-invariant. For all $g\in G$ define a scalar product on $g.V_1$ by $$ (gv_1 \vert gw_1 )_{gV_1} = (v_1 \vert w_1 ) $$ It does not depend on the choice of $g$. Let $<-\vert ->$ be the scalar product on ${\mathbb C}^d$ given by the orthogonal sum of the $(-\vert -)_{gV_1}$, $g\in G/G_1$. It is clearly $G$-invariant. Now since the representation ${\mathbb C}^d$ is irreducible the scalar products $(-\vert -)$ and $<-\vert ->$ are proportional and we are done. N.B. Of course Schur's lemma is hidden in my proof.

$\endgroup$
1
  • $\begingroup$ Paul, thanks for this. In fact, I think it's exactly the same proof, and in fact when I first came up with the proof I formulated it this way. Maybe it's more natural. Schur's lemma is hidden in the last step and I think < x, y > = (phi(x), y) with my notation. I don't know how you got to this argument, but for me it was via the fact that imprimitive => induced from G_1, and then look at how induced representations are given a unitary structure. $\endgroup$
    – Ben Green
    Jan 10, 2018 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.