\(SL_4/\mu_2\) and a mod \(8\) congruence

This is a continuation of a previous post.  Recall that we wanted to prove the following claim:

Claim.  Let \(\rho\) be a representation of \(SL_4\) such that no irreducible subrepresentation of \(\rho\) descends to \(SL_4/\mu_2\).  Then if \(\rho\) is self-dual, we have that $$8\mid\dim \rho.$$

We first translate this into the language of weights, so we can use the Weyl dimension formula.  Recall that irreducible representations of \(SL_4\) are indexed by \(4\)-tuples of non-negative integers $$\lambda_1\geq \lambda_2\geq \lambda_3\geq \lambda_4=0.$$  We let \(\lambda\) be the \(n\)-tuple \((\lambda_1, \lambda_2, \lambda_3, \lambda_4)\), and denote the representation corresponding to \(\lambda\) by \(S^\lambda\).  So for example $$S^{(n, 0, 0, 0)}=\text{Sym}^n(V),$$ where \(V\) is the standard representation of \(SL_4\).  We first show:

Lemma 1.  Let \(\rho\) be an irreducible representation of \(SL_4\) which does not descend to \(SL_4/\mu_2\).  Then \(4\mid \dim\rho\).

Proof.  We have \(\rho=S^\lambda\) for some \(\lambda\); the hypothesis is equivalent to the statement that $$\sum_{i=1}^4 \lambda_i \equiv 1\bmod 2.$$  In other words, three of the \(\lambda_i\) have the same parity.

Now the Weyl dimension formula (or equivalently, standard results on specializations of Schur polynomials) tells us that $$\dim \rho =\prod_{1\leq i<j\leq 4}\frac{\lambda_j-\lambda_i+j-i}{j-i}.$$  The denominator of this product is \(12\), so we must show that the numerator is divisible by \(16\).  Now of the four pairs \(i, j\) with \(j-i\) odd, at least two must have that \(\lambda_i-\lambda_j\) is odd as well.  Thus $$4\mid\prod_{1\leq i<j\leq 4, j-i\text{ odd}} \lambda_j-\lambda_i+j-i.$$  For the two pairs with \(j-i=2\), at least one must have that \(\lambda_i-\lambda_j\) is even, so $$2\mid \prod_{1\leq i<j\leq 4, j-i\text{ even}} \lambda_j-\lambda_i+j-i.$$

Now some annoying casework lets us eke out one more factor of two; if I come up with a slick way of doing it, I will write it down...

Multiplying gives the desired result.  \(\blacksquare\)

Now we analyze the case where \(\rho\) is self-dual.

Lemma 2.  Let \(\rho\) be as in Lemma 1.  Then  \(\rho\) is not self-dual.

Proof.  Let \(\lambda\) be the \(4\)-tuple of integers corresponding to \(\rho\).  Then \(\rho\) is self-dual iff $$(\lambda_1, \lambda_2, \lambda_3, \lambda_4)=(-\lambda_4+\lambda_1, -\lambda_3+\lambda_1, -\lambda_2+\lambda_1, -\lambda_1+\lambda_1).$$  In other words, we have $$\lambda_2=\lambda_1-\lambda_3.$$  But recall that $$\lambda_1+\lambda_2+\lambda_3=2\lambda_1$$ had to be odd in the setting of Lemma 1.  Contradiction. \(\blacksquare\)

Proof of Claim.  Let \(\rho\) be as in the claim.  Then it is a direct sum of irreducibles, none of which descend to \(SL_4/\mu_2\).  Thus there exist a collection of irreducibles \(V_i\) (none of which are self-dual, by Lemma 2, such that $$\rho=\bigoplus_i V_i \oplus V_i^{\vee}.$$  But by Lemma 1, each \(V_i\) has dimension divisible by \(4\), so $$8\mid \dim \rho$$ as desired. \(\blacksquare\)

Applying the arguments from my previous post on this topic, we get a version of the Auel-First-Williams result for the stack \(B(SL_4/\mu_2)\) (as well as a version of their theorem for symplectic involutions of Azumaya algebras).  In a sequel post, I might explain how to deduce a weak version of their actual results from this computation.