$SL_4/\mu_2$ 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.