# $$SL_4/\mu_2$$ 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.