## Virtual Congruence Betti Numbers

Suppose that $G$ is a real semisimple group and that $X = \Gamma \backslash G/K$ is a compact arithmetic locally symmetric space. Let us call a cohomology class tautological if it is invariant under the group $G$. For example, if $X$ is a 3-manifold, then the tautological classes are all multiples of either the trivial class in $H^0$ or the fundamental class in $H^3$. We say that $X$ has positive Betti number if there exist any non-tautologial classes in the cohomology of $X$. One can pose the following question:

Problem: Show that there exists a finite congruence cover $\widetilde{X} \rightarrow X$ such that $\widetilde{X}$ has positive Betti number.

An automorphic way of phrasing this question is as follows: do there exist any automorphic forms besides the trivial representation for the $\mathbf{Q}$-group $\mathbb{G}$ associated to $\Gamma$. If $G$ admits discrete series, then the result is obvious for automorphic reasons (from the trace formula, by de George-Wallach). If $X$ has non-zero Betti number Euler characteristic, then the result is obvious for topological reasons. In fact, as I leant from Gross one day at tea, these two situations coincide (this certainly follows from Borel-Wallach, even in the stronger form that the contribution from each $\pi$ via Matsushima’s formula has zero Euler characteristic if it is not a discrete series; I’m not sure if there’s a slicker argument).

The problem is obviously related to the virtual positive Betti number theorem of Agol, but there are a few important subtle differences. The first is that we insist that the cover $\widetilde{X}$ is congruence. Hence, the problem remains open for a general arithmetic 3-manifold. Second, we also allow (as we must) cohomology in any degree. Another example to consider is $G = U(2,1)$. In this case, $X$ is a compact complex hyperbolic manifold. It is an open problem whether such manifolds have virtual positive first Betti number. In contrast, by a theorem of Rogawski, they certainly don’t have virtual positive first Betti number in congruence covers, although they clearly do have virtual positive Betti number in congruence covers for the two equivalent reasons given above.

What I want to do in this post is discuss a related problem, namely, can one find arbitrarily large congruence covers $\widetilde{X}$ which all fail to have positive Betti number? Specific examples of this kind (for a compact arithmetic 3-manifold $X$) were given in my paper with Dunfield (conditional on local-global compatibility of certain Galois representations, now known), and Boston-Ellenberg shortly thereafter found a different (unconditional) argument using group theory (which applied to the same example). I want to explain how to generalize these results to higher dimension, contingent on computations which might be hard to carry out explicitly.

Choose:

• An imaginary quadratic field $F$.
• A prime $p$ which splits as $\mathfrak{p} \overline{\mathfrak{p}}$ in $F$.
• A central simple algebra $D/F$ with local invariants $1/N$ and $-1/N$ at the primes dividing $p$.

Associated to $D$ is a maximal lattice $\Gamma$ in $G/K = \mathrm{SL}_N(\mathbf{C})/\mathrm{SU}_N(\mathbf{C})$ whose quotient is a compact finite volume orbifold of real dimension $N^2 - 1$. For sufficiently large $n$, the congruence covers $X(\mathfrak{p}^n)$ are manifolds which are $K(\pi,1)$ spaces with fundamental group $\Gamma(\mathfrak{p}^n)$. When $F = \mathbf{Q}(\sqrt{-2})$, $N = 2$, and $p = 3$, one recovers the manifolds considered in my paper with Nathan.

Let me now make another definition. Let $F_S$ be the maximal pro-p extension of $F(\zeta_p)$ unramified outside the primes dividing $p$.

Definition: The prime $p$ is very regular in $F$ if the map:

$\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p) \rightarrow D_v \subset \mathrm{Gal}(F_S/F)$

is surjective for either $v|p$.

The notion of very regular primes arose in my latest paper on $K$-theory and completed cohomology in the stable range, but more on that later. One last definition: say that an ideal $\mathfrak{m}$ of a Hecke algebra $\mathbf{T}$ is Eisenstein if the image of any Hecke operator $T$ in $\mathbf{T}/\mathfrak{m}$ coincides with multiplication by the degree $\deg(T)$. This is how $\mathbf{T}$ acts on the trivial representation. We then have the following:

Conditional Theorem: Suppose that $p$ is very regular, and that $\mathfrak{m}$ is an Eisenstein maximal ideal. Then for all $n$ there is an equality:

$H^*(X(\mathfrak{p}^n),\mathbf{Z}_p)_{\mathfrak{m}} \otimes \mathbf{Q}_p = H^*(\mathrm{SU}(N),\mathbf{Q}_p)$

In particular, if the only maximal ideals of $\mathbf{T}$ on $H^*(X(\mathfrak{p}),\mathbf{Z}_p)$ are Eisenstein, then all the $X(\mathfrak{p}^n)$ are rational $SU(N)$-homology spaces.

Example: The prime $p = 3$ is strongly regular for $F = \mathbf{Q}(\sqrt{-2})$, and — by a computation — the only maximal ideals of $\mathbf{T}$ on $H^*(X(\mathfrak{p}),\mathbf{Z}_p)$ are Eisenstein. Of course, a rational $SU(2)$-homology space is a homology 3-sphere.

Proof: Suppose that there is exists a non-trivial class in the cohomology of $X(\mathfrak{p}^n)$. It will give rise to an automorphic representation $\pi$ which is tempered, because $X$ are Shimura varieties manifolds for which we can show (reference?) have no endoscopic forms. Hence, by HLTT or Scholze, there exists a corresponding Galois representation

$r(\pi): G_{F} \rightarrow \mathrm{GL}_n(\mathbf{Q}_p)$

that is unramified away from $p$. We now assume (this may be proved soon, but this is the reason for the “conditional” in the statement) that we know enough about local-global compatibility to deduce that this representation is also ordinary at the prime $\mathfrak{p}'$. Note that the reason it should be ordinary is that the level is prime to $\mathfrak{p}'$, and since the quaternion algebra is ramified at this prime we know that $\pi_{\mathfrak{p}'}$ is Steinberg. We deduce that $r(\pi)$ is completely reducible after restriction to $D_v$ for $v = \mathfrak{p}'$. The Eisenstein assumption and the ramification assumption imply that $\overline{r(\pi)}$ and hence $r(\pi)$ factor through $\mathrm{Gal}(F_S/F)$. Hence, using the fact that $p$ is very regular, we immediately deduce that $r(\pi)$ itself is reducible and ordinary. It follows that, after semisimplification, $r(\pi)$ is a direct sum of characters, which leads to an easy contradiction.

Experts will recognize this argument as a generalized and more streamlined version of what appears in my paper with Nathan. One may naturally ask whether there is a generalization of the Boston-Ellenberg argument as well. Emerton and I already explained that the correct way to view that argument was as follows. What one really wants to prove is that the partially completed cohomology groups:

$\displaystyle{\widetilde{H}^*(\mathfrak{p}) = \lim_{\rightarrow} H^*(X(\mathfrak{p}^n),\mathbf{F}_p)}$

all vanish identically outside degree zero. For 3-manifolds, it suffices to prove this for $\widetilde{H}^1$. For what $X$ might one be able to prove such vanishing? As Matt and I explained in our paper on 3-manifolds, for all these groups to vanish there has to be a delicate balancing act between the dimension of the group acting on completed cohomology and the dimension of the manifold. For example, it is crucial that there is an equality

$\dim(G/K) = \dim(\prod_{S} G(F_v))$

where one partially completes at primes $S$ above $p$. (Otherwise one obtains an immediate contradition by Hochschild-Serre.) In the case at hand, this inequality is satisfied, since:

$\dim(G/K) = \dim(\mathrm{SL}_N(\mathbf{C})) - \dim(\mathrm{SU}_N(\mathbf{C})) = N^2 - 1 = \dim(\mathrm{SL}_N(\mathbf{Z}_p))$

Hence, it is really possible that all the completed cohomology groups may vanish in this case. In fact, if one instead considers the split group $\mathrm{GL}(N)/F$, then the partially completed cohomology groups do vanish in the stable range exactly for very regular primes. (This is where the definition of very regular primes comes from.) By Nakayama’s Lemma, one can explicitly compute at some finite level to determine whether the $\widetilde{H}^*(\mathfrak{p})$ vanish or not. In fact, it suffices to compute that the maps:

$H^*(G(p),\mathbf{F}_p) \rightarrow H^*(X(\mathfrak{p}),\mathbf{F}_p)$

are isomorphisms, where $G(p)$ is the congruence subgroup of $\mathrm{SL}_n(\mathbf{Z}_p)$.
If one wanted to find an explicit example where these theorems applied for $N \ge 3$, the first place to look would probably be to take $F = \mathbf{Q}(\sqrt{-2})$, $p = 3$, and $N = 3$. One would then have to compute the cohomology of a certain $8$-dimensional manifold! (The resulting manifolds would potentially all be rational $SU(3)$-homology space = rational $S^5 \times S^3$-homology space). This computation is within the realms of plausibility. To rule out characteristic zero representations, we can pass by functoriality to the split side. So, if there is a characteristic zero class which is not Eisenstein mod-p, that residual representation also has to occur at low(ish) level inside the cohomology of $\mathrm{GL}_3(\mathbf{Z}[\sqrt{-2}])$. This is the sort of cohomology that people like Gunnells might almost be able to compute!

This entry was posted in Mathematics and tagged , , , , , , , , . Bookmark the permalink.

### 8 Responses to Virtual Congruence Betti Numbers

1. AV says:

Guess you meant “Euler characteristic” right after you mention dG-W? Interesting question and thoughts. What if you ask about the integral question, not just rational?

• Corrected! When you mean integral, however, I’m not entirely sure what question you have in mind. Showing the partially p-adically completed cohomology groups vanish is an integral question. Perhaps you mean:

(i). Can one say anything about l-torsion for l prime to p?

(ii). Can one show that the completed cohomology groups vanish using a Galois theoretic argument?

For (ii), I think what one gets from the Galois argument (now assuming Scholze+ and Local-Global compatibility) is that, after localizing at an Eisenstein maximal ideal $\mathfrak{m}$, the completed cohomology groups are themselves totally Eisenstein (in the strong sense). But then you are left with the problem of showing that any such Eisenstein class must come from a K-theoretic class, which is hard.

2. AV says:

Yeah, sorry, I had in mind (i), l-torsion for l prime to p (and assuming that rank(G)-rank(K) is at least 2.) I guess it’s probably inaccessible…

3. Wholesome Breakfast says:

A small correction: the X’s are not Shimura varieties.

4. Dear GR,

I think there’s a typo in the discussion of Rogawski’s result. You write that it’s an open question whether the SU(2,1) manifolds have virtual postive first Betti number, but then assert that they don’t (by Rogawski). Is it that Rogawski rules out the virtual congruence case, but the non-congruence case is open (related perhaps to not knowing the status of the congruence sg. property in this case)?

Cheers,

Matt

• That’s right. Benson knows something about the non-congruence case, btw.

5. Douglas Knight says:

What do you mean by a “rational X-homology sphere”? Do you mean a space whose rational homology is the same as that of X? Usually the term is a “rational homology X,” since there is nothing spherical about it.

• The extra “spheres” were just typos for “spaces”, now fixed.