When does a locally symmetric space have no odd degree Betti numbers?

Question

Let $G$ be a semisimple real lie group, $K$ be a maximal compact subgroup of $G$, $\Gamma$ be a torsion-free cocompact discrete subgroup. The Betti number the locally symmetric space $X_{\Gamma}:=\Gamma \backslash G/K$ can be expressed in term of unitary representations of $G$ by Matsushima's formula.

The formula looks quite complicated in general, the question is: when does $X_{\Gamma}$ have all vanishing odd degree Betti numbers? I am interested in some examples and some necessary conditions.

Example: Consider the toroidal compactification of Siegel threefold with $3$ level structure, its Betti numbers are $1, 0, 61, 0, 61, 0, 1$, see "The Siegel modular variety of degree two and level three" by J. William Hoffman and Steven H. Weintraub (Trans AMS 2001).

Answer 1

Here is an answer of sorts.

For simplicity, I restrict to cocompact torsion-free lattices. Then $b_{odd}(X_\Gamma)=0$ implies that $X$ is even-dimensional. The situation is clear if $dim(X)\le 2$, so the first interesting case is when $dim(X)=4$. There are (essentially) three 4-dimensional symmetric spaces of noncompact type (up to rescaling the metric on de Rham factors): The real-hyperbolic space, ${\mathbb H}^4$ (when $G=PO(4,1)$), the complex-hyperbolic space ${\mathbb H}^2_{\mathbb C}$ (when $G=PU(2,1)$) and the product of two hyperbolic planes ${\mathbb H}^2\times {\mathbb H}^2$ (when $G=PO(2,1)\times PO(2,1)$), up to commensuration of Lie groups.

1. $X={\mathbb H}^4$. I do not know if there are examples with $b_1=0$. However, it is known that for every arithmetic uniform lattice $\Gamma< PO(4,1)$ there exists a congruence-subgroup $\Gamma_1<\Gamma$ such that $b_1(\Gamma_1)= b_1(X_{\Gamma_1})>0$ (Millson). I do not think anybody knows what happens with non-arithmetic manifolds.

2. $X={\mathbb H}^2_{\mathbb C}$. Arithmetic lattices come in two types: All lattices of type I again have congruence-subgroups $\Gamma_1$ with $b_1(\Gamma_1)>0$ (Kazhdan). However, for all congruence-subgroups $\Gamma_1$ in type II arithmetic lattice, $b_1(\Gamma_1)=b_3(\Gamma_1)=0$. So all odd Betti numbers vanish. The question of positivity of 1st Betti number of finite-index non-congruence subgroups is wide-open. The non-arithmetic case is mostly open. (There is one nonvanishing result, due to Yeung.)

3. $X={\mathbb H}^2\times {\mathbb H}^2$. The case of reducible lattices is (essentially) clear, so I will consider irreducible lattices. For all such lattices $\Gamma$, $b_1(\Gamma)=b_3(\Gamma)=0$. (This follows from Margulis' super-rigidity, but might have been known earlier.) So all odd Betti numbers vanish.

I will not attempt to analyze the case when $dim(X)\ge 6$. One gets vanishing results for $b_1$, but the situation with higher odd Betti numbers are very much unclear, unless you know something about odd Betti numbers of the dual compact metric space: Nonvanishing of some of those implies nonvanishing of $b_{odd}(\Gamma)$. There are vanishing theorems going back to Matsushima, but they do not deal with all odd Betti numbers.