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

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).

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.