Manifolds with trivial mapping class group and large $H^1$? Are there smooth closed manifolds $M^n$ in every dimension $n \geq 3$ with trivial mapping class groups and with $H^1(M^n;\mathbb{Z}/2\mathbb{Z})$ arbitrarily large?
I am under the impression that "generically" a manifold will have no mapping class group (but maybe I am totally mistaken).  There are presumably lots of constructions of such manifolds with trivial mapping class groups.  So I guess I'm hoping to hear of some such construction where $H^1(M^n;\mathbb{Z}/2\mathbb{Z})$ can get very large.
 A: I think that the construction in Belolipetsky--Lubotzky, Finite Groups and Hyperbolic Manifolds (https://arxiv.org/abs/math/0406607) provides a more algebraic construction of such manifolds for any $n$.
One of the results in this paper (Theorem 3.1) is the following : given groups $\Gamma \triangleright \Delta \triangleright M$ with $\Gamma/\Delta$ finite and $F = \Delta/M$ a nonabelian free group, there exists $\Gamma \ge D \ge \Delta$ and $B \le D$ of finite index with $N_\Gamma(B) = B$ (in fact they prove that for any finite group $G$ there are infinitely many $B$ with $N_\Gamma(B)/B$ isomorphic to $G$).
They use this to construct closed hyperbolic manifolds with trivial isometry group ; by Mostow rigidity the mapping class group of $X$ is isomorphic to the isometry group of $X$ (every homeomorphism is isotopic to an isometry), so these manifolds will have trivial mapping class group. Without too much details their construction uses a specific lattice $\Gamma$ in the isometry group of hyperbolic $n$-space $\mathbb H^n$, and subgroups $D, \Delta, M, B$ as in the previous paragraph ; the manifold is the quotient of $\mathbb H^n$ by $B$ (their construction ensures that $B$ is torsion-free though $\Gamma$ will likely not be).
Now to look at Betti numbers, for which we need to dive a bit more into the details of this paper. Since $X$ is aspherical we have
$$b_1(X, \mathbb Q) = b_1(B, \mathbb Q) \ge b_1(D, \mathbb Q).$$
On the other hand $b_1(D, \mathbb Q)$ is the dimension of the fixed subspace of $D/\Delta$ in $H^1(\Delta, \mathbb Q)$ (conjugation action on the normal subgroup $\Delta$ and its morphisms to $\mathbb Z$). By the definition of $D$ in the paper of Belolipetsky--Lubotzky (p. 4 in the arxiv version) we have that it acts trivially on the part of the cohomology coming from the surjective morphism $\Delta \to F$. So we have that $b_1(X, \mathbb Q)$
is larger than the rank of $F$. It remains to remark that in the data for Theorem 3.1 the quotient $\Delta/M$ can be taken to have arbitrarily large rank (since a non-abelian free group contains free subgroups of finite index with arbitrarily large rank), so we get examples with $b_1(B, \mathbb Q)$ arbitrarily large.
A: Here is one such construction in dimension three. Suppose that $G$ is a finite, connected, simple graph without symmetries. We build a manifold $M^3$ by taking manifolds with boundary for the vertices of $G$ and gluing them in pairs “along” the edges.  We arrange matters so that the vertex manifolds are hyperbolic and with totally geodesic boundary. We also require that the vertex manifolds have no reflection symmetries. Finally we glue boundary components in pairs according to the edges of the graph, using high powers of pseudo-Anosov maps.
If the power is high enough, then any mapping class of $M$ sends vertex manifolds to each other, and so preserves the graph structure. Thus the manifold $M$ has no nontrivial mapping classes.
We can arrange also for each vertex manifold to contribute to homology.  Thus the rank of the first (co)homology is linear in the number of vertex manifolds.
Note that, with a bit more work, such manifolds are hyperbolic.  Thus we know that their mapping class groups are finite.  The messing about (with high powers of pseudo-Anosov maps) is needed to enforce “rigidity”: that is, to reduce the analysis from symmetries of $M$ to symmetries of $G$.
