Let ${\mathcal M}\_S$ be the moduli space of (a family of) minimal smooth algebraic surfaces. (A precise definition does not matter here.) Denote $M=\dim {\mathcal M}\_S$.
 Is it true that 
$$M\le b\_2-p\_g-2,$$
where $b\_2$ is the second Betti number and $p\_g$ is the geometric genus?
(Excluding the case when RHS is $-1$. I am not referring to a fancy dimension, so LHS cannot be negative.) 


I arrived at this looking for surfaces with large moduli spaces. The inequality is sharp for Abelian surfaces and K3 surfaces, but in other cases there seems to be a margin.



A bound for the number of moduli of a surface?
Let $k$ be a field of characteristic prime to $5$. Denote $\text{Proj}\ k[x\_0,x\_1,x\_2,x\_3]$ with $\text{deg}(x\_i)=1$ by $\mathbb{P}^3\_k.$ Denote by $G$ the following copy of $\mu\_5$ inside $\textbf{PGL}\_4$, $$\rho:\mu\_5 \to \textbf{PGL}\_4, \ \ \zeta\cdot[x\_0,x\_1,x\_2,x\_3] = [\zeta^0x\_0,\zeta^1x\_1,\zeta^2x\_2,\zeta^3x\_3].$$ The following quintic Fermat surface $Y$ is invariant under this action, $$Y = \text{Zero}(x\_0^5+x\_1^5+x\_2^5+x\_3^5) \subset \mathbb{P}^3\_k.$$ The induced action of $G$ on $Y$ is free. Denote the quotient by this action as follows, $$f:Y\to X.$$ This is a finite, étale morphism of degree $5$. The surface $X$ is a **Godeaux surface**. The dimensions $h^q(X,\Omega^p\_{X/k})$ are as follows, $$ \begin{array}[ccccc] & & & 1 & & \\ & 0 & & 0 & \\ 0 & & 9 & & 0 \\ & 0 & & 0 & \\ & & 1 & & \end{array}.$$ In particular, in characteristic $0$, the second Betti number equals $9$ and the geometric genus equals $0$, so that $b\_2-p\_g-2$ equals $9-0-2$, i.e., $7$.


On the other hand, by the Griffiths residue calculus (or other means), the group $H^1(X,T\_X) = H^1(Y,T\_Y)^G$ is the $G$-invariant vector subspace of the degree $5$ graded piece of the Jacobian ring, $$k[x\_0,x\_1,x\_2,x\_3]/\text{Jac}(Y) = k[x\_0,x\_1,x\_2,x\_3]/\langle x\_0^4,x\_1^4,x\_2^4,x\_3^4 \rangle.$$ The $G$-invariant subspace has basis consisting of monomials $x\_0^{e\_0}x\_1^{e\_1}x\_2^{e\_2}x\_3^{e\_3}$ such that $e\_0+e\_1+e\_2+e\_3=5$, such that every $e\_i \leq 3$, and such that $0e\_0+1e\_1+2e\_2+3e\_3$ is divisible by $5$. It is straightforward to check that this defines the following $8$ monomials,
$$\{ x\_3^3x\_1x\_0, x\_3^2x\_2^2x\_0,x\_3^2x\_2x\_1^2,x\_3x\_2^3x\_1,x\_3x\_2x\_0^3,x\_3x\_1^2x\_0^2, x\_2^2x\_1x\_0^2,x\_2x\_1^3x\_0\}.$$ Thus, $h^1(X,T\_X)$ equals $8$. Also, by Serre duality, $$h^3(\mathbb{P}^3\_k,T\_{\mathbb{P}^3}(-\underline{Y})) = h^0(\mathbb{P}^3\_k,\omega\_{\mathbb{P}^3/k}\otimes \Omega\_{\mathbb{P}^3/k}(\underline{Y})) = h^0(\mathbb{P}^3\_k, \Omega\_{\mathbb{P}^3/k}(1)).$$ Using the Euler sequence, this last cohomology group is zero. Thus, also $H^2(Y,T\_Y)$ equals $0$ (maybe there is a faster computation of that), so that $H^2(Y,T\_Y)^G$ is zero. In other words, $H^2(X,T\_X)$ is zero. Therefore, the infinitesimal deformations of $X$ are unobstructed. Finally, $H^0(Y,T\_Y)$ equals $0$, so that also $H^0(X,T\_X)=H^0(Y,T\_Y)^G$ is zero. 


Altogether, the versal deformation space of $X$ is smooth of dimension $8$, without any infinitesimal automorphisms. So the integer $M$ equals $8$. Yet $8$ is strictly greater than $7$.


