This is a weaker version of my previous (unanswered) question MO429574.
Let us start with a smooth, ample divisor $X$ in an abelian threefold $A$. It is a surface of general type such that $\Omega_X$ is globally generated, and so $\operatorname{Sym}^n \Omega_X$ is globally generated for all $n \geq 0$.
By [1, Proposition 13], for all $n \geq 0$ the restriction $$H^0(A, \, \operatorname{Sym}^n \Omega_A) \longrightarrow H^0(X, \, \operatorname{Sym}^n \Omega_X)$$ is an isomorphism, in particular we get $h^0(X, \, \operatorname{Sym}^n \Omega_X) = \frac{1}{2}(n+2)(n+1)$.
Let us consider now an arbitrary surface $X$ of general type with $\operatorname{Sym}^n \Omega_X$ globally generated. In all the examples that I am able to compute, the dimension of $H^0(X, \, \operatorname{Sym}^n \Omega_X)$ is never smaller than the number obtained above. So, let me ask the following
Question. Let $X$ be a surface of general type with $\operatorname{Sym}^n \Omega_X$ globally generated. Is it true that $$h^0(X, \, \operatorname{Sym}^n \Omega_X) \geq \frac{1}{2}(n+2)(n+1)?$$ Otherwise, what is a counterexample?
Every answer of reference to the relevant literature will be greatly appreciated.
Bibliography
1 Debarre, Olivier, Varieties with ample cotangent bundle, Compos. Math. 141, No. 6, 1445-1459 (2005); corrigendum ibid. 149, No. 3 (2013). ZBL1086.14038.
