Let $X^4$ be a closed spin 4-manifold with $b_2^+\geq 2$ and let $$ SW_X:\mathrm{Spin}^c(X)\to \mathbb{Z} $$ be the Seiberg-Witten map. If either $\hat{A}(X)\neq 0$ or $SW_X\not \equiv 0$, then $X$ admits no metric of positive scalar curvature. (See any reference on SW equations, e.g. Morgan or Moore.)
My question is whether in dimension 4, the Seiberg-Witten invariants provide a "stronger" obstruction to positive scalar curvature. More precisely,
Question: Let $X^4$ be a closed spin 4-manifold with $b_2^+\geq 2$ and at least one SW basic class, i.e. $SW_X\not\equiv 0$. Then must we have $\hat{A}(X)\neq 0$?
My idea: For spin 4-manifolds, $\hat{A}(X)$ is related to the signature: $$ \hat{A}(X)=-\frac{1}{8}\sigma(X). $$ So the answer to the question is negative if and only if there exists a closed spin 4-manifold satisfying
- $b_2^+\geq 2$,
- $\sigma(X)=0$, and
- $SW_X\not \equiv 0$.
Akhmedov and Park have constructed exotic copies of $S^2\times S^2$ with nontrivial SW invariants, but $b_2^+(S^2\times S^2)=1$, so this doesn't give a counterexample.
Maybe combining the formula for the dimension of the SW moduli space with the index formula of the Dirac operator $D^+_A$ would imply that $\sigma(X)$ cannot vanish if $SW_X\neq 0$.
Thanks for your help.