Converse to Lichnerowicz Vanishing Theorem?

Question

The Lichnerowicz vanishing theorem says that if on a compact 4-dimensional spin manifold there exists a metric whose scalar curvature $R>0$, then there are no harmonic spinors; $$D\psi=0 \implies \psi=0$$

This follows from the Bochner-type formula $$D^2=\nabla^*\nabla + \displaystyle\frac{R}{4}$$ by applying it to a spinor field $\psi$ and pairing with $\psi$ to obtain $$\int \, \rvert D\psi\rvert^2 \,dV = \int \, \vert \nabla\psi \rvert^2 \, dV + \displaystyle\frac{R}{4}\int \, \rvert \psi \rvert^2 \, dV$$ from which the statement follows directly.

My question is about whether there's a sort of converse to this reasoning. That is, suppose we have a compact Riemannian spin manifold and we know that there are no solutions to $D\psi = 0$ except the trivial one $\psi=0$. Does it follow that $R\ge 0$ (hopefully with $R>0$ at some point)?

My vague intuition for the plausibility of this is the following: Staying away from the zero section, we know that as you range over all spinor fields, the LHS of the integral formula above is strictly positive. Suppose $R<0$ at some point. Is there enough "flexibility" in our choice of spinor field to localize near this negative value and make the RHS negative (by making the second term dominate the first in absolute value) so as to generate a contradiction?

Answer 1

I think the answer is No. You are essentially asking the following: If $0$ is not an eigenvalue of the Dirac operator $D$ on a compact Riemannian manifold, then does the underlying Riemannian metric have non-negative scalar curvature?

Well, pick some Riemannian manifold whose scalar curvature is very much not non-negative (in any way that you like). If $0$ is not an eigenvalue of $D$, you have your counter example. If $0$ happens to be an eigenvalue of $D$, then perturb the metric slightly. The scalar curvature will remain very much not non-negative, but each eigenvalue of $D$ will (generically) move slightly. Since $D$ is an elliptic operator on a compact manifold, its spectrum is discrete. So the perturbed spectrum will avoid $0$ and hence give the desired counter example. The extra conditions in your question, being $4$-dimensional and spin, shouldn't affect this reasoning.

Answer 2

This is far from true! For a generic metric on a spin manifold of dimension at least 3, the kernel of the Dirac operator will be as small as it can be, subject to the index theorem. This was proved by Ammann-Dahl-Humbert (Surgery and harmonic spinors. Adv. Math. 220 (2009), no. 2, 523–539). In particular, the index of the Dirac operator on a 4-dimensional spin 4-manifold with signature $0$ is $0$. But there are many such $4$-manifolds that don't admit a metric of non-negative scalar curvature with positive curvature somewhere. The $4$-torus is an example, as is $S^1$ times any hyperbolic $3$-manifold; one can find similar examples in higher dimensions.

In dimension $4$, there are even simply-connected spin manifolds with signature $0$ that don't admit metrics of positive scalar curvature; this can be detected by the Seiberg-Witten invariants.