Lichnerowicz vanishing theorem

Remarks

Let $\psi$ be a spinor with $D\psi = 0$. Integrating the Lichnerowicz formula $D^2 = \nabla^* \nabla + \frac{R}{4}$ against $\psi$ gives $0 = \int_M \abs{\nabla \psi}^2 + \frac{1}{4} \int_M R \abs{\psi}^2$. Both terms are nonnegative and $R > 0$, so $\psi = 0$. By the Atiyah–Singer index theorem, $\hat{A}(M)$ is the index of the positive part of the Dirac operator, $\dim \ker D^+ - \dim \ker D^-$, which is then $0$. The argument works as well when $R \ge 0$ everywhere and $R > 0$ at some point.

Example: a K3 surface is a closed spin $4$-manifold with signature $-16$, so $\hat{A} = -\frac{\sigma}{8} = 2 \ne 0$. Hence a K3 surface has no metric of positive scalar curvature.

The spin hypothesis is necessary. The complex projective plane $\C P^2$ with the Fubini–Study metric has positive scalar curvature, but $\hat{A}(\C P^2) = -\frac{1}{8}$, computed from its Pontryagin number $p_1 = 3$ by $\hat{A} = -\frac{p_1}{24}$. This is consistent because $\C P^2$ is not spin.

Going deeper

For a student

This is the first and basic obstruction to metrics of positive scalar curvature. Hitchin refined it to a $KO$-theoretic invariant. Gromov and Lawson extended the method to Dirac operators twisted by bundles, which shows for example that tori admit no metric of positive scalar curvature. For simply connected manifolds of dimension at least $5$, the refined invariant decides the question for spin manifolds.