Notes

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Mark has a proof of vanishing for $\mathrm{Spin}(7)$. I think it's correct. Here's a careful statement and an attempt at a proof.

Theorem: Suppose $(E,A) \to X$ is a (hermitian) vector bundle with (compatible) instanton connection over a $\mathrm{Spin}(7)$ base (that is, $F^7 = 0$). There are no harmonic spinors in $\Gamma(S_7^+ \otimes E)$. That is, if $\psi \in \Gamma(S_7^+ \otimes E)$ and $D\psi = 0$, then $\psi = 0$.

Through the proof it suffices to work in a nice frame $\{e_1, \ldots, e_8\}$ of $TM$ around some point $x$ chosen so that at $x$, Clifford multiplication by the standard basis $\gamma^i$ of $\Lambda^2_7$ is covariant constant. That is, at $x$ $$[c(\gamma^i), \nabla] = 0. \tag{1}$$ We define $$D^7 := \frac{1}{28} \sum_{j=1}^7 \sum_{b=1}^8 c(\gamma^j e_b) c(\gamma^j) \nabla_b \tag{2}$$ $$D^{21} := \frac{1}{56} \sum_{1 \le j < k \le 7} \sum_{b=1}^8 c(\gamma^{jk} e_b) c(\gamma^{jk}) \nabla_b \tag{3}$$

Lemma 1: $D = D^7 + D^{21}$

Lemma 2: If $\psi \in \Gamma(S_7^+ \otimes E)$, there exist 7 sections $\psi_i \in \Gamma(S_1^+ \otimes E)$ such that $\psi = c(\gamma^i) \psi_i$.

Proof: Essentially, this is a consequence of the following orthogonality relation: for unit $\psi \in S^+_1$,

\begin{align*} \langle c(\gamma^j) \psi, c(\gamma^k)\psi \rangle &= \frac{1}{4} \langle [c(\gamma^{jk}), c(\gamma^k)] \psi, c(\gamma^k) \psi \rangle \\ &= \frac{1}{4}\langle c(\gamma^{jk})c(\gamma^k) \psi, c(\gamma^k) \psi \rangle \\ &= 0 \end{align*} since, for example, Clifford multiplication is skew-adjoint. Note we used that $c(\gamma^{jk}) \psi = 0$.

Moreover, note that $|c(\gamma^j) \psi|^2 = |\psi|^2 = 1$, so the $c(\gamma^j)\psi$ are an orthogonal set of the right dimension inside $S^+_7$.

Lemma 3: $\ker D = \ker D^7 \cap \ker D^{21}$, the kernels being in $\Gamma(S^+ \otimes E)$.

Lemma 4: If $D^7 \psi = 0$