Notes

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 $\psi = c(\gamma^i) \psi_i$. Moreover, $c(\gamma^j) c(\gamma^m) \psi = -16 \delta_{jm} \psi$.

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

Suppose $j \neq k$. Then \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 $c(\gamma^{jk}) \psi = 0$. Skew-adjointness gives that $c(\gamma^j) c(\gamma^m) \psi = 0$ if $j \neq m$.

For $j = k$ one requires an eigendecomposition of $\psi$. It is a fact that the eigenvalues of $c(\gamma^k)$ acting on $S^+$ are $\pm 4i$ with multiplicity 1 for each sign and 0 with multiplicity 6. Thus $\psi = \psi_{+} + \psi_{-} + \psi_0$ splits as a sum of eigenvectors for a fixed $k$. However, $\psi_0 = 0$;

Thus $\psi = \psi_+ + \psi_-$ and $c(\gamma^k)^2 \psi = (4i)^2 \psi_+ + (-4i)^2 \psi_{-} = -16\psi$.

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

We have $|D\psi|^2 = |D^7 \psi|^2 + |D^{21}\psi|^2 + 2\langle D^7\psi, D^{21}\psi\rangle$. It suffices (by the assumption that Clifford multiplication commutes with derivatives) to consider $\psi \in S^+_1$; but then the cross term vanishes a priori

Lemma 4: If $D^7 \psi = 0$ then $c(\gamma^j e_a) \nabla_a \psi_j = 0$ for all $j$ and $a$. If $D^{21} \psi = 0$ then $c(\gamma^{jk} e_a) \nabla_a \psi_k = 0$ for all $j, k$ and $a$.

Proof: Compute:

\begin{align*} 0 &= \frac{28^2}{16^2}|D^7 \psi|^2 \\ &= \frac{1}{16^2} \langle c(\gamma^j e_a) c(\gamma^j) \nabla_a c(\gamma^m) \psi_m, c(\gamma^k e_b) c(\gamma^k) \nabla_b c(\gamma^\ell) \psi_\ell \rangle \\ &= \langle c(\gamma^j e_a) \nabla_a \psi_j, c(\gamma^k e_b) \nabla_b \psi_k \rangle \\ &= |c(\gamma^j e_a) \nabla_a \psi_j|^2 \end{align*}