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--- spin7_vanishing [2023/05/11 17:30] – spencer
+++ spin7_vanishing [2023/05/15 10:32] (current) – spencer
@@ Line 12: / Line 12: @@
 $$D^7 := \frac{1}{28} \sum_{j=1}^7 \sum_{b=1}^8 c(\gamma^j e_b) c(\gamma^j) \nabla_b \tag{2}$$
 $$D^{21} := \frac{1}{56} \sum_{1 \le j < k \le 7} \sum_{b=1}^8 c(\gamma^{jk} e_b) c(\gamma^{jk}) \nabla_b \tag{3}$$
+<WRAP center round info 60%>
+Here we are working with the convention that for a skew operator $A$, we have $c(A) = \frac{1}{2} A_{ab} c_a c_b$.
+</WRAP>
 **Lemma 1:** $D = D^7 + D^{21}$
 <WRAP center round todo 60%>
-**Proof:** Elided, for now
+**Proof:** We have
+\begin{align*}
+D &= c_b \nabla_b \\
+&= -\frac{1}{8} c_a c_a c_b \nabla_b \\
+&= \frac{1}{8} D - \frac{1}{8} \sum_{a \neq b} c_a c_a c_b \nabla_b
+\end{align*}
+so
+$$D = -\frac{1}{7} \sum_{a \neq b} c_a c_a c_b \nabla_b$$
 </WRAP>
 **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 \\
-&= \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 $j \neq m$.
-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 $S^+_7$.
+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, $\psi_0 = 0$;
+<WRAP center round todo 60%>
+Finish proof that zero part vanishes
+</WRAP>
+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
 <WRAP center round todo 60%>
-**Proof:** Elided, for now
+do this more carefully later
 </WRAP>
-**Lemma 4:** If $D^7 \psi = 0$
+**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*}
+&= \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*}
+<WRAP center round tip 60%>
+We know $\tau_{aj} := \nabla_a \psi_j \in S_1^+$ since $\nabla$ preserves the decomposition.
+If it is non-zero, then $c(\gamma^k) \tau_{aj}$ span $S_7^+$ by Lemma 2.
+So idea is $c(\gamma^j e_a)$, $c(\gamma^{jk} e_a)$ should span everything and in particular one of them should contain something in $S_7^+$, so this is our contradiction.
+</WRAP>