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--- spin7_vanishing [2023/05/11 20:28] – 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 $\psi = c(\gamma^i) \psi_i$.
@@ Line 47: / Line 58: @@
 **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 $D^7 \psi = 0$ 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: