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math214:04-06 [2020/04/06 09:00] pzhou |
math214:04-06 [2020/04/06 10:35] (current) pzhou [Bianchi Identity] |
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===== Bianchi Identity ===== | ===== Bianchi Identity ===== | ||
- | If $T \in \Omega^p(M, \End(E))$, then we have $$\Phi_T: \Omega^k(M, E) \to \Omega^{k+p}(M, | + | $$ \nabla^{\End(E)}(F_\nabla) = 0 $$ |
+ | |||
+ | (1) Formal proof. | ||
$$ \Phi_{\nabla(T)} = [\nabla, \Phi_T] = \nabla \Phi_T - (-1)^p \Phi_T \nabla $$ | $$ \Phi_{\nabla(T)} = [\nabla, \Phi_T] = \nabla \Phi_T - (-1)^p \Phi_T \nabla $$ | ||
that is, for a section $u \in \Omega^k(M, E)$, we have | that is, for a section $u \in \Omega^k(M, E)$, we have | ||
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Now, take $T = F_\nabla \in \Omega^2(M, \End(E))$, we need to show that $\nabla(F_\nabla) = [\nabla, \nabla^2] = 0$. done. | Now, take $T = F_\nabla \in \Omega^2(M, \End(E))$, we need to show that $\nabla(F_\nabla) = [\nabla, \nabla^2] = 0$. done. | ||
- | This seems too easy, I might have a sign wrong somewhere | + | This seems too easy, did I miss a sign? (no..) |
+ | |||
+ | (2) Try again, using local presentation | ||
+ | $$ F = (d + A)^2 = dA + A \wedge A = dA + (1/2) [A, A] $$ | ||
+ | where in the last expression $\End(E)$ is viewed as a Lie algebra. | ||
+ | |||
+ | Then | ||
+ | $$ \nabla(F) = [\nabla, F] = [d+A, dA + (1/2) [A, A]] = (1/2) d[A, A] + [A, dA] + (1/2) [A, [A, A]] = (1/2) [A, [A, A]] $$ | ||
+ | The last quantity is zero, by Jacobi identity, to be more explicity, we write | ||
+ | $$A = \sum_i dx^i \ot \Gamma_i, \quad \Gamma_i \in M_n(\R)$$ | ||
+ | Then | ||
+ | $$[A, [A, A]] = \sum_{i, | ||
+ | for example, the term with $dx^1 \wedge dx^2 \wedge dx^3$ has (2 times) | ||
+ | $$ [\Gamma_1, [\Gamma_2, \Gamma_3]] + [\Gamma_2, [\Gamma_3, \Gamma_1]] + [\Gamma_3, [\Gamma_1, \Gamma_2]] = 0. $$ | ||
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