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math214:04-06 [2020/04/06 09:05]
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 $$\Phi_T: \Omega^k(M, E) \to \Omega^{k+p}(M, E)$$ a $C^\infty(M)$-linear map. The exterior covariant derivative $\nabla (T) \in \Omega^{p+1}(M, \End(E))$ satisfies+$$ \nabla^{\End(E)}(F_\nabla) = 0 $$ 
 + 
 +(1) Formal proof. If $T \in \Omega^p(M, \End(E))$, then we have $$\Phi_T: \Omega^k(M, E) \to \Omega^{k+p}(M, E)$$ a $C^\infty(M)$-linear map. The exterior covariant derivative $\nabla (T) \in \Omega^{p+1}(M, \End(E))$ satisfies
 $$ \Phi_{\nabla(T)} = [\nabla, \Phi_T] = \nabla \Phi_T - (-1)^p \Phi_T \nabla $$ $$ \Phi_{\nabla(T)} = [\nabla, \Phi_T] = \nabla \Phi_T - (-1)^p \Phi_T \nabla $$
 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, did I miss a sign? ....+This seems too easy, did I miss a sign? (no..)
  
-Try again, using local presentation+(2) Try again, using local presentation
 $$ F = (d + A)^2 = dA + A \wedge A = dA + (1/2) [A, A] $$ $$ F = (d + A)^2 = dA + A \wedge A = dA + (1/2) [A, A] $$
 where in the last expression $\End(E)$ is viewed as a Lie algebra.  where in the last expression $\End(E)$ is viewed as a Lie algebra. 
  
 Then Then
-$$ \nabla(F) = [\nabla, F] = [d+A, dA + (1/2) [A, A]] = (1/2) d[A, A] + [A, dA] + [A, [A, A]] = [A, [A, A]] $$ +$$ \nabla(F) = [\nabla, F] = [d+A, dA + (1/2) [A, A]] = (1/2) d[A, A] + [A, dA] + (1/2) [A, [A, A]] = (1/2) [A, [A, A]] $$ 
-The last quantity is zero, by Jacobi identity. +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 
 +$$[A, [A, A]] = \sum_{i,j,k} dx^i \wedge dx^j \wedge dx^k [\Gamma_i, [\Gamma_j, \Gamma_k]] = 0$$ 
 +for example, the term with $dx^1 \wedge dx^2 \wedge dx^3$ has (2 times) 
 +$$ [\Gamma_1, [\Gamma_2, \Gamma_3]] + [\Gamma_2, [\Gamma_3, \Gamma_1]] + [\Gamma_3, [\Gamma_1, \Gamma_2]] = 0. $$ 
 + 
 + 
  
-===== 
  
  
  
math214/04-06.1586189153.txt.gz · Last modified: 2020/04/06 09:05 by pzhou

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