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--- math214:04-06 [2020/04/06 08:28]
pzhou
+++ math214:04-06 [2020/04/06 10:35] (current)
pzhou [Bianchi Identity]
@@ Line 39: / Line 39: @@
 $$  \d_t u^\alpha(t) = - [\Gamma_1]^\alpha_\beta(0, t) u^\beta(t) $$
 Approximately, we have
-$$ [u](\epsilon) \approx (1 + \epsilon \Gamma_1(0,0)) [u](0). $$
+$$ [u](\epsilon) \approx (1 - \epsilon \Gamma_1(0,0)) [u](0). $$
-The parallel transport along the first segment is $$P_1 \approx 1 + \epsilon \Gamma_1(0,0),$$
+The parallel transport along the first segment is $$P_1 \approx 1 - \epsilon \Gamma_1(0,0),$$
 Similarly, we have
-$$ P_2 \approx 1 + \delta \Gamma_2(\epsilon, 0), \quad P_3 =  1 - \epsilon \Gamma_1(0,\delta) \quad P_4 \approx 1 - \delta \Gamma_2 (0,0) $$
+$$ P_2 \approx 1 - \delta \Gamma_2(\epsilon, 0), \quad P_3 =  1 + \epsilon \Gamma_1(0,\delta) \quad P_4 \approx 1 + \delta \Gamma_2 (0,0) $$
 Using Taylor expansion for $\Gamma$ at $(0,0)$,
-$$ P_4 P_3 P_2 P_1 \approx (1 - \delta \Gamma_2 ) (1 - \epsilon \Gamma_1  - \epsilon\delta \d_2 \Gamma_1 ) (1 + \delta \Gamma_2 + \delta \epsilon \d_1 \Gamma_2) (1+\epsilon \Gamma_1)|_{(0,0)} $$
+$$ P_4 P_3 P_2 P_1 \approx (1 + \delta \Gamma_2 ) (1 + \epsilon \Gamma_1  - \epsilon\delta \d_2 \Gamma_1 ) (1 - \delta \Gamma_2 - \delta \epsilon \d_1 \Gamma_2) (1 - \epsilon \Gamma_1)|_{(0,0)} $$
+$$  \approx 1 - \epsilon \delta (\d_1 \Gamma_2 - \d_2 \Gamma_1 + \Gamma_1 \Gamma_2 - \Gamma_2 \Gamma_1)|_{(0,0)}. $$
+Hence we are done. See also [Ni] 3.3 for a more rigorous derivation.
+===== Bianchi Identity =====
+$$ \nabla^{\End(E)}(F_\nabla) = 0 $$
+(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
+$$ \Phi_{\nabla(T)} = [\nabla, \Phi_T] = \nabla \Phi_T - (-1)^p \Phi_T \nabla $$
+that is, for a section $u \in \Omega^k(M, E)$, we have
+$$ [\nabla(T)](u) = \nabla (T \wedge u) - (-1)^p T \wedge (\nabla u). $$
+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? (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
+$$[A, [A, A]] = \sum_{i,j,k} dx^i \wedge dx^j \wedge dx^k [\Gamma_i, [\Gamma_j, \Gamma_k]] = 0$$
+for example, the term with $dx^1 \wedge dx^2 \wedge dx^3$ has (2 times)
+$$ [\Gamma_1, [\Gamma_2, \Gamma_3]] + [\Gamma_2, [\Gamma_3, \Gamma_1]] + [\Gamma_3, [\Gamma_1, \Gamma_2]] = 0. $$

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