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--- math214:04-06 [2020/04/06 07:47]
pzhou [Geometric Meaning of Curvature]
+++ math214:04-06 [2020/04/06 10:35] (current)
pzhou [Bianchi Identity]
@@ Line 2: / Line 2: @@
 $$\gdef\xto\xrightarrow \gdef\End{\z{ End}}$$
-==== Geometric Meaning of Curvature ====
+==== Summary of Connection and Curvature ====
 Let $E$ be a vector bundle over a smooth manifold. The connection $\nabla$ is given as such
 $$ \Omega^0(M, E) \xto{ \nabla} \Omega^1(M, E) \xto{ \nabla}  \Omega^2(M, E) \cdots.  $$
@@ Line 10: / Line 10: @@
 Locally, if we have a base coordinates $x_i$, and local trivialization $e_\alpha$ of $E$, then we may write the connection as
-$$ \nabla = d +  \Gamma_i^\alpha_\beta dx^i \otimes e_\alpha \otimes \delta^\beta $$
+$$ \nabla = d +  [\Gamma_i]^\alpha_\beta dx^i \otimes e_\alpha \otimes \delta^\beta $$
 where $\{\delta^\alpha\}$ is the dual basis to $\{e_\alpha\}$, and $e_\alpha \otimes \delta^\beta$ is a local section of $\End(E)$.
 $$ F_\nabla = \nabla^2 = (F_{ij})^\alpha_\beta dx^j \wedge d x^i \otimes e_\alpha \otimes \delta^\beta$$
@@ Line 16: / Line 16: @@
 $$ [F_{ij}]^\alpha_\beta = \d_i [\Gamma_j]^\alpha_\beta -  \d_j [\Gamma_i]^\alpha_\beta + [\Gamma_i]^\alpha_\gamma [\Gamma_j]^\gamma_\beta - [\Gamma_j]^\alpha_\gamma [\Gamma_i]^\gamma_\beta $$
 If we view $[--]^\alpha_\beta$ as $(\alpha, \beta)$ entry of an $n \times n$ matrix, then the above can be written as an equation matrix valued forms
-$$ F_{ij} = \d_i \Gamma_j - \d_j \Gamma_i + \Gamma_i \Gamma_j - \Gamma_j \Gamma_i $$
+$$ F_{ij} = \d_i \Gamma_j - \d_j \Gamma_i + \Gamma_i \Gamma_j - \Gamma_j \Gamma_i. $$
+==== Holonomy Around the Loop ====
+Recall that given a path $\gamma: [0,1] \to M$, we can define parallel transport
+$$ P_\gamma: E_{\gamma(0)} \to E_{\gamma(1)}.$$
+In particular, if the path is a loop, ie., $\gamma(0)=\gamma(1)$, then we have an endormorphism
+$ P_\gamma \in \End(E_{\gamma(0)})$. This is called the holonomy of the loop.
+==== Geometric Picture ====
+What is the geometric meaning of curvature?
+Suppose we are given a local presentation of $\nabla, F_\nabla$ as above. For simplicity, assume $M=\R^2$,  then we claim that
+$$ F_{12} = - \lim_{\epsilon, \delta \to 0} \frac{1}{\epsilon \delta} P_\gamma $$
+where $\gamma$ is the loop around the boundary of the small square $[0, \epsilon] \times [0, \delta]$ counter-clockwise.
+To see this, we break down the loop into four segments, and denote the parallel transports on the four segments as $P_i, i=1, \cdots, 4$, and we work out the expansion modulo $\epsilon^2, \delta^2$ terms
+From $(0,0)$ to $(\epsilon,0)$, say we have path $\gamma_1(t) = (t, 0)$ for $t \in [0, \epsilon]$.  We need to solve equation
+$$ \d_t u^\alpha(t) + [\Gamma_i]^\alpha_\beta \dot \gamma_1^i(t) u^\beta(t) = 0 $$
+since $\dot \gamma_1^i(t) = 1$ only if $i=1$, thus
+$$  \d_t u^\alpha(t) = - [\Gamma_1]^\alpha_\beta(0, t) u^\beta(t) $$
+Approximately, we have
+$$ [u](\epsilon) \approx (1 - \epsilon \Gamma_1(0,0)) [u](0). $$
+The parallel transport along the first segment is $$P_1 \approx 1 - \epsilon \Gamma_1(0,0),$$
+Similarly, we have
+$$ P_2 \approx 1 - \delta \Gamma_2(\epsilon, 0), \quad P_3 =  1 + \epsilon \Gamma_1(0,\delta) \quad P_4 \approx 1 + \delta \Gamma_2 (0,0) $$
+Using Taylor expansion for $\Gamma$ at $(0,0)$,
+$$ P_4 P_3 P_2 P_1 \approx (1 + \delta \Gamma_2 ) (1 + \epsilon \Gamma_1  - \epsilon\delta \d_2 \Gamma_1 ) (1 - \delta \Gamma_2 - \delta \epsilon \d_1 \Gamma_2) (1 - \epsilon \Gamma_1)|_{(0,0)} $$
+$$  \approx 1 - \epsilon \delta (\d_1 \Gamma_2 - \d_2 \Gamma_1 + \Gamma_1 \Gamma_2 - \Gamma_2 \Gamma_1)|_{(0,0)}. $$
+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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