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--- math214:04-15 [2020/04/15 00:15]
pzhou created
+++ math214:04-15 [2020/04/15 00:33]
pzhou
@@ Line 15: / Line 15: @@
 $$ [X_\gamma, X_\sigma] = \nabla_{X_\gamma} X_\sigma - \nabla_{X_\sigma} X_\gamma = \omega_\sigma^\beta(X_\gamma) X_\beta - \omega_\gamma^\beta(X_\sigma) X_\beta$$
 Hence
-$$- \theta^{\alpha}([X_\gamma, X_\sigma]) = - \omega_\sigma^\alpha(X_\gamma)  + \omega_\gamma^\alpha(X_\sigma)$$
+$$LHS = - \theta^{\alpha}([X_\gamma, X_\sigma]) = - \omega_\sigma^\alpha(X_\gamma)  + \omega_\gamma^\alpha(X_\sigma)$$
 And on RHS
 $$RHS = \theta^\beta \wedge \omega_\beta^\alpha ( X_\gamma, X_\sigma) = \theta^\beta(X_\gamma)\omega_\beta^\alpha(X_\sigma) - \theta^\beta(X_\sigma)\omega_\beta^\alpha(X_\gamma)= \omega_\gamma^\alpha(X_\sigma) - \omega_\sigma^\alpha(X_\gamma)$$
+How to interpret this $\omega^\alpha_\beta$? A choice of frame $X_\alpha$ induces a local trivialization
+of $E$, so we have a local trivial connection $d$, it acts on local section $u \In \Gamma(U, TM)$ as
+$$u = u^\alpha X_\alpha, \quad d(u) = d(u^\alpha) \otimes X_\alpha \in \Omega^1(U, TM) $$
+That is, $d X_\alpha = 0$, $d$ views $X_\alpha$ as constant section. Thus
+$$ \nabla = d + A $$
+where $A$ is a matrix valued (or $End(TM)$ valued) 1-form, we get
+$$ \nabla X_\alpha = A (X_\alpha) = \omega_\alpha^\beta X_\beta $$
+$A$ is valued not just in $End(E)$, but in $\mathfrak{so}(E)$, since the connection preserves length (inner product), that explains why $\omega_\alpha^\beta$ is anti-symmetric in the indices.
+Given this understanding, we have a local matrix-valued 2-form
+$$ R^\alpha_\beta = d \omega^\alpha_\beta + \omega^\alpha_\gamma \wedge \omega^\gamma_\beta. $$
+==== Application: Orthogonal Coordinate ====
+Suppose we have coordinate that looks like the following
+$$ g_{ij} = h_i^2 \delta_{ij} $$
+then, we may choose the orthonormal frame $X_i$ as (no summation of repeated indices here)
+$$X_i = \frac{1}{h_i} \d_{x^i}, \theta^i = h_i dx^i $$
+Then, we can get $\omega_i^j$ by solving
+$$ d \theta^i = d(h_i) \wedge dx^i $$
+Now, one need to figure out what is $\omega_i^j$ in each specific cases. Once that is done, one can easily get the curvature.

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