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2020-04-15, Wednesday

Cartan's Moving Frame

Pick any Orthonormal Frame $X_\alpha$ of $TM$, and choose its dual frame $\theta^\alpha$ of $T^*M$. Introduce a collection of 1-forms using covariant derivatives $$ \nabla X_\alpha = \omega_\alpha^\beta X_\beta $$

Proposition : $$ \omega_\alpha^\beta = - \omega_\beta^\alpha $$ $$ d \theta^\alpha = \theta^\beta \wedge \omega_\beta^\alpha $$ Proof: (1) For any vector field $Y$, we have $$ \la \nabla_Y X_\alpha, X_\beta \ra + \la X_\alpha, \nabla_Y X_\beta \ra = 0. $$ (2) Evaluate on $X_\gamma, X_\sigma$, we get $$ LHS=d \theta^\alpha(X_\gamma, X_\sigma) = X_\gamma(\theta^\alpha (X_\sigma)) - X_\sigma(\theta^\alpha(X_\gamma)) - \theta^{\alpha}([X_\gamma, X_\sigma]) = - \theta^{\alpha}([X_\gamma, X_\sigma])$$ Then, we use $$ [X_\gamma, X_\sigma] = \nabla_{X_\gamma} X_\sigma - \nabla_{X_\sigma} X_\gamma = \omega_\sigma^\beta(X_\gamma) X_\beta - \omega_\gamma^\beta(X_\sigma) X_\beta$$ Hence $$LHS = - \theta^{\alpha}([X_\gamma, X_\sigma]) = - \omega_\sigma^\alpha(X_\gamma) + \omega_\gamma^\alpha(X_\sigma)$$ 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.

Aside: Ehresmann connection, Connection on Principal Bundle

Another more geometric notion of connection is called Ehresmann connection, it applies to fiber bundle. Let $\pi: E \to M$ be a fiber bundle. A connection is a splitting of $$T_p E = T_p E_h \oplus T_p E_v $$ where $h$ stands for horizontal and $v$ for vertical. $TE_v$ is canonical, it is the tangent space to the fiber. $T_p E_h$ contains information.

If $P$ is a principal $G$-bundle, then we want the choice of horizontal subspace to be invariant under the (right) action of $G$.

Parallel transport along a path $\gamma: [0,1] \to M$ on $M$ gives a diffeomorphism $$ P_\gamma: E_{\gamma(0)} \to E_{\gamma(1)}. $$ In the case of $G$-bundle, this diffeomorphism commute with the right $G$-action on the fiber.

If $V$ is a represenation of $G$, we may form the associated bundle $P \times_G V$, where each fiber over $b \in M$ is $$ P_b \times_G V = \{ (p , v) \in P_b \times V \} / (p g, v) \sim (p, g \cdot v) \cong V $$ since $G$ acts on $P_b$ freely and transitively ($P_b$ is said to be a $G$-torsor). Then, $P_\gamma$ also induces parallel transport on $P \times_G V$.

math214/04-15.txt · Last modified: 2020/04/15 00:55 by pzhou