This shows you the differences between two versions of the page.
Next revision | Previous revision | ||
math214:04-06 [2020/04/06 07:44] pzhou created |
math214:04-06 [2020/04/06 10:35] (current) pzhou [Bianchi Identity] |
||
---|---|---|---|
Line 2: | Line 2: | ||
$$\gdef\xto\xrightarrow \gdef\End{\z{ End}}$$ | $$\gdef\xto\xrightarrow \gdef\End{\z{ End}}$$ | ||
- | ==== Geometric Meaning | + | ==== Summary |
Let $E$ be a vector bundle over a smooth manifold. The connection $\nabla$ is given as such | 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^0(M, E) \xto{ \nabla} \Omega^1(M, E) \xto{ \nabla} | ||
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 | Locally, if we have a base coordinates $x_i$, and local trivialization $e_\alpha$ of $E$, then we may write the connection as | ||
- | $$ \nabla = d + \Gamma_i^\alpha_\beta dx^i \otimes e_\alpha \otimes \delta^\beta $$ | + | $$ \nabla = d + |
where $\{\delta^\alpha\}$ is the dual basis to $\{e_\alpha\}$, | where $\{\delta^\alpha\}$ is the dual basis to $\{e_\alpha\}$, | ||
$$ F_\nabla = \nabla^2 = (F_{ij})^\alpha_\beta dx^j \wedge d x^i \otimes e_\alpha \otimes \delta^\beta$$ | $$ F_\nabla = \nabla^2 = (F_{ij})^\alpha_\beta dx^j \wedge d x^i \otimes e_\alpha \otimes \delta^\beta$$ | ||
where | where | ||
- | $$ (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 $$ | + | $$ [F_{ij}]^\alpha_\beta = \d_i [\Gamma_j]^\alpha_\beta - \d_j [\Gamma_i]^\alpha_\beta + [\Gamma_i]^\alpha_\gamma |
+ | 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 | ||
+ | $$ F_{ij} = \d_i \Gamma_j - \d_j \Gamma_i + \Gamma_i \Gamma_j - \Gamma_j \Gamma_i. $$ | ||
+ | |||
+ | |||
+ | ==== 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., $\gamma(0)=\gamma(1)$, | ||
+ | $ 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 $M=\R^2$, | ||
+ | $$ F_{12} = - \lim_{\epsilon, | ||
+ | where $\gamma$ is the loop around the boundary of the small square $[0, \epsilon] \times [0, \delta]$ counter-clockwise. | ||
+ | |||
+ | To see this, we break down the loop into four segments, and denote the parallel transports on the four segments as $P_i, i=1, \cdots, 4$, and we work out the expansion modulo $\epsilon^2, | ||
+ | |||
+ | From $(0,0)$ to $(\epsilon, | ||
+ | $$ \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, | ||
+ | Approximately, | ||
+ | $$ [u](\epsilon) \approx (1 - \epsilon \Gamma_1(0, | ||
+ | The parallel transport along the first segment is $$P_1 \approx 1 - \epsilon \Gamma_1(0, | ||
+ | Similarly, we have | ||
+ | $$ P_2 \approx 1 - \delta \Gamma_2(\epsilon, | ||
+ | Using Taylor expansion for $\Gamma$ at $(0,0)$, | ||
+ | $$ P_4 P_3 P_2 P_1 \approx (1 + \delta \Gamma_2 ) (1 + \epsilon \Gamma_1 | ||
+ | $$ \approx 1 - \epsilon \delta (\d_1 \Gamma_2 - \d_2 \Gamma_1 + \Gamma_1 \Gamma_2 - \Gamma_2 \Gamma_1)|_{(0, | ||
+ | Hence we are done. See also [Ni] 3.3 for a more rigorous derivation. | ||
+ | |||
+ | ===== Bianchi Identity ===== | ||
+ | $$ \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, | ||
+ | $$ \Phi_{\nabla(T)} = [\nabla, \Phi_T] = \nabla \Phi_T - (-1)^p \Phi_T \nabla $$ | ||
+ | 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 | ||
+ | $$[A, [A, A]] = \sum_{i, | ||
+ | for example, the term with $dx^1 \wedge dx^2 \wedge dx^3$ has (2 times) | ||
+ | $$ [\Gamma_1, [\Gamma_2, \Gamma_3]] + [\Gamma_2, [\Gamma_3, \Gamma_1]] + [\Gamma_3, [\Gamma_1, \Gamma_2]] = 0. $$ | ||
+ | |||
+ | |||
+ | |||
+ | |||