math214:04-15
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math214:04-15 [2020/04/15 00:29] pzhou |
math214:04-15 [2020/04/15 00:33] pzhou |
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| $$X_i = \frac{1}{h_i} \d_{x^i}, \theta^i = h_i dx^i $$ | $$X_i = \frac{1}{h_i} \d_{x^i}, \theta^i = h_i dx^i $$ | ||
| Then, we can get $\omega_i^j$ by solving | Then, we can get $\omega_i^j$ by solving | ||
| - | $$ d \theta^i = -\d_j(h_i) dx^i \wedge dx^j $$ | + | $$ d \theta^i = d(h_i) \wedge |
| + | 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. | ||
math214/04-15.txt · Last modified: 2020/04/15 00:55 by pzhou
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