Timeline for Finding covariant derivative of a riemanian submanifold
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jul 26, 2010 at 20:47 | answer | added | Willie Wong | timeline score: 0 | |
| Jul 26, 2010 at 17:37 | comment | added | tamir | OK I understand this, but I'm still not sure how do I demonstrate that $ g^N $ would be diagonal. Intuitively - Suppose I have a "vertical vector" $ a \in M \subset N $ then in the local coordinates you showed me $ g^N * a $ as a matrix operating on a vector, should give me a vector which is in $ T_a N $ but not in $ T_a M $ And this shows that the matrix elements "*" would have to be zero. But again, this is intuitive, from linear algebra. Can you please help me to understand this delicate point? thanks for the time, Tamir | |
| Jul 26, 2010 at 0:29 | answer | added | Greg Kuperberg | timeline score: 1 | |
| Jul 26, 2010 at 0:22 | comment | added | Willie Wong | (Of course when you do this you have to be careful when you compute the curvature; beware of taking additional vertical derivatives!) | |
| Jul 26, 2010 at 0:19 | comment | added | Willie Wong | You don't need * to be all zero. You just need it to be zero along $M$. So just take an arbitrary local coordinate on $M$ to start. The metric defines along $M$ the normal direction. Choose a field of normal vectors, extend the field arbitrarily in a thickened slab around $M$. Flow $M$ along the vector field. Then the flow $t$ gives the "vertical coordinate". The lie transport of the local coordinates on $M$ gives the coordinate on the slab. And along $M$ the total metric $g^N$ is block diagonal. | |
| Jul 26, 2010 at 0:06 | history | asked | tamir | CC BY-SA 2.5 |