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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Nov 23, 2018 at 22:05 history edited Ali Taghavi
I add a tag.
Aug 16, 2015 at 5:53 vote accept Ali Taghavi
Aug 9, 2015 at 4:52 history edited Ali Taghavi CC BY-SA 3.0
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Aug 9, 2015 at 4:45 history edited Ali Taghavi CC BY-SA 3.0
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Aug 6, 2015 at 18:25 answer added David E Speyer timeline score: 5
Aug 6, 2015 at 18:04 comment added David E Speyer In other words, yes, I think the manifold has to be parallelizable to find a global solution.
Aug 6, 2015 at 18:02 comment added David E Speyer Maybe I'm missing something but, if the $X_i$ became linearly dependent at some $p \in M$, it seems to me that the symbol of $\sum \partial^2/(\partial X_i)^2$ at $p$ would be a degenerate quadratic form, so $\Delta \neq \sum \partial^2/(\partial X_i)^2$.
Aug 6, 2015 at 17:58 answer added Robert Bryant timeline score: 22
Aug 6, 2015 at 8:44 comment added Ali Taghavi @Andrew Do you think any such vector field should be a frame hence manifold is parallelizable?
Aug 6, 2015 at 8:37 history edited Ali Taghavi
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Aug 6, 2015 at 8:35 comment added Ali Taghavi @Andrew but they do not have necearilly a common singularity, hence this is not an obstruction, right?
Aug 6, 2015 at 8:30 comment added Andrew Globally continuous vector fields on some manifolds, say on a sphere, should vanish somewhere.
Aug 6, 2015 at 7:28 history edited Ali Taghavi CC BY-SA 3.0
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Aug 5, 2015 at 7:29 answer added Raziel timeline score: 11
Aug 5, 2015 at 6:22 history edited Ali Taghavi CC BY-SA 3.0
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Aug 5, 2015 at 3:56 history asked Ali Taghavi CC BY-SA 3.0