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Aug 10, 2015 at 9:14 history edited Raziel CC BY-SA 3.0
Added explicit example on the sphere.
Aug 10, 2015 at 7:29 history edited Raziel CC BY-SA 3.0
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Aug 10, 2015 at 7:28 comment added Raziel Correct, that's sufficient. So it seems the problem it's even easier (at least with this relaxed condition on the ''sum of squares''). I will edit my post.
Aug 10, 2015 at 2:29 comment added Robert Bryant @Raziel: Actually, the equation $\sum_{i=1}^N\mathrm{div}(X_i)X_i=0$ is at most $n$ equations, not $N>n$ equations, since you are only trying to get the vanishing of a single vector field, not setting $\mathrm{div}(X_i) = 0$ for $i=1,\ldots,N$.
Aug 9, 2015 at 9:30 history edited Raziel CC BY-SA 3.0
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Aug 9, 2015 at 9:23 history edited Raziel CC BY-SA 3.0
Fixed typos, improved notation for consistency
Aug 9, 2015 at 9:22 comment added Raziel As I explained the formula is valid for N>n under the conditions on the $W_I$ that I described. This indeed put constraints on the $W_I$. For example all $\|W_I\| \leq 1$ (and at least some of them is strictly $<1$ as soon as $N > n$. You can test it explicitly with the example on the $2$-sphere that I suggested. As a matter of fact, it's a matter of linear algebra: you can write a positive definite matrix in $n$-dimension as $\sum_{I=1}^N w_I w_I^*$ for an arbitrarily large number of vectors $w_I \in \mathbb{R}^n$.
Aug 9, 2015 at 3:53 comment added Ali Taghavi @Raziel Thanks again for your answer. You wrote the above formula works for N>n. Am I mistaken to think the formula is sensitive to orthonotmality?So not valid for N>n?
Aug 8, 2015 at 8:41 history edited Raziel CC BY-SA 3.0
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Aug 7, 2015 at 8:27 history edited Raziel CC BY-SA 3.0
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Aug 7, 2015 at 8:12 history edited Raziel CC BY-SA 3.0
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Aug 7, 2015 at 8:07 history edited Raziel CC BY-SA 3.0
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Aug 7, 2015 at 8:01 history edited Raziel CC BY-SA 3.0
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Aug 7, 2015 at 7:48 history edited Raziel CC BY-SA 3.0
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Aug 7, 2015 at 7:35 history edited Raziel CC BY-SA 3.0
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Aug 6, 2015 at 18:03 history edited Raziel CC BY-SA 3.0
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Aug 6, 2015 at 9:43 history edited Raziel CC BY-SA 3.0
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Aug 6, 2015 at 9:35 comment added Raziel Thanks for the comment. Indeed I believe the best I can do is build such a frame where the fields are divergence-free at a given point $x_0$. The requirement on a full neighborhood leads to a system of PDEs. The integrability conditions give then the local obstruction.
Aug 6, 2015 at 9:25 comment added Jean Van Schaftingen Can you explain how can you construct in general a local frame of orthogonal and divergence-free vector fields?
Aug 5, 2015 at 7:29 history answered Raziel CC BY-SA 3.0