bio | website | math.poly.edu/~yang |
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location | New York, New York | |
age | ||
visits | member for | 5 years, 1 month |
seen | 6 mins ago | |
stats | profile views | 12,257 |
Nov 15 |
comment |
When is a `1-form' with continuous coefficients exact?
Doesn't the Poincare lemma hold for distributions? And if the weak derivatives of $u$ are continuous, it follows that $u$ is $C^1$. |
Nov 9 |
answered | Which are the recommended books for an introductory study of complex manifolds? |
Nov 9 |
awarded | Good Answer |
Nov 4 |
awarded | Popular Question |
Nov 2 |
comment |
Is there a complex structure on the 6-sphere?
That's too bad. |
Nov 2 |
comment |
Lower regularity version of Moser's theorem on volume elements
If you can somehow make the flow also satisfy either a parabolic PDE or a transverse elliptic PDE, then you would achieve the regularity you want. |
Nov 2 |
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Lower regularity version of Moser's theorem on volume elements
I suspect Robert is correct about the regularity. When you integrate a flow, you gain a derivative in the time direction only and not in any transverse ("spatial") directions. This can be seen from the proof of the existence and uniqueness of parameterized ODE's. |
Oct 28 |
comment |
Riemann's formula for the metric in a normal neighborhood
Of the modern proofs, I like the Jacobi field approach the best. You start by using the distance from a given point as one co-ordinate function and showing that you can extend the angular variables from the tangent space at the origin. This is most easily done using Jacobi fields $J_1, \dots, J_n$,. The metric in these co-ordinates is given by $g_{ij} = J_i\cdot J_j$, so its Taylor expansion is easily calculated using the formal solution to the Jacobi equation. |
Oct 27 |
comment |
Riemann's formula for the metric in a normal neighborhood
Martin, I find Christoffel symbols highly unenlightening, so that's not a simple definition to me. It seems to me that both normal co-ordinates (up to second order only) and the Riemann curvature tensor arise pretty naturally if you search for co-ordinates that simplify the 2nd order Taylor expansion of the metric as possible. But since the Hessian of the metric is a 4th order tensor, that's still a bit tricky. |
Oct 27 |
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Riemann's formula for the metric in a normal neighborhood
It seems to me that it's first necessary to find a "simple" definition of the Riemann curvature tensor. I'm not sure how Levi-Civita defined it. |
Oct 27 |
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Uniqueness of scalar curvature
Robert, oops. You're right. I once knew stuff like that. |
Oct 27 |
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Uniqueness of scalar curvature
I don't have a proof, but it seems to me that more is true. Namely, any scalar function defined using a Riemannian metric and its derivatives up to second order that is equivariant under diffeomorphisms has to be a function of the scalar curvature. |
Oct 23 |
comment |
Why do we teach calculus students the derivative as a limit?
Depends on what you mean. If you need it for your work, you'll learn it properly through what you do. That's always the best way, because you understand why it's needed. If you just want to learn it, you succeed either by having a really good teacher or studying it on your own (and not being satisfied until you understand it inside out). |
Oct 18 |
awarded | Yearling |
Oct 16 |
comment |
Regularity of the Minkowski functionnal of a convex
Use polar coordinates. |
Oct 16 |
comment |
Source needed (at final-year undergrad level) for the double cover of SO(3) by SU(2)
Yes, but without discussing the topological aspect. Uses quaternions. |
Oct 16 |
answered | Source needed (at final-year undergrad level) for the double cover of SO(3) by SU(2) |
Oct 15 |
comment |
How is this transformation related to the Legendre transform?
If you apply the Legendre transform to $-xf(x,y)$ in the $x$ variable and for fixed $y$, then you should get $s$ back. And, yes, it appears that $f$ is concave with respect to $y$ since it's obtained as an infimum over a family of concave functions. |
Oct 15 |
comment |
How is this transformation related to the Legendre transform?
It looks to me like that, for each $y$, $-xf(x,y)$ is the Legendre transform of $-s(t,y)$. |
Oct 11 |
awarded | Custodian |