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1d
comment Extremal functions for Gagliardo-Nirenberg inequality
For $\mathbb{R}^n$ or any other noncompact Riemannian manifold, I'm not sure. It's basically a question of showing that an appropriately normalized minimizing sequence of functions is compact with respect to some topology and that the limit is nonzero. The main issue here is making sure that the "mass" of the function does not leak out to infinity. If the domain is a closed Riemannian manifold, then the answer is yes.
2d
revised Extremal functions for Gagliardo-Nirenberg inequality
added 469 characters in body
2d
answered Extremal functions for Gagliardo-Nirenberg inequality
Jun
25
comment Complex structure on $S^6$ gets published in Journ. Math. Phys
Misha, I don't really want to push this back towards you, but there probably aren't a lot of other people who are in a better position than you to judge the correctness of this paper. Especially since you already looked at an earlier version.
Jun
22
comment Taylor expansions of Riemannian exponential map and Jacobi fields?
Oops. I didn't see that you already posted this on math.stackexchange.com. But I stand by my other advice. If you dig around MathOverflow, I believe you'll also find other answers to this question.
Jun
22
comment Taylor expansions of Riemannian exponential map and Jacobi fields?
Yes, please migrate to math.exchange.com. Two hints: use Jacobi equation and describe differential of exponential in terms of Jacobi fields. Solid exercise for a differential geometry student. Might take you a while but you'll learn more than just reading about it.
Jun
20
comment Proving a differential inequality without performing iteration
This is pretty straightforward. It's more appropriate for math.stackexchange.com.
Jun
17
comment Measurement of “symmetry” of a convex body
I don't like to say that the simplex is the least symmetric body, but there are a lot of geometric invariants, where one extremal value is achieved by the ball and the other by the regular simplex. The simplest invariant that is conjectured to have this property is $V(K)V(K^*)$, where $K$ is a convex body and $K^*$ the polar body.
Jun
12
comment Blow-up as polar coordinates?
I think it's reasonable that the idea of blowing up a point came before higher dimensions, and, whether it was explicitly viewed as polar co-ordinates or not, it's likely that it was somewhere in the back of the mind of the person(s) who discovered the idea of blowing up.
Jun
11
comment Extension of a smooth function from a convex set
Stein's book, "Singular Integrals and Differentiability Properties of Functions" shows how to extend a smooth function on a domain with boundary (including any convex domain) to an open neigbborhood of the domain. This is easily extended to a manifold with boundary using partitions of unity.
Jun
8
comment Is this differential identity known?
I think posting such a simple proof is well justified here, even though it doesn't answer the exact question asked.
Jun
7
comment Examples of Smooth, Compact and Non-rigid Manifolds that Bound a Finite, Non-zero Volume
Manfred, you have to be careful about what you're saying is smooth or not. Kuiper's theorem says that given a smooth Riemannian 2-manifold, there is an infinite family of $C^1$ isometric embeddings of the surface into $\mathbb{R}^3$. So the Riemannian manifold is smooth, but as a surface in $\mathbb{R}^3$, it's not.
Jun
7
comment Examples of Smooth, Compact and Non-rigid Manifolds that Bound a Finite, Non-zero Volume
In mathoverflow.net/questions/1975/…, "smooth" is used only in terms of the regularity of the embedding and not in terms of the dependence of the embedding in terms of the parameter.
Jun
5
comment Examples of Smooth, Compact and Non-rigid Manifolds that Bound a Finite, Non-zero Volume
mathoverflow.net/questions/1975/…
Jun
5
comment Examples of Smooth, Compact and Non-rigid Manifolds that Bound a Finite, Non-zero Volume
I'm under the impression that there is no known example of a smooth closed surface in $\mathbb{R}^2$ that can be deformed isometrically as smooth surfaces. I believe it's a conjecture that in fact there are none and all smooth closed surfaces are rigid in the smooth category. And things only get more rigid in higher dimensions. This is in sharp contrast to polyhedral surfaces, where Connelly constructed a counterexample (and there is a metal model of it at IHES).
Jun
2
comment The existence of differential operator of the form $AB=0$
Vit, I'm not familiar with the details, but it seems to me that the cohomology associated with any elliptic complex of differential operators, such as Dolbeault cohomology as well as operators on vector bundles, depend on more than the topology of the domain.
Jun
1
comment The existence of differential operator of the form $AB=0$
One possible question might be whether "local exactness" holds, which means at any point in the domain there exists a (sufficiently small) neighborhood of that point on which the operators are exact.
Jun
1
comment The existence of differential operator of the form $AB=0$
As the comments above, there are a lot of issues with your question as posed. For example, a simple one is that whether exactness, as you define it, holds depends on the domain of the operator. For example, the exterior derivative of differential forms is exact if and only if the corresponding deRham cohomology group vanishes. For other operators, the exactness depends on even more than just the domain. In general, exactness, as you define it, is a rather subtle phenomenon that is understood fully only for special operators.
Jun
1
comment The existence of differential operator of the form $AB=0$
$AB = 0$ implies only that the image of $B$ is a subset of the kernel of $A$. They don't have to be equal.
May
29
comment John Nash's Mathematical Legacy
c) The h-principle is the abstract version of the arguments used in Nash's second isometric embedding theorem. As far as I know, it has nothing to do with the Nash-Moser implicit function theorem. However, see arxiv.org/abs/1111.2700 for a explanation why I might be wrong about this.