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2d
comment Symmetries of non-Riemannian curvature tensor
Do you have any thoughts about what kind of condition you're looking for?
2d
comment Symmetries of non-Riemannian curvature tensor
Ben, this doesn't look like Ricci to me. The result is a skew-symmetric tensor and vanishes if the connection is the Levi-Civita connection.
Jul
26
comment Fundamental solution for a parabolic PDE with constant coefficents
No, you're right. I stand corrected. The Malgrange-Ehrenpreis theorem does not make any assumptions about the top order term at all.
Jul
26
comment Fundamental solution for a parabolic PDE with constant coefficents
Could you write out the precise form of the PDE you want to solve? Is it $u_t = a^{ij}\partial_i\partial_ju$?
Jul
26
comment Fundamental solution for a parabolic PDE with constant coefficents
I could be wrong since I'm relying on distant memories, but I believe that the theorems of Malgrange and Ehrenpreis apply only to PDE's where the top order term is nondegenerate (i.e., PDE's whose real analytics solutions can be obtained via Cauchy-Kovalevski). Without this assumption there might not be any solutions at all. The heat equation has a degenerate top order term, so I'm not sure that the theorems you cite apply.
Jul
18
comment Oriented volume and determinants: Circularity
Jochen, to quibble with this, in my definition of orientation the number of possible orientations is the number of connected components of $GL(n)$. Also, you can use the polar decomposition of invertible matrices to show that the number of components is 2. It's just that using the determinant is much easier.
Jul
17
comment Oriented volume and determinants: Circularity
But it's way easier just to define orientation directly using the determinant of the change-of-basis matrix and observing that it's well-defined.
Jul
17
comment Oriented volume and determinants: Circularity
It suffices to distinguish between the two different orientations. An orientation is associated with a basis of the vector space, commonly known as a frame. So start with a frame and declare it to have positive orientation. Any other frame has positive orientation, if there exists a continuous path in the space of all frames joining the first frame to the other and negative orientation otherwise. In other words, orientation identifies the two connected components of $GL(n)$. The fact that there are two separate components and that orientation is well-defined is proved using the determinant.
Jul
17
comment Is this has anything to do with Riesz representation?
Condition (2) is needed to show that the operator $\beta \mapsto \beta'$ is continuous in the Frechet topology of the space of distributions. And, yes, $\nu$ is in general not a measure. It is something weaker than that. Condition (1) holds, if interpreted appropriately. Since $\nu$ is not a measure, it cannot hold since it isn't even well-defined.
Jul
17
comment Is this has anything to do with Riesz representation?
To be more precise, a distribution is a continuous linear functional on, say, $C^\infty_c(I)$ with respect to the Frechet topology. The derivative of a distribution is a distribution. In this example, the distributions $\alpha: \phi \rightarrow \int \phi\,d\mu$ and $\beta: \phi\rightarrow \int \phi'\,d\mu$ satisfy $\beta' = -\alpha$, where the operator $\beta \mapsto \beta'$ is commonly known as weak differentiation and is defined using integration by parts, as you have described.
Jul
17
comment Is this has anything to do with Riesz representation?
$\nu$ is not a measure. It is a distribution (as defined by Laurent Schwartz). See, for example, en.wikipedia.org/wiki/Distribution_(mathematics)
Jul
17
comment Is this has anything to do with Riesz representation?
This defines a distribution ("generalized function") that is the derivative of the measure $\mu$ (which is itself a distribution.
Jul
13
comment Method of characteristics
This question really should be migrated to math.stackexchange.com
Jul
7
comment Elliptic operators corresponds to non vanishing vector fields
I just want to reiterate what Liviu Nicolaescu has already said: Any PDO defined using only the vector field $X$ and no other PDEO is never elliptic. The best way to understand this is to look up the most general definition of an elliptic partial differential operator and test it against examples such as $\nabla_X$, $[X,\cdot]$, and any other example you can think of. Any PDO defined using only $X$ is essentially an ODE along the integral curves of $X$. If such an operator is Fredholm, it is due to the global dynamics of the operator and not a local property of the PDO such as ellipticity.
Jul
7
awarded  Nice Answer
Jul
3
comment $C^1$ regularity of harmonic functions on Riemannian manifolds
I'm not sure if this works (I'm too lazy to do the tensor calculation), but if you differentiate $\Delta f = 0$ twice, you should get something like $\Delta \nabla^2f + A\nabla^2 f = \nabla\cdot(B\nabla f)$, where $A$ and $B$ depend on the curvature tensor (and maybve its covariant derivative). If so, then a straightforward Moser iteration argument gives an $L^p$ bound on $\nabla^2f$ in terms of the Sobolev constant, curvature (and maybe its covariant derivative). This in turn implies a Holder bound on $\nabla f$, where the constant can also be bounded in terms of geometric invariants.
Jun
28
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.
Jun
27
revised Extremal functions for Gagliardo-Nirenberg inequality
added 469 characters in body
Jun
27
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.