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location  Rehovot, Israel  
age  24  
visits  member for  2 years, 5 months 
seen  23 hours ago  
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I'm a PhD student in Weizmann Institute, specializing in Probability.
2d

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Does exterior product commute functor Hom?
@DinhVanHoang: My comment still applies. For infinitedimensional vector spaces, for instance, $\mathrm{Hom}(\wedge^2 M, R)$ consists of all skewsymmetric matrices, while $\wedge^2 \mathrm{Hom}(M, R)$ only contains those of finite rank. 
2d

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Results true in a dimension and false for higher dimensions
Contraction in Hilbert space, I suppose? 
2d

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Does exterior product commute functor Hom?
I strongly suspect that it's false even for infinitedimensional vector spaces over fields. Did you check that? 
Aug 12 
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Is the ideal of functions vanishing at a set complementable in $C(X)$?
@SergeiAkbarov: Actually, it is. $\mathbb{N} \subset \beta \mathbb{N}$ consists of isolated points, so it's open. And in any case, your ideal $I$ only depends on the closure of the set. 
Aug 12 
answered  Is the ideal of functions vanishing at a set complementable in $C(X)$? 
Aug 7 
awarded  Popular Question 
Aug 4 
revised 
When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function?
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Aug 4 
revised 
When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function?
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Aug 4 
answered  When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function? 
Jul 12 
comment 
Equivalent Gaussian measures
Shouldn't it be "HilbertSchmidt" instead of "trace class"? Anyway, I believe that for most $v$ your $I  G^{1/2} G^\prime G^{1/2}$ is not even bounded. 
Jul 2 
awarded  Curious 
Jun 18 
awarded  Nice Answer 
Jun 17 
comment 
Why differential forms are important?
Another thing you can integrate is a measure. So I wouldn't call differential forms "the" natural way to express "the" idea of integration. :) 
Jun 17 
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Why differential forms are important?
@LeeMosher: Right, but that requires a fullblown tensor calculus, not just differential forms. 
Jun 17 
answered  Why differential forms are important? 
May 30 
answered  Does every operator from a Hilbert space to $L^0$ factor through a canonical one? 
May 22 
answered  The borel $\sigma$algebra of the set of probability measures 
May 6 
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smooth Luzin theorem
Any such function must have an "$L^0$ derivative", in the sense that $\frac{1}{t}(f(\cdot + t)  f(\cdot))$ has a limit in measure as $t \to 0$. 
Apr 23 
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How to define the distributional Hessian for a convex function on a $C^0$ Riemannian manifold?
My comment would only apply if the manifold and the metric were smooth, in which case $f \mapsto \nabla^2 f$ would be a smooth differential operator. In your nonsmooth case I don't know how to make sense of this. 
Apr 23 
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How to define the distributional Hessian for a convex function on a $C^0$ Riemannian manifold?
Any differential operator with smooth coefficients acts on distributions, 