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location Rehovot, Israel
age 24
visits member for 2 years, 5 months
seen 23 hours ago

I'm a PhD student in Weizmann Institute, specializing in Probability.


2d
comment Does exterior product commute functor Hom?
@DinhVanHoang: My comment still applies. For infinite-dimensional vector spaces, for instance, $\mathrm{Hom}(\wedge^2 M, R)$ consists of all skew-symmetric matrices, while $\wedge^2 \mathrm{Hom}(M, R)$ only contains those of finite rank.
2d
comment Results true in a dimension and false for higher dimensions
Contraction in Hilbert space, I suppose?
2d
comment Does exterior product commute functor Hom?
I strongly suspect that it's false even for infinite-dimensional vector spaces over fields. Did you check that?
Aug
12
comment 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 "Hilbert-Schmidt" 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
comment Why differential forms are important?
@LeeMosher: Right, but that requires a full-blown 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
comment 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
comment 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
comment 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,