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Aug
21
answered Method of characteristics with constant PDE
Aug
13
answered Definition of the rank of the isometry group of a manifold.
Aug
10
comment Are there n-th roots of differential operators?
This technique depends sensitively on both the given differential operator, as well as the space of functions on which it acts. In the case of the Dirac operator, it is not the Laplacian on functions that admits an algebraic square root, but rather the Laplacian on spinors. This also works for vector-valued functions, since we can decompose vectors into products of spinors. These kinds of $n$th roots are completely different than what I described in my answer, but both can be useful depending on the situation.
Aug
10
comment Are there n-th roots of differential operators?
Yes, if you write $\sqrt[1/2]{D_y} \sqrt[1/2]{D_y} g(y)$ as a multiple integral, there will be a term like $\int e^{2\pi i y(k-k')} dy$ which is $\delta(k-k')$ and everything takes care of itself. This is what makes the pseudodifferential calculus work.
Aug
10
answered Are there n-th roots of differential operators?
Aug
10
comment Computation of determinant of a matrix with elements from an arbitrary commutative ring
I don't know the answer to this particular question, but there is a general phenomenon in algebraic geometry that computational complexity of certain algebraic problems is (inversely) related to the "niceness" of an associated geometric object. The presence of zero divisors definitely complicates the geometry. So I wouldn't find it completely surprising if cofactor expansion really is the most efficient method for an arbitrary commutative ring, without any additional assumptions.
Aug
3
comment The boundedness of an integral
After integrating by parts and playing around a little bit, you can easily get an estimate like this provided you make an assumption like $|a| > a_0$, $|b| > b_0$. If you want an estimate that makes no such assumption, then I'm not sure there is any reason to expect such an estimate to exist.
Aug
3
revised Asymptotic behavior of a sequence based on a subsequence.
deleted 177 characters in body
Aug
3
comment Asymptotic behavior of a sequence based on a subsequence.
I didn't realize that $m$ is supposed to be fixed. I'll update my answer when I get a chance.
Aug
3
answered Asymptotic behavior of a sequence based on a subsequence.
Jun
24
awarded  Yearling
May
1
comment What is the Lie algebra of the ``indefinite orthogonal group''?
Just to add--these are all non-compact real forms of the corresponding complex groups $O(n, \mathbb{C})$, etc. So from the point of view of their complex representations, the signature $(p,q)$ does not really play a role. But the signature does play a role in the real representation theory (see e.g. the classification of spinors, which exhibits mod 8 periodicity).
Apr
28
answered Approximating the Hessian
Apr
28
awarded  Supporter
Apr
27
answered An isomorphism in relative De Rham cohomology
Apr
27
awarded  Commentator
Apr
27
comment An isomorphism in relative De Rham cohomology
I'm a bit confused. If $E \to M$ is a vector bundle, then $M$ is a retract of $E$ and they have the same cohomology. It seems to me that the relative cohomology $H^\ast(E, E^o)$ should be isomorphic to the (reduced?) cohomology of the Thom space of $E$, which is certainly not isomorphic to $H^\ast(M)$ in general.
Apr
26
answered Scalar product on manifold.
Apr
25
comment Why is $W(V)\simeq D(k[X_1,\dots,X_n])$?
@Kally what is your working definition of polynomial differential operators, if not the algebra generated by $x_i, \partial_j$ subject to the above relations? There are several equivalent definitions, so the answer depends on which you take as fundamental.
Apr
25
answered Reference request for “Hodge Theorem”