Jonathan
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 Aug 22 answered Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? Aug 21 answered relation between integral and summation 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.