# Kofi

 1,159 Reputation 573 views

## Registered User

 Name Kofi Member for 1 year Seen 1 hour ago Website Location Age
 3h comment Dirac measures dense in space of measures?What an easy argument! Thanks! 5h asked Dirac measures dense in space of measures? 1d revised The first eigenvalue of the Schrödinger operator is simple.deleted 2 characters in body; edited body; added 20 characters in body 1d comment The first eigenvalue of the Schrödinger operator is simple.The argument in the last paragraph also works with $C^2$ instead of $C^\infty$, and the solutions are always $C^2$. You should check out some book about PDE (e.g. the mentioned one by Evans or Gilbarg-Trudinger) to get the exact statements about regularity (especially at the border), but in principle, the same argument should work in most related cases. 2d answered The first eigenvalue of the Schrödinger operator is simple. May16 comment Special kind of operatorsThank You for the reference. It turned out, such an operator is (2,1,2) nuclear. May15 asked Special kind of operators May14 comment Absolutely 2-summable operator on a Hilbert spaceWow, that is really interesting! Thank you for your answer! May13 asked Absolutely 2-summable operator on a Hilbert space May13 comment Triangle area on surfaces of constant curvatureI don't understand what you mean. I have a Riemannian metric of constant curvature on a Surface. This gives me curvature and a volume density, so I know both what curvature and area is and I can in theory calculate what the area of a geodesic triangle is. Where do I need an axiom? May12 asked Triangle area on surfaces of constant curvature May9 comment Fourier transform of a bounded functionI suppose, continuity of the function does not help either? May9 asked Fourier transform of a bounded function May5 comment zeta(3) in terms of derivatives of zeta at 1/2 and piWhat do you mean with "the last equality holds to precision 10^-4"? If the result was true, shouldn't the equality hold up to any precision? May2 comment Gap between first two nonzero Laplacian eigenvalues on closed compact surface? Probably one should skip the "nonzero" requirement as in this case, the gap is always positive. Apr28 answered Linearization of vector fields Apr17 asked Heat Kernel Asymptotics with low regularity Mar31 asked Difference between parallel transport and derivative of the exponential map Mar22 comment sections of tensor product bundle ( tensor product of two vector bundles )First, how is your actual question related to connections? Secondly, on any bundle, you can construct a section by patching it together locally with partitions of unity! Feb26 revised Heat Kernel Asymptotics on Manifold with Boundaryadded 255 characters in body; added 4 characters in body; deleted 39 characters in body Feb26 asked Heat Kernel Asymptotics on Manifold with Boundary Feb26 awarded ● Popular Question Feb7 comment From Topological to Smooth and Holomorphic Vector BundlesI think what you want to hear in relation to (C) is the word "Thom isomorphism" Jan17 comment About writing of mathematical papersI almost never use $\Longrightarrow$, $\exists$ or $\forall$. Jan14 asked Asymptotic Expansion of the Schrödinger kernel? Jan8 comment Trace formula for PSDOsCan you elaborate on this? I don't really know what to make of it. Jan7 asked Trace formula for PSDOs Jan2 comment Solution of a PDE and its uniquenessI may be wrong, but at first sight I would say this looks pretty hopeless to answer. Dec29 comment Math for a cakeThis should be two, not zero, shouldn't it? Dec19 comment Excellent mathematical explanationsI think that whether a proof is explanatory is highly individual and depends on the knowledge you already have. Some people may be satisfied with some argument while others still don't have a feeling of understanding. Dec13 asked Heat kernel proof of Poincaré Hopf Dec9 revised Combinatorics: Product Rules.deleted 10 characters in body Dec8 asked Combinatorics: Product Rules. Dec5 asked Index theorems and orientability Dec2 comment Fourier Transform, for entire functionYour statement about defining the Fourier transform by duality does not make sense... As you pointed out yourself, the fourier transform of a function in D is not in D again, but in S. Hence taking the dual operator gives again only an operator from S' to D'. Dec2 answered Fourier Transform, for entire function Dec1 comment What are the Dirac operators on $S^1$?Hopefully, my answer at stackexchange is answering your question. Dec1 comment An invariant method of stationary phaseSo the constant depends on $t$? Nov28 comment An invariant method of stationary phaseYes, this is basically what I wrote in my edit to the above post, except that I considered the real case (I think the $e^{it\psi(x)}$ is wrong in your definition of $c$?). But the higher coefficients always make use of a chart. Nov27 revised An invariant method of stationary phaseadded 2 characters in body; added 2 characters in body Nov27 comment An invariant method of stationary phaseYes, this is the standard way. It works just as well in the case that $\phi$ is purely real (and gets even simpler). However, it describes the asymptotic expansion via differential operators given in Morse coordinates. And there is a vast amount of Morse coordinates for a given function. Of course, the term cannot depend on this choice, however when looking at its definition, this is not obvious at all. I am looking for a description of these terms such that one can see their invariance of the choice of Morse chart by its definition. Nov27 revised An invariant method of stationary phaseadded 37 characters in body Nov27 revised An invariant method of stationary phaseadded 736 characters in body Nov27 comment An invariant method of stationary phaseI know the statement in Duistermaat, and I know that one can determine the sj more specifically using local coordinates. In fact, it is just a Laplacian when using Morse coordinates. However, it should be possible to figure out the $s_j$ without using any coordinates, just by using invariance theory. Nov27 asked An invariant method of stationary phase