Kofi

1,159
Reputation
573 views
Is this your account?

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.
May
16
comment Special kind of operators
Thank You for the reference. It turned out, such an operator is (2,1,2) nuclear.
May
15
asked Special kind of operators
May
14
comment Absolutely 2-summable operator on a Hilbert space
Wow, that is really interesting! Thank you for your answer!
May
13
asked Absolutely 2-summable operator on a Hilbert space
May
13
comment Triangle area on surfaces of constant curvature
I 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?
May
12
asked Triangle area on surfaces of constant curvature
May
9
comment Fourier transform of a bounded function
I suppose, continuity of the function does not help either?
May
9
asked Fourier transform of a bounded function
May
5
comment zeta(3) in terms of derivatives of zeta at 1/2 and pi
What 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?
May
2
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.
Apr
28
answered Linearization of vector fields
Apr
17
asked Heat Kernel Asymptotics with low regularity
Mar
31
asked Difference between parallel transport and derivative of the exponential map
Mar
22
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!
Feb
26
revised Heat Kernel Asymptotics on Manifold with Boundary
added 255 characters in body; added 4 characters in body; deleted 39 characters in body
Feb
26
asked Heat Kernel Asymptotics on Manifold with Boundary
Feb
26
awarded  Popular Question
Feb
7
comment From Topological to Smooth and Holomorphic Vector Bundles
I think what you want to hear in relation to (C) is the word "Thom isomorphism"
Jan
17
comment About writing of mathematical papers
I almost never use $\Longrightarrow$, $\exists$ or $\forall$.
Jan
14
asked Asymptotic Expansion of the Schrödinger kernel?
Jan
8
comment Trace formula for PSDOs
Can you elaborate on this? I don't really know what to make of it.
Jan
7
asked Trace formula for PSDOs
Jan
2
comment Solution of a PDE and its uniqueness
I may be wrong, but at first sight I would say this looks pretty hopeless to answer.
Dec
29
comment Math for a cake
This should be two, not zero, shouldn't it?
Dec
19
comment Excellent mathematical explanations
I 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.
Dec
13
asked Heat kernel proof of Poincaré Hopf
Dec
9
revised Combinatorics: Product Rules.
deleted 10 characters in body
Dec
8
asked Combinatorics: Product Rules.
Dec
5
asked Index theorems and orientability
Dec
2
comment Fourier Transform, for entire function
Your 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'.
Dec
2
answered Fourier Transform, for entire function
Dec
1
comment What are the Dirac operators on $S^1$?
Hopefully, my answer at stackexchange is answering your question.
Dec
1
comment An invariant method of stationary phase
So the constant depends on $t$?
Nov
28
comment An invariant method of stationary phase
Yes, 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.
Nov
27
revised An invariant method of stationary phase
added 2 characters in body; added 2 characters in body
Nov
27
comment An invariant method of stationary phase
Yes, 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.
Nov
27
revised An invariant method of stationary phase
added 37 characters in body
Nov
27
revised An invariant method of stationary phase
added 736 characters in body
Nov
27
comment An invariant method of stationary phase
I 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.
Nov
27
asked An invariant method of stationary phase