Timeline for Is the heat kernel more spread out with a smaller metric?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 24, 2017 at 8:28 | history | edited | Tom Price | CC BY-SA 3.0 |
The image no longer loads, so I removed the parts of the question that refer to it.
|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Feb 18, 2015 at 6:38 | vote | accept | Tom Price | ||
| Feb 18, 2015 at 2:48 | answer | added | Noah Stephens-Davidowitz | timeline score: 9 | |
| Dec 12, 2014 at 9:17 | comment | added | Delio Mugnolo | @Paul Siegel: I am surprised by your assertion. The heat semigroup typically has infinite speed of propagation, that is, as soon as the initial data are positive but not identically 0, the solution of the heat equation is instanteneously strictly larger than 0 at any point of the domain/manifold. This can be e.g. proved by irreducibility of the semigroup, see e.g. Ouhabaz' 2005 book. Am I missing something? | |
| Dec 12, 2014 at 5:01 | comment | added | Paul Siegel | The heat operator has the "finite propagation speed" property which allows one to estimate the support of $e^{t\Delta}f$ in terms of the support of $f$. There are estimates of the propagation speed involving metric data, but I don't know off the top of my head if they are sharp enough to get what you want. | |
| Dec 11, 2014 at 23:06 | history | edited | Tom Price | CC BY-SA 3.0 |
Explained a proof of a special case I recently found.
|
| Nov 8, 2014 at 9:04 | comment | added | Sebastian | The only thing I know about the $\theta$'s is that they can be expressed in terms of connection data along the diagonal, see for example the book of John Roe: Elliptic operators,... . But if I remember correctly, the series $(\theta_0(p,q)+t\theta_1(p,q)+...)$ converges for small t and nearby $p,q$ to a smooth map which is the identity for $t=0$ and $p=q.$ Therefore, you can get the inequality in a neighbourhood of the diagonal times the $t=0$ slice. | |
| Nov 7, 2014 at 16:54 | comment | added | Tom Price | @Sebastian No I wasn't aware, thanks for pointing that out. I see that the $h_t$ functions will satisfy the inequality, but I don't see how that is necessarily preserved after multiplying by the series on the right without more information about these $\theta_n$. | |
| Nov 7, 2014 at 8:12 | comment | added | Sebastian | Are you aware of the asymptotic expansion for the heat kernel in terms of $t:$ $H_t(p,q)= h_t(p,q)(\theta_0(p,q)+t\theta_1(p,q)+...),$ where $h_t(p,q)=\frac{1}{(4 \pi t)^{dim M/2}}exp(-d(p,q)^2/4t)$ and $\theta_0(p,p)=1.$ At least for small $t$ and nearby points $p,q$ this should give your inequality. | |
| Nov 7, 2014 at 0:16 | history | edited | Tom Price |
edited tags
|
|
| Nov 7, 2014 at 0:09 | history | asked | Tom Price | CC BY-SA 3.0 |