The idea to construct a heat kernel is first construct a parametric in a small neighbourhood. Then use a bump function to extend it. And do convolution iteratively. (Reference: Laplacian on a Riemannian manifold [ROSENBERG].)

My question is will it spread all over the whole manifold? It seems that the bump function cut it off. But this should not happen for a heat kernel that it just vanish outside a neighbourhood. Because physically, the heat should be able to conduct to everywhere.

  • $\begingroup$ It certainly should be true that the heat kernel on a connected Riemannian manifold is strictly positive everywhere at all positive times. Can you explain in more detail why you think it wouldn't be? Perhaps you could include a scan of the relevant page from the book. $\endgroup$ – Nate Eldredge Sep 4 '18 at 22:05
  • $\begingroup$ @NateEldredge Just by the way of construction: it is constructed locally by a parametrix times a bump function. So it vanishes outside that neighbourhood. Will the last step - the infinite iteration makes it spread all over the whole manifold? cambridge.org/core/books/laplacian-on-a-riemannian-manifold/… chapter 3 $\endgroup$ – Upc Sep 4 '18 at 22:31
  • $\begingroup$ The book is paywalled so I can't read it. But I think that's the issue. When you take the convolution of two compactly supported functions, the result is still compactly supported but the support is bigger. When you take a limit of such functions, the limit need not be compactly supported at all. $\endgroup$ – Nate Eldredge Sep 4 '18 at 23:11
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    $\begingroup$ @NateEldredge, the book is freely available on the authors website here: math.bu.edu/people/sr. (A small plug, I find this is an excellent text for students, and its accessibility is also a mark in its favor). $\endgroup$ – Hadrian Quan Sep 5 '18 at 14:42

This bump function is not supported on an arbitrary neighborhood, but one very specific to the construction of $H(t,x,y)$. Rosenberg's argument proceeds by constructing the parametrix $H_k(t,x,y)$ in a neighborhood of the diagonal $M_{\text{diag}}\subset M\times M$. Namely the neighborhood $U_\epsilon=\{(x,y)\in M\times M:d(x,y)<\epsilon\}$.

The idea is that the heat kernel $H(t,x,y)$ vanishes for small $t$, provided that $x\neq y$. In fact, you should expect it vanish like $t^{-m/2}\text{exp}\left(-\frac{d(x,y)^2}{4t}\right)$, when comparing it with the Euclidean heat kernel. For this reason, you don't expect points in $(0,\infty)_t\times (M\times M\setminus U_\epsilon)$ to contribute much to the value of the Heat kernel, and the parametrix construction relies on this to some degree. The parametrix $H_k$ does not agree pointwise with the true heat kernel $H$, but instead differs by some small error (see theorem 3.22)

What he proves is that $H(t,x,y)=H_k(t,x,y)+Q_k\ast H_k(t,x,y)$, where the error function $Q_k$ satisfies $|Q_k|<Ct^{k-m/2}$, and that $Q_k\in C^l$ as long as $k>l+\frac{n}{2}$. In particular, on a closed manifold, convolution with $Q_k$ defines a compact operator (it is in this sense that the error is "small").

  • $\begingroup$ Also, presumably $Q_k$ is not compactly supported, and so this wouldn't imply that $H$ is compactly supported even if $H_k$ is. $\endgroup$ – Nate Eldredge Sep 5 '18 at 1:06
  • $\begingroup$ Indeed it is not. $H_k$ is compactly supported on $M\times M$ for the silly reason that $M$ is compact without boundary. However, the smooth cutoff we are multiplying by to obtain our $H_k$ is not even defined on $\mathbb{R}^+$, so it does nothing to the support of $H_k$ or $H$ on this factor. $\endgroup$ – Hadrian Quan Sep 5 '18 at 14:47
  • $\begingroup$ What you said are correct. I was reminded of a task i did in first year analysis: convolve a mountain function (0 on $(-\infty,-1)\cup(1,+\infty)$, $x+1$ on $[-1,0)$ and $1-x$ on $[0,1]$)with itself. Gradually, the support gets bigger and bigger and spread on the whole real line eventually. If we choose the normalize factor nicely, eventually, we will get the Gaussian curve due to central limit theorem. Gaussian curve is also the shape of heat kernel of $\mathbb{R}^n$. And it really doesn't matter what shape of the mountain is as long as it converges. I thinkg that is idea of the construction. $\endgroup$ – Upc Sep 5 '18 at 15:01
  • $\begingroup$ But the problem is: can one be sure that it always spreads on the whole manifold by this way of construction no matter how does the manifold look like? It looks like a magic that you only work locally, then you get a something global? If I had read carefully, during the construction, I did not see anywhere we need the assumption that the manifold is connect although this was said at the beginning of this chapter. But if we want to construct such kernel on two disjoint $S^2$. Assume we just pick a small neighborhood of a point on one $S^2$. We can get the heat kernel. But I dont think $\endgroup$ – Upc Sep 5 '18 at 15:09
  • $\begingroup$ this will be conducted onto another one. So it is always be zero on the other $S^2$ but it is absurd.. $\endgroup$ – Upc Sep 5 '18 at 15:11

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