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Let $(M,g)$ be a compact Riemannian manifold. It is known there exist Gaussian estimates of the heat kernel and its derivatives acting on functions on $M$.

The kind of estimate I'm looking for could be found in the first page of the following article : http://www.math.uni-bielefeld.de/~grigor/grad.pdf

My question is : it is possible to extends this kind of result at the level of the $k$-forms in $M$ ?

More specifically I would be interested by inequalities of the following kind, for small value of $t$ :

\begin{equation*} ||\nabla_y H^1(x,y,t)|| \le C_1(t) \exp({\frac{- c_2 d(x,y)^2}{t}}) \end{equation*}

where $H^1(x,y,t)$ is the heat kernel for $1$-forms viewed as a $(1,1)$ form on the product $M \times M \verb+\+ \Delta$, if $t$ is freeze. $\Delta$ denote the diagonal and $|| \cdot ||$ the punctual norm induced by $g$. And $C_1(t)$ a well controlled function of $t$, some rational fraction for instance. I've read a lot of stuff about functions but nothing about the forms. Someone suggests me to look at the Kato's inequalities but I don't really see how to use it because this one compare the solutions of different elliptic equations, not the kernel.

Any help and references will be appreciated.

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There should be better references, but for a beginning, how about the following: Look at Lemme 1 in Thierry Bouche's paper Convergence de la metrique de Fubini-Study d'un fibre lineaire positif, Ann. Inst. Grenoble 1990. Here is the link to the paper:

http://archive.numdam.org/ARCHIVE/AIF/AIF_1990__40_1/AIF_1990__40_1_117_0/AIF_1990__40_1_117_0.pdf .

The proof of Lemme 1 on page 122 tells you how to apply Kato's inequality to read the desired result from the leading term in the Minakshisundaram-Pleijel asymptotic expansion (the case of functions).

The paper is about complex projective manifolds, but $(M,g)$ in section 1 is an arbitrary compact Riemannian manifold. Consider just $E = M \times \mathbb{C}$ (trivial), $F = T^*M$, $k = 1$, and $V = 0$.

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  • $\begingroup$ Thanks for the reference, do you think that it is easy to extend the result to the first derivative ? $\endgroup$
    – user50806
    Jan 7, 2015 at 17:36
  • $\begingroup$ @AdrienBoulanger: I think so, it should be easy to extend it to all derivatives. $\endgroup$ Jan 7, 2015 at 19:30

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