In $\mathbb{R}^n$, if $\gamma$ is a line segment between $x_0 = \gamma (0)$ and $x = \gamma (t)$, one has the following formula:

$$\frac {\mathbb{e}^{- \frac{1}{4} \int_0^t <\dot{\gamma}, \dot{\gamma}> \mathbb{d}s}}{\sqrt {4 \pi t}^n} = h(t,x,x_0) ,$$

where $h$ is the kernel of the heat equation centered in $x_0$ at time $0$.

Now, if $(M,g)$ is a Riemann manifold (with the appropriate conditions for the heat kernel to exist), is there any curve $\gamma$ that could play a similar role to the line segment in the above formula (possibly with a different denominator)?

The only formula that I know of is

$$\mathbb{e}^{- \frac{1}{4} \int_0^t <\dot{\gamma}, \dot{\gamma}> \mathbb{d}s} = \mathbb{e}^ {- \frac{d(x,x_0)^2}{4 t}} ,$$

where $d$ is the distance associated to $g$ and $\gamma$ is a minimizing geodesic. This is nice, but quite far from the heat kernel.

  • 1
    $\begingroup$ There exists such formula in a stochastic way. Choose a measure on the space of curves. Take the integral on the space of curves will give the heat kernel. The first formula you mentioned is just take the dirac measure with support of the curve connecting $x_0$ and $x$. Google Brownian motion will give you more results. $\endgroup$
    – shu
    Feb 16, 2015 at 19:28
  • $\begingroup$ This link[en.wikipedia.org/wiki/Heat_kernel] gives a similar result. $\endgroup$
    – Semsem
    Feb 16, 2015 at 21:00

1 Answer 1


Actually, I think your second formula might indeed relate the distance on a Riemannian manifold to the heat kernel - see this paper by Varadhan, in which it is shown that if $p(t,x,y)$ is a solution to

$\frac{\partial p}{\partial t} = \frac{1}{2} \sum_{i,j=1}^k a_{ij}(x) \frac{\partial^2 p}{\partial x^i \partial x^j}$

such that $p(t,x,y) \to \delta_x(y)$ as $t \to 0$, then

$\lim_{t \to 0} -2t \log p(t,x,y) = d(x,y)^2,$

where $d$ is distance in the Riemannian metric derived from the coefficients $\left\{ a_{ij}(x) \right\}$.

(the link below is to a preview of the paper, but the relevant equation (what I gave above) is right on the front page):


I found this from another paper by Saloff-Coste (available in full here):


which in turn I found by some strategic googling because it looked to me as though you might be able to get somewhere by viewing the inner product $\langle \dot{\gamma}, \dot{\gamma} \rangle$ in your exponent as a Dirichlet form (see e.g. Dirichlet Forms and Symmetric Markov Processes by Fukushima, Oshima, and Takeda).

Hope this helps!

  • $\begingroup$ Indeed, it is helpful, and hopefully setting me on some right track. This is why I have upvoted your answer. Nevertheless, it is not the answer I was exactly looking for. $\endgroup$
    – Alex M.
    Feb 18, 2015 at 17:12

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