2 Heat Kernel for the Laplacian

In the previous section, our solutions $f(x, t)$ had the property that $f(-, t) \to f_0$ in $L^2$. One might hope that, for example, if $f_0$ is continuous, then this converges in $C^0$. Another shortcoming of our previous approach is that we have no control over what $H_t$ looks like as a function as the whole construction is manifestly global.

In this section, we will construct the heat kernel via an alternative method. This method involves starting with an initial “guess” of what the heat kernel should be. We then show that if our initial guess is “good enough”, then there is an iterative procedure that gives us the actual heat kernel. Moreover, this iterative procedure is sufficiently explicit that we can relate the asymptotic behaviour of $H_t$ as $t \to 0$ back to that of our original guess.

We will now restrict to the case of the Laplace–Beltrami operator, since we can use the explicit solution for $\mathbb {R}^d$ as an inspiration for our initial guess. We fix a $p$ and consider the Laplace–Beltrami operator on $\Omega ^p$.

To motivate why one should care about the convolution, notice that Duhamel's principle essentially says that the heat kernel $H$ satisfies

$$\left(\frac{\mathrm{d}}{\mathrm{d}t} + \Delta _x\right) H * F = F$$

for all $F$. The condition that the convolution is $C^2$ is a technical point that will be useful later on.

Write $r(x, y)$ for the geodesic distance between $x, y \in M$. The very first naive guess for $K_t(x, y)$ might just be

$$\frac{1}{(4\pi t)^{d/2}} e^{-r^2/4t}.$$

This definitely satisfies (i). To see how good an attempt we made, we compute $\frac{\partial }{\partial t} + \Delta _x$ of this.

To do so, we fix a $y \in M$ and express $x$ in geodesic polar coordinates about $y$. Then there is a formula

$$\Delta _x (F(r) \alpha ) = \left(\frac{\mathrm{d}^2 F}{\mathrm{d}r^2} + \frac{d - 1}{r} \frac{\mathrm{d}F}{\mathrm{d}r} + \frac{1}{2g} \frac{\mathrm{d}g}{\mathrm{d}r} \frac{\mathrm{d}F}{\mathrm{d}r}\right)\alpha + \frac{2}{r} \frac{\mathrm{d}F}{\mathrm{d}r} \nabla _{r \frac{\partial }{\partial r}} \alpha + F \Delta _x \alpha$$

where $g$ is the determinant of the metric (note that $r\frac{\partial }{\partial r}$ is defined at $r = 0$, but $\frac{\partial }{\partial r}$ is not, hence the funny notation). So we find that

$$\left(\frac{\partial }{\partial t} + \Delta _x\right) \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} = -\frac{r}{4g t} \frac{\mathrm{d}g}{\mathrm{d}r} \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}}.$$

If we were in flat space, then $\frac{\mathrm{d}g}{\mathrm{d}r} = 0$, and so this is in fact a legitimate fundamental solution, as one would expect. However, when $\frac{\mathrm{d}g}{\mathrm{d}r} \not= 0$, then we can't quite bound this as $t \to 0$.

To solve this, we introduce some correction terms, which are small enough to preserve the property (i). We shall find $u_i(x, y) \in \Gamma (\Omega ^p \boxtimes (\Omega ^p)^*)$ with $f_0(x, x) = I$ such that

$$K_{N, t} (x, y) = \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} \left(\sum _{i = 0}^N t^i u_i(x, y)\right)$$

satisfies the equality

$$\left(\frac{\partial }{\partial t} + \Delta _x\right) K^N_t(x, y) = \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}}\, t^N \Delta _x u_N(x, y).$$

The precise shape of the right-hand side does not matter. It is just what happens to be convenient. The point is that the exponent of $t$ is $N - \frac{d}{2}$, which is eventually positive. Then we simply have to take $K = K_N$ for large enough $N$ (note that this $N$ is different from the exponent $N$ in (ii) by $\frac{d}{2}$). Actually, we will only succeed in finding $u_N$ in a neighbourhood of the diagonal, since we work in geodesic polars. But we can simply take $K = K_N \varphi (x, y)$, where $\varphi$ is a bump function supported near the diagonal. ((iii) is left as an exercise)

To find the $u_i$, we assume $K_N$ is of the given form, and compute

$$\left(\frac{\partial }{\partial t} - \Delta _x\right) K_{N, t}(x, y) = \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} \sum _{i = 0}^N \left[\left( \frac{i}{t} + \frac{r}{4gt} \frac{\mathrm{d}g}{\mathrm{d}r}\right) t^i u_i(x, y) + t^{i - 1} \nabla _{r \frac{\partial }{\partial r}} u_i + t^i \Delta _x u_i(x, y)\right].$$

To obtain the desired result, we need the coefficient of $\frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} t^{i - 1}$ to vanish for $i \leq N$. So we need to inductively solve

$$\left(i + \frac{r}{4g} \frac{\mathrm{d}g}{\mathrm{d}r}\right) u_i(x, y) + \nabla _{r \frac{\partial }{\partial r}} u_i(x, y) = \Delta _x u_{i - 1}(x, y).$$

The diligent reader will observe that the left-hand side is exact, so we can write this as

$$\nabla _{r\frac{\partial }{\partial r}} \left(r^i g^{1/4} u_i(x, y)\right) = r^i g^{1/4} \Delta _x u_{i - 1}(x, y).$$

The existence and uniqueness of a local solution then follows easily (one has to be slightly careful about $r = 0$, where the original equation gives an initial condition $i u_i = \Delta _x u_{i - 1}$).

For future reference, we note that $\nabla _{r \frac{\partial }{\partial r}} = 0$ when $r = 0$, so

$$K_t(x, x) = \text{const} \cdot \sum _{i = 0}^N t^{i - d/2} u_i(x, x),$$

where $u_i(x, x)$ is a rational function in $g^{ij}$ and its derivatives.

We will now fix an $N$ large enough and take $K$ to be the function given by the previous lemma.

To find the actual fundamental solution, we have to understand the convolution a bit better. Recall that we expect a genuine fundamental solution $H$ to satisfy $\left(\frac{\mathrm{d}}{\mathrm{d}t} + \Delta _x\right) (H * F) = F$. Using the fact that $K_t(x, y) \to \delta (x, y)$ as $t \to 0$, we can calculate

So at least formally, the heat kernel should be given by

$$H = \sum _{m = 0}^\infty (-1)^m K * \left[\left(\frac{\partial }{\partial t} + \Delta _x\right)K\right]^{*m}.$$

Barring convergence issues, the formula above also shows that $\left(\frac{\mathrm{d}}{\mathrm{d}t} + \Delta _x\right) H = 0$.

To check convergence, the naive bound

$$\| F * G\| _{L^\infty } \leq C t \operatorname{vol}(M) \| F\| _{L^\infty } \| G\| _{L^\infty }$$

is not quite enough. We set

$$G = \left(\frac{\partial }{\partial t} + \Delta _x \right) K.$$

We then have $\| G_t\| _{L^\infty } \leq C t^N$. We claim that

$$\| G^{*(j + 1)}_t\| _{L^\infty } \leq \frac{C^{j + 1} \operatorname{vol}(M)^j}{(N + j) \cdots (N + 2)(N + 1)} t^{N(j + 1) + j}.$$

The terms in the denominator are the interesting bits that come from doing the bound more carefully. They ensure the sum above converges by the ratio test. This is a standard inductive calculation

$$\begin{aligned} |G^{*(j + 1)}_t(x, y)| & \leq \int _0^t \mathrm{d}s\int _M \mathrm{d}z\; |G_{t - s}(x, z)| |G_s^{*j}(z, y)|\\ & \leq \int _0^t \mathrm{d}s\int _M \mathrm{d}z\; C t^N \cdot \frac{C^j \operatorname{vol}(M)^{j - 1}}{(N + j - 1) \cdots (N + 1)} s^{Nj + j - 1}\\ & \leq \frac{C^{j + 1} \operatorname{vol}(M)^j}{(N + j - 1) \cdots (N + 1)} t^{Nj} \int _0^t s^{N + j - 1}\; \mathrm{d}s\\ & = \frac{C^{j + 1} \operatorname{vol}(M)^j}{(N + j) \cdots (N + 1)} t^{N(j + 1) + j}. \end{aligned}$$

Adding these up and using the ratio test, we find that

$$H = K + K * \sum _{m = 1}^\infty (-1) G^{*m}$$

converges, and we have the following theorem:

From this, we can read off the leading term of $H_t(x, x)$ as $t \to 0$, which is, unsurprisingly, $\frac{e^{-r^2/4t}}{(4\pi t)^{d/2}}$. As a recap, to produce this estimate, we used this as an initial approximation to the heat kernel, and then perform some procedures to turn this into an actual heat kernel. We then observe that all the correction terms we add are of higher order, so we are happy.

Note that all we used about $K_t$ was the properties stated in Lemma 2.1, and the procedure is pretty general. One can interpret the success of this procedure as testifying to the locality of the heat kernel. The properties of $K_t$ we required were always local conditions, namely that $K_t$ looks locally like a heat kernel for small $t$. We then get to produce an actual heat kernel $H_t$, whose difference from $K_t$ is something we can control very well, and in particular vanishes in the limit $t \to 0$.

In general, if our manifold is written as a union of two open submanifolds, and we have produced heat kernels on each submanifold (suitably interpreted), then we can attempt to glue them together using a partition of unity. This will not give us an actual heat kernel, but the failure to be a heat kernel decreases exponentially as $t \to 0$ (since $H_t(x, y)$ falls exponentially as $t \to 0$ whenever $x \not= y$). The above argument then lets us produce a heat kernel whose difference from the “fake” one vanishes to all orders as $t \to 0$.