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  "body": "\n## Definition\n\nLet $(M, g)$ be a closed [[Riemannian manifold|7CBT4SM83Y]] with volume measure $dv$, and let $\\Delta = -\\operatorname{div} \\operatorname{grad}$ be its [[Laplace–Beltrami operator|F4NWZNTRMD]], with the sign that makes it nonnegative.\nThe *heat kernel* of $M$ is the function $K \\colon (0, \\infty) \\times M \\times M \\to \\R$ such that:\n\n1. $K$ is continuous, and it is smooth in $t$ and $x$;\n2. $\\partial_t K(t, x, y) + \\Delta_x K(t, x, y) = 0$ for all $t \u003e 0$ and $x, y \\in M$;\n3. for every continuous $f \\colon M \\to \\R$ and every $x \\in M$, $\\int_M K(t, x, y) f(y) \\, dv(y) \\to f(x)$ as $t \\to 0^+$.\n\nSuch a function exists and is unique.\n\n## Remarks\n\nIf $\\phi_0, \\phi_1, \\ldots$ is an orthonormal basis of $L^2(M)$ of eigenfunctions with $\\Delta \\phi_j = \\lambda_j \\phi_j$, then\n$$K(t, x, y) = \\sum_{j \\ge 0} e^{-t \\lambda_j} \\phi_j(x) \\phi_j(y),$$\nthe series converging uniformly for $t \\ge t_0 \u003e 0$.\nSo $K(t, \\cdot, \\cdot)$ is the integral kernel of the operator $e^{-t\\Delta}$, and $K$ is smooth, symmetric in $x$ and $y$, and positive.\nThe same definition, with $\\Delta$ replaced by a Laplace-type operator on a vector bundle, gives heat kernels of operators such as the Hodge Laplacian.\nOn a complete noncompact manifold one takes the minimal positive fundamental solution; for $\\R^n$ it is the Gaussian [[heat kernel|KD404MF042]].\n\nExample: on $S^1 = \\R / 2\\pi\\Z$ with $\\Delta = -d^2/d\\theta^2$, the eigenvalues are $k^2$ with eigenfunctions $e^{ik\\theta}$, and\n$$K(t, \\theta, \\theta') = \\frac{1}{2\\pi} \\sum_{k \\in \\Z} e^{-k^2 t} e^{ik(\\theta - \\theta')}.$$\nEach term solves the heat equation.\nFor fixed $t$, $K(t, \\theta, \\cdot)$ is the sum over $m \\in \\Z$ of the Gaussian kernel of $\\R$ translated by $2\\pi m$.\nIt is a positive kernel of total mass $1$ that concentrates at $\\theta$ as $t \\to 0^+$, which gives clause 3.\n\nNon-example: on $S^1$ the constant $K \\equiv 1/(2\\pi)$ satisfies clauses 1 and 2, but $\\int K f \\to \\frac{1}{2\\pi}\\int f$, not $f(x)$, so clause 3 fails.\n",
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