[proofplan] The proof is a pointwise change of variables between the positive function $u$ and its logarithmic potential $f$. We differentiate $u=(4\pi\tau)^{-n/2}e^{-f}$ in time, using the time derivative equation for $\tau$, and compute its spatial Laplacian using the product rule for the Laplace-Beltrami operator. Substituting these two identities into the conjugate [heat equation](/page/Heat%20Equation) and dividing by the positive function $u$ gives the stated equation for $f$; the reverse implication follows by the same algebra in reverse. [/proofplan]

[step:Differentiate the logarithmic density in time] By hypothesis, $f:M\times I\to\mathbb{R}$ and $\tau:I\to(0,\infty)$ are smooth, and $n=\dim M$. Define the time-dependent scalar factor $a: I \to (0,\infty)$ by \begin{align*} a(t) = (4\pi\tau(t))^{-n/2}. \end{align*} Then $u(x,t)=a(t)e^{-f(x,t)}$. By the ordinary Chain Rule in $t$ and the time derivative equation for $\tau$, \begin{align*} \frac{a'(t)}{a(t)} = -\frac{n}{2}\frac{\tau'(t)}{\tau(t)} = \frac{n}{2\tau(t)}. \end{align*} Therefore, for every $(x,t)\in M\times I$, \begin{align*} \partial_t u(x,t) = a'(t)e^{-f(x,t)} - a(t)e^{-f(x,t)}\partial_t f(x,t) = u(x,t)\left(\frac{n}{2\tau(t)}-\partial_t f(x,t)\right). \end{align*} [/step]

[step:Compute the spatial Laplacian of the logarithmic density]Fix $t\in I$. By the smoothness hypotheses on $f:M\times I\to\mathbb{R}$ and the Ricci flow $(M,g(t))_{t\in I}$, and since $a(t)>0$, the functions $f(\cdot,t):M\to\mathbb{R}$ and $u(\cdot,t):M\to(0,\infty)$ are smooth on the Riemannian manifold $(M,g(t))$. Since $a(t)$ is independent of the spatial variable $x$, the spatial Chain Rule gives the gradient of $u(\cdot,t)$ with respect to $g(t)$: \begin{align*} \nabla u = -u\nabla f. \end{align*} Using the definition of the Laplace-Beltrami Operator, $\Delta_{g(t)}h=\operatorname{div}_{g(t)}(\nabla h)$, and the Product Rule for Divergence, \begin{align*} \Delta_{g(t)}u = \operatorname{div}_{g(t)}(-u\nabla f). \end{align*} The product rule gives \begin{align*} \operatorname{div}_{g(t)}(-u\nabla f) = -g(t)(\nabla u,\nabla f)-u\Delta_{g(t)}f. \end{align*} Substituting $\nabla u=-u\nabla f$ gives \begin{align*} \Delta_{g(t)}u = u|\nabla f|_{g(t)}^2-u\Delta_{g(t)}f. \end{align*}[/step]

[guided]Fix a time $t\in I$. The theorem assumes that $f:M\times I\to\mathbb{R}$ is smooth, that $(M,g(t))_{t\in I}$ is a smooth Ricci flow, and that $\tau:I\to(0,\infty)$ is smooth. Since $a(t)=(4\pi\tau(t))^{-n/2}>0$, the functions $f(\cdot,t):M\to\mathbb{R}$ and $u(\cdot,t):M\to(0,\infty)$ are smooth on the Riemannian manifold $(M,g(t))$. Thus the spatial chain rule and the divergence product rule apply to these functions and vector fields. The operator $\Delta_{g(t)}$ differentiates only in the spatial variable $x$, while the factor $(4\pi\tau(t))^{-n/2}$ depends only on time. Thus, when taking the spatial gradient, this prefactor behaves as a constant. From \begin{align*} u(x,t)=(4\pi\tau(t))^{-n/2}e^{-f(x,t)}, \end{align*} the Chain Rule for the spatial gradient gives \begin{align*} \nabla u = -u\nabla f. \end{align*} Now apply the Laplace-Beltrami Operator with the convention $\Delta_{g(t)}h=\operatorname{div}_{g(t)}(\nabla h)$. Since $\nabla u=-u\nabla f$, we have \begin{align*} \Delta_{g(t)}u = \operatorname{div}_{g(t)}(-u\nabla f). \end{align*} Let $\mathfrak{X}(M)$ denote the space of smooth vector fields on $M$. The Product Rule for Divergence says that for a smooth scalar function $\phi: M\to\mathbb{R}$ and a smooth vector field $X\in\mathfrak{X}(M)$, \begin{align*} \operatorname{div}_{g(t)}(\phi X) = g(t)(\nabla \phi,X)+\phi\,\operatorname{div}_{g(t)}X. \end{align*} We apply this with $\phi=-u$ and $X=\nabla f$. This gives \begin{align*} \Delta_{g(t)}u = -g(t)(\nabla u,\nabla f)-u\operatorname{div}_{g(t)}(\nabla f). \end{align*} Because $\operatorname{div}_{g(t)}(\nabla f)=\Delta_{g(t)}f$, this becomes \begin{align*} \Delta_{g(t)}u = -g(t)(\nabla u,\nabla f)-u\Delta_{g(t)}f. \end{align*} Substituting $\nabla u=-u\nabla f$ into the first term yields \begin{align*} -g(t)(\nabla u,\nabla f) = -g(t)(-u\nabla f,\nabla f) = u|\nabla f|_{g(t)}^2. \end{align*} Therefore \begin{align*} \Delta_{g(t)}u = u|\nabla f|_{g(t)}^2-u\Delta_{g(t)}f. \end{align*}[/guided]

[step:Substitute the identities into the conjugate heat equation] Assume first that $u$ satisfies \begin{align*} \partial_t u = -\Delta_{g(t)}u + S_{g(t)}u. \end{align*} Using the two identities above, we obtain \begin{align*} u\left(\frac{n}{2\tau}-\partial_t f\right) = -u\left(|\nabla f|_{g(t)}^2-\Delta_{g(t)}f\right)+S_{g(t)}u. \end{align*} Collecting the terms multiplied by $u$ gives \begin{align*} u\left(\frac{n}{2\tau}-\partial_t f\right) = u\left(\Delta_{g(t)}f-|\nabla f|_{g(t)}^2+S_{g(t)}\right). \end{align*} Since $u(x,t)>0$ for every $(x,t)\in M\times I$, division by $u$ gives \begin{align*} \frac{n}{2\tau}-\partial_t f = \Delta_{g(t)}f-|\nabla f|_{g(t)}^2+S_{g(t)}. \end{align*} Rearranging terms gives \begin{align*} \partial_t f = -\Delta_{g(t)}f+|\nabla f|_{g(t)}^2-S_{g(t)}+\frac{n}{2\tau}. \end{align*} [/step]

[step:Reverse the algebra to prove the converse] Conversely, assume that $f$ satisfies \begin{align*} \partial_t f = -\Delta_{g(t)}f+|\nabla f|_{g(t)}^2-S_{g(t)}+\frac{n}{2\tau}. \end{align*} Then \begin{align*} \frac{n}{2\tau}-\partial_t f = \Delta_{g(t)}f-|\nabla f|_{g(t)}^2+S_{g(t)}. \end{align*} Multiplying by $u$ and using the time derivative identity gives \begin{align*} \partial_t u = u\left(\Delta_{g(t)}f-|\nabla f|_{g(t)}^2+S_{g(t)}\right). \end{align*} Using the Laplacian identity \begin{align*} \Delta_{g(t)}u = u|\nabla f|_{g(t)}^2-u\Delta_{g(t)}f, \end{align*} we rewrite the right-hand side as \begin{align*} u\left(\Delta_{g(t)}f-|\nabla f|_{g(t)}^2+S_{g(t)}\right) = -\Delta_{g(t)}u+S_{g(t)}u. \end{align*} Hence \begin{align*} \partial_t u = -\Delta_{g(t)}u+S_{g(t)}u, \end{align*} so $u$ satisfies the conjugate heat equation. This proves both implications. [/step]