[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](/page/Chain%20Rule) 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](/page/Chain%20Rule) 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](/page/Laplace-Beltrami%20Operator), $\Delta_{g(t)}h=\operatorname{div}_{g(t)}(\nabla h)$, and the [Product Rule](/page/Product%20Rule) for [Divergence](/page/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*}
[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](/page/Chain%20Rule) for the spatial gradient gives
\begin{align*}
\nabla u = -u\nabla f.
\end{align*}
Now apply the [Laplace-Beltrami Operator](/page/Laplace-Beltrami%20Operator) 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](/page/Product%20Rule) for [Divergence](/page/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]
[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]
