[proofplan]
The proof compares the time derivative of $|\nabla u|_{g(t)}^2$ with the static Bochner formula for the fixed metric $g(t)$. The time derivative has two contributions: one from differentiating $u$ and one from differentiating the inverse metric, the latter producing $2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u)$. The fixed-time Bochner identity produces the same Ricci term with the opposite sign after subtracting $\Delta_{g(t)}|\nabla u|^2$. The remaining first-order term is $2\langle \nabla u,\nabla(\partial_tu-\Delta_{g(t)}u)\rangle_{g(t)}$, which vanishes by the [heat equation](/page/Heat%20Equation).
[/proofplan]
[step:Differentiate the gradient norm along the evolving metric]
Fix $t \in I^\circ$ and $p \in M$. Let $(U,\varphi)$ be a coordinate chart around $p$ with coordinates $(x_1,\dots,x_n)$, and write
\begin{align*}
g(t) &= \sum_{i,j=1}^n g_{ij}(\cdot,t)\,dx_i \otimes dx_j, &
g(t)^{-1} &= \sum_{i,j=1}^n g^{ij}(\cdot,t)\,\partial_{x_i}\otimes \partial_{x_j}.
\end{align*}
For the scalar function $u$, define the coordinate first derivatives
\begin{align*}
u_i: U \times I &\to \mathbb{R}, &
u_i(q,s) &= \partial_{x_i}u(q,s).
\end{align*}
Then
\begin{align*}
|\nabla u|_{g(t)}^2
=
\sum_{i,j=1}^n g^{ij}u_i u_j.
\end{align*}
Differentiating the identity $\sum_{k=1}^n g^{ik}g_{kj}=\delta_{ij}$ in time gives
\begin{align*}
\partial_t g^{ij}
=
-\sum_{a,b=1}^n g^{ia}(\partial_t g_{ab})g^{bj}.
\end{align*}
Since $\partial_t g_{ab}=-2(\operatorname{Ric}_{g(t)})_{ab}$, this becomes
\begin{align*}
\partial_t g^{ij}
=
2\sum_{a,b=1}^n g^{ia}(\operatorname{Ric}_{g(t)})_{ab}g^{bj}.
\end{align*}
Therefore the product rule gives
\begin{align*}
\partial_t |\nabla u|_{g(t)}^2 = \sum_{i,j=1}^n (\partial_t g^{ij})u_i u_j + 2\sum_{i,j=1}^n g^{ij}u_i\partial_{x_j}(\partial_tu).
\end{align*}
Using the preceding formula for \partial_t g^{ij}, this becomes
\begin{align*}
\partial_t |\nabla u|_{g(t)}^2 = 2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u) + 2\langle \nabla u,\nabla(\partial_tu)\rangle_{g(t)}.
\end{align*}
[/step]
[step:Compute the fixed-time Laplacian by the Bochner identity]
At the fixed time $t$, regard $g(t)$ as a single Riemannian metric. We prove the pointwise identity
\begin{align*}
\Delta_{g(t)}|\nabla u|_{g(t)}^2 = 2|\operatorname{Hess}_{g(t)}u|_{g(t)}^2 + 2\langle \nabla u,\nabla\Delta_{g(t)}u\rangle_{g(t)} + 2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u).
\end{align*}
Choose [normal coordinates](/theorems/2713) for $g(t)$ at $p$. Let $\nabla^{g(t)}$ denote the Levi-Civita connection of $g(t)$, and define the Christoffel symbols $\Gamma_{ij}^k: U \to \mathbb{R}$ by
\begin{align*}
\nabla^{g(t)}_{\partial_{x_i}}\partial_{x_j}
=
\sum_{k=1}^n \Gamma_{ij}^k\,\partial_{x_k}.
\end{align*}
At $p$, the metric satisfies $g_{ij}=\delta_{ij}$ and the Christoffel symbols satisfy $\Gamma_{ij}^k=0$. Define
\begin{align*}
u_{ij}: U \times I &\to \mathbb{R}, &
u_{ij}(q,s) &= (\operatorname{Hess}_{g(s)}u)_{ij}(q,s).
\end{align*}
At the point $(p,t)$, the frame $\{\partial_{x_1},\dots,\partial_{x_n}\}$ is $g(t)$-orthonormal and the Christoffel symbols vanish. Thus the tensorial contraction defining the Laplacian at $p$ gives
\begin{align*}
\Delta_{g(t)}|\nabla u|_{g(t)}^2(p,t)
=
\sum_{k=1}^n \nabla_k\nabla_k\left(|\nabla u|_{g(t)}^2\right)(p,t)
=
\sum_{k=1}^n \nabla_k\nabla_k\left(\sum_{i=1}^n u_i^2\right)(p,t).
\end{align*}
The final equality is asserted only at the base point $p$, where $g^{ij}(p,t)=\delta_{ij}$; it is not a neighbourhood identity for $|\nabla u|_{g(t)}^2$. Applying the product rule to each summand gives
\begin{align*}
\Delta_{g(t)}|\nabla u|_{g(t)}^2 = 2\sum_{i,k=1}^n u_{ik}^2 + 2\sum_{i=1}^n u_i\sum_{k=1}^n \nabla_k\nabla_k\nabla_i u.
\end{align*}
We use the Laplace-Beltrami sign convention
\begin{align*}
\Delta_{g(t)}f
=
\operatorname{tr}_{g(t)}\operatorname{Hess}_{g(t)}f
\end{align*}
for scalar functions $f: M \to \mathbb{R}$.
[claim:Commute the rough Laplacian of the gradient]
With the Riemann curvature tensor convention
\begin{align*}
R(X,Y)Z
=
\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,
\end{align*}
and Ricci curvature obtained by tracing the first and third slots, which is the curvature convention used throughout this proof, the following identity holds at $(p,t)$:
\begin{align*}
\sum_{k=1}^n \nabla_k\nabla_k\nabla_i u
=
\nabla_i\Delta_{g(t)}u
+
\sum_{\ell=1}^n(\operatorname{Ric}_{g(t)})_{i\ell}u_\ell.
\end{align*}
[/claim]
[proof]
At $p$, the chosen coordinates are normal for $g(t)$, so covariant second derivatives of the scalar function $u(\cdot,t): M \to \mathbb{R}$ commute. Thus
\begin{align*}
\sum_{k=1}^n \nabla_k\nabla_k\nabla_i u
=
\sum_{k=1}^n \nabla_k\nabla_i\nabla_k u.
\end{align*}
Commuting the two covariant derivatives acting on the covector field $\nabla u$ gives
\begin{align*}
\sum_{k=1}^n \nabla_k\nabla_i\nabla_k u
=
\sum_{k=1}^n \nabla_i\nabla_k\nabla_k u
+
\sum_{k,\ell=1}^n (R_{g(t)})_{k i k \ell}u_\ell.
\end{align*}
The first term is $\nabla_i\Delta_{g(t)}u$ by the sign convention for $\Delta_{g(t)}$, and the curvature trace satisfies $\sum_{k=1}^n (R_{g(t)})_{k i k \ell}=(\operatorname{Ric}_{g(t)})_{i\ell}$ by the stated Ricci convention. This proves the claimed commutation identity.
[/proof]
Substituting this into the previous formula yields the displayed Bochner identity at $(p,t)$. Since both sides are tensorial, the identity holds independently of the chosen normal coordinates.
[guided]
We now compute the spatial Laplacian while freezing the time variable $t$. This means $g(t)$ is treated as an ordinary Riemannian metric on $M$, and the time derivative of the metric plays no role in this step.
Choose normal coordinates for $g(t)$ at the fixed point $p$. Let $\nabla^{g(t)}$ denote the Levi-Civita connection of $g(t)$, and define the Christoffel symbols $\Gamma_{ij}^k: U \to \mathbb{R}$ by
\begin{align*}
\nabla^{g(t)}_{\partial_{x_i}}\partial_{x_j}
=
\sum_{k=1}^n \Gamma_{ij}^k\,\partial_{x_k}.
\end{align*}
In these coordinates, at $p$ we have $g_{ij}=\delta_{ij}$ and $\Gamma_{ij}^k=0$. These coordinates simplify the calculation at the point, but the quantities being computed are tensorial, so the final identity does not depend on this choice.
For the scalar function $u$, the first covariant derivative agrees with the coordinate derivative, so $u_i=\partial_{x_i}u$. The Hessian components are
\begin{align*}
u_{ij}
=
(\operatorname{Hess}_{g(t)}u)_{ij}
=
\nabla_i\nabla_j u.
\end{align*}
At $(p,t)$, the squared gradient is
\begin{align*}
|\nabla u|_{g(t)}^2
=
\sum_{i=1}^n u_i^2.
\end{align*}
At the point $p$, applying the Laplace-Beltrami operator means taking the trace
\begin{align*}
\Delta_{g(t)}
=
\sum_{k=1}^n\nabla_k\nabla_k
\end{align*}
in this normal frame. Therefore
\begin{align*}
\Delta_{g(t)}|\nabla u|_{g(t)}^2(p,t)
=
\sum_{k=1}^n\nabla_k\nabla_k\left(|\nabla u|_{g(t)}^2\right)(p,t)
=
\sum_{k=1}^n\nabla_k\nabla_k\left(\sum_{i=1}^n u_i^2\right)(p,t).
\end{align*}
The equality $|\nabla u|_{g(t)}^2=\sum_i u_i^2$ is being used only at $p$, where the normal frame is orthonormal; away from $p$, the expression includes the components $g^{ij}$.
Applying the product rule to each term in the sum gives
\begin{align*}
\Delta_{g(t)}|\nabla u|_{g(t)}^2 = 2\sum_{i,k=1}^n(\nabla_k u_i)^2 + 2\sum_{i=1}^n u_i\sum_{k=1}^n\nabla_k\nabla_k u_i.
\end{align*}
Because $u_i=\nabla_i u$ and $\nabla_k u_i=\nabla_k\nabla_i u=u_{ki}$, the first term is exactly
\begin{align*}
2\sum_{i,k=1}^n u_{ik}^2 = 2|\operatorname{Hess}_{g(t)}u|_{g(t)}^2.
\end{align*}
The second term is where curvature enters. We use the Laplace-Beltrami sign convention
\begin{align*}
\Delta_{g(t)}f
=
\operatorname{tr}_{g(t)}\operatorname{Hess}_{g(t)}f
\end{align*}
for scalar functions $f: M \to \mathbb{R}$. We also use the Riemann curvature tensor convention
\begin{align*}
R(X,Y)Z
=
\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,
\end{align*}
with Ricci curvature obtained by tracing the first and third slots. This convention determines the sign of the curvature term in the commutation formula below.
Let us verify the commutation formula rather than treating it as a black box. At $p$, the coordinates are normal for $g(t)$, and covariant second derivatives of the scalar function $u(\cdot,t): M \to \mathbb{R}$ commute. Therefore
\begin{align*}
\sum_{k=1}^n\nabla_k\nabla_k\nabla_i u
=
\sum_{k=1}^n\nabla_k\nabla_i\nabla_k u.
\end{align*}
Now the derivatives are acting on the covector field $\nabla u$, whose components are $u_i$. Commuting the two covariant derivatives on this covector field gives
\begin{align*}
\sum_{k=1}^n\nabla_k\nabla_i\nabla_k u
=
\sum_{k=1}^n\nabla_i\nabla_k\nabla_k u
+
\sum_{k,\ell=1}^n(R_{g(t)})_{k i k \ell}u_\ell.
\end{align*}
The first term is $\nabla_i\Delta_{g(t)}u$ by the sign convention for $\Delta_{g(t)}$, while the curvature trace is the Ricci tensor component $\sum_{k=1}^n(R_{g(t)})_{k i k \ell}=(\operatorname{Ric}_{g(t)})_{i\ell}$. Hence
\begin{align*}
\sum_{k=1}^n\nabla_k\nabla_k\nabla_i u
=
\nabla_i\Delta_{g(t)}u
+
\sum_{\ell=1}^n(\operatorname{Ric}_{g(t)})_{i\ell}u_\ell.
\end{align*}
Substituting this identity into the second term produces
\begin{align*}
2\sum_{i=1}^n u_i\sum_{k=1}^n\nabla_k\nabla_k\nabla_i u = 2\sum_{i=1}^n u_i\nabla_i\Delta_{g(t)}u + 2\sum_{i,\ell=1}^n(\operatorname{Ric}_{g(t)})_{i\ell}u_i u_\ell.
\end{align*}
By the coordinate expression for the metric [inner product](/page/Inner%20Product) and for the Ricci tensor pairing, this is
\begin{align*}
2\sum_{i=1}^n u_i\sum_{k=1}^n\nabla_k\nabla_k\nabla_i u = 2\langle \nabla u,\nabla\Delta_{g(t)}u\rangle_{g(t)} + 2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u).
\end{align*}
Combining the Hessian term with these two terms gives
\begin{align*}
\Delta_{g(t)}|\nabla u|_{g(t)}^2 = 2|\operatorname{Hess}_{g(t)}u|_{g(t)}^2 + 2\langle \nabla u,\nabla\Delta_{g(t)}u\rangle_{g(t)} + 2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u).
\end{align*}
[/guided]
[/step]
[step:Subtract the Bochner identity and use the heat equation]
Combining the time derivative formula with the fixed-time Bochner identity gives, at $(p,t)$,
\begin{align*}
(\partial_t-\Delta_{g(t)})|\nabla u|_{g(t)}^2 = 2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u) + 2\langle \nabla u,\nabla(\partial_tu)\rangle_{g(t)} - 2|\operatorname{Hess}_{g(t)}u|_{g(t)}^2 - 2\langle \nabla u,\nabla\Delta_{g(t)}u\rangle_{g(t)} - 2\operatorname{Ric}_{g(t)}(\nabla u,\nabla u).
\end{align*}
Canceling the two Ricci terms and combining the two first-order terms gives
\begin{align*}
(\partial_t-\Delta_{g(t)})|\nabla u|_{g(t)}^2 = -2|\operatorname{Hess}_{g(t)}u|_{g(t)}^2 + 2\langle \nabla u,\nabla(\partial_tu-\Delta_{g(t)}u)\rangle_{g(t)}.
\end{align*}
Since $u$ satisfies $\partial_tu=\Delta_{g(t)}u$, the function
\begin{align*}
\partial_tu-\Delta_{g(t)}u: M\times I &\to \mathbb{R}
\end{align*}
is identically zero. Hence its spatial gradient is zero, and the preceding identity reduces to
\begin{align*}
(\partial_t-\Delta_{g(t)})|\nabla u|_{g(t)}^2
=
-2|\operatorname{Hess}_{g(t)}u|_{g(t)}^2.
\end{align*}
Because $p \in M$ and $t \in I^\circ$ were arbitrary, the identity holds on all of $M\times I^\circ$.
[/step]
