[proofplan]
We compute the evolution at a fixed spacetime point using a time-dependent orthonormal frame, so that the fibre metric used to take tensor norms is fixed at that point. In that frame, the curvature tensor satisfies a [heat equation](/page/Heat%20Equation) with a quadratic reaction term. The Bochner identity converts the diffusion term into $\Delta |\operatorname{Rm}|^2 - 2|\nabla \operatorname{Rm}|^2$, and the remaining quadratic reaction term is bounded by a dimension-dependent multiple of $|\operatorname{Rm}|^3$.
[/proofplan]
[step:Freeze the norm computation in a moving orthonormal frame]
Fix a point $(p,t_0) \in M \times (0,T)$. Since $g(t)$ is a smooth Ricci flow, choose $\varepsilon>0$ such that $(t_0-\varepsilon,t_0+\varepsilon) \subset (0,T)$ and choose a smooth local frame $(e_1(t),\dots,e_n(t))$ near $p$ which is orthonormal for $g(t)$ at the point $p$ for times $t \in (t_0-\varepsilon,t_0+\varepsilon)$. We take the frame to satisfy the moving-frame transport equation
\begin{align*}
\partial_t e_i(t) = \operatorname{Ric}_{g(t)}^{\sharp}(e_i(t))
\end{align*}
at $p$, where $\operatorname{Ric}_{g(t)}^{\sharp}:T_pM \to T_pM$ is the endomorphism obtained from the Ricci tensor by raising one index using $g(t)$. This transport equation gives
\begin{align*}
\partial_t g(t)(e_i(t),e_j(t))=0,
\end{align*}
so the frame remains $g(t)$-orthonormal at $p$. In this frame, the fibre [inner product](/page/Inner%20Product) on tensor components at $p$ is the Euclidean inner product on the fixed [vector space](/page/Vector%20Space) of algebraic curvature tensors on $\mathbb{R}^n$.
Let $\mathcal{C}_n$ denote the finite-dimensional real vector space of algebraic curvature tensors on $\mathbb{R}^n$. Define the component curve $R:(t_0-\varepsilon,t_0+\varepsilon)\to \mathcal{C}_n$ by declaring $R(t)$ to be the component vector of $\operatorname{Rm}(p,t)$ in the moving orthonormal frame $(e_i(t))$.
We use the Ricci flow curvature evolution formula in a moving orthonormal frame at $(p,t_0)$:
\begin{align*}
\partial_t R = \Delta R + Q(R).
\end{align*}
Here $\Delta R$ denotes the component vector of the rough Laplacian $\Delta_{g(t_0)}\operatorname{Rm}:=\operatorname{tr}_{g(t_0)}(\nabla\nabla\operatorname{Rm})$, so in a $g(t_0)$-orthonormal frame
\begin{align*}
(\Delta_{g(t_0)}\operatorname{Rm})_{ijkl}=\sum_{a=1}^{n}(\nabla_a\nabla_a\operatorname{Rm})_{ijkl}.
\end{align*}
The map
\begin{align*}
Q: \mathcal{C}_n &\to \mathcal{C}_n
\end{align*}
is Hamilton's universal homogeneous quadratic curvature reaction term in the moving-frame curvature evolution equation. Its coefficients depend only on the dimension $n$.
[/step]
[step:Convert the diffusion term into a negative gradient term]
At $(p,t_0)$, the Bochner identity for the tensor $\operatorname{Rm}$ gives
\begin{align*}
\Delta_{g(t_0)}|\operatorname{Rm}|_{g(t_0)}^2
=
2(\Delta \operatorname{Rm},\operatorname{Rm})_{g(t_0)}
+
2|\nabla \operatorname{Rm}|_{g(t_0)}^2.
\end{align*}
The moving-frame evolution equation gives
\begin{align*}
\partial_t |\operatorname{Rm}|_{g(t_0)}^2
=
2(\partial_t \operatorname{Rm},\operatorname{Rm})_{g(t_0)}
=
2(\Delta \operatorname{Rm},\operatorname{Rm})_{g(t_0)}
+
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}.
\end{align*}
Subtracting the Bochner identity from this equality yields
\begin{align*}
(\partial_t-\Delta_{g(t_0)})|\operatorname{Rm}|_{g(t_0)}^2
=
-2|\nabla \operatorname{Rm}|_{g(t_0)}^2
+
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}.
\end{align*}
[guided]
We work at the fixed spacetime point $(p,t_0)$ and use the moving $g(t)$-orthonormal frame $(e_1(t),\dots,e_n(t))$ chosen in the previous step. The frame is transported by
\begin{align*}
\partial_t e_i(t)=\operatorname{Ric}_{g(t)}^{\sharp}(e_i(t))
\end{align*}
at $p$, so the component inner product on the fixed vector space $\mathcal{C}_n$ is independent of $t$ at the point. This is the reason the time derivative of the squared curvature norm has no extra metric-variation term in these components: it is the ordinary derivative of the Euclidean squared norm of the curve $R:(t_0-\varepsilon,t_0+\varepsilon)\to\mathcal{C}_n$. Hence
\begin{align*}
\partial_t |\operatorname{Rm}|_{g(t_0)}^2
=
2(\partial_t \operatorname{Rm},\operatorname{Rm})_{g(t_0)}.
\end{align*}
The Ricci flow curvature evolution formula in the moving orthonormal frame applies because the Ricci flow and the moving frame are smooth near $(p,t_0)$. It gives, at $(p,t_0)$,
\begin{align*}
\partial_t \operatorname{Rm}
=
\Delta_{g(t_0)} \operatorname{Rm}
+
Q(\operatorname{Rm}),
\end{align*}
where $\Delta_{g(t_0)}\operatorname{Rm}:=\operatorname{tr}_{g(t_0)}(\nabla\nabla\operatorname{Rm})$ is the rough Laplacian of the curvature tensor and $Q(\operatorname{Rm})$ is Hamilton's universal quadratic curvature reaction term. Substituting this equation into the derivative of the squared norm gives
\begin{align*}
\partial_t |\operatorname{Rm}|_{g(t_0)}^2
=
2(\Delta \operatorname{Rm},\operatorname{Rm})_{g(t_0)}
+
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}.
\end{align*}
We now compare this time derivative with the spatial Laplacian of the squared norm. The Bochner identity for the rough Laplacian applies to the smooth tensor field $\operatorname{Rm}$ at time $t_0$ and states
\begin{align*}
\Delta_{g(t_0)}|\operatorname{Rm}|_{g(t_0)}^2
=
2(\Delta \operatorname{Rm},\operatorname{Rm})_{g(t_0)}
+
2|\nabla \operatorname{Rm}|_{g(t_0)}^2.
\end{align*}
The hypotheses are the smoothness of the tensor field and the use of the Levi-Civita connection of $g(t_0)$; both hold because $g(t)$ is a smooth Ricci flow. The same rough-Laplacian pairing $2(\Delta \operatorname{Rm},\operatorname{Rm})_{g(t_0)}$ appears in the time derivative and in the Bochner identity. Subtracting the Bochner identity from the time derivative cancels that pairing and leaves
\begin{align*}
(\partial_t-\Delta_{g(t_0)})|\operatorname{Rm}|_{g(t_0)}^2
=
-2|\nabla \operatorname{Rm}|_{g(t_0)}^2
+
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}.
\end{align*}
The negative gradient term is the dissipative contribution from the heat-operator part of the curvature evolution, while the only remaining term is the quadratic curvature reaction paired with $\operatorname{Rm}$.
[/guided]
[/step]
[step:Bound the quadratic reaction term by a cubic curvature norm]
Since $Q:\mathcal{C}_n \to \mathcal{C}_n$ is a homogeneous quadratic polynomial map on a finite-dimensional [normed vector space](/page/Normed%20Vector%20Space) and its coefficients depend only on $n$, there exists a constant $c_n>0$ depending only on $n$ such that
\begin{align*}
|Q(A)| \leq c_n |A|^2
\end{align*}
for every $A \in \mathcal{C}_n$. Applying this with $A=\operatorname{Rm}(p,t_0)$ and then using the [Cauchy-Schwarz inequality](/theorems/432) for the fibre inner product gives
\begin{align*}
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}
\leq
2|Q(\operatorname{Rm})|_{g(t_0)}|\operatorname{Rm}|_{g(t_0)}.
\end{align*}
The quadratic bound on $Q$ then gives
\begin{align*}
2|Q(\operatorname{Rm})|_{g(t_0)}|\operatorname{Rm}|_{g(t_0)}
\leq
2c_n|\operatorname{Rm}|_{g(t_0)}^3.
\end{align*}
Define
\begin{align*}
C_n := 2c_n.
\end{align*}
Then
\begin{align*}
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}
\leq
C_n|\operatorname{Rm}|_{g(t_0)}^3.
\end{align*}
[/step]
[step:Conclude the pointwise inequality at every spacetime point]
Combining the identity from the diffusion computation with the reaction estimate gives the identity
\begin{align*}
(\partial_t-\Delta_{g(t_0)})|\operatorname{Rm}|_{g(t_0)}^2
=
-2|\nabla \operatorname{Rm}|_{g(t_0)}^2
+
2(Q(\operatorname{Rm}),\operatorname{Rm})_{g(t_0)}.
\end{align*}
Using the bound on the reaction term, this implies
\begin{align*}
(\partial_t-\Delta_{g(t_0)})|\operatorname{Rm}|_{g(t_0)}^2
\leq
-2|\nabla \operatorname{Rm}|_{g(t_0)}^2
+
C_n|\operatorname{Rm}|_{g(t_0)}^3.
\end{align*}
The point $(p,t_0)\in M\times(0,T)$ was arbitrary, and the constant $C_n$ depends only on the dimension. Hence the stated pointwise inequality holds on all of $M\times(0,T)$.
[/step]
