[proofplan]
We use Hamilton's tensor maximum principle for sections of a vector bundle over a compact manifold. The curvature operator evolves by a heat-type equation whose reaction term is the quadratic expression $\mathcal R^2+\mathcal R^{\#}$. Thus the proof reduces to the algebraic fact that this reaction field points inward along the boundary of the cone of nonnegative curvature operators. At a boundary point, the kernel vector annihilates the $\mathcal R^2$ term, and Hamilton's Lie-algebraic identity gives nonnegativity of the $\mathcal R^{\#}$ term; the maximum principle then propagates the cone condition for all time.
[/proofplan]
[step:Place the curvature operator in the bundle maximum principle setting]
For each $t \in [0,T]$, define the vector bundle
\begin{align*}
E_t := \Lambda^2T^*M
\end{align*}
with the fiber [inner product](/page/Inner%20Product) induced by $g(t)$. For $p \in M$, let
\begin{align*}
\mathcal C_{p,t} := \{A \in \operatorname{Sym}(E_{p,t}) : (A\omega,\omega)_{g(t)} \ge 0 \text{ for all } \omega \in E_{p,t}\}
\end{align*}
be the closed convex cone of nonnegative self-adjoint endomorphisms of $E_{p,t}$.
We now pass to the standard Uhlenbeck gauge for Ricci flow. Let $h := g(0)$ be the fixed initial metric on $TM$. Let
\begin{align*}
P_t: (TM,h) \to (TM,g(t))
\end{align*}
denote the time-dependent bundle isometry determined by the Uhlenbeck gauge equation, with $P_0 = \operatorname{id}_{TM}$. Its induced exterior-square isometry is
\begin{align*}
\Lambda^2P_t^*: \Lambda^2T^*M \to \Lambda^2T^*M,
\end{align*}
where the domain is equipped with the metric induced by $h$ and the target is equipped with the metric induced by $g(t)$. Thus positivity of the curvature operator with respect to $g(t)$ is equivalent to positivity of the pulled-back endomorphism on the fixed Euclidean bundle
\begin{align*}
E := \Lambda^2T^*M
\end{align*}
with the $h$-induced fiber inner product.
The Levi-Civita connection of $g(t)$ pulls back through $P_t$ to a time-dependent metric connection $\nabla^{t}$ on $E$. Under this identification the curvature operator becomes a smooth section
\begin{align*}
\mathcal R: M \times [0,T] \to \operatorname{Sym}(E),
\end{align*}
and it satisfies the curvature-operator evolution equation
\begin{align*}
\partial_t \mathcal R = \Delta_t \mathcal R + \mathcal R^2 + \mathcal R^{\#}.
\end{align*}
Here $\Delta_t := \operatorname{tr}_{g(t)}(\nabla^t\nabla^t)$ is the connection Laplacian on $\operatorname{Sym}(E)$, $\mathcal R^2$ is composition of endomorphisms, and $\mathcal R^{\#}$ is the quadratic Lie-algebraic term determined by the bracket on $\Lambda^2T_p^*M \cong \mathfrak{so}(T_pM,g(t))$ through the metric-induced skew-adjoint endomorphism associated to a two-form.
This is the Ricci Flow Curvature Operator Evolution Equation in Uhlenbeck gauge.
[/step]
[step:Reduce preservation of the cone to an algebraic boundary inequality]
We apply the Hamilton Tensor Maximum Principle for compact base manifolds without boundary. The base manifold $M$ is compact and has no boundary by the theorem statement, the section $\mathcal R$ is smooth on $M \times [0,T]$, and the equation has the required form
\begin{align*}
\partial_t \mathcal R = \Delta_t\mathcal R + \Phi(\mathcal R),
\qquad
\Phi(A) := A^2 + A^{\#},
\end{align*}
where $\Delta_t$ is the connection Laplacian of the metric connection $\nabla^t$ and $\Phi$ is fiberwise.
For each $(p,t)$, the set $\mathcal C_{p,t}$ is closed because it is the intersection of the closed half-spaces
\begin{align*}
\{A \in \operatorname{Sym}(E_{p,t}) : (A\omega,\omega)_{g(t)} \ge 0\},
\qquad \omega \in E_{p,t},
\end{align*}
and it is convex because each of those half-spaces is convex. Since $\nabla^t$ is metric, parallel transport by $\nabla^t$ is a fiber isometry; consequently it carries nonnegative self-adjoint endomorphisms to nonnegative self-adjoint endomorphisms by conjugation, so the cone field is invariant under $\nabla^t$-parallel transport.
Thus Hamilton's maximum principle reduces preservation of $\mathcal C$ to the following fiberwise tangency condition: whenever $A \in \mathcal C_{p,t}$ and $\omega \in E_{p,t}$ satisfy
\begin{align*}
(A\omega,\omega)_{g(t)} = 0,
\end{align*}
one must have
\begin{align*}
((A^2 + A^{\#})\omega,\omega)_{g(t)} \ge 0.
\end{align*}
[guided]
The maximum principle is being used in the vector-bundle form rather than the scalar form. The scalar maximum principle tracks one real-valued function, while the condition here is that an entire self-adjoint endomorphism have nonnegative spectrum. We encode that condition fiberwise by the cone
\begin{align*}
\mathcal C_{p,t} = \{A \in \operatorname{Sym}(E_{p,t}) : (A\omega,\omega)_{g(t)} \ge 0 \text{ for all } \omega \in E_{p,t}\}.
\end{align*}
We verify the hypotheses of the Hamilton Tensor Maximum Principle. First, the base manifold is compact and has no boundary by the theorem statement. Second, the section
\begin{align*}
\mathcal R: M \times [0,T] \to \operatorname{Sym}(E)
\end{align*}
is smooth because $g(t)$ is a smooth Ricci flow and the Uhlenbeck gauge is smooth in space and time. Third, the evolution equation has the required connection-heat form
\begin{align*}
\partial_t \mathcal R = \Delta_t\mathcal R + \Phi(\mathcal R),
\qquad
\Phi(A) := A^2 + A^{\#},
\end{align*}
where $\Delta_t$ is the connection Laplacian of the metric connection $\nabla^t$ on $\operatorname{Sym}(E)$ and $\Phi$ is a fiberwise map on self-adjoint endomorphisms.
Now we check the cone hypotheses. For fixed $(p,t)$, the cone $\mathcal C_{p,t}$ is closed because it is the intersection, over all $\omega \in E_{p,t}$, of the closed half-spaces
\begin{align*}
\{A \in \operatorname{Sym}(E_{p,t}) : (A\omega,\omega)_{g(t)} \ge 0\}.
\end{align*}
It is convex because if $A,B \in \mathcal C_{p,t}$ and $\lambda \in [0,1]$, then for every $\omega \in E_{p,t}$,
\begin{align*}
((\lambda A + (1-\lambda)B)\omega,\omega)_{g(t)}
=
\lambda(A\omega,\omega)_{g(t)} + (1-\lambda)(B\omega,\omega)_{g(t)}
\ge 0.
\end{align*}
Finally, $\mathcal C$ is invariant under $\nabla^t$-parallel transport. Indeed, parallel transport for a metric connection is a fiber isometry; if $P: E_{p,t} \to E_{q,t}$ denotes such parallel transport and $A \in \mathcal C_{p,t}$, then for every $\eta \in E_{q,t}$,
\begin{align*}
((PAP^{-1})\eta,\eta)_{g(t)}
=
(A(P^{-1}\eta),P^{-1}\eta)_{g(t)}
\ge 0.
\end{align*}
Thus $PAP^{-1} \in \mathcal C_{q,t}$.
The remaining hypothesis of the theorem is the boundary tangency condition for the reaction term. A boundary point of the cone of nonnegative [self-adjoint operators](/page/Self-Adjoint%20Operators) is detected by a nonzero vector $\omega$ with
\begin{align*}
(A\omega,\omega)_{g(t)} = 0.
\end{align*}
Therefore the exact condition needed is
\begin{align*}
((A^2 + A^{\#})\omega,\omega)_{g(t)} \ge 0
\end{align*}
for every $A \in \mathcal C_{p,t}$ and every nonzero $\omega \in E_{p,t}$ with $(A\omega,\omega)_{g(t)}=0$. This is the algebraic inequality proved in the next step; once it is known, Hamilton's maximum principle propagates membership in the cone from time $0$ to every later time in $[0,T]$.
[/guided]
[/step]
[step:Verify the inward tangency of the quadratic reaction term]
Fix $p \in M$, $t \in [0,T]$, and write $V := T_pM$ with the inner product $g(t)_p$. Let
\begin{align*}
A: \Lambda^2V^* \to \Lambda^2V^*
\end{align*}
be a nonnegative self-adjoint algebraic curvature operator. Let $\omega \in \Lambda^2V^*$ satisfy
\begin{align*}
(A\omega,\omega)_{g(t)} = 0.
\end{align*}
Since $A$ is nonnegative and self-adjoint, the spectral theorem gives $A\omega=0$. Therefore
\begin{align*}
(A^2\omega,\omega)_{g(t)}
= (A\omega,A\omega)_{g(t)}
= 0.
\end{align*}
It remains to check the $\#$ term. Identify $\Lambda^2V^*$ with the Lie algebra $\mathfrak{so}(V,g(t)_p)$ by the metric-induced skew-adjoint map associated to a two-form. The Hamilton Algebraic Identity for the Curvature Operator Cone states that, for every nonnegative algebraic curvature operator $A$ and every $\omega \in \ker A$,
\begin{align*}
(A^{\#}\omega,\omega)_{g(t)} \ge 0.
\end{align*}
Its hypotheses apply here because $A$ is nonnegative by assumption and the preceding paragraph proved $\omega \in \ker A$.
Combining the two estimates gives
\begin{align*}
((A^2 + A^{\#})\omega,\omega)_{g(t)}
=
(A^2\omega,\omega)_{g(t)}
+
(A^{\#}\omega,\omega)_{g(t)}
\ge 0.
\end{align*}
Thus the reaction field is tangent inward along the boundary of the cone of nonnegative curvature operators.
[/step]
[step:Apply the tensor maximum principle to preserve nonnegativity]
By hypothesis,
\begin{align*}
\mathcal R(p,0) \in \mathcal C_{p,0}
\end{align*}
for every $p \in M$. The previous step verifies the inward-pointing condition required by Hamilton's tensor maximum principle for the reaction term
\begin{align*}
\Phi(A) = A^2 + A^{\#}.
\end{align*}
Therefore
\begin{align*}
\mathcal R(p,t) \in \mathcal C_{p,t}
\end{align*}
for every $p \in M$ and every $t \in [0,T]$.
Unwinding the definition of $\mathcal C_{p,t}$, this says that
\begin{align*}
(\mathcal R_{p,t}\omega,\omega)_{g(t)} \ge 0
\end{align*}
for every $p \in M$, every $t \in [0,T]$, and every $\omega \in \Lambda^2T_p^*M$. Hence the curvature operator of $g(t)$ is nonnegative at every point for all $t \in [0,T]$.
[/step]
