[proofplan]
We apply Hamilton's vector bundle maximum principle to the vector bundle of symmetric two-tensors. The relevant fiberwise closed convex set is the cone of nonnegative symmetric bilinear forms. We verify that this cone is invariant under parallel transport and that the null-vector hypothesis is precisely the condition that the reaction term points into the cone along its boundary. The bundle maximum principle then shows that the solution cannot leave the cone once it starts inside it.
[/proofplan]
[step:Define the preserved cone of nonnegative symmetric two-tensors]
Let
\begin{align*}
E := \operatorname{Sym}^2(T^*M)
\end{align*}
be the vector bundle over $M$ whose fiber over $x \in M$ is
\begin{align*}
E_x := \operatorname{Sym}^2(T_x^*M).
\end{align*}
For each $x \in M$, define the cone
\begin{align*}
K_x := \{A \in \operatorname{Sym}^2(T_x^*M) : A(v,v) \ge 0 \text{ for every } v \in T_xM\}.
\end{align*}
Thus $h(t) \ge 0$ means precisely that
\begin{align*}
h(x,t) \in K_x
\end{align*}
for every $x \in M$.
Each $K_x$ is closed because it is the intersection of the closed half-spaces
\begin{align*}
\{A \in E_x : A(v,v) \ge 0\}, \qquad v \in T_xM.
\end{align*}
It is convex because if $A,B \in K_x$ and $\lambda \in [0,1]$, then for every $v \in T_xM$,
\begin{align*}
\bigl(\lambda A+(1-\lambda)B\bigr)(v,v)
=
\lambda A(v,v)+(1-\lambda)B(v,v)
\ge 0.
\end{align*}
It is also fiberwise closed under multiplication by nonnegative scalars, hence is a closed convex cone.
[/step]
[step:Verify that the cone is invariant under parallel transport]
Fix $t \in [0,T]$. Let $\gamma : [0,1] \to M$ be a smooth curve, and let
\begin{align*}
P_\gamma^{0,1}: T_{\gamma(0)}M \to T_{\gamma(1)}M
\end{align*}
denote parallel transport along $\gamma$ with respect to the Levi-Civita connection of $g(t)$. The induced parallel transport on $E=\operatorname{Sym}^2(T^*M)$ is the map
\begin{align*}
\mathcal{P}_\gamma^{0,1}: \operatorname{Sym}^2(T_{\gamma(0)}^*M) \to \operatorname{Sym}^2(T_{\gamma(1)}^*M)
\end{align*}
defined by sending each $A \in \operatorname{Sym}^2(T_{\gamma(0)}^*M)$ to the symmetric [bilinear form](/page/Bilinear%20Form)
\begin{align*}
\mathcal{P}_\gamma^{0,1}(A)=\Bigl[(w_1,w_2) \mapsto A\bigl((P_\gamma^{0,1})^{-1}w_1,(P_\gamma^{0,1})^{-1}w_2\bigr)\Bigr].
\end{align*}
If $A \in K_{\gamma(0)}$ and $w \in T_{\gamma(1)}M$, then $(P_\gamma^{0,1})^{-1}w \in T_{\gamma(0)}M$, and therefore
\begin{align*}
\bigl(\mathcal{P}_\gamma^{0,1}A\bigr)(w,w)
=
A\bigl((P_\gamma^{0,1})^{-1}w,(P_\gamma^{0,1})^{-1}w\bigr)
\ge 0.
\end{align*}
Hence
\begin{align*}
\mathcal{P}_\gamma^{0,1}(K_{\gamma(0)}) \subset K_{\gamma(1)}.
\end{align*}
Applying the same argument to the reversed curve gives the reverse inclusion, so the family $(K_x)_{x \in M}$ is invariant under parallel transport.
[/step]
[step:Identify the null-vector condition with inward pointing reaction]
Fix $x \in M$, $t \in [0,T]$, and $A \in \partial K_x$. We show that the hypothesis on $N$ says that the reaction vector $N(A,t) \in E_x$ lies in the tangent cone to $K_x$ at $A$. Since $K_x$ is a closed convex cone in the finite-dimensional [vector space](/page/Vector%20Space) $E_x$, we use the supporting-functional characterization of the tangent cone:
\begin{align*}
T_{K_x}(A):=\{Y\in E_x : L(Y)\ge 0 \text{ for every linear } L:E_x\to\mathbb{R} \text{ with } L(B)\ge 0 \text{ for all } B\in K_x \text{ and } L(A)=0\}.
\end{align*}
Thus it is enough to prove that every supporting functional $L$ at $A$ satisfies $L(N(A,t))\ge 0$.
Since $A \in \partial K_x$, there exists a nonzero vector $v \in T_xM$ with
\begin{align*}
A(v,v)=0.
\end{align*}
For such a vector $v$, define the linear functional $\ell_v: \operatorname{Sym}^2(T_x^*M) \to \mathbb{R}$ by
\begin{align*}
\ell_v(B)=B(v,v).
\end{align*}
The cone $K_x$ lies in the closed half-space $\{B \in E_x : \ell_v(B) \ge 0\}$, and $A$ lies on the boundary hyperplane $\{B \in E_x : \ell_v(B)=0\}$. Thus $\ell_v$ is a supporting functional for $K_x$ at $A$. The hypothesis gives
\begin{align*}
\ell_v\bigl(N(A,t)\bigr)=N(A,t)(v,v) \ge 0.
\end{align*}
It remains to justify that these rank-one functionals test every supporting direction of the cone. Fix a $g(t)_x$-[orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_n)$ of $T_xM$, and use it to identify $E_x$ with the vector space of real symmetric $n \times n$ matrices. Let $L:E_x\to\mathbb{R}$ be any supporting functional for $K_x$ at $A$, meaning that $L(B)\ge 0$ for every $B\in K_x$ and $L(A)=0$. By the self-duality of the positive semidefinite cone with respect to the Frobenius trace pairing, there is a positive semidefinite symmetric matrix $C$ such that
\begin{align*}
L(B)=\operatorname{tr}(CB)
\end{align*}
for every $B\in E_x$. Diagonalize $C$ in the chosen orthonormal basis as
\begin{align*}
C=\sum_{k=1}^n \mu_k u_k\otimes u_k,
\end{align*}
where $u_1,\dots,u_n\in T_xM$ are $g(t)_x$-orthonormal eigenvectors and $\mu_k\ge 0$. Since $A\in K_x$, each $A(u_k,u_k)\ge 0$, and the supporting condition gives
\begin{align*}
0=L(A)=\operatorname{tr}(CA)=\sum_{k=1}^n \mu_k A(u_k,u_k).
\end{align*}
Thus whenever $\mu_k>0$, one has $A(u_k,u_k)=0$. Applying the null-vector hypothesis to each such $u_k$ gives
\begin{align*}
L\bigl(N(A,t)\bigr)=\operatorname{tr}\bigl(CN(A,t)\bigr)=\sum_{k=1}^n \mu_k N(A,t)(u_k,u_k)\ge 0.
\end{align*}
Therefore $N(A,t)$ pairs nonnegatively with every supporting functional of $K_x$ at $A$, which is exactly the tangent-cone boundary condition. Hence the reaction ordinary differential equation
\begin{align*}
\frac{dA}{dt}=N(A,t)
\end{align*}
cannot point strictly outside $K_x$ at any boundary point.
[guided]
The maximum principle for vector bundles does not require us to solve the reaction equation. It requires a boundary test: whenever the solution reaches the boundary of the allowed fiberwise set, the non-diffusive part of the equation must not point outside that set. We now verify that boundary test for the cone
\begin{align*}
K_x=\{A \in E_x : A(w,w)\ge 0 \text{ for every } w \in T_xM\},
\end{align*}
where $E_x=\operatorname{Sym}^2(T_x^*M)$.
Because $K_x$ is a closed convex cone in the finite-dimensional vector space $E_x$, the tangent cone at a boundary point $A\in \partial K_x$ can be tested by supporting functionals. In this proof, this means
\begin{align*}
T_{K_x}(A):=\{Y\in E_x : L(Y)\ge 0 \text{ for every linear } L:E_x\to\mathbb{R} \text{ with } L(B)\ge 0 \text{ for all } B\in K_x \text{ and } L(A)=0\}.
\end{align*}
So to show that $N(A,t)$ points inward, we must prove $L(N(A,t))\ge 0$ for every such supporting functional $L$.
A boundary point $A \in \partial K_x$ is a nonnegative bilinear form with at least one null vector. Thus there exists $v \in T_xM$, $v \ne 0$, such that
\begin{align*}
A(v,v)=0.
\end{align*}
Define the linear functional $\ell_v:E_x\to\mathbb{R}$ by
\begin{align*}
\ell_v(B)=B(v,v).
\end{align*}
This functional cuts out a supporting half-space for $K_x$: every $B\in K_x$ satisfies
\begin{align*}
\ell_v(B)=B(v,v)\ge 0,
\end{align*}
and the boundary tensor satisfies
\begin{align*}
\ell_v(A)=A(v,v)=0.
\end{align*}
For this supporting hyperplane, inward-pointing means
\begin{align*}
\ell_v\bigl(N(A,t)\bigr)\ge 0.
\end{align*}
By the definition of $\ell_v$, this is exactly
\begin{align*}
N(A,t)(v,v)\ge 0,
\end{align*}
which is the null-vector hypothesis.
We also need to know that no other supporting functional gives an additional condition. Fix a $g(t)_x$-orthonormal basis $(e_1,\dots,e_n)$ of $T_xM$, and use it to identify $E_x$ with the vector space of real symmetric $n\times n$ matrices. Let $L:E_x\to\mathbb{R}$ be any supporting functional for $K_x$ at $A$, so $L(B)\ge 0$ for all $B\in K_x$ and $L(A)=0$. The positive semidefinite cone is self-dual for the Frobenius trace pairing, so there is a positive semidefinite symmetric matrix $C$ such that
\begin{align*}
L(B)=\operatorname{tr}(CB)
\end{align*}
for every $B\in E_x$.
Diagonalize $C$ as
\begin{align*}
C=\sum_{k=1}^n \mu_k u_k\otimes u_k,
\end{align*}
where $u_1,\dots,u_n\in T_xM$ are $g(t)_x$-orthonormal eigenvectors and $\mu_k\ge 0$. Since $A\in K_x$, each $A(u_k,u_k)\ge 0$. The equation $L(A)=0$ gives
\begin{align*}
0=\operatorname{tr}(CA)=\sum_{k=1}^n \mu_k A(u_k,u_k).
\end{align*}
Every summand is nonnegative, so whenever $\mu_k>0$ we must have $A(u_k,u_k)=0$. Applying the theorem's null-vector hypothesis to those vectors gives
\begin{align*}
N(A,t)(u_k,u_k)\ge 0
\end{align*}
whenever $\mu_k>0$. Therefore
\begin{align*}
L\bigl(N(A,t)\bigr)=\operatorname{tr}\bigl(CN(A,t)\bigr)=\sum_{k=1}^n \mu_k N(A,t)(u_k,u_k)\ge 0.
\end{align*}
Thus $N(A,t)$ pairs nonnegatively with every supporting functional at $A$, so the reaction term satisfies the full tangent-cone boundary condition for $K_x$.
[/guided]
[/step]
[step:Apply Hamilton's vector bundle maximum principle]
We now invoke [Hamilton Vector Bundle Maximum Principle](/theorems/TEMP-28) for parabolic equations on compact manifolds, applied to the bundle
\begin{align*}
E=\operatorname{Sym}^2(T^*M)
\end{align*}
with the time-dependent connection induced by $g(t)$ and to the fiberwise family of closed convex sets $(K_x)_{x \in M}$. This is the time-dependent-connection formulation of Hamilton's principle, so the evolving Levi-Civita connection of the Ricci flow metric $g(t)$ is part of the theorem's allowed setting.
The hypotheses of that principle are satisfied. The base manifold $M$ is compact. The local Lipschitz regularity of $N$ in the tensor variable, together with the stated time-dependence of $N(h,t)$ along the smooth family $h$, is the regularity required for the reaction term in the vector bundle maximum principle. The section
\begin{align*}
h : M \times [0,T] \to E
\end{align*}
is smooth. The equation has the required heat-type form
\begin{align*}
\partial_t h=\Delta_t h+N(h,t),
\end{align*}
where $\Delta_t$ is the connection rough Laplacian. The fiberwise sets $K_x$ are closed and convex by the first step, and they are invariant under parallel transport by the second step. Finally, the reaction term satisfies the required inward-pointing boundary condition by the third step.
Hamilton's vector bundle maximum principle therefore implies that if
\begin{align*}
h(x,0)\in K_x
\end{align*}
for every $x \in M$, then
\begin{align*}
h(x,t)\in K_x
\end{align*}
for every $x \in M$ and every $t \in [0,T]$.
[/step]
[step:Translate cone preservation back into tensor nonnegativity]
By the definition of $K_x$, the conclusion
\begin{align*}
h(x,t)\in K_x
\end{align*}
means that for every $v \in T_xM$,
\begin{align*}
h(x,t)(v,v)\ge 0.
\end{align*}
Thus $h(t)$ is nonnegative as a symmetric two-tensor on $M$ for every $t \in [0,T]$. This is exactly the asserted preservation of the inequality $h(t)\ge 0$.
[/step]
