[proofplan] We prove preservation by applying Hamilton's tensor maximum principle to the Ricci tensor. The only point to check is the null-vector condition for the reaction term in the Ricci evolution equation. In dimension three, the Riemann curvature tensor is determined algebraically by the Ricci tensor, and evaluating the reaction term on a null Ricci eigendirection gives $(\mu-\nu)^2\geq 0$. This verifies inward-pointingness of the evolution at the boundary of the cone of nonnegative symmetric two-tensors, so compactness of $M$ allows the tensor maximum principle to preserve nonnegative Ricci curvature on the whole time interval. [/proofplan]

[step:Write the Ricci evolution as a heat equation with an algebraic reaction term] For each $t\in[0,T]$, let $\nabla^{g(t)}$ denote the Levi-Civita connection of $g(t)$, let $\Delta_{g(t)}$ denote the rough Laplacian on covariant two-tensors induced by $\nabla^{g(t)}$, and let $R_{ikjl}(t)$ and $R_{ij}(t)$ denote respectively the components of the Riemann curvature tensor and Ricci tensor of $g(t)$ in a local $g(t)$-orthonormal frame. Along Ricci flow, the covariant Ricci tensor satisfies the evolution equation \begin{align*} \partial_t R_{ij} = \Delta_{g(t)} R_{ij} + Q_{ij}, \qquad Q_{ij} = 2\sum_{k,l=1}^{3} R_{ikjl}R_{kl} - 2\sum_{k=1}^{3}R_{ik}R_{kj}. \end{align*} This is the standard Ricci tensor evolution equation under Ricci flow (citing a result not yet in the wiki: Ricci tensor evolution equation under Ricci flow). Let $\mathcal{C}_{p,t}\subset S^2T_p^*M$ denote the closed convex cone of symmetric two-tensors nonnegative with respect to $g(t)$: \begin{align*} \mathcal{C}_{p,t} = \{A\in S^2T_p^*M: A(v,v)\geq 0\text{ for every }v\in T_pM\}. \end{align*} [Hamilton's tensor maximum principle for symmetric two-tensors](/theorems/5987) says that, on a compact manifold, this cone is preserved by an equation of the form $\partial_t A=\Delta_{g(t)}A+Q(A)$ if the algebraic reaction term satisfies the null-vector condition: \begin{align*} A\in\mathcal{C}_{p,t},\quad A(v,v)=0 \quad\Longrightarrow\quad Q(A)(v,v)\geq 0 \end{align*} for every $p\in M$, $t\in[0,T]$, and $v\in T_pM$ (citing a result not yet in the wiki: Hamilton tensor maximum principle for symmetric two-tensors). Thus it remains to verify this condition for $A=\operatorname{Ric}_{g(t)}$. [/step]

[step:Verify the null-vector condition by diagonalising the Ricci tensor]Fix $t\in[0,T]$ and $p\in M$. Assume that $\operatorname{Ric}_{g(t)}\in\mathcal{C}_{p,t}$ and that $v\in T_pM$ satisfies \begin{align*} \operatorname{Ric}_{g(t)}(v,v)=0. \end{align*} If $v=0$, then $Q(v,v)=0$. Suppose $v\neq 0$, and define $e_1:=v/|v|_{g(t)}$. Since $\operatorname{Ric}_{g(t)}$ is a symmetric nonnegative [bilinear form](/page/Bilinear%20Form), there is a $g(t)$-[orthonormal basis](/page/Orthonormal%20Basis) $(e_1,e_2,e_3)$ of $T_pM$ diagonalising $\operatorname{Ric}_{g(t)}$. Write its eigenvalues in this basis as \begin{align*} R_{11}=0,\qquad R_{22}=\mu,\qquad R_{33}=\nu, \end{align*} where $\mu,\nu\geq 0$. Because $R_{1k}=0$ for every $k\in\{1,2,3\}$ in this basis, the quadratic Ricci term vanishes in the null direction: \begin{align*} \sum_{k=1}^{3}R_{1k}R_{k1}=0. \end{align*} Therefore \begin{align*} Q_{11} = 2\sum_{k,l=1}^{3}R_{1k1l}R_{kl} = 2R_{1212}\mu+2R_{1313}\nu. \end{align*} In dimension three, the curvature operator is determined by the Ricci eigenvalues. In the above $g(t)$-orthonormal Ricci eigenbasis, the sectional curvature components satisfy \begin{align*} R_{1212} = \frac{R_{11}+R_{22}-R_{33}}{2} = \frac{\mu-\nu}{2}, \qquad R_{1313} = \frac{R_{11}+R_{33}-R_{22}}{2} = \frac{\nu-\mu}{2}. \end{align*} Substituting these identities gives \begin{align*} Q_{11}=2\left(\frac{\mu-\nu}{2}\right)\mu+2\left(\frac{\nu-\mu}{2}\right)\nu. \end{align*} Combining the two products gives \begin{align*} Q_{11}=\mu(\mu-\nu)+\nu(\nu-\mu). \end{align*} Expanding and collecting terms gives \begin{align*} Q_{11}=\mu^2-2\mu\nu+\nu^2. \end{align*} Factoring the square gives \begin{align*} Q_{11}=(\mu-\nu)^2\geq 0. \end{align*} Since $v=|v|_{g(t)}e_1$, bilinearity gives \begin{align*} Q(v,v)=|v|_{g(t)}^2Q_{11}\geq 0. \end{align*} Thus the reaction term satisfies the null-vector condition.[/step]

[guided]The boundary of the cone of nonnegative symmetric two-tensors consists of tensors that are nonnegative but have at least one null direction. We therefore fix a time $t\in[0,T]$, a point $p\in M$, and a tangent vector $v\in T_pM$ such that \begin{align*} \operatorname{Ric}_{g(t)}(w,w)\geq 0\quad\text{for every }w\in T_pM, \qquad \operatorname{Ric}_{g(t)}(v,v)=0. \end{align*} If $v=0$, then $Q(v,v)=0$, so assume $v\neq 0$ and set $e_1:=v/|v|_{g(t)}$. Because $\operatorname{Ric}_{g(t)}$ is symmetric with respect to the [inner product](/page/Inner%20Product) $g(t)_p$, the spectral theorem gives a $g(t)$-orthonormal eigenbasis $(e_1,e_2,e_3)$ after extending the null eigenvector $e_1$. The eigenvalue corresponding to $e_1$ is $0$: nonnegativity and $\operatorname{Ric}_{g(t)}(e_1,e_1)=0$ force $\operatorname{Ric}_{g(t)}(e_1,w)=0$ for every $w\in T_pM$ by applying nonnegativity to $e_1+sw$ and viewing the resulting quadratic polynomial in $s\in\mathbb{R}$. Thus in this basis the Ricci tensor has diagonal entries \begin{align*} R_{11}=0,\qquad R_{22}=\mu,\qquad R_{33}=\nu, \end{align*} with $\mu,\nu\geq 0$. Now evaluate the reaction term in the null direction $e_1$. The second part of the reaction term is \begin{align*} -2\sum_{k=1}^{3}R_{1k}R_{k1}. \end{align*} Every summand is zero because $R_{1k}=0$ for all $k$. Hence \begin{align*} Q_{11} = 2\sum_{k,l=1}^{3}R_{1k1l}R_{kl}. \end{align*} Since the Ricci tensor is diagonal in the chosen basis, $R_{kl}=0$ when $k\neq l$, and $R_{11}=0$. Therefore only the $(k,l)=(2,2)$ and $(k,l)=(3,3)$ terms remain: \begin{align*} Q_{11} = 2R_{1212}\mu+2R_{1313}\nu. \end{align*} The special feature of dimension three now enters. In a three-dimensional Riemannian [vector space](/page/Vector%20Space), the full curvature tensor is determined by the Ricci tensor. For a $g(t)$-orthonormal Ricci eigenbasis with eigenvalues $0,\mu,\nu$, the sectional curvature components are \begin{align*} R_{1212} = \frac{0+\mu-\nu}{2} = \frac{\mu-\nu}{2}, \qquad R_{1313} = \frac{0+\nu-\mu}{2} = \frac{\nu-\mu}{2}. \end{align*} Substituting these two identities into the expression for $Q_{11}$ gives \begin{align*} Q_{11}=2\left(\frac{\mu-\nu}{2}\right)\mu+2\left(\frac{\nu-\mu}{2}\right)\nu. \end{align*} Combining the two products gives \begin{align*} Q_{11}=\mu(\mu-\nu)+\nu(\nu-\mu). \end{align*} Expanding and collecting terms gives \begin{align*} Q_{11}=\mu^2-2\mu\nu+\nu^2. \end{align*} Factoring the square gives \begin{align*} Q_{11}=(\mu-\nu)^2. \end{align*} This quantity is nonnegative for all [real numbers](/page/Real%20Numbers) $\mu$ and $\nu$, and in particular for the nonnegative Ricci eigenvalues here. Finally, since $v=|v|_{g(t)}e_1$, bilinearity of $Q$ as a two-tensor gives \begin{align*} Q(v,v)=|v|_{g(t)}^2Q_{11}=|v|_{g(t)}^2(\mu-\nu)^2\geq 0. \end{align*} This is exactly the null-vector condition required by the tensor maximum principle.[/guided]

[step:Apply the tensor maximum principle to preserve the Ricci cone] The initial hypothesis states that \begin{align*} \operatorname{Ric}_{g(0)}\in\mathcal{C}_{p,0} \end{align*} for every $p\in M$. The preceding step proves that the reaction term in the Ricci evolution equation points into the cone $\mathcal{C}_{p,t}$ at every boundary point of the cone. Since $M$ is compact and $g(t)$ is a smooth Ricci flow on the closed interval $[0,T]$, Hamilton's tensor maximum principle applies to $\operatorname{Ric}_{g(t)}$ and yields \begin{align*} \operatorname{Ric}_{g(t)}\in\mathcal{C}_{p,t} \end{align*} for every $p\in M$ and every $t\in[0,T]$. By the definition of $\mathcal{C}_{p,t}$, this means \begin{align*} \operatorname{Ric}_{g(t)}(v,v)\geq 0 \end{align*} for every $t\in[0,T]$, every $p\in M$, and every $v\in T_pM$. This is the asserted preservation of nonnegative Ricci curvature in dimension three. [/step]