[proofplan] We prove the contrapositive. If the full curvature tensor stays uniformly bounded on $M \times [0,T)$, then Shi derivative estimates give uniform bounds for all covariant derivatives of curvature on every terminal subinterval $[\tau,T)$. These estimates, together with the Ricci flow equation, imply that the metrics $g(t)$ converge smoothly as $t \uparrow T$ to a smooth Riemannian metric $g_T$ on $M$. Hamilton short-time existence starting from $g_T$ then extends the flow beyond $T$, contradicting the assumed maximality of $[0,T)$. [/proofplan]

[step:Assume bounded curvature and derive uniform equivalence of the metrics]Assume, toward a contradiction, that there is a constant $K \in [0,\infty)$ such that \begin{align*} |\operatorname{Rm}(g(t))|_{g(t)}(x) \leq K \end{align*} for every $(x,t) \in M \times [0,T)$. Let $n = \dim M$. Since $\operatorname{Ric}(g(t))$ is the trace of $\operatorname{Rm}(g(t))$, there is a dimensional constant $c_n > 0$ such that \begin{align*} |\operatorname{Ric}(g(t))|_{g(t)}(x) \leq c_n K \end{align*} for every $(x,t) \in M \times [0,T)$. Fix $p \in M$ and $v \in T_pM$. Define the scalar function $a_{p,v}: [0,T) \to [0,\infty)$ by setting $a_{p,v}(t) = g(t)_p(v,v)$ for $t \in [0,T)$. The Ricci flow equation gives \begin{align*} \frac{d}{dt}a_{p,v}(t) = -2\operatorname{Ric}(g(t))_p(v,v). \end{align*} By the tensor norm estimate for a symmetric [bilinear form](/page/Bilinear%20Form), \begin{align*} \left|\operatorname{Ric}(g(t))_p(v,v)\right| \leq |\operatorname{Ric}(g(t))|_{g(t)}(p)\, g(t)_p(v,v) \leq c_nK\, a_{p,v}(t). \end{align*} Hence \begin{align*} -2c_nK\,a_{p,v}(t) \leq \frac{d}{dt}a_{p,v}(t) \leq 2c_nK\,a_{p,v}(t). \end{align*} Integrating this differential inequality on $[s,t] \subset [0,T)$ gives \begin{align*} e^{-2c_nK(t-s)}g(s)_p(v,v) \leq g(t)_p(v,v) \leq e^{2c_nK(t-s)}g(s)_p(v,v). \end{align*} Taking $s=0$ and using $t<T$, we obtain the uniform equivalence \begin{align*} e^{-2c_nKT}g(0)_p(v,v) \leq g(t)_p(v,v) \leq e^{2c_nKT}g(0)_p(v,v). \end{align*} Thus all metrics $g(t)$, for $t \in [0,T)$, are uniformly equivalent to the fixed metric $g(0)$.[/step]

[guided]The first point is that a uniform curvature bound also controls how quickly the metric itself can change. Assume that there is a constant $K \in [0,\infty)$ such that \begin{align*} |\operatorname{Rm}(g(t))|_{g(t)}(x) \leq K \end{align*} for every $(x,t) \in M \times [0,T)$. Let $n = \dim M$. Since the Ricci tensor is obtained by tracing the Riemann curvature tensor, its pointwise norm is bounded by a dimensional multiple of the curvature norm. Thus there is a constant $c_n > 0$, depending only on $n$, such that \begin{align*} |\operatorname{Ric}(g(t))|_{g(t)}(x) \leq c_nK \end{align*} for every $(x,t) \in M \times [0,T)$. Now fix a point $p \in M$ and a tangent vector $v \in T_pM$. To measure how the metric changes in this fixed direction, define the scalar function $a_{p,v}: [0,T) \to [0,\infty)$ by $a_{p,v}(t) = g(t)_p(v,v)$ for $t \in [0,T)$. The Ricci flow equation says that \begin{align*} \frac{\partial g}{\partial t}(t) = -2\operatorname{Ric}(g(t)). \end{align*} Evaluating this identity on the fixed pair $(v,v)$ gives \begin{align*} \frac{d}{dt}a_{p,v}(t) = -2\operatorname{Ric}(g(t))_p(v,v). \end{align*} The norm of a symmetric bilinear form controls its value on a vector twice, so \begin{align*} \left|\operatorname{Ric}(g(t))_p(v,v)\right| \leq |\operatorname{Ric}(g(t))|_{g(t)}(p)\, g(t)_p(v,v). \end{align*} Using the Ricci bound, this becomes \begin{align*} \left|\operatorname{Ric}(g(t))_p(v,v)\right| \leq c_nK\,a_{p,v}(t). \end{align*} Therefore \begin{align*} -2c_nK\,a_{p,v}(t) \leq \frac{d}{dt}a_{p,v}(t) \leq 2c_nK\,a_{p,v}(t). \end{align*} Integrating this scalar differential inequality from $s$ to $t$, where $0 \leq s \leq t < T$, gives \begin{align*} e^{-2c_nK(t-s)}g(s)_p(v,v) \leq g(t)_p(v,v) \leq e^{2c_nK(t-s)}g(s)_p(v,v). \end{align*} In particular, with $s=0$ and $t<T$, \begin{align*} e^{-2c_nKT}g(0)_p(v,v) \leq g(t)_p(v,v) \leq e^{2c_nKT}g(0)_p(v,v). \end{align*} This uniform equivalence is essential: it prevents the metric from degenerating while the curvature remains bounded.[/guided]

[step:Use Shi estimates to control all curvature derivatives]Fix $\tau \in (0,T)$. The manifold $M$ is closed, hence compact and complete for each metric $g(t)$. The assumed bound gives a uniform estimate $|\operatorname{Rm}(g(t))|_{g(t)} \leq K$ on the whole spacetime region $M \times [0,T)$. Therefore Shi’s derivative estimates for Ricci flow, for complete Ricci flows with uniformly bounded curvature, applied on the time interval $[0,T)$ and then restricted to the terminal subinterval $[\tau,T)$ away from the initial time, imply that for every integer $m \geq 0$ there is a constant \begin{align*} C_m = C_m(n,m,K,\tau,T,g(0)) > 0 \end{align*} such that \begin{align*} \sup_{(x,t) \in M \times [\tau,T)} |\nabla_{g(t)}^m \operatorname{Rm}(g(t))|_{g(t)}(x) \leq C_m. \end{align*} Here $\nabla_{g(t)}^m\operatorname{Rm}(g(t))$ denotes the $m$-fold covariant derivative of the Riemann curvature tensor with respect to the Levi-Civita connection of $g(t)$. Since $\tau$ was arbitrary, these bounds hold uniformly on every compact terminal subinterval $[\tau,T)$ with $\tau>0$.[/step]

[guided]We now use the standard local smoothing estimate for Ricci flow, namely Shi’s derivative estimates for Ricci flow. Its hypotheses are satisfied in this setting. First, $M$ is closed, so $M$ is compact; in particular each Riemannian manifold $(M,g(t))$ is complete. Second, the contradiction hypothesis gives the uniform spacetime curvature bound \begin{align*} |\operatorname{Rm}(g(t))|_{g(t)}(x) \leq K \end{align*} for every $(x,t) \in M \times [0,T)$. Third, we only ask for derivative bounds on $[\tau,T)$ with $\tau>0$, so the estimates are used away from the initial time, where their constants may depend on $\tau$. Shi’s derivative estimates therefore give the following conclusion: for each integer $m \geq 0$, there is a constant \begin{align*} C_m = C_m(n,m,K,\tau,T,g(0)) > 0 \end{align*} such that \begin{align*} \sup_{(x,t) \in M \times [\tau,T)} |\nabla_{g(t)}^m \operatorname{Rm}(g(t))|_{g(t)}(x) \leq C_m. \end{align*} Here $\nabla_{g(t)}^m\operatorname{Rm}(g(t))$ means the $m$-fold covariant derivative of the Riemann curvature tensor using the Levi-Civita connection of the metric $g(t)$. These estimates supply the spatial regularity needed for smooth continuation at the terminal time.[/guided]

[step:Apply the smooth continuation compactness criterion to obtain the terminal metric]We now use Hamilton’s smooth continuation criterion for Ricci flow: if $g(t)$ is a Ricci flow on a closed manifold on $[0,T)$, the metrics are uniformly equivalent to a fixed smooth background metric, and the curvature tensor together with all of its $g(t)$-covariant derivatives is uniformly bounded on every terminal subinterval $[\tau,T)$ with $\tau>0$, then $g(t)$ converges in $C^\infty$ as $t \uparrow T$ to a smooth Riemannian metric on $M$. The hypotheses have already been verified. The manifold $M$ is closed by assumption. The first step proves that $g(t)$ is uniformly equivalent to $g(0)$ on $M \times [0,T)$. The preceding step supplies the required bounds for $\nabla_{g(t)}^m\operatorname{Rm}(g(t))$ on $M \times [\tau,T)$ for every integer $m \geq 0$ and every $\tau \in (0,T)$. For completeness, we record the Cauchy estimate contained in the criterion. Fix a coordinate chart $(U,\varphi)$ with $\varphi: U \to \varphi(U) \subseteq \mathbb{R}^n$, fix a compact subset $L_U \subset U$, and write the coordinate components of the metric as functions $g_{ij}(\cdot,t): U \to \mathbb{R}$ for $1 \leq i,j \leq n$. The continuation criterion gives, for every integer $r \geq 0$, a constant $B_{r,L_U}>0$ such that \begin{align*} \sup_{t \in [\tau,T)}\sum_{i,j=1}^n\|\operatorname{Ric}_{ij}(g(t))\|_{C^r(L_U)} \leq B_{r,L_U}. \end{align*} Since the Ricci flow equation in this chart is \begin{align*} \frac{\partial}{\partial t}g_{ij}(\cdot,t) = -2\operatorname{Ric}_{ij}(g(t)), \end{align*} we obtain, for $\tau \leq s \leq t < T$, \begin{align*} \sum_{i,j=1}^n\|g_{ij}(\cdot,t)-g_{ij}(\cdot,s)\|_{C^r(L_U)} \leq 2B_{r,L_U}(t-s). \end{align*} Thus the coordinate components are Cauchy in every $C^r(L_U)$ norm as $s,t \uparrow T$. The criterion identifies the resulting local limits as the coordinate components of a smooth symmetric $(0,2)$-tensor $g_T \in \Gamma(S^2T^*M)$, and the lower metric bound \begin{align*} g(t)_p(v,v) \geq e^{-2c_nKT}g(0)_p(v,v) \end{align*} passes to the limit for every $p \in M$ and every $v \in T_pM$. Hence $g_T$ is positive definite, so $g_T$ is a smooth Riemannian metric, and \begin{align*} g(t) \to g_T \end{align*} smoothly on $M$ as $t \uparrow T$.[/step]

[guided]The delicate point is that curvature bounds alone are not coordinate derivative bounds for the metric in an arbitrary chart. We therefore use the precise continuation compactness statement for Ricci flow rather than trying to infer coordinate estimates directly from the curvature tensor. Hamilton’s smooth continuation criterion for Ricci flow says that a Ricci flow on a closed manifold has a smooth terminal limit provided three inputs hold: a fixed background metric uniformly equivalent to $g(t)$, uniform bounds for all covariant derivatives of curvature on terminal subintervals, and compactness of the underlying manifold. Each input is available here. The manifold $M$ is closed, so it is compact. The first step gives the uniform equivalence \begin{align*} e^{-2c_nKT}g(0)_p(v,v) \leq g(t)_p(v,v) \leq e^{2c_nKT}g(0)_p(v,v) \end{align*} for all $p \in M$, all $v \in T_pM$, and all $t \in [0,T)$. The Shi estimates from the preceding step give, for every integer $m \geq 0$ and every $\tau \in (0,T)$, \begin{align*} \sup_{(x,t) \in M \times [\tau,T)} |\nabla_{g(t)}^m \operatorname{Rm}(g(t))|_{g(t)}(x) \leq C_m. \end{align*} Thus the continuation criterion applies and produces a smooth limit tensor $g_T \in \Gamma(S^2T^*M)$ with $g(t) \to g_T$ in $C^\infty$ as $t \uparrow T$. Let us also spell out why the convergence is genuinely Cauchy in local coordinates. Fix a chart $(U,\varphi)$, a compact subset $L_U \subset U$, and write $g_{ij}(\cdot,t): U \to \mathbb{R}$ for the coordinate components of $g(t)$. The continuation criterion supplies constants $B_{r,L_U}>0$ such that \begin{align*} \sup_{t \in [\tau,T)}\sum_{i,j=1}^n\|\operatorname{Ric}_{ij}(g(t))\|_{C^r(L_U)} \leq B_{r,L_U} \end{align*} for every integer $r \geq 0$. Since the Ricci flow equation gives \begin{align*} \frac{\partial}{\partial t}g_{ij}(\cdot,t) = -2\operatorname{Ric}_{ij}(g(t)), \end{align*} integrating from $s$ to $t$ yields \begin{align*} \sum_{i,j=1}^n\|g_{ij}(\cdot,t)-g_{ij}(\cdot,s)\|_{C^r(L_U)} \leq 2B_{r,L_U}(t-s) \end{align*} whenever $\tau \leq s \leq t < T$. Hence the coordinate components converge in every $C^r(L_U)$ norm. Finally, the positive lower bound from uniform equivalence passes to the limit, so $g_T$ is not merely a smooth symmetric tensor: it is positive definite and therefore a smooth Riemannian metric.[/guided]

[step:Restart the flow from the limiting metric and contradict maximality] By Hamilton’s short-time existence theorem for Ricci flow on closed manifolds, applied to the smooth closed manifold $M$ and the smooth Riemannian metric $g_T$, there exist $\varepsilon > 0$ and a smooth Ricci flow \begin{align*} \tilde{g}: [T,T+\varepsilon) \to \Gamma(S^2T^*M) \end{align*} such that \begin{align*} \tilde{g}(T) = g_T. \end{align*} Define the tensor-valued map $\hat{g}: [0,T+\varepsilon) \to \Gamma(S^2T^*M)$ by requiring \begin{align*} \hat{g}(t) = g(t) \quad \text{for } 0 \leq t < T. \end{align*} and \begin{align*} \hat{g}(t) = \tilde{g}(t) \quad \text{for } T \leq t < T+\varepsilon. \end{align*} Because $g(t) \to g_T$ smoothly as $t \uparrow T$, the zeroth-order tensors match at $t=T$. The time derivatives also match: the first derivative is determined by $\partial_t g=-2\operatorname{Ric}(g)$, and higher time derivatives are obtained recursively by differentiating the Ricci flow equation, so each is a universal smooth expression in $g(t)^{-1}$ and finitely many spatial derivatives of $g(t)$. The smooth convergence from the preceding step therefore implies that the left time derivatives at $T$ agree with the corresponding derivatives of the solution starting from $g_T$. Hence the piecewise tensor field $\hat{g}$ is smooth at $t=T$ and solves the Ricci flow equation on $[0,T+\varepsilon)$. This extends $g$ beyond $T$, contradicting the assumption that $[0,T)$ is the maximal time interval of existence. Therefore the bounded-curvature assumption is false. Hence \begin{align*} \sup_{(x,t) \in M \times [0,T)} |\operatorname{Rm}(g(t))|_{g(t)}(x) = \infty. \end{align*} This proves the theorem. [/step]