[proofplan] We first turn the uniform curvature bound into uniform control of the geometry on every terminal slab $[\tau,T)$, using Shi derivative estimates for Ricci flow. These estimates give smooth convergence of $g(t)$ as $t \uparrow T$ to a smooth Riemannian metric $g_T$ on the compact manifold $M$. Hamilton's short-time existence theorem applied to $g_T$ then produces a Ricci flow starting at time $T$, and the original flow and the new flow patch smoothly because the Ricci flow equation determines all time derivatives from the spatial jets of the metric. [/proofplan]

[step:Use the curvature bound to obtain uniform equivalence of the metrics] Let $n := \dim M$ denote the dimension of the compact smooth manifold $M$. Let $K \in [0,\infty)$ denote the finite curvature bound \begin{align*} K := \sup_{M \times [0,T)} |\operatorname{Rm}(g(t))|_{g(t)}. \end{align*} For each $t \in [0,T)$, let $\operatorname{Ric}(g(t))$ denote the Ricci tensor of $g(t)$. The Ricci tensor is obtained by contracting one pair of indices of $\operatorname{Rm}(g(t))$ against a $g(t)$-[orthonormal basis](/page/Orthonormal%20Basis). Hence the standard finite-dimensional tensor contraction estimate gives $|\operatorname{Ric}(g(t))|_{g(t)} \leq n|\operatorname{Rm}(g(t))|_{g(t)} \leq nK$. The Ricci flow equation $\partial_t g(t) = -2\operatorname{Ric}(g(t))$ implies that for every $p \in M$ and every $v \in T_pM$, \begin{align*} \left|\frac{d}{dt} g(t)_p(v,v)\right| \leq 2nK\, g(t)_p(v,v). \end{align*} Gronwall's inequality on $[0,t]$ gives \begin{align*} e^{-2nKt} g(0)_p(v,v) \leq g(t)_p(v,v) \leq e^{2nKt} g(0)_p(v,v). \end{align*} Because $T<\infty$, the metrics $g(t)$ are uniformly equivalent to the fixed background metric $g(0)$ on $[0,T)$. [/step]

[step:Apply Shi derivative estimates on compact subintervals below $T$]Fix $\tau \in (0,T)$. For every $T_0 \in (\tau,T)$, the restriction of $g(t)$ to $M \times [0,T_0]$ is a smooth Ricci flow on a compact manifold with \begin{align*} \sup_{M \times [0,T_0]} |\operatorname{Rm}(g(t))|_{g(t)} \leq K. \end{align*} The hypotheses of the compact interior form of the Shi derivative estimates for Ricci flow are therefore satisfied on each compact subinterval $[0,T_0]$. We use the interior-in-time compact version: if a smooth Ricci flow on a compact $n$-manifold satisfies $|\operatorname{Rm}|\leq K$ on $M\times[0,T_0]$, then for every integer $m\geq0$ and every $\tau<T_0$ there is a constant depending only on $n$, $m$, $K$, $\tau$, and an upper bound for $T_0$ such that $|\nabla^m\operatorname{Rm}|$ is bounded on $M\times[\tau,T_0]$. Since every interval used here has $T_0<T$, the upper-time argument may be bounded by the fixed number $T$, so the constants are independent of the particular choice of $T_0$. Thus, for every integer $m \geq 0$, there exists a constant $C_m = C_m(n,m,K,\tau,T) < \infty$ such that \begin{align*} \sup_{M \times [\tau,T)} |\nabla^{m}\operatorname{Rm}(g(t))|_{g(t)} \leq C_m, \end{align*} where $\nabla^{m}\operatorname{Rm}(g(t))$ denotes the $m$th covariant derivative of the Riemann curvature tensor with respect to $g(t)$.[/step]

[guided]We want bounds not only for curvature but also for all spatial derivatives of curvature as $t$ approaches $T$. This is exactly the role of Shi's derivative estimates. We fix a positive time $\tau \in (0,T)$ because Shi's estimates may lose control at the initial time, while the extension problem only concerns the behaviour as $t \uparrow T$. There is a small endpoint issue to handle carefully: the original flow is defined on the half-open interval $[0,T)$, not on a closed slab ending at $T$. To apply Shi's estimates, choose an arbitrary $T_0 \in (\tau,T)$. Then $g(t)$ restricted to $M \times [0,T_0]$ is a smooth Ricci flow on a compact manifold, and the curvature bound gives \begin{align*} \sup_{M \times [0,T_0]} |\operatorname{Rm}(g(t))|_{g(t)} \leq K. \end{align*} Thus the hypotheses of Shi's interior-in-time derivative estimate for compact Ricci flows are satisfied on the closed time interval $[0,T_0]$. This version of the estimate says that a curvature bound $|\operatorname{Rm}|\leq K$ on $M\times[0,T_0]$ controls $|\nabla^m\operatorname{Rm}|$ on $M\times[\tau,T_0]$, with constants depending on $n$, $m$, $K$, $\tau$, and the upper time parameter. Because every $T_0$ under consideration satisfies $T_0<T$, we replace that upper time parameter by $T$ and obtain a constant $C_m=C_m(n,m,K,\tau,T)$ independent of $T_0$. Hence \begin{align*} \sup_{M \times [\tau,T_0]} |\nabla^{m}\operatorname{Rm}(g(t))|_{g(t)} \leq C_m. \end{align*} Since $T_0 \in (\tau,T)$ was arbitrary, taking the supremum over all such $T_0$ yields \begin{align*} \sup_{M \times [\tau,T)} |\nabla^{m}\operatorname{Rm}(g(t))|_{g(t)} \leq C_m. \end{align*} These estimates are the analytic input that prevents the geometry from oscillating in higher derivatives while the curvature itself remains bounded.[/guided]

[step:Take the smooth limit metric as $t$ approaches $T$]Let $g_0 := g(0)$ be the fixed background metric on $M$. For each integer $m \geq 0$, let $C^m(M;T^*M \otimes T^*M)$ denote the [Fréchet space](/page/Fr%C3%A9chet%20Space) of $m$-times continuously differentiable covariant $2$-tensor fields on $M$, measured using any finite atlas and a subordinate [partition of unity](/page/Partition%20of%20Unity); since $M$ is compact, different choices give equivalent norms. The uniform equivalence from the first step gives a constant $A = e^{2nKT}$ such that \begin{align*} A^{-1} g_0 \leq g(t) \leq A g_0 \end{align*} for all $t \in [0,T)$. The Ricci bound and uniform equivalence imply that the map $t \mapsto g(t)$ is Cauchy in $C^0(M;T^*M \otimes T^*M)$ as $t \uparrow T$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Indeed, for $0 \leq s < t < T$ and $p \in M$, \begin{align*} |g(t)_p - g(s)_p|_{g_0} \leq \int_s^t |\partial_r g(r)_p|_{g_0}\,d\mathcal{L}^1(r) \leq 2nKA\,(t-s). \end{align*} Thus there exists a continuous positive-definite symmetric $2$-tensor $g_T$ such that $g(t) \to g_T$ uniformly as $t \uparrow T$. We next upgrade the convergence to smooth convergence without introducing a moving gauge. Fix $\tau \in (0,T)$ and let $\nabla^\tau$ denote the Levi-Civita connection of the fixed metric $g(\tau)$. For $t \in [\tau,T)$, let $B(t) := \nabla^{g(t)} - \nabla^\tau$ denote the difference tensor between the Levi-Civita connection of $g(t)$ and $\nabla^\tau$, so $B(t)$ is a smooth section of $T^*M\otimes T^*M\otimes TM$. Along Ricci flow the connection evolves by \begin{align*} \partial_t \Gamma_{ij}^{k}(g(t)) = -g(t)^{k\ell}\left(\nabla_i \operatorname{Ric}(g(t))_{j\ell}+\nabla_j \operatorname{Ric}(g(t))_{i\ell}-\nabla_\ell \operatorname{Ric}(g(t))_{ij}\right), \end{align*} where the displayed formula is written in any coordinate chart and $\nabla$ denotes the Levi-Civita connection of $g(t)$. The uniform metric equivalence and the Shi estimate with $m=1$ therefore bound $\partial_t B(t)$ uniformly on $M\times[\tau,T)$. Integrating from $\tau$ to $t$ gives a uniform bound for $B(t)$ with respect to $g(\tau)$. For each integer $r\geq1$, applying $(\nabla^\tau)^r$ to the connection evolution formula expresses $\partial_t(\nabla^\tau)^r B(t)$ as a universal finite sum formed from $g(t)^{-1}$, $B(t),\dots,(\nabla^\tau)^rB(t)$, and covariant derivatives of $\operatorname{Rm}(g(t))$ of order at most $r+1$. The already obtained lower-order bounds, the uniform metric equivalence, and the Shi estimates imply inductively that $(\nabla^\tau)^rB(t)$ is uniformly bounded on $M\times[\tau,T)$ for every $r\geq0$. Since \begin{align*} \partial_t g(t)=-2\operatorname{Ric}(g(t)), \end{align*} applying $(\nabla^\tau)^r$ and using the bounds for $B(t)$ together with the Shi estimates gives a uniform bound for $(\nabla^\tau)^r\partial_t g(t)$ on $M\times[\tau,T)$ for every $r\geq0$. The [fundamental theorem of calculus](/theorems/632) in time then shows that $g(t)$ is Cauchy in every $C^m(M;T^*M\otimes T^*M)$ norm as $t\uparrow T$. Let $g_T$ denote the resulting smooth limit. This smooth limit agrees with the previously obtained uniform limit, so \begin{align*} g(t) \to g_T \quad \text{in } C^m(M;T^*M \otimes T^*M) \text{ as } t \uparrow T \end{align*} for every integer $m \geq 0$.[/step]

[guided]The first goal is to get an actual metric at the terminal time. Define $g_0 := g(0)$. From the first step, with $A=e^{2nKT}$, we have \begin{align*} A^{-1}g_0 \leq g(t) \leq Ag_0 \end{align*} for every $t\in[0,T)$. This inequality has two roles: it prevents degeneration of the limiting tensor, and it lets us compare tensor norms taken with respect to $g(t)$ to tensor norms taken with respect to the fixed background metric $g_0$. For precision, for each integer $m\geq0$ we define $C^m(M;T^*M\otimes T^*M)$ to be the space of $m$-times continuously differentiable covariant $2$-tensor fields on $M$, equipped with any standard $C^m$ norm coming from a finite atlas and a partition of unity. Compactness of $M$ ensures that all such choices produce equivalent norms, so convergence in this space is intrinsic. Now we prove [uniform convergence](/page/Uniform%20Convergence) first. Since \begin{align*} \partial_t g(t)=-2\operatorname{Ric}(g(t)) \end{align*} and $|\operatorname{Ric}(g(t))|_{g(t)}\leq nK$, the fixed-background norm satisfies $|\partial_t g(t)|_{g_0}\leq2nKA$. Therefore, for $0\leq s<t<T$ and $p\in M$, \begin{align*} |g(t)_p-g(s)_p|_{g_0} \leq \int_s^t |\partial_r g(r)_p|_{g_0}\,d\mathcal{L}^1(r) \leq 2nKA(t-s). \end{align*} This shows that $g(t)$ is Cauchy in $C^0(M;T^*M\otimes T^*M)$ as $t\uparrow T$, so it converges uniformly to a continuous symmetric $2$-tensor $g_T$. The lower bound $A^{-1}g_0\leq g(t)$ passes to the limit, so $g_T$ is positive definite. The remaining question is why the convergence is smooth, not only uniform. We avoid a circular DeTurck argument by working directly with a fixed connection after a positive time. Fix $\tau\in(0,T)$ and let $\nabla^\tau$ be the Levi-Civita connection of the fixed metric $g(\tau)$. For $t\in[\tau,T)$, define \begin{align*} B(t):=\nabla^{g(t)}-\nabla^\tau. \end{align*} This is a tensor because the difference of two connections is tensorial; more precisely, $B(t)$ is a smooth section of $T^*M\otimes T^*M\otimes TM$. Controlling $B(t)$ and its $\nabla^\tau$-derivatives is the same as controlling the Christoffel symbols of $g(t)$ relative to the fixed background connection. The connection evolution formula along Ricci flow is \begin{align*} \partial_t \Gamma_{ij}^{k}(g(t)) = -g(t)^{k\ell}\left(\nabla_i \operatorname{Ric}(g(t))_{j\ell}+\nabla_j \operatorname{Ric}(g(t))_{i\ell}-\nabla_\ell \operatorname{Ric}(g(t))_{ij}\right), \end{align*} where the formula is written in any coordinate chart and $\nabla$ is the Levi-Civita connection of $g(t)$. The right-hand side contains one covariant derivative of the Ricci tensor. Since Ricci is a contraction of the Riemann curvature tensor, the Shi estimate with $m=1$ bounds this term uniformly on $M\times[\tau,T)$. The uniform equivalence of $g(t)$ to $g_0$ controls the inverse metric factors. Hence $\partial_t B(t)$ is uniformly bounded. Integrating from $\tau$ to $t$ gives a uniform bound for $B(t)$. We now repeat this after applying spatial derivatives with respect to the fixed connection $\nabla^\tau$. For each integer $r\geq1$, applying $(\nabla^\tau)^r$ to the connection evolution formula expresses $\partial_t(\nabla^\tau)^rB(t)$ as a universal finite sum involving $g(t)^{-1}$, the tensors $B(t),\dots,(\nabla^\tau)^rB(t)$, and covariant derivatives of $\operatorname{Rm}(g(t))$ of order at most $r+1$. The lower-order $B$ terms have already been bounded by induction, and the curvature terms are bounded by Shi's estimates. Therefore $(\nabla^\tau)^rB(t)$ is uniformly bounded on $M\times[\tau,T)$ for every $r\geq0$. Finally, the Ricci flow equation gives \begin{align*} \partial_t g(t)=-2\operatorname{Ric}(g(t)). \end{align*} Applying $(\nabla^\tau)^r$ to this equation and rewriting $\nabla^\tau$-derivatives in terms of $\nabla^{g(t)}$-derivatives and the bounded tensors $B(t),\dots,(\nabla^\tau)^rB(t)$ gives a uniform bound for $(\nabla^\tau)^r\partial_tg(t)$ for every $r\geq0$. Therefore, for $s<t<T$, the fundamental theorem of calculus in time gives Cauchy convergence in every fixed $C^m$ norm. The limit is a smooth metric, and it agrees with the uniform limit already called $g_T$. Thus \begin{align*} g(t)\to g_T \quad \text{in } C^m(M;T^*M\otimes T^*M) \text{ as } t\uparrow T \end{align*} for every integer $m\geq0$.[/guided]

[step:Start a new Ricci flow from the limiting metric]Let $\operatorname{Sym}^2(T^*M)$ denote the bundle of symmetric covariant $2$-tensors on $M$, and let $\Gamma(\operatorname{Sym}^2(T^*M))$ denote its space of smooth sections. Since $M$ is compact and $g_T$ is a smooth Riemannian metric on $M$, the hypotheses of Hamilton's compact short-time existence theorem for Ricci flow are satisfied: the initial manifold is compact and the initial metric is smooth. Hence there exist $\varepsilon>0$ and a smooth Ricci flow $\hat g: [0,\varepsilon) \to \Gamma(\operatorname{Sym}^2(T^*M))$. For each $s\in[0,\varepsilon)$, the value $\hat g(s)$ is a smooth Riemannian metric on $M$. This flow satisfies \begin{align*} \hat g(0)=g_T, \qquad \partial_s \hat g(s) = -2\operatorname{Ric}(\hat g(s)) \end{align*} for $s \in [0,\varepsilon)$.[/step]

[guided]We now restart the equation from the terminal metric. The input theorem is Hamilton's compact short-time existence theorem for Ricci flow: on a compact smooth manifold, every smooth Riemannian metric is the initial metric of a smooth Ricci flow for some positive time. Its hypotheses are met because $M$ is compact and the previous step proved that $g_T$ is a smooth Riemannian metric on $M$. Therefore there are $\varepsilon>0$ and a smooth map \begin{align*} \hat g:[0,\varepsilon)&\to\Gamma(\operatorname{Sym}^2(T^*M)) \end{align*} such that each $\hat g(s)$ is a smooth Riemannian metric on $M$ and \begin{align*} \hat g(0)=g_T, \qquad \partial_s\hat g(s)=-2\operatorname{Ric}(\hat g(s)) \end{align*} for every $s\in[0,\varepsilon)$. This produces the piece of the extended flow that begins exactly at the limiting metric.[/guided]

[step:Patch the old and new flows smoothly at time $T$]Define a time-dependent symmetric $2$-tensor $\tilde g: [0,T+\varepsilon) \to \Gamma(\operatorname{Sym}^2(T^*M))$ by setting $\tilde g(t) := g(t)$ for $0 \leq t < T$ and $\tilde g(t) := \hat g(t-T)$ for $T \leq t < T+\varepsilon$. The smooth convergence $g(t) \to g_T = \hat g(0)$ shows that $\tilde g$ is spatially smooth at $t=T$. Moreover, the Ricci flow equation expresses each time derivative $\partial_t^j g(t)$ as a universal polynomial in the inverse metric and finitely many spatial derivatives of $g(t)$; the same formula holds for $\hat g(s)$. Since the spatial jets of $g(t)$ converge to the spatial jets of $g_T=\hat g(0)$, the left and right limits of every mixed space-time derivative of $\tilde g$ agree at $t=T$. Thus $\tilde g$ is smooth on $M \times [0,T+\varepsilon)$ and satisfies \begin{align*} \partial_t \tilde g(t) = -2\operatorname{Ric}(\tilde g(t)) \end{align*} for every $t \in [0,T+\varepsilon)$. This is a smooth Ricci flow extension of the original $g(t)$, proving the theorem.[/step]

[guided]We must check more than continuity at the gluing time. Define $\tilde g: [0,T+\varepsilon) \to \Gamma(\operatorname{Sym}^2(T^*M))$ as follows: for $0\leq t<T$, set $\tilde g(t)=g(t)$, and for $T\leq t<T+\varepsilon$, set $\tilde g(t)=\hat g(t-T)$. The equality $\hat g(0)=g_T$ and the smooth convergence $g(t)\to g_T$ imply that the spatial derivatives of $\tilde g$ from the left at $T$ agree with the spatial derivatives of $\hat g(0)$ from the right. It remains to check time derivatives. The Ricci flow equation gives the first time derivative as \begin{align*} \partial_t g(t)=-2\operatorname{Ric}(g(t)). \end{align*} The Ricci tensor is a smooth expression in the inverse metric, the first spatial derivatives of the metric, and the second spatial derivatives of the metric in local coordinates. Therefore the smooth convergence $g(t)\to g_T$ implies \begin{align*} \lim_{t\uparrow T}\partial_t g(t)=-2\operatorname{Ric}(g_T)=\partial_s \hat g(0). \end{align*} For higher derivatives, differentiate the Ricci flow equation repeatedly in time. Inductively, each $\partial_t^j g(t)$ is a universal smooth expression in the inverse metric and finitely many spatial derivatives of $g(t)$. The same universal expression applies to $\hat g(s)$ because $\hat g$ satisfies the same Ricci flow equation. Since all spatial jets of $g(t)$ converge to the corresponding spatial jets of $g_T=\hat g(0)$, every mixed space-time derivative has the same left and right limit at $t=T$. Thus $\tilde g$ is smooth on $M\times[0,T+\varepsilon)$ and satisfies \begin{align*} \partial_t\tilde g(t)=-2\operatorname{Ric}(\tilde g(t)) \end{align*} on both sides of $T$, and also at $T$ by the derivative matching above. Hence $\tilde g$ is a smooth Ricci flow extension of the original flow $g(t)$.[/guided]