Finite-time extinction of Ricci flow with surgery

Step 1 (a dimension-3 Cauchy–Schwarz inequality). At a point of a 3-manifold, the Ricci tensor is a symmetric $3\times3$ matrix with eigenvalues $\lambda_1,\lambda_2,\lambda_3$, and $R=\lambda_1+\lambda_2+\lambda_3$. The Cauchy–Schwarz (power-mean) inequality on three numbers gives $(\lambda_1+\lambda_2+\lambda_3)^2\le3(\lambda_1^2+\lambda_2^2+\lambda_3^2)$, i.e. $|\mathrm{Ric}|^2\ge R^2/3$, with equality exactly at Einstein points ($\mathrm{Ric}=\tfrac{R}{3}g$). This inequality is special to dimension 3 and is exactly what lets the argument close using only the scalar curvature $R$.

**Step 2 (a differential inequality for $R$).** Substituting into Hamilton's evolution formula $\partial_tR=\Delta R+2|\mathrm{Ric}|^2$ gives $\partial_tR=\Delta R+2|\mathrm{Ric}|^2\ge\Delta R+\tfrac23R^2$.

Step 3 (maximum principle at the minimum). Let $R_{\min}(t)=\min_{x\in M}R(x,t)$. At a point where the spatial minimum of $R$ is attained, $\Delta R\ge0$ (second-derivative test), so the differential inequality of Step 2 forces $\frac{d}{dt}R_{\min}\ge\frac23R_{\min}^2$ in the barrier sense (Hamilton's maximum principle for evolving minima under Ricci flow). Surgery only grafts in metrics whose curvature is at least as large as the surgery threshold, so it can never lower $R_{\min}$ below where it already was; hence this differential inequality persists across the finitely many surgery times.

Step 4 (solving the model ODE). Solve $y'=\tfrac23y^2$, $y(0)=r_0>0$, by separation of variables: $\int y^{-2}dy=\int\tfrac23dt$ gives $-\tfrac1y=\tfrac23t-\tfrac1{r_0}$, i.e. $\tfrac1{y(t)}=\tfrac1{r_0}-\tfrac23t$, i.e. $y(t)=\dfrac{r_0}{1-\frac23 r_0 t}$. This is exactly the scalar curvature of a shrinking round metric sphere of matching initial size, and $y(t)\to+\infty$ precisely as the denominator vanishes, at $t=T$, where $T=\dfrac{3}{2r_0}$.

Step 5 (comparison). Since $R_{\min}(t)$ satisfies $\frac{d}{dt}R_{\min}\ge\tfrac23R_{\min}^2$ while $y(t)$ satisfies this ODE with equality and the same initial value $r_0$, the ODE comparison principle gives $R_{\min}(t)\ge y(t)$ for as long as both remain defined. Because $y(t)\to+\infty$ as $t\to T^-$, $R_{\min}(t)$ must already have escaped to $+\infty$ at or before $T$: the flow cannot be smoothly continued past $T$ without curvature blowing up, so surgery must intervene no later than $T$.

Step 6 (from blow-up to extinction). A component on which $R_{\min}\to\infty$ is, by comparison with the model shrinking sphere of Step 4, shrinking to zero volume everywhere at least that fast; there is no room for it to persist past $T$. Because $M$ is simply connected, geometrization allows no non-positively-curved piece ($\mathbb{H}^3$, $\mathrm{Nil}$, $\mathrm{Sol}$, $\mathbb{H}^2\times\mathbb{R}$, $\widetilde{\mathrm{SL}_2\mathbb{R}}$ pieces would each force a nontrivial fundamental group on a compact 3-manifold), so no surviving geometric piece can be left over once the positively-curved pieces have shrunk away: every connected component present at time $t$ is forced to disappear by surgery no later than time $T=\tfrac{3}{2r_0}$, i.e. the flow with surgery is extinct by $T$. $\blacksquare$