Monotonicity of Perelman's $\mathcal{F}$-entropy

Step 1 (conserved measure). Under Ricci flow, $\partial_t\,dV=-R\,dV$ (since $\partial_t\log\sqrt{\det g}=\tfrac12g^{ij}\partial_tg_{ij}=-R$). Combining this with the evolution $\partial_t f = -\Delta f + |\nabla f|^2 - R$ for $f$ gives $\partial_t(e^{-f}dV)=(-\partial_tf-R)e^{-f}dV=(\Delta f-|\nabla f|^2)e^{-f}dV=-\Delta(e^{-f})\,dV$, using $\Delta(e^{-f})=e^{-f}(|\nabla f|^2-\Delta f)$. Integrating over the closed manifold $M$, the right side vanishes by the divergence theorem, so $\int_M e^{-f}dV$ is constant in time: $e^{-f}dV$ is a conserved measure carried along by the coupled flow.

**Step 2 (reduce to an integral of $\partial_t\varphi-\Delta\varphi$).** Write $\varphi=R+|\nabla f|^2$, so $\mathcal F(g,f)=\int_M\varphi\,e^{-f}dV$. Differentiating under the integral sign and using $\partial_t(e^{-f}dV)=-\Delta(e^{-f})\,dV$ from Step 1, $\frac{d}{dt}\mathcal F=\int_M(\partial_t\varphi)e^{-f}dV-\int_M\varphi\,\Delta(e^{-f})\,dV$. Since the ordinary Laplacian is self-adjoint with respect to $dV$ on the closed manifold $M$, $\int_M\varphi\,\Delta(e^{-f})\,dV=\int_M(\Delta\varphi)e^{-f}dV$, so $\frac{d}{dt}\mathcal F=\int_M\big[\partial_t\varphi-\Delta\varphi\big]e^{-f}dV$.

**Step 3 (the $R$ part).** Hamilton's evolution formula for scalar curvature is $\partial_tR=\Delta R+2|\mathrm{Ric}|^2$, hence $\partial_tR-\Delta R=2|\mathrm{Ric}|^2$ directly. This already contributes $2|\mathrm{Ric}|^2$ to the bracket in Step 2.

**Step 4 (the $|\nabla f|^2$ part).** Since $\partial_tg^{ij}=2R^{ij}$, the chain rule gives $\partial_t|\nabla f|^2=2\mathrm{Ric}(\nabla f,\nabla f)+2\langle\nabla f,\nabla(\partial_tf)\rangle$, and substituting $\partial_tf=-\Delta f+|\nabla f|^2-R$ together with the identity $\langle\nabla f,\nabla|\nabla f|^2\rangle=2\nabla^2f(\nabla f,\nabla f)$ yields $\partial_t|\nabla f|^2=2\mathrm{Ric}(\nabla f,\nabla f)-2\langle\nabla f,\nabla\Delta f\rangle+4\nabla^2f(\nabla f,\nabla f)-2\langle\nabla f,\nabla R\rangle$. Bochner's formula states $\Delta|\nabla f|^2=2|\nabla^2f|^2+2\langle\nabla f,\nabla\Delta f\rangle+2\mathrm{Ric}(\nabla f,\nabla f)$; subtracting cancels both copies of $\mathrm{Ric}(\nabla f,\nabla f)$ and leaves $\partial_t|\nabla f|^2-\Delta|\nabla f|^2=-2|\nabla^2f|^2+4\nabla^2f(\nabla f,\nabla f)-4\langle\nabla f,\nabla\Delta f\rangle-2\langle\nabla f,\nabla R\rangle$.

Step 5 (weighted Bochner and a vanishing divergence). Introduce the Bakry–Émery ($f$-weighted) Laplacian $\Delta_f h = \Delta h - \langle\nabla f,\nabla h\rangle$; it satisfies $\int_M(\Delta_fh)e^{-f}dV=0$ for every function $h$, since $\Delta_fh\,e^{-f}=\mathrm{div}(e^{-f}\nabla h)$ is a pure divergence. Writing $\Delta f=\Delta_ff+|\nabla f|^2$ turns $\langle\nabla f,\nabla\Delta f\rangle$ into $\langle\nabla f,\nabla\Delta_ff\rangle+2\nabla^2f(\nabla f,\nabla f)$, and the weighted Bochner–Weitzenböck identity $\tfrac12\Delta_f|\nabla f|^2=|\nabla^2f|^2+\langle\nabla f,\nabla\Delta_ff\rangle+(\mathrm{Ric}+\nabla^2f)(\nabla f,\nabla f)$ isolates $\langle\nabla f,\nabla\Delta_ff\rangle$. Substituting all of this back into the Step 4 expression, every term built from $\Delta_f|\nabla f|^2$ integrates to zero against $e^{-f}dV$ by the vanishing-divergence property above, and what survives is exactly $\int_M\big[2|\nabla^2f|^2+4\mathrm{Ric}(\nabla f,\nabla f)-2\langle\nabla f,\nabla R\rangle\big]e^{-f}dV$.

Step 6 (completing the square via the contracted Bianchi identity). Integrating $\langle\mathrm{Ric},\nabla^2f\rangle$ by parts and using the contracted second Bianchi identity $\nabla^iR_{ij}=\frac12\nabla_jR$ gives exactly $4\int_M\langle\mathrm{Ric},\nabla^2f\rangle e^{-f}dV=\int_M\big[4\mathrm{Ric}(\nabla f,\nabla f)-2\langle\nabla f,\nabla R\rangle\big]e^{-f}dV$ — precisely the cross term left over from Step 5. Substituting back and adding the $2|\mathrm{Ric}|^2$ from Step 3, the integrand assembles into a perfect square: $2|\mathrm{Ric}|^2+4\langle\mathrm{Ric},\nabla^2f\rangle+2|\nabla^2f|^2=2|\mathrm{Ric}+\nabla^2f|^2$.

Conclusion. Therefore $\frac{d}{dt}\mathcal{F}(g,f)=2\int_M|R_{ij}+\nabla_i\nabla_jf|^2e^{-f}dV\ge0$, a sum of squares against the positive weight $e^{-f}dV$, so $\mathcal{F}$ is non-decreasing along the coupled flow, and it is constant on an interval exactly when $R_{ij}+\nabla_i\nabla_jf\equiv0$ there, i.e. exactly on steady gradient Ricci solitons. $\blacksquare$