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Sepideh Bakhoda
Sepideh Bakhoda
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In study of Ricci flow, for making Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$. So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know that $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. Why don't we define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? In this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Is notDoesn't this make a better definition?

In study of Ricci flow, for making Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$. So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know that $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. Why don't we define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? In this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Is not this better?

In study of Ricci flow, for making Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$. So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know that $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. Why don't we define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? In this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Doesn't this make a better definition?

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Sepideh Bakhoda
Sepideh Bakhoda
  • 1.3k
  • 1
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In study of Ricci flow, for makemaking Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$,. So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know that $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. why weWhy don't we define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? inIn this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Is not this better?

In study of Ricci flow, for make Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$, So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. why we don't define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? in this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Is not this better?

In study of Ricci flow, for making Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$. So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know that $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. Why don't we define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? In this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Is not this better?

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Sepideh Bakhoda
Sepideh Bakhoda
  • 1.3k
  • 1
  • 15
  • 23

Ricci flow as a gradient flow and its Lyapunov function

In study of Ricci flow, for make Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$, So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. why we don't define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? in this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Is not this better?