# Perelman's entropy functional computation problem.

I was reading Hopper and Andrews's book on Ricci flow in Riemannian Geometry. I came across the following proposition on the monotonicity of Perelman's $$\mathcal{W}$$-functional.

Let $$(g(t),f(t), \tau(t))$$ evolve by \begin{align*} \frac{\partial g}{\partial t} &= -2\text{Ric} \\ \frac{\partial f}{\partial t} &= -|\nabla f|^{2} + \Delta f - \text{Scal} + \frac{n}{2\tau} \\ \frac{d \tau}{d t} &= -1. \end{align*}

Consider the function $$w = (\tau(\text{Scal} + 2\Delta f - |\nabla f|^{2}) + f - n)u,$$ where $$u = (4\pi \tau)^{-n/2}e^{-f}$$.

Also, consider the operator $$\Box^{*} = -\frac{\partial}{\partial t} - \Delta + \text{Scal}.$$

We have to show that $$\Box^{*}w = -2\tau \left | \text{Ric} + \text{Hess}{f} - \frac{g}{2\tau} \right|^{2} u.$$

I'm having trouble proving this. The book refers to Peter Topping's lecture notes on Ricci flow where the computation is done.

In the first line of their proof, I found the following which I find problematic. They write

$$\Box^{*}w = \Box^{*}(u) \frac{w}{u} - \left(\frac{\partial}{\partial t} + \Delta \right) \left(\frac{w}{u}\right) - 2 \left \langle \nabla \left( \frac{w}{u} \right) , \nabla u \right \rangle.$$

I believe that the term comes from the following intermediate step, $$\Box^{*}w = \Box^{*}(u) \left( \frac{w}{u} \right) + u \Box^{*}\left( \frac{w}{u} \right).$$ The last term in the above expression troubles me. What should have been simply

$$u \text{Scal} \frac{w}{u}$$ is written as $$-2\left \langle \nabla \frac{w}{u}, \nabla u \right \rangle.$$ I don't understand how are these equal. When I tried to solve the derivative myself I found exactly these terms which I couldn't wish away.

I found the same problem as an exercise in Chow, Lu, and Ni's book on Hamilton's Ricci flow as well.

I would be happy to provide more details if needed.

• I don't even see how $\Box^{*}w = \Box^{*}(u) \left( \frac{w}{u} \right) + u \Box^{*}\left( \frac{w}{u} \right)$ is true. Commented Jul 27, 2020 at 14:58
• This is just the derivative rule over the product. $w$ is written as $u . w/u$. Commented Jul 27, 2020 at 15:25
• Oh, I see. The product rule doesn't hold for $\Delta$. My bad. Should I delete the question, having found the problem with my issue? Commented Jul 27, 2020 at 15:28
• I think your question is still a legit one and should be kept. Especially when the computation is not shown in the book. If you later find the answer you may consider answering this question. Commented Jul 27, 2020 at 15:29
• I think I should still edit the question to be factually correct. In topping's lecture notes, they actually write $\Box^{*}w =$ the final expression I wrote. I added the middle expression as my interpretation for reaching there. That interpretation was the actual issue. Commented Jul 27, 2020 at 15:33

First, as noted by Arctic Char, $$\Box^{*}(uv) = v \Box^{*}u + u \Box^{*}v$$ is not true. This only holds for first-order derivatives like $$\partial/\partial t$$. This doesn't hold for $$\Delta$$ and $$\text{Scal}$$. So we need to find appropriate product rules for these operators and then apply them here.
Computation shows that $$\Delta(uv) = u\Delta v + v \Delta u + 2\langle \nabla u, \nabla v \rangle$$ and $$\text{Scal}(uv) = \text{Scal}(u) v$$.
Combining these three we get that \begin{align*} \Box^{*}w &= \left( -\frac{\partial}{\partial t} - \Delta +\text{Scal}\right) \left( u \frac{w}{u}\right)\\ &= -\frac{w}{u}\frac{\partial u}{\partial t} - u \frac{\partial}{\partial t}\left( \frac{w}{u} \right) - u \Delta\left( \frac{w}{u}\right) - \frac{w}{u}\Delta u - 2 \langle \nabla \left( \frac{w}{u} \right), \nabla u \rangle + \frac{w}{u} \text{Scal}(u)\\ &= \frac{w}{u} \Box^{*}u - u\left( \frac{\partial}{\partial t} + \Delta \right)\left(\frac{w}{u}\right) - 2 \langle \nabla \left( \frac{w}{u} \right), \nabla u \rangle. \end{align*}