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.
Arctic Char's comment correctly finds my issue hence practically answers the question. Here I intend to provide the details of the computation which shows why the expression in Peter Topping's book is indeed correct.
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*}