## Normalized Ricci Flow

Calculation of Matrix

Let $A(t)$ be a smooth family of $n\times n$ matrix. The determinant of $A$ is given as
$\det A=\sum_{\sigma}\text{sgn}(\sigma)a_{1\sigma(1)\cdots a_{n\sigma(n)}}$.
We can calculate the derivative of the determinant of a time-dependent matrix as following.
$\frac{d\det A}{dt}=\sum_{\sigma}\text{sgn}(\sigma)\sum_{k}a_{1\sigma(1)}\cdots\frac{d a_{k\sigma(k)}}{dt}\cdots a_{n\sigma(n)}$
$\hskip1cm=\sum_{ij}\frac{d a_{ij}}{dt}\sum_{\sigma(i)=j}\text{sgn}(\sigma)a_{1\sigma(1)}\cdots\widehat{a_{i\sigma(i)}} \cdots a_{n\sigma(n)}$
$\hskip1cm=\sum_{ij}\frac{d a_{ij}}{dt}A^{ij}$ where $(A^{ij})$ is the adjoint matrix of $A$.

Let $M$ be a $n$-dimensional smooth manifold and $g_t$ be a smooth family of Riemannian metrics on $M$. The corresponding volume form is given as
$\mathrm{Vol}_{g}(x)=\sqrt{\det(g_{ij}(x))}dx^1\wedge\cdots\wedge dx^n$.
The following differential equation is known as Ricci Flow: $\frac{\partial g_{ij}}{\partial t}=-2Ric_{ij}$ where $Ric_{ij}$ is Ricci curvature of $(M,g)$. The Scalar Curvature of $(M,g)$ is given as $S=g^{ij}Ric_{ij}$.
Then applying above calculation we have
$\frac{d\mathrm{Vol}_g}{dt}=\frac{d\sqrt{\det(g_{ij})}}{dt}dx^1\wedge\cdots\wedge dx^n=\frac{\sqrt{\det g}}{2} g^{ij}\frac{d g_{ij}}{dt}dx^1\wedge\cdots\wedge dx^n$
$\qquad=\frac{1}{2}g^{ij}\frac{d g_{ij}}{dt}\mathrm{Vol}_g$.

Consider the following Normalized Ricci Flow $\frac{\partial g_{ij}}{\partial t}=-2 Ric_{ij}+\frac{2}{n}\frac{\int_M Sd\mathrm{Vol}_g}{\mathrm{Vol}_g(M)}g_{ij}$. — ($\star$)
For this flow the evolution of the Volume is:
$\frac{d\mathrm{Vol}_g(M)}{\partial t}=\int_M \frac{d\mathrm{Vol}_g(M)}{\partial t}=\int_M g^{ij}\frac{\partial g_{ij}}{\partial t}d\mathrm{Vol}_g(M)$
$=-2\int_M g^{ij} Ric_{ij}d\mathrm{Vol}_g(M)+\frac{2}{n}\frac{\int_M Sd\mathrm{Vol}_g}{\mathrm{Vol}_g(M)}\int_M g^{ij}g_{ij}d\mathrm{Vol}_g(M)=0$.

So we see the volume is preserved under the normalized Ricci Flow ($\star$).

Given a solution $g(t)$ for $t\in[0,\epsilon)$ of the Ricci Flow $\frac{\partial g_{ij}}{\partial t}=-2Ric_{ij}$, the metrics $\tilde{g}(t')=c(t)g(t)$ where
$c(t)=\exp\left(\frac{2}{n}\int_0^t\frac{2}{n}\frac{\int_M Sd\mathrm{Vol}_g}{\mathrm{Vol}_g(M)}ds\right)$ and $t'=\int_0^t c(s)ds$
are a solution of the Normalized Ricci Flow with $\tilde{g}(0)=g(0)$. Hence solutions of the normalized Ricci flow differ from solutions of the Ricci flow only by rescalings in space and time.

Under Ricci Flow,
regions of positive curvature tends to shrink;
regions of negative curvature tends to extend.

Under Normalized Ricci Flow,
regions which has more positive curvature than average tends to shrink;
regions which has more negative curvature than average tends to extend.

Let $\phi^t_s:SM\rightarrow SM$ be the geodesic flow induced by the metric $g(t)$, and $h_{\text{top}}(g(t))$ be the topological entropy of that geodesic flow.
Canonically there is a Liouville measure on the bundle $SM$ associated to $g(t)$ and the associated metric entropy $h_{\text{Liou}}(g(t))$.

Under some assumption,
the entropy $h_{\text{top}}(g(t))$ is decreasing under the flow (at least for a short time), and the minimum of this enropy among the negative sect. curv. with fixed volume is attained by the constant nega. curv..

For Liouville measure the following behavior is more possible, under some assumption:
the entropy $h_{\text{Liou}}(g(t))$ is increasing under the flow (at least for a short time), and the maximum of this enropy among the negative sect. curv. with fixed volume is attained by the constant nega. curv..

This is strange.