[proofplan]
We differentiate the total Riemannian volume along the metric family. The [first variation formula](/theorems/2728) for the Riemannian volume form expresses the derivative of $d\operatorname{vol}_{g(t)}$ in terms of the $g(t)$-trace of $\partial_t g(t)$. Substituting the normalized Ricci flow equation gives an integrand $-R_{g(t)} + r(t)$, whose integral vanishes exactly because $r(t)$ is the average scalar curvature. Hence the derivative of total volume is zero for every time, so the volume is constant on the interval.
[/proofplan]
[step:Differentiate the total volume using the variation formula for the volume form]
Define the volume function $V: I \to \mathbb{R}$ by
\begin{align*}
V(t)=\int_M 1\,d\operatorname{vol}_{g(t)}(x).
\end{align*}
Since $(g(t))_{t \in I}$ is a smooth one-parameter family of Riemannian metrics on the closed manifold $M$, the Riemannian volume form depends smoothly on $t$, and differentiation under the integral is valid. The first variation formula for the Riemannian volume form gives the following identity of time-dependent top-degree forms on $M$:
\begin{align*}
\partial_t d\operatorname{vol}_{g(t)}=\frac{1}{2}\operatorname{tr}_{g(t)}(\partial_t g(t))\,d\operatorname{vol}_{g(t)}.
\end{align*}
Therefore, for every $t$ in the interior of $I$,
\begin{align*}
V'(t)=\frac{1}{2}\int_M \operatorname{tr}_{g(t)}(\partial_t g(t))\,d\operatorname{vol}_{g(t)}(x).
\end{align*}
[guided]
We begin by converting the geometric statement into a derivative computation for a real-valued function. Define the volume function $V: I \to \mathbb{R}$ by
\begin{align*}
V(t)=\int_M 1\,d\operatorname{vol}_{g(t)}(x).
\end{align*}
Because $M$ is closed, $M$ is compact, so this integral is finite for every $t \in I$. Because $g(t)$ is a smooth one-parameter family of Riemannian metrics, the Riemannian volume form $d\operatorname{vol}_{g(t)}$ depends smoothly on $t$. Compactness of $M$ then permits differentiation under the integral sign for every $t$ in the interior of $I$.
The first variation formula for the Riemannian volume form states that, for a smooth metric variation $h(t)=\partial_t g(t)$, the corresponding variation of the volume form is
\begin{align*}
\partial_t d\operatorname{vol}_{g(t)}=\frac{1}{2}\operatorname{tr}_{g(t)}(h(t))\,d\operatorname{vol}_{g(t)}.
\end{align*}
Here $h(t)$ is a symmetric $(0,2)$-tensor on $M$, and $\operatorname{tr}_{g(t)}(h(t))$ denotes its trace with respect to the metric $g(t)$. Applying this formula with $h(t)=\partial_t g(t)$ gives the derivative formula
\begin{align*}
V'(t)=\frac{1}{2}\int_M \operatorname{tr}_{g(t)}(\partial_t g(t))\,d\operatorname{vol}_{g(t)}(x).
\end{align*}
This step isolates the only quantity that remains to compute: the $g(t)$-trace of the normalized Ricci flow equation.
[/guided]
[/step]
[step:Compute the trace of the normalized Ricci flow equation]
For each $t \in I$, let $R_{g(t)}: M \to \mathbb{R}$ denote the scalar curvature of $g(t)$. Define the average scalar curvature $r: I \to \mathbb{R}$ by
\begin{align*}
r(t)=\frac{1}{\operatorname{Vol}_{g(t)}(M)}\int_M R_{g(t)}(x)\,d\operatorname{vol}_{g(t)}(x),
\end{align*}
where $\operatorname{Vol}_{g(t)}(M)=\int_M 1\,d\operatorname{vol}_{g(t)}(x)$. The denominator is positive because $M$ is nonempty and $g(t)$ is a Riemannian metric. Taking the $g(t)$-trace of the normalized Ricci flow equation
\begin{align*}
\partial_t g(t)=-2\operatorname{Ric}_{g(t)}+\frac{2}{n}r(t)g(t)
\end{align*}
gives the first identity
\begin{align*}
\operatorname{tr}_{g(t)}(\partial_t g(t))=-2\operatorname{tr}_{g(t)}(\operatorname{Ric}_{g(t)})+\frac{2}{n}r(t)\operatorname{tr}_{g(t)}(g(t)).
\end{align*}
Since $\operatorname{tr}_{g(t)}(\operatorname{Ric}_{g(t)})=R_{g(t)}$ and $\operatorname{tr}_{g(t)}(g(t))=n$, this becomes
\begin{align*}
\operatorname{tr}_{g(t)}(\partial_t g(t))=-2R_{g(t)}+2r(t).
\end{align*}
Substituting this identity into the formula for $V'(t)$ yields
\begin{align*}
V'(t)=\int_M \left(-R_{g(t)}(x)+r(t)\right)\,d\operatorname{vol}_{g(t)}(x).
\end{align*}
[/step]
[step:Use the definition of average scalar curvature to make the derivative vanish]
By the definition of $r(t)$ from the preceding step,
\begin{align*}
r(t)\operatorname{Vol}_{g(t)}(M)=\int_M R_{g(t)}(x)\,d\operatorname{vol}_{g(t)}(x).
\end{align*}
Since
\begin{align*}
\operatorname{Vol}_{g(t)}(M)=\int_M 1\,d\operatorname{vol}_{g(t)}(x),
\end{align*}
we obtain
\begin{align*}
V'(t)=-\int_M R_{g(t)}(x)\,d\operatorname{vol}_{g(t)}(x)+r(t)\operatorname{Vol}_{g(t)}(M).
\end{align*}
By the defining identity for $r(t)$, this right-hand side is $0$. Thus $V'(t)=0$ for every $t$ in the interior of $I$. Since $(g(t))_{t \in I}$ solves a flow, its time domain $I$ is an interval in $\mathbb{R}$. If $I$ is a singleton, then the conclusion is immediate. Otherwise, the interior of the interval $I$ is connected. For any two points $a,b$ in the interior of $I$ with $a<b$, the compact interval $[a,b]$ is contained in the interior of $I$, so the [Mean Value Theorem](/theorems/186) applied to $V|_{[a,b]}$ gives $V(b)-V(a)=V'(c)(b-a)=0$ for some $c \in (a,b)$. Hence $V$ is constant on the interior of $I$. Since $V$ is continuous on all of $I$, if an endpoint $e \in I$ is present and $(t_j)_{j \in \mathbb{N}}$ is any sequence of interior points with $t_j \to e$, then $V(e)=\lim_{j \to \infty}V(t_j)$ equals the same interior constant. Therefore $\operatorname{Vol}_{g(t)}(M)$ is independent of $t$.
[/step]
