[guided]We now identify the rigidity mechanism. The quantity that detects shrinking solitons is Perelman's reduced volume, but there is an important endpoint issue. The spacetime point $(p,T)$ is not a regular point of the original flow, and after rescaling it corresponds to $(p,1)$, while $g_j$ is defined only for $s<1$. Thus we cannot simply apply the ordinary reduced-distance construction with a regular basepoint. We instead use the [Type I Singular Reduced Distance and Reduced Volume Convergence Theorem](/theorems/6021), which defines the singular reduced distance by taking regular base times tending to the Type I endpoint, proves monotonicity for the resulting singular reduced volume, and proves convergence of these quantities under pointed Type I blow-up.
For each rescaled flow $g_j$, let $\ell_j: M\times(-\lambda_jt_j,1)\to\mathbb{R}$ denote this singular reduced distance with boundary base point $(p,1)$ in the rescaled time coordinate. This is the same spacetime endpoint as $(p,T)$ for the original flow, because $s=\lambda_j(t-t_j)$ sends $t=T$ to $s=1$.
The backward-time parameter from the boundary base time $1$ to a time $s<1$ is not $-s$; it is
\begin{align*}
\tau_j(s)=1-s.
\end{align*}
Thus the corresponding singular reduced volume is the function $\widetilde V_j: (-\lambda_jt_j,1)\to\mathbb{R}$ defined by
\begin{align*}
\widetilde V_j(s)=\int_M (4\pi(1-s))^{-n/2}e^{-\ell_j(x,s)}\,d\operatorname{vol}_{g_j(s)}(x),
\end{align*}
where $n=\dim M$ and $d\operatorname{vol}_{g_j(s)}$ is the Riemannian volume measure of the metric $g_j(s)$. This is the point where the base-time bookkeeping matters: the factor $(4\pi(1-s))^{-n/2}$ is forced by the boundary base point $(p,1)$.
The hypotheses of the [Type I Singular Reduced Distance and Reduced Volume Convergence Theorem](/theorems/6021) are satisfied. Each $g_j$ is smooth and complete on compact time subintervals, because $M$ is compact and parabolic rescaling preserves smoothness and completeness. The Type I curvature estimate gives the required endpoint curvature control as $s\uparrow 1$. The theorem therefore gives monotonicity in the backward-time variable $\tau_j=1-s$ and the bound $0\leq \widetilde V_j\leq 1$.
Why should the limit have constant reduced volume? Fix $s_1<s_2<0$. Under the inverse rescaling, these times correspond to
\begin{align*}
t_{j,k}:=t_j+\frac{s_k}{\lambda_j}, \qquad k\in\{1,2\}.
\end{align*}
The corresponding backward times from $T$ are
\begin{align*}
T-t_{j,k}=\frac{1-s_k}{\lambda_j}, \qquad k\in\{1,2\},
\end{align*}
and both tend to $0$ as $j\to\infty$. Since the original singular reduced volume based at $(p,T)$ is monotone and bounded in $[0,1]$, it has a finite limit as the backward time tends to $0$. Parabolic scaling leaves singular reduced distance and singular reduced volume unchanged with this matching boundary base time, so the drop satisfies
\begin{align*}
\lim_{j\to\infty}\left(\widetilde V_j(s_2)-\widetilde V_j(s_1)\right)=0.
\end{align*}
It remains to justify passage of the global integral to the pointed limit. Local smooth Cheeger-Gromov convergence alone only gives convergence on fixed compact subsets, so we invoke the compactness and convergence part of the [Type I Singular Reduced Distance and Reduced Volume Convergence Theorem](/theorems/6021). Its assumptions are verified as follows: the basepoints are the fixed points $p$ and converge to $p_\infty$ in the pointed limit; the Type I estimate gives uniform curvature bounds on compact negative time intervals; Perelman's noncollapsing gives the local volume lower bounds; and the Gaussian lower estimates for singular reduced distance give uniform integrable tails for $(4\pi(1-s))^{-n/2}e^{-\ell_j}$ outside large pointed balls. The theorem yields local convergence of $\ell_j$ to the limiting singular reduced distance $\ell_\infty: M_\infty\times(-\infty,0)\to\mathbb{R}$ based at $(p_\infty,1)$, and the tail estimate upgrades local integral convergence to convergence of the full reduced volumes. Passing to the limit in the zero-drop identity gives
\begin{align*}
\widetilde V_\infty(s_1)=\widetilde V_\infty(s_2)
\end{align*}
for every $s_1<s_2<0$. Thus the singular reduced volume of the limit is constant on $(-\infty,0)$.[/guided]