[proofplan] Because the Ricci flow is smooth on the compact spacetime $M \times [0,T]$, its curvature and injectivity radius are uniformly controlled. We choose a uniform radius below the injectivity radius on which all geodesic polar-coordinate volume densities are at least one half of their Euclidean values. This gives a uniform lower bound for the volume of every sufficiently small geodesic ball, so the required $\kappa$-noncollapsing estimate follows without using the additional curvature hypothesis. [/proofplan]

[step:Extract uniform geometric bounds from compact spacetime] For each $t \in [0,T]$, let $\operatorname{inj}_{g(t)}(x)$ denote the injectivity radius of the Riemannian metric $g(t)$ at $x \in M$. Since $g(t)$ is smooth on the compact set $M \times [0,T]$, the curvature norm and the injectivity-radius function are uniformly controlled. Define \begin{align*} K_0 := \sup_{(x,t) \in M \times [0,T]} |\operatorname{Rm}_{g(t)}(x)|_{g(t)}. \end{align*} Then $K_0 < \infty$. Define also \begin{align*} i_0 := \inf_{(x,t) \in M \times [0,T]} \operatorname{inj}_{g(t)}(x). \end{align*} The smooth compact family of metrics has $i_0 > 0$: each metric $g(t)$ has positive injectivity radius on compact $M$, and the injectivity radius varies continuously under smooth variation of the metric on a compact manifold. [/step]

[step:Choose a uniform radius where polar volume density is Euclidean from below]Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R^n$, let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $(0,\infty)$, and let $\mathcal{H}^{n-1}$ denote the induced hypersurface measure on the unit sphere $S^{n-1}$. Let $\alpha_n := \mathcal{L}^n(B(0,1))$ denote the Euclidean volume of the unit ball in $\mathbb{R}^n$. For each $(x,t) \in M \times [0,T]$, geodesic polar coordinates for $g(t)$ are valid on all radii $0 < s < i_0$. In these coordinates the Riemannian volume measure has the form $J_{x,t}(s,\omega)\, d\mathcal{L}^1(s)\, d\mathcal{H}^{n-1}(\omega)$, where $\omega \in S^{n-1}$ and $J_{x,t}(s,\omega)$ is the polar Jacobian density. Smooth dependence of the metric and the exponential map on $(x,t)$ gives a number $\rho_0 \in (0,i_0)$ such that \begin{align*} J_{x,t}(s,\omega) \geq \frac{1}{2}s^{n-1} \end{align*} for every $(x,t) \in M \times [0,T]$, every $\omega \in S^{n-1}$, and every $0 < s < \rho_0$.[/step]

[guided]We need a lower volume bound that is uniform in the centre point and in time. The reason compact finite time is enough is that small balls in a smooth Riemannian manifold have Euclidean first-order volume density, and this approximation is uniform over the compact parameter set $M \times [0,T]$. Let $\mathcal{L}^n$ be $n$-dimensional Lebesgue measure, let $\mathcal{L}^1$ be one-dimensional Lebesgue measure in the radial coordinate, and let $\mathcal{H}^{n-1}$ be hypersurface measure on $S^{n-1}$. Let $\alpha_n := \mathcal{L}^n(B(0,1))$ be the Euclidean volume of the unit ball in $\mathbb{R}^n$. From the previous step, $i_0>0$ is a uniform lower bound for the injectivity radii. Therefore, for every $(x,t) \in M \times [0,T]$, the exponential map of $g(t)$ at $x$ gives geodesic polar coordinates for radii $0<s<i_0$. In those coordinates the Riemannian volume measure is \begin{align*} J_{x,t}(s,\omega)\, d\mathcal{L}^1(s)\, d\mathcal{H}^{n-1}(\omega), \end{align*} where $\omega \in S^{n-1}$ and $J_{x,t}(s,\omega)$ denotes the polar Jacobian density. At $s=0$, the polar density has the Euclidean leading term $s^{n-1}$. Equivalently, $J_{x,t}(s,\omega)s^{1-n} \to 1$ as $s \downarrow 0$. This convergence is uniform in $(x,t,\omega)$ because the metrics $g(t)$ and their exponential maps vary smoothly and the parameter set $M \times [0,T] \times S^{n-1}$ is compact. Hence there exists $\rho_0 \in (0,i_0)$ such that \begin{align*} J_{x,t}(s,\omega) \geq \frac{1}{2}s^{n-1} \end{align*} for every $(x,t) \in M \times [0,T]$, every $\omega \in S^{n-1}$, and every $0<s<\rho_0$. This is the precise point where compactness in space and finite time are used.[/guided]

[step:Integrate the uniform density bound to obtain noncollapsing] Define \begin{align*} \rho := \rho_0 \end{align*} and \begin{align*} \kappa := \frac{1}{2}\alpha_n. \end{align*} Let $x_0 \in M$, $t_0 \in [0,T]$, and $0<r<\rho$. The geodesic ball $B_{g(t_0)}(x_0,r)$ lies inside the injectivity-radius domain because $r<\rho_0<i_0$. Applying the polar-coordinate formula and the lower bound for $J_{x_0,t_0}$ gives \begin{align*} \operatorname{Vol}_{g(t_0)}(B_{g(t_0)}(x_0,r)) \geq \int_{S^{n-1}}\int_0^r \frac{1}{2}s^{n-1}\, d\mathcal{L}^1(s)\, d\mathcal{H}^{n-1}(\omega). \end{align*} Since $\mathcal{H}^{n-1}(S^{n-1}) = n\alpha_n$, the right-hand side equals \begin{align*} \frac{1}{2}n\alpha_n\int_0^r s^{n-1}\, d\mathcal{L}^1(s) = \frac{1}{2}\alpha_n r^n = \kappa r^n. \end{align*} This lower bound holds for every $x_0$, every $t_0$, and every $0<r<\rho$. In particular it holds for balls satisfying the additional curvature bound $|\operatorname{Rm}_{g(t_0)}| \leq r^{-2}$, so $g(t)$ is $\kappa$-noncollapsed at every scale $r<\rho$ on such balls for every $t \in [0,T]$. [/step]