This is a comment on the Ricci flow aspect of Terence Tao's answer. Let
$\frac{\partial g}{\partial\tau}=2\operatorname{Ric}$ be a solution to the
backward Ricci flow on $M\times(0,T)$. For $N\gg1$ define the potentially
infinite space-time metric $\tilde{g}\doteqdot g+\tilde{R}d\tau^{2}$, where
$\tilde{R}=R+\frac{N}{2\tau}$. This metric is dual (i.e., it is modeled on
shrinkers instead of expanders) to the one defined by Sun-Chin Chu (see
Section 4 of arXiv:0211349) and is Perelman's metric without the potentially
infinite dimensional $S^{N}$ factor. Perelman wrote in 6.4 of arXiv:0211159
that this is "a potentially degenerate Riemannian metric [on
$M\times\mathbb{R}$], which potentially satisfies the Ricci soliton
equation."

Let $\nabla$ and $\tilde{\nabla}$ be the Levi-Civita connections of $g$ and
$\tilde{g}$, respectively. Let $T=\frac{\partial}{\partial\tau}$ and let other
capital letters denote space vectors. Then $\tilde{R}^{-1}=O(N^{-1})$ and
\begin{gather*}
\tilde{\nabla}_{X}Y=\nabla_{X}Y-\frac{\operatorname{Rc}(X,Y)T}{\tilde{R}
},\quad\;\tilde{\nabla}_{X}T=\operatorname{Rc}(X)+\frac{\left\langle \nabla
R,X\right\rangle T}{2\tilde{R}},\\
\tilde{\nabla}_{T}T=-\frac{\nabla R+\frac{T}{\tau}}{2}+\frac{(\frac{\partial
R}{\partial\tau}+\frac{R}{\tau})T}{2\tilde{R}},
\end{gather*}
where $\frac{\partial R}{\partial\tau}=-\Delta R-2\left\vert \operatorname{Rc}
\right\vert ^{2}$. The terms with $\frac{1}{\tilde{R}}$ factors comprise the
dual of Hamilton's trace Harnack. The space-time Riemann curvature tensor is
(Gauss equations):
\begin{align*}
\widetilde{\operatorname{Rm}}(X,Y,Z,W) & =\operatorname{Rm}(X,Y,Z,W)+\frac
{\operatorname{Rc}(X,W)\operatorname{Rc}(Y,Z)-\operatorname{Rc}%
(X,Z)\operatorname{Rc}(Y,W)}{\tilde{R}},\\
\widetilde{\operatorname{Rm}}(\cdot,T,Z,W) & =(\nabla_{W}\operatorname{Rc}
)(Z)-(\nabla_{Z}\operatorname{Rc})(W)+\frac{Z(R)\operatorname{Rc}
(W)-W(R)\operatorname{Rc}(Z)}{2\tilde{R}},\\
\widetilde{\operatorname{Rm}}(\cdot,T,T,\cdot) & =\Delta_{L}\operatorname{Rc}
-\frac{\nabla^{2}R}{2}+\operatorname{Rc}^{2}-\frac{\operatorname{Rc}}{2\tau
}+\frac{1}{\tilde{R}}(\frac{\nabla R\otimes\nabla R}{4}+(\frac{\partial
R}{\partial\tau}+\frac{R}{\tau})\frac{\operatorname{Rc}}{2}),
\end{align*}
where $\cdot$ denotes a slot for a space vector. The space-time geometry has
some sort of divergence structure in the sense that taking the space
divergence of $\widetilde{\operatorname{Rm}}$ is essentially the same as
taking one of the components to be $T$. One already sees this in Hamilton's
matrix Harnack quadratic. The space-time second Bianchi identity can be used
to efficiently compute the evolution of $\widetilde{\operatorname{Rm}}$,
leading to an alternate proof of Hamilton's matrix Harnack estimate.

Let $\bar{\nabla}=\lim_{N\rightarrow\infty}\tilde{\nabla}$, $\overline
{\operatorname{Rm}}=\lim_{N\rightarrow\infty}\widetilde{\operatorname{Rm}}$,
and $\overline{\operatorname{Ric}}=\lim_{N\rightarrow\infty}
\widetilde{\operatorname{Ric}}$. Define $\bar{V}=\frac{\partial}{\partial\tau
}$. Then we obtain the potentially Ricci soliton equation $\overline
{\operatorname{Ric}}-\bar{\nabla}\bar{V}=\frac{1}{2\tau}d\tau\otimes
\frac{\partial}{\partial\tau}$. Furthermore, we have
\begin{align*}
(\bar{\nabla}_{X}\overline{\operatorname{Ric}})(Y,T)-(\bar{\nabla}
_{Y}\overline{\operatorname{Ric}})(X,T) & =0,\\
(\bar{\nabla}_{X}\overline{\operatorname{Ric}})(T,T)-(\bar{\nabla}
_{T}\overline{\operatorname{Ric}})(X,T) & =-\frac{1}{4\tau}\left\langle
\nabla R,X\right\rangle ,\\
(\bar{\nabla}_{X}\overline{\operatorname{Ric}})(Y,T)-(\bar{\nabla}
_{T}\overline{\operatorname{Ric}})(X,Y) & =\overline{\operatorname{Rm}
}(X,T,T,Y)+\frac{1}{2\tau}\overline{\operatorname{Ric}}(X,Y).
\end{align*}
This is analogous to the gradient Ricci soliton equation $(\nabla
_{X}\operatorname{Ric})(Y,Z)-(\nabla_{Y}\operatorname{Ric}
)(X,Z)=\operatorname{Rm}(X,Y,Z,\nabla f)$.

We see some nice properties of the space-time metric; but for applications to
Ricci flow, following Perelman one should expand in $N$ the path length
functional (here, one may do without the $S^{N}$ factor, i.e., potentially
infinite metric versus dimension). Let $\tilde{\gamma}(\tau)\doteqdot
(\gamma(\tau),\tau)$, $\tau\in\lbrack0,\bar{\tau}]$, be a path. Then
Perelman's $\mathcal{L}$-geometry arises from
$$
\operatorname{L}_{\tilde{g}}\left( \tilde{\gamma}\right) =\int_{0}
^{\bar{\tau}}\sqrt{\tilde{g}(\tilde{\gamma}^{\prime}(\tau),\tilde{\gamma
}^{\prime}(\tau))}\,d\tau=\sqrt{2N\bar{\tau}}(1+\frac{\mathcal{L}(\gamma
)}{2\sqrt{\bar{\tau}}}\,N^{-1}+O\left( N^{-2}\right) ),
$$
where $\mathcal{L}(\gamma)=\int_{0}^{\bar{\tau}}\sqrt{\tau}(R\left(
\gamma(\tau),\tau\right) +|\gamma^{\prime}(\tau)|_{g\left( \tau\right)
}^{2})d\tau$. Another strong motivation for not only this, but also Hamilton's matrix Harnack estimate, is the work of Peter Li and
Shing-Tung Yau (see Section 3 of Acta Math. 1986).