[proofplan] The proof has two ingredients. First, because the diffeomorphism $\phi$ is independent of time, differentiating the pullback metric commutes with taking the pullback. Second, Ricci curvature is natural under diffeomorphisms: the Ricci tensor of $\phi^*g$ is the pullback of the Ricci tensor of $g$. Combining these two identities with the Ricci flow equation for $g(t)$ gives the Ricci flow equation for $\tilde g(t)$. [/proofplan]

[step:Differentiate the pulled-back metric in time]Fix $t \in I$, $p \in M$, and tangent vectors $X,Y \in T_pM$. By definition of the pullback metric, \begin{align*} \tilde g(t)_p(X,Y)=(\phi^*g(t))_p(X,Y)=g(t)_{\phi(p)}(d\phi_p(X),d\phi_p(Y)). \end{align*} The point $\phi(p)$ and the tangent vectors $d\phi_p(X),d\phi_p(Y) \in T_{\phi(p)}M$ are independent of $t$, since $\phi$ is fixed. Define the scalar map $a:I\to \mathbb{R}$ by sending $s \in I$ to $a(s)=g(s)_{\phi(p)}(d\phi_p(X),d\phi_p(Y))$. Differentiating $a$ at $s=t$ gives \begin{align*} (\partial_t \tilde g(t))_p(X,Y)=(\partial_t g(t))_{\phi(p)}(d\phi_p(X),d\phi_p(Y)). \end{align*} Since this holds for every $p \in M$ and every $X,Y \in T_pM$, we have the tensor identity \begin{align*} \partial_t \tilde g(t)=\phi^*(\partial_t g(t)). \end{align*}[/step]

[guided]We want to compute the time derivative of $\tilde g(t)=\phi^*g(t)$. The only possible subtlety is that pullback involves the map $\phi$, but here $\phi$ is fixed in time. Thus the only quantity depending on $t$ is the metric $g(t)$ itself. Fix $t \in I$, $p \in M$, and $X,Y \in T_pM$. The pullback metric is defined by \begin{align*} \tilde g(t)_p(X,Y)=g(t)_{\phi(p)}(d\phi_p(X),d\phi_p(Y)). \end{align*} The point $\phi(p)$ is fixed, and the vectors $d\phi_p(X)$ and $d\phi_p(Y)$ are fixed vectors in $T_{\phi(p)}M$. Therefore, when we differentiate with respect to time, we are differentiating only the [bilinear form](/page/Bilinear%20Form) $g(t)_{\phi(p)}$. Hence \begin{align*} (\partial_t \tilde g(t))_p(X,Y)=(\partial_t g(t))_{\phi(p)}(d\phi_p(X),d\phi_p(Y)). \end{align*} The right-hand side is precisely the pullback of the tensor $\partial_t g(t)$ evaluated on $X$ and $Y$. Since the equality holds at every point and on every pair of tangent vectors, it gives \begin{align*} \partial_t \tilde g(t)=\phi^*(\partial_t g(t)). \end{align*}[/guided]

[step:Show that Ricci curvature is natural under the diffeomorphism] Let $h$ be an arbitrary Riemannian metric on $M$, and define the pullback metric $\hat h:=\phi^*h$. Let $\mathfrak{X}(M):=\Gamma(TM)$ denote the space of smooth vector fields on $M$. Let $\nabla^h$ and $\nabla^{\hat h}$ denote the Levi-Civita connections of $h$ and $\hat h$, respectively. Define the pushforward map on vector fields $\phi_*:\mathfrak{X}(M)\to\mathfrak{X}(M)$ by sending each $X\in\mathfrak{X}(M)$ to the vector field $\phi_*X\in\mathfrak{X}(M)$, where, for each $q \in M$, \begin{align*} (\phi_*X)_q=d\phi_{\phi^{-1}(q)}(X_{\phi^{-1}(q)}). \end{align*} Define its inverse map $\phi_*^{-1}:\mathfrak{X}(M)\to\mathfrak{X}(M)$ by sending each $W\in\mathfrak{X}(M)$ to the vector field $\phi_*^{-1}W\in\mathfrak{X}(M)$, where, for each $p \in M$, \begin{align*} (\phi_*^{-1}W)_p=d(\phi^{-1})_{\phi(p)}(W_{\phi(p)}). \end{align*} Let $[\cdot,\cdot]:\mathfrak{X}(M)\times\mathfrak{X}(M)\to\mathfrak{X}(M)$ denote the Lie bracket of smooth vector fields. For smooth vector fields $X,Y \in \mathfrak{X}(M)$, define the $\phi$-pulled connection \begin{align*} \nabla^\phi_XY := \phi_*^{-1}\left(\nabla^h_{\phi_*X}(\phi_*Y)\right). \end{align*} We verify that $\nabla^\phi$ is a connection on $TM$. Let $f\in C^\infty(M)$ and $X_1,X_2,Y_1,Y_2\in\mathfrak{X}(M)$. Since $\phi_*(fX_1)=(f\circ\phi^{-1})\phi_*X_1$, the $C^\infty(M)$-linearity of $\nabla^h$ in its first argument gives \begin{align*} \nabla^\phi_{fX_1+X_2}Y_1=f\nabla^\phi_{X_1}Y_1+\nabla^\phi_{X_2}Y_1. \end{align*} The $\mathbb{R}$-linearity of $\phi_*$, $\phi_*^{-1}$, and $\nabla^h$ gives $\mathbb{R}$-linearity in the second argument. Finally, since $\phi_*(fY_1)= (f\circ\phi^{-1})\phi_*Y_1$, the Leibniz rule for $\nabla^h$ gives \begin{align*} \nabla^\phi_{X_1}(fY_1)=X_1(f)Y_1+f\nabla^\phi_{X_1}Y_1. \end{align*} Thus $\nabla^\phi$ satisfies the connection axioms. The connection $\nabla^\phi$ is torsion-free because \begin{align*} \nabla^\phi_XY-\nabla^\phi_YX=\phi_*^{-1}\left(\nabla^h_{\phi_*X}\phi_*Y-\nabla^h_{\phi_*Y}\phi_*X\right)=\phi_*^{-1}([\phi_*X,\phi_*Y])=[X,Y]. \end{align*} It is compatible with $\hat h$ because, for all $X,Y,Z \in \mathfrak{X}(M)$, \begin{align*} X(\hat h(Y,Z))=X\left(h(\phi_*Y,\phi_*Z)\circ \phi\right)=\left((\phi_*X)h(\phi_*Y,\phi_*Z)\right)\circ \phi. \end{align*} Metric compatibility of $\nabla^h$ with $h$ gives \begin{align*} \left((\phi_*X)h(\phi_*Y,\phi_*Z)\right)\circ \phi=\left(h(\nabla^h_{\phi_*X}\phi_*Y,\phi_*Z)+h(\phi_*Y,\nabla^h_{\phi_*X}\phi_*Z)\right)\circ \phi. \end{align*} By the definition of $\nabla^\phi$ and $\hat h=\phi^*h$, this last expression equals \begin{align*} \hat h(\nabla^\phi_XY,Z)+\hat h(Y,\nabla^\phi_XZ). \end{align*} The connection $\nabla^\phi$ is a connection on $TM$, and the preceding two paragraphs verify that it is torsion-free and compatible with the Riemannian metric $\hat h$. By the [Fundamental Theorem of Riemannian Geometry](/theorems/1552), the torsion-free metric-compatible connection for $\hat h$ is unique. Hence $\nabla^\phi=\nabla^{\hat h}$. Let $R^h$ and $R^{\hat h}$ denote the curvature tensors of $\nabla^h$ and $\nabla^{\hat h}$, with convention \begin{align*} R^h(A,B)C=\nabla^h_A\nabla^h_BC-\nabla^h_B\nabla^h_AC-\nabla^h_{[A,B]}C. \end{align*} Using $\nabla^{\hat h}=\nabla^\phi$ and the definition of curvature gives \begin{align*} R^{\hat h}(X,Y)Z = \phi_*^{-1}\left(R^h(\phi_*X,\phi_*Y)\phi_*Z\right). \end{align*} Now fix $p \in M$. Let $e_1,\dots,e_n \in T_pM$ be a $\hat h_p$-[orthonormal basis](/page/Orthonormal%20Basis), where $n=\dim M$. Then $d\phi_p(e_1),\dots,d\phi_p(e_n)$ is an $h_{\phi(p)}$-orthonormal basis because $\hat h=\phi^*h$. Therefore the definition of Ricci curvature as the trace of the curvature tensor gives \begin{align*} \operatorname{Ric}(\hat h)_p(X,Y)=\sum_{i=1}^n \hat h_p\left(R^{\hat h}(e_i,X)Y,e_i\right). \end{align*} Using the curvature identity and the defining formula for the pullback metric, this equals \begin{align*} \sum_{i=1}^n h_{\phi(p)}\left(R^h(d\phi_p(e_i),d\phi_p(X))d\phi_p(Y),d\phi_p(e_i)\right). \end{align*} Since $d\phi_p(e_1),\dots,d\phi_p(e_n)$ is an $h_{\phi(p)}$-orthonormal basis, the last expression is \begin{align*} \operatorname{Ric}(h)_{\phi(p)}(d\phi_p(X),d\phi_p(Y))=(\phi^*\operatorname{Ric}(h))_p(X,Y). \end{align*} Thus \begin{align*} \operatorname{Ric}(\phi^*h)=\phi^*\operatorname{Ric}(h). \end{align*} [/step]

[step:Substitute the Ricci flow equation and conclude] Apply the naturality identity from the previous step with $h=g(t)$. Since $\tilde g(t)=\phi^*g(t)$, we have \begin{align*} \operatorname{Ric}(\tilde g(t))=\operatorname{Ric}(\phi^*g(t))=\phi^*\operatorname{Ric}(g(t)). \end{align*} Using the time-derivative identity and the Ricci flow equation for $g(t)$, \begin{align*} \partial_t\tilde g(t)=\phi^*(\partial_t g(t))=\phi^*(-2\operatorname{Ric}(g(t)))=-2\phi^*\operatorname{Ric}(g(t))=-2\operatorname{Ric}(\tilde g(t)). \end{align*} This holds for every $t \in I$, so $\tilde g$ satisfies the Ricci flow equation on $I$. Hence $\tilde g(t)=\phi^*g(t)$ is a Ricci flow. [/step]