[proofplan] We construct the self-similar flow by integrating the time-dependent vector field $c(t)^{-1}\nabla_g f$. Differentiating the ansatz $g(t)=c(t)\varphi_t^*g$ gives one term from the homothetic scale factor and one term from the Lie derivative along the generating vector field. The choice of coefficient $c(t)^{-1}$ makes the pullback term equal to $2\varphi_t^*(\operatorname{Hess}_g f)$, and the soliton equation converts the resulting expression into $-2\varphi_t^*\operatorname{Ric}_g$. Finally, Ricci curvature is unchanged by constant positive scaling and natural under pullback, so this is exactly $-2\operatorname{Ric}_{g(t)}$. [/proofplan]

[step:Construct the diffeomorphisms from the soliton vector field]Let $X:M\to TM$ denote the smooth section of the tangent bundle defined by $X(p)=\nabla_g f(p)$ for every $p\in M$; this is the gradient vector field of $f$ with respect to $g$. For $t\in I_\lambda$, define the time-dependent vector field $X_t:M\to TM$ by $X_t(p)=c(t)^{-1}X(p)$ for every $p\in M$. Since $c(t)>0$ on $I_\lambda$, each $X_t$ is a smooth vector field on $M$. We use the completeness theorem for gradient soliton vector fields: if $(M,g,f)$ is a complete gradient Ricci soliton, then the smooth vector field $\nabla_g f$ is complete. The hypotheses match the present theorem because $(M,g,f)$ is assumed complete and satisfies the gradient Ricci soliton equation. Let $\Phi_s:M\to M$ denote the complete flow of $X=\nabla_g f$, with flow parameter $s\in\mathbb{R}$ and initial condition $\Phi_0=\operatorname{id}_M$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Define $a:I_\lambda\to\mathbb{R}$ by $a(t)=c(t)^{-1}$. Define the reparametrisation $\tau:I_\lambda\to\mathbb{R}$ by \begin{align*} \tau(t)=\int_0^t c(s)^{-1}\,d\mathcal{L}^1(s). \end{align*} For each fixed $t\in I_\lambda$, the closed interval between $0$ and $t$ is compactly contained in $I_\lambda$, so $c(s)^{-1}$ is bounded on that interval and the integral defining $\tau(t)$ is finite. Set $\varphi_t:=\Phi_{\tau(t)}$. Then $\varphi_t:M\to M$ is a diffeomorphism, $\varphi_0=\Phi_0=\operatorname{id}_M$, and the chain rule gives \begin{align*} \frac{d}{dt}\varphi_t(p)=\tau'(t)X(\Phi_{\tau(t)}(p))=c(t)^{-1}\nabla_g f(\varphi_t(p))=X_t(\varphi_t(p)), \end{align*} for every $p\in M$.[/step]

[guided]We first isolate the only global existence input in the proof. Define $X:M\to TM$ by $X(p)=\nabla_g f(p)$ for every $p\in M$; this is the smooth gradient vector field of $f$ with respect to $g$. The desired diffeomorphisms are meant to be the flow of a time-dependent multiple of $X$. More precisely, for each $t\in I_\lambda$ define $X_t:M\to TM$ by $X_t(p)=c(t)^{-1}X(p)$ for every $p\in M$. This is well-defined because $c(t)>0$ on $I_\lambda$, so $c(t)^{-1}$ is a smooth real-valued function of time. The noncompactness of $M$ means that a smooth vector field need not have a globally defined flow for all times. This is exactly where completeness of the gradient Ricci soliton is used. We invoke the completeness theorem for gradient soliton vector fields: if $(M,g,f)$ is a complete gradient Ricci soliton, then the vector field $\nabla_g f$ is complete. The hypotheses match the present setting because the theorem assumes completeness of $(M,g)$ and the soliton equation, both of which are part of the statement. Let $\Phi_s:M\to M$ denote the complete flow of $X=\nabla_g f$, with flow parameter $s\in\mathbb{R}$ and initial condition $\Phi_0=\operatorname{id}_M$. Now we convert the autonomous complete flow $\Phi_s$ into the desired time-dependent flow. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define $a:I_\lambda\to\mathbb{R}$ by $a(t)=c(t)^{-1}$, and define $\tau:I_\lambda\to\mathbb{R}$ by \begin{align*} \tau(t)=\int_0^t c(s)^{-1}\,d\mathcal{L}^1(s). \end{align*} For a fixed $t\in I_\lambda$, the interval with endpoints $0$ and $t$ is compactly contained in $I_\lambda$. Since $c(s)^{-1}$ is continuous on that compact interval, it is bounded there, so the integral defining $\tau(t)$ is finite. This finiteness is the point of using the maximal open interval $I_\lambda$: the coefficient may blow up at an endpoint of $I_\lambda$, but it is harmless at every interior time. Set $\varphi_t:=\Phi_{\tau(t)}$. Because each $\Phi_s$ is a diffeomorphism of $M$, each $\varphi_t:M\to M$ is a diffeomorphism. Also $\tau(0)=0$, so $\varphi_0=\Phi_0=\operatorname{id}_M$. Finally, the chain rule and the defining equation for the flow $\Phi_s$ give \begin{align*} \frac{d}{dt}\varphi_t(p)=\tau'(t)X(\Phi_{\tau(t)}(p))=c(t)^{-1}\nabla_g f(\varphi_t(p)). \end{align*} Thus the family required in the statement exists on the whole maximal interval $I_\lambda$.[/guided]

[step:Differentiate the self-similar metric ansatz]Let $\Gamma(S^2T^*M)$ denote the space of smooth symmetric covariant $2$-tensor fields on $M$. Define $g_\ast:I_\lambda\to \Gamma(S^2T^*M)$ by $g_\ast(t)=c(t)\varphi_t^*g$ for every $t\in I_\lambda$. We will show that $g_\ast$ satisfies Ricci flow. Since \begin{align*} c'(t)=-2\lambda, \end{align*} the product rule gives \begin{align*} \partial_t g_\ast(t) = -2\lambda\,\varphi_t^*g + c(t)\,\partial_t(\varphi_t^*g). \end{align*} Let $\mathcal{L}_{X_t}g$ denote the Lie derivative of the metric tensor $g$ along the vector field $X_t$. The derivative of a pullback by the flow of the time-dependent vector field $X_t$ is the pullback of the Lie derivative: \begin{align*} \partial_t(\varphi_t^*g)=\varphi_t^*(\mathcal{L}_{X_t}g). \end{align*} Therefore \begin{align*} \partial_t g_\ast(t) = -2\lambda\,\varphi_t^*g + c(t)\,\varphi_t^*(\mathcal{L}_{X_t}g). \end{align*}[/step]

[guided]We differentiate $g_\ast(t)=c(t)\varphi_t^*g$ as a curve in the [vector space](/page/Vector%20Space) $\Gamma(S^2T^*M)$ of smooth symmetric covariant $2$-tensor fields. The scalar factor satisfies $c'(t)=-2\lambda$, so the product rule gives \begin{align*} \partial_t g_\ast(t) = -2\lambda\,\varphi_t^*g + c(t)\,\partial_t(\varphi_t^*g). \end{align*} The family $\varphi_t$ is generated by the time-dependent vector field $X_t$, as established in the previous step. The standard pullback differentiation formula therefore applies: the derivative of $\varphi_t^*g$ is the pullback of the Lie derivative of $g$ along the generating vector field. Thus \begin{align*} \partial_t(\varphi_t^*g)=\varphi_t^*(\mathcal{L}_{X_t}g). \end{align*} Substituting this identity into the product-rule formula yields \begin{align*} \partial_t g_\ast(t) = -2\lambda\,\varphi_t^*g + c(t)\,\varphi_t^*(\mathcal{L}_{X_t}g). \end{align*}[/guided]

[step:Convert the Lie derivative term into the Hessian of the potential]For every $t\in I_\lambda$, the scalar $c(t)^{-1}$ is constant on $M$. Since \begin{align*} X_t=c(t)^{-1}\nabla_g f, \end{align*} linearity of the Lie derivative in the vector field gives \begin{align*} \mathcal{L}_{X_t}g=c(t)^{-1}\mathcal{L}_{\nabla_g f}g. \end{align*} Let $\mathfrak{X}(M)$ denote the space of smooth vector fields on $M$. Let $\operatorname{Hess}_g f\in\Gamma(S^2T^*M)$ denote the Hessian of $f$ with respect to the Levi-Civita connection $\nabla$ of $g$. For vector fields $Y,Z\in\mathfrak{X}(M)$, the Lie derivative of $g$ along $\nabla_g f$ satisfies \begin{align*} (\mathcal{L}_{\nabla_g f}g)(Y,Z)=g(\nabla_Y\nabla_g f,Z)+g(Y,\nabla_Z\nabla_g f)=2\operatorname{Hess}_g f(Y,Z), \end{align*} where the last equality uses the symmetry of the Hessian of a smooth function. Hence \begin{align*} \mathcal{L}_{X_t}g=2c(t)^{-1}\operatorname{Hess}_g f. \end{align*} Substituting this into the differentiated ansatz gives \begin{align*} \partial_t g_\ast(t)=-2\lambda\,\varphi_t^*g+c(t)\,\varphi_t^*\left(2c(t)^{-1}\operatorname{Hess}_g f\right). \end{align*} Since pullback is linear on covariant tensor fields and $c(t)c(t)^{-1}=1$, this becomes \begin{align*} \partial_t g_\ast(t)=\varphi_t^*(-2\lambda g+2\operatorname{Hess}_g f). \end{align*}[/step]

[guided]The purpose of this step is to rewrite the Lie derivative term in geometric quantities already present in the soliton equation. For a fixed $t\in I_\lambda$, the factor $c(t)^{-1}$ is a real number, not a function on $M$. Since $X_t=c(t)^{-1}\nabla_g f$, linearity of the Lie derivative in the vector field gives \begin{align*} \mathcal{L}_{X_t}g=c(t)^{-1}\mathcal{L}_{\nabla_g f}g. \end{align*} Let $\operatorname{Hess}_g f\in\Gamma(S^2T^*M)$ be the Hessian of $f$ with respect to the Levi-Civita connection $\nabla$ of $g$. For arbitrary vector fields $Y,Z\in\mathfrak{X}(M)$, the formula for the Lie derivative of a metric gives \begin{align*} (\mathcal{L}_{\nabla_g f}g)(Y,Z)=g(\nabla_Y\nabla_g f,Z)+g(Y,\nabla_Z\nabla_g f). \end{align*} By the definition of the Hessian and its symmetry for a smooth real-valued function, \begin{align*} g(\nabla_Y\nabla_g f,Z)=\operatorname{Hess}_g f(Y,Z), \qquad g(Y,\nabla_Z\nabla_g f)=\operatorname{Hess}_g f(Z,Y)=\operatorname{Hess}_g f(Y,Z). \end{align*} Therefore \begin{align*} \mathcal{L}_{\nabla_g f}g=2\operatorname{Hess}_g f, \qquad \mathcal{L}_{X_t}g=2c(t)^{-1}\operatorname{Hess}_g f. \end{align*} Substituting this into the differentiated ansatz from the previous step gives \begin{align*} \partial_t g_\ast(t)=-2\lambda\,\varphi_t^*g+c(t)\,\varphi_t^*\left(2c(t)^{-1}\operatorname{Hess}_g f\right). \end{align*} Pullback is linear on covariant tensor fields, and $c(t)c(t)^{-1}=1$, so this simplifies to \begin{align*} \partial_t g_\ast(t)=\varphi_t^*(-2\lambda g+2\operatorname{Hess}_g f). \end{align*}[/guided]

[step:Use the soliton equation to obtain the pulled-back Ricci tensor]The soliton equation is \begin{align*} \operatorname{Ric}_g+\operatorname{Hess}_g f=\lambda g. \end{align*} Rearranging it gives \begin{align*} -2\lambda g+2\operatorname{Hess}_g f=-2\operatorname{Ric}_g. \end{align*} Therefore \begin{align*} \partial_t g_\ast(t) = -2\varphi_t^*\operatorname{Ric}_g. \end{align*}[/step]

[guided]We now use the defining equation of the gradient Ricci soliton. In the notation of this proof, the equation is \begin{align*} \operatorname{Ric}_g+\operatorname{Hess}_g f=\lambda g. \end{align*} Solving this identity for $\operatorname{Hess}_g f$ gives \begin{align*} \operatorname{Hess}_g f=\lambda g-\operatorname{Ric}_g. \end{align*} Substitute this expression into the tensor obtained in the previous step: \begin{align*} -2\lambda g+2\operatorname{Hess}_g f = -2\lambda g+2(\lambda g-\operatorname{Ric}_g) = -2\operatorname{Ric}_g. \end{align*} Since pullback by $\varphi_t$ is linear on covariant tensor fields, the identity \begin{align*} \partial_t g_\ast(t)=\varphi_t^*(-2\lambda g+2\operatorname{Hess}_g f) \end{align*} becomes \begin{align*} \partial_t g_\ast(t)=-2\varphi_t^*\operatorname{Ric}_g. \end{align*}[/guided]

[step:Identify the pulled-back Ricci tensor with the Ricci tensor of $g(t)$]For each $t\in I_\lambda$, define the positive constant \begin{align*} a_t:=c(t). \end{align*} The metric under consideration is \begin{align*} g_\ast(t)=a_t\varphi_t^*g. \end{align*} Ricci curvature is natural under diffeomorphism pullback, so \begin{align*} \operatorname{Ric}_{\varphi_t^*g}=\varphi_t^*\operatorname{Ric}_g. \end{align*} Also, multiplication of a Riemannian metric by a positive constant leaves the Levi-Civita connection unchanged, because the Koszul formula has the same constant factor on both sides. Hence the $(0,2)$ Ricci tensor is unchanged under constant positive scaling: \begin{align*} \operatorname{Ric}_{a_t\varphi_t^*g} = \operatorname{Ric}_{\varphi_t^*g}. \end{align*} Combining these two facts yields \begin{align*} \operatorname{Ric}_{g_\ast(t)} = \operatorname{Ric}_{a_t\varphi_t^*g} = \operatorname{Ric}_{\varphi_t^*g} = \varphi_t^*\operatorname{Ric}_g. \end{align*} Thus \begin{align*} \partial_t g_\ast(t) = -2\varphi_t^*\operatorname{Ric}_g = -2\operatorname{Ric}_{g_\ast(t)}. \end{align*} Since $g_\ast(t)=c(t)\varphi_t^*g$ was the family denoted $g(t)$ in the statement, the metrics $g(t)$ solve Ricci flow on $I_\lambda$.[/step]

[guided]It remains to identify the tensor $\varphi_t^*\operatorname{Ric}_g$ with the Ricci tensor of the metric that is actually evolving. Fix $t\in I_\lambda$ and define the positive constant $a_t:=c(t)$. Then \begin{align*} g_\ast(t)=a_t\varphi_t^*g. \end{align*} Ricci curvature is natural under diffeomorphism pullback: because $\varphi_t:M\to M$ is a diffeomorphism, the Ricci tensor of the pulled-back metric is \begin{align*} \operatorname{Ric}_{\varphi_t^*g}=\varphi_t^*\operatorname{Ric}_g. \end{align*} Next, multiplying a Riemannian metric by a positive constant does not change its Levi-Civita connection. The Koszul formula has the same factor $a_t$ on both sides, so the connection coefficients are unchanged. Therefore the $(0,2)$ Ricci tensor is also unchanged under this constant scaling: \begin{align*} \operatorname{Ric}_{a_t\varphi_t^*g}=\operatorname{Ric}_{\varphi_t^*g}. \end{align*} Combining naturality under pullback with invariance under positive constant scaling gives \begin{align*} \operatorname{Ric}_{g_\ast(t)} = \operatorname{Ric}_{a_t\varphi_t^*g} = \operatorname{Ric}_{\varphi_t^*g} = \varphi_t^*\operatorname{Ric}_g. \end{align*} Using the differential identity from the previous step, \begin{align*} \partial_t g_\ast(t) = -2\varphi_t^*\operatorname{Ric}_g = -2\operatorname{Ric}_{g_\ast(t)}. \end{align*} Since $g_\ast(t)=c(t)\varphi_t^*g$ is precisely the family denoted $g(t)$ in the theorem statement, this proves that $g(t)$ solves Ricci flow on $I_\lambda$.[/guided]