[proofplan] We differentiate the gradient Ricci soliton equation and use the contracted [second Bianchi identity](/theorems/1541) together with the Ricci commutation formula for the Hessian of a function. These identities show that the [exterior derivative](/theorems/1525) of the scalar curvature is twice the Ricci tensor with one slot filled by the gradient of the potential function. We then differentiate Hamilton's scalar quantity, substitute the soliton equation, and obtain cancellation of the Ricci and metric terms. A smooth function with vanishing differential is constant on each connected component. [/proofplan]

[step:Derive the scalar curvature identity from the soliton equation]Let $\nabla$ denote the Levi-Civita connection of $g$, let $\operatorname{Ric}_g$ denote the Ricci curvature of $g$, let $R_g$ denote the scalar curvature of $g$, let $\operatorname{div}_g$ denote the divergence induced by $g$, and let $\operatorname{Hess}_g f$ denote the Hessian of $f$ with respect to $g$. Let $C^\infty(M)$ denote the real [vector space](/page/Vector%20Space) of smooth maps $u:M\to\mathbb{R}$, let $\Omega^1(M)$ denote the real vector space of smooth one-forms on $M$, and let $d:C^\infty(M)\to\Omega^1(M)$ denote the exterior derivative. Define the Laplace-Beltrami scalar $\Delta_g f \in C^\infty(M)$ by \begin{align*} \Delta_g f := \operatorname{tr}_g(\operatorname{Hess}_g f). \end{align*} Taking the $g$-trace of the soliton equation gives \begin{align*} R_g + \Delta_g f = n\lambda, \end{align*} where $n := \dim M$. Since $\lambda$ and $n$ are constant, applying the exterior derivative $d$ to this identity gives the one-form identity \begin{align*} d(\Delta_g f) = -dR_g. \end{align*} Now take the divergence of the soliton equation. Since $\nabla g = 0$ by metric compatibility and $\lambda$ is constant, $\operatorname{div}_g(\lambda g)=0$. Hence \begin{align*} \operatorname{div}_g(\operatorname{Ric}_g) + \operatorname{div}_g(\operatorname{Hess}_g f) = 0. \end{align*} The contracted second Bianchi identity gives \begin{align*} \operatorname{div}_g(\operatorname{Ric}_g)=\frac{1}{2}dR_g. \end{align*} With the curvature sign convention used here, the Ricci commutation identity for the Hessian of a smooth function gives \begin{align*} \operatorname{div}_g(\operatorname{Hess}_g f)=d(\Delta_g f)+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Substituting both identities into the divergence equation gives \begin{align*} 0=\frac{1}{2}dR_g + d(\Delta_g f)+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Using $d(\Delta_g f)=-dR_g$, we obtain \begin{align*} 0=\frac{1}{2}dR_g - dR_g+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Combining the two scalar-curvature terms gives \begin{align*} 0=-\frac{1}{2}dR_g+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Therefore \begin{align*} dR_g = 2\operatorname{Ric}_g(\nabla f,\cdot). \end{align*}[/step]

[guided]The goal of this step is to convert the soliton equation into an identity involving the first derivative of the scalar curvature. The soliton equation is a tensor identity, \begin{align*} \operatorname{Ric}_g+\operatorname{Hess}_g f=\lambda g, \end{align*} so we first extract two scalar or one-form consequences from it. Let $\nabla$ be the Levi-Civita connection of $g$, let $\operatorname{Ric}_g$ be the Ricci curvature of $g$, let $R_g$ be the scalar curvature of $g$, let $\operatorname{div}_g$ be the divergence induced by $g$, and let $\operatorname{Hess}_g f$ be the Hessian of $f$ with respect to $g$. Let $C^\infty(M)$ denote the real vector space of smooth maps $u:M\to\mathbb{R}$, let $\Omega^1(M)$ denote the real vector space of smooth one-forms on $M$, and let $d:C^\infty(M)\to\Omega^1(M)$ denote the exterior derivative. Define the Laplace-Beltrami scalar $\Delta_g f \in C^\infty(M)$ by \begin{align*} \Delta_g f := \operatorname{tr}_g(\operatorname{Hess}_g f). \end{align*} Taking the trace of the soliton equation with respect to $g$ gives \begin{align*} R_g+\Delta_g f=n\lambda, \end{align*} where $n:=\dim M$. The right-hand side is constant because both $n$ and $\lambda$ are constant. Therefore applying the exterior derivative $d$ gives \begin{align*} d(\Delta_g f)=-dR_g. \end{align*} Next we take the divergence of the original soliton equation. Metric compatibility gives $\nabla g=0$, and since $\lambda$ is constant this implies \begin{align*} \operatorname{div}_g(\lambda g)=0. \end{align*} Thus \begin{align*} \operatorname{div}_g(\operatorname{Ric}_g)+\operatorname{div}_g(\operatorname{Hess}_g f)=0. \end{align*} We now use two standard differential identities. The contracted second Bianchi identity applies because $(M,g)$ is a smooth Riemannian manifold, and it says \begin{align*} \operatorname{div}_g(\operatorname{Ric}_g)=\frac{1}{2}dR_g. \end{align*} The Ricci commutation identity for the Hessian of a smooth function applies because $f$ is smooth. With the curvature sign convention used here, it says \begin{align*} \operatorname{div}_g(\operatorname{Hess}_g f)=d(\Delta_g f)+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} This is the point where curvature enters the calculation: the divergence of the Hessian is not merely $\nabla(\Delta_g f)$; commuting the covariant derivatives produces the Ricci term. Substituting these two identities into the divergence equation gives \begin{align*} 0 &= \frac{1}{2}dR_g+d(\Delta_g f)+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Using the trace consequence $d(\Delta_g f)=-dR_g$, this becomes \begin{align*} 0=\frac{1}{2}dR_g-dR_g+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Combining the two scalar-curvature terms gives \begin{align*} 0=-\frac{1}{2}dR_g+\operatorname{Ric}_g(\nabla f,\cdot). \end{align*} Rearranging yields the scalar curvature identity \begin{align*} dR_g=2\operatorname{Ric}_g(\nabla f,\cdot). \end{align*}[/guided]

[step:Differentiate Hamilton's quantity and substitute the soliton equation] Define the smooth function $H: M \to \mathbb{R}$ by \begin{align*} H(p)=R_g(p)+|\nabla f|_g^2(p)-2\lambda f(p) \end{align*} for $p \in M$. Let $\mathfrak{X}(M)$ denote the real vector space of smooth vector fields on $M$. For every vector field $X \in \mathfrak{X}(M)$, metric compatibility gives \begin{align*} X(|\nabla f|_g^2)=2\operatorname{Hess}_g f(X,\nabla f). \end{align*} Since the Levi-Civita Hessian of a smooth function is symmetric, the one-form $X\mapsto 2\operatorname{Hess}_g f(X,\nabla f)$ equals $2\operatorname{Hess}_g f(\nabla f,\cdot)$. Therefore \begin{align*} dH=dR_g+2\operatorname{Hess}_g f(\nabla f,\cdot)-2\lambda df. \end{align*} Using $dR_g=2\operatorname{Ric}_g(\nabla f,\cdot)$, $df=g(\nabla f,\cdot)$, and $\operatorname{Hess}_g f=\lambda g-\operatorname{Ric}_g$, we get \begin{align*} dH=2\operatorname{Ric}_g(\nabla f,\cdot)+2(\lambda g-\operatorname{Ric}_g)(\nabla f,\cdot)-2\lambda g(\nabla f,\cdot). \end{align*} Expanding the middle term gives \begin{align*} dH=2\operatorname{Ric}_g(\nabla f,\cdot)+2\lambda g(\nabla f,\cdot)-2\operatorname{Ric}_g(\nabla f,\cdot)-2\lambda g(\nabla f,\cdot). \end{align*} The Ricci terms and the metric terms cancel, so \begin{align*} dH=0. \end{align*} Thus the differential of $H$ vanishes everywhere on $M$. [/step]

[step:Conclude constancy on connected components] Let $C \subset M$ be a connected component. Since $dH=0$, the differential $dH_p:T_pM\to\mathbb{R}$ is the zero [linear map](/page/Linear%20Map) for every $p\in C$. Hence $H|_C$ is locally constant. Because $C$ is connected, every locally constant real-valued function on $C$ is constant. Therefore $H$ is constant on $C$. Since $C$ was arbitrary, $H$ is constant on each connected component of $M$. [/step]