[proofplan] The family is a scalar multiple of the initial Einstein metric, so the metric condition reduces to positivity of the scalar factor. The main geometric point is that multiplying a Riemannian metric by a positive constant leaves its Levi-Civita connection, curvature endomorphism, and Ricci tensor as a $(0,2)$-tensor unchanged. After proving this invariance for each fixed time, the Ricci flow equation becomes the scalar ordinary differential equation $a'(t)=-2\kappa$, whose solution with $a(0)=1$ is $a(t)=1-2\kappa t$. [/proofplan]

[step:Verify that the homothetic family is a smooth family of metrics] Define the scalar function $a: I \to (0,\infty)$ by $a(t):=1 - 2\kappa t$ for $t \in I$. By the hypothesis on $I$, $a(t)>0$ for every $t \in I$. Hence, for each $t \in I$, the tensor $g(t)=a(t)g_0$ is symmetric and positive definite because $g_0$ is symmetric and positive definite. Since $a$ is a smooth function of $t$, the map $t \mapsto g(t)$ is a smooth family of Riemannian metrics. Also, \begin{align*} g(0) = (1 - 2\kappa \cdot 0)g_0 = g_0. \end{align*} [/step]

[step:Show that constant scaling leaves the Ricci tensor unchanged]Fix $t \in I$ and set $c:=a(t)>0$. Define the scaled metric $\widetilde g := c g_0$. Let $\nabla$ denote the Levi-Civita connection of $g_0$, and let $\widetilde\nabla$ denote the Levi-Civita connection of $\widetilde g$. Let $\mathfrak{X}(M)$ denote the [vector space](/page/Vector%20Space) of smooth vector fields on $M$. For smooth vector fields $X,Y,Z \in \mathfrak{X}(M)$, we use the Koszul formula for the Levi-Civita connection, which characterises the Levi-Civita connection by its metric compatibility and torsion-free properties. Applied to $\widetilde g$, it gives \begin{align*} 2\widetilde g(\widetilde\nabla_XY,Z)=X\widetilde g(Y,Z)+Y\widetilde g(X,Z)-Z\widetilde g(X,Y)+\widetilde g([X,Y],Z)-\widetilde g([X,Z],Y)-\widetilde g([Y,Z],X). \end{align*} Since $\widetilde g = c g_0$ and $c$ is constant on $M$, the right-hand side equals \begin{align*} c\bigl(Xg_0(Y,Z)+Yg_0(X,Z)-Zg_0(X,Y)+g_0([X,Y],Z)-g_0([X,Z],Y)-g_0([Y,Z],X)\bigr). \end{align*} Applying the Koszul formula for $g_0$, this is \begin{align*} 2c\, g_0(\nabla_XY,Z)=2\widetilde g(\nabla_XY,Z). \end{align*} Because $\widetilde g$ is nondegenerate, $\widetilde\nabla_XY=\nabla_XY$ for all $X,Y \in \mathfrak{X}(M)$. Let $R$ denote the curvature endomorphism of $g_0$, defined by \begin{align*} R(X,Y)Z:=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z, \end{align*} and let $\widetilde R$ denote the curvature endomorphism of $\widetilde g$, defined by the same formula with $\widetilde\nabla$ in place of $\nabla$. Since $\widetilde\nabla=\nabla$, the curvature endomorphisms of $\widetilde g$ and $g_0$ agree: \begin{align*} \widetilde R(X,Y)Z = R(X,Y)Z \end{align*} for all $X,Y,Z \in \mathfrak{X}(M)$. Let $p \in M$, and let $n:=\dim M$. Let $(e_1,\dots,e_n)$ be a $g_0$-[orthonormal basis](/page/Orthonormal%20Basis) of $T_pM$. Then $(c^{-1/2}e_1,\dots,c^{-1/2}e_n)$ is a $\widetilde g$-orthonormal basis of $T_pM$. For $X,Y \in T_pM$, the Ricci contraction gives \begin{align*} \operatorname{Ric}(\widetilde g)(X,Y)=\sum_{i=1}^n \widetilde g(\widetilde R(c^{-1/2}e_i,X)Y,c^{-1/2}e_i)=\sum_{i=1}^n c\, g_0(c^{-1/2}R(e_i,X)Y,c^{-1/2}e_i)=\sum_{i=1}^n g_0(R(e_i,X)Y,e_i)=\operatorname{Ric}(g_0)(X,Y). \end{align*} Thus \begin{align*} \operatorname{Ric}(g(t))=\operatorname{Ric}(g_0)=\kappa g_0. \end{align*}[/step]

[guided]The only point needing care is the sentence “constant scaling leaves Ricci unchanged.” We prove it directly. Fix $t \in I$ and write $c:=a(t)>0$, so the metric at time $t$ is $\widetilde g:=c g_0$. Let $\nabla$ be the Levi-Civita connection of $g_0$, and let $\widetilde\nabla$ be the Levi-Civita connection of $\widetilde g$. Let $\mathfrak{X}(M)$ denote the vector space of smooth vector fields on $M$. We compare the two connections using the Koszul formula for the Levi-Civita connection. This formula applies to the Levi-Civita connection of a Riemannian metric, and both $g_0$ and $\widetilde g$ are Riemannian metrics. For smooth vector fields $X,Y,Z \in \mathfrak{X}(M)$, the formula gives \begin{align*} 2\widetilde g(\widetilde\nabla_XY,Z)=X\widetilde g(Y,Z)+Y\widetilde g(X,Z)-Z\widetilde g(X,Y)+\widetilde g([X,Y],Z)-\widetilde g([X,Z],Y)-\widetilde g([Y,Z],X). \end{align*} Since $\widetilde g=cg_0$ and $c$ is a constant on $M$, every term on the right has a common factor $c$. Therefore \begin{align*} 2\widetilde g(\widetilde\nabla_XY,Z)=c\bigl(Xg_0(Y,Z)+Yg_0(X,Z)-Zg_0(X,Y)+g_0([X,Y],Z)-g_0([X,Z],Y)-g_0([Y,Z],X)\bigr). \end{align*} The expression in parentheses is $2g_0(\nabla_XY,Z)$ by the Koszul formula for $g_0$. Hence \begin{align*} 2\widetilde g(\widetilde\nabla_XY,Z)=2c\,g_0(\nabla_XY,Z)=2\widetilde g(\nabla_XY,Z). \end{align*} Since $\widetilde g$ is nondegenerate, equality of inner products against every $Z$ implies \begin{align*} \widetilde\nabla_XY=\nabla_XY. \end{align*} The curvature endomorphism is built only from the connection and Lie brackets: \begin{align*} R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. \end{align*} Because the two connections agree, their curvature endomorphisms agree: \begin{align*} \widetilde R(X,Y)Z=R(X,Y)Z. \end{align*} It remains to check that the Ricci tensor as a $(0,2)$-tensor is unchanged, not merely rescaled. Let $p \in M$, and let $n:=\dim M$. Let $(e_1,\dots,e_n)$ be a $g_0$-orthonormal basis of $T_pM$. Since $\widetilde g= c g_0$, the basis $(c^{-1/2}e_1,\dots,c^{-1/2}e_n)$ is $\widetilde g$-orthonormal. We now use the standard orthonormal-frame contraction formula in the definition of Ricci curvature. Therefore, for $X,Y \in T_pM$, \begin{align*} \operatorname{Ric}(\widetilde g)(X,Y)=\sum_{i=1}^n \widetilde g(\widetilde R(c^{-1/2}e_i,X)Y,c^{-1/2}e_i)=\sum_{i=1}^n c\, g_0(c^{-1/2}R(e_i,X)Y,c^{-1/2}e_i)=\sum_{i=1}^n g_0(R(e_i,X)Y,e_i)=\operatorname{Ric}(g_0)(X,Y). \end{align*} Thus constant positive scaling leaves the Ricci tensor unchanged as a $(0,2)$-tensor. Applying this with $\widetilde g=g(t)$ gives \begin{align*} \operatorname{Ric}(g(t))=\operatorname{Ric}(g_0)=\kappa g_0. \end{align*}[/guided]

[step:Differentiate the family and verify the Ricci flow equation] Since \begin{align*} g(t)=a(t)g_0 \end{align*} and $a'(t)=-2\kappa$, differentiation with respect to $t$ gives \begin{align*} \frac{\partial g}{\partial t}(t)=a'(t)g_0=-2\kappa g_0. \end{align*} From the preceding step, \begin{align*} \operatorname{Ric}(g(t))=\kappa g_0. \end{align*} Therefore \begin{align*} \frac{\partial g}{\partial t}(t)=-2\kappa g_0=-2\operatorname{Ric}(g(t)). \end{align*} This proves that $g(t)=(1-2\kappa t)g_0$ solves the Ricci flow equation on every interval $I$ on which $1-2\kappa t>0$, with initial metric $g(0)=g_0$. [/step]