[proofplan] The proof has two separate scaling computations. First, the time derivative of $g_\lambda$ cancels the factor $\lambda$ in the time change, and constant metric scaling leaves the Ricci tensor as a $(0,2)$-tensor unchanged. Second, the Levi-Civita connection is unchanged under constant scaling, so the $(1,3)$ curvature tensor is unchanged, while the fully covariant curvature tensor gains one factor of $\lambda$; the four inverse metrics appearing in its norm then give the pointwise scaling factor $\lambda^{-1}$. [/proofplan]

[step:Differentiate the rescaled metric in the new time variable] Define the affine time-change map $\tau: I_\lambda \to I$ by \begin{align*} \tau(s)=t_0+s/\lambda. \end{align*} By the definition of $I_\lambda$, this map takes every $s \in I_\lambda$ to a time $\tau(s) \in I$. Since $g$ is a smooth curve in the [vector space](/page/Vector%20Space) $\Gamma(S^2T^*M)$ of smooth symmetric $(0,2)$-tensors on $M$, and since $\tau$ is a smooth affine map with $\tau'(s)=1/\lambda$, the chain rule applied to the smooth tensor-field curve $s \mapsto g(\tau(s))$ gives the following identity at interior points of $I_\lambda$, and the same identity with one-sided derivatives at endpoints of $I_\lambda$: \begin{align*} \partial_s g_\lambda(s) = \partial_s\bigl(\lambda g(\tau(s))\bigr) = \lambda \cdot \frac{1}{\lambda}\,\partial_t g(\tau(s)) = \partial_t g(\tau(s)). \end{align*} Since $g$ is a Ricci flow, \begin{align*} \partial_s g_\lambda(s) = -2\operatorname{Ric}_{g(\tau(s))}. \end{align*} [/step]

[step:Show that constant metric scaling preserves the Ricci tensor]Let $h$ be a Riemannian metric on $M$, and define the scaled metric $\tilde h := \lambda h$. We prove \begin{align*} \operatorname{Ric}_{\tilde h} = \operatorname{Ric}_h \end{align*} as $(0,2)$-tensors. Let $(U,\varphi)$ be a coordinate chart with coordinates $(x_1,\dots,x_n)$. Write $h_{ij}$ for the components of $h$ in this chart, $\tilde h_{ij}$ for the components of $\tilde h$, and $h^{ij}$, $\tilde h^{ij}$ for the components of the inverse matrices. Since $\tilde h_{ij}=\lambda h_{ij}$ and $\lambda$ is constant, the inverse metric coefficients satisfy \begin{align*} \tilde h^{ij}=\lambda^{-1}h^{ij}. \end{align*} The spatial derivatives of the metric coefficients satisfy \begin{align*} \partial_{x_k}\tilde h_{ij}=\lambda\,\partial_{x_k}h_{ij}. \end{align*} By the coordinate formula for the Levi-Civita connection, the Christoffel symbols of the Levi-Civita connection of $\tilde h$ satisfy \begin{align*} \tilde\Gamma_{ij}^k = \frac{1}{2}\sum_{\ell=1}^n\tilde h^{k\ell}\left(\partial_{x_i}\tilde h_{j\ell}+\partial_{x_j}\tilde h_{i\ell}-\partial_{x_\ell}\tilde h_{ij}\right). \end{align*} Substituting the scaling identities for $\tilde h^{k\ell}$ and $\partial_{x_m}\tilde h_{pq}$ gives \begin{align*} \tilde\Gamma_{ij}^k=\frac{1}{2}\sum_{\ell=1}^n\lambda^{-1}h^{k\ell}\left(\lambda\partial_{x_i}h_{j\ell}+\lambda\partial_{x_j}h_{i\ell}-\lambda\partial_{x_\ell}h_{ij}\right)=\Gamma_{ij}^k. \end{align*} Thus the Levi-Civita connections of $h$ and $\tilde h$ are equal. Since the Riemann curvature tensor as a $(1,3)$-tensor is defined from the connection and its first derivatives, the two $(1,3)$ curvature tensors are equal. Since the Ricci curvature is obtained by tracing this $(1,3)$ curvature tensor in the first and third slots, we get \begin{align*} \operatorname{Ric}_{\tilde h} = \operatorname{Ric}_h. \end{align*} Applying this with $h=g(\tau(s))$ and $\tilde h=g_\lambda(s)=\lambda g(\tau(s))$, we obtain \begin{align*} \operatorname{Ric}_{g_\lambda(s)} = \operatorname{Ric}_{g(\tau(s))}. \end{align*} Because $\lambda>0$, each tensor $g_\lambda(s)=\lambda g(\tau(s))$ is positive definite whenever $g(\tau(s))$ is positive definite. Since $g$ is smooth on $I$ and $\tau: I_\lambda \to I$ is smooth, the curve $s \mapsto g_\lambda(s)$ is smooth on $I_\lambda$ as a curve of Riemannian metrics. Together with the previous step, \begin{align*} \partial_s g_\lambda(s) = -2\operatorname{Ric}_{g_\lambda(s)}. \end{align*} Hence $g_\lambda$ is a Ricci flow on $I_\lambda$.[/step]

[guided]The point of this step is to justify the statement that Ricci curvature is unchanged by multiplying the metric by a positive constant. This is not because every curvature object is unchanged; it is specifically true for the Ricci tensor viewed as a $(0,2)$-tensor. Let $h$ be a Riemannian metric on $M$, and define the scaled metric $\tilde h := \lambda h$. We compare the two Levi-Civita connections in a coordinate chart $(U,\varphi)$ with coordinates $(x_1,\dots,x_n)$. Write $h_{ij}$ and $\tilde h_{ij}$ for the metric coefficients, and write $h^{ij}$ and $\tilde h^{ij}$ for the entries of the inverse metric matrices. Since $\tilde h=\lambda h$, we have \begin{align*} \tilde h_{ij}=\lambda h_{ij}. \end{align*} Inverting the matrix multiplies by the reciprocal scalar, so \begin{align*} \tilde h^{ij}=\lambda^{-1}h^{ij}. \end{align*} Because $\lambda$ is constant in space, differentiating the metric coefficients gives \begin{align*} \partial_{x_k}\tilde h_{ij} = \lambda\,\partial_{x_k}h_{ij}. \end{align*} Now compute the Christoffel symbols using the coordinate formula for the Levi-Civita connection. For $\tilde h$, \begin{align*} \tilde\Gamma_{ij}^k = \frac{1}{2}\sum_{\ell=1}^n\tilde h^{k\ell}\left(\partial_{x_i}\tilde h_{j\ell}+\partial_{x_j}\tilde h_{i\ell}-\partial_{x_\ell}\tilde h_{ij}\right). \end{align*} Substituting $\tilde h^{k\ell}=\lambda^{-1}h^{k\ell}$ and $\partial_{x_m}\tilde h_{pq}=\lambda\partial_{x_m}h_{pq}$ gives \begin{align*} \tilde\Gamma_{ij}^k=\frac{1}{2}\sum_{\ell=1}^n\lambda^{-1}h^{k\ell}\left(\lambda\partial_{x_i}h_{j\ell}+\lambda\partial_{x_j}h_{i\ell}-\lambda\partial_{x_\ell}h_{ij}\right)=\Gamma_{ij}^k. \end{align*} The factor $\lambda^{-1}$ from the inverse metric cancels exactly with the factor $\lambda$ from differentiating the scaled metric coefficients. Therefore the Levi-Civita connections of $h$ and $\tilde h$ are the same. Define $\operatorname{Rm}^{(1,3)}_h$ and $\operatorname{Rm}^{(1,3)}_{\tilde h}$ to be the Riemann curvature tensors of $h$ and $\tilde h$ with one contravariant and three covariant slots. Since this tensor is defined from the connection and its first derivatives, equal connections give equal $(1,3)$ curvature tensors: \begin{align*} \operatorname{Rm}^{(1,3)}_{\tilde h} = \operatorname{Rm}^{(1,3)}_h. \end{align*} The Ricci curvature is the trace of this $(1,3)$ curvature tensor in the first and third slots, so the trace is also unchanged: \begin{align*} \operatorname{Ric}_{\tilde h} = \operatorname{Ric}_h. \end{align*} Now take $h=g(\tau(s))$ and $\tilde h=g_\lambda(s)=\lambda g(\tau(s))$. The scaling result gives \begin{align*} \operatorname{Ric}_{g_\lambda(s)} = \operatorname{Ric}_{g(\tau(s))}. \end{align*} Because $\lambda>0$, multiplying the positive definite metric $g(\tau(s))$ by $\lambda$ gives another positive definite metric. Also, $g$ is smooth as a curve of metrics on $I$, and $\tau$ is a smooth affine map from $I_\lambda$ into $I$, so the composite-rescaled curve $s \mapsto g_\lambda(s)=\lambda g(\tau(s))$ is smooth on $I_\lambda$. It remains to connect this Ricci-scaling identity with the rescaled time variable. Define the affine map $\tau: I_\lambda \to I$ by $\tau(s)=t_0+s/\lambda$. Since $g$ is a smooth curve in $\Gamma(S^2T^*M)$ and $\tau'(s)=1/\lambda$, the chain rule, interpreted one-sidedly at endpoints of $I_\lambda$, gives \begin{align*} \partial_s g_\lambda(s) = \partial_s\bigl(\lambda g(\tau(s))\bigr) = \lambda\cdot\frac{1}{\lambda}\,\partial_t g(\tau(s)) = \partial_t g(\tau(s)). \end{align*} Because $g$ is a Ricci flow on $I$, this becomes \begin{align*} \partial_s g_\lambda(s) = -2\operatorname{Ric}_{g(\tau(s))}. \end{align*} Substituting the Ricci scaling identity yields \begin{align*} \partial_s g_\lambda(s) = -2\operatorname{Ric}_{g_\lambda(s)}. \end{align*} Thus $g_\lambda$ satisfies the Ricci flow equation on the rescaled interval $I_\lambda$.[/guided]

[step:Compute the scaling of the curvature norm] Fix $x \in M$ and $s \in I_\lambda$. Set $h := g(\tau(s))$. Then set $\tilde h := g_\lambda(s)=\lambda h$. Let $(U,\varphi)$ be a coordinate chart around $x$ with coordinates $(x_1,\dots,x_n)$. Write $(\operatorname{Rm}_h)_{ijk\ell}$ and $(\operatorname{Rm}_{\tilde h})_{ijk\ell}$ for the components of the fully covariant Riemann curvature tensors. Since the $(1,3)$ curvature tensors are equal and lowering the first index uses the metric, we have \begin{align*} (\operatorname{Rm}_{\tilde h})_{ijk\ell} = \lambda(\operatorname{Rm}_h)_{ijk\ell}. \end{align*} Also $\tilde h^{ij}=\lambda^{-1}h^{ij}$. Therefore the squared pointwise norm satisfies The squared pointwise norm is the full contraction of the two covariant curvature tensors using four copies of the inverse metric. In coordinates, this is \begin{align*} |\operatorname{Rm}_{\tilde h}|_{\tilde h}^2(x) = \sum_{i,j,k,\ell=1}^n\sum_{a,b,c,d=1}^n \tilde h^{ia}\tilde h^{jb}\tilde h^{kc}\tilde h^{\ell d} (\operatorname{Rm}_{\tilde h})_{ijk\ell}(\operatorname{Rm}_{\tilde h})_{abcd}. \end{align*} Substituting $\tilde h^{pq}=\lambda^{-1}h^{pq}$ and $(\operatorname{Rm}_{\tilde h})_{ijk\ell}=\lambda(\operatorname{Rm}_h)_{ijk\ell}$ into this norm formula gives \begin{align*} |\operatorname{Rm}_{\tilde h}|_{\tilde h}^2(x)=\lambda^{-2}|\operatorname{Rm}_h|_h^2(x). \end{align*} Taking the non-negative square root gives \begin{align*} |\operatorname{Rm}_{\tilde h}|_{\tilde h}(x) = \lambda^{-1}|\operatorname{Rm}_h|_h(x). \end{align*} Substituting back $h=g(t_0+s/\lambda)$ and $\tilde h=g_\lambda(s)$ proves \begin{align*} |\operatorname{Rm}_{g_\lambda(s)}|_{g_\lambda(s)}(x) = \lambda^{-1} |\operatorname{Rm}_{g(t_0+s/\lambda)}|_{g(t_0+s/\lambda)}(x). \end{align*} This is the desired curvature scaling identity. [/step]