[proofplan] The key point is that multiplying the metric by a positive constant does not change the Levi-Civita connection: the Koszul formula acquires the same factor $\lambda$ on both sides. Therefore the curvature operator, viewed as a $(1,3)$-tensor, is unchanged. The $(0,4)$ curvature tensor gains one factor of $\lambda$ because it is obtained by lowering an index with the metric, Ricci curvature is the metric-independent trace of the unchanged curvature operator, scalar curvature gains one inverse-metric factor, and the pointwise norm of the $(0,4)$ curvature tensor combines one tensor factor $\lambda$ with four inverse-metric factors. [/proofplan]

[step:Show that the Levi-Civita connection is unchanged by constant rescaling]Let $\Gamma(TM)$ denote the space of smooth sections of the tangent bundle $TM$, that is, the space of smooth vector fields on $M$. Let $\mathbb R$ denote the [real numbers](/page/Real%20Numbers), and let $C^\infty(M)$ denote the algebra of smooth real-valued functions on $M$. Let $\nabla$ and $\tilde\nabla$ denote the Levi-Civita connections of $g$ and $\tilde g$, respectively. Let $X,Y,Z \in \Gamma(TM)$ be smooth vector fields. We use the convention that the curvature operator of a connection $\nabla$ is $\mathcal R(X,Y)Z:=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$, the Riemann curvature tensor as a $(0,4)$-tensor is obtained by lowering the final vector slot with the metric, Ricci curvature is the contraction $\operatorname{Ric}_p(X,Y)=\sum_{i=1}^n \operatorname{Rm}_p(e_i,X,Y,e_i)$ in any [orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_n)$ of $T_pM$, and scalar curvature is the metric trace of Ricci curvature. The Koszul formula for $\tilde g$ gives \begin{align*} 2\tilde g(\tilde\nabla_XY,Z)=X\tilde g(Y,Z)+Y\tilde g(Z,X)-Z\tilde g(X,Y)+\tilde g([X,Y],Z)-\tilde g([Y,Z],X)+\tilde g([Z,X],Y). \end{align*} Since $\tilde g=\lambda g$ and $\lambda$ is constant, the right-hand side is \begin{align*} \lambda\bigl( Xg(Y,Z)+Yg(Z,X)-Zg(X,Y) +g([X,Y],Z)-g([Y,Z],X)+g([Z,X],Y) \bigr). \end{align*} Applying the Koszul formula for $g$, this equals $2\lambda g(\nabla_XY,Z)$. Hence \begin{align*} 2\tilde g(\tilde\nabla_XY,Z)=2\lambda g(\nabla_XY,Z)=2\tilde g(\nabla_XY,Z). \end{align*} Because $\tilde g$ is nondegenerate at every point and $Z$ was arbitrary, the equality $\tilde g(\tilde\nabla_XY-\nabla_XY,Z)=0$ for all smooth vector fields $Z$ implies $(\tilde\nabla_XY)_p=(\nabla_XY)_p$ for every $p\in M$. Hence $\tilde\nabla_XY=\nabla_XY$. Thus $\tilde\nabla=\nabla$.[/step]

[guided]We prove first that the connection does not change. This is the structural reason all the scaling laws follow. Let $\nabla$ be the Levi-Civita connection of $g$, and let $\tilde\nabla$ be the Levi-Civita connection of $\tilde g$. For smooth vector fields $X,Y,Z \in \Gamma(TM)$, where $\Gamma(TM)$ is the space of smooth sections of the tangent bundle $TM$, the Koszul formula for the metric $\tilde g$ states that \begin{align*} 2\tilde g(\tilde\nabla_XY,Z)=X\tilde g(Y,Z)+Y\tilde g(Z,X)-Z\tilde g(X,Y)+\tilde g([X,Y],Z)-\tilde g([Y,Z],X)+\tilde g([Z,X],Y). \end{align*} Here $\mathbb R$ denotes the real numbers and $C^\infty(M)$ denotes the algebra of smooth real-valued functions on $M$; these are the codomains used below for tensor-valued evaluations. Now use $\tilde g=\lambda g$. Because $\lambda$ is constant, differentiating $\tilde g(Y,Z)=\lambda g(Y,Z)$ gives $X\tilde g(Y,Z)=\lambda Xg(Y,Z)$, and the same applies to the other derivative terms. Therefore the whole right-hand side factors as \begin{align*} \lambda\bigl( Xg(Y,Z)+Yg(Z,X)-Zg(X,Y) +g([X,Y],Z)-g([Y,Z],X)+g([Z,X],Y) \bigr). \end{align*} The expression in parentheses is $2g(\nabla_XY,Z)$ by the Koszul formula for $g$. Hence \begin{align*} 2\tilde g(\tilde\nabla_XY,Z)=2\lambda g(\nabla_XY,Z)=2\tilde g(\nabla_XY,Z). \end{align*} Thus $\tilde g(\tilde\nabla_XY-\nabla_XY,Z)=0$ for every smooth vector field $Z$. To pass from this to equality of vector fields, fix a point $p\in M$. Since the value $Z_p\in T_pM$ can be chosen arbitrarily by choosing a smooth vector field with that value at $p$, and since $\tilde g_p$ is a nondegenerate [inner product](/page/Inner%20Product) on $T_pM$, we get $(\tilde\nabla_XY)_p=(\nabla_XY)_p$. Because $p$ was arbitrary, $\tilde\nabla_XY=\nabla_XY$. Because $X$ and $Y$ were arbitrary, $\tilde\nabla=\nabla$.[/guided]

[step:Deduce that the curvature operator is unchanged]For $X,Y,Z \in \Gamma(TM)$, define the curvature operator $\mathcal R_g: \Gamma(TM)\times\Gamma(TM)\times\Gamma(TM)\to\Gamma(TM)$ by \begin{align*} \mathcal R_g(X,Y)Z:=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. \end{align*} Define the curvature operator $\mathcal R_{\tilde g}: \Gamma(TM)\times\Gamma(TM)\times\Gamma(TM)\to\Gamma(TM)$ by \begin{align*} \mathcal R_{\tilde g}(X,Y)Z:=\tilde\nabla_X\tilde\nabla_YZ-\tilde\nabla_Y\tilde\nabla_XZ-\tilde\nabla_{[X,Y]}Z. \end{align*} Since $\tilde\nabla=\nabla$, each term in these two definitions agrees. Therefore \begin{align*} \mathcal R_{\tilde g}(X,Y)Z=\mathcal R_g(X,Y)Z \end{align*} for all $X,Y,Z \in \Gamma(TM)$.[/step]

[guided]The curvature operator is built only from the connection and the Lie bracket of vector fields. For the metric $g$, define \begin{align*} \mathcal R_g(X,Y)Z:=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. \end{align*} For the metric $\tilde g$, define \begin{align*} \mathcal R_{\tilde g}(X,Y)Z:=\tilde\nabla_X\tilde\nabla_YZ-\tilde\nabla_Y\tilde\nabla_XZ-\tilde\nabla_{[X,Y]}Z. \end{align*} The previous step proved $\tilde\nabla=\nabla$. Substituting this equality into each of the three terms in the definition of $\mathcal R_{\tilde g}$ gives \begin{align*} \mathcal R_{\tilde g}(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z=\mathcal R_g(X,Y)Z. \end{align*} Thus the curvature operator is unchanged as a $(1,3)$-tensor.[/guided]

[step:Lower one index to obtain the scaling of the $(0,4)$ curvature tensor]For $X,Y,Z,W \in \Gamma(TM)$, define the $(0,4)$ curvature tensor $\operatorname{Rm}(g):\Gamma(TM)^4\to C^\infty(M)$ by \begin{align*} \operatorname{Rm}(g)(X,Y,Z,W):=g(\mathcal R_g(X,Y)Z,W). \end{align*} Define the $(0,4)$ curvature tensor $\operatorname{Rm}(\tilde g):\Gamma(TM)^4\to C^\infty(M)$ by \begin{align*} \operatorname{Rm}(\tilde g)(X,Y,Z,W):=\tilde g(\mathcal R_{\tilde g}(X,Y)Z,W). \end{align*} Using $\mathcal R_{\tilde g}=\mathcal R_g$ and $\tilde g=\lambda g$, we get \begin{align*} \operatorname{Rm}(\tilde g)(X,Y,Z,W)=\tilde g(\mathcal R_g(X,Y)Z,W)=\lambda g(\mathcal R_g(X,Y)Z,W)=\lambda\operatorname{Rm}(g)(X,Y,Z,W). \end{align*} Since this holds for all smooth vector fields $X,Y,Z,W$, we have \begin{align*} \operatorname{Rm}(\tilde g)=\lambda\operatorname{Rm}(g) \end{align*} as $(0,4)$-tensors.[/step]

[guided]The curvature operator itself did not change, but the $(0,4)$ curvature tensor also uses the metric to lower the final vector slot. For $X,Y,Z,W \in \Gamma(TM)$, define \begin{align*} \operatorname{Rm}(g)(X,Y,Z,W):=g(\mathcal R_g(X,Y)Z,W) \end{align*} and define \begin{align*} \operatorname{Rm}(\tilde g)(X,Y,Z,W):=\tilde g(\mathcal R_{\tilde g}(X,Y)Z,W). \end{align*} Using the previously proved identity $\mathcal R_{\tilde g}=\mathcal R_g$ and the metric rescaling $\tilde g=\lambda g$, we compute \begin{align*} \operatorname{Rm}(\tilde g)(X,Y,Z,W)=\tilde g(\mathcal R_g(X,Y)Z,W)=\lambda g(\mathcal R_g(X,Y)Z,W)=\lambda\operatorname{Rm}(g)(X,Y,Z,W). \end{align*} Because this equality holds for every quadruple of smooth vector fields, it is an equality of $(0,4)$-tensors: \begin{align*} \operatorname{Rm}(\tilde g)=\lambda\operatorname{Rm}(g). \end{align*}[/guided]

[step:Take traces to obtain the Ricci and scalar curvature scaling laws]Fix $p \in M$. Let $T_pM$ denote the tangent space of $M$ at $p$, let $n:=\dim T_pM$, let $g_p:T_pM\times T_pM\to\mathbb R$ denote the inner product induced by $g$ at $p$, and let $\tilde g_p:T_pM\times T_pM\to\mathbb R$ denote the inner product induced by $\tilde g$ at $p$. Choose a $g_p$-orthonormal basis $(e_1,\dots,e_n)$ of $T_pM$. Then $(\lambda^{-1/2}e_1,\dots,\lambda^{-1/2}e_n)$ is a $\tilde g_p$-orthonormal basis. For $X,Y \in T_pM$, the Ricci tensor of $\tilde g$ satisfies \begin{align*} \operatorname{Ric}(\tilde g)_p(X,Y)=\sum_{i=1}^n \operatorname{Rm}(\tilde g)_p(\lambda^{-1/2}e_i,X,Y,\lambda^{-1/2}e_i)=\sum_{i=1}^n \lambda^{-1}\operatorname{Rm}(\tilde g)_p(e_i,X,Y,e_i)=\sum_{i=1}^n \operatorname{Rm}(g)_p(e_i,X,Y,e_i)=\operatorname{Ric}(g)_p(X,Y). \end{align*} Thus $\operatorname{Ric}(\tilde g)=\operatorname{Ric}(g)$. Let $(\tilde g^{ij})$ and $(g^{ij})$ denote the inverse metric matrices in any local coordinate chart. Since $\tilde g_{ij}=\lambda g_{ij}$, we have $\tilde g^{ij}=\lambda^{-1}g^{ij}$. Therefore \begin{align*} R(\tilde g)=\sum_{i,j=1}^n \tilde g^{ij}\operatorname{Ric}(\tilde g)_{ij}=\sum_{i,j=1}^n \lambda^{-1}g^{ij}\operatorname{Ric}(g)_{ij}=\lambda^{-1}R(g). \end{align*}[/step]

[guided]Fix $p \in M$. The Ricci tensor is defined by tracing the $(0,4)$ curvature tensor over a metric-orthonormal basis. Let $T_pM$ be the tangent space at $p$, let $n:=\dim T_pM$, and choose a $g_p$-orthonormal basis $(e_1,\dots,e_n)$ of $T_pM$. Since $\tilde g_p=\lambda g_p$, the vectors $\tilde e_i:=\lambda^{-1/2}e_i$ satisfy \begin{align*} \tilde g_p(\tilde e_i,\tilde e_j)=\lambda g_p(\lambda^{-1/2}e_i,\lambda^{-1/2}e_j)=g_p(e_i,e_j), \end{align*} so $(\tilde e_1,\dots,\tilde e_n)$ is a $\tilde g_p$-orthonormal basis. Now compute the trace defining Ricci curvature for $\tilde g$. Using multilinearity in the first and fourth slots, and using the already proved identity $\operatorname{Rm}(\tilde g)=\lambda\operatorname{Rm}(g)$ as a $(0,4)$-tensor, we obtain \begin{align*} \operatorname{Ric}(\tilde g)_p(X,Y)=\sum_{i=1}^n \operatorname{Rm}(\tilde g)_p(\lambda^{-1/2}e_i,X,Y,\lambda^{-1/2}e_i)=\sum_{i=1}^n \lambda^{-1}\operatorname{Rm}(\tilde g)_p(e_i,X,Y,e_i)=\sum_{i=1}^n \operatorname{Rm}(g)_p(e_i,X,Y,e_i)=\operatorname{Ric}(g)_p(X,Y). \end{align*} This proves $\operatorname{Ric}(\tilde g)=\operatorname{Ric}(g)$ as a $(0,2)$-tensor. For scalar curvature, take the metric trace of Ricci curvature. In a local coordinate chart, let $(g_{ij})$ and $(\tilde g_{ij})$ denote the metric matrices, and let $(g^{ij})$ and $(\tilde g^{ij})$ denote their inverse matrices. Since $\tilde g_{ij}=\lambda g_{ij}$, inversion gives $\tilde g^{ij}=\lambda^{-1}g^{ij}$. Combining this inverse-metric scaling with $\operatorname{Ric}(\tilde g)=\operatorname{Ric}(g)$ gives \begin{align*} R(\tilde g)=\sum_{i,j=1}^n \tilde g^{ij}\operatorname{Ric}(\tilde g)_{ij}=\sum_{i,j=1}^n \lambda^{-1}g^{ij}\operatorname{Ric}(g)_{ij}=\lambda^{-1}R(g). \end{align*} The inverse factor appears here because scalar curvature is the trace of Ricci curvature with respect to the metric, and the inverse metric scales oppositely to the metric.[/guided]

[step:Compute the pointwise norm scaling of the curvature tensor]Fix $p \in M$, and let $(e_1,\dots,e_n)$ be a $g_p$-orthonormal basis of $T_pM$. Define $\tilde e_i:=\lambda^{-1/2}e_i$ for each $i \in \{1,\dots,n\}$. Then $(\tilde e_1,\dots,\tilde e_n)$ is a $\tilde g_p$-orthonormal basis. By the definition of the pointwise norm of a $(0,4)$-tensor, \begin{align*} |\operatorname{Rm}(\tilde g)|_{\tilde g,p}^2=\sum_{i,j,k,l=1}^n\left|\operatorname{Rm}(\tilde g)_p(\tilde e_i,\tilde e_j,\tilde e_k,\tilde e_l)\right|^2. \end{align*} Using multilinearity of $\operatorname{Rm}(\tilde g)$ and $\operatorname{Rm}(\tilde g)=\lambda\operatorname{Rm}(g)$, we obtain \begin{align*} \operatorname{Rm}(\tilde g)_p(\tilde e_i,\tilde e_j,\tilde e_k,\tilde e_l)=\lambda^{-2}\operatorname{Rm}(\tilde g)_p(e_i,e_j,e_k,e_l)=\lambda^{-2}\lambda\operatorname{Rm}(g)_p(e_i,e_j,e_k,e_l)=\lambda^{-1}\operatorname{Rm}(g)_p(e_i,e_j,e_k,e_l). \end{align*} Hence \begin{align*} |\operatorname{Rm}(\tilde g)|_{\tilde g,p}^2=\lambda^{-2}\sum_{i,j,k,l=1}^n\left|\operatorname{Rm}(g)_p(e_i,e_j,e_k,e_l)\right|^2=\lambda^{-2}|\operatorname{Rm}(g)|_{g,p}^2. \end{align*} Taking square roots and using $\lambda>0$ gives \begin{align*} |\operatorname{Rm}(\tilde g)|_{\tilde g,p} = \lambda^{-1}|\operatorname{Rm}(g)|_{g,p}. \end{align*} Since $p \in M$ was arbitrary, the pointwise norm identity holds on all of $M$. This completes the proof.[/step]

[guided]Fix $p \in M$, and choose a $g_p$-orthonormal basis $(e_1,\dots,e_n)$ of $T_pM$. Define $\tilde e_i:=\lambda^{-1/2}e_i$ for each $i \in \{1,\dots,n\}$. Since $\tilde g_p=\lambda g_p$, the basis $(\tilde e_1,\dots,\tilde e_n)$ is orthonormal for $\tilde g_p$. The pointwise norm of a $(0,4)$-tensor is computed by summing the squares of its components in an orthonormal basis for the metric in question. Therefore \begin{align*} |\operatorname{Rm}(\tilde g)|_{\tilde g,p}^2=\sum_{i,j,k,l=1}^n\left|\operatorname{Rm}(\tilde g)_p(\tilde e_i,\tilde e_j,\tilde e_k,\tilde e_l)\right|^2. \end{align*} Now substitute $\tilde e_i=\lambda^{-1/2}e_i$ in each of the four tensor slots. Multilinearity contributes the factor $(\lambda^{-1/2})^4=\lambda^{-2}$, and the curvature tensor identity $\operatorname{Rm}(\tilde g)=\lambda\operatorname{Rm}(g)$ contributes one factor of $\lambda$. Hence \begin{align*} \operatorname{Rm}(\tilde g)_p(\tilde e_i,\tilde e_j,\tilde e_k,\tilde e_l)=\lambda^{-2}\operatorname{Rm}(\tilde g)_p(e_i,e_j,e_k,e_l)=\lambda^{-1}\operatorname{Rm}(g)_p(e_i,e_j,e_k,e_l). \end{align*} Squaring and summing over all four indices gives \begin{align*} |\operatorname{Rm}(\tilde g)|_{\tilde g,p}^2=\lambda^{-2}\sum_{i,j,k,l=1}^n\left|\operatorname{Rm}(g)_p(e_i,e_j,e_k,e_l)\right|^2=\lambda^{-2}|\operatorname{Rm}(g)|_{g,p}^2. \end{align*} Because $\lambda>0$, taking square roots gives \begin{align*} |\operatorname{Rm}(\tilde g)|_{\tilde g,p}=\lambda^{-1}|\operatorname{Rm}(g)|_{g,p}. \end{align*} Since $p$ was arbitrary, this proves the pointwise norm identity on all of $M$.[/guided]