Constant metric rescaling preserves the connection and Ricci tensor and inversely scales scalar and sectional curvature

(Symbols of the rescaled metric.) In a chart, the matrix of ${\gamma }_{\lambda }$ is $\lambda ({\gamma }_{ij})$ with inverse ${\lambda }^{-1}({\gamma }^{ij})$, and its coordinate derivatives are ${\mathrm{\partial }}_i(\lambda {\gamma }_{j\mathrm{\ell }})=\lambda {\mathrm{\partial }}_i{\gamma }_{j\mathrm{\ell }}$. Substituting into [L1], ${\mathrm{\Gamma }}_{\lambda }{}^k{}_{ij}=\frac{1}{2}{\lambda }^{-1}{\gamma }^{k\mathrm{\ell }}\lambda ({\mathrm{\partial }}_i{\gamma }_{j\mathrm{\ell }}+{\mathrm{\partial }}_j{\gamma }_{i\mathrm{\ell }}-{\mathrm{\partial }}_{\mathrm{\ell }}{\gamma }_{ij})={\mathrm{\Gamma }}^k{}_{ij}$: the symbols of ${\gamma }_{\lambda }$ and $\gamma$ agree in every chart.

(Connection equality.) Fix a chart. By [L2] a vector field on its domain is ${\sum }_i{f}^i{\mathrm{\partial }}_i$ with smooth $f^i$, and the symbols describe the connection on coordinate fields, ${\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_j={\sum }_k{\mathrm{\Gamma }}^k{}_{ij}{\mathrm{\partial }}_k$. Since $\mathrm{\nabla }$ is function-linear in its first argument and satisfies the Leibniz rule in its second, ${\mathrm{\nabla }}_{X}Y=X^a{\mathrm{\partial }}_a(Y^b){\mathrm{\partial }}_b+X^a{Y}^b{\mathrm{\nabla }}_{{\mathrm{\partial }}_a}{\mathrm{\partial }}_b$ for smooth fields; hence two affine connections with the same symbols in every chart agree on all smooth vector fields. By step 1.1 the Levi-Civita connections of ${\gamma }_{\lambda }$ and $\gamma$ have the same symbols in every chart, so they are equal.

(Curvature operator equality.) The curvature operator is defined from the connection alone by $R(X,Y)Z={\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_{[X,Y]}Z$; substituting the common connection of step 2.1 gives $R_{\lambda }=R$ on smooth fields. Equivalently, the coordinate formula [L3] expresses the components through the common symbols and their derivatives, so the components agree in every chart and [L3] makes the operators equal as $(1,3)$-tensors.

(Ricci equality.) By the definition recalled in [L4], $\mathrm{Ric}(X,Y)$ is the trace of the endomorphism $Z\mapsto R(Z,X)Y$ of $T_p\mathrm{\Sigma }$; this trace is computed from the curvature endomorphism and the vector-space structure alone and uses no metric. Since $R_{\lambda }=R$ by step 3.1, the endomorphisms coincide at every point and so do their traces: $\mathrm{Ric}({\gamma }_{\lambda })=\mathrm{Ric}(\gamma )$. In particular no factor of $\lambda$ appears.

(Lowered curvature, scalar curvature and sectional curvature.) At a point, ${\mathrm{Rm}}_{{\gamma }_{\lambda }}(X,Y,Z,W)=(\lambda \gamma )(R_{\lambda }(X,Y)Z,W)=\lambda \gamma (R(X,Y)Z,W)=\lambda {\mathrm{Rm}}_{\gamma }(X,Y,Z,W)$ by [L4], step 3.1 and bilinearity of $\gamma$. Taking the metric trace of step 4.1, $S({\gamma }_{\lambda })={\sum }_{i,j}(\lambda \gamma {)}^{ij}\mathrm{Ric}({\gamma }_{\lambda }{)}_{ij}={\lambda }^{-1}{\sum }_{i,j}{\gamma }^{ij}\mathrm{Ric}(\gamma {)}_{ij}={\lambda }^{-1}S(\gamma )$, since the inverse metric of $\lambda \gamma$ is ${\lambda }^{-1}{\gamma }^{-1}$. Finally, for an independent pair $(u,v)$ the numerator of the sectional curvature of ${\gamma }_{\lambda }$ is ${\mathrm{Rm}}_{{\gamma }_{\lambda }}(u,v,v,u)=\lambda {\mathrm{Rm}}_{\gamma }(u,v,v,u)$ while the Gram determinant of $(u,v)$ with respect to ${\gamma }_{\lambda }$ is ${\lambda }^2$ times the Gram determinant with respect to $\gamma$; the ratio is therefore multiplied by ${\lambda }^{-1}$, $K_{{\gamma }_{\lambda }}(\sigma )={\lambda }^{-1}{K}_{\gamma }(\sigma )$. This proves the four claims.