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Sorry if this is a silly question. I recently read the post: Relating the Ricci curvatures of two Riemannian metrics and wonder if there are any ways to relate the scalar curvature of two Riemannian metrics?

I tried to take the trace of the following equation $$R_{ij}=\bar R_{ij}+(\bar \nabla_kW_{ij}^k-\bar\nabla_jW_{ki}^k)+(W_{kl}^kW_{ij}^l-W_{jl}^kW_{ik}^l)$$ i.e., $$R_g=g^{ij}R_{ij}=g^{ij}\bar R_{ij}+g^{ij}(\bar \nabla_kW_{ij}^k-\bar\nabla_jW_{ki}^k)+g^{ij}(W_{kl}^kW_{ij}^l-W_{jl}^kW_{ik}^l).$$ But I am not sure how to proceed from here. Are there any well-known relation between two scalar curvatures with respect to two different metrics $g$ and $\bar g$ on the same manifold $M$?

ldgo
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    I don't think a nice formula exist for general $g,\bar{g}$. On the other hand, if $g,\bar{g}$ are conformal then you have a conformal frame that helps a lot with finally contracting $i,j$ with respect to both $g$ and $\bar{g}$ (or computing the difference of their Christoffel symbols). – user10354138 Oct 13 '23 at 15:12
  • @user10354138 Thank you for your answer! I don't know if I should ask a new question in a new post or just comment on it here. Here is another situation: is it possible to relate the scalar curvatures between two manifolds, say, $(M,g)$ and $(N,h)$ in general? Suppose $f:(M,g)\to (N,h)$ is a diffeomorphism, are the scalar curvatures related somehow? – ldgo Oct 13 '23 at 15:19
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    In general no. For example, Aubin (1970) proved that in dimension $\geq 3$ every closed manifold admits a negative scalar curvature metric. – user10354138 Oct 13 '23 at 15:29
  • @user10354138 Thank you very much for your help! Much appreciated! – ldgo Oct 13 '23 at 15:42

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