Let $(M,g)$ be a Riemannian manifold with Ricci curvature $R_{ij}$. After using some conditions, I have the following equation $$R_{ij}=\lambda g^2_{ij},$$ where $g^2_{ij}=g(\partial_i,\partial_j)g(\partial_i,\partial_j)$ is the square of the coefficient of the metric tensor $g$. Now if I take trace in both sides of the above equation, the right hand side becomes scalar curvature but I am not able to find trace of the left hand side. I suppose that it will be $n$ (the dimension of the manifold) but don't have enough confidence.
Please help me. Thank you.

  • 3
    $\begingroup$ If I understand your notations correctly, $(g^2)_{ij}=g_{ik}g_{lj}g^{kl}$ (summation over repeated indices is implied). In this case $(g^2)_i^i=(g^2)_{ij}g^{ij}=g_{ik}g_{lj}g^{kl}g^{ij}=\delta_i^l\delta_l^i=\delta_i^i=n$ and $R=n\lambda$ $\endgroup$
    – Svyatoslav
    Mar 27 at 14:08
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    $\begingroup$ The right side has to be a tensor. So Syvatoslav’s interpretation is the most reasonable one. $\endgroup$
    – Deane
    Mar 27 at 14:43
  • $\begingroup$ Thank you very much. $\endgroup$ Mar 27 at 18:28
  • 2
    $\begingroup$ So what @Svyatoslav has suggested is very different from what you wrote in your question. Are you agreeing that you meant to have what he said? You specifically said "the square of the coefficient of the metric tensor," and, as Deane said, that can't make sense. $\endgroup$ Mar 27 at 22:24
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    $\begingroup$ No. How do you think $g_{ij}^2$ defines a tensor? $\endgroup$ Mar 28 at 20:24

I think the comments mislead the OP. In my opinion the only way to make it meaningful is that by $\mathrm{R}_{ij}=\lambda g^2_{ij}$ you probably meant $\mathrm{Ric}=\lambda h$ where $h=(g_{ij})^2dx^i\otimes dx^j$ i.e., $h_{ij}=(g_{ij})^2=g_{ij}\times g_{ij}$ (no sum assumed). In this case its trace is $=\sum_{i}{(g_{ii})^2}=(g_{11})^2+(g_{22})^2+\dots+(g_{nn})^2$ which cannot be simplified further or rewrite it using $tr(g)$.

NOTE: In comments some commenters thought that $g_{ij}^2=(g^2)_{ij}$ and some others considered it as $S=g_{ij}g_{ij}dx^i\otimes dx^j\otimes dx^i\otimes dx^j=g_{11}^2dx^1\otimes dx^1\otimes dx^1\otimes dx^1+g_{12}^2dx^1\otimes dx^2\otimes dx^1\otimes dx^2+\dots$ which is not a tensor if one consider it as rank 2-tensor wrongly i.e. $S(X,Y)=g(X,Y)g(X,Y)$.


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