# Trace of the square of the metric tensor

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.
• 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 $R=n\lambda$ Mar 27 at 14:08
• No. How do you think $g_{ij}^2$ defines a tensor? 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. $$S(X,Y)=g(X,Y)g(X,Y)$$.