What is the significance of defining the trace of a tensor as $g^{\alpha\beta} R_{\alpha\beta}$ instead of $R_{\alpha\alpha}$ on a Riemannian manifold?

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    $\begingroup$ Invariant under coordinate transformation $\endgroup$ – RoderickLee Jul 19 '16 at 13:36

The components of $\text{Ric}$ transform during coordinate change $x^\mu\mapsto \tilde{x}^\mu$ as $\tilde{R}_{\mu\nu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\nu}R_{\sigma\rho}$. This is just the usual transformation rule for coordinate-components of tensors.

Contracting over the two indices gives $$ \tilde{R}_{\mu\mu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\mu}R_{\sigma\rho}, $$ and there is no way to simplify this expression further, therefore this contraction is not coordinate-independent.

On the other hand for $R^\mu_{\ \ \nu}=g^{\mu\sigma}R_{\sigma\nu}$, the transformation rule is $$ \tilde{R}^\mu_{\ \ \nu}=\frac{\partial\tilde{x}^\mu}{\partial x^\sigma}\frac{\partial x^\rho}{\partial\tilde{x}^\nu}R^\sigma_{\ \ \rho}, $$ and contracting over this gives $$ \tilde{R}=\tilde{R}^\mu_{\ \ \mu}=\frac{\partial\tilde{x}^\mu}{\partial x^\sigma}\frac{\partial x^\rho}{\partial\tilde{x}^\mu}R^\sigma_{\ \ \rho}=\delta^\rho_\sigma R^\sigma_{\ \ \rho}=R^\sigma_{\ \ \sigma}=R. $$ Thatis why contractions may only be performed over mixed indices. The only thing the metric tensor does in that expression is to turn the two lower indices into one upper and one lower index.

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  • $\begingroup$ How does the inability of the simplification of the expression translate to coordinate variance? Intuitively $R_{\mu\mu}$ is the trace of a matrix which is variant, but can this be proved mathematically? $\endgroup$ – Geeth Chandra Apr 13 at 15:44
  • $\begingroup$ @GeethChandra Because the inability to simplify means that $\tilde R_{\mu\mu}\neq R_{\mu\mu}$, which is variance? $\endgroup$ – Bence Racskó Apr 13 at 15:57
  • $\begingroup$ Does that mean $R_{\mu\mu}$ is not a (legal) tensor? Because a(any) tensor at its very core is coordinate invariant. $\endgroup$ – Geeth Chandra Apr 15 at 14:20

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