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8

A physicist would write your first equation $x^a = x^\mu e_\mu^a$. The notation $x^a$ is invariant in your terminology. The $a$ is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that $x$ is a vector (i.e., rank 1,0 tensor.) Similarly for each $\mu$, $e^a_\mu$ is ...

4

The Ricci tensor is built adding up some of the components of the Riemann tensor, as the definition specifies: $$R_{\mu\nu} = R^{\lambda}_{\ \mu\lambda\nu}$$ The repetition of two indices above and below means that all these components must be summed: $$R^{\lambda}_{\ \mu\lambda\nu} = R^{0}_{\ \mu 0\nu} + \dots + R^{3}_{\ \mu 3\nu}$$ This operation ...

3

Comments to the question (v5): In General Relativity (GR), the notation $x^{\mu}$, $\mu=0,1,2,3,$ usually denotes some (local) coordinates of a (spacetime) manifold $M$. Note that $x^{\mu}$ does in general not transform as a $(1,0)$ (contravariant) tensor in the sense that  x^{\prime \nu}~=~\frac{\partial x^{\prime \nu}}{\partial x^{\mu}} x^{\mu} ...

1

It is a tensor product. (At least it always was when I encountered such notations, I can't speak with authority about SPDC specifically) Let $\mathcal{H}_1$ be the Hilbert space of polarization states for a single photon. Then the space of states for a two photon system is $\mathcal{H}_2 = \mathcal{H}_1 \otimes \mathcal{H}_1$, and the state you consider in ...

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