Purpose of the Einstein Tensor I was reading "Tensor Analysis and Elementary Differential Geometry for Physicists and Engineers" by Hung Nguyen-Schäfer and Jan-Philip Schmidt and in the chapter 2 named "Tensor Analysis" they introduce the Einstein Tensor:

$$ G_j^i \equiv R_j^i - \frac 1 2 \delta_j^i R$$

With $R_j^i$ the second-kind Ricci tensor, $\delta_j^i$ the usual and friendly Kronecker tensor and $R$ the Ricci curvature.
Then, few properties of this tensor were shown on this subsection dedicated to this tensor:


*

*Symmetry of the Einstein tensor

*Divergence of the Einstein tensor is always 0


The chapter ends with the second property with:

This result is very important and has been often used in the general relativity theories and other relativity fields.

This leads me into thinking that the property of 0-divergence was the goal of this construction.
My questions are:


*

*Is the purpose of the Einstein tensor to have a linear combination of Ricci Tensor and it being a symmetric tensor so that it has the divergence equal to zero? Or is it for another purpose and the null divergence is just a handy property that has nothing to do with the construction of this tensor? To a greater extent, what were the objectives of the Einstein tensor?

*I am unsure if this question is more physics or mathematics, but why is it " used in the general relativity theories and other relativity fields" to have a tensor with a divergence of zero? For a vector field, it would mean that the "quantity" that goes inside the infinitesimal domain equals the "quantity" that exits the same domain. However, I am not sure what it means for a tensor or if it's useful for certain results in mathematics and physics.
Please apologize me if this belongs more in physics SE.
 A: General relativity combines differential geometry with a stationary action principle that gives rise to gravity. The action $S$ is the definite time integral of the Lagrangian $L$, which in turn is the definite space integral of the Lagrangian density $\mathcal{L}$, which in turn is $\sqrt{|g|}$ times a scalar sometimes called the scalar Lagrangian density, where $g$ is the determinant of the index-lowering metric tensor. (The terminology is a bit confusing because $\mathcal{L}$ is a weight-1 "scalar density", whereas the SLD is a true scalar, i.e. is weight-$0$.)
To get general relativity we start with an action in which gravity is "turned off", and the scalar Lagrangian density depends only on the matter fields. For example, a universe with one matter field, a minimally coupled real scalar field $\phi$ of mass $M$, would have $\mathcal{L}=\frac12\sqrt{|g|}(\partial^a\phi\partial_a\phi-(M^2+\xi R)\phi^2)$ for some parameter $\xi\in\Bbb R$. Note that $R$ is allowed to appear in this result, thereby introducing a dependence on the metric tensor and its first two derivatives, and we scale by $\sqrt{|g|}$ at the end anyway. Let's call the no-gravity action $S_\text{matter}$. The Euler-Lagrange equation we get by varying with respect to $g^{ab}$ can be written as $T_{ab}=0$, where in terms of functional derivatives the stress-energy tensor $T_{ab}:=\frac{-2}{\sqrt{|g|}}\frac{\delta S_\text{matter}}{\delta g^{ab}}$.
To turn gravity on, we add a new term to $\mathcal{L}$, namely $\dfrac{\sqrt{|g|}}{2\kappa}(R-2\Lambda)$, with $\Lambda$ the cosmological constant and $\kappa=\frac{8\pi G}{c^4}$ if spacetime is $4$-dimensional. We have added the Einstein-Hilbert action to proceedings. A fairly lengthy derivation shows that, in the $\Lambda=0$ case, $\frac{2}{\sqrt{g}}\frac{\delta S_\text{EH}}{\delta g^{ab}}$ is $\kappa^{-1}$ times the Einstein tensor, so the equation of motion obtained from varying $g^{ab}$ is $G_{ab}=\kappa T_{ab}$. More generally, we get $G_{ab}+\Lambda g_{ab}=\kappa T_{ab}$. This equation has geometry on the left-hand side and matter on the right-hand side; it explains how space tells matter how to move, and how matter tells space how to curve. (You can interpret the $\Lambda g_{ab}$ term as material by moving it to the right-hand side, but that's probably best left for another time.)
If $X^{ab}$ is symmetric and divergenceless, and $\xi_b$ is a vector field for which $\nabla_a\xi_b$ is antisymmetric (we say $\xi_b$ is a Killing vector), $$\nabla_a(X^{ab}\xi_b)=X^{ab}\nabla_a\xi_b=0,$$so $X^{ab}\xi_b$ is a conserved current. KVs are notable for providing a vector space of such currents, whose dimension is proportional to the dimension of the vector space of KVs (which is finite, but that's not our topic today). But divergenceless symmetric rank-$2$ tensors are the other ingredient for generating such symmetries, and $G_{ab}$ is one such tensor. But Lovelock proved something interesting: in 4D, it's the only divergenceless rank-2-tensor-valued function of $g_{ab}$ and at most its first two derivatives. That, in a way, is a hint at its role in any Lagrangian account of how geometry could beget gravity. (Don't try going to even higher-order derivatives in a Lagrangian field theory; it doesn't end well.)
As an added bonus, $T_{ab}$ is therefore also a symmetric divergenceless tensor on-shell, because $\Lambda$ is spacetime-constant. This means $T_{ab}\xi^b$ is conserved for any KV $\xi^b$. This ensures local four-momentum conservation.
