# Doubt about the “kernel of Einstein's equations”

From a coordinate-free point of view, we can rewrite the Einstein Field Equations $$R^{\mu}\hspace{0.5mm}_{\nu} - \frac{1}{2}R \delta^{\mu}\hspace{0.5mm}_{\nu}=:G^{\mu}\hspace{0.5mm}_{\nu} = 8\pi T^{\mu}\hspace{0.5mm}_{\nu}$$ like:

$$\textbf{Ric} - \frac{1}{2}S\hspace{0.5mm}\textbf{Id}=: \textbf{G} = 8\pi \textbf{T} \tag{1}$$

Now, suppose we define the following tensor:

$$R^{\mu}\hspace{0.5mm}_{\nu} - \frac{1}{2}R\delta^{\mu}\hspace{0.5mm}_{\nu} - 8\pi T^{\mu}\hspace{0.5mm}_{\nu} \equiv G^{\mu}\hspace{0.5mm}_{\nu} - 8\pi T^{\mu}\hspace{0.5mm}_{\nu} = 0 \tag{2}$$

Then, we can (naively) define:

$$\textbf{Ric} - \frac{1}{2}S\hspace{0.5mm}\textbf{Id} - 8\pi \textbf{T} \equiv \textbf{G} - 8\pi \textbf{T} =: \textbf{E} = \textbf{0} \tag{3}$$

Which is a mixed tensor and in coordinate free notation we have:

$$\textbf{E}: T_{p}M \times (T_{p}M)^{*} \to \mathbb{R}$$

Now, a $$(1,1)-$$tensor defines a Linear Transformation $$[1]$$ and due to that, we can talk about the definition of kernel of linear transformation:

$$Ker(\textbf{E}) =: \Big\{X \in T_{p}M : \textbf{E}(X) = \textbf{0} \Big\}$$

Now, what suppose to mean (physically) the kernel of Einstein's Equations? $$* * *$$

$$[1]$$ ROMANO.A. Classical Mechanics with Mathematica . Birkhäuser. page 26. 2012

• Since $\mathbf E$ is the zero tensor, isn’t its kernel the entire tangent space? If you apply the all-zero linear transformation to any vector, you get zero. – G. Smith Nov 2 '19 at 4:10