I am currently writing an expository paper on gauge theory and gravity and throughout the gauge theory part I have been able to mostly stay away from coordinates unless I wished to
provide specific examples. I am trying to continue this practice through the section on general relativity but am struggling at writing a coordinate free proof of the variation of the Einstein-Hilbert action.


Ideally what I am looking for is a way to take the Einstein-Hilbert action :
$$
S[g]=\int\_M\langle g,Rc\rangle\_g\text{dvol}\_g
$$
(where $\langle\cdot,\cdot\rangle$ is the induced inner product on the bundle of symmetric tensors, and $\text{dvol}\_g$ is the volume form) and then vary it with a section of the symmetric tensor bundle $h$:
$$
\frac{d}{dt}\Big|\_{t=0}S[g+th]=0
$$


A similar question was asked [here](https://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-act) many years ago, and I have tried to follow up on their recommended sources, namely Besse's Einstein Manifolds and I am honestly struggling to figure out how they are obtaining the results they are obtaining as they do not provide many details.


For concrete examples of where I am struggling, take the variation of the volume form. Besse states that:
$$
\begin{align}
 \frac{d}{dt}\Big|\_{t=0}\text{dvol}\_{g+th}=\frac{1}{2}\text{tr}\_g(h)\text{dvol}\_g
\end{align}
$$
and I agree, but I can only get there in coordinates by writing:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{dvol}\_{g+th}=&\frac{d}{dt}\Big|\_{t=0} \sqrt{\det(g+th)}dx^1\wedge \cdots 
\wedge dx^n\\
=& \frac{1}{2}\frac{1}{\sqrt{\det(g)}}\frac{d}{dt}\Big|\_{t=0}\det(g+th)dx^1\wedge \cdots dx^n\\
=&\frac{1}{2}\frac{1}{\sqrt{\det(g)}}\det(g)\text{tr}(g^{-1}h)dx^1\wedge \cdots \wedge dx^n\\
=&\frac{1}{2}\sqrt{\det(g)}g^{ij}h\_{ij}dx^1\wedge \cdots \wedge dx^n\\
=&\frac{1}{2}\text{tr}\_g(h)\text{dvol}\_g
\end{align}
$$


Furthermore, when varying the scalar curvature, $s=\langle g,Rc\rangle\_g$ we writes that this most easily found by noting that $s=\text{tr}\_g(Rc)$ and then applying the product rule:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{tr}\_{g+th}(Rc)=\langle h , Rc\rangle \_g+\text{tr}\_g\left(\frac{d}{dt}\Big|\_{t=0}Rc\_{g+th}\right)
\end{align}
$$
I can see how the second term comes about as the trace of $Rc$ with respect to $g$ is just $g^{-1}\lrcorner Rc$, but the first term is harder for me to come by without coordinates. With coordinates I can write that:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{tr}\_{g+th}(Rc)=&\frac{d}{dt}\Big|\_{t=0}(g+th)^{ij}Rc\_ij\\
=&-g^{ik}h\_{kl}g^{jl}Rc\_{ij}+g^{ij}\frac{d}{dt}\Big|\_{t=0}Rc\_ij\\
=&\langle h , Rc\rangle \_g+\text{tr}\_g\left(\frac{d}{dt}\Big|\_{t=0}Rc\_{g+th}\right)
\end{align}
$$
But is there any way to see this without coordinates? Perhaps some notation that allows one to write the inner product in terms of $g$ without reference to coordinates? I feel like if I can past my initial uneasiness here I'll be able to follow the rest of the argument in the book, albeit with some work, so any help would be appreciated.



Variation of the Einstein Hilbert action in a coordinate-free way
This is really just a long commentary about your question. First, it is always possible to write everything without using coordinates, because the indices can refer to a (moving) frame of tangent vectors. If I understand correctly, your main goal is to not use indices.


I've always preferred index-free formulas over ones with indices. But usually only for the final formula. It is unusual in Riemannian geometry to be able to write full calculations without indices.


If you want to compute the variation of something, call it $\mathrm{Blah}$, with respect to something else, say $\mathrm{Else}$, then you need a precise definition of $\mathrm{Blah}$ with respect to $\mathrm{Else}$. Gauge theory is in a sense less nonlinear than Riemannian geometry. So it is often possible to write rigorous formulas without indices and that can be differentiated with respect to a variation in the connection relatively easily.


However, this simply isn't true in Riemannian geometry, and your two examples demonstrate this well. The concepts of determinant and trace (with respect to a Riemannian metric) are awkward to define rigorously using index-free notation, i.e., without using a basis of tangent vectors.


So before you can even differentiate the volume form without indices, you need a formula for it that does not use indices.


Another awkward issue is when you want to contract a tensor of higher order with a lower order tensor. Compare
$$ g^{jk}\nabla\_jR\_{kl}\,dx^l $$
to
$$ g^{jk}\nabla\_lR\_{jk}\,dx^l. $$
Each is a contraction of the tensors $g^{-1}$ and $\nabla R$. So what notation can you use do distinguish between these two possibilities without using indices?


It might be possible to invent notation that allows you to do calculation without using indices, but I don't know of any successful effort to do this. The closest I know of is [Penrose's abstract index notation](https://en.wikipedia.org/wiki/Abstract_index_notation).


$\newcommand\Hom{\operatorname{Hom}}$ ADDED: You can of course define everything functorially.


In particular, if $T = T\_xM$, then $g \in S^2T^\*$ defines functorially a map $g: T \rightarrow T^\*$, which induces functorially a map $$\det g: \Lambda^nT \rightarrow \Lambda^nT^\*.$$ This means that $\det g \in \Lambda^nT^\*\otimes\Lambda^nT^\*$. If $g$ is positive definite (i.e., $g(v,v) > 0$ for any $v \ne 0$), then for any nonzero $\omega \in \Lambda^nT$, $$(\det g)(\omega,\omega) > 0.$$ Since $\dim \Lambda^nT^\* = 1$, this implies that there exists, unique up to sign, $\omega \in \Lambda^nT^\*$ such that $$\omega\otimes \omega = \det g.$$ Then $$dV\_g = \omega.$$
I'm pretty sure you could do all the variation calculations using this definition of $dV\_g$, but, as far as I can tell, it isn't worth the trouble.


