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 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.

  • 2
    $\begingroup$ You might check Topping's lectures on Ricci flow, there are a lot of derivations of the change in various quantities as the metric varies (in general not just for Ricci flow). See homepages.warwick.ac.uk/~maseq/topping_RF_mar06.pdf section 2.3. I am not sure if your exact concern is addressed there or not. $\endgroup$ Dec 17, 2022 at 4:30

2 Answers 2


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.

$\newcommand\Hom{\operatorname{Hom}}$ 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.


I agree with Deane that it isn't worth it, but nontheless it is possible to make an index-free derivation.

As indicated, the main culprit is the volume form $\mu_g$. This is defined uniquely by $\mu_g(u_1,\dots,u_m)=1$ when $u_1,\dots,u_m\in T_xM$ is a positively oriented orthonormal basis. Let $g_t$ be a smooth deformation of the metric (with $g_0=g$), then we get a smooth family $\mu_{g_t}$ of volume forms. Since volume forms provide a one-element basis for $\Omega^m(M)$, there is a smooth family of smooth functions $f_t$ such that $\mu_{g_t}=f_t\mu_g$. Then we have $ f_t(x)=\mu_{g_t}(u_1,\dots,u_m)$, where the vectors are $g$-orthonormal (rather than $g_t$-orthonormal). Then $\delta\mu_g=\delta f\mu_g$, where $\delta F=(dF_t/dt)|_{t=0}$.

Given the deformation $g_t$ and an orthonormal basis $u_1,\dots,u_m\in T_xM$ of $g$, there is also a (non-unique) family $(u_{1,t},\dots,u_{m,t})$ of orthonormal bases such that $\mu_{g_t}(u_{1,t},\dots,u_{m,t})=1$. Then there is a smooth family $A_t:T_xM\rightarrow T_xM$ of endomorphisms with positive determinant such that $u_{i,t}=A_tu_i$. We have $$ \mu_{g_t}(u_{1,t},\dots,u_{m,t})=\det(A_t)\mu_{g_t}(u_1,\dots,u_m)=1. $$Clearly $A_0=\mathrm{id}$, so differentiating at $t=0$, we get $$0=\delta \det A+\delta\mu_{g}(u_1,\dots,u_m)=\delta \det A+\delta f(x). $$

On the other hand, we have $g_t(A_tu,A_tv)=g(u,v)$ for any $u,v\in T_xM$, so differentiating at $t=0$ we get $$ 0=\delta g(u,v)+g(\delta Au,v)+g(u,\delta Av). $$ Now $\delta A:T_xM\rightarrow T_xM$ is an endomorphism, but let $(\delta A)^\flat:T_xM\otimes T_xM\rightarrow\mathbb R$ be the corresponding bilinear form. What the above equation says is that $$ \delta g(u,v)=-(\delta A)^\flat(u,v)-(\delta A)^\flat(v,u). $$

Then we recall Jacobi's formula, that $\delta\det A=\det(A_{t=0})\mathrm{Tr}\left(A^{-1}_{t=0}\delta A\right)=\mathrm{Tr}(\delta A)$ (this can be proven coordinate-free using exterior algebra). Then the preceding equation says that $$ \mathrm{Tr}_g(\delta g)(x)=-2\mathrm{Tr}(\delta A)=-2\delta \det A=2\delta f(x), $$ so putting things together, we get $$ \delta\mu_g=\frac{1}{2}\mathrm{Tr}_g(\delta g)\mu_g. $$

Now that we have the volume form, the rest of the derivation is fairly easy. If $g_t$ is a smooth deformation, let $\nabla_t$ be the smooth family of Levi-Civita connections. There is then a tensor field $\Delta:\Gamma(TM)\rightarrow\Gamma(\mathrm{End}(TM))$ given by $$ \Delta(X)Y=\frac{d}{dt}\nabla_{t,X}Y|_{t=0}. $$

We can thus construct the variation of the curvature tensor: $$ R_t(X,Y)Z=\nabla_{t,X}\nabla_{t,Y}Z-\nabla_{t,Y}\nabla_{t,X}Z-\nabla_{t,[X,Y]}Z, $$ so differentiating at $t=0$, we get $$ \delta R(X,Y)Z=(\nabla_X\Delta)(Y)Z-(\nabla_Y\Delta)(X)Z. $$

Then since we have $\mathrm{Ric}(X,Y)=\mathrm{Tr}(R(-,X)Y)$, we get $$ \delta\mathrm{Ric}(X,Y)=\mathrm{Tr}\left[(\nabla_{(-)}\Delta)(X)Y-(\nabla_X\Delta)(-)Y\right]. $$

Now the following gets completely nuts without index notation (in my opinion, this is worse than the volume form). We take the metric trace of this, and traces commute with each other as well as covariant derivatives (including metric traces). So let $\mathrm{Tr}_g(\Delta)$ the vector field given by taking the trace on the arguments $X,Y$ in $\Delta(X)Y$, and let $\mathrm{Tr}(\Delta)(X)=\mathrm{Tr}(\Delta(X))=\mathrm{Tr}(\Delta(-)X)$ be the ordinary trace ($\Delta$ is symmetric in the two vector arguments). Then we can write this as $$ \delta\mathrm{Ric}(X,Y)=\mathrm{Div}(\Delta)(X,Y)-\nabla_X[\mathrm{Tr}(\Delta)](Y), $$ where $\mathrm{Div}(\Delta)$ is the covariant divergence taken with respect to the obvious part (i.e. not the vector arguments, but the implicit one-form argument).

We can now take the metric trace as $$ \langle g,\delta\mathrm{Ric}\rangle=\mathrm{Div}(\mathrm{Tr}_g(\Delta))-\mathrm{Div}(\mathrm{Tr}(\Delta)^\sharp), $$ where $\sharp$ is index raising. Either way, this is a divergence, so we have that $\langle g,\delta\mathrm{Ric}\rangle=\mathrm{Div}(Q)$ for some appropriate vector field $Q$.

Putting these together, the Lagrangian is $L=S\mu_g$ ($S=\mathrm{Tr}_g\mathrm{Ric}=\langle g, \mathrm{Ric}\rangle$), so we get $$ \delta L=\delta S\mu_g+S\delta\mu_g=\delta S\mu_g+\frac{1}{2}S\mathrm{Tr}_g(\delta g)\mu_g, $$ and if $g^\ast$ denotes the dual of the metric, we have $S=\langle g^\ast, \mathrm{Ric}\rangle,$ where now the angular brackets denote duality pairing, rather than inner product, hence $$ \delta S=\delta\langle g^\ast,\mathrm{Ric}\rangle=\langle \delta g^\ast,\mathrm{Ric}\rangle+\langle g^\ast,\mathrm{Ric}\rangle=-\langle \delta g,\mathrm{Ric}\rangle_g+\langle g,\delta\mathrm{Ric}\rangle_g \\ =\mathrm{Div}(Q)-\langle \delta g,\mathrm{Ric}\rangle_g. $$

This gives $$ \delta L=\frac{1}{2}S\langle g,\delta g\rangle_g\mu_g-\langle \delta g,\mathrm{Ric}\rangle_g\mu_g +\mathrm{Div}(Q)\mu_g. $$ Throwing away the divergence and "canceling" the contraction with $\delta g$ and the multiplication by the volume, we get that $$E=-\mathrm{Einstein}=\frac{1}{2}Sg-\mathrm{Ric}.$$

Hopefully this computation illustrates why us caveman relativists still use index notation :) . Some notation is probably a bit inconsistent and I skipped some steps. Still, this calculation should nonetheless illustrate that the variation of the EH action can be computed index-free, just it shouldn't be.

  • $\begingroup$ This is brilliant. I had long forgotten about this question, but I am glad to see your answer. Thank you very much $\endgroup$
    – Chris
    Feb 24 at 12:35

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