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It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional is given by the Einstein field equation (for a statement, see previous question). The standard derivation of this is through Koszul's formulae either in coordinates (for example wikipedia), or in abstract index notation (for example, in Wald's General Relativty), or in coordinate-free notation (for example, as pointed out by Thomas Richard in Besse - Einstein manifold). This approach is mainly algebraic by using the definition in terms of Koszul's formulae and then calculus in various notations. Essentially the derivation is a direct calculation without the need to even mention the manifold.

I am wondering if there is a way to derive/interprete the statement refered to at the beginning using an alternative method which is more geometric, ie. using parallel transport or alike. The criterion for "geometric" being (a) a direct reference to the manifold is necessary; or (b) a picture, at least in principle a mental picture, can be drawn to at least carry the main idea of the derivation (of course, pictures of formulae don't count).

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This seems very hard to do. Can you cite an example of what you're looking for for a simpler variational problem? Also, you could take a look at the book "Gravitation" by Misner, Thorne, and Wheeler. They try very hard to explain everything using pictures. –  Deane Yang Sep 10 '12 at 22:36
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For example, can the equation for a geodesic in a Riemannian manifold be derived geometrically? I'd love to know about this, if it can. –  Deane Yang Sep 10 '12 at 22:37
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It is also not clear to me what you mean by "direct reference to the manifold". To even begin to contemplate Einstein's equation you have to have a smooth manifold with its associated tangent bundle. If the fact that the manifold itself is required for whatever you are computing to be even defined is not sufficient as a reference, can you give an example of an argument where "direct reference to the manifold is necessary"? Also, I don't quite understand the distinction you are trying to draw between parallel transport and connection, can you explain more? –  Willie Wong Sep 11 '12 at 10:54
    
Thank you for the comments. @Deane Yang - I am not that optimistic about it as well. But the situation for the geodesic equation is somewhat different. The equation itself has the Christoffel symbol in it. So I don't think there is much you can do about it. The meaning of the various curvatures appeared here are beyond mere calculation tools. Hilbert action is the total curvature (integration of the 2nd Lipschitz-Killing curvature), which at least for simple objects have a very clear interpretation other than the formulae itself. The Einstein tensor has geometric characterizations too. –  Lizao Li Sep 11 '12 at 15:17
    
But I don't really see how things fit in here. Also, it seems even in MTW - Gravitation they just give a direct derivation. –  Lizao Li Sep 11 '12 at 15:18

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Comment: the following is a somewhat convoluted way of deriving the Euler-Lagrange equation using Clairaut's theorem for the volume functional and some standard, albeit not simpler, variation formulas (all is $C^{\infty}$ in the following). Let $g_{0}$ be a Riemannian metric and let $v$ be a symmetric $2$-tensor. Let $g_{0,s}$ satisfy $\frac{\partial}{\partial s}|_{s=0} g_{0,s}=v$ and $g_{0,0}=g_{0}$. With $g_{0,s}$ as initial data, let $g_{t,s}$, $t\in\lbrack0,\varepsilon)$, solve the Ricci-Hamilton-DeTurck flow $\frac{\partial}{\partial t}g_{t,s}=-2\operatorname{Ric}{}_{g_{t,s} }+\mathcal{L}_{W_{t,s}}g_{t,s}$, where $W_{t,s}=\operatorname{tr}^{1,2} {}_{g_{t,s}}(\nabla_{g_{t,s}}-\nabla_{g_{t,0}})$. Note that $\frac{\partial }{\partial t}g_{t,0}=-2\operatorname{Ric}{}_{g_{t,0}}$. Let $v_{t,s} =\frac{\partial}{\partial s}g_{t,s}$. We have $\frac{\partial^{2}}{\partial s\partial t}|_{s=0}g_{t,s}=\Delta_{L}v_{t,0}$, which equals $\frac {\partial^{2}}{\partial t\partial s}|_{s=0}g_{t,s}=\frac{\partial}{\partial t}v_{t,0}$ (Lichnerowicz Laplacian heat equation). We compute $\frac{\partial }{\partial t}\operatorname{Vol}(g_{t,s})=\frac{1}{2}\int\operatorname{tr} {}_{g_{t,s}}(\frac{\partial}{\partial t}g_{t,s})d\mu_{g_{t,s}}=-\int R_{g_{t,s}}d\mu_{g_{t,s}}$ since $\int\operatorname{tr}{}_{g_{t,s} }(\mathcal{L}_{W_{t,s}}g_{t,s})d\mu_{g_{t,s}}=0$ by the divergence theorem. Now \begin{align*} \frac{\partial}{\partial s}|_{s=0}\int R_{g_{t,s}}d\mu_{g_{t,s}} & =-\frac{\partial^{2}}{\partial s\partial t}|_{s=0}\operatorname{Vol} (g_{t,s})=-\frac{\partial^{2}}{\partial t\partial s}|_{s=0}\operatorname{Vol} (g_{t,s})\\ & =-\frac{1}{2}\frac{\partial}{\partial t}\int\operatorname{tr}{}_{g_{t,0} }(v_{t,0})d\mu_{g_{t,0}}\\ & =\int\langle-\operatorname{Ric}{}_{g_{t,0}}+\frac{g_{t,0}}{2}R_{g_{t,0} },v_{t,0}\rangle_{g_{t,0}}d\mu_{g_{g_{t,0}}} \end{align*} since $\int\operatorname{tr}{}_{g_{t,0}}(\frac{\partial}{\partial t} v_{t,0})d\mu_{g_{t,0}}=\int\operatorname{tr}{}_{g_{t,0}}(\Delta_{L} v_{t,0})d\mu_{g_{t,0}}=\int\Delta_{g_{t,0}}(\operatorname{tr}{}_{g_{t,0} }(v_{t,0}))d\mu_{g_{t,0}}=0$. Finally, take $t=0$.

December 18, 2013. The notion of volume, in various guises, occurs throughout the study of Ricci flow, especially in Perelman's work. Now, per unit increase in scale $t$, the volume form of a metric changes with velocity $\frac {\partial}{\partial t}d\mu=-Rd\mu$. By Clairaut's theorem, the variation of $-Rd\mu$ is equal to the change per unit increase in scale of the variation of the volume form, i.e., $$ \frac{\partial}{\partial t}(\frac{\operatorname{tr}_{g}v}{2}d\mu_{g})=\left( \langle\operatorname{Ric}-\frac{1}{2}Rg,v\rangle+\operatorname{div} (\frac{\nabla\operatorname{tr}v}{2})\right) d\mu. $$

In the $f$-warped or entropy version of this, we have $\frac{\partial }{\partial t}(fe^{-f}d\mu)=(-R-\Delta f)e^{-f}d\mu$ under $\frac{\partial }{\partial t}g=-2(\operatorname{Ric}+\nabla^{2}f)$ and $\frac{\partial f}{\partial t}=-R-\Delta f$. Integrating this yields that $\mathcal{N} \doteqdot\int_{\mathcal{M}}fe^{-f}d\mu$ satisfies $-\frac{d\mathcal{N}} {dt}=\mathcal{F}\doteqdot\int(R+|\nabla f|^{2})e^{-f}d\mu$. If $\frac {\partial}{\partial s}g=v$ and $\frac{\partial f}{\partial s}=\frac {\operatorname{tr}_{g}v}{2}$, then the variation of the energy integrand is \begin{align*} & \frac{\partial}{\partial s}((-R-\Delta f)e^{-f}d\mu)\\ & =\left( (-L(v,\nabla f)+2\langle\operatorname{Ric}+\nabla^{2} f,v\rangle)e^{-f}+\operatorname{div}(e^{-f}\{\frac{\nabla\operatorname{tr} v}{2}-v(\nabla f)\})\right) d\mu\doteqdot A, \end{align*} where $L(v,X)=\operatorname{div}^{2}v+\langle\operatorname{Ric},v\rangle -2\langle\operatorname{div}v,X\rangle+v(X,X)$ is the linear trace Harnack quadratic. On the other hand, $\frac{\partial}{\partial s}(fe^{-f}d\mu )=\frac{\operatorname{tr}_{g}v}{2}e^{-f}d\mu$. So Perelman's version is $\frac{\partial}{\partial t}(\frac{\operatorname{tr}_{g}v}{2}e^{-f}d\mu )=\frac{\partial^{2}}{\partial s\partial t}(fe^{-f}d\mu)=A$, using Clairaut's theorem. Note that integration by parts gives $\int L(v,\nabla f)e^{-f} d\mu=\int\langle\operatorname{Ric}+\nabla^{2}f,v\rangle e^{-f}d\mu$, from which one obtains Perelman's energy variation formula.

In Section 6.2 of arXiv:0211159 Perelman argues that the $\mathcal{W}$-entropy (i.e., $\mathcal{F}$ with scaling) integrand is a warped scalar curvature. So, without scaling (i.e., $\tau$), we would have the correspondences $\mathcal{N}\sim\operatorname{Vol}$ and $\mathcal{F}\sim\int Rd\mu$, which is also clear from taking $f=\operatorname{const}\neq0$ as a special case. However, in 6.2, Perelman's volume is essentially $\int e^{-f}d\mu$, which is constant under the above variations.

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