# Equations of motion for action with differential forms/Hodge star

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I'm trying to compute the equations of motion for an action, but I'm not really familiar with the notation and so I'm not entirely sure what to do. It's a non-linear sigma model, describing maps $X: \Sigma \to M$ where $\Sigma$ is two-dimensional, given by $$S[X] = \frac{1}{2} \int_{\Sigma} g_{ij}(X) \, dX^i \wedge \star dX^j$$ I'm used to seeing actions of the form $S = \int g_{ij}(x) \partial_{\mu} X^i \, \partial^{\mu}X^j$, and then getting the equations of motions from the Euler-Lagrange equations, but I don't know what the Euler-Lagrange equations look like in this notation.

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asked Oct 19, 2016

## 1 Answer

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Lets begin by defining more precisely the objects which we are dealing with there. Let $\phi=(X^\mu):\Sigma\longrightarrow \mathcal M$ an embedding of our world-sheet $\Sigma$ in spacetime $\mathcal M$ with metric $\eta$, $h = \phi^*g\in \mathrm{sec}T^{(0,2)}(\Sigma )$ the metric induced on the world-sheet $\Sigma$ by the background metric $\eta$, and let $\gamma = \mathrm{sec}T^{(0,2)}(\Sigma )$ be a metric on the world-sheet. Observe that $h$ and $\gamma$ are unrelated: the former is induced by the extrinsic geometry, the latter defines the intrinsic world-sheet geometry. The world-sheet is oriented by the 2-form $\omega \in \bigwedge^2 (\Sigma)$ defining our element of volume. The Hodge dual $\star_\gamma$ is then the duality operator define on the oriented pseudo-Riemannian surface $(\Sigma,\gamma,\omega)$ (and not the Hodge dual induced on the bundle of differential forms of the background metric!). We may then define the action functional

$S[X^\mu,\gamma]=\frac{1}{2}\int_\Sigma g_{\mu \nu} [X] dX^\mu \wedge \star_\gamma dX^\nu$

where $(g_{\mu \nu})$ is family of functions living on $\Sigma$ depending on the non-linear sigma model you are dealing with. (For a bosonic string moving in a curved background $\mathcal M$ with curved metric $G_{\mu\nu}$, we let $g_{\mu\nu}=G_{\mu \nu}$. To study the interaction with gravitons, let $G_{\mu\nu}=\eta_{\mu\nu}+\xi_{\mu\nu}$ for a small metric pertubation $\xi$ . See Polchinski 3.7 for details).

Since you only want to know how to study the calculus of variations with this formalism, lets choose the linear sigma model for which $g_{\mu\nu}=\eta_{\mu\nu}$. Observe that from the calculus of exterior differential forms, the component version of this action is actually

$S[X^\mu,\gamma]=\frac{1}{2}\int_\Sigma \eta_{\mu \nu} \gamma(dX^\mu, dX^\nu)\omega =\frac{1}{2}\int_\Sigma \eta_{\mu \nu} \gamma^{ab} \partial _aX^\mu \partial_bX^\nu|\gamma|^{1/2}d\sigma \wedge d\tau$

This is just the Polyakov action. The fact is that $\alpha \wedge \star_\gamma \beta = \gamma(\alpha,\beta)\omega$ for all $\alpha,\beta\in \bigwedge\Sigma$ of the same degree. So, in the enunciation of this question, the author forgot to include the volume element, and the corresponding metric determinant, in the second line of the second paragraph).

To do the variation using geometric formalism, consider the variation $X^\mu \mapsto X^\mu+\bar{\delta}X^\mu$. (The bar on the delta will distinguish the variation symbol to the coderivative, to be introduced below). The action becomes, after using the Leibnitz rule for the exterior derivative, Stokes theorem and ignoring boundary ($\in \partial \Sigma$) terms,

$\bar{\delta}_XS[X,\gamma]=-\int_\Sigma \eta_{\mu \nu} \bar{\delta}X^{\mu}\wedge d \star_\gamma d X^\nu=0$

Recalling that, for surfaces of Lorentz signatures the coderivative is $\delta=-\star_\gamma d \star_\gamma$ we get the equation of motion $\delta d X^\mu=0$. If $\square = d\delta + \delta d$ denotes the Laplace-Beltrami operator on the world-sheet, our equations of motion are just the wave-equations

$\square X^\mu=0$

since $\delta X=0$ for all 0-forms $X\in \bigwedge^0\Sigma$.

To carry the world-sheet metric variation now, I will introduce the notation $\langle T|V \rangle$ for the contraction of the tensors $T,V$ of the same degree, and $\alpha \lrcorner \beta$ for the contraction of the forms $\alpha,\beta$ of the same degree. Observe that by varying the world-sheet metric $\gamma$, one is changing both the volume element $\omega$ and the Hodge star $\star_\gamma$. The variation of terms involving the Hodge dual is not so trivial, and in fact, can be proven (assuming $\mathrm{deg}(\alpha)=\mathrm{deg}(\beta)=1$) to be just

$\bar{\delta}(\alpha \wedge \star_\gamma \beta)=\big\langle \frac{1}{2}(\alpha\lrcorner\beta)\gamma^{-1}-\alpha\otimes\beta\big|\bar{\delta}\gamma\big\rangle\omega$

where by $\gamma^{-1}$ I actually denote the inverse metric of $\gamma$ (namely, whose elements are $(\gamma^{ab})$ in world-sheet coordinates). This formula can be used to perform the variation of the Einstein-Hilbert action in differential forms which I commented in another question here and is based on Thirring's course on mathematical physics. I have also seen it been derived in  Göckeler & Schücker and in this paper. If someone is really interested, I can provide a more easier derivation of this result upon request. Anyway, applying this to our equation gives:

$\bar{\delta}_\gamma S[X,\gamma]=-\int_\Sigma \big\langle \frac{1}{2}(dx^\mu\lrcorner dX_\mu)\gamma^{-1}-dX^\mu\otimes dX_\mu\big|\bar{\delta}\gamma\big\rangle\omega$

and the equation of motion completely fix our (auxiliary) world-sheet metric as

$\frac{1}{2} (dX^\mu\lrcorner dX_\mu) \gamma^{-1}=dX^\mu\otimes dX_\mu$

By recalling that $h_{ab}=\partial_aX^\mu \partial_bX_\mu$ is our induced metric on the world-sheet, the above equation just becomes the well-known formula

$\frac{1}{2}\gamma^{cd}\gamma^{ab}h_{ab}=h^{cd}$

that you may grab your copy of Polchinski and see Eq.(1.2.16) (with slight changes in index positioning). So, to summarize, in differential forms, the equations of motion for the bosonic Polyakov action are just

$\square X^\mu=0 \\ \frac{1}{2} (dX^\mu\lrcorner dX_\mu) \gamma^{-1}=h$

answered Dec 8, 2016 by (550 points)
edited Apr 8, 2019 by Igor Mol

I may be missing something trivial, but it seems that the formula you gave for $\delta\langle\alpha\wedge\star_\gamma\beta\rangle$ only works if $\alpha,\beta\in\Omega^{1}(\mathcal{M})$, as $\langle\alpha\otimes\beta|\delta\gamma\rangle$ is only defined if $\alpha$ and $\beta$ are forms of degree one.

Dear @BobKnighton,

Yes, the equation for the variation of $\alpha\wedge\star_\gamma \beta$ with respect to the metric field $\gamma$ given in my answer is indeed specific to the case where $\mathrm{deg}(\alpha)=\mathrm{deg}(\beta)=1$, which suffices for the solution of the problem raised in this question since here we only need $\alpha=dX^\mu$ and $\beta=dX_\mu$ (where $(X^\mu):\Sigma\longrightarrow \mathcal{M}$ denotes the embedding of the world-sheet into spacetime). I added a parenthesis in my answer to specify this point, thank you.

If you are interested, cf. Eq.(33) of this paper for a general formula of the variation $\bar\delta (\star_\gamma\phi)$ of the Hodge dual $\star_\gamma$ (induced by the metric $\gamma$) of a p-form $\phi$.

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