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 motion from the Euler-Lagrange equations, but I don't know what the Euler-Lagrange equations look like in this notation.


You can write this action in components if you want, and then proceed as you are used to.

The Hodge star here is a two-dimensional Hodge star (because $\Sigma$ is two-dimensional, so $* dX^j$ must be a one-form). Remember that for any function $f$ on the surface $\Sigma$, if you choose coordinates $(\sigma^1,\sigma^2)$ on $\Sigma$, then you can write $$df = (\partial_a f) d \sigma^a $$ where $a=1,2$. You can do it here for the functions $X^i$: $$dX^i = (\partial_a X^i) d \sigma^a $$

Finally, the Hodge star is obtained using the totally antisymmetric tensor $\epsilon_{ab}$: $$*dX^j = \epsilon_{ab} (\partial^b X^j) d \sigma^a $$

So your action reads $$S[X] ~=~ \frac{1}{2}\int_{\Sigma} g_{ij} (X) \, (\partial_a X^i) \epsilon_{cb} (\partial^b X^j) d \sigma^a \wedge d \sigma^c$$ Using the volume form $\omega = \epsilon_{ac} d \sigma^a \wedge d \sigma^c$, this reduces to $$S[X] ~=~ \frac{1}{2}\int_{\Sigma} g_{ij} (X) \, \epsilon^{ab} (\partial_a X^i) (\partial_b X^j)\omega $$ which probably sounds familiar. Note that I have not been careful about factors of $2$ or $1/2$ that may appear depending on the normalization you use for the tensor $\epsilon^{ab}$.

Note also that a more concise way to get rid of the Hodge star is to use the abstract definition, which gives immediately $$dX^i \wedge \star \,dX^j = \langle dX^i , dX^j \rangle \omega . $$

  • $\begingroup$ Brilliant! Thanks! I guess if $\Sigma$ is Lorentzian then that changes the form of $\star dX^j$? At the moment I care about Euclidean signature, but I'd like to understand this better. $\endgroup$ – Mark B Oct 20 '16 at 19:44
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    $\begingroup$ Yes, the definition of the Hodge star depends on the symmetric bilinear form on the variety, through the definition $\alpha \wedge \star d \beta = \langle \alpha , \beta \rangle \omega$. So if the signature in the orthonormal basis $(e_1,e_2)$ is $(-,+)$ you have $e_1 \wedge \star e_1 = - e_1 \wedge e_2$ (here we have made a choice of orientation for the volume form $\omega = e_1 \wedge e_2$), hence $\star e_1 = - e_2$. You can also show that $\star e_2 = - e_1$. $\endgroup$ – Antoine Oct 21 '16 at 8:23
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    $\begingroup$ For comparison, if the signature is $(+,+)$ then $\star e_1 = e_2$ and $\star e_2 = - e_1$. You see that : for Euclidean signature, $\star^2$ applied on 1-forms is $-1$, while for Lorentzian signature, $\star^2$ applied on 1-forms is $+1$. For the general result, see en.wikipedia.org/wiki/Hodge_dual#Duality . $\endgroup$ – Antoine Oct 21 '16 at 8:26

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