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

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 $$dX^i \wedge \star \,dX^j = \langle dX^i , dX^j \rangle \omega .$$

• 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. – Mark B Oct 20 '16 at 19:44
• 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 $\star e_2 = - e_1$. – Antoine Oct 21 '16 at 8:23
• 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 . – Antoine Oct 21 '16 at 8:26