An intrinsic supergeometric description of the Green–Schwarz supersymmetric action

Question

The Green-Schwarz action is a natural supersymmetric extension of the Polyakov action (with a $B$-field which I will omit in what follows since it is not relevant to the question).

For a morphism $X:\Sigma\to M$ from a worldsheet $X$ endowed with a (pseudo-)Riemannian metric $h$ to a stacetime $M$ endowed with a (pseudo-)Riemannian metric $g$, the Polyakov action is usually written in Physics textbooks as $$ S_P[X,h]=\int_\Sigma d\sigma^2 \sqrt{h} h^{ab}\partial_aX^\mu\partial_bX^\nu g_{\mu\nu} $$ It is very simple to rewrite this in a completely intrinsic way. Namely, the differential of the smooth map $X$ is a morphism $dX:T\Sigma\to X^*TM$ of vector bundles over $\Sigma$, while the metric $h$ induces a canonical volume form $dvol_\Sigma$ and a cometric $\gamma_h$, which is a section of $T\Sigma\otimes T\Sigma$. Finally, $X^*g$ is a morphism $X^*(TM)\otimes X^*(TM)\to \mathbb{R}_\Sigma$ and the Polyakov action is intrinsically given by $$ S_P[X,h]=\int_\Sigma dvol_\Sigma \, X^*g((dX\otimes dX)(\gamma_h)) $$ Equivalently, and possibly more nicely, one can rewrite this in terms of the Hodge star operator $\star_h$ for the metric $h$ as $$ S_P[X,h]=\int_\Sigma X^*g(dX\wedge\star_h dX) $$ The differential $dX$, as an element in $\Omega^1(\Sigma;X^*TM)$ can actually be seen as the pullback via $X$ of the canonical 1-form $E$ in $\Omega^1(M,TM)$ corresponding to the identity morphism of $TM$ over $M$. That is, $dX=X^*(E)$ and the action becomes $$ S_P[X,h]=\int_\Sigma dvol_\Sigma \, X^*(g(E\otimes E))(\gamma_h)= \int_\Sigma X^*g(X^*(E)\wedge\star_h X^*(E)) $$ Written this way, the action is immediately translated in what physicists call the vielbein formalism: if $\{e_1,e_2,\dots\}$ is a local orthonormal frame for $TM$ then $X^*E$ is written locally as $E^\mu_\nu e_\mu dX^\nu$ so that if one adopts the shorthand notation $E^\mu_a$ for $E^\mu_\nu \partial_a X^\mu$ the action reads $$ S_P[X,h]=\int_\Sigma d\sigma^2 \sqrt{h} h^{ab}E_a^\mu E_b^\nu \eta_{\mu\nu} $$ which is another of the forms one often finds the action written in textbooks.

Then the action is promoted to its supersymmetric version simply by replacing the vielbein $E_a^\mu$ with the "supervielbein" $\tilde{E}_a^\mu$ (whatever it is: this is something I have not clearly understood, yet).

My question is: what is the supergeometry behind this?

I would expect one has some supermanifold $\tilde{\Sigma}$ with underlying classical manifold $\Sigma$ and $\tilde{M}$ with underlying classical manifold $M$, as well as a map $\tilde{X}: \tilde{\Sigma}\to \tilde{M}$ coming into play, together with supermetrics $\tilde{h}$ and $\tilde{g}$, but I'm unable to find a reference presenting the action directly in these geometric terms, nor to guess exactly which the precise supergeometrical picture should be from the supervielbein description of the action (the only thing I'm reasonably sure about is that $\tilde{M}$ should be the supermanifold associated with some spinor bundle over $M$).

This post imported from StackExchange MathOverflow at 2015-11-21 09:31 (UTC), posted by SE-user domenico fiorenza

Answer 1

Maybe I have solved: looking at the description of the coordinates above equation (1) in Superstrings in D=10 from supermembranes in D=11 by Duff-Howe-Inami-Stelle, it is manifest that there is no $\tilde{\Sigma}$ or, more precisely, that $\tilde{\Sigma}=\Sigma$ is just an ordinary manifold. This was actually the point most confusing me: what $\tilde{\Sigma}$ should be, and instead it was the most trivial one. Once this is settled, everything else follows easily: one takes as $\tilde{M}$ the supermanifold corresponding to a spinor bundel $\mathbb{S}\to M$ together with its canonical supermetric $\tilde{g}$, and considers a morphism of supermanifolds $X\colon \Sigma\to \tilde{M}$; one denotes by $\tilde{E}$ the canonical 1-form on $\tilde{M}$ with coefficients in $T\tilde{M}$ corresponding o the identity of $T\tilde{M}$ and one is done: the action is $$ S_{GS}[X,h]=\int_\Sigma dvol_\Sigma X^*(\tilde{g}(\tilde{E}\otimes\tilde{E}))(\gamma_h)= \int_\Sigma X^*(\tilde{g})(X^*\tilde{E}\wedge \star_h X^*\tilde{E}) $$ which in super-coordinates precisely reads $$ S_{GS}[X,h]=\int_\Sigma d\sigma^2\sqrt{h} h^{ab} \tilde{E}_a^\mu\tilde{E}_b^\mu \eta_{\mu\nu} $$ Now that I write it I see the answer was the most obvious and I don't understund how this can have been puzzling me so much. I will let the admins decide whether this question and its answer may be worth be left here for the sake of possibly interested readers or it is better to just delete them.

This post imported from StackExchange MathOverflow at 2015-11-21 09:31 (UTC), posted by SE-user domenico fiorenza