7

The WZW model on the sphere $S^2$ with group $G$ and level $k$ is described by the action for a $G$-valued field $g : S^2\to G$ (see these notes by Lorenz Eberhardt):

$$S[g]=\dfrac{1}{4\lambda^2}\int_{S^2}d^2z\ \operatorname{tr}(g^{-1}\partial_\mu g \ g^{-1}\partial^\mu g)-\dfrac{ik}{12\pi}\int_Bd^3y\ \epsilon_{\alpha\beta\gamma}\operatorname{tr}(g^{-1}\partial^\alpha g\ g^{-1}\partial^\beta g\ g^{-1}\partial^\gamma g).\tag{1}$$

On the other hand, the Polyakov action for the bosonic string in a background spacetime $(M,G)$ is $$S_P=\dfrac{T}{2}\int d^2\sigma\sqrt{h}\ h^{ab}G_{\mu\nu}\partial_a X^\mu \partial_b X^\nu\tag{2}.$$

Now, I've heard that string theory on ${\rm AdS}_3$ is somehow equivalent to a WZW model with group ${\rm SL}(2,\mathbb{R})$. In particular this is briefly mentioned in the notes I have linked and is discussed by Maldacena and Ooguri in arXiv:0001053. The thing is that I can't see how is the string theory in ${\rm AdS}_3$ a WZW model.

The authors seem to argue that as a manifold ${\rm SL}(2,\mathbb{R})\simeq {\rm AdS}_3$. In that case I do understand that if we consider string theory in ${\rm AdS}_3$ we would have fields in a two-dimensional surface (the worldsheet), taking values in ${\rm AdS}_3$ (the target space), which turns out to be as a manifold the same as the group under consideration.

But at the level of the actions it seems things are quite different. I see no way in which (1) and (2) end up describing the same theory in the worldsheet. And perhaps this is not even the point for this equivalence.

So what is behind this equivalence between string theory in ${\rm AdS}_3$ and the ${\rm SL}(2,\mathbb{R})$ WZW model on the worldsheet?

Qmechanic
  • 220,844
Gold
  • 38,087
  • 19
  • 112
  • 289

1 Answers1

6

  1. Preliminary: $\mathrm{AdS_3}$ can be embedded in $\mathbb R^{2,2}$ subject to the hyperboloid constraint $-\eta_{\mu\nu}x^\mu x^\nu=L^2$, $\eta_{\mu\nu}=\mathrm{diag}(-1,-1,1,1)$. This can additionally be parameterised by $$ \chi=\frac{1}{L^2}\pmatrix{x^0-x^2 & -x^1+x^3 \\ x^1+x^3 & x^0+x^2}\overset!\in\mathrm{SL}(2,\mathbb R)\tag{1} $$ as the hyperboloid constraint imposes $\det\chi=1$.

  2. The flat spacetime Polyakov action can be converted into a $\mathrm U(1)^D$ model: $$ S\sim\int_\Sigma\mathrm d^2\sigma\ \sqrt{-h}h^{\alpha\beta}\eta_{\mu\nu}\partial_\alpha X^\mu\partial_\beta X^\nu \\\sim\int\mathrm d^2\sigma\ \mathrm{Tr}\left(\partial g\ \bar\partial g^{-1}\right) $$ $$= -\int\mathrm d^2\sigma\ \mathrm{Tr}\left(g^{-1}\partial g\ g^{-1}\bar\partial g\right)\tag{2} $$ where $\mathrm U(1)^D\ni g = \exp(t_i X^i), [t_i,t_j]=0, \mathrm{Tr}(t_i t_j)=\eta_{ij}$. Note that $g$ is not the metric, but rather the space of maps from the 2D worldsheet $\Sigma$ into the target (group) manifold.

  3. The action $(2)$ is easy to generalise to non-abelian semisimple Lie algebras, yielding the corresponding metric in the non-linear sigma model. In particular, choosing the (simple) Lie algebra $\mathfrak{sl}(2, \mathbb R)$ results in a non-linear sigma model over $\mathrm{AdS_3}$, from $(1)$. Note that $g^{-1}\partial g\in\Omega^1(M)\otimes\mathfrak{Lie}(G)$, cf. the Maurer-Cartan form. In particular, choosing the Maurer-Cartan form as the connection on $G$ leads to $R=0$, via the Cartan structure equation $\mathrm d\omega+\frac12[\omega, \omega]=0$. But this precisely encodes the constraints imposed upon worldsheet fields by the one-loop beta functions!

  4. However, unlike for the $\mathrm U(1)^D$ model, the group manifold action for the non-linear sigma model does not automatically satisfy the worldsheet beta function constraints for a general connection. Enter the Witten term: $$ S_W=\int_\mathcal{Y} W=\int_\mathcal{Y}\mathrm{Tr}(g^{-1}\mathrm dg\ g^{-1}\mathrm dg\ g^{-1}\mathrm dg),\qquad \partial\mathcal Y=\Sigma \\\Omega^3(\mathcal Y)\ni W\overset!=\mathrm dB\Rightarrow\int_\mathcal{Y} W = \int_\Sigma B\tag{3} $$ This term corresponds to the topological NS-NS B-field.

  5. Ordinarily, this term would have a quantized prefactor in the action, to ensure conformal invariance at the quantum level. However, for $\mathrm{SL}(2, \mathbb R)$, this is unnecessary as the NS-NS three-form field vanishes.

Nihar Karve
  • 8,672