Orbifolds of the $c =1$ Bosonic theory on a circle

Question

For a $c=1$ Boson on a circle at the self-dual rdius, we get an enhanced gauge symmetry $\hat{SU}(2)_1$. It is said that we can orbifold this model by any finite subgroup of $SU(2)$ since $SU(2)$ is a symmetry of the model. But the Lagrangian of a $c=1$ Boson does not have an $SU(2)$ symmetry even at the self-dual radius which is

$$ L=\sqrt{2}\int d^2 z\ \partial \phi \bar{\partial} \phi $$

right?

I know that at the self-dual radius we get extra marginal operators which close among each other to form the OPE of $\hat{SU}(2)_1$ but is there another form of the above Lagrangian which makes the $SU(2)$ symmetry manifest?

Is the $c=1$ Boson at the self-dual radius equivalent to a level 1 WZW model? Thanks.

Answer 1

Yes, a $c=1$ boson at the self-dual radius is exactly equivalent to the SU(2) WZW model at $k=1$. The latter makes the $SU(2)$ symmetry manifest. When one gets used to this and similar equivalences and to the extra marginal momentum/winding operators in the boson description, the $SU(2)$ symmetry becomes "manifest" in both formalisms. There is always some degree of psychology or subjective judgement in what is "manifest".

If you want to be sure about all the $c=1$ CFTs, pages 261-262 of Joe Polchinski's book, volume I, may be helpful. They approximately look like this:

Special message for Joe: if you wonder why I use an electronic version of the book, it's because Nima Arkani-Hamed borrowed and lost my paper edition. Well, it was actually Volume II that disappeared and I still own this Volume I, but let's ignore those details. ;-)