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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

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


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.

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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:

enter image description here

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. ;-)

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Right, in particular you don't really need the Lagrangian for pretty much anything in these models. –  user566 Mar 8 '11 at 5:55
Good reminder. And even when the Lagrangian may be written, like for the free boson, there often exist the operators and effects - like the $\exp(ik\cdot X)$ operator here - whose behavior is very different from the classical counterpart. In particular, the classical dimension of the exponential would be zero. But quantum mechanically, it's one: a big difference. This difference means that the classical approximation - and therefore the Lagrangian approach itself - is not enough to see all important operators and effects. –  Luboš Motl Mar 8 '11 at 6:15
However, I would probably agree that the non-Abelian orbifolds (like the icosahedron's symmetry) of this model are more doable in the WZW formalism. –  Luboš Motl Mar 8 '11 at 6:24
@Luboš Motl: Thanks for your clarifications. In WZW models like the SUˉ(2)k there is always a truncation of the spectrum, i.e, j≤k/2, where j is the spin of reps. if the c=1 Boson is equivalent to a k=1 WZW, then how can I see the truncation of the spectrum in the Boson side? Also, SUˉ(2)1 describes the propagation of a string on the group manifold of SU(2) which is S3 with a radius ≈1 and the above truncation means there is a bound on the internal energy of a string moving on S3, right? What is the equivalence of this in the Boson side? Thanks. –  Axion Mar 9 '11 at 3:01
Dear Axion, first, your last question: there is no bound on energy of a particle or string moving on a compact manifold. Obviously, the motion may always be faster and faster, which means more energy. In the first question, the truncation of the spectrum, you probably mean that the Fourier modes of the $j^\pm(\sigma)$ current are truncated - which means that $:(j^\pm)^l:$ vanishes for $l>k$. This fact may be derived in any realization of the SU(2) current algebra. At any rate, you may check the spectra level-by-level to see that the two theories are equivalent. –  Luboš Motl Mar 9 '11 at 7:16

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