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It seems that the $SU(2)_1$ and $SO(3)_1$ Wess-Zumino-Witten models are quite different despite the Lie algebras being identical. The $SO(3)_1$ model has central charge 3/2 and is equivalent to 3 free Majorana fermions. The $SU(2)_1$ model has central charge 1, and can be expressed in terms of a compactified free boson (see for instance section 15.6 in Di Francesco et al's CFT textbook).

So unless I'm misunderstanding something, through ordinary bosonization the $SU(2)_1$ model should be equivalent to 2 Majorana fermions and thus equivalent to the $SO(2)_1$ model rather than $SO(3)_1$.

This situation seems very strange to me. Can someone point out where the global difference between $SO(3)$ and $SU(2)$ leads to a loss of a Majorana fermion?

Note that in this related question the brief answers claim the $SU(2)_1$ and $SO(2)_1$ WZW models are not equivalent, but frankly I don't see why that is the case. So perhaps my confusion with that question is related to this one.

octonion
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1 Answers1

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The $G$-WZW model depends not only on the group $G$, but also on a number $k$ called the level. The symmetry algebra is an affine Lie algebra, and it also depends on $k$. Both $SU(2)$ and $SO(3)$ have the same affine Lie algebra, and the central charge is $$ c = \frac{3k}{k+2} $$ where $k\in \mathbb{N}$ for $SU(2)$ and $k\in 2\mathbb{N}$ for $SO(3)$. It seems you are considering the $SU(2)$ WZW model at level $k=1$ (so $c=1$), and the $SO(3)$ WZW model at level $k=2$ (so $c=\frac32$). Their symmetry algebras differ because their levels differ.

Even at the same level, the $SO(3)$ and $SU(2)$ WZW models would differ. They would have the same symmetry algebra, but different spectrums. (Diagonal for $SU(2)$, non-diagonal for $SO(3)$.)

(I am currently trying to improve the Wikipedia page on WZW models. Help and suggestions are welcome.)

Sylvain Ribault
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