# $SU(2)$ and $SO(3)$ WZW models

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.

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 $$SO(3)$$.)
• No I'm considering it for $k=1$ in both cases. Sorry for not making that clear in the notation of my post. I won't downvote but I don't think your answer is correct. The formula for central charge differs between the $SU(N)$ and $SO(N)$ cases. I think you are showing here the formula for $SU(N)$. Witten's original paper on the subject treats $SO(N)$ and shows the level $k=1$ to be equivalent to $N$ free Majorana fermions, thus for $N=3$ it is $3/2$. Aug 26 '19 at 22:08
• FWIW, $c(G_k)=\frac{k}{k+h}\dim(\mathfrak g)$, where $h=N$ for $SU(N)$ and $h=N-2$ for $SO(N)$. Therefore, $c(SU(2)_k)=\frac{3k}{k+2}$ and $c(SO(3)_k)=\frac{3k}{k+1}$. There are some low-rank exceptions though: e.g., $SO(3)$ doesn't really "exist" as a group, and it is customary to define $SO(3)_k:=PSU(2)_{2k}$. Note that the level has a different normalisation. So in this case, $c(SO(3)_k)=\frac{2k}{2k+2}(2^2-1)\equiv\frac{3k}{k+1}$, as expected. So I believe I agree with octonion: your central charge doesn't seem to be correct for $SO(3)$. Aug 26 '19 at 22:32
• The problem seems to be that we have different definitions of the level. I am using the definition of di Francesco et al, who state (page 620) that the level has to be an even integer for $SO(3)$, so your case should be $k=2$ by that definition. Anyway, the orginal question was about two models whose central charges differ, and whose symmetry algebras must therefore also differ. Aug 27 '19 at 10:10
• Ok, I see what you are getting at. The volume of the Lie group is half as big in the $SO(3)$ case so keeping the same normalization in the action the level needs to be restricted to even integers. Thanks for your answer Aug 27 '19 at 22:22