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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.

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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.)

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  • $\begingroup$ 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 $3/2$. $\endgroup$
    – octonion
    Commented Aug 26, 2019 at 22:08
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    $\begingroup$ 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 $SO(3)$. $\endgroup$
    – AccidentalFourierTransform
    Commented Aug 26, 2019 at 22:32
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    $\begingroup$ 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. $\endgroup$
    – Sylvain Ribault
    Commented Aug 27, 2019 at 10:10
  • $\begingroup$ 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 $\endgroup$
    – octonion
    Commented Aug 27, 2019 at 22:22

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