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Normalization of the Chern-Simons level in $SO(N)$ gauge theory

+ 8 like - 0 dislike
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In a 3d SU(N) gauge theory with action $\frac{k}{4\pi} \int \mathrm{Tr} (A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$, where the generators are normalized to $\mathrm{Tr}(T^a T^b)=\frac{1}{2}\delta^{ab}$, it is well known the Chern-Simons level $k$ is quantized to integer values, i.e. $k \in \mathbb{Z}$.

My question is about the analogous quantization in $SO(N)$ gauge theories (A more standard normalization in this case would be $\mathrm{Tr}(T^a T^b)=2\delta^{ab}$ ). Some related subtleties are discussed in a (rather difficult) paper by Dijkgraaf and Witten Topological Gauge Theories and Group Cohomology, but I am not sure about the bottom line.

Does anyone know how to properly normalize the Chern-Simons term in $SO(N)$ gauge theories, or know a reference where this is explained?

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Chern
asked Oct 30, 2011 in Theoretical Physics by anonymous [ no revision ]

2 Answers

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Let me normalize the action as $$S=\frac{k}{4\pi}\int\langle A\wedge dA + \frac{1}{3} A\wedge[A\wedge A]\rangle$$ for $\langle,\rangle$ being the Killing form. This coincides with your normalization for $SU(N)$.

Variation of the Chern-Simons action under a gauge transformation $g:M\rightarrow G$ is given by $$S\rightarrow S + \frac{k}{24\pi}\int_{g_*[M]} \langle\theta\wedge[\theta\wedge\theta]\rangle,$$ where $\theta\in\Omega^1(G;\mathfrak{g})$ is the Maurer-Cartan form (Proposition 2.3 in http://arxiv.org/abs/hep-th/9206021). The last term is also called the Wess-Zumino term. Therefore, $\exp(iS)$ is invariant if $$\frac{k}{24\pi}\int_{[C]} \langle\theta\wedge[\theta\wedge\theta]\rangle\in2\pi\mathbf{Z}$$ for $[C]$ the generator of $H_3(G;\mathbf{Z})$.

For $G=SO(N)$, the homology is generated by $SO(3)\subset SO(N)$, and that term can be computed as follows. As you say, $$\frac{1}{24\pi}\int_{SU(2)} \langle\theta\wedge[\theta\wedge\theta]\rangle=2\pi,$$ but $SU(2)\rightarrow SO(3)$ is a 2:1 local diffeomorphism, so $$\frac{1}{24\pi}\int_{SO(3)} \langle\theta\wedge[\theta\wedge\theta]\rangle=\pi.$$

Therefore, the level $k$ in this case has to be even. See also appendix 15.A in the conformal field theory book by Di Francesco, Mathieu and Senechal.

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Pavel Safronov
answered Oct 30, 2011 by anonymous [ no revision ]
Right... Here you are carefully distinguishing e.g. $SO(3)=SU(2)/Z_2$ and $SU(2)$, right? They have different normalizations i.e. different allowed $k$, don't they?

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Luboš Motl
That is correct. Moreover, if you want to consider nontrivial bundles (and write corresponding actions), you have to consider $k$ divisible by 4. This statement appears in Dijkgraaf-Witten, section 4.3.

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Pavel Safronov
@Pavel: We are struggling to understand why "k is divisible by 4" as mentioned in Dijkgraaf-Witten. Could you kindly expand/explain your statement "if you want to consider nontrivial bundles (and write corresponding actions), you have to consider k divisible by 4"?

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Xiao-Gang Wen
+ 1 like - 0 dislike

Can we simply comment that the $\text{Tr}[T^a_rT^b_r]\equiv C(r) \delta^{ab}$ depends on the representation. For the case of SU(2) and SO(3), we can relate this to the spin-S representation. By the manner that for SU(2) group is in a spin 1/2 representation and SO(3) group is in a spin 1 representation. One can write down the relation of spin operators as: $$ S_x^2+S_y^2+S_z^2=S(S+1) \;\mathbb{I}_{2s+1}. $$ ($\hbar=1$). And $$ \sum_a(S^a)^2=S_x^2+S_y^2+S_z^2=\sum_{a=x,y,z}(T^a)^2=3 (T^b)^2 $$ here $b$ can be $x,y,z$. So, combine the two relations above: $$ \frac{1}{2}\text{Tr}[T^a_rT^b_r]=\frac{1}{2}\frac{S(S+1)}{3}\text{Tr}[\mathbb{I}_{2s+1}]=\frac{S(S+1)(2S+1)}{6} $$ For SU(2), spin-1/2 representation, we have: $$ \text{Tr}[T^a_rT^b_r]=2\frac{S(S+1)(2S+1)}{6}|_{S=1/2}=1/2 $$ For SO(3), spin-1 representation, we have: $$ \text{Tr}[T^a_rT^b_r]=2\frac{S(S+1)(2S+1)}{6}|_{S=1}=2 $$.

For spin-3/2 representation, we have: $$ \text{Tr}[T^a_rT^b_r]=2\frac{S(S+1)(2S+1)}{6}|_{S=3/2}=5, $$ etc. Shall we say the level k quantization of SU(2) C-S and SO(3) C-S are thus related by a factor of: $$(1/2)/2=1/4.$$

And this quantization value presumably is a measurable quantized value for the spin-Hall conductance. See for example, the discussion in this paper: Symmetry-protected topological phases with charge and spin symmetries: response theory and dynamical gauge theory in 2D, 3D and the surface of 3D: arXiv-1306.3695v2, in Eq(26) and its p.7 right column and in p.8 left column. See also this Phys Rev B paper.

This way of interpretation simplifies Dijkgraaf-Witten or Moore-Seiberg's mathematical argument to a very physical level of the spin $S$ property. Would you agree?

Any further thoughts/comments?

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Idear
answered Dec 14, 2013 by wonderich (1,400 points) [ no revision ]

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