Consider a Chern-Simons system with gauge group $G$ and level $k$. Such a system can be used to model anyons, where the latter are identified with the integrable representations of $G$.

One of the most important properties of anyons is that they provide a non-trivial representation of the braiding group. This braiding is encoded in the so-called braiding matrix $B$. For the particular case of $G=\mathrm{SU}(2)$, ref.1 gives the expression for the $B$ matrix: $$ B=q^{T^a_j\otimes T^a_j},\qquad q\equiv \mathrm e^{\frac{4\pi i}{k+2}}\tag{7.74} $$

I have two questions:

  1. First, one of notation. I don't really know what the author means by $T^a_j\otimes T^a_j$, and I'd be glad if someone could translate this into a more precise mathematical language. In particular, given two highest-weight representations of $G_k$, $\hat\mu,\hat\nu$, what is their brading $B_{\hat\mu\hat\nu}\in\mathbb C$?

  2. Second, I'd like to know how the formula above is generalised to an arbitrary group $G$. I think the correct answer is that $q\to\exp\frac{4\pi i}{k+g}$, where $g$ is the dual Coxeter number of $G$, but I haven't been able to find this formula anywhere.


  1. Pachos, J.K. - Introduction To Topological Quantum Computation.

2 Answers 2


The configuration space $\mathcal {M}$ of $N$ non-Abelian anyons on the $2-$ sphere $S^2$ consists of $N$ copies of $S^2$ excluding the collision points and in addition $N$ irreducible representations of a non-Abelian Lie group $G$ corresponding to the highest weight vectors $j_k$, $k= 1, .,.,.,N$.

The corresponding (physical) Hilbert space is given by: $$H = \mathrm{inv}\left(\bigotimes_{k=1}^{N} H^{(j_k)}\right)$$ Where, $H^{(j_k)}$ is the Hilbert space of the $k-$th anyon carrying a representation $j_k$ of a non-Abelian Lie group $G$. ($j_k$ is the highest weight vector of the representation).

The physical Hilbert space includes only the scalar representations in the tensor product . The braiding matrix is a non-Abelian holonomy $B$ of a flat connection (which will be described below) on the trivial bundle $\mathcal {P} = \mathcal {M} \otimes G$, corresponding to a closed path in the configuration space in which one anyon say the $m-$th one encircles a single other anyon say the $n-$th. (The exact trajectory is immaterial since the connection is flat). In this case the braiding matrix $B = q^t$ is given by: (Here I am writing explicitly the short hand notation of Pachos):

$$t = \sum_{a=1}^{\dim(G)} I_1 \otimes . . . \otimes T^a_{j_m}\otimes . . . \otimes T^a_{j_n}\otimes ... \otimes I_k$$

The sum in the braiding matrix is over an orthonormal basis of the Lie algebra $\mathfrak{g}$ of $G$. $I_k$ are unit matrices, so the braiding matrix acts only on the tensor product of the local Hilbert spaces corresponding to the anyons $m$ and $n$.

Please notice that due to fact that $t$ is invariant under $G$, the braiding works within the physical Hilbert space of invariants.

The braiding matrix can be obtained from a quantum mechanical model of the Anyon system. I am following here Oh and Verlinde . A system of particles in $2+1$ dimensions acquires anyon statistics when it is coupled to a Chern-Simons term. In the case of non-Abelian Anyons, the particles need to have internal structure and to be coupled to a non-Abelian Chern Simons term. Oh works with the following theory of non-relativistic Anyons whose action (for $SU(2)$ anyons is given by:

$$ I = \int dt \sum_{k=1}^N \left( \frac{m_k}{2} \dot{q_k}^2 - Q_k^a (\theta_k, \phi_k)\left(A_i^a (q_k, t)\dot{ q_k }^i - A_0^a(q_k, t)\right) +j_k\cos(\theta_k)\dot{\phi_k}\right ) + \quad k \int dt d^2x \epsilon^{\mu \nu \lambda} (A_{\mu }\partial_{\nu} A_{\lambda } + \frac{2}{3} A_{\mu }A_{\nu} A_{\lambda } )$$ This is equation (2) in the article with a slight change of notation. $j_k$ is the spin of the $k-$th anyon; the second term in the action is the non-Abelian Lorentz term interaction. The third term is the symplectic potential of the two-spheres consisting of the classical phase spaces of the spin degrees of freedom. $ Q_k^a$ are the classical non-Abelian charges, which due to the fact that the group is $SU(2)$ are given by the components of the angular momentum (The Poisson algebra of the sphere): $$(Q_k^1, Q_k^2, Q_k^3) = j_k \left(\cos(\phi_k) \sin(\theta_k), \sin(\phi_k) \sin(\theta_k), \cos(\theta_k) \right)$$ The Gauss' law (equations of motion of $A_0^a$) takes the form: $$k F_{12}^a = \sum_{k=1}^N Q_k^a \delta^2(x-x_k)$$ Except at the particle points the gauge field is flat, and the flat connection solving the Gauss Law is given in the holomorphic gauge ($z = x^1+ix^2$) $$A_{z}^a = \sum_{k=1}^N Q_k^a \frac{1}{z-z_k}$$ The charges $ Q_k^a $ quantize into the representation matrices $T_k^a$ of the $\mathfrak{su}(2)$ algebra. The holonomy of the connection becomes: $$B = Pe^{\frac{1}{2 \pi k}\int_{\Gamma} \sum_k dz^k \sum_{l\ne k} dz^l T_k^a \otimes T_l^a \frac{1}{z_k-z_l}}$$ In the two-body case the exponent becomes self commuting over time and the path ordering becomes unnecessary and we get the final result:

$$B = e^{\frac{i}{2k} T_1^a \otimes T_2^a}$$


For your first question, the notation is that $j$ labels the representation of $SU(2)$, while the index $a$ runs over the generators of the algebra, and should be summed. So once you choose a given representation $j$, the corresponding matrix $B$ is a $(2j+1)^2 \times (2j+1)^2$ matrix, and not a scalar as you seem to suggest. In the textbook by Pachos, you have the explicit example $j=1/2$ treated in section 7.3.

For your second question, I think you are right. See for instance page 4 in this paper.

  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files $\endgroup$
    – Qmechanic
    Commented Jun 25, 2018 at 14:35
  • $\begingroup$ Hi @Antoine, thank you for your answer. The case of $\mathrm{SU}(2)$ is rather simple, because representations are indexed by a single number, $j$. But for more general groups, a representation is specified by a tuple of numbers (Dynkin labels). Given two highest-weight representations $\hat\mu,\hat\nu$, their braiding should be given by a $\dim(\hat\mu)\times\dim(\hat\nu)$ matrix, which I denote by $B_{\hat\mu\hat\nu}$ (and, indeed, $\not\in\mathbb C$, that was typo). What is the explicit formula for this matrix? Something of the form $B_{\hat\mu\hat\nu}=q^{R(\hat\mu)\otimes R(\hat\nu)}$? $\endgroup$ Commented Jun 25, 2018 at 21:26

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