In QFT and the Jones polynomial, Witten claims that it is possible to perform symplectic reduction from the distributional Poisson bracket on the unconstrained phase space to a symplectic structure on the finite-dimensional constrained moduli space of flat connections. How can this be done explicitly?

More details:

Take the Chern-Simons action (with no charges for simplicity) with compact Lie group $G$. Fix the time gauge: $$ A_0^{a} = 0. $$

The action becomes quadratic, and the Gauss constraint reads $$ f = da + a \wedge a = 0,$$

where $a$ is the pull-back of $A$ on the 2d surface $\Sigma$.

Witten argues that in this situation it is more illuminating to first impose the constraint on the classical theory and then quantize. The constrained phase space is just the moduli space $M$ of flat connections on $\Sigma$, which is compact and finite-dimensional.

Then(page 18) Witten says that (quote) "On general grounds, $M$ inherits a symplectic structure from the symplectic structure present on $M_0$",

where $M_0$ is the unconstrained, infinite-dimensional phase space of the field theory, with the distributional Poisson bracket:

$$ \left\{ A_u^a(x), A_v^b(y) \right\} = \frac{4\pi}{k} \varepsilon_{uv} \delta^{ab} \delta^{(2)}(x, y). $$

Question: I don't understand how this could be done. Here's my thoughts on this:

Moduli space of flat connections $M$ can be obtained by first restricting $M_0$ to a space of flat connections, and then to a space of gauge orbits on it (generated by global gauge transformations). To define symplectic form on $M$ an ordinary symplectic reduction can be used, provided that a symplectic form is defined on the space of flat connections. However, I don't know how to explicitly pull-back the symplectic form from the space of all connections to the space of flat connections.

Bonus question: If it is possible to explicitly define symplectic structure on the finite-dimensional compact $M$, is it also possible to obtain the Chern-Simons quantum Hilbert space

$$H_{CS} = \text{Inv}_q \left( R^{\otimes g} \right) $$ (where $g$ is the genus of $\Sigma$, $R$ is an irrep of $SU(2)_q$ corresponding to a single handle of $\Sigma$, and $\text{Inv}_q$ is the invariant part of the tensor product of irreps of the quantum group $SU(2)_q$) by applying the Kontsevich deformation quantization formula and passing to the GNS representation of the $C^{*}$ algebra obtained via deformation quantization?

Why I think it may be possible: flat connections are basically homomorphisms from $\pi_1(\Sigma)$ to $G$. Classical observables are functions over the phase space, hence functions over those homomorphisms and (very roughly) functions over some copies of $G$.

Deformation quantization of this phase space seems closely related to the approach of non-commutative geometry to quantum groups. By deforming the algebra of functions over $G$ we get an "algebra of functions over $G_q$".


2 Answers 2


You can write the symplectic form on the large phase space as

$$\Omega(\delta_i A,\delta_j A) = \frac{k}{4\pi} \int_\Sigma \langle\delta_i A \wedge \delta_j A\rangle.$$

Here $\langle, \rangle$ is the Killing form on $\mathfrak{g}$, where the $\delta_j A$ are valued. This formula is derived by studying the variational problem of the Chern-Simons action on a 3-manifold with boundary $\Sigma$. It's discussed here in a rigorous formulation (pdf) in section 4.2. They also go on to discuss Hamiltonian reduction to the finite dimensional phase space. The idea is that this small phase space is a $G$-quotient of a level set of the moment map $$\mu = \frac{-k}{2\pi} F$$ which generates the action of gauge transformations. You can use this to construct the symplectic form, since now you can simply pull back $\Omega$ to $\mu^{-1}(\epsilon)$ and evaluate it on $G$-invariant vector fields there, which are equivalent to vector fields on the true phase space $\mu^{-1}(\epsilon)/G$. So basically you use the same formula as $\Omega$.

Edit: As requested, if we're studying the moduli space of flat connections on a torus, we can think of this as a pair of commuting elements $g_1, g_2 \in G$ defined up to simultaneous conjugation $(g_1,g_2) \mapsto (hg_1h^{-1},hg_2h^{-1})$. This is a subquotient of $G \times G/G$, but the quotient $G$ acts by conjugation, so it's not the same as $G$, for example with $G = U(1)$ it's $U(1) \times U(1)$.

We can write $g_j = \exp(2\pi i t_j)$ and the connection form $A = t_1 dx_1 + t_2 dx_2$, where $x_1,x_2$ are $2\pi$ periodic coordinates on the torus. This way, $\exp i \int_{j} A = g_j$ and $dA + [A,A] = 0$.

The tangent space at $A$ is given by all 1-forms $\delta A$ such that $A + \epsilon\delta A$ is flat to order $\epsilon^2$. This means $d\delta A + [A,\delta A] = 0$. You can solve this equation using the structure constants of the Lie algebra. Then the symplectic form is evaluated as $\Omega$ above with $\Sigma = T^2$.

For instance, the tangent space at the zero sections is just the space of closed $\mathfrak{g}$-valued 1-forms, and we use the Killing form to evaluate $\Omega$.

  • $\begingroup$ I don't understand your notation. What is $\left< \delta_i A \delta_j A \right>$? (In fact, the only symbol I understand in this expression is $A$ :) ). $\endgroup$ Mar 11, 2018 at 5:29
  • $\begingroup$ Ah that $\langle \rangle$ was superfluous with the symplectic form $\Omega$, which is a 2-form on the space of flat connections, whose tangent space is given by 1-forms $\delta_i A$. I got rid of it. $\endgroup$ Mar 11, 2018 at 19:58
  • $\begingroup$ Thank you for this answer! I have one more question, after which I'm gonna accept your answer. I now understand how symplectic reduction works abstractly, but have no idea of how to do this explicitly (say, for a torus). Could you update your answer with an explicit computation of $\Omega$ reduced to the moduli space on the torus (which is I believe just $G$, since $\pi_1$ is generated by two elements any of which can be sent to an arbitrary element in $G$, and we have a global $G$ invariance to get rid of one copy of $G$)? $\endgroup$ Mar 12, 2018 at 14:46
  • $\begingroup$ @SolenodonParadoxus Now I remembered why the brackets were there! And what you say about the moduli space on the torus is not quite right. It's not isomorphic to G because G doesn't act freely on the pairs of commuting elements. $\endgroup$ Mar 12, 2018 at 19:10
  • $\begingroup$ PS. I don't know how to answer your bonus question because I don't understand deformation quantization in these weird settings very well, but many people have written about obtaining the Chern-Simons Hilbert space from geometric quantization. For instance, T. Kohno has a whole book called CFT and Topology about this. $\endgroup$ Mar 12, 2018 at 20:14

This question has many sub-questions. I'll try to give you brief explanations and good references.

How to perform symplectic reduction in "Quantum mechanics"

Symplectic reduction is basically equivalent to Dirac's first class constraints, i.e., reduction of gauge symmetry. The prototype model used in QFT obtained by symplectic reduction is the $CP^{N}$ model, which is a symplectic reduction of $C^{N+1}$ by $U(1)$:

We start from $N+1$ scalar fields $\phi_i$ with its standard Lagrangian. (Everything can be performed in $0+1$ space-time dimensions i.e., quantum mechanics, however, this is not obligatory, we can work in any dimension). This model has $U(N)$ global symmetry. If we want to reduce the $U(1)$ symmetry $\phi_i \rightarrow e^{i\theta} \phi_i$, we perform two operations:

  • first, we minimally couple the model to a non-dynamical gauge field $A_{\mu}$.

  • Secondly, we observe that the Noether charge of this symmetry is $\phi_i \phi_i$ (Einstein's convention - this is the analog of the Gauss' law in electrodynamics). This charge must be a constant. Thus we add a constraint term to the Lagrangian: $\lambda(\phi_i \phi_i-\mathrm{const.}) $.

Now, we solve the classical equations of motion for the gauge field and the Noether charge constraint and substitute the solutions and voilà, we get the $CP^N$ sigma model.

Please see BKMU, (section 5.1.) for the explicit construction. The solution of the gauge field removes one real parameter and the constraint removes an additional real parameter, thus the resulting model target space is one complex dimension less.

How to perform symplectic reduction and quantize the 2+1 D Chern Simons theory

In the Abelian Chern-Simons theory, (in the Hamiltonian form with $A_0 = 0$). The number of degrees of freedom before the reduction is $2 $ , ($A_1$ and $A_2$) and as above the reduction removes two real degrees of freedom, thus we are left with zero degrees of freedom and this means that even though we started with a field theoretical model, the residual model can be at most quantum mechanical (in $0+1$ dimensions). Said differently, the moduli space is finite dimensional. Thus the above method does not directly work. The same reasoning works also for the non-Abelian case.

What we can do, is to substitute the most general classical solution with the residual quantum mechanical degrees of freedom (i.e., the moduli space), compatible with the boundary conditions. All the other degrees of freedom, but the moduli space, will disappear from the action due to the gauge symmetry (and the Gauss law) and we will reach a quantum mechanical model.

This is what was exactly performed by EMSS, who obtained a lot of exact solutions to the Chern Simons theory with and without sources, in particular they identified the phase space and the resulting Hilbert space. Please see other equivalent methods Bos and Nair and Dunne.

On the origin of the quantum group symmetry

The finite dimensional moduli spaces which emerge from the symplectic reduction of the Chern-Simons theory are symplectic.( In finite number of dimensions, It was proved by Marsden and Weinstein that a symplectic reduction of a symplectic manifold results a symplectic space. In our case where the initial space is infinite dimensional (Chern-Simons), there is no such a general result, but it can be proved case by case). The symplectic form on the reduced theory can be considered as a magnetic field (Closed but non-exact $2-$ form).

It is a general result that the reduced quantum mechanical theory of the Chern-Simons theory always describe the restricted to the lowest Landau level dynamics of a nonrelativistic particle on the moduli space in the presence of a magnetic field equal to $k$ (the level) times a basic symplectic structure (generator of $H^2(M, \mathbb{Z})$) of the moduli space.

The quantum group symmetry is the algebra satisfied by the magnetic finite translation operators, having the form: $$e^{i \alpha^i (p_i-A_i)}$$

Please see the following article by Sato with a concise explanation of this fact.

The solution of the theory on the moduli space, thus for the full Chern-Simons model will be given by a representation of the quantum group. In particular, it fixes the dimension of the Hilber sapce of the theory.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.