5
$\begingroup$

With a $SU(N)_k$ or $U(N)_{k,k}$ Chern-Simons term written as $$\mathcal{L} = \frac{k}{4\pi} \left(AdA + \frac{2i}{3} A^3\right)$$ it is often stated that $k\in \mathbb{Z}$. Usually, I see this coming out as a result of flux quantization, though to do this they place the theory on a closed manifold (such as $S^1\times S^2$).

  1. Under what precise conditions is $k\in \mathbb{Z}$ required? If I have an open manifold, say $\mathbb{R}^3$, must $k$ still be quantized?

  2. I have also heard that the particulars of the level quantization depends on if the manifold is spin or non-spin (or $\text{spin}_c$). Why does this matter?

$\endgroup$
1
$\begingroup$

Comments to OP's first subquestion (v3):

  • The Chern-Simons (CS) action $S[A]$ is always invariant under infinitesimal gauge transformations. If we require the CS Boltzmann factor to be invariant under large gauge transformations $g:M\to G$, then the level $k$ has to be quantized. See also e.g. G.V. Dunne, Aspects of Chern-Simons Theory, arXiv:hep-th/9902115, eq. (58).

  • One should ensure that the CS Lagrangian density is integrable, so that the action $S[A]$ is well-defined and finite. This in turn put restrictions on allowed gauge potentials $A$ and allowed gauge transformations $g$, in particular if the 2+1D spacetime $M$ is a non-compact manifold. Typically one would impose that $A$ and $g$ should vanish sufficiently fast at "infinity", i.e. essentially one-point compactify the manifold.

$\endgroup$
0
$\begingroup$

Chern-Simon action is classically not gauge invariant. Under $A_\mu \rightarrow A^g_\mu=g A_\mu g^{-1} - \partial_\mu g g^{-1}$, the Lagrangian transforms as

$\mathcal{L}_{CS} \rightarrow \mathcal{L}_{CS} + \frac{k}{4\pi} \epsilon^{\mu\nu\alpha} \partial_\mu\ Tr(\partial_\nu g g ^{-1} A_\alpha) + \frac{k}{12\pi} \epsilon^{\mu\nu\alpha}\ Tr(g^{-1}\partial_\mu g g^{-1}\partial_\nu g g^{-1}\partial_\alpha g)$.

The second term is a total derivative that vanishes. The last term however, does not vanish. Up to a constant, the integral of this term is called the winding number $\omega(g)$, given by

$\omega(g)=\frac{1}{24\pi^2}\int d^3x\ \epsilon^{\mu\nu\alpha}\ Tr(g^{-1}\partial_\mu g g^{-1}\partial_\nu g g^{-1}\partial_\alpha g)$.

$\omega(g)$ is an integer called the winding number. Now, we can write

$S_{CS}(A) \rightarrow S_{CS}(A^g) = S_{CS}(A) + 2\pi k \omega(g)$.

Chern-Simons action is classically not gauge invariant but it can be made gauge invariant at the quantum level by constraining $k$ to be an integer. In that case, the weight of the path integral $e^{iS_{CS}}$ does not change, thus the theory becomes gauge invariant. The integer $k$ is usually referred to as the "level number" of Chern-Simons theory.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.