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Let us consider abelian and non-abelian 3d quantum Chern-Simons theory path integrals:

  1. abelian Chern-Simons theory on non-spin manifolds --- $$ \int [DA]\exp(i \frac{k}{2\pi} \int_X (A \wedge dA )) $$

  2. abelian Chern-Simons theory on spin manifolds --- $$ \int [DA]\exp(i \frac{k}{4\pi} \int_X (A \wedge dA )) $$

  3. non-abelian Chern-Simons theory --- $$ \int [DA]\exp(i \frac{k}{4\pi} \int_X \mathrm{Tr}_{} (A \wedge dA + \frac{2}{3} A \wedge A \wedge A)) $$ where $A$ takes values in the Lie algebra valued $\mathcal{G}$ 1-form. So does the Tr take the matrix representations in the Lie algebra $\mathcal{G}$.

What are the correct and rigorous ways to argue the quantization of values of $k$?

I think there are three possible helpful ideas:

  • extend 3-manifolds $X$ to 4-manifolds $Y$?

  • large gauge transformation.

  • Use Wess Zumino Witten like terms.

Could any expert demonstrate these line by line?

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  • $\begingroup$ None of the actions you write are well-defined. Once you pick a proper definitions, the action will be valued in $\mathbb{R}/\mathbb{Z}$. The quantization follows accordingly. You can find one definition in the appendix to Freed’s arxiv.org/abs/0808.2507. $\endgroup$ – Aaron Bergman Aug 9 '20 at 3:01
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Our recent paper https://arxiv.org/abs/2008.02613 provides an answer. It reveals that how quantization of chiral central charge and Hall conductance depend on the groundstate degeneracy on Riemannian surfaces. Note that groundstate degeneracy is not even defined for an arbitrary spacetime manifold.

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  • $\begingroup$ Do Riemannian surfaces include all orientable 2d spatial manifolds already? What is the emphasis of the sentence: " groundstate degeneracy is not even defined for an arbitrary spacetime manifold?" $\endgroup$ – annie marie heart Aug 8 '20 at 23:18
  • $\begingroup$ Riemannian surfaces do include all orientable 2d spatial manifolds. Groundstate degeneracy is defined for any Riemannian surfaces, but is not defined for an arbitrary spacetime which is a 3d manifold. In your question, you only have 3d spacetime manifold. $\endgroup$ – Xiao-Gang Wen Aug 9 '20 at 0:15

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