6
$\begingroup$

I was trying to understand formula(2.21) in Witten's paper "Quantum Field Theory and Jones Polynomial"(link: https://projecteuclid.org/euclid.cmp/1104178138) (Page 360).

There, it was mentioned, the gravitational Chern Simons action:

$$I(g)=\frac{1}{4\pi} \int_M Tr(\omega d \omega + \frac{2}{3} \omega^3)$$

depends on the choice of framing, i.e. trivialization of tangent bundle of M, in the way that $I(g)\rightarrow I(g)+2\pi s$, where s is how many "units" you twisted the framing.

My question is: how to imagine the twist of framing on this 3 manifold, and further compute the number 's' for given two tangent bundle?

A related, and might be more interesting question is about (2.25) of the same paper:

$$Z\rightarrow Z \exp(2\pi i s \frac{c}{24})$$

Which says, if you twist the frame by s unit, the total contribute to the partition function of Chern Simons theory of gauge group G at level k will be a phase proportion to $c/24$, where c is the central charge of the corresponding current algebra.

Again, how to under to understand this formula, e.g., is there a CFT derivation of this phase shift?

I also post this question in physics stack exchange: https://physics.stackexchange.com/questions/191362/choice-of-framing-in-gravitational-chern-simons

For mathematicians, I want to mention that $\Omega^{fr}_3=\mathbb{Z}/24$, is there any hint that this 24 might related to the $1/24$ in the above formula?

$\endgroup$
1
  • $\begingroup$ I was told the generator of $\Omega^{fr}_3$ is a $S^3=SU(2)$ with Lie group framing on it, so I am wondering how to see 24 copy of it is null-cobordant? $\endgroup$ – Yingfei Gu Jun 26 '15 at 3:06

Your Answer

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

Browse other questions tagged or ask your own question.