# Choice of framing in Gravitational Chern-Simons

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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?

This post imported from StackExchange Physics at 2015-06-26 10:40 (UTC), posted by SE-user Yingfei Gu

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If you have two (spin) framings, the difference between them is an $Spin(3)$ transformation. As you go around the 3-manifold, you get an element of $H_3 Spin(3) = \mathbb{Z}$. This is $s$ that Witten is talking about. So to visualize all this in an example, I would take $X = SU(2)$ starting with the framing determined by the group structure. As far as cobordism is concerned, this is the most intricate of all the 3-manifold framings: twisting a framing by 1 unit means taking a connect sum with this guy. So, try imagining a connect sum of two of these SU(2) things.

There is of course a WZW calculation of the phase shift. Take the SU(2) example and decompose it as two solid balls. Since the Hilbert space associated to a sphere is one dimensional, the partition function is a phase. It is in fact the twisting phase by the connect sum argument above. Now, decomposing it as two solid tori glued by the mapping class group element S, we see this phase is the modular anomaly associated with the S transformation on the CFT on a 2-torus.

answered Jun 27, 2015 by (1,895 points)

Thanks, it really helps. And I have a following up question concerning about $\Omega^{fr}_3=\mathbb{Z}_{24}$. There seems to be a homomorphism from $H_3 Spin(3)$ to $\Omega^{fr}_3$? First I hope to understand why 24 twists are null-cobordant.

Then this cobordism group seems to suggest an independent argument about why there is a shift when change the framing. I was thinking about a Dijgraaf-Witten(treatment of CS with topological group) type of argument:

Now if I have a frame bundle on 3manifold and try to write Gravitational Chern-Simons action assoicated with this, i should make it well defined by finding a 4manifold bounding it, carrying the frame consist with it at boundary, and evaluate $c \int R \wedge R$ in the bulk as the definition. However, non-trivial $\Omega_3^{fr}$ suggest we can not do it naively, and we need 24 copies instead.

Then following the standard argument about the independence of choice of bounding manifold, we will see the GCS action is well-defined modulo $c/24$(times 2 $\pi$ in some convention).

This topological argument seems to be an independent calculation of the dependence on framing in the partition function, which gives the same result. And I hope to learn the deep connection between these two, (one from modular property of CFT, one from topological argument).

Btw, I like your SPT paper.

That 24 is quite interesting to think about. I think you can take K3 minus 24 points as the nullbordism of 24 SU(2)s.

There are two ways to fix the framing anomaly. One is to frame the 3-manifold, the other is to choose a bounding spin 4-manifold. A nice way to think about this is via the exact sequence

0 -> spin closed 4-manifolds -> spin 4-manifolds w framed boundary -> framed closed 3-manifolds -> 0,

all modulo bordism.

Thanks for the readership and feedback (:

Why do we need spin 4manifold in the sequence? I thought it could be arbitrary.

It has to do with K3 generating the spin bordism group and K3 with 24 punctures being the nullbordism for 24 SU(2)s.

There seems to be a contradiction in the two ways of fixing the framing: (1) 3-manifold: a twist will induce a phase shift $c/24 \times 2 \pi$; (2) bounding 4-manifold: a change of bounding manifold will induce a phase-shift $c/24 \times 2 \pi \times 3$(3 from $p_1=3\sigma$, and for spin 4-mfd, 3 is replaced by $3\times 16$). So there is a discrepancy of factor 3. Do you have any idea on how to understand it?

The signature of a spin 4-manifold is divisible by 16. In particular, $\frac{1}{2}p_1(K3) = 24$.

Yes, I know. Maybe I didn't say it clearly, the discrepancy is about the minimal phase shift you can get. For 3mfd argument, the minimal shift is $c/24\times 2 \pi$; while for the 4mfd argument, the minimal shift is $c/8 \times 2 \pi$ or $2c \times 2\pi$(if assuming spin).

I think you're just missing a 1/2 somewhere?

What is $H_{3}$?

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