The existence of the extension of a non-trivial line bundle

In Three Dimensional Gravity Revisted, Witten studied the Abelian Chern-Simons theory in three dimensions.

Let $$W$$ be a three dimensional manifold. Let $$\mathcal{L}$$ be a non-trivial line-bundle over $$W$$. On page 11, Witten claims that one can pick up a four dimensional manifold $$M$$ such that $$W$$ is the boundary of $$M$$ and the line-bundle $$\mathcal{L}$$ can be extended over $$M$$. Such a four dimensional manifold $$M$$ always exists.

Is there any proof of the above statements? They seem quite non-trivial to me.

I am also interested in the case of $$SL(2,\mathbb{R})$$ Chern-Simons theory. Witten claims that the $$SL(2,\mathbb{R})$$ case can be reduced to the $$U(1)$$ case because $$U(1)$$ and $$SL(2,\mathbb{R})$$ are homotopy equivalent.

How to prove that the extension of the non-trivial principal $$SL(2,\mathbb{R})$$ over $$M$$ really exists?

This is a bordism problem, and as such can be answered using algebraic topology. I'll answer in the unoriented setting, then indicate how to modify things if $$M$$ and $$W$$ are required to be oriented.

Complex line bundles $$\mathcal{L}$$ over $$W$$ are classified by maps $$f:W\to BU(1)\simeq \mathbb{C}P^{\infty}$$. We want to decide if there is a $$4$$-manifold $$M$$ with a map $$F:M\to BU(1)$$ such that $$\partial M=W$$ and $$F|_{\partial M}=f$$.

We can define an equivalence relation called bordism on the set of pairs $$(W,f)$$ where $$W$$ is a closed $$3$$-manifold and $$f:W\to BU(1)$$ is a continuous map. Two such pairs $$(W_0,f_0)$$ and $$(W_1,f_1)$$ are bordant if there is a pair $$(M,F)$$ consisting of a compact $$4$$-manifold $$M$$ with $$\partial M = W_0\sqcup W_1$$ and a map $$F:M\to BU(1)$$ satisfying $$F|_{\partial M} = f_0\sqcup f_1$$.

The set of equivalence classes $$[W,f]$$, denoted $$\mathfrak{M}_3(BU(1))$$, becomes an abelian group under the operation of disjoint union. The zero element is represented by the empty $$3$$-manifold. This group is a homotopy invariant, and so $$\mathfrak{M}_3(BU(1))\cong \mathfrak{M}_3(\mathbb{C}P^\infty)$$.

All of this is fairly standard, and can of course be generalised. A classic reference is Conner and Floyd's Differentiable periodic maps.

Eventually we see that Witten's claim is equivalent to the group $$\mathfrak{M}_3(BU(1))$$ being trivial. There may be more elementary ways to see this, but an algebraic topologist would use the following spectral sequence argument. Let $$\mathfrak{M}_q$$ denote the group of closed $$q$$-manifolds up to bordism (the same equivalence relation as above, but without the maps to $$BU(1)$$). These groups have been computed by Thom and others. All we need to know here is that $$\mathfrak{M}_q\cong \mathbb{Z}/2,0,\mathbb{Z}/2,0$$ for $$q=0,1,2,3$$.

There is a spectral sequence, called the Atiyah-Hirzebruch spectral sequence or the unoriented bordism spectral sequence, whose $$E^2$$-term is $$E^2_{p,q} = H_p(BU(1),\mathfrak{M}_q)$$ and which converges to $$\mathfrak{M}_{p+q}(BU(1))$$. Since $$BU(1)\simeq \mathbb{C}P^\infty$$ has homology concentrated in even degrees, we see that the groups $$H_p(BU(1),\mathfrak{M}_q)$$ are zero for $$p+q=3$$, and it follows that $$\mathfrak{M}_3(BU(1))\cong 0$$ as claimed.

The same argument works for oriented bordism, since the low dimensional oriented bordism groups are $$\Omega_q\cong \mathbb{Z}, 0 , 0, 0$$ for $$q=0,1,2,3$$. It also works for oriented rank $$2$$ real bundles, since $$BSL(2,\mathbb{R})\simeq BU(1)\simeq \mathbb{C}P^\infty$$.

• Thank you so much for your help. As a physics student I don't have any background knowledge in algebraic topology. I will try to digest your answer. – Libertarian Monarchist Bot Dec 11 '18 at 14:03
• @TheLastKnightofSilkRoad: No problem. I assume Witten knows this stuff, but also suspect he has a more "physical" way of seeing it! – Mark Grant Dec 11 '18 at 14:05
• As I recall, the original argument used bordism, but I can’t remember where it is. It is mentioned in Dijkgraaf and Witten for example. projecteuclid.org/… – Aaron Bergman Dec 11 '18 at 19:19
• @AaronBergman Thank you Sir. I am studying that paper recently. – Libertarian Monarchist Bot Dec 14 '18 at 21:48
• For what it’s worth, you can check out various papers by Freed for rigorous definitions. The appendix of arxiv.org/abs/0808.2507 for example. – Aaron Bergman Dec 15 '18 at 1:27