Witten's QFT and Jones Poly paper - MathOverflow

Witten's QFT and Jones Poly paper

2011-02-21T06:48:16Z

Data: $M$ is an oriented 3-dim manifold, $E$ is a $G$-bundle over $M$, with $G$ compact simple Lie group.

Question: How does $\pi_3(G)\cong \mathbb{Z}$ imply that there exists non-trivial gauge transformations (i.e., continuous maps $M\rightarrow G$ which are not homotopic to the trivial map)?

If anyone would like to read it from the source, check the paragraph leading up to equation 1.4.

Answer by Kelly Davis for Witten's QFT and Jones Poly paper

2011-02-21T07:11:58Z

First consider the case $M = S^3$. Generalizing, consider the connected sum of a generic M with a sphere $M = M \# S^3$

Edit Here's what I was thinking (Still not sure if it's all correct, but it seems closer to the spirit of Witten's paper than the obstruction arguments.)

Consider a gauge transform $f': M \rightarrow G$. Also, consider a gauge transformation $g' : S^3 \rightarrow G$ not homotopic to the identity. Continuity allows us to change $f'$ to a map $f$ homotopic to $f'$ such that in a neighborhood $U$ of $p \in M$ the map $f$ maps to the identity of $G$. We can define a map $g$ to have similar properties in a neighborhood $V$ of $q \in S^3$.

Do the connected sum around $p$ and $q$ and obtain $M \# S^3 = M$ as well as a gauge transform $h$ on $M \# S^3 = M$ obtained by joining $f$ and $g$. Now, assume $h$ is homotopic to the identity.

The homotopy taking $h$ to the identity can be used to construct a homotopy of $g$ to the identity. (Here we use the fact that $\pi_2(G)$ is trivial to continue the homotopy over the ball removed from $S^3$.)

But, no such homotopy of $g$ to the identity exists. Thus, $h$ is not homotopic to the identity. Hence, $\pi_3(G) = \mathbf{Z}$ implies there exist continuous maps $M \rightarrow G$ not homotopic to the identity.

Answer by Konrad Waldorf for Witten's QFT and Jones Poly paper

2011-02-21T14:21:15Z

I think the important information is that $H^3(G,\mathbb{R}) \neq 0$.

By the way, every $G$-bundle over $M$ is trivializable, under the conditions you have mentioned. That's why a gauge transformation can be regarded as a (smooth) map $g: M \to G$.

Now you look at the behaviour of the Chern-Simons 3-form $CS(A)$ of a connection $A$ on $E$ under a gauge transformation $g$. The formula is $$ CS(g^*A) = CS(A) + g^*H + \text{exact terms}. $$ where $H$ is the canonical 3-form of $G$ that represents a non-trivial element of $H^3(G,\mathbb{R})$. Now you can find a gauge transformation $g$ such that $g^*H$ is not exact. In that sense you have non-trivial gauge transformations.

EDIT: The comment that every $G$-bundle is trivializable is only true if $G$ is additionally assumed to be simply-connected, sorry. So you either assume that (so does Witten) or you must see gauge transformations as maps $g:P \to G$, rather, and the Chern-Simons form $CS(A)$ as a form on $P$, not on $M$.

Answer by Peter Woit for Witten's QFT and Jones Poly paper

2011-02-21T21:14:24Z

As Konrad Waldorf noted, in this case G-bundles are trivializable (since $\pi_2(G)$ is trivial). So gauge transformations are just maps $$\phi:M\rightarrow G$$

and these have a homotopy invariant that can be non-trivial, the degree of the map. One way to compute this is as $$\int_M \phi^*\omega_3$$

where $\omega_3$ is a generator of $H^3(G)$. Or, as usual for a degree, just pick an element of G, and count points (with sign) in the inverse image.

Answer by Paul for Witten's QFT and Jones Poly paper

2011-02-22T00:26:28Z

@kwl1026. Gauge transformations are sections of the Ad bundle $P\times_{Ad} g$ where $P\to M$ is the principal $G$ bundle; $g$ the lie algebra. When $G$ is abelian the adjoint action is trivial so, e.g. the $U(1)$ gauge group is always $Map(M,U(1))$ whether or not $P$ is trivial. Its homotopy classes are then $[M, U(1)]= H^1(M;Z)$, which is zero (for $M$ a closed 3-manifold) if and only if $M$ is a rational homology sphere.

An elementary answer to your original question for $SU(2)=S^3$ is that obstruction theory shows that the primary obstruction gives an isomorphism $[M,S^3]\to H^3(M;Z)$. An induction using the fibration $SU(n)\to SU(n+1)\to S^{2n+1}$ and cellular approximation shows that $[M,SU(n)]=[M,SU(2)]$. Other tricks can get you there for other $G$. It is true that the differnence in Chern-Simons invariants (suitably normalized) coincides with this isomorphism (composed with $H^3(M;Z)\to Z$), as indcated by Konrad. For $SU(2)$ it also agrees with the degree, as mentioned by Peter.

If $P$ is non-trivial you have to work a little harder, since you are asking what is the set of homotopy classes of sections of the fiber bundle $P\times_{Ad} g$. A useful reference is Donaldson's book on Floer homology.