10

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.

Kevin Wray
  • 1,689
  • If you like rational homotopy theory you could use the fact that rationally all these Lie groups look like products of odd-spheres... Then you would be analyzing maps from an algebra \wedge v_2i+1 \to C*(M) all. The map which sends v_3 \to a generator in H^3(M) should be the one... This should all be okay even in the non-simply connected case since pi_1 acts trivially on the higher homotopy groups for Lie groups. – Daniel Pomerleano Feb 21 '11 at 10:43

4 Answers4

5

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 $M$.

Konrad Waldorf
  • 4,304
  • But, isn't true that $H^3\big(U(1),\mathbb{R}\big)\neq 0$? If so, then shouldn't we also have nontrivial gauge transformations in this case? However, we know that $k$ is not quantized for $U(1)$ theories - quantization comes directly from nontrivial gauge transformations.

    I'm sure that I am mixing something up here...just not sure what.

     – Kevin Wray Feb 21 '11 at 18:36
  • @ klw1026 - $U(1)$ is the circle which is $1$-dimensional, whence $H^3(U(1);\mathbb{R})=0$. – Somnath Basu Feb 21 '11 at 19:14
  • Sorry, I didn't mean to say $U(1)$. – Kevin Wray Feb 21 '11 at 19:19
  • @ klw1026: $U(1)$ is not simple. It's a subtle terminology, but simple contains the assumption of being non-abelian. Also, I think we need the assumption of simply-connectedness, see my edit. – Konrad Waldorf Feb 21 '11 at 20:07
  • By the way, Chern-Simons theory for $G=U(1)$ is non-trivial. That's because $H^4(BU(1),\mathbb{Z})$ is non-trivial. – Konrad Waldorf Feb 21 '11 at 20:16
  • @ Konrad: Yes, but there is no "quantization" condition on $k$ for the $U(1)$ theory, right? And this has to do with $\pi_3\big(U(1)\big)=0$, right?

    Also, are you saying: If $G$ is compact and simple, then we know that $\pi_3(G)\cong \mathbb{Z}$. Hence, using the Hurewicz isom., we get that $H_3(G)\cong \mathbb{Z}$. Therefore, $H^3(G;\mathbb{R})$ must be nontrivial.

     – Kevin Wray Feb 21 '11 at 20:54
  • Further, we care about elements of $H^n(G;\mathbb{R})$ where $n=3$ since we are integrating over the $3$-manifold $M$, and no other reason; i.e., why not care about $\pi_1(G)$ or $\pi_2(G)$ (I know they are trivial in these two cases, but still would like a nice description for why we don't care about them)? – Kevin Wray Feb 21 '11 at 20:55
4

@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.

Paul
  • 4,304
  • Yes, I agree with what everyone has said. I am just trying to understand Witten's reasoning when he says that since $\pi_3(G) \cong \mathbb{Z})$ (i.e., when $\pi_3(G)$ is non-trivial) we have nontrivial gauge transformations. I understand Konrad's description (since $\pi_1(G)=\pi_2(G) =0$ there is no obstruction to a global section - giving gauge transformations as maps $M\rightarrow G$), but he is starting with a cohomological statement. So, is it that Witten is just thinking about the Hurewicz isom. to got to homology, then cohom., or is it straightforward from $pi_3(G)\neq 0$? – Kevin Wray Feb 22 '11 at 00:40
  • 1
    For any (simply connected) space $G$, the obstructions to nulhomotoping $f:M\to G$ lie in $H^i(M;\pi_i(G))$. If $\pi_1(G)=0=\pi_2(G)$, then since $M$ is 3-dimensional there is only one obstruction in $H^3(M;\pi_3(G))$. So $\pi_3(G)=Z$ means the only (primary) obstruction is in $H^3(M;Z)$.

    Although I'm citing obstruction theory, this is elementary since you can easily nullhomotop the 2-skeleton of $M$, so the map factors through $M/2-skeleton=S^3$ for an appropriate cell structure.

    Incidentally, for $G=Spin(4)$, $\pi_3(G)=Z\oplus Z$ since $Spin(4)=SU(2)\times SU(2)$.

     – Paul Feb 22 '11 at 00:55
  • Ok, this I like! For some reason though I was thinking that the obstruction (extending over the $3$-skeleton) was an element in $H^3\big(M;\pi_2(G)\big)$, not in $H^3\big(M;\pi_3(G)\big)$. Maybe that is the obstruction to extending a section over the $G$-bundle? – Kevin Wray Feb 22 '11 at 01:30
  • 1
    This is a homotopy obstruction, not an extension obstruction. Of course they are related: Think of it as a relative extension problem for $(M\times I, M\times{0,1})$, then use the suspension iso $H^i(M\times I, M\times{0,1})=H^{i-1}(M)$. This explains the shift. – Paul Feb 22 '11 at 13:10
  • @Paul: Yes, I realized this last night. Everything is ok now. Thanks for the explanation. – Kevin Wray Feb 23 '11 at 00:13
3

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.

Peter Woit
  • 927
0

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.

Michael Albanese
  • 18,817
Kelly Davis
  • 1,897
  • Yes, I understand the $M=S^3$ case, but not following what you are implying by considering the connected sum $M# S^3$ – Kevin Wray Feb 21 '11 at 07:23
  • Presumably you should be able to take a map which is constant on $M\backslash D^3$ (and on the connecting tube) and essential on $(S^3\backslash D^3,\partial D^3) \simeq (S^3,pt)$? Then we could prove this is essential too by looking at the induced map on $H_3$ or something like that. (Or maybe there's an easier direct argument proving that such a construction can't be nullhomotopic.) – Aaron Mazel-Gee Feb 21 '11 at 09:00
  • Yeah at least in the simply connected case, that's probably what she means. We can just collapse the two skeleton of M and look at the map S^3 \to G given by a generator of pi_3. By Hurewicz and the fact that the collapsing map induces an iso on H_3, the map must be non-trivial on homology. In the non-simply connected case, pass to the universal cover and note that the map $H_3(\tilde(G))\to H_3(G)$ is multiplication by some non-zero number. – Daniel Pomerleano Feb 21 '11 at 10:57
  • the weird tilde thing is supposed to be the universal cover but it's late and i'm to lazy to fix it... – Daniel Pomerleano Feb 21 '11 at 10:58