Is the following reasoning correct?

  • The classifying space of the 1-torus $\mathbb T$ is $\mathbb{CP^\infty}$.

  • Hence isomorphism classes of $\mathbb T$-principal bundles over $\mathbb T$ are in bijection with homotopy classes of continuous maps $\mathbb T\to\mathbb{CP^\infty}$, or equivalently, since $\mathbb{CP^\infty}$ is connected, with $\pi_1(\mathbb{CP^\infty})$.

  • But the $\mathbb T$-principal bundle $S^\infty\to\mathbb{CP^\infty}$ gives an exact (in the middle) sequence $\pi_1(S^\infty)\to\pi_1(\mathbb{CP^\infty})\to\pi_0(\mathbb T)$.

  • The groups $\pi_0(\mathbb T)$ and $\pi_1(S^\infty)$ vanish since $\mathbb T$ is connected and $S^\infty$ is contractible. Hence $\pi_1(\mathbb{CP^\infty})$ vanishes as well.

  • In conclusion, every $\mathbb T$-principal bundle over $\mathbb T$ is trivial.

This suprises me, because I feel like I could get a non-trivial bundle by "turning the fiber around once during one circulation."

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    $\begingroup$ Looks fine to me. Another way to see this is that $\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z}, 2)$ (for example because it is the infinite symmetric product of $S^2$) and $H^2(\mathbb{T}; \mathbb{Z})$ vanishes. $\endgroup$ – Qiaochu Yuan May 2 '11 at 15:59
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    $\begingroup$ "turning the fiber around once during one circulation" rather shows that the group of automorphisms of the trivial $\mathbb{T}$-bundle over $\mathbb{T}$ is not connected (which is the same as $\pi_1(\mathbb{T})$ being non-trivial) $\endgroup$ – user8268 May 2 '11 at 16:06
  • $\begingroup$ @user8268: thanks, your comment is very helpful. $\endgroup$ – Rasmus May 2 '11 at 16:34
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    $\begingroup$ @Grigory: well, I know very little about these matters (beyond what I've already said), and I can't really resolve the surprise part of the question, so I would prefer that someone more knowledgeable write something more detailed. $\endgroup$ – Qiaochu Yuan May 2 '11 at 19:30
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    $\begingroup$ (Re: last par) maybe you will find en.wikipedia.org/wiki/Dehn_twist interesting (cf. user8268's comment) $\endgroup$ – Grigory M May 2 '11 at 20:28

Your argument is fine - there is a unique principal $\mathbb{T}$ bundle over $\mathbb{T}$, namely, the trivial product.

Your intuition about turning the fiber around, I think, comes from confusing principal bundles with general bundles.

In fact, there are two fiber bundles over a circle whose fiber is a circle - the two dimensional torus and the Klein bottle (at least when the structure group is finite dimensional - though I believe the diffeomorphism group of $S^1$ retracts onto $O(2)$, so perhaps the finite dimensionality hypothesis is superfluous.)

In this case, one can prove that the structure group reduces down to $Gl(2,\mathbb{R})$. Now, such a bundle over $S^1$ is trivial on the "northern" half and trivial on the "southern" half and must be glued together at the "east" and "west" poles (the points $(\pm 1,0)$ in $\mathbb{R}^2$). The gluing is given by a (homotopy class of) map $\phi:S^0\rightarrow Gl(2,\mathbb{R})$ and we may assume wlog that $\phi$ of the basepoint of $S^0$ (whichever point we select to be the basepoint) is sent to the identity. The map $\phi$ is known as the clutching function.

Then, because $GL(2,\mathbb{R})$ has 2 components (corresponding to those matrices of negative and positive determinant), there are exactly two homotopy class of such $\phi$ depending on which component $\phi$ of the non-basepoint is.

This proves there are two bundles over $S^1$ with fiber $S^1$. It's not too hard to pictorially draw the nontrivial bundle. Draw the Klein bottle as a square with sides identified. Identify the upper side with lower orientation preservingly and identify the sides orientation reversingly. Then the map $\pi$ defining the fiber bundle structure sends a point in the square to it's projection along the bottom line. The fibers are the vertical lines (circles after identification).

So we have two bundles, the trivial bundle and the Klein bottle. You have already proven that the trivial bundle is a principal bundle (and is the only one), but let me give some intuition for why that the Klein bottle is not.

Pick a fiber of $K$. I'm thinking of the fiber going through the center of the square. In order to be a principal bundle, $\mathbb{T}$ must act simply transitively on this fiber. There are precisely two such actions, namely $z* w = zw$ or $z*w = \overline{z}w$. Intuitively, these actions corresponding to the circle acting on itself by either spinning one direction or the opposite direction. In our square model of $K$, this corresponds to moving points either up or down along the vertical line containing them. I'll assume we're pick "moving up", the other case being basically identical.

By continuity of the action, nearby fibers must also move up. In fact, one can extend this everywhere - by continuity, $\mathbb{T}$ must act on all the fibers by moving them up. But if you now look at how the left edge and right edge are identified, you'll see you've now specified contradictory actions on the identified left/right edge. This shows $K$ is not a principal bundle, even though it is a perfectly good fiber bundle.

The moral of the story is that "principal $G$-bundle" and "fiber bundle with fiber $G$" are different ideas. The second needs a consistent choice of "multiplication by G" on all fibers in order to become the first and sometimes this consistent choice can't be made.


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