Principal bundles with connections

[holy_toaster]

Suppose (P,M,\pi,G) is a G-principal bundle. With this I mean a locally trivial fibration (G acts freely on P) over M=P/G with total space P and typical fibre G, as well as a differentiable surjective submersion \pi\colon P\to M. In this case M is nearly a manifold, but may be non-Hausdorff.

Now it is known that every principal bundle admits a connection if the base M is paracompact (this is the case if it is Hausdorff).

My question is now if the converse does also hold. If I have a G-principal bundle with a G-invariant splitting of the tangent spaces of P into a vertical and horizontal part (or equivalently a connection one-form), does this imply that the base M must be Hausdorff?

Any ideas how one could prove that? Or is it not true?

 

[quasar987]

If you do not assume that the base M is a smooth manifold, then how do you define "differentiable surjective submersion" for the map pi?

 

[holy_toaster]

Well, differentiable structures can be defined on non-Hausdorff manifolds in the same way as on Hausdorff ones. As it is for example done in the book by N.J. Hicks: Note on Differential Geometry (Van Nostrand Reynhold, 1965). The only thing one does not have at hand in this case is paracompactness, hence a countable cover with a subordinate partition of unity. This partition of unity is crucial in the proof of the existence of a connection if the base is indeed a smooth manifold. Starting from that I was wondering about proving the opposite direction...

 

[quasar987]

Oh I see. Sorry I can't help you.

 

[lavinia]

Well, differentiable structures can be defined on non-Hausdorff manifolds in the same way as on Hausdorff ones. As it is for example done in the book by N.J. Hicks: Note on Differential Geometry (Van Nostrand Reynhold, 1965). The only thing one does not have at hand in this case is paracompactness, hence a countable cover with a subordinate partition of unity. This partition of unity is crucial in the proof of the existence of a connection if the base is indeed a smooth manifold. Starting from that I was wondering about proving the opposite direction...

What about a trivial bundle over a non-Hausdorff space? It has a connection.

 

[holy_toaster]

What about a trivial bundle over a non-Hausdorff space? It has a connection.

Are you sure? Can you provide an example? I searched for something like that but was not able to find or construct one.

 

[holy_toaster]

I think I figured out a proof of this: Assume M is non-Hausdorff and P admits a connection. Let x,y be two points in M which cannot be separated by disjoint open sets in M. Then there are two curves c_1 and c_2 in M, c_1 closed at x and c_2 connecting x and y, which coincide everywhere but at x respectively y. As there is a connection, there is a unique horizontal lift of the two curves given some point u in the fiber over x as the initial point of the lifts. Then the horizontal lift of c_1 terminates at some point w in the fiber over x. But by continuity one concludes then the lift of c_2 should also terminate in the fiber over x. As this holds for all given points u in the fiber over x, it follows that the fibers over x and y coincide, in contradiction to the non-Hausdorff condition.

Does that make sense?