Subject: Re: two postings From: Kevin Iga Date: Mon, 6 Sep 2004 14:46:31 PDT To: dmd1@lehigh.edu (Don Davis) >> >> Two postings: A question and a conf update............DMD >> __________________________________________________________ >> >> From: David Roberts >> Date: Mon, 6 Sep 2004 15:16:52 +0930 >> >> I'm trying to find a non-orientable principal S1-bundle (E) over an >> orientable >> base space M of 'low' dimension (say 2-4). Here's the tricky bit: >> >> Want: H2(M) nontrivial - so we can twist the bundle. Are you sure you mean "principal"? By definition, a principal bundle with group S1 has fibre S1, structure group S1, with the group S1 acting on the fiber by left multiplication. Now, this action is orientation-preserving on S1, and therefore every principal S1 bundle over any space is orientable. (This is true when S1 is replaced by any connected Lie group, too.) But sometimes when people say "non-orientable bundle" they don't mean that the bundle is non-orientable, but that the total space E is a non-orientable manifold. In case you meant this, note that it is also true that if the base space is orientable, the total space of any principal S1 bundle will also be orientable, and here is the proof: at every point in the n-dimensional base space M, we have an n-form \omega that defines the orientation. Define the 1-form d\theta on S1 that is the standard orientation on S1. Note that left-multiplication by elements in S1 preserves this 1-form, and so this 1-form also exists in the total space of the bundle. Wedge \omega\wedge d\theta and we now have a non-vanishing n+1 form on the total space E that serves as the orientation there. So under these two possible interpretations of your question, there are no non-orientable principal S1 bundles over an orientable base space M, whether or not H_2(M) is nontrivial. In case you don't really mean "principal", but simply that the fibre is S1, here's an example of what you want: Consider the "twisted" 3-torus given below: E=R3/(x,y,z)~(x,y+1,z)~(x,y,z+1)~(x+1,y,-z) This is non-orientable since dx^dy^dz, when translated by (1,0,0), and associated via (x,y,z)~(x+1,y,-z), becomes -dx^dy^dz. Or you could also notice that E is a Klein bottle cross a circle (the middle coordinate is the circle in that case). Map this space E to the 2-torus using the projection function p(x,y,z)=(x,y). This is well-defined. The preimage of a point is a circle. It is straightforward to check that this is a fibre bundle with fibre S1 (and structure group Z/2, in case you care). The base space is the 2-torus, which of course is orientable with H2(M) nontrivial, and in the dimension range you want. More generally M may be chosen to be any orientable manifold with nontrivial H1(M;Z/2). That is, assume the abelianization of the fundamental group has at least one free summand or at least one Z/2^n summand. This guarantees the existence of a subgroup of pi_1(M) of index 2. Find a subgroup K of pi_1(M) of index 2. This defines a double-cover of M, called M~, where loops in M remain loops in M~ if and only if they correspond to elements of K. The projection p:M~ -> M is defined. For every point x in M, we have preimages p(x_1)=x=p(x_2) in M~. Let E~ = M~ x S1. Then let E be E~/(x_1,a)~(x_2,-a). (Here, S1 is considered as R/Z so that negation is an orientation-reversing action). Now E will be non-orientable, and any loop in K demonstrates this non-orientability. If M also has the other properties you want (is low-dimensional and has non-trivial H2) then you are done. Examples of such M in 2 dimensions include any closed Riemann surface other than the sphere. In 3 dimensions, we can let M be any orientable closed manifold other than a rational homology sphere. Kevin Iga Pepperdine University