While reading Poincare Duality a new idea of orientability of manifold came in my mind.I dont know wheather this idea is new or not, or even true or false.

My idea is following... A $n$-dim manifold $X$ is called non-orientable if there exists an embedding $i: M\times \mathbb{D^{n-2}} \to X$, where $M$ is a mobius strip. Otherwise $X$ is called an orientable manifold.

For an example if $n=2$, then by using of classification theorem of compact $2$-manifold we can see that my definition works for 2-dim compact manifold...But I got stuck while doing this generalization...Till now I've knowladge of algebraic topology only upto Poincare Duality of Hatcher's book...

I want to know whether my this general idea is true or not, and in case if it is true then how do I prove it?? I need some HINTS for that...and if it is a well known result then please give me some reference where I can study this properly.

  • $\begingroup$ At least for a smooth manifold, this idea is equivalent to the usual notion of orientability. The idea of the proof uses normal bundles of embedded closed curves. $\endgroup$ Aug 29, 2015 at 21:49
  • $\begingroup$ @JasonDeVito: This should be true in the topological case. You just need a curve that 'demonstrates' non-orientability and has a disc bundle over it. For $n=4$ noncompact manifolds are smoothable; for $n \geq 6$ decent subcomplexes have regular neighborhoods, so you can take the regular neighborhood and by Mazur's obstruction theory you have that it's PL; in this PL manifold take the star of our curve; this is a disc bundle. Compact topological disc bundles are either the Mobius bundle or trivial. I dunno about $n=5$. $\endgroup$
    – user98602
    Aug 29, 2015 at 22:21
  • $\begingroup$ This is probably overkill but the only thing I know about topological manifold theory is the part that lets me turn it into PL or smooth manifold theory. $\endgroup$
    – user98602
    Aug 29, 2015 at 22:22
  • $\begingroup$ @Mike: You know more about topological manifolds than I do! $\endgroup$ Aug 30, 2015 at 0:30

1 Answer 1


This is correct, and a fun idea. Here's why. I'll talk about smooth manifolds first, but it's true in full generality. All my manifolds have dimension $n \geq 3$.

First off, we need to classify $n$-dimensional vector bundles over the circle. There are precisely two: one is trivial, the other is non-orientable, with total space homeomorphic to the (open) Mobius band times the open disc $D^n$. This is because bundles over the circle are classified by $\pi_0 GL_n(\Bbb R) = \mathbb Z/2$; the nontrivial bundle is the Mobius line bundle plus a bunch of trivial line bundles.

Given a smooth manifold $M$, $M$ is orientable if and only if every embedded circle $S^1 \hookrightarrow M$ has trivial normal bundle. (If it's orientable, because the circle is orientable, its normal bundle is orientable, hence trivial; conversely if it's not orientable its orientation double cover is not trivial, and hence there is some loop that doesn't lift as a loop to the double cover. This is the desired loop with nontrivial normal bundle.) Now by the classification of line bundles above, we can pick a tubular neighborhood and then a closed disc bundle in that neighborhood to give the desired embedding of the "thickened Mobius strip".

The same technique works for arbitrary topological manifolds, as the above equivalence still works: a manifold is orientable if and only if there is some loop in the manifold that does not lift to the orientation double cover. Because the dimension is at least 3, you can homotope this loop "by hand" to be a locally flat embedding.

What we want to do now is find a closed disc bundle around this loop. Then, by the classification of disc bundles (there are again precisely two, because $\pi_0 \text{Diff}(D^n) = \Bbb Z/2$, and they are what you think), we have the desired result.

The one trouble is that disc bundles are hard to find. I'm told by Remark 4 in this paper by Edwards that as long as the dimension of my manifold is at least 5, my loop as constructed above has one. I don't have a precise reference for you. We can get around this problem in small dimensions: for $n = 3$ all manifolds are smoothable, so this is encapsulated in the first case, and for $n=4$ it's part of Freedman's work that noncompact 4-manifolds are smoothable, so just puncture it at a point not in the loop. This does not destroy nonorientability.

If there's a way to avoid Freedman's result here I'd be glad to know.


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