Prove that a manifold is not orientable

Question

I have found a proposition who says: A manifold M is not orientable if it contains a Moebius band. How can I prove this?

More hypothesis required. Can you check the the exact wording? — Martín-Blas Pérez Pinilla, Jan 31 '14 at 09:25
If a Manifold isn't orientable then it contains an open subset diffeomorphic to the Moebius band.. Is this the right claim? — andreasvr, Jan 31 '14 at 09:31
Definitely true for surfaces. Maybe true for embedded submanifolds (http://en.wikipedia.org/wiki/Submanifold#Embedded_submanifolds). I think that the oriented atlas of $M$ restricted to the submanifold works. — Martín-Blas Pérez Pinilla, Jan 31 '14 at 10:20

score 4 · Answer 1 · answered Jan 31 '14 at 08:43

4

If $M$ were orientable, the Moebius band will inherit the orientation of $M$.

answered Jan 31 '14 at 08:43

Martin thanks for your answer. If i want to prove that the Moebius band isn't orientable, without using the vector normal, only with the definition of oriented atlas, what i might to do? I've seen there is a similar post but it don't answered completely this question. Thanks. – andreasvr Jan 31 '14 at 08:56
Thank you for your comment! – user89987 Jan 31 '14 at 09:05
See http://math.stackexchange.com/questions/15602/why-is-the-mobius-strip-not-orientable. And maybe Differential Forms and Applications (do Carmo) has another argument that I can't remember. – Martín-Blas Pérez Pinilla Jan 31 '14 at 09:06
I think we can´t use this argument because the mobius strip is not open. We have: If M is a orientable manifold, then every open subset of M is a orientable manifold. – Luiz Apr 23 '16 at 14:48

score 2 · Answer 2 · answered Jan 31 '14 at 09:04

2

The claim is incorrect. Euclidean space $M=\mathbb{R}^3$ contains the Mobius band but $M$ is orientable.

answered Jan 31 '14 at 09:04

Mikhail Katz

Is a submanifold? – Martín-Blas Pérez Pinilla Jan 31 '14 at 09:07
The usual picture of the Mobius band in Euclidean space represents it as a submanifold. A parametrisation can be found in most elementary differential geometry textbooks. Of course if $M$ is 2-dimensional the claim is true. – Mikhail Katz Jan 31 '14 at 09:09
This is the exercice: I have the Moebius band M, the quotient of the square $[0,1]x\times [0,1]$ with the projection map $\pi \colon [0,1]\times [0,1]\to M$. After prove that the Moebius band isn't orientable conseider $M^°=\pi([0,1]\times (0,1)$. Show that $M^°$ and every manifold which contains an open set diffeomorphic to $M^°$ is not orientable. – andreasvr Jan 31 '14 at 09:13
So @user72694 the claim is true only in 2-dimensional manifolds? How we can generalize it for prove that a manifold isn't orientable? thanks – andreasvr Jan 31 '14 at 09:16
@user72694 the Moebius band is contained in ${\Bbb R}^3$ like ${\Bbb R}^2$ in ${\Bbb R}^3$? There are several definitions of submanifolds with adjectives. In any case, you are right, the simple condition of subset isn't enough. More is required. – Martín-Blas Pérez Pinilla Jan 31 '14 at 09:22
@Martín-BlasPérezPinilla however, what is the right claim? what hyp. needs? – andreasvr Jan 31 '14 at 09:33
@user72694, please, can you check if my comment about embedded submanifolds makes sense? – Martín-Blas Pérez Pinilla Jan 31 '14 at 10:21

score 1 · Accepted Answer · answered Apr 23 '16 at 17:41

We have:

1) The mobius strip is not orientable.

2) let $f:M\rightarrow$ N be is a diffeo. M is orientable iff N is orientable.

3) Every open subset of a orientable manifold is a orientable manifold.

Let M be the mobius strip, U subset of N and $f:U\rightarrow$ M diffeomorphism.

Suppose that N is orientable. Then by 3) we have U is a orientable manifold.

Since $f:U \rightarrow$ M is a diffeo by 2) M is orientable. Contradition.

3 Answers3