# Prove that a manifold is not orientable

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

• Interesting question: the converse implication! – Martín-Blas Pérez Pinilla Jan 31 '14 at 8:54
• @Martín-BlasPérezPinilla interesting remark! Is it true? – andreasvr Jan 31 '14 at 8:58
• More hypothesis required. Can you check the the exact wording? – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:25
• 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 9:31
• Open is impossible if ${\rm dim M}>2$. – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:34

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.

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

• 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 8:56
• Thank you for your comment! – user89987 Jan 31 '14 at 9:05
• See math.stackexchange.com/questions/15602/…. And maybe Differential Forms and Applications (do Carmo) has another argument that I can't remember. – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:06
• 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

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

• Is a submanifold? – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:07
• 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 9:09
• 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 9:13
• 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 9:16
• @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 9:22