# Orientation of manifold in topological sense

What do we mean by orientability of a topological manifold? How do we orient two dimensional Euclidean space and why is Moebius band non-orientable? And it would be a great favour to me if you can tell me a good book that explains this idea. I read "Algebraic topology: An Introduction" by W. S. Massey but have a confusion.

-

To answer your question about Euclidean two-space first: If $V$ is a vector space and $e_i$ and $f_i$ are two bases then they are said to have the same orientation if the determinant of the matrix expressing one basis in terms of the other is positive that is, $>0$.

The relation you define in such a way on the set of all bases of $V$ is an equivalence relation and consists of two equivalence classes.

Hence you may define the orientation of a vector space as a choice of an equivalence class of bases.

See e.g. Boothby on page 207/208.

Why is the Moebius band not orientable? In the case of topological manifold one can choose a normal vector. Say you start at a point $a$ on the Moebius band and pick a vector pointing upwards. If you slide this vector along the middle circle of the band one time around it will point the opposite way. A manifold where you can continuously move around a point from "one orientation" into another different one is non-orientable.

See e.g. this post which explains it nicely.

Another way of defining it is given in Hatcher starting on page 233.

-
"A topological manifold is orientable if it does not contain the Moebius band as a submanifold." This is a completely surrealistic definition. I challenge you to use it to prove that $\mathbb P^3(\mathbb R)$ is orientable. Good luck! – Georges Elencwajg Dec 16 '12 at 10:12
Dear Matt, it would be false (and circular): $\mathbb R^4$ contains the non-orientable Möbius strip as a closed submanifold, but is of course orientable. – Georges Elencwajg Dec 16 '12 at 10:22
Dear Matt, I certainly don't want to pick on you, but the implication arrows in your preceding comment go in the wrong direction (look at $\mathbb R^4$ again). – Georges Elencwajg Dec 16 '12 at 10:57
I think what is true and sounds similar to what you want to state, is that a smooth manifold (or more general a vectorbundle possibly different from the tangent bundle) is orientable if and only if the restriction along any embedded $S^1$ does not contain the moebius strip as a subbundle. – mland Dec 16 '12 at 11:01
A two-dimensional manifold is non-orientable iff it has a Möbius-strip submanifold, but this is not true for higher dimensions unless we replace the "Möbius strip" with an appropriate higher analogue (which I don't know what is called). – Henning Makholm Dec 16 '12 at 12:13