Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let be $M$ a compact $3$-manifold. If $\Sigma$ is a embedded surface in $M$, such that $\Sigma$ is homeomorphic to $\mathbb{RP}^2$.

If $i: \pi_1(\Sigma) \longrightarrow \pi_1(M)$ is not injective, then $\Sigma$ is non-orientable?

Note that $\pi_1(\Sigma) = \mathbb{Z}/2$.

This is utilized in the proof of Proposition 3 in http://arxiv.org/abs/0909.1665.

Thank you!

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The claim they're making is not about whether or not $\Sigma$ is orientable (it can't be since it's homeomorphic to $\mathbb{R}P^2$), but whether or not $TM$, when restricted to $\Sigma$ (that is, $i^* TM$) is an orientable vector bundle.

A vector bundle is orientable is the transition maps can be chosen to lie in $Gl^+$ (orientation preserving linear isomorphisms) instead of just in $Gl$ (all linear isomorphisms).

Alternatively, a vector bundle is classified by a (homotopy class of) map $\Sigma\rightarrow BGl$ and orientability is precisely the condition that there is a lift of this map to $BGl^+$.

Alternatively again, (and this is the characterization I'll use), a bundle $\xi$ is orientable iff the first Stiefel-Whitney class $w_1(\xi)\in H^1(\Sigma; \mathbb{Z}/2)$ is $0$. (I honestly forgot how to prove the equivalence between the three notions, if I ever knew it. I'm thinking Lawson's book on spin geometry has a proof of the equivalence, but I could be misremembering).

The tangent bundle over $\Sigma$ is not orientable (since $\mathbb{R}P^2$ isn't), but other vector bundles over $\Sigma$ may be. For example, the trivial bundle (of any rank) is always orientable.

Anyway, to see why their claim is true, note that if $\pi_1(\Sigma)\rightarrow \pi_1(M)$ is not injective, it's actually the $0$ map. This implies the induced map $H_1(\Sigma;\mathbb{Z}/2)\rightarrow H_1(M;\mathbb{Z}/2)$ is the $0$ map, and this in turn implies the induced map $H^1(M;\mathbb{Z}/2)\rightarrow H^1(\Sigma, \mathbb{Z}/2)$ is the $0$ map.

By naturality of characteristic classes, we have $w_1(i^* TM) = i^*w_1(TM) = 0$ since $i^*$ is the $0$ map, so the bundle $TM$, when restricted to $\Sigma$ is orientable.

Now they use the fact that $i^*TM = T\Sigma \oplus \nu$ where $\nu$ is the normal bundle to argue that $\nu$ is nontrivial. If it were trivial, it'd be orientable, so $T\Sigma\oplus \nu$ wouldn't be, but $i^*TM$ is, giving a contradiction.

share|improve this answer
    
Thank you @JasonDeVito! I saw Lawson's book on spin geometry. Also, I noticed, that Σ is non-orientable. But, I not understand two facts. First, why they claim that $T\Sigma$ is non-orientable? (see exercise 2 of M. P. do Carmo, Riemannian Geometry, Birkhauser, 1992). Second, since $\Sigma$ is non-orientable, then $\nu\Sigma$ is non-trivial (see Problem 13-9 of J. M. Lee, Introduction to Smooth Manifolds, Springer, 2002). Thus, why prove that $i^*TM$ is orientable for conclude that $v\Sigma$ is non-trivial? Thank you, again! –  Kelson Vieira Nov 1 '12 at 4:53
1  
If $\Sigma$ is orientable, then the "usual" transitions between trivializations on $\Sigma$ will be orientation preserving, so if $\Sigma$ is orientable, then $T\Sigma$ is (as a bundle) as well. If $T\Sigma$ is orientable (as a bundle), then using the usual trivializations will show that $\Sigma$ is as well. So, orientability of $\Sigma$ and of $T\Sigma$ (as a bundle) are equivalent. Note that there are two notions of orientability here. One, $T\Sigma$ is a manifold, so you can ask if it's orientable as a manifold. The answer to that is YES (as in Do Carmo). (continued)... –  Jason DeVito Nov 1 '12 at 13:18
1  
Second, $T\Sigma$ is a bundle, and there's a notion of a vector bundle being orientable as a bundle. It is this notion of orientability that I'm using in my answer. As to your second question if $T\Sigma$ is nonorientable and $\nu\Sigma$ is orientable, then $T\Sigma \oplus \nu\Sigma$ will be *non*orientable. But $T\Sigma\oplus\nu\Sigma = i^*TM$ is orientable, so $\nu\Sigma$ must be nonorientable. In particular, $\nu\Sigma$ cannot be trivial. (I loaned my copy of Lee out to a student, so I can't see the exercise you're referring to.) –  Jason DeVito Nov 1 '12 at 13:20
    
Ok @JasonDeVito! The problem 13-9 claim that: If $M$ is a smooth orientable Riemannian manifold and $\S\subset M$ is an immersed or embedded submanifold, then $S$ is orientable when $S$ has trivial normal bundle. –  Kelson Vieira Nov 1 '12 at 14:11
1  
Well, I didn't read the .pdf that carefully, but I don't think we know that $M$ is orientable. The argument in their proof works regardless of whether or not $M$ is orientable. –  Jason DeVito Nov 1 '12 at 14:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.