# How show that a surface embedded is non-orientable?

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!

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## 1 Answer

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.

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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
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
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 nonorientable. 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
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