# Orientability and trivialization of the tangent bundle over the 1-skeleton

I was reading the following post on mathoverflow: https://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds in an answer there is written that

Orientability means the tangent bundle trivializes over a 1-skeleton. Dually you could think of that as saying the complement of a co-dimension 2 subcomplex has a trivial tangent bundle.

As Lee Mosher suggests in the comment section, this means that the tangent bundle of an $n$ dimensional $M$ (homeomorphic to a simplicial complex) is trivial when restricted to the $1$-skeleton of $M$.

Unfortunately, I do not manage to prove that this is equivalent to the definition of orientability that I am used to (for smooth manifolds). Namely that exists a never vanishing section of the bundle of $n$-forms over an $n$ dimensional manifold. Can someone explain it to me?

Question: Show that the above definition of orientability (in the smooth setting) is equivalent to the existence of a volume form. Also I wonder why we have that "dual" definition.

• Your remark that "any embedding $\gamma$... is such that the pull-back of the tangent bundle ... is a trivial bundle" is confusing; under what hypotheses are you making this assertion? There are only two $\mathbb{R}^n$ bundles over $\mathbb{S}^1$ up to isomorphism, the trivial bundle $\mathbb{S}^1 \times \mathbb{R}^n$ and a nontrivial example. If you consider just a single continuous map $\gamma : \mathbb{S}^1 \to M^n$, triviality of $\gamma^* TM$ is equivalent to saying that $\gamma$ preserves orientation. – Lee Mosher Jan 12 '18 at 17:06
• Do you know the fact that $M$ is orientable iff every loop on $M$ is orientation preserving? Meaning that you can find a finite partition of $I$ ($0=t_0<t_1...<t_n=1$) and charts $U_i$ such that the loop maps $[t_1, t_{i+1}]$ to $U_i$ and on each overlap $U_i\cap U_{i+1}$ (with the convention that $U_{n+1}=U_0$) the jacobian of the transition maps are positive. – Ahr Jan 12 '18 at 17:06
• @Warlock: Forget about $n$-forms per se, and just think about the line bundle $\mathscr L = \bigwedge^n T^*M$. $M$ is orientable if and only if $\mathscr L$ is trivial. Can you convince yourself that a line bundle on $M$ is trivial if and only if its restriction to the $1$-skeleton is trivial? (Think about the usual differential topology proof that a simply connected manifold is orientable.) – Ted Shifrin Jan 12 '18 at 18:41
• The meaning of the statement "The tangent bundle trivializes over the 1-skeleton $M^{(1)}$" is that $$TM \bigm| M^{(1)} \cong M^{(1)} \times \mathbb{R}^n$$ – Lee Mosher Jan 12 '18 at 19:16
• As Lee suggested, up to isomorphism, on a circle, there are only two vector bundles of rank $n$ [because $\pi_1(SO(n)) \cong \Bbb Z_2$]. One is an orientable vector bundle; the other is non-orientable. And of course the top exterior power detects precisely this: For a vector bundle $E$ on any reasonable space, $E$ is an orientable bundle of rank $n$ if and only if $\bigwedge^n E$ is the trivial line bundle. – Ted Shifrin Jan 12 '18 at 19:51

A manifold is orientable if and only if its first Stiefel-Whitney class vanishes, which is equivalent to say that this class vanishes on the $1$-skeleton.
Basically, by taking a good cover $(U_i)$ that is such that $(U_i\cap U_j)$ is connected of $M$, you can associate a $1$-Cech cocycle which takes its values in $\mathbb{Z}$, it is the sign of the determinant of the coordinate change. This $1$-cocycle is the first Stiefel-Whitney class and is trivial if the manifold is oriented. And $1$-cocycle is trivial if it is trivial on the $1$-skeleton.
• If the class vanishes on the skeleton it implies that the tangent bundle is trivial over the $1$-skeleton because of the dimension. – Tsemo Aristide Jan 12 '18 at 17:37
• You can also remark that an orientable vector bundle over $S^1$ is defined by a classifying map $S^1\rightarrow BSO(n)$, the homotopy exact sequence of Serre applied to the universal fibration $ESO(n)\rightarrow BSO(n)$ shows that $\pi_1(BSO(n))=1$ since $ESO(n)$ is contractible and $SO(n)$is connected, this implies that the bundle is trivial. – Tsemo Aristide Jan 12 '18 at 19:25