What I know:
- A tangent bundle $TM$ of a manifold $M$ is orientable if and only if the manifold $M$ is orientable.
- If the first Stiefel-Whitney class $w_1(E) \neq 0$, the vector bundle $E$ is non-orientable.
- The first Stiefel-Whitney class $w_1 \in H^1(M;\mathbb{Z}_2)$, where $M$ is the base space.
I read an Answer: If a manifold $M$ embedded in Euclidean space is non-orientable, its normal bundle must also be non-orientable. (from here is normal bundle of a manifold trivial?)
My question: How to prove the answer above using $w_1(NM)\neq 0$?
Best Answer
Suppose $M$ is embedded into the standard inner product space $\Bbb{R}^n$.
Since $TM \oplus NM \cong M \times \Bbb{R}^n$, by the Whitney sum formula, $w(TM) \cdot w(NM)=w(TM \oplus NM)=w(M \times \Bbb{R}^n)=1 \in H^0(M;\Bbb{Z}_2)$ where $w$ denotes the total Stiefel-Whitney class.
Specifically, $w_1(TM) + w_1(NM)=w_1(M \times \Bbb{R}^n)=0 \in H^1(M;\Bbb{Z}_2)$, which means $w_1(NM) = w_1(TM) \ne 0$.