Problem with normal bundle to sphere $\mathbb S^n\subset \mathbb R^{n+1}$

Question

I'm in trouble for understanding the normal bundle to $\mathbb S^n\subset\mathbb R^{n+1}$. By definition $$\nu(\mathbb S^n)=\bigcup_{p\in\mathbb S^n}\nu_p(\mathbb S^n),$$ where $\nu_p(\mathbb S^n)=\mathbb T_p\mathbb R^{n+1}/\mathbb T_p\mathbb S^n$. I have some questions about it:

(i) What do the elements of $\nu(\mathbb S^n)$ look like?

(ii) I know $T_\mathbb R^{n+1}\simeq \mathbb R^{n+1}$ whereas $T_p\mathbb S^n\simeq \mathbb R^n$, is it true that $T_p\mathbb R^{n+1}/\mathbb R^n\simeq \mathbb R^{n+1}/\mathbb R^n?$

(iii) Finally how can I stablish and isomorphism between $\nu(\mathbb S^n)$ and $\mathbb S^n\times \mathbb R$?

Thanks..

Answer 1

i) I think of elements as an ordered pair $(v,tv)$, where $|v|=1$ and $t$ is a real number. The normal bundle (when we're thinking about $S^n$ under its usual embedding in $\mathbb{R}^{n+1}$) is just the vectors that are orthogonal to $S^n$, so we can think of each point as lying on a line through the origin, so to nail down a a point we just need to know where this line pierces the sphere and how far along the line we are.

ii) I'm not sure what you mean by $\mathbb{R}^{n+1}/S^n$, I don't imagine that you mean that quotient space? If you meant to write $T_p \mathbb{R}^{n+1}/T_p S^n \cong \mathbb{R}^{n+1}/\mathbb{R}^n$, that is correct and sort of tautological.

iii) Do you know the result that a $1$ dimensional vector bundle is trivial if and only if it admits a nonvanishing section?

Answer 2

(i) Normal bundle over $S^n$ looks like a porcupine rolled into a ball: there's an orthogonal needle (1-D line) sticking out of every point on the sphere.

(ii) It's better to picture $T_pS^n$ not as an abstract $R^n$, but as the tangent plane to $S^n$ at point $p$. What matters here is how $T_pS^n$ changes as you move $p$ along $S^n$.

(iii) Your assertion is wrong for even $n$. Here's how you can show that: the tautological sum of the tangent bundle and normal bundle would be the trivial bundle of $R^{n+1}$ over each point on $S^n$: you can identify it with the $R^{n+1}$ containing the sphere. For the proposed triviality of the normal bundle to hold the tangent bundle would need to be trivial as well. This is not the case for even $n$ because Euler characteristic of $S^n$ is non-zero. Alternatively, one can deny triviality of the tangent bundle by using fixed point theorem to show that even-dimensional spheres cannot admit non-vanishing vector fields.