Problem 2-A

Question

Show that the unit sphere $S^n$ admits a vector field which is nowhere zero, providing that $n$ is odd. Show that the normal bundle of $S^n \subset \mathbf{R}^{n+1}$ is trivial for all $n$.

Amit Awasthi · Accepted Answer

\begin{proof}
Let $S^n \subset \mathbf{R}^{n+1}$ be the standard embedding and let $(x^1,\ldots,x^{n+1})$ be the standard coordinates on $\mathbf{R}^{n+1}$. For each $\vec{x} = (x^1,\ldots,x^{n+1}) \in S^n$, define $v(\vec{x}) := (x^2, -x^1, \ldots, x^{n+1}, -x^n) \in \mathbf{R}^{n+1}$. This function $v \colon S^n \to TS^n$ is a vector field, since 
$$
v(\vec{x}) \cdot \vec{x}  = (x^1 x^2 - x^2 x^1 ) + \ldots + (x^{n} x^{n+1} - x^{n+1} x^n) = 0 \textrm{ for all $\vec{x} \in S^n$.}
$$
It is clear that $v$ does not vanish.

To each $\vec{x} \in S^n$, let $s(\vec{x}) = (\vec{x},\vec{x})$, where the first coordinate denotes the basepoint and the second coordinate denotes the point on the line through the origin and $\vec{x}$. As the normal bundle of $S^n \subset \mathbf{R}^{n+1}$ has rank 1, the existence of the nowhere vanishing section $s$ implies that the bundle is trivial, by Theorem 2.2.

\end{proof}

Problem 2-A

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Problem 2-A

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