Problem 2-B

Question

If $S^n$ admits a vector field which is nowhere zero, show that the identity map of $S^n$ is homotopic to the antipodal map. For $n$ even, show that the antipodal map of $S^n$ is homotopic to the reflection 
$$
r(x^1,\ldots,x^{n+1}) = (-x^1,x^2,\ldots,x^{n+1}),
$$
and therefore has degree $-1$. Combining these facts, show that $S^n$ is not parallelizable for $n$ even, $n \geq 2$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Let $v \colon S^n \to \mathbf{R}^{n+1}$ be a nonwhere zero vector field, then for $t \in [0,\pi]$,
  $$
    (\vec{x},t) \mapsto H(\vec{x},t) := \cos(t) \vec{x} + \sin(t) \frac{v(\vec{x})}{\| v(\vec{x}) \|^2 }
  $$
  defines a homotopy from the identity map to the antipodal map (remark that $v$ is needed so that the homotopy stays on $S^n$: indeed, for any fixed $t$, $H(\vec{x},t) \cdot H(\vec{x},t) = 1$ since $\vec{x} \cdot v(\vec{x}) = 0$.)

Middle proof still to be added.

Let $n \geq 2$ be even and assume that $S^n$ is parallelizable, then there exist $n$ nowhere dependent (hence, nowhere zero) sections of $S^n$. Thus, the identity map is homotopic to the antipodal map, which is in turn homotopic to the above reflection; taking degrees, we find that $1= \deg(id) = \deg(-id) = \deg(r) = -1$, a contradiction.
\end{proof}

Problem 2-B

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Problem 2-B

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