Show that $\Bbb R^{2n+1}$ is not a division algebra over $\Bbb R$ for $n>0$

Question

This is an exercise from Hatcher's Algebraic Topology (exercise 2.B.8). Here is the problem statement:

Show that, for $n>0$, $\Bbb R^{2n+1}$ is not a division algebra over $\Bbb R$ by showing that if it were, then for nonzero $a \in \Bbb R^{2n+1}$ the map $S^{2n} \to S^{2n}$ defined by $x \mapsto \dfrac{ax}{|ax|}$ would be homotopic to $x \mapsto -\dfrac{ax}{|ax|}$, but these maps have different degrees.

I found a duplicate of this question here where a suggestion was to find a path between $a$ and $-a$ that does not pass through $0$. I am struggling to do this and I would appreciate if someone could give me some direction here. I don't have much experience with algebras or division algebras, so I may be missing something simple.

I am also somewhat confused about how to find the degrees of the maps $x \mapsto {ax\over |ax|}$ and $x \mapsto {-ax\over |ax|}$. As I understand it a map $S^{n} \to S^{n}$ induces a homomorphism $H_n(S^n) \to H_n(S^n)$, and $H_n(S^n)$ is isomorphic to $\Bbb Z$, so the induced homomorphism is of the form $x \mapsto nx$ for some $n \in \Bbb Z$. Then that $n$ is the degree of the original map. I'm not sure how to actually find the induced maps on homology in this problem (or if there is another way to find the degree).

Edit: As anomaly has pointed out, the second part of my question is simpler than I thought.

Answer 1

Too long for a comment:

It is nice to prove that there are no nontrivial odd-dimensional real division algebras using algebraic topology. However, it should be noted that there are, of course, far more elementary ways of proving such a statement. One of my favorites is the following (I think I learnt this one from Alain Hillion).

Suppose that the real vector space $\mathbf R^{m}$ carries the structure of a real division algebra, where $m$ is odd. We prove that $m=1$. Let $a\in \mathbf R^m$. Consider the multiplication-on-the-left-by-$a$ map from $\mathbf R^m$ into itself, which we denote again by $a$. Its characteristic polynomial $\det(\lambda I_m-a)$ is a polynomial of degree $m$ with real coefficients. Since $m$ is odd, it has a real root $\lambda$. This means that that there is a nonzero $v\in\mathbf R^m$ such that $(a-\lambda)v=0$. Since $\mathbf R^m$ is a division algebra and $v\neq0$, one has $a-\lambda=0$, i.e., $a=\lambda\in\mathbf R\cdot 1$ in $\mathbf R^m$. This proves that $\mathbf R^m=\mathbf R\cdot 1$, i.e., $m=1$.

Answer 2

For the first part, you want a path from $a \in \mathbb{R}^{2n+1}$ to $-a$ that doesn't go through zero. There are plenty of ways to do this; for example, let $b$ be some point which is not colinear with $0$ and $a$ (this exists because $2n+1 \ge 2$, draw a picture to convince yourself). Then concatenate the line segment from $a$ to $b$ and the line segment from $b$ to $-a$ to get a path $\gamma$ from $a$ to $-a$ that doesn't go through zero, i.e. $\gamma(0) = a$, $\gamma(1) = -a$ and $\gamma(t) \neq 0$ for all $t$.

This path then defines a homotopy $H : [0,1] \times S^{2n} \to S^{2n}$ given by $$H(t,x) = \frac{\gamma(t)x}{\|\gamma(t) x\|}.$$ Since the two maps are homotopic, they have the same degree.

But as anomaly mentions in the comments, $x \mapsto \frac{ax}{\|ax\|}$ has nonzero degree (say $n \neq 0$), and since $2n$ is even, $x \mapsto -x$ has degree $-1$; thus $x \mapsto \frac{-ax}{\|-ax\|}$ has degree $-n \neq n$. This contradicts the fact that they are homotopic.

Answer 3

Another answer that uses the hint Hatcher suggests:

Let $T_a : \mathbb R^{2n+1} \to \mathbb R^{2n+1}$ be the linear map $T_a(x) = ax$. By examining coordinates, since the multiplication map $(x,y) \mapsto xy$ is bilinear, one can show multiplication is continuous. In particular, if $\gamma : [0,1] \to \mathbb R^{2n+1} \setminus \{0\}$ is a path with $\gamma(0) = a$ and $\gamma(1) = -a$, then the map $$t \mapsto \det(T_{\gamma(t)})$$ is continuous. And, since $\mathbb R^{2n+1}$ is odd-dimensional and $T_{-a} = -T_a$, $$ \det(T_{-a}) = \det(-T_a) = -\det(T_a). $$ So $\det(T_{\gamma(1)}) = -\det(T_{\gamma(0)})$. By the intermediate value theorem, $\det(T_{\gamma(t_0)}) = 0$ for some $t_0 \in [0,1]$. Letting $b = \gamma(t_0)$, this means $b$ has zero divisors. This is impossible in a division algebra.