Here is a bit from Hatcher's book:

enter image description here

I don't understand part (f); why is the antipodal map the composition of $n+1$ reflections? Even if I accept that, I still don't know why does it have degree $(-1)^{n+1}$. What would that mean for $S^2$? We fix the equator $S^1$ and take every other point $x$ to $-x$? And then we are doing this for $S^1$? That sounds a little pointless. In general if $\operatorname{deg}f=x$ and $\operatorname{deg}g=y$ then what is $\operatorname{deg}f\circ g$? It's $xy$ I guess.

  • 3
    $\begingroup$ On $S^2$. You take the map $f_x \colon S^2 \to S^2 \,\, f_x(x,y,z) = (-x,y,z)$ and define $f_y$ and $f_z$ in a similar way. Then observe that the antipodal map $-\mathbb{1}$ is just $-\mathbb{1} = f_x \circ f_y \circ f_z$. By (e), each of the $f_i$ has degree $-1$, so by (d) you have $\mathrm{deg}\, (-\mathbb{1}) = (-1)(-1)(-1) = -1$. Just do the same things on $S^n$. $\endgroup$ – Ivo Oct 16 '13 at 13:12
  • $\begingroup$ possible duplicate of The degree of antipodal map. $\endgroup$ – Noah Snyder Dec 5 '14 at 4:06
  • $\begingroup$ which page of Hatcher? $\endgroup$ – Idonotknow Nov 3 '18 at 18:23

Consider $S^n$ as a subspace of $\mathbb{R}^{n+1}$. Then a point $x$ in $S^n$ is given by an $(n+1)$--tuple $(x_0,\dotsc,x_n)$, hence the antipodal map is given by $$ (x_0,\dotsc,x_n)\mapsto (-x_0,\dotsc,-x_n) $$ we can get this map by composing all $n+1$ reflections with respect to the hyperplanesplanes orthogonal to the coordinate axes. (I.e. the maps $$ (x_0,\dotsc, x_k,\dotsc,x_n)\mapsto (x_0,\dotsc,-x_k,\dotsc,x_n) $$ Furthermore, since homology is functorial, we have $H_n(f\circ g,\mathbb{Z}) = H_n(f,\mathbb{Z})\circ H_n(g,\mathbb{Z})$. Thus given $f,g\colon S^n\to S^n$ with $$ H_n(f,\mathbb{Z})=f_*\colon H_n(S^n,\mathbb{Z})\to H_n(S^n,\mathbb{Z}) $$ $$x\mapsto\alpha x$$ and $$ H_n(g,\mathbb{Z})=g_*\colon H_n(S^n,\mathbb{Z})\to H_n(S^n,\mathbb{Z}) $$

$$x\mapsto\beta x $$ we have $(f\circ g)_*(x) = f_*\circ g_*(x) = f_*(\beta x) = \alpha\beta x$, so as you assumed $\operatorname{deg}(f\circ g) = \operatorname{deg}(f)\operatorname{deg}(g)$. And combining both results, we get the degree $(-1)^{n+1}$ for the antipodal map on $S^n$.

  • $\begingroup$ If it is $S^\infty$ how to compute the deg ? For example, $E$ is infinity dimensional Banach space,$\Omega={x\in E:||x||=1}$,$L:\Omega\rightarrow\Omega ,L(x)=-x$ $\endgroup$ – lanse2pty Mar 2 '16 at 2:01

If $u$ is a unit vector in $\mathbf{R}^{n+1}$ and $H = u^\perp$ is the hyperplane orthogonal to $u$, then reflection in $H$ is the linear transformation $R_u(v) = v - 2\langle u, v\rangle u$.

For example, if $u = \mathbf{e}_i$ is the $i$th standard basis vector (so $H$ is the hyperplane $\{x_i = 0\}$), then $$ R_u(x_1, \dots, x_{i-1}, x_i, x_{i+1}, \dots, x_{n+1}) = (x_1, \dots, x_{i-1}, -x_i, x_{i+1}, \dots, x_{n+1}). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.