A map without fixed points - two wrong approaches

For the unit sphere $S^n \subset \mathbb{R}^{n+1}$ let $f : S^n \to S^n$ be the map reversing the signs of all but one coordinate, $$f(x_0, x_1, \dots, x_n) = (x_0, -x_1, \dots, -x_n):$$

(a) Compute the Lefschetz number $L(f)$.

My attempt to this question is Lefschetz number.

(b) For which values of $n$ is $f$ homotopic to a map without fixed points?

First consider

Poincare-Hopf Index Theorem. If $\vec{v}$ is a smooth vector field on the compact, oriented manifold $X$ with only finitely many zeros, then the global sum of the indices of $\vec{v}$ equals the Euler characteristic of $X$.

The Euler characteristic of an $n$-sphere is $1 + (-1)^n$. Hence, the degree of $f$ is $2$ if $k$ is even, and $0$ otherwise by Poincare-Hopf Index Theorem.

The Hopf Degree Theorem. Two maps of a compact, connected, oriented $k$-manifold $X$ into $S^k$ are homotopic if and only if they have the same degree.

We attempt a homotopy with $f$ and the antipodal map. They both are compact, connected, oriented $k$-manifold $X$ into $S^k$. So they are homotopic if and only if they have the same degree.

Based on Jared's absolutely worth reading answer The degree of antipodal map. is $(-1)^{k+1}$.

So, I am completely wrong here!

Alternatively,

Consider the homotopy $f_t = f + t(-2x_1)$, so that $f_0 = f$, but $f_1$ is the antipodal map.

So it is irrelevant to $n$! This also can't be right..

Solution 1:

To compute the Lefschetz number, you just look at the induced map on homology. $H_0(S^n)=H_n(S^n)=\mathbb Z$ and all other homology groups are zero. The induced map $H_0(f)\colon \mathbb Z\to\mathbb Z$ is the identity, as is always the case for a map from a connected space to itself. Since $f$ is invertible, $H_n(f)=\pm\mathrm{id}$. To figure out which one, we need to figure out whether it preserves or reverses orientation. If you write down charts and an orientation in these charts, you can see that $f$ preserves orientation iff $n$ is even. (In fact, $f$ is the suspension of the antipodal map on $S^{n-1}$.) So $H_n(f)=(-1)^n\mathrm{id}$. Thus the Lefschetz number is $$L(f)=(-1)^0Tr(H_0(f))+(-1)^nTr(H_n(f))=1+(-1)^n(-1)^n=2.$$

Solution 2:

WishingFish, beware $f_t$ is not an homotopy $f_t(x)\not\in S^n$ (for $t\neq0,1$).

The two fixed points of $f$ are : $A=(1,0,...,0)$ and $B=(-1,0,...,0)$.

Let $e_i,\ (i=1,...,n+1)$ denotes the canonical basis of $\mathbb{R}^{n+1}$.

Since $T_AS^n=<e_2,...,e_{n+1}>$, if $\gamma$ is a smooth curve in $S^n$ passing through $A$, such that $\gamma'(0)=e_i$ then: $$T_Af(e_i)=\frac{d}{dt}\big|_{t=0}f(\gamma(t))=-e_i$$ so that $D_Af=-\mathrm{Id}$ and also $D_Bf=-\mathrm{Id}$. The Lefschetz number is: $$L(f)=\mathrm{sign}\det(T_Af-\mathrm{Id})+\mathrm{sign}\det(T_Bf-\mathrm{Id})=(-1)^n+(-1)^n=2(-1)^n.$$