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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..

Community
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WishingFish
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  • A fixed point of a function/map $f\colon S\to S$ is an element $x\in S$ such that $f(x)=x$. That should answer your question about what a map without fixed points would be. – dfeuer Jul 25 '13 at 17:48
  • Oh yes, I know the definition, but I am trying to figure out which map I should think of. I actually think $f$ itself is... – WishingFish Jul 25 '13 at 17:51
  • $f(1,0,\dots,0)=(1,0,\dots,0)$ – dfeuer Jul 25 '13 at 17:55
  • Thanks @dfeuer, I've thought about this, but it is not a map without fixed points... It is a map completely are fixed points.. – WishingFish Jul 25 '13 at 17:58
  • Huh? Your $f$ has exactly one fixed point. – dfeuer Jul 25 '13 at 17:59
  • $f$ is just identity map - or your are implying restrict the domain to be $(1,0,\dots, 0)$? Thank you @dfeuer. – WishingFish Jul 25 '13 at 18:04
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    I mean exactly two fixed points. – dfeuer Jul 25 '13 at 18:10
  • Well if it is homotopic to any map without fixed points, it's likely homotopic to the antipodal map, so for certain $n$ try to construct a homotopy between $f$ and the antipodal map. – PVAL-inactive Jul 25 '13 at 18:33
  • Yes Indeed @PVAL, but I think $f$ is just an antipodal map with first coordinate fixed...? – WishingFish Jul 26 '13 at 17:02
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    @WishingFish, the point I was trying to make was that though obviously this map has fixed points, the antipodal map does not. So if you can show it is homotopic to the antipodal map then it is homotopic to a "map with no fixed points". I have seen this problem stated as when is $f$ is homotopic to the antipodal map. I am really not sure why that is enough (Maybe by removing a point and applying Brouwer's Fixed Point Theorem), but I do think that it is. – PVAL-inactive Jul 27 '13 at 19:36
  • Oh, so you mean do not invoke the degree theorem, but instead show $f$ is homotopic to antipodal map @PVAL? – WishingFish Jul 27 '13 at 19:46
  • Oh no, I tried as updated... my attempt is incorrect, please help @PVAL.. – WishingFish Jul 27 '13 at 20:16
  • Are you really sure that your $f_t$ is really a homotopy between maps on the sphere? I mean, if you think about sphere $S^k$ as $\sum_{i=1}^{k+1} x^2_i = 1$, $f_0$ and $f_1$ are obviously maps of sphere, but all the intermediate maps doesn't seem to be. – Evgeny Aug 01 '13 at 14:39

2 Answers2

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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 $$L(f)=(-1)^0Tr(H_0(f))+(-1)^nTr(H_n(f))=1+(-1)^n(-1)^n=2.$$

Cheerful Parsnip
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  • When I use the Guillemin and Pollack definition of Lefschetz number I get the same answer as @amine. But I follow this answer and it makes sense to me too. What gives? Do these definitions only agree up to sign? – Max Jul 30 '16 at 17:25
  • @Max: looks like the definitions agree up to a factor of $(-1)^n$ in general. – Cheerful Parsnip Jul 30 '16 at 20:12
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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: $$L(f)=\mathrm{sign}\det(T_Af-\mathrm{Id})+\mathrm{sign}\det(T_Bf-\mathrm{Id})=(-1)^n+(-1)^n=2(-1)^n.$$

amine
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