Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be the wedge sum of two $2$-dimensional spheres and $f$ a continuous function from $X$ into $X$. Does $f$ have either a fixed point or a periodic point of order 2? Thanks

share|improve this question
    
Is there an example where $f$ has no fixed point? –  Noam D. Elkies Jul 9 '13 at 6:33
4  
@Noam: Yes. On one sphere, send each point to its antipode. Send everything else to the antipode of the base point. –  Will Sawin Jul 9 '13 at 6:37
4  
By the Lefschetz trace formula, if $M$ is the $2 \times 2$ matrix that gives the action of $f$ on $H^2$, then $tr(M)=-1$ and $tr(M^2)=-1$, so $tr(M)^2-2 det (M) = -1$, so $det(M)=1$. –  Will Sawin Jul 9 '13 at 6:45
    
@Will I do not know what are you saying, yes or no? –  Pedro Perez Jul 9 '13 at 7:12
2  
@PedroPerez Will's argument says that if $f$ is a self-map such that neither $f$ nor $f^2$ has a fixed point, then $f$ must act with determinant one on $H^2(X)\simeq\mathbb{Z}^2$. In particular, $f$ must be a homotopy equivalence. I don't think that Will is making any claim about whether such a map $f$ exists. –  Neil Strickland Jul 9 '13 at 7:43
add comment

1 Answer

Will's comment above shows that the action of $f$ on $H^2$ is of order 3 (the eigenvalues are the nontrivial third roots of unity) and something like $\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$. So I thought one should try a construction using third roots of unity realizing this.

Write the wedge as $S^1 \times [0,2] / \sim $ where one identifies $(s,0)\sim (s',0), (s,1)\sim (s',1), (s,2)\sim (s',2)$.

Let $\zeta$ be a third root of unity, and define

$f: S^1 \times [0,2] / \sim\ \to \ S^1 \times [0,2] / \sim$

by

$f(s,t)=(\zeta s, 2-2t)$ for $t\le 1$ and $f(s,t)=(\zeta s, t-1)$ for $t\ge 1$

Note that the three points where $t=0,t=1,t=2$ are permuted cyclically. This implies $f$ and $f^2$ have no fixed points (for all other points the $\zeta$ respectively $\zeta^2$ factor works).

Note also that multiplication by $\zeta$ is not really necessary, almost any rotation of the circle would be fine. Multiplication by $\zeta$ allows $f^3$ to be close to the identity. If one uses a different rotation, the three points with $t=0,t=1,t=2$ are the only fixed points.

share|improve this answer
5  
Let me connect the dots of what nsrt and @WillSawin are saying: If $f$ and $f^2$ are fixed point free, then $Tr(M) = Tr(M^2) = -1$. Let the eigenvalues of $M$ be $\lambda$ and $\mu$. Then $\lambda+\mu = \lambda^2 + \mu^2 = -1$. The only solution to these equations is $(\lambda, \mu) = (\zeta, \zeta^{-1})$ for $\zeta$ a primitive cube root of unity. In particular, $M^3$ is the identity, so $f^3$ has fixed points and, if they are isolated smooth points of the spheres, then there are $3$ of them. –  David Speyer Jul 9 '13 at 13:21
    
I want to make a very small remark. The rotation by $\zeta$ is not necessary in the second branch ($t\geq 1$) of the definition of $f$. So the following would also work: $f(s,t)=(\zeta s, 2-2t)$ for $t\leq 1$, and $f(s,t)=(s, t-1)$ for $t\geq 1$. Here, and also in the answer above, $\zeta$ can actually be any self-map of the circle such that $\zeta$ and $\zeta^2$ do not have fixed points: for example, $\zeta$ can be any rotation of the circle which is neither the identity nor the antipodal map. –  Ricardo Andrade Jul 9 '13 at 21:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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