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
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$\begingroup$ Is there an example where $f$ has no fixed point? $\endgroup$– Noam D. ElkiesCommented Jul 9, 2013 at 6:33
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5$\begingroup$ @Noam: Yes. On one sphere, send each point to its antipode. Send everything else to the antipode of the base point. $\endgroup$– Will SawinCommented Jul 9, 2013 at 6:37
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4$\begingroup$ 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 $det(M)=1$. $\endgroup$– Will SawinCommented Jul 9, 2013 at 6:45
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$\begingroup$ @Will I do not know what are you saying, yes or no? $\endgroup$– Pedro PerezCommented Jul 9, 2013 at 7:12
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2$\begingroup$ @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. $\endgroup$– Neil StricklandCommented Jul 9, 2013 at 7:43
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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.
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5$\begingroup$ 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. $\endgroup$ Commented Jul 9, 2013 at 13:21
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$\begingroup$ 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. $\endgroup$ Commented Jul 9, 2013 at 21:09