Does there exist any non-trivial space (i.e not deformation retract onto a point) in $\mathbb R^n$ such that any continuous map from the space onto itself has a fixed point. I highly suspect that the quasi circle on $\mathbb R^2$ is an example. Yet I've not written down the (dirty) proof. But in this case all its homotpy groups are trivial. So if I assume my space as a manifold, then (QUESTION:) does this fixed point property force it to become a contractible manifold? I read somewhere that there exists a contractible compact manifold which does not satisfy this fixed point property. So does there exist any non-contractible manifold (compact) where this property follows? Or otherwise can anyone please provide an outline of how to prove that such a manifold is contractible?

  • 4
    $\begingroup$ There are many examples of orientable closed manifolds (hence not contractible since the top dimensional homology group is non zero) in that question: mathoverflow.net/questions/265743/… $\endgroup$
    – coudy
    Oct 24, 2017 at 17:04
  • $\begingroup$ @coudy Thanks, that's a nice source. $\endgroup$ Oct 24, 2017 at 17:08

2 Answers 2


Take the space $\mathbb{CP}^2$. Its cohomology ring is given by $\mathbb{Z}[a]/a^3$, where $a$ has degree $2$. A map $f:\mathbb{CP}^2\rightarrow \mathbb{CP}^2$ induces a map on the second cohomology group with $f^*(a)=k a$ with $k\in \mathbb{Z}$. From this you can compute the action on (co)homology on the other degrees. In degree zero it is the identity and on the fourth degree it is given by multiplying with $k^2$. Then the Lefschetz number of this map is seen to be $L(f)=k^2+k+1$. This number is never zero. A non-zero Lefschetz number implies a fixed point by the Lefschetz fixed point Theorem.

  • $\begingroup$ Wow!! Nice argument. I was trying to use Lefschetz fixed point theorem, but this one didn't come in my mind. $\endgroup$ Oct 24, 2017 at 14:15
  • 11
    $\begingroup$ The same argument shows that $\mathbb{CP}^{2n}$ has the fixed point property for all $n$. Incidentally, this implies the fundamental theorem of algebra, because it implies that every complex matrix has an eigenvalue. $\endgroup$ Oct 24, 2017 at 17:54
  • 1
    $\begingroup$ @QiaochuYuan I believe that it's fixable. If you assume that M is invertible (which is fine, otherwise 0 is an eigenvector) then it induces a self-map that has to send $x \in H^2$ to $k x$. When $n$ is odd the above argument works; when $n$ is even, $x^{n-1}$ is sent to $k^{n-1} x^{n-1}$, but then the constraint that M induces an orientation-preserving diffeomorphism (due to the complex structure) forces $k=1$, and so the Lefschetz number is $n$. $\endgroup$ Oct 24, 2017 at 21:07
  • 2
    $\begingroup$ I think that quaternions projective space $\mathbb{H}P^n$ gives more examples. Regarding $\mathbb{C}P^{2n}$ there is an alternative proof of FPP without using Lefschetz number as follows: If $f$ is a fixed point free map on $\mathbb{C}P^{2n}$ then $f^{*}(\ell)$ is transverse to $\ell$ where $\ell$ is the tautological line bundle. We consider both bundles $\ell$ and $f^{*}(\ell)$ as subbundles of the trivial $(n+1)$ dimensional bundle $\epsilon_{n+1}$. $\endgroup$ Oct 25, 2017 at 3:44
  • 1
    $\begingroup$ @AliTaghavi I think your argument also shows that any self-map of $\mathbb{C}P^m$ has fixed points even when $m$ is odd. In that case we can take $f([z])=[\overline{z}]$ and we find that $f^*(x)=-x$ in cohomology and the Lefschetz number is zero. The fixed point set is $\mathbb{R}P^m$ which is nonempty but has Euler characteristic zero, consistent with the Lefschetz theorem. A nice example! $\endgroup$ Oct 25, 2017 at 10:29

Another nice example is $\mathbb RP^{2n}$. Here is a simple proof...

If $f:\mathbb RP^{2n} \to \mathbb RP^{2n}$ is a continuous function then look at the lift of the map $\bar{f} :S^{2n}\to S^{2n}$. A fixed point of $f$ is equivalent to a point $x\in S^{2n}$ s.t. $\bar{f}(x)=x \ or \ -x$. If such a point doesn't exist, then we can have a homotopy between Identitity and Antipodal map via $H(x,t)=cos(t)x+sin(t)\bar{f}(x)$.[This is well defined map from sphere to sphere]. But for even dimensional sphere antipodal map has degree $-1$. That's a contradiction.

  • 4
    $\begingroup$ This should also follow from the Lefschetz fixed point theorem, since $\Bbb{RP}^{2n}$ has the rational homology of a point. (Compact simplicial complexes whose homology is all torsion fall to the same argument.) $\endgroup$
    – mme
    Oct 26, 2017 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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