Fixed-point property

This article defines a property of topological spaces: a property that can be evaluated to true/false for any topological space|View a complete list of properties of topological spaces

Definition

A topological space is said to have the fixed-point property if every continuous map (not necessarily a self-homeomorphism) from the topological space to itself has a fixed point.

Relation with other properties

Stronger properties

acyclic compact polyhedron (nonempty)
rationally acyclic compact polyhedron (nonempty)

Examples

Manifolds without boundary

Manifold or family of manifolds	Dimension in terms of parameter	Does it satisfy the fixed-point property?	Proof/explanation
sphere $S^{n}, n \geq 1$	$n$	No	The antipodal map is a fixed-point-free self-map (in fact, it's a self-homeomorphism without fixed points).
real projective space $P^{n} (R), n \geq 1$	$n$	Even $n$ : Yes Odd $n$ : No	For $n$ even, follows from rationally acyclic compact polyhedron has fixed-point property. For $n$ odd, we use the mapping $[a_{1} : a_{2} : a_{3} : a_{4} : \dots : a_{n} : a_{n + 1}] \mapsto [- a_{2} : a_{1} : - a_{4} : a_{3} : \dots : - a_{n + 1} : a_{n}]$ . Note that the map has no fixed points because solving the fixed point condition gives $a_{1}^{2} + a_{2}^{2} = a_{3}^{2} + a_{4}^{2} = \dots = a_{n}^{2} + a_{n + 1}^{2} = 0$ , forcing $a_{1} = a_{2} = \dots = a_{n} = a_{n + 1} = 0$ .
complex projective space $P^{n} (C), n \geq 1$	$2 n$	Even $n$ : Yes Odd $n$ : No	For $n$ odd, we use the mapping $[a_{1} : a_{2} : a_{3} : a_{4} : \dots : a_{n} : a_{n + 1}] \mapsto [- \bar{a_{2}} : \bar{a_{1}} : - \bar{a_{4}} : \bar{a_{3}} : \dots : - \bar{a_{n + 1}} : \bar{a_{n}}]$ . Note that the map has no fixed points because solving the fixed point condition gives $\| a_{1} \|^{2} + \| a_{2} \|^{2} = \| a_{3} \|^{2} + \| a_{4} \|^{2} = \dots = \| a_{n} \|^{2} + \| a_{n + 1} \|^{2} = 0$ , forcing $a_{1} = a_{2} = \dots = a_{n} = a_{n + 1} = 0$ .
compact orientable surface of genus $g \geq 0$	2	$g = 0, 1$ : No $g \geq 2$ : Yes (need to verify)
torus $T^{n}, n \geq 1$ , product of $n$ circles	$n$	No	Follows from the fact that it's a nontrivial Lie group, so multiplication by a non-identity element. Also, we can use the coordinate-wise antipodal map.
product of spheres $S^{m_{1}} \times S^{m_{2}} \times \dots S^{m_{r}}$	$m_{1} + m_{2} + \dots + m_{r}$	No	Take the antipodal map for each coordinate sphere.

Manifolds with boundary

Facts

Fixed-point theorems

In general, we combine the Lefschetz fixed-point theorem with the structure of the cohomology ring of the space to determine whether or not it has the fixed-point property. For instance, we can show that complex projective space in even dimensions has the fixed-point property, by combining the Lefschetz fixed-point theorem with the fact that the trace on the $(2 k)^{t h}$ homology is $d^{k}$ where $d$ is the trace on the second homology.

Products

The product of any topological space that does not satisfy the fixed-point property with any nonempty topological space gives a space that does not satisfy the fixed-point property.

Metaproperties

Retract-hereditariness

This property of topological spaces is hereditary on retracts, viz if a space has the property, so does any retract of it
View all retract-hereditary properties of topological spaces

Every retract of a space with the fixed-point property also has the fixed-point property. Further information: fixed-point property is retract-hereditary