# Closed manifolds with the fixed point property

The real projective plane ${\bf P}^2({\bf R})$ is the only closed surface with the fixed point property: all continuous maps from the plane to itself has a fixed point. The projective spaces ${\bf P}^{2k}({\bf R})$ and ${\bf P}^{2k}({\bf C})$ also have this property; this follows from the Lefschetz fixed point formula. There seems to be few examples of manifolds with this property though. I think that odd dimensional orientable manifolds always have a vector field without zeros, and a map without fixed points can be obtained by integrating such a vector field.

Are there any other examples of compact manifolds without boundary with the fixed point property? Please explain briefly how the fixed point property is obtained.

Is there some hope to get a complete list of such manifolds?

## 3 Answers

According to this paper, there are various such manifolds constructed as cartesian products and connected sums of real, complex and quaternion projective spaces.

For example, it is shown that $\mathbb{R}P^4 \# \mathbb{R}P^4\# \mathbb{R}P^4$ has the fixed-point property. The paper also mentions that products such as $\mathbb{R}P^{2n} \times \mathbb{C}P^{2m}$ have that property.

• Nice. For completeness, the paper also mentions $({\bf P}^{2}({\bf C}))^n$ and ${\bf P}^{2m}({\bf C})\times {\bf P}^{n}({\bf H})$, ${\bf P}^{2m}({\bf R})\times {\bf P}^{n}({\bf H})$. – coudy Mar 28 '17 at 9:38

Products do not always have the property, but products with "disjoint" rational homology (in other words $X\times Y$ where $H_i(X, \mathbb{Q}) \neq 0$ implies that $H_i(Y, \mathbb{Q}) = 0)$ do. For more, see the Monthly(!) paper by Brown.

Brown, Robert F., The fixed point property and Cartesian products, Am. Math. Mon. 89, 654-668, 677-678 (1982). ZBL0499.54028.

• The article of Brown indeed contains two examples of 56-dimensional manifolds with the fixed point property whose product does not have the property. The construction is quite tricky. – coudy Mar 28 '17 at 16:59

The Lefschetz argument you mentioned, of course, also works on ${\bf P}^{2k}({\bf H})$ and ${\bf P}^2({\bf O})$.

What is a bit more surprising, is that ${\bf P}^k({\bf H})$ has the fixed point for any $k > 1$. (Of course, ${\bf P}^1({\bf H})\cong S^4$ does not have the fixed point property: the antipodal map has no fixed points).

This is proven in Hatcher's Algebraic Topology book (The end of example 4L.4, pg 493 in my edition) as an application of Steenrod powers with $p=3$.