Suppose we are given a triangulable topological space $X$. If $X$ has the fixed point property (FPP), then obviously for every triangulation $K$ of $X$ and every simplicial map $f:K\to K$ a simplex $\sigma\in K$ exists such that $f(\sigma)=\sigma$. This will be called the fixed simplex property (FSP).

One can give examples of simplicial complexes (included below) with FSP whose geometric realizations do not have FPP. Moreover, triangulations of spaces without FPP exist whose iterated barycentric subdivisions all have FSP.

Is it true that if $X$ is a triangulable space without FPP, then a triangulation of $X$ exists that does not have FSP? A related question: given a triangulation $K$ of $X$, can we find a (non-barycentric) subdivision of $K$ that does not have FSP?

Example (taken from J.A. Barmak's thesis (p. 101), who in turn cites K.Baclawski and A.Björner "Fixed points in partially ordered sets" (Adv. Math. 1979)):

Consider the regular CW-complex (square + 4 triangles) $C$ which is the border of a pyramid with square base. $C$ is homeomorphic to $S^2$, so it doesn't have FPP. Let $K$ be the following subdivision of $C$: divide each of the 3 triangular sides into 3 triangles by adding a vertex in the middle and divide the bottom square into 4 triangles the same way.

Name the top vertex of the pyramid $x$. Let $f:K\to K$ be simplicial. If $f$ is onto (on the vertices), then, by finiteness of $K$, $f$ is an automorphism. Since $x$ is the uniqe vertex that belongs to exactly eight 1-simplices, it is a fixed point. If $f$ is not onto, then $|f|$ is nullhomotopic and thus the Lefschetz number $\lambda(f)=1$. $|f|$ has a fixed point, so $f$ fixes a simplex.

The same argument (just count the 1-simplices that $x$ and other vertices belong to) applies to arbitrary barycentric subdivisions of $K$.

However, if in $C$ we subdivide only the square base into 4 triangles by adding a vertex in the middle, then the obtained simplicial complex doesn't have FSP.

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    $\begingroup$ Can you actually give us the examples you mention in your second paragraph? $\endgroup$ Jan 25, 2010 at 22:25
  • $\begingroup$ Example is now included. Hopefully without mistakes. $\endgroup$ Jan 25, 2010 at 23:58
  • $\begingroup$ I corrected the example. The barycentric subdivision of C has a 2nd vertex with 8 incident edges in the middle of the square, which is no good for our argument. $\endgroup$ Jan 26, 2010 at 8:55

1 Answer 1


EDITED. The arugment related to Mostov rigidity is completed according to a nice suggestion of Tom Church

The answer to the first question is no. There exsit manifolds of dimension 3 such that every simlicial map of the manfiold to itself (for any simplicial decomposition) has a fixed point (and hence a fixed simplex). At the same time every 3-mafiold adimits a smooth self-map without fixed points.

Namely, take $M^3$ with vanishing first and second homology ($H_1(M^3,R)=H_2(M^3,R)=0$) and that is a hyperbolic 3-manifold. Moreover take such $M^3$ that does not have isometries. Existence of such manifolds is a standard result of 3-dimensional hyperbolic geometry. Let us prove that every such manifold gives us an example.

Proof. All compact 3-manifolds have zero Euler characteristics, so on $M^3$ there is a non-vanishing vector field $v$. Take the flow $F_t$ generated by $v$ in small time $t$. This will give us a family of diffeo $F_t$ of $M^3$ that don't have fixed points for small $t$. So $M^3$ in not FPP.

Now, let us show that $M^3$ has FSP. Take any simplicial decomposition of $M^3$. First we state a simple lemma (without a a proof)

Lemma. Consider a simplicial decomposition of a compact orienable manifold. Suppose we have a simpicial map from it to itself, that send simplexes of highest dimensions to simplexes of higher dimentions (i.e. don't collapse them) and don't indentify them. Then this is an automorphism of finite order.

Corollary. Every non-identical simplicial map $\phi$ from $M^3$ to itself either collapses a simplex of dimension 3 or identifies two such simplexes. In paricular the generator of $H^3(M^3,\mathbb Z)$ is sent to zero by this map.

This corollary together with Lefshetz fixed point theorem implies immediately that $\phi$ has a fixed point, and so it proves FSP for $M^3$ (we use that $H_1(M^3)=H_2(M^3)=0$).

Proof of corollary. If $\phi$ does not collapse 3 simplexes of $M^3$ or identify them, then it is a homeomorphism of $M^3$ of finite order (Lemma above). From Mostov rigidity it follows that this automorphism is homotopic to the identity. In order to show that it IS in fact the identity we need to use a more involved statement suggseted by Tom Church below. Namely, a paritial case of Proposition 1.1 in http://www.math.uchicago.edu/~farb/papers/hidden.pdf
says that for a hyperbolic 3-manifold the group of isometries of any Riemanninan metric on it is isomorphic to a subgroup of the group of hyperbolic isometries. By our choice the group of hyperbolic isometries of $M^3$ is trivial. It is clear that $\phi$ preserves a Riemannian metric on $M^3$. So by Prop 1.1 it is the identity.

From this it immediately follows that $\phi$ sends $H^3(M^3, Z)$ to zero (since the volume is contacted). End of proof.

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    $\begingroup$ Applying Mostow rigidity gives only that the finite-order homeomorphism phi is homotopic to the identity (since although phi preserves some metric, it need not be a hyperbolic isometry). To conclude that phi is the identity, you need a further argument. One way to see this is to apply Smith theory to a lift of phi acting on hyperbolic space; for the details, plus two other proofs of this fact, see Section 2 of Farb-Weinberger, "Hidden symmetries and arithmetic manifolds" (Geometry, spectral theory, groups, and dynamics, p111-119, math.uchicago.edu/~farb/papers/hidden.pdf) $\endgroup$
    – Tom Church
    Jan 26, 2010 at 22:41
  • $\begingroup$ Tom, many thanks for your comment and your refference! Indded my argument with Mostov rigidity was not complete. I incorporated your remark an the answer :) Thanks! $\endgroup$ Jan 26, 2010 at 23:14
  • $\begingroup$ Thanks for the answer, though it looks like I'm going to have to learn a lot in order to understand it. :) $\endgroup$ Jan 28, 2010 at 12:27

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