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I want to prove a statement about the fixed points of simplicial maps.

If $f: |K|\to |K|$ is a simplicial map prove that the set of fixed points of $f$ is the polyhedron of a subcomplex of $K^1$ (where $K^1$ denotes the first barycentric subdivision of the simplicial complex $K$), though not necessarily of a subcomplex of $K$.

My ideas:

If a fixed point is a vertex of $K$ it is also a barycentre of a $0$-simplex in $K$.

If a fixed point lies in the interior of a simplex $S$ of $K$ then $f$ must take that simplex to itself. Since this induces a permutation on the vertices of $S$ I can prove that the barycentre $A_S$ is a fixed point too.

I can collect all the barycenters in a set $M$. Then $M$ is a subset of the vertices of $K^1$.

If I define $L$ to be the simplex that has the elements of $M$ as its vertices then each element of $L$ is a fixed point of $f$.

My problems with this construction are:

i) I don't think I can prove that each fixed point of $f$ is also in $L$.

ii) I don't see how to prove that $L$ is a simplex of $K^1$. If $M=\{A_0,...,A_k\}$ and $S_0,...,S_k$ are the corresponding simplexes I would have to prove that there is a permutation $\sigma$ of the integers $0,1,2,...,k$ such that

$S_{\sigma(0)} \lt S_{\sigma(1)} \lt ... \lt S_{\sigma(k)}$ where the notation $A \lt B$ means that the simplex $A$ has to be a face of the simplex $B$.

If I succeed in proving i) and ii) I can define the set $H=\{L\}$. By the above results this would be a subcomplex of $K^1$ such that the polyhedron $|H|$ is the set of fixed points of $f$. But I do not know how to prove i) and ii).

blancket
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Polymorph
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1 Answers1

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Let $x$ be a fixed point of $f$. Then $x$ is an interior point of a simplex $\Delta^{\prime k}$ of $K_1$ that is lying in a simplex $[v_0, \dots, v_{n} ]$ of $K.$

Let's prove that the whole $\Delta^{\prime k}$ lies in the set $A$ of fixed points of $f$.

Proof

Let $S_{0} \lt S_{1} \lt ... \lt S_{k}$ be the chain of faces, corresponding to $\Delta^{\prime k}$, $d_i = \operatorname{dim}(S_{i})$. We can re-enumerate $\{v_i \}$ so that $S_{i} = [v_0, \dots, v_{d_i} ].$ Note that the sequence $\{ d_i \}$ is increasing. Let $\{w_i\}_{i=0}^k$ be barycenters of $\{ S_i \} $: \begin{equation} w_i = \frac{1}{d_i + 1} \sum_{l=0}^{d_i} v_l . \end{equation} There are $\{ \alpha_i \}, i = 0\dots k,$ such that $ x = \Sigma \alpha_i w_i, \Sigma \alpha_i = 1, \alpha_i \gt 0. $

By substituting $w_i$ we get $\{ a_i \}$ such that: $ x = \sum_{i = 0}^{n} a_i v_i $ and $ \forall i $ $ a_i \ge a_{i + 1}. $ It's easy to see from the definition of $\{ w_i \}$ that $a_i \gt a_{i + 1} $ iff $ i = d_l$ for some $l$.

$$ f(x) = \sum_{i = 0}^{n} a_i f(v_i) = \sum_{i = 0}^{n} a_{\sigma (i)} v_i, $$ here $\sigma$ is the inverse permutation to $f$ on indices of $\{v_i \}$. Thus, $ \forall i $ $a_{\sigma(i)} = a_i,$ so $f(v_j) = v_k $ only if $a_j = a_k.$ By using the properties of ${a_i}$ above, we see that each $S_i$ is invariant under action of $f.$ This proves that each $w_i$ is in $A.$ $f$ is a simplicial map, thus any point of $\Delta^{\prime k}$ is in $A.$

QED

By using this fact we get that $A$ is a union of simplices of $K_1.$

Here's an example, where $A$ is not a subcomplex of $K:$ $$ K = \Delta^2 = [v_0, v_1, v_2], $$ $$ f(v_0) = v_0, $$ $$ f(v_1) = v_2, f(v_2) = v_1. $$

Igor Ernst
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