Mesh, Barycentric Subdivision and Simplicial Approximation Theorem · DL Ferrario's Test Web Page

# Mesh, Barycentric Subdivision and Simplicial Approximation Theorem

## Mesh of a euclidean simplicial complex.

Let $|K| \subset \RR^k$ an euclidean simplicial complex (i.e., an abstract simplicial complex, where the set of vertices actually is a set of points in $\RR^k$, such that $|K|$ is the disjoint union of the interiors of the convex hulls of the sets of vertices ranging over the set of simplices).

A vertex $P$ of $K$ is therefore a point of $\RR^k$.

For each simplex $\sigma \subset K$, the diameter of $\sigma$ (or $|\sigma|$ more precisely) is the maximum distance $|x-y|$ as $x,y$ vary in $|\sigma|$.

(1)
The diameter of $|\sigma| = [P_ 0, \ldots, P_ n]$ is the maximum of the distances $|P_ i - P_ j|$ for $i,j=0,\ldots, n$.

Proof: If $x,y \in \sigma$, then in barycentric coordinates $$x = \sum_ {j=0} ^n x_ j P_ j,$$ $$y = \sum_ {i=0} ^n y_ i P_ i,$$ with $x_ j \geq 0$, $y_ j \geq 0$ and $\sum_ j x_ j = \sum_ j y_ j = 1$. Hence \begin{aligned} x-y & = \sum_ {j=0}^ n x_ j P_ j - \sum_ {i=0} ^n y_ iP_ i \\ &= \sum_ {j=0}^n \left( \sum_ {i=0}^n y_ i \right) x_ j P_ j - \sum_ {i=0} ^n \left( \sum_ {j=0}^n x_ j \right) y_ iP_ i \\ &= \sum_ {i,j=0}^n x_ j y_ i \left( P_ j - P_ i \right) \\ & \implies \\ |x-y| & \leq \sum_ {i,j=0} ^n x_ j y_ i \max_ {i,j} |P_ i - P_ j| \\ &= \left( \sum_ {i,j=0} ^n x_ j y_ i \right) \max_ {i,j} |P_ i - P_ j| = \max_ {i,j} |P_ i - P_ j| \end{aligned} /qed/


Now, define the mesh of $K$, denoted by $\mesh K$, as the maximum diameter of all simplices of $K$.

## Barycentric subdivision of a euclidean simplicial complex.

Let $K$ be as above, and $K'$ the euclidean simplicial complex with as vertices the simplices of $K$, with coordinates the barycenter of the simplex $$b([P_ 0, \ldots, P_ n]) = \dfrac{1}{n+1} \sum_ {j=0}^n P_ j.$$ A simplex in $K'$ is an (strictly) ascending chain of simplices of $K$, $\sigma_ 0 < \sigma_ 1 \ldots, \sigma_ n$.

(2)
(Mesh Lemma) The following inequality holds: $$\mesh K' \leq \dfrac{\dim K}{\dim K + 1} \mesh K.$$

Proof: Let $\sigma_ 0$ and $\sigma_ 1$ be two vertices of a simplex in $K'$ $\implies$ $\sigma_ 0 \subset \sigma_ 1$, and therefore $$\sigma_ 0 = [P_ 0, \ldots, P_ n]$$ $$\sigma_ 1 = [P_ 0, \ldots, P_ n, Q_ 0,\ldots, Q_ m ].$$ Hence \begin{aligned} b(\sigma_ 0 ) - b ( \sigma_ 1) & = \dfrac{1}{n+1} \sum_ {j=0} ^n P_ j - \sum \dfrac{1}{n+m+2} \left( \sum_ {j=0} ^n P_ j - \sum_ {i=0} ^ m Q_ i \right) \\ &= \left( \dfrac{1}{n+1} - \sum \dfrac{1}{n+m+2} \right) \sum_ {j=0} ^n P_ j - \sum \dfrac{1}{n+m+2} \sum_ {i=0} ^ m Q_ i \\ &= \left( \dfrac{m+1}{n+m+2} \right) \left( b([P_ 0, \ldots, P_ n]) - b([Q_ 0, \ldots, Q_ m]) \right) . \end{aligned} Now, note that if $0 < k < k'$, then $\dfrac{1}{k'} > \dfrac{1}{k}$, $\implies$ $1 + \dfrac{1}{k'} > 1+ \dfrac{1}{k}$ $\implies$ $\dfrac{k'+1}{k} > \dfrac{k+1}{k}$ $\implies$ $\dfrac{k}{k+1} < \dfrac{k'}{k'+1}$. Hence from $n+m+1 \leq \dim K$ the following inequality follows $$\dfrac{m+1}{n+m+2} \leq \dfrac{\dim K}{\dim K + 1},$$ and since $$\left| b([P_ 0, \ldots, P_ n]) - b([Q_ 0, \ldots, Q_ m]) \right| \leq \mesh K$$ one deduces that $$\mesh K' \leq \dfrac{\dim K}{\dim K + 1} \mesh K .$$ /qed/

## Simplicial Approximation Theorem

Recall that, for $x\in |K|\subset \RR^k$, we denote by $\supp x$ the support of $x$, i.e. the (unique) set of vertices of $K$ with positive barycentric coordinates of $x$, i.e. if $x\in |K|$, then $$x = \sum_ {j=0} ^n x_ j P_ j$$ with $x_ j >0$, $\sum_ j x_ j = 1$, and $\supp x = \{ P_ 0, \ldots, P_ n \}$. The support of $P$ is a simplex of $K$. Let $|x|$ denote its geometric realization $|x|\subset |K|$.

For each vertex $v$ of $K$, let $A(v)$ denote the set of all points $P\in |K|$ such that $v \in \supp P$. It is an open subset of $|K|$, such that :

(3)
$[P_ 0, \ldots, P_ n]$ is a simplex of $K$ if and only if $\bigcap_ {j=0} ^n A(P_ j) \neq \emptyset$. Moreover, the diameter of $A(P)$ is less than or equal to $\mesh K$.

Let $f \from |K| \to |L|$ be a continuous map. Let $\delta$ be the Lebesgue number of the open covering $$\{ f^{-1} A ( Q) \}_ {Q}$$ where $Q$ ranges over all vertices of $L$ and $A(Q) \subset |L|$ is defined as above. There is an iterated barycentric subdivision $K'$ such that $\mesh K' \leq \epsilon/2$. Hence for each vertex $P'$ of $K'$, the diameter of $A(P')$ is at most $2 \mesh K' \leq \epsilon$. Define the map $\varphi \from K' \to L$ as follows: for each $P'$ vertex of $K'$, $A(P')$ is an open subset of $|K'| \equiv |K|$ of diameter $\leq \epsilon$, and hence it is contained in at least one of the $f^{-1} A(Q)$. Choose one of the $Q$ with such property, and set $\varphi (P') = Q$.

It happens that $\varphi \from K' \to K$ is a simplicial function: If $[P'_ 0, \ldots, P'_ n]$ is a simplex in $K'$, then by (3) $\bigcap_ {j=0} ^n A(P'_ j) \neq \emptyset$, and hence $\bigcap_ {j=0} ^n f ( A(P'_ j) ) \neq \emptyset$, $\implies$ $\bigcap_ {j=0} ^n A(\varphi (P'_ j) ) \neq \emptyset$, which implies that $[\varphi P'_ 0, \ldots, \varphi P'_ n]$ is an $n$-simplex in $L$.

Moreover, for any $x \in |K'|$, the geometric realization of $\varphi$ sends $$x = \sum_ {j=0} ^n x_ j P'_ j \mapsto |\varphi|(x) = \sum_ {j=0} ^n x_ j \varphi(P'_ j) = \sum_ {j=0} ^n x_ j Q_ j \in |L|$$ with the property that $f( A(P'_ j) ) \subset A(Q_ j)$ (assume all $x_j >0$). This implies that for any $x\in |K'|$, $|\varphi|(x) \in | \supp f(x)|$.

A simplicial function $\varphi \from K' \to L$ such that for each $x\in |K'|=|K|$ the equality $$|\varphi|(x) \in | \supp f(x) |$$ holds is termed a simplicial approximation function of $f$.

(4)
Given a continuous map $f\from |K| \to |L|$, there exists a suitable iterated barycentric subdivision $K' = K^{(r)}$ and a simplicial map $\varphi \from K' \to L$ such that $\varphi$ is a simplicial approximation of $f$. Furthermore $|\varphi|\from |K'|=|K| \to |L|$ is homotopic to $f$.

## Hauptvermutung and homeomorphisms

Let $K$ and $L$ be two simplicial complexes.

(H)
(Question) If $|K| \approx |L|$, then is it true that there are subdivisions $K'$ and $L'$ (of $K$ and $L$ respectively) such that $K'=L'$?

Radó (1920s): Yes, in dimension 2.

Moise (1950s): Yes, in dimension 3.

Milnor (1961, Fields Medal 1962, Wolf Prize 1989, Abel Prize 2011): Answer is "No". Counterexample in dimension $\geq 6$. {LINK}

(Question) If $M$ is a $C^k$ manifold, is it true that there is a simplicial complex $K$ such that $M\approx |K|$? In other words, is a $C^k$ manifold triangulable?

Radó (1925): Any $C^0$ $2$-manifold is combinatorially triangulable.

Moise (1952): Any $C^0$ $3$-manifold is combinatorially triangulable.

Cairns (1935); Whitehead (1940): Every smooth manifold is (combinatorially) triangulable.

Kirby-Siebenmann (1969): No, for topological manifolds of dimension $\geq 5$.

Freedman (1982): No, there is a $4$-dimensional $C^0$-manifold which has no combinatorial triangulation, the $E_ 8$ manifold.

Casson (1990): Freedman's $E_ 8$ manifold is not triangulable.

Manolescu (2013): No, for any dimension $\geq 5$ there are $C^0$ manifolds without combinatorial triangulations.

{LINK (PDF)} from CIPRIAN MANOLESCU webpage.