Topological Dimension of Polyhedra

Michel Coornaert

doi:10.1007/978-3-319-19794-4_3

Topological Dimension and Dynamical Systems pp 49-67 | Cite as

Topological Dimension of Polyhedra

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First Online: 21 June 2015

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Abstract

In this chapter, we introduce the notion of a simplicial complex of $\mathbb {R}^n$ and that of a polyhedron. A simplicial complex is a finite assembly of simplices and a polyhedron is a topological space that is homeomorphic to some simplicial complex..

Keywords

Simplicial Complex Topological Dimension Combinatorial Dimension Geometric Realization Open Star

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Notes

The Lebesgue lemma (cf. Lemma 3.5.2) about the finite open covers of the n-cube was stated by Lebesgue in [66]. It was used by Lebesgue in order to show that $[0,1]^n$ and $[0,1]^m$ are not homeomorphic for $n \not = m$. The proof of Lebesgue’s lemma given in [66] contains an error as was pointed out by Brouwer in [19]. A corrected proof appeared in [67].

Let $D^n$ (resp. $\mathbb {S}^{n-1}$) denote the closed unit ball (resp. the unit sphere) in the Euclidean space $\mathbb {R}^n$. The Brouwer fixed point theorem, which says that every continuous map $f :D^n \rightarrow D^n$ admits at least one fixed point, may be proved by using tools from algebraic topology (see for example [101]) in the following way.

Suppose for contradiction that there exists a continuous map $f :D^n \rightarrow D^n$ without fixed points ($n \ge 2$). Consider the map $g :D^n \rightarrow S^{n-1}$ that sends every point $x \in D^n$ to the intersection point of the half-line starting from f(x) and passing through x with the boundary sphere $\mathbb {S}^{n-1}$ (see Fig. 3.4). Clearly g is continuous. On the other hand, the map g fixes every point belonging to $\mathbb {S}^{n-1}$, that is, it satisfies

$$\begin{aligned} g \circ h = i, \end{aligned}$$

(3.6.1)

where h is the inclusion map $\mathbb {S}^{n-1} \rightarrow D^n$ and i is the identity map on $\mathbb {S}^{n-1}$. On the level of $n-1$-dimensional real homology, the maps g, h and i induce linear maps $g_* :H_{n-1}(D^n) \rightarrow H_{n-1}(\mathbb {S}^{n-1})$, $h_* :H_{n-1}(\mathbb {S}^{n-1}) \rightarrow H_{n-1}(\mathbb {S}^{n-1})$ and $i_* = {{\mathrm{Id}}}:H_{n-1}(\mathbb {S}^{n-1}) \rightarrow H_{n-1}(\mathbb {S}^{n-1})$, where ${{\mathrm{Id}}}$ is the identity map on $H_{n-1}(\mathbb {S}^{n-1})$. Now it follows from (3.6.1) that

$$ g_* \circ h_* = I_* = {{\mathrm{Id}}}, $$

which is impossible since $H_{n-1}(D^n) = 0$ while $H_{n-1}(\mathbb {S}^{n-1}) \cong \mathbb {R}\not = 0$. Actually, there are many other proofs of the Brouwer fixed point theorem (see [116] and the references therein). The one presented by Milnor in [76] is elementary an especially clever.

Fig. 3.4
The map $g :D^n \rightarrow S^{n-1}$

Exercises

3.1

Show that, up to homeomorphism, there are only countably many polyhedra.
3.2

Let X be a non-empty topological space. Show that X is a polyhedron with $\dim (X) = 0$ if and only if X is finite and discrete.
3.3

Show that a polyhedron has only finitely many connected components.
3.4

A topological space X is called path-connected if, given any two points $x,y \in X$, there exists a continuous map $\gamma :[0,1] \rightarrow X$ such that $\gamma (0) = x$ and $\gamma (1) = y$. Show that every connected polyhedron is path-connected.
3.5

Show that neither the Cantor set nor the Hilbert cube are polyhedra.
3.6

Find a proof of Lemma 3.5.1 that does not require Baire’s theorem. Hint: proceed by contradiction and use induction on the dimension n (consider a suitable translate of one of the hyperplanes for going from n to $n - 1$).
3.7

(Lebesgue’s lemma for closed coverings). Let $\alpha $ be a finite closed cover of $[0,1]^n$ such that no element of $\alpha $ meets two opposite faces of $[0,1]^n$. Show that one has ${{\mathrm{ord}}}(\alpha ) \ge n$. Hint: use Proposition 1.6.3 and Lemma 3.5.2.
3.8

Show that the unit sphere $\mathbb {S}^{n - 1} \subset \mathbb {R}^n$, defined by
$$ \mathbb {S}^{n - 1} := \{ (x_1,\dots ,x_n) \in \mathbb {R}^n \mid \; \sum _{k = 1}^n x_k^2 = 1 \}, $$
satisfies $\dim (\mathbb {S}^{n - 1}) = n - 1$.
3.9

Let X be a polyhedron and n a non-negative integer. Show that one has $\dim (X) \ge n$ if and only if X contains a subset homeomorphic to $[0,1]^ n$.
3.10
Let $\Gamma = (V,\Sigma )$ and $\Gamma ' = (V',\Sigma ')$ be two abstract simplicial complexes. Equip each of the sets V and $V'$ with some total ordering. Let $W := V \times V'$ denote the Cartesian product of the sets V and $V'$. Consider the set $\Lambda $ consisting of all subsets $\lambda \subset W$ satisfying the following condition: there exist an integer $n \ge 0$, vertices $v_0,v_1,\dots ,v_n \in V$ and $v_0',v_1',\dots ,v_n' \in V'$ such that $v_0 \le v_1 \le \dots \le v_n$, $v_0' \le v_1' \le \dots \le v_n'$, $\{v_0,v_1,\dots ,v_n\} \in \Sigma $, $\{v_0',v_1',\dots ,v_n'\} \in \Sigma '$, and
$$ \lambda = \{(v_0,v_0'),(v_1,v_1'), \dots , (v_n,v_n') \}. $$
1. (a)
  
  Show that $\Pi := (W,\Lambda )$ is an abstract simplicial complex.
2. (b)
  
  Let P (resp. $P'$, resp. Q) denote the support of the geometric realization of $\Gamma $ (resp. $\Gamma '$, resp. $\Pi $). Show that Q is homeomorphic to the topological product $P \times P'$.
3.11

Let X and Y be polyhedra. Show that the product space $X \times Y$ is a polyhedron. Hint: use Exercise 3.10.

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Michel Coornaert
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1.Institut de Recherche Mathématique AvancéeUniversity of StrasbourgStrasbourgFrance

Topological Dimension of Polyhedra

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