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The symmetric group acts continuously and freely on $F_{n} (X)$ by permuting labels

Statement

Let $n \in N$ and let $X$ be a topological space (Ordered configuration spaces $F_{n} (X)$ ). Give $S_{n}$ the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and $F_{n} (X)$ the subspace topology of its definition. Then

$S_{n} \times F_{n} (X) ⟶ F_{n} (X), (σ, x) ⟼ σ \cdot x, (σ \cdot x)_{i} : = x_{σ^{- 1} (i - 1) + 1} (1 \leq i \leq n),$

is a continuous left action of $S_{n}$ on $F_{n} (X)$ , and it is free (A free group action has no nonidentity element fixing a point): $σ \cdot x = x$ forces $σ = id$ . The cases $n = 0$ and $n = 1$ are included, $S_{0}$ and $S_{1}$ being the trivial group, and so is the case $F_{n} (X) = \emptyset$ , where the action is continuous and free vacuously.

Facts & Assumptions

Given: A natural number $n$ , a topological space $X$ , the ordered configuration space $F_{n} (X)$ with its label convention, and the symmetric group $S_{n}$ acting on the label set ${1, \dots, n}$ through $κ (i) = i - 1$ .

[F1]

Points of $F_{n} (X)$ are the tuples $(x_{1}, \dots, x_{n}) \in X^{n}$ with $x_{i} \neq x_{j}$ for $i \neq j$ , carrying the subspace topology, and the label $i$ names the coordinate of index $i - 1$ under the identification $κ (i) = i - 1$ of ${1, \dots, n}$ with $n = {0, \dots, n - 1}$ (Ordered configuration spaces $F_{n} (X)$ ).

[L2]

$S_{n} = Sym (n)$ is a group under composition, with $(σ τ) (i) = σ (τ (i))$ for $i \in n$ , so that $(σ τ)^{- 1} = τ^{- 1} σ^{- 1}$ ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

[L3]

A left action of a group $G$ on a set $X$ is a map $(g, x) \mapsto g \cdot x$ with $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ , and it is free when $g \cdot x = x$ implies $g = e$ (Left group actions, transitive actions, and faithful actions, A free group action has no nonidentity element fixing a point).

[L4]

A map into a product is continuous if and only if each of its components is; the projections are continuous (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[L5]

A function on a space is continuous if its restriction to each member of an open cover is continuous, and composites and restrictions of continuous maps are continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally).

[L6]

A set with the discrete topology has every subset open, and a finite group such as $S_{n}$ carries the discrete topology here (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

Proof

technique · direct

1.1

The formula is well defined: for $i \in {1, \dots, n}$ the index $σ^{- 1} (i - 1)$ lies in $n$ , so $σ^{- 1} (i - 1) + 1$ is a label of ${1, \dots, n}$ , and the resulting tuple in $X^{n}$ has the coordinates of $x$ reindexed along the bijection $σ^{- 1} \circ κ$ : writing $y (k) : = x_{k + 1}$ for $k \in n$ , its $i$ -th coordinate is $y (σ^{- 1} (κ (i)))$ . A reindexing of pairwise distinct coordinates is again pairwise distinct, so $σ \cdot x \in F_{n} (X)$ .

F1L2algebra

1.2

The assignment is a left action. The identity of $S_{n}$ gives $(id \cdot x)_{i} = x_{i - 1 + 1} = x_{i}$ , so $id \cdot x = x$ ; and for $σ, τ \in S_{n}$ and every label $i$ , $((σ τ) \cdot x)_{i} = x_{(σ τ)^{- 1} (i - 1) + 1} = x_{τ^{- 1} (σ^{- 1} (i - 1)) + 1} = (τ \cdot x)_{σ^{- 1} (i - 1) + 1} = (σ \cdot (τ \cdot x))_{i},$ using $(σ τ)^{- 1} = τ^{- 1} σ^{- 1}$ and the composition convention $(σ τ) (k) = σ (τ (k))$ . Since coordinates determine a tuple, $(σ τ) \cdot x = σ \cdot (τ \cdot x)$ .

F1L2L3algebra

1.3

Each slice map $x \mapsto σ \cdot x$ is continuous: its $i$ -th component is the map $x \mapsto x_{σ^{- 1} (i - 1) + 1}$ , the composite of the coordinate projection $π_{σ^{- 1} (i - 1)} ⁣ : X^{n} \to X$ with the inclusion $F_{n} (X) ↪ X^{n}$ , and both are continuous; the characteristic property of the product therefore gives continuity of the slice map into $X^{n}$ , and its values lie in $F_{n} (X)$ , so it is continuous into $F_{n} (X)$ .

F1L4L5

1.4

The action is free. Suppose $σ \cdot x = x$ for some $σ \in S_{n}$ and $x \in F_{n} (X)$ , and put $y (k) : = x_{k + 1}$ for $k \in n$ . Comparing coordinates gives $y (σ^{- 1} (k)) = y (k)$ for every $k \in n$ , and replacing $k$ by $σ (k)$ gives $y (k) = y (σ (k))$ for every $k$ . The coordinates of $x$ are pairwise distinct, so $y$ is injective, hence $σ (k) = k$ for every $k \in n$ and $σ = id$ . Thus no nonidentity element fixes a point of $F_{n} (X)$ .

F1L2L3algebra

2.1

The action map $S_{n} \times F_{n} (X) \to F_{n} (X)$ is continuous. Since $S_{n}$ is discrete, each ${σ} \times F_{n} (X)$ is open in the product and these sets cover it; the restriction of the action map to ${σ} \times F_{n} (X)$ is, after the evident identification with $F_{n} (X)$ , the continuous slice map of step 1.3. Continuity is local on an open cover, so the action map is continuous.

step 1.3L5L6

3.1

Steps 1.2 and 2.1 give a continuous left action and step 1.4 gives freeness in the sense of the definition, which is the assertion.

step 1.2step 2.1step 1.4L3∎

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Juan Gonzalez-Meneses, Basic results on braid groups, sections 1.1 and 1.3, printed pp. 3-6 (standard reference, not scraped)

The symmetric group acts continuously and freely on Fn(X)Fn​(X) by permuting labels