The Metrizability of Symmetric Products of Metric Spaces

Question

The (infinite) symmetric product of a based topological space $(X,e)$, denoted by $\mathrm{SP}(X,e)$, can be viewed as the topological space of ''multisets'' in $X$ containing the base point $e$ infinitely many times (please see http://en.wikipedia.org/wiki/Infinite_symmetric_product for the precise definition). The following is my question:

Provided that $X$ is metrizable, can we say that $\mathrm{SP}(X,e)$ is metrizable in general?

Due to my poor knowledge I have absolutely no idea as to how the symmetric product of simple spaces look like. However, I suppose I can explain how I come up with this question at least. Let $(X,d)$ be a metric space with a base point $e$. An arbitrary element $S$ in $\mathrm{SP}(X,e)$ admits a representation $S = [s_1,s_2,\ldots]$, where $(s_i)_{i \in \mathbb{N}}$ is a sequence in $X$ with the property that all but finitely many terms of the sequence are the base point $e$. We can now define a metric on $\mathrm{SP}(X,e)$ by $$ \mathrm{dist}([s_1,s_2,\ldots],[t_1,t_2,\ldots]) := \inf_{\pi} \sum_i d(s_i,t_{\pi(i)}), $$ where the inf is taken over all permutations $\pi$. In my masters thesis (on functional analysis) I had to show that the fundamental group of $\mathrm{SP}(X,e)$, equipped with the above metric topology, is the first homology group of $X$ (provided that $X$ is both path connected and locally simply connected). I have met one algebraic-topologist who made a somewhat interesting remark that this result is analogous to the Dold-Thom theorem (please see http://en.wikipedia.org/wiki/Dold%E2%80%93Thom_theorem). This is how I started finding a relationship between the topology induced by this metric and the standard topology on $\mathrm{SP}(X,e)$. The Dold-Thom theorem seems pretty famous, and so I thought the metrizability of the symmetric product of a metric space might have been well studied. Sadly, I know nothing about topology, so this is how I ended up using this website.

Cheers.

Answer 1

The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$.

Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^n(X,e)\setminus SP^{n-1}(X,e)$ for each $n>0$. By first countability, we can find elements $x_n\in SP^n(X,e)\setminus SP^{n-1}(X,e)$ such that the sequence $(x_n)$ converges to $[e]$ in $SP(X,e)$. But the set $\{x_n\}$ has finite and hence closed intersection with each $SP^n(X,e)$, so the entire set $\{x_n\}$ is closed in $SP(X,e)$. This contradicts the assumption that $(x_n)$ converges to $[e]$. More generally, this argument shows that if a $T_1$ space $A$ is a colimit of subspaces $A_0\subset A_1\subset \dots$ such that the interiors of the $A_n$ do not cover $A$, then $A$ cannot be first countable.

As a final note, let $SP_d(X,e)$ be the symmetric product with the topology induced by your metric. Then the identity $i:SP(X,e)\to SP_d(X,e)$ is continuous, and in fact is a homeomorphism when restricted to each $SP^n(X,e)$. Since every compact subset of $SP(X,e)$ is contained in some $SP^n(X,e)$ (by essentially the same argument as the previous paragraph), the map $i$ is a weak equivalence (i.e., induces an isomorphism on homotopy groups). In particular, when $(X,e)$ is a pointed metric space that is homotopy equivalent to a connected CW-complex (relative to the basepoint), the Dold-Thom theorem implies that $\pi_n(SP_d(X,e))=\pi_n(SP(X,e))=H_n(X,e)$.

Answer 2

NOTATION:   $\ \mathbb Z_+:=\{0\ 1\ \ldots\}\ $ is the set of the non-negative integers.

Let me propose a metric closely related to the goal. One may check how well it fits the problem (for what class of spaces this is a perfect fit; and how the topology induced by this metric deviates from the respective quotient topology). Perhaps this metric should be useful for similar purposes even without referring to the quotient topology.

Let $\ \mathbf X:=(X\ e)\ $ be a pointed set, where $\ e\in X,\ $and let $\ T\ $ be an arbitrary non-empty set. The symmetric power $\ S^\infty(\mathbf X)\ $ can be defined as the set of all functions

$$\ f:X\setminus\{e\}\rightarrow\mathbb Z_+$$

such that the set $\ X\setminus f^{-1}(0)\ $ is finite. This is a pointed set after we defined the point $\ e^\infty: X\setminus\{e\}\rightarrow\mathbb Z_+\ \ $ as the zero function.

Now let's consider a metric $\ d\ $ in $\ X.\ $ Thus first let's define the left Hausdorff metric $\ \ell_\infty\ $ in the symmetric power $\ S^\infty(\mathbf X),\ $ i.e. between every $\ f\ g\ \in S^\infty(\mathbf X)$:

$$\ell_\infty(f\ g)\ :=\ \inf_{s:X\rightarrow S^T(\mathbf X)} \{\ell_s(f\ g): \sigma s\le f\ \ and\ \ \rho s\le g\}$$

where:

$$\sigma s(x)\ :=\ \sum_{y\in X\setminus\{e\}} (s(x))(y)$$

and

$$\rho s(y)\ :=\ \sum_{x\in X\setminus\{e\}} (s(x))(y) $$

and

$$\ell_s(f\ g)\ :=\\ \sup\{d(x\ e): \sigma s(x)<f(x)\}\ +\ \sup\{d(x\ y):\ f(x)\cdot((s(x))(y)>0\}$$

Now let me define the Hausdorff distance in $\ S^\infty(\mathbf X)$:

$$\forall_{f\ g\in S^\infty(\mathbf X)}\ d_T(f\ g)\ :=\ \max(\ell_\infty(f\ g)\ \ \ell_\infty(g\ f))$$

The pointed set $\ (S^\infty(\mathbf X)\,\ e^\infty)\ $ is simply the free monoid.