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Let $\mathcal M_1(S)$ be the (compact, metrizable) space of probability Borel measures on the circle $S=\{z\in\mathbb C: |z|=1\}$ with its weak $*$ topology, so $\mu_n\to\mu$ if and only if $$ \int_S f\, d\mu_n\to \int_S f\, d\mu $$ for all $f\in C(S)$.

Is $\mathcal M_1(S)\times \mathcal M_1(S)$ homeomorphic to $\mathcal M_1(S)$?

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    $\begingroup$ If instead you considered a norm topology on the space of measures, it seems that the answer is yes, since there is a Borel bijection between $S$ and $S\cup S$. $\endgroup$ Commented Dec 6 at 21:42
  • $\begingroup$ Yes, I understand, just wrote it for the people who might've thought of the other problem. $\endgroup$ Commented Dec 7 at 0:24
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    $\begingroup$ @AlekseiKulikov: We can clearly not preserve the linear structure since $\mathcal M_1(S)$ has extreme points $\{ \delta_x\}\cong S$, which is not homeomorphic to $S\sqcup S$. $\endgroup$ Commented Dec 7 at 0:37
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    $\begingroup$ @AlekseiKulikov I’m not disputing that $\mathcal{M}_1(S)$ and $\mathcal{M}_1(S\sqcup S)$ are isometric under the total variation norm, which seems to be what you have shown. But why is the latter space homeomorphic to $\mathcal{M}_1(S)\times\mathcal{M}_1(S)$? They are clearly not affinely homeomorphic since $\mathcal{M}_1(S\sqcup S)$ has extreme points $S\sqcup S$ while $\mathcal{M}_1(S)\times\mathcal{M}_1(S)$ has extreme points $S\times S$. $\endgroup$
    – David Gao
    Commented Dec 7 at 3:41
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    $\begingroup$ @AlekseiKulikov Agreeing with David Gao, try thinking about the two-point space instead of $S$, where $\mathcal{M}_1(2) \times \mathcal{M}_1(2)$ is a square, but $\mathcal{M}_1(2 + 2)$ is a tetrahedron. In general $\mathcal{M}_1(X) \times \mathcal{M}_1(Y)$ is not a space of probability measures but the state space of the coproduct order-unit space $C(X) \oplus C(Y)$. $\endgroup$ Commented Dec 7 at 11:42

1 Answer 1

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The answer is yes, following from Klee's extension of Keller's theorem on homeomorphisms of infinite dimensional compact convex sets:

Klee, V. L., Some topological properties of convex sets, Trans. Am. Math. Soc. 78, 30-45 (1955). ZBL0064.10505.

Specifically, combining (1.1) with the theorem of Keller:

If $K$ is an infinite-dimensional compact convex subset of $E$, a topological vector space, such that there is a countable set $F \subseteq E^*$ separating the points of $K$, then $K$ is homeomorphic to the Hilbert cube.

(For the avoidance of doubt, I use $E^*$ to mean the space of continuous linear functionals throughout)

For $\mathcal{M}_1(S)$, consider it inside $C(S)^*$ with the weak-* topology and take $F$ to be your favourite countable norm-dense subset of $C(S)$, considered as weak-* continuous maps $C(S)^* \rightarrow \mathbb{R}$. For $\mathcal{M}_1(S) \times \mathcal{M}_1(S)$, consider it inside the product topological vector space $C(S)^* \times C(S)^*$ with the product of weak-* topologies (the underlying Banach space is the $\ell^\infty$-direct sum). Define $$ F' = \{ \phi \circ \pi_1 \mid \phi \in F \} \cup \{ \phi \circ \pi_2 \mid \phi \in F \} $$ where $\pi_1 : E \times F \rightarrow E$ and $\pi_2 : E \times F \rightarrow F$ are the usual projection maps. Then $F'$ is a countable subset of $(C(S)^* \times C(S)^*)^*$ separating the points of $\mathcal{M}_1(S) \times \mathcal{M}_1(S)$, so this is also homeomorphic to the Hilbert cube and therefore to $\mathcal{M}_1(S)$.

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    $\begingroup$ Thank you, this is very useful for me. In the last part, we could also just use that (obviously) $H\times H\cong H$ for the Hilbert cube $H=[0,1]^{\mathbb N}$. $\endgroup$ Commented Dec 7 at 15:18

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