I would like to ask a question which may look strange at the first sight nevertheless I find it interesting. Let $H$ be a separable Hilbert space: for any separable $C^*$-algebra $A$ one can embed $A$ into $B(H)$. Very often one can do even better and many nonseparable $C^*$-algebras still embed into $B(H)$, for example $\ell^{\infty}, L^{\infty}[0,1]$ or $B(H)$ itself. However it is not always possible even for quotients of such algebras, $\ell^{\infty}/c_0$ being an example. One can view $\ell^{\infty}$ as $C(\beta\mathbb{N})$ where $\beta\mathbb{N}$ is the Stone-Cech compactification of $\mathbb{N}$ (which is of cardinality $2^{\mathfrak{c}}$) and $\ell^{\infty}/c_0$ as $C(\beta\mathbb{N} \setminus \mathbb{N})$ (still of cardinality $2^{\mathfrak{c}}$). So here comes my question:

Let $X$ be a compact nonmetrizable Hausdorff topological space of cardinality $\mathfrak{c}$. Is it true that $C(X)$ embeds into $B(H)$ for separable $H$?

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    $\begingroup$ I don't know if useful: maximal abelian subalgebras of $B(H)$ seem to be known (see Tomiyama, JFA 1971): these are isomorphic to $C(Y)$ for some class of hopefully identifiable spaces $Y$, say $Y\in\mathcal{Y}$. Then the spaces $X$ are the continuous images of the spaces $Y$ when $Y$ ranges over$\mathcal{Y}$. $\endgroup$
    – YCor
    Apr 3, 2021 at 17:53

1 Answer 1


No, not necessarily. If $C(X)$ embeds into $B(H)$ then $X$ must be ccc, but there exist compact Hausdorff spaces of cardinality $\mathfrak{c}$ which are not ccc. (One example is to take $\mathfrak{c}$ with its discrete topology and form the one-point compactification $\mathfrak{c} \cup \{\infty\}$; then all the singletons except $\infty$ are disjoint open sets. Another is the ordinal $\mathfrak{c}+1$ with its order topology; all the successor singletons $\{\alpha+1\}$, $\alpha < \mathfrak{c}$, are disjoint open sets.)

To see this, suppose that $X$ is not ccc, so that there is an uncountable family $\{U_\alpha\}$ of disjoint nonempty open sets. By the appropriate version of Urysohn's lemma, for each $\alpha$ there is a nonzero real-valued $f_\alpha \in C(X)$ supported inside $U_\alpha$; in particular, $f_\alpha f_\beta = 0$ for $\alpha \ne \beta$. Now if there is an injective *-homomorphism $\Phi : C(X) \to B(H)$, then each $\Phi(f_\alpha)$ is nonzero, so we may find $h_\alpha \in H$ so that $\|\Phi(f_\alpha) h_\alpha\|=1$. Also, $\Phi(f_\alpha)$ is self-adjoint, so for $\alpha \ne \beta$ we have $$\langle \Phi(f_\alpha) h_\alpha, \Phi(f_\beta) h_\beta\rangle = \langle h_\alpha, \Phi(f_\alpha)\Phi(f_\beta) h_\beta\rangle = \langle h_\alpha, \Phi(f_\alpha f_\beta) h_\beta\rangle = 0$$ so that $\{\Phi(f_\alpha) h_\alpha\} \subset H$ is an uncountable orthonormal set, which is impossible if $H$ is separable.

This is fairly close to optimal because every separable compact Hausdorff $X$ does have $C(X)$ embeddable into $B(H)$ (recall that separable implies ccc but not conversely). Indeed, if $\{x_n\}$ is a countable dense subset of $X$, then identifying $f \in C(X)$ with the sequence $(f(x_1), f(x_2), \dots)$ gives an isometric embedding of $C(X)$ into $\ell^\infty$, which as we know embeds in $B(\ell^2)$. Presumably there is a necessary and sufficient condition somewhere in between, but I don't know what it is.

  • $\begingroup$ Nice, thank you! $\endgroup$
    – truebaran
    Apr 4, 2021 at 1:16

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