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:



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> 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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Embedding of $C(X)$ into $B(H)$ where $H$ is separable
No, not necessarily. If $C(X)$ embeds into $B(H)$ then $X$ must be [ccc](https://en.wikipedia.org/wiki/Countable_chain_condition), 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.




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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.


