# Embedding of $C(X)$ into $B(H)$ where $H$ is separable

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 $$H$$?

• 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$\mathcal{Y}$. – YCor Apr 3 at 17:53

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.