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The original title of this question was "Is there only one (up to homeomorphism) zero-dimensional homogeneous dyadic space of weight $\mathfrak{c}$?". I changed it with the hope of getting a bit more attention.

So I´m interested in the case $\kappa=\mathfrak{c}$ of the following question, asked originally by Efimov in 1965: Is there a zero-dimensional homogeneous dyadic space of weight $\kappa$ that is not homeomorphic to $2^\kappa$?

The only answers I know of to Efimov´s question are:

Pashenkov (1974): Yes, if $\kappa=2^\mu$ for some uncountable $\mu$.

Shapiro (1993) and Bell (1994): No, if $\kappa=\aleph_1$.

So the answer to the question in the original title is Yes when $\mathfrak{c}=\aleph_1$ and it is No when $\mathfrak{c}=2^{\aleph_1}$. Then the real question (perhaps still open) is:

Suppose that $\aleph_1 < \mathfrak{c} < 2^{\aleph_1}$. Is there a zero-dimensional homogeneous dyadic space of weight $\mathfrak{c}$ that is not homeomorphic to $2^\mathfrak{c}$?

Maybe it is more reasonable to replace $\mathfrak{c}$ by $\aleph_2$ in this question. I would also like to know what happens in general with a $\kappa > \aleph_1$ which is not a power of $2$.

Space = Hausdorff topological space.
$X$ is homogeneous if for all $x,y \in X$ there is an autohomeomorphism of $X$ sending $x$ to $y$.
A space is dyadic if it is the continuous image of some Cantor cube $2^\kappa $.
A space is zero-dimensional if it has a base of clopen subsets.
The weight of a space is the least size of a base for its topology.

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I cooked up a quotient construction in the same spirit of Pashenkov's in an old paper of mine. When I have time, I'd like to try to apply it to this question. But maybe someone else can get there first: the idea is to wisely pick points $(p^i:i<\aleph_2)$ in $2^{\aleph_1}$ and form the quotient $Q$ of $(2^{\aleph_1})^{\aleph_2}$ where $x\equiv y$ iff $x_0=y_0$ and, for all $i<\aleph_2$, $x_{1+i}=y_{1+i}$ or $x_0=p^i$. (P.S. Readers less familiar with this area might be interested to know that Shchepin proved that every 0-dimensional compact group is homeomorphic to a Cantor cube.) – David Milovich Feb 7 '12 at 19:21
Thank you David. What´s your paper about? To completely put Schepin´s result in this context, it may be worth mentioning that any compact group is dyadic (by Kuzminov) and of course any group is homogeneous. You may think that this is what motivated Efimov´s question, but Schepin´s result appeared much later. Unfortunately I don´t have access to either of the original papers. – Ramiro de la Vega Feb 8 '12 at 11:55
The paper is on my website and has MR #2298632. The part of the paper that your question reminded me of is a ZFC construction of a homogeneous compactum that is not homeomorphic to any product where each factor is dyadic or first-countable. However, the proof is a connectedness argument, and your question is about zero-dimensional spaces. – David Milovich Feb 8 '12 at 16:03

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