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Is there a group homeomorphic to but not homeomorphically isomorphic to the circle group?

Let $Circ$ be the topological group

$(\{z\in \mathbb{C} : \overline{z}\cdot z = 1\},\cdot , \{U\in 2^{\{z\in \mathbb{C} : \; \overline{z}\cdot z \, = \, 1\}} : \{z\in \mathbb{C} : \overline{z}\cdot z = 1\}-U$ is closed in $\mathbb{C}\})$.

Let $(G,\star,T)$ be a topological group such that $(G,T)$ is homeomorphic to $(\{z\in \mathbb{C} : \overline{z}\cdot z = 1\}, \{U\in 2^{\{z\in \mathbb{C} : \; \overline{z}\cdot z \, = \, 1\}} : \{z\in \mathbb{C} : \overline{z}\cdot z = 1\}-U$ is closed in $\mathbb{C}\})$.

1. $(G,\star)$ is isomorphic to $(\{z\in \mathbb{C} : \overline{z}\cdot z = 1\},\cdot)$?
2. $(G,\star,T)$ is homeomorphically isomorphic to $Circ$?
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What a complicated way of saying "with its usual topology"! – Mariano Suárez-Alvarez Oct 28 2010 at 5:06
-1 for the complicated formulation. – Qfwfq Oct 28 2010 at 9:34
Note that it's not true for $Circ \times Z_2$, which has three group structures, as pointed out to me by Greg Kuperberg. – Allen Knutson Oct 28 2010 at 18:28

It is a theorem of Scheinberg that two compact connected abelian groups are topologically isomorphic whenever they are homeomorphic.

Now, by the solution of Hilbert's 5th problem, a topological group $G$ homeomorphic to $S^1$ is a Lie group, which is necessarily of dimension $1$. It follows that $G$ is abelian. Afirmative answers to your two questions follow.

More generally, since A compact solvable Lie group is abelian, the same reasoning shows that a topological group homeomorphic to a 2-torus is isomorphic to a 2-torus

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 Hilbert's 5th is a heavy stuff: there is an elementary proof that takes a fraction of time it would cost to do Hilbert's 5th... I will have a go. – Bugs Bunny Oct 28 2010 at 9:00 OK, so what is an example of two locally compact (not discrete) groups that are homeomorphic, but not isomorphic as topological groups? By this, they are not compact connected abelian; but can they have two of those three properties? – Gerald Edgar Oct 28 2010 at 18:13 @Gerald: a countable product of copies of the discrete group $\mathbb Z_2$ is homemorphic to a countable product of copies of the discrete group $\mathbb Z_3$, but they are not isomorphic as groups. So the groups can be compact, non-discrete and abelian. – Mariano Suárez-Alvarez Oct 28 2010 at 21:42 If now you do direct productsof those two groups with a discrete $\mathbb Z$, I thing you have an example of what you want. – Mariano Suárez-Alvarez Oct 28 2010 at 22:25

Here is also a direct argument (which only works in dimension $1$): The universal cover of $G$ is a group homeomorphic to ${\mathbb R}$. Now, it is enough to show that the only topological group homeomorphic to ${\mathbb R}$ is the additive group of reals. To show this, first note that each right or left translation $$\lambda_g: x \mapsto g \cdot x , \quad \rho_g: x \mapsto x \cdot g$$ is continuous and bijective, hence it has to be monotone. Now, the ones that are increasing form a subgroup of index at most $2$, and since ${\mathbb R}$ is connected, it follows that every left or right translation is strictly increasing. Also $\lambda_g$ has no fixed point, which implies that either $\lambda_g(x)>x$ or $\lambda_g (x)0$, (here $0$ is the identity) then $\lim_{n \to \infty} f^n (x)=\infty$ (since $f$ is increasing and bijective) so for large enough $n$, we have $f^n (x)>g(x)$, so $f^n >g$. Now, by Hölder's theorem, $G$ (as a group) is isomorphic to the group of real numbers, and hence Abelian. Now, for each integer $n>0$, the map $g \mapsto g^n$ is also increasing and proper, which implies that every element has an $n$-th root. So, you can define $g^q$ for any rational number $q$, and then for any real $q$. This allows you to construct a topological isomorphism from the additive group to $G$.

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