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Suppose $G$ is a compact $T_2$ group. Can there be other compact $T_2$ topologies on $G$ which also turn $G$ into a topological group? ($T_2$ refers to the Hausdorff separation axiom)

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The topology of a compact Hausdorff space is maximal compact and minimal Hausdorff; that is, no finer topology is compact, and no coarser topology is compact. So if you have another compact Hausdorff topology, then it is neither finer nor coarser to the original one. – Mariano Suárez-Alvarez Feb 15 '12 at 9:00
It may be worth stating that if you pick a topology once and for all and ask about uniqueness of smooth structures (if it has any at all!), then the answer is yes. – Jason DeVito Feb 15 '12 at 13:49
up vote 7 down vote accepted

Take the circle group $G=S^1=\mathbb R/\mathbb Z$. Any non-continuous automorphism of $\mathbb R$ which fixes pointwise the subgroup $\mathbb Z$ passes to the quotient and gives an automorphism $f$ of the abstract group $G$, which is not continuous. Now define a topology on $G$ so that a set $U$ is open iff $f(U)$ is open in the usual topology. This new topology is of course Hausdorff and compact, but it is different to the usual topology.

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Interesting, the resulting topological group is isomorphic (as topological group) to the original topological group, still the topology on the set $G$ is different. But I wonder more whether I could somehow prevent the implicit use of the axiom of choice. One idea would be to prescribe the Borel $\sigma$-algebra of a second-countable space and only allow topologies whose open sets belong to that $\sigma$-algebra. – Thomas Klimpel Feb 15 '12 at 15:50

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