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edited May 5, 2017 at 7:06

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A simple way to come up with non-complete topologies on $\text{SO}(3,\mathbb{R})$ is by embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C})$$ and then using a discontinuous automorphism of $\mathbb{C}$ to produce a discontinuous automorphism of $\text{SL}(3,\mathbb{C})$. Knowing all the (not so many) closed subgroups of $\text{SL}(3,\mathbb{C})$, it is easy to deduce that the image of $\text{SO}(3,\mathbb{R})$ under such an automorphism is not closed, hence the induced topology not complete.

Another nice way to finish the argument is by making an identification of $\mathbb{C}$ with another algebraically closed field of the same cardinality, carrying a different topological structure. For this one can use the "$p$-adic complex field", $\mathbb{C}_p$. One gets the embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C}) \to \text{SL}(3,\mathbb{C}_p).$$ Finally, to cope with Andreas' remark, note that $\text{SL}(3,\mathbb{C}_p)$ has a countable permutation action. Indeed, this Polish group has an open subgroup, namely $\text{SL}(3,\mathcal{O})$, where $\mathcal{O}<\mathbb{C}_p$ is the ring of integers.

A simple way to come up with non-complete topologies on $\text{SO}(3,\mathbb{R})$ is by embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C})$$ and then using a discontinuous automorphism of $\mathbb{C}$ to produce a discontinuous automorphism of $\text{SL}(3,\mathbb{C})$. Knowing all the (not so many) closed subgroups of $\text{SL}(3,\mathbb{C})$, it is easy to deduce that the image of $\text{SO}(3,\mathbb{R})$ under such an automorphism is not closed, hence not complete.

Another nice way to finish the argument is by making an identification of $\mathbb{C}$ with another algebraically closed field of the same cardinality, carrying a different topological structure. For this one can use the "$p$-adic complex field", $\mathbb{C}_p$. One gets the embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C}) \to \text{SL}(3,\mathbb{C}_p).$$ Finally, to cope with Andreas' remark, note that $\text{SL}(3,\mathbb{C}_p)$ has a countable permutation action. Indeed, this Polish group has an open subgroup, namely $\text{SL}(3,\mathcal{O})$, where $\mathcal{O}<\mathbb{C}_p$ is the ring of integers.

A simple way to come up with non-complete topologies on $\text{SO}(3,\mathbb{R})$ is by embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C})$$ and then using a discontinuous automorphism of $\mathbb{C}$ to produce a discontinuous automorphism of $\text{SL}(3,\mathbb{C})$. Knowing all the (not so many) closed subgroups of $\text{SL}(3,\mathbb{C})$, it is easy to deduce that the image of $\text{SO}(3,\mathbb{R})$ under such an automorphism is not closed, hence the induced topology not complete.

Another nice way to finish the argument is by making an identification of $\mathbb{C}$ with another algebraically closed field of the same cardinality, carrying a different topological structure. For this one can use the "$p$-adic complex field", $\mathbb{C}_p$. One gets the embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C}) \to \text{SL}(3,\mathbb{C}_p).$$ Finally, to cope with Andreas' remark, note that $\text{SL}(3,\mathbb{C}_p)$ has a countable permutation action. Indeed, this Polish group has an open subgroup, namely $\text{SL}(3,\mathcal{O})$, where $\mathcal{O}<\mathbb{C}_p$ is the ring of integers.

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created May 5, 2017 at 6:59

11.6k
2
37
60

A simple way to come up with non-complete topologies on $\text{SO}(3,\mathbb{R})$ is by embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C})$$ and then using a discontinuous automorphism of $\mathbb{C}$ to produce a discontinuous automorphism of $\text{SL}(3,\mathbb{C})$. Knowing all the (not so many) closed subgroups of $\text{SL}(3,\mathbb{C})$, it is easy to deduce that the image of $\text{SO}(3,\mathbb{R})$ under such an automorphism is not closed, hence not complete.

Another nice way to finish the argument is by making an identification of $\mathbb{C}$ with another algebraically closed field of the same cardinality, carrying a different topological structure. For this one can use the "$p$-adic complex field", $\mathbb{C}_p$. One gets the embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C}) \to \text{SL}(3,\mathbb{C}_p).$$ Finally, to cope with Andreas' remark, note that $\text{SL}(3,\mathbb{C}_p)$ has a countable permutation action. Indeed, this Polish group has an open subgroup, namely $\text{SL}(3,\mathcal{O})$, where $\mathcal{O}<\mathbb{C}_p$ is the ring of integers.