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Oct 18, 2011 at 4:23 comment added Theo Johnson-Freyd "discrete" here means "is discrete with the induced topology from the ambient Lie group". If you think about the irrational angle, you'll see that what you've done is given a homomorphism from the discrete group $\mathbb Z$ into $\mathrm{SO}(2)$, but the image is not discrete as a topological space.
Oct 18, 2011 at 3:41 comment added Mariano Suárez-Álvarez Your question is off-topic here, as explained in the FAQ. math.stackexchange.com is a good place, to ask, though.
Oct 18, 2011 at 3:40 comment added Mariano Suárez-Álvarez That is not a counterexample, because the subgroup is not discrete.
Oct 18, 2011 at 3:39 history asked ka9q CC BY-SA 3.0