Let $G$ be an algebraic group, i.e., a topological group homeomorphic to an algebraic variety. In particular, $G$ is a topological manifold. Let $H\subset G$ be a subgroup which is a closed submanifold in the manifold topology, not Zariski topology.


  1. How far is $H$ from being Zariski-closed, i.e., an algebraic subgroup?

  2. How far is the homogeneous space $G/H$ from being an algebraic variety? Maybe projective variety?

It is known that a Lie group can be defined as a smooth group, but then the group structure forces it to be necessarily analytic. Because the multiplicative law in $H$ is still algebraic (that of $G$), I thought maybe this forces $H$ to be Zariski closed.

Note that I am neither an algebraist nor a geometer, so a very basic language would be appreciated. Thank you.

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    $\begingroup$ You seem to have an unusual definition of an algebraic group. Usually one also requires the operations to be morphisms of varieties. $\endgroup$ – Tobias Kildetoft Apr 1 '17 at 17:23
  • $\begingroup$ Thanks for the comment. The algebraic variety is taken with its Zariski topology, so if the topological group is homeomorphic to it, then its group operations are continuous with respect to the Zariski topology. Isn't that wat is needed? I can't tell what a morphism of varieties is. $\endgroup$ – Bedovlat Apr 1 '17 at 17:38
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    $\begingroup$ One of the small examples is $H=\Bbb{Z}$ as a subgroup of the additive group $G=\Bbb{C}$. Because $H$ is infinite its Zariski closure is all of $\Bbb{C}$. Here $G/H$ is isomorphic as a Lie group to $\Bbb{C}^*$ via the mapping $\phi:z+H\mapsto e^{2\pi i z}$. But even though $\Bbb{C}^*$ is also an algebraic group, $\phi$ is not a morphism of varieties because $z\mapsto e^{2\pi iz}$ isn't. $\endgroup$ – Jyrki Lahtonen Apr 1 '17 at 18:07
  • $\begingroup$ $\mathbb{C}^*$ as an algebraic group has a different topology than $\mathbb{C}^*$ as a Lie group. $\phi$ is a homeomorphism between $G/H$ on one hand, and on the other hand $\mathbb{C}^*$ as a Lie group, and not as an algebraic group. If it were a homeomorphism with an algebraic group taken with the topology of an algebraic group, it would be what you call a morphism of varieties, I guess. $\endgroup$ – Bedovlat Apr 1 '17 at 19:28
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    $\begingroup$ @Bedovlat Being a morphism of varieties is a very strong condition. It's not merely topological, there's algebraic stuff going on. $\endgroup$ – Matt Samuel Apr 1 '17 at 21:54

The easiest example I can think of where both fail miserably is $S^1\subseteq\mathbb C-\{0\}$ under multiplication, where we treat $\mathbb C-\{0\}$ as a complex variety. $S^1$ is uncountably infinite, whereas the proper Zariski closed subsets are finite. The quotient is the group of positive reals, which is not algebraic over the complex numbers.

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  • $\begingroup$ Hmm, thanks! Things are much more complicated than they seem at first. $\endgroup$ – Bedovlat Apr 1 '17 at 18:03
  • $\begingroup$ @Bedovlat No problem. $\endgroup$ – Matt Samuel Apr 1 '17 at 18:08

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