Given an algebraic group $G$ over an algebraically closed field $K$, if $H$ is a subvariety of $G$, then is $H$ a subgroup of $G$? This seems rather strong. If it is indeed false, is there a geometric characterization of when $H$ is a subgroup of $G$?
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4An arbitrary subvariety need not even contain whatever element is the identity of the group. – Tobias Kildetoft May 15 '13 at 16:57
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1I doubt a geometric characterization is possible, precisely because $G$ has automorphisms (say, multiplication by a non-identity element) that don't preserve the identity. More basically, you can't tell geometrically which point is the trivial subgroup. – mdp May 15 '13 at 17:19
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If we're allowed to know the non-geometric information of which point is the identity, then a potentially interesting question is whether every subvariety of $G$ can be mapped to a subgroup by multiplying by a group element. Thinking about this question in reverse, it becomes "is every subvariety of $G$ a coset of some subgroup $H$?". – mdp May 15 '13 at 17:37
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Also, the subvariety $H$ may not be an algebraic group at all, for instance if $H$ fails to be nonsingular. – Corey Harris May 15 '13 at 17:44
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Just to get this off the unanswered list.
No, it is not true that a subvariety of an algebraic group is an algebraic group. Moreover, there is no real geometric quality that guarantees this. Namely, note that $\mathbb{G}_a^n$ is an algebraic group, and the subvarieties of this (as $n$ varies) are precisely the affine varieties. There is no geometric condition that guarantees that an arbitrary affine variety is even a group scheme.
Alex Youcis
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