Is every connected solvable algebraic group a Borel subgroup of a reductive group? If a counterexample exists, I would ideally like it to be over $\Bbb C$.
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4$\begingroup$ Yes, every unimodular solvable algebraic that is not a torus is not a Borel subgroup of any reductive group (a torus being Borel of itself). Also most non-unimodular solvable 3-dimensional algebraic groups are not Borel either. $\endgroup$– YCorCommented Nov 8, 2020 at 21:40
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$\begingroup$ What’s a unimodular solvable group? Also, would you be able to explain why these are counterexamples? $\endgroup$– Avi SteinerCommented Nov 8, 2020 at 21:43
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4$\begingroup$ For an algebraic group, the modular character is given by $g\mapsto\det(\mathrm{Ad}(g))$. Unimodular means that this character has finite image (for connected, it means it is trivial). $\endgroup$– YCorCommented Nov 8, 2020 at 21:52
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2$\begingroup$ (And why that lets us recognise counterexamples is that the modular character of the Borel subgroup of a reductive group is $2\rho$ on a maximal torus, where $\rho$ is the half-sum of positive roots; and this is non-$0$ unless the group equals its torus.) But note that @YCor's comment means "No, not every connected soluble algebraic group is a Borel subgroup of a reductive group". $\endgroup$– LSpiceCommented Nov 8, 2020 at 22:11
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1$\begingroup$ Yes, but I'd say: you have a classification of simple complex groups (and the 1-dimensional torus), and the classification follows (taking direct products of such things), maybe modulo finite central subgroups. I don't really see what an intrinsic condition would be, except that being Borel is very rare, and there are many obvious necessary conditions in terms of geometry of weights). $\endgroup$– YCorCommented Nov 9, 2020 at 18:04
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