Let $G$ be a semisimple Lie group and $T$ be its maximal torus. Can we say that $G/T$ is a projective variety?. Is there any proof or counterexample for it?
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4$\begingroup$ In the theory of algebraic groups subgroups $H$ such that $G/H$ is a complete variety are called parabolic subgroups. Borel subgroup are characterized as being minimal parabolic subgroup, so in general maximal tori won't be parabolic. See e.g. Springer, Linear Algebraic Groups. $\endgroup$– Denis NardinCommented Feb 1, 2014 at 19:11
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$\begingroup$ We can not define Borel or parabolic subgroups on non-algebraic groups?. also complex semisimple lie groups can not be viewed as complex linear algebraic groups? $\endgroup$– user21574Commented Feb 1, 2014 at 19:20
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1$\begingroup$ I am not sure what do you mean by "is a projective variety". If your group is algebraic there is a natural algebraic variety structure on its quotients, otherwise I do not know. I think you are correct that any complex semisimple Lie group has an algebraic group structure but I'm far from an expert of the subject and I don't want to say something that might be false or misleading. $\endgroup$– Denis NardinCommented Feb 1, 2014 at 20:26
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4$\begingroup$ @Hassan: It's easy to answer your question if you make clear at the outset whether your "Lie group" has a structure of algebraic variety (say over the complex field). For a semisimple algebraic group $G$ over $\mathbb{C}$, the algebraic variety $G/T$ for a maximal algebraic torus $T$ is definitely not projective but instead affine. (Bernstein-Gelfand-Gelfand wrote a long paper about this variety.) On the other hand, the real manifold $G/T$ for a compact Lie group $G$ is certainly compact, and is related indirectly to a complex projective flag variety $G/B$ as in the Borel-Weil theorem. $\endgroup$– Jim HumphreysCommented Feb 1, 2014 at 23:14
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1$\begingroup$ @JimHumphreys Do you have a precise reference for this paper of Bernstein, Gelfand and Gelfand? $\endgroup$– Daniel PomerleanoCommented Feb 1, 2014 at 23:42
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2 Answers
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Suppose that $G$ is compact, connected, and semisimple. Let $T\subseteq G$ be a maximal torus. Take the complexification $G_{\mathbb{C}}$ of $G$, and choose a Borel subgroup $B\subseteq G_{\mathbb{C}}$ containing $T$. Using the Iwasawa decomposition $G_{\mathbb{C}}=GB$, we see that $G$ acts transitively on $G_{\mathbb{C}}/B$. Also, the stabilizer of the identity coset is $T$, giving us a $G$-equivariant diffeomorphism $G/T\cong G_{\mathbb{C}}/B$.
The thing to note is that $G_{\mathbb{C}}/B$ naturally carries the structure of a complex projective variety. Hence, $G/T$ inherits the projective variety structure for which the isomorphism $G/T\cong G_{\mathbb{C}}/B$ is an isomorphism of projective varieties. Note, however, that the projective variety structure of $G/T$ depends on the choice of $B$ containing $T$, or equivalently, the choice of positive roots for the adjoint representation of $T$ on $\mathfrak{g}_{\mathbb{C}}$.
answered Feb 1, 2014 at 19:38
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$\begingroup$ Peter thanks for your answer, You said $G/T$ inherits the projective variety structure while Ben found an counterexample. The point is the additional conditions $G$ must be compact, connected? $\endgroup$– user21574Commented Feb 1, 2014 at 21:54
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1$\begingroup$ You're welcome! Compactness is definitely required for this construction to work. $\endgroup$ Commented Feb 1, 2014 at 22:05
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4$\begingroup$ Peter's setting is a real compact group and it's complexification. If you will take complex group and it's maximal torus it won't be compact for sure. And any complex connected compact group is an abelian variety, so it's definitely not you are asking for. If you consider an algebraic setting, then $G/T$ is NEVER projective unless it's a point. But it is true that for a compact group $G$ $G/T$ is homeomorphic to the $G_\mathbf{C} / B$, which is projective and called the generalized flag variety. $\endgroup$ Commented Feb 2, 2014 at 13:24
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Try $G=SL(2,\mathbb{R})$; the $T$ is the stabilizer of a metric, so $G/T$ is the set of metrics on $\mathbb{R}^2$ with unit volume, certainly not projective, because they are just positive definite symmetric $2 \times 2$ matrices with determinant 1, an affine hypersurface. Projective varieties are compact.
answered Feb 1, 2014 at 19:09