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Suppose $T$ is a maximal torus of $G$ with dimension = $n$. If there is another torus $H \subset G$ of the same dimension, could I then conclude that $H$ is also a maximal torus? In other words once you know the dimension of a maximal torus you can just use the dimension to judge whether another torus is maximal or not.

PhysicsMath
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Yes. By definition, $H$ is contained in some maximal torus, and it's a theorem that all maximal tori are conjugate, and in particular have the same dimension. So $H$ is contained in a conjugate of $T$. Now show that if a torus is contained in another torus and they have the same dimension then they must in fact be the same.

Qiaochu Yuan
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  • Thanks a lot for the reply Qiaochu! I totally agree with your argument and actually what I wish to see is only this step: "show that if a torus is contained in another torus and they have the same dimension then they must in fact be the same". Do you need to pass to Lie algebra and use the dimension argument for vector space? – PhysicsMath May 20 '16 at 19:03
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    @Physics: yes, that's one possible approach. You can show that any inclusion of tori induces an inclusion of Lie algebras. If they have the same dimension, this inclusion of Lie algebras must be an isomorphism, which implies that the corresponding map of tori is surjective (since it's surjective around the origin and the target is connected). – Qiaochu Yuan May 20 '16 at 19:12
  • Many thanks for your confirmation Qiaochu! But there is any way to show that without appealing to Lie algebra i.e. a purely group theoretical approach? By the way why don't you append this discussion to your answer so that more people would find your answer useful and give you credit? – PhysicsMath May 20 '16 at 19:22