Talk:Maximal torus

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"Given a maximal torus T in G, every element g ∈ G is conjugate to an element in T." What if G is not compact? Does this hold iff g is semisimple? —The preceding unsigned comment was added by 84.165.180.8 (talk • contribs) 18:20, 23 December 2006 (UTC)

No, the compact condition is necessary. Take SL(2,R) for instance. This group is semisimple, noncompact, with maximal torus equal to the rotation subgroup SO(2). Conjugacy must preserve the trace of a matrix and every matrix in SO(2) has a trace between −2 and 2. Therefore any matrix in SL(2,R) with |Tr| > 2 (the so-called hyperbolic elements) cannot be conjugate to a rotation. Likewise, the parabolic elements (the nonidentity elements with Tr^2 = 4) are not conjugate to rotations. -- Fropuff 01:55, 24 December 2006 (UTC)Reply

The introduction states that a maximal torus in a Lie group is "not to be confused with the mathematical torus".

Whoever wrote that appears not to have been aware that a maximal torus in a Lie group is indeed a "mathematical" torus. Just not necessarily the 2-dimensional torus usually denoted by T².

In the general case, any compact connected abelian Lie group is topologically an n-torus Tⁿ for some integer n. (In fact as a Lie group it must be the quotient Rⁿ / L of Euclidean n-space Rⁿ by a n-dimensional lattice L. (That is, L is a subgroup of Rⁿ that is isomorphic to Zⁿ and whose span over the reals is all of Rⁿ. In fact if we are discussing complex Lie groups, then the Rⁿ in question is in fact the complex vector space C^k where 2k = n, and in this case C^k / L is isomorphic as a complex Lie group to the maximal torus in question.)108.245.209.39 (talk) 03:38, 24 June 2017 (UTC)Reply

Yes, of course. The phrase "not to be confused with" was removed sometime in the past. linas (talk) 18:20, 4 April 2026 (UTC)Reply