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So I'm looking to find the maximal abelian subgroup of SL(3,C). I know that if a maximal torus for SL(3,C) exists that said torus is the maximal abelian subgroup. Is it enough to know that since SU(3) is a subgroup that they have the same maximally abelian subgroup (namely, a maximal torus of SU(3))? Is there a simple way to go about showing that they share this maximal abelian subgroup?

Apologies if the wording lacks precision, feel free to guide me toward clarifications.

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If I am not being mistaken $sl_{n}$'s complexification should be $su_{n}$. –  Kerry Apr 24 '12 at 1:45
    
@ChangweiZhou: $su_n$ is not a complex Lie algebra. $su_n \otimes \mathbb C \simeq sl_n(\mathbb C)$. –  Eric O. Korman Apr 24 '12 at 1:51
    
@Eric: Glad to know I was being mistaken. –  Kerry Apr 24 '12 at 1:59
    
Why do you write "the" maximal abelian subgroup. Why should there be a unique such subgroup (presumably you mean up to conujugacy)? –  Derek Holt Apr 24 '12 at 7:33
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2 Answers

up vote 2 down vote accepted

Surely the subgroup of diagonal matrices with determinant 1, and the subgroup of matrices of the form $\left(\begin{array}{ccc}a&b&c\\0&a&0\\0&0&a\end{array}\right)$ with $a^3=1$ are both self-centralizing, and hence they are maximal abelian subgroups of ${\rm SL}_3(\mathbb{C})$?

Added later: two more maximal abelian subgroups are matrices of the form $\left(\begin{array}{ccc}a&b&c\\0&a&b\\0&0&a\end{array}\right)$ with $a^3=1$, and matrices of form $\left(\begin{array}{ccc}a&b&0\\0&a&0\\0&0&a^{-2}\end{array}\right)$ with $a \ne 0$.

That makes at least four distinct conjugacy classes of maximal abelian subgroups. Perhaps they are the only ones.

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Much thanks, sir. Might you educate me with regard to proving how rudely self-centered they are? The case of the diagonal matrices particularly... –  AsinglePANCAKE Apr 25 '12 at 1:35
    
Ah, considering the matrix with i, -i, 1 on the diagonal in SL(3,C). If a matrix A commutes with anything in our diagonal torus, it must certainly commute with that one. Respective left and right multiplications show that A was in our torus already. Boom, roasted...methinks. –  AsinglePANCAKE Apr 25 '12 at 4:03
    
I would do it this way. The subgroup $H$ of diagonal matrices fixes exactly three 1-dimensional subspace in its action on ${\mathbb C}^3$, namely the multiples of the natural basis elements. So the normalizer $N$ of $H$ subgroup must permute these three spaces, and $N$ is a split extension of $H$ and the group $S$ of permutation matrices, which has order 6. It is easy to see that no nonidentity element of $S$ centralizes $H$, so $H$ is self-centralizing in $N$ and hence in ${\rm SL}_3({\mathbb C})$. –  Derek Holt Apr 25 '12 at 7:45
    
Nice. Seems to throw a lot more machinery at it than I'm assumed to know, but more mature solutions are always fun to know. –  AsinglePANCAKE Apr 25 '12 at 14:35
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The Lie group $SL_{3}\mathbb{C}$ is simply connected, thus the maximal torus is in one to one correspondence with the Cartan subalgebra of $sl_{3}\mathbb{C}$. We can identify it as generated by $e_{1,1}-e_{2,2}$, $e_{2,2}-e_{3,3}$. This coincide with the Cartan subalgebra of $su_{3}$. Thus I believe the maximal torus of them are the same.

Another (possibly wrong) way of thinking is the maximal torus of $SL_{3}\mathbb{C}$ has rank $2$ because it has two generators(the third uniquely determined by the previous two since it is traceless). We know for $SU_{3}$ the maximal torus is the intersection of $U_{n}$ with $T^{n}$, which cuts dimension by 1. Thus both maximal torus has rank 2. Since $SU_{3}\subset SL_{3}\mathbb{C}$ the two maximal torus should be conjugate to each other or can be identified to be the same.

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Hmmm, while I'm sure your answer is awesome, it's a little (way) over my head. I don't actually know what a Cartan subalgebra is... –  AsinglePANCAKE Apr 24 '12 at 2:18
    
Let me try to argue it in other ways. –  Kerry Apr 24 '12 at 2:18
    
What do you mean by «the maximal torus is in one to one correspondence with the Cartan subalgebra»? –  Mariano Suárez-Alvarez Apr 24 '12 at 2:28
    
@MarianoSuárez-Alvarez: I mean that by exponentiating it we should get the corresponding maximal torus. –  Kerry Apr 24 '12 at 2:30
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