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A subalgebra which is a Semisimple Lie algebra with the 2 properties

  • The subalgebra is maximal abelian
  • All elements are diagonalizable

is called Cartan subalgebras. The most common example is the algebra of all diagonalizable matrices but I don't quite understand why these matrices are maximal abelian since I believe that maximal abelian means $[X,Y] = 0$ for all $X,Y \in J$ with J being the subalgebra.

For my understanding maximal abelian is not guaranteed for that specific example. I would be interested in an explanation why diagonalizable matrices are (seemingly) a Cartan subalgebra. Can someone think of other examples?

gamma
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    I think you've confused diagonalizable with diagonal: diagonal matrices do indeed form an abelian subalgebra. – lisyarus Jan 07 '21 at 15:20
  • Your definition is imprecise. It should read "a subalgebra of a complex semisimple Lie algebra ...". -- Also, $[X,Y]= 0$ for all $X,Y \in J$ means nothing more or less than that $J$ is abelian. To be maximal abelian means that there is no bigger abelian subalgebra which contains $J$, or equivalently, $J$ is its own centraliser. – Torsten Schoeneberg Jan 07 '21 at 18:04
  • That the diagonal matrices in $\mathfrak{sl}_n(\mathbb C)$ are a CSA is a standard exercise. Show a) its elements are $ad$-diagonalisable (should be easy), b) they all commute with each other (meaning they are an abelian subalgebra, should also be easy if not known from linear algebra) and c) (to show maximal abelian) that for every matrix $X$ which has a non-zero entry off the diagonal there exists some (traceless) diagonal matrix which does not commute with $X$. Also, cf. https://math.stackexchange.com/q/3968855/96384, https://math.stackexchange.com/q/2095754/96384 and links therein. – Torsten Schoeneberg Jan 07 '21 at 18:12
  • As @lisyarus points out, the set of all diagonalisable matrices (in $\mathfrak{sl}_n(\mathbb C)$, say) is not a CSA, indeed it is not even a subalgebra (exercise: find two diagonalisable matrices in $\mathfrak{sl}_2$ whose sum is not diagonalisable). As said, the subalgebra of all diagonal matrices is one, and that is "the most common example". – Torsten Schoeneberg Jan 07 '21 at 18:18
  • Thanks for your help. Obviously, I've been a little confused with the terminus. I'll try your exercise and ask again is this post if I'm stuck. – gamma Jan 08 '21 at 10:27

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