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The notes I am following for my class on Lie algebra introduced the Cartan subalgebra for a complex semisimple Lie algebra $L$ as the set $$H=\{x : \mathrm{ad}_x=\mathrm{semisimble}\}$$ and proceeded in proving that this set is actually a Lie subalgebra. the following are what I read in the notes that have been provided to me..

We know that if $L$ is semisimple every element have the abstract Jordan decomposition; and that since $L$ is semisimple there must be some elements of $L$ with non zero semisimple part.

Let $x,y \in H$ then $\mathrm{ad}_x , \mathrm{ad}_y$ are diagonalizable. This means that $L$ decomposes on a direct sum of generalized eigenspases of the two endomorphisms and the endomorphisms act as a scalar multiple of the unit endomorphism on every generalized eigenspase.

Symbolically, $$ L = \bigoplus_{\lambda}L_{\lambda} = \bigoplus_{\mu}L_{\mu} $$ where $\lambda$ and $\mu$ range over the eigenvalues of $\mathrm{ad}_x$ and $\mathrm{ad}_y$ respectively. And $\mathrm{ad}_xz=\lambda z$ if $z\in L_{\lambda}$ ; and $\mathrm{ad}_yz=\mu z$ if $z\in L_{\mu}$.

Finally sinse $L_{\lambda}$ are disjoint when $\lambda$ ranges, and so are $L_{\mu}$ we can write $L$ as a direct sum of the subspaces $L_{\lambda \mu}=L_{\lambda}\cap L_{\mu}$, $L = \bigoplus_{\lambda, \mu} L_{\lambda \mu}$

From the definition of adjoint representation $\mathrm{ad}_xz=[x,z]$ we have that $$\mathrm{ad}_{x+y}z = \mathrm{ad}_x z + \mathrm{ad}_y z$$ and $$\mathrm{ad}_{[x,y]} z = [\mathrm{ad}_x z,y] + [x,\mathrm{ad}_y z]$$ so if $z \in L_{\lambda \mu}=L_{\lambda}\cap L_{\mu}$ we have that $\mathrm{ad}_{x+y}z= (\lambda + \mu)z$ and $\mathrm{ad}_{[x,y]}z= (\lambda + \mu)z$.

The last observation together with the fact that $L = \bigoplus_{\lambda, \mu} L_{\lambda \mu}$ gives that $\mathrm{ad}_{x+y}$ and $\mathrm{ad}_{[x,y]}$ are diagonalizable; hence, since $L$ is complex Lie algebra the are semisimple endomorphisms.

So, the set $H \subseteq L$ is a subalgebra of $L$.

my question

I feel that the above argument of the decomposition of $L$ on the subspases $L_{\lambda \mu}$ works just fine for every pair of diagonazible endomorphisms of a complex vector space which means that the sum of two diagonizable matrices is again diagonizable, which is a falce fact.

So, are there something that I missing? Are the notes using some fact that could only exist in the context of complex semisimple Lie algebras? or the above argument is false, and something is wrong with the notes?

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karhas
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1 Answers1

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There is no such thing as "the" Cartan subalgebra of a Lie algebra: They generally contain many Cartan subalgebras. This already shows that this definition of Cartan subalgebras is wrong. There certainly are various definitions for Cartan subalgebras in various contexts, and for some of them it's non-trivial to show that they are equivalent (cf. Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?, Is Cartan subalgebra of Complex semisimple Lie algebra the maximal Abelian subalgebra? I found two places give the different answers., Equivalence of Two Cartan Subalgebra Definitions in Semi-Simple Lie Algebra, Definition of Cartan subalgebra in Erdmann-Wildon), but none of them matches the one in your first paragraph.

Rather, in a semisimple Lie algebra $L$, one can define a Cartan subalgebra as a subalgebra $H$ which

  1. consists of semisimple elements,

  2. is abelian, and

  3. is maximal among the subalgebras which satisfy conditions 1 and 2.

So in particular your definition misses the "abelian" condition, which solves your issue: Namely, any two elements of a CSA commute, and the sum of two commuting diagonalisable endomorphisms is indeed diagonalisable.

Torsten Schoeneberg
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