Existence of Cartan subalgebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:54:51Z http://mathoverflow.net/feeds/question/79417 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79417/existence-of-cartan-subalgebra Existence of Cartan subalgebra CEOandVIP 2011-10-28T17:47:39Z 2011-10-30T18:15:23Z <p>I am reading Helgason's book. In Chapter 3 he proved the existence of Cartan subalgebra for a semisimple Lie algebra $\mathfrak g$ (definition: a Cartan subalgebra is a maximal abelian subalgebra all whose element $H$ satisfies $\text{ad}_H$ is semisimple). </p> <p>It seems to me the proof is quick: if $H\in {\mathfrak g}$, then $\text{ad}_H$ is automatically semisimple because $K(\text{ad}_H X, Y)+K(X, \text{ad}_H Y)=0$, where the Killing form $K$ is nondegenerate (since $\mathfrak g$ is semisimple) - thus any $\text{ad}_H$ invariant subspace has an invariant complementary subspace. So any maximal abelian subalgebra in a semisimple Lie algebra is a Cartan subalgebra. </p> <p>My question is, is the above argument valid? I am confused since Helgason spent more than 3 pages proving the existence of Cartan subalgebra in the semisimple case - of course his proof contains a lot of information. </p> http://mathoverflow.net/questions/79417/existence-of-cartan-subalgebra/79421#79421 Answer by David Loeffler for Existence of Cartan subalgebra David Loeffler 2011-10-28T18:03:20Z 2011-10-28T18:03:20Z <p>This argument does not work because the Killing form is not generally positive definite, so the orthogonal of a subspace wrt the Killing form is not necessary a complement of the subspace.</p> http://mathoverflow.net/questions/79417/existence-of-cartan-subalgebra/79457#79457 Answer by Salvatore Siciliano for Existence of Cartan subalgebra Salvatore Siciliano 2011-10-29T09:52:50Z 2011-10-30T18:15:23Z <p>Let me just add some remarks. In general, if $L$ is a finite dimensional Lie algebra over an arbitrary field $F$ then a subalgebra $H$ of $L$ is called a Cartan subalgebra if $H$ is nilpotent and self-normalising in $L$. If $L$ is semisimple and $F$ has characteristic zero (as in the case asked by the OP) then the Cartan subalgebras of $L$ are precisely the maximal tori of $L$. (A torus of $L$ is an abelian subalgebra consisting of semisimple elements). Note that the existence of a Cartan subalgebra is always assured whenever the ground field has more than $\dim_F L$ elements. In particular, finite dimensional Lie algebras over infinite field always have Cartan subalgebras. Moreover, the Cartan subalgebras coincides with the minimal Engel subalgebras of $L$. (A subalgebra of $L$ is called an Engel subalgebra if it is the null Fitting component of $L$ with respect to $ad x$ for some $x\in L$.) See the paper </p> <p>R.E. Barnes: On Cartan Subalgebras of Lie Algebras, Math. Z. 101 (1967), 350-355.</p> <p>On the other hand, the existence of Cartan subalgebras of Lie algebras defined over small fields still remains an OPEN problem. </p> <p>It is also worth to mention that solvable Lie algebras always have Cartan subalgebras.</p> <p>Finally, if $L$ is a finite dimensional restricted Lie algebra over a field of characteristic $p>0$, then $H$ is a Cartan subalgebra of $L$ if and only if is the centralizer of a maximal torus of $L$. (Here a torus is an abelian subalgebra consisting of semisimple elements; an element $x$ of $L$ is semisimple it $x$ is in the restricted subalgebra generated by $x^{[p]}$). </p>