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I was looking at the proof of the conjugacy of Cartan subalgebras from Carter's Lie Algebra's of Finite and Affine Type. An important part of the proof is to show that every Cartan subalgebra $H$ has a regular element. To show this, the proof constructs a polynomial function $f$ such that $$f(x) = T_x(x)$$ where $T_x$ is in the group of inner automorphisms, $G$, and then shows that the range of $f$ is large. Is there a more explicit way to prove the presence of a regular element in $H$? Why not consider the set of all $G$ translates of $H$? I am also interested in knowing if there is a simpler proof if we allow conjugacy by automorphisms which are not inner.

  • What definition of regular are you using? $x$ is regular if $L_0(x)$ is a Cartan subalgebra of $L$, see here. – Dietrich Burde Feb 24 '18 at 20:04
  • The definition I'm using is: "$x$ is regular if the characteristic polynomial $p(t)$ of $ad x$ is divisible by as small a power of $t$ as possible." –  Feb 25 '18 at 08:18

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For the proof of the conjugacy of Cartan subalgebras in $L$ over an algebraically closed field one needs to show that the regular elements form a non-empty Zariski open set in $L$, invariant under all automorphisms of $L$; and, that for a regular element $h_0$ the Lie algebra $H=L_0(h_0)$ is nilpotent, and equal to its own normaliser. Then the Theorem follows easily. Proving "Zariski-open" has to use that the condition $\dim L_0(h_0)>rank (L)$ can be expressed by the vanishing of polynomial functions as above, given the definition of regular.

If you want a more explicit proof of the conjugacy, then follow the proof given in the lecture notes Lie algebras - Harvard Mathematics Department by Sternberg, Theorem $10$, page $76$ till page $81$.

Dietrich Burde
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  • That the regular elements (as defined above) form a a Zariski open set is clear from the definition (the characteristic polynomial's coefficients are polynomials on the Lie algebra, and the regular elements are those for which one of those polynomials does not vanish). Again, it is clear to me that the regular elements are invariant under automorphisms of L, as is the fact that $L_0(h_0)$ is a Cartan algebra. I fail to see how the theorem then follows easily. –  Feb 25 '18 at 14:51
  • The proof in Sternberg's notes goes through BSAs and I don't find it particularly direct (but of course, this is a matter of opinion). My question is about finding regular elements in a given Cartan subalgebra, which neither your answer, nor Sternberg's notes seem to answer. –  Feb 25 '18 at 14:52
  • Actually Sternberg's proof is quite detailed and nice, but of course this depends on your taste. You will have to find your way then. This is anyway the best method to learn something. A Cartan subalgebra is just a $L_0(h_0)$ with any regular element from $L$. So just take one $h_0$. – Dietrich Burde Feb 25 '18 at 17:40
  • Taking one $h_0$ yields a Cartan subalgebra. But how do we know that every Cartan subalgebra is of this form? Am I missing something? Or why does the Zariski open set of regular elements have to intersect every Cartan subalgebra (post facto it does, of course)? –  Feb 25 '18 at 19:41
  • This is the point. All Cartan subalgebras arise as $L_0(h_0)$. This is certainly explained in Carter? It is Lemma $9.4.7$ here, and the proof consists only of a few lines, again using Zariski-dense. – Dietrich Burde Feb 25 '18 at 19:54
  • Yes, this is what my question is about, and it is explained in Carter. The method in Carter is the indirect construction I mention in my question, which I want a replacement for. The method in your comment is different but the important part is in Lemma 9.4.6 I think. Lemma 9.4.7 is just Zariski-dense, as you say. It is still unclear to me whether there is any way to do this for automorphisms without using the 'inner' property. –  Feb 25 '18 at 21:46