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For a Lie algebra $L$ of dimension $n$ over a field ${\mathbb F}$ we denote by $\beta(L)$ the maximal dimension of abelian ideals of $L$. In general, $\beta(L)$ is not preserved under extensions of the ground field (see e.g. Example 2.7 in http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf). Do you know any example in which $\beta(L)<\beta(L\otimes_{\mathbb F} \bar{{\mathbb F}})=n-1$, where $\bar{\mathbb F}$ is the algebraic closure of ${\mathbb F}$? (In other words, is it possible that $L\otimes_{\mathbb F} \bar{{\mathbb F}}$ contains an abelian ideal of codimension 1 and $L$ has no abelian ideal of codimension 1?)

I am mainly interested in the case where $L$ is a restricted Lie algebra over a field of characteristic $p>0$.

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It seems risky to jump into prime characteristic here (including finite fields) when the existing literature is mostly oriented toward characteristic 0. But the question asked does make sense. –  Jim Humphreys Mar 10 '12 at 16:59
    
Professor Humphreys: I agree that jumping in positive characteristic could seem risky, but I faced with this problem when I was dealing with restricted enveloping algebras satisfying certain polynomial identities. This is the real motivation of my question. –  Salvatore Siciliano Mar 10 '12 at 17:34
    
What I find strange is that in the paper you linked, Burde and Ceballos formulate Proposition 3.1 only for the case when the field has characteristic zero. In my opinion, his proof (modulo one trivial typo: "Rescaling $e_1$" should be "Rescaling $e_2$") works equally well in every characteristic. –  darij grinberg May 5 '12 at 0:19
    
darij: I agree with you. –  Salvatore Siciliano May 5 '12 at 12:11
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1 Answer

up vote 4 down vote accepted

There's no such example.

Since this is convenient, I denote by $L$ the Lie algebra over the algebraic closure. Let $A$ be a codimension 1 abelian ideal and let us show that some (possibly other) abelian ideal $A'$ is defined over the ground field, i.e. is a hyperplane that can be defined by a linear equation with coefficients in $K$.

Since $L/A$ is abelian, we have $[L,L]$ contained in $A$. In particular, $A$ is contained in the centralizer of $[L,L]$. In case $A$ is equal to the centralizer of $[L,L]$, this is defined over the ground field and thus we are done. So I now assume that the centralizer of $[L,L]$ is all of $L$ (so $L$ is nilpotent of step 2).

The case when $L$ is abelian is trivial. If the derived subalgebra of $L$ is 1-dimensional, then the Lie algebra law can be viewed as an alternate form. Since $A$ is a codimension 1 isotropic subspace for this form, it is easy to check that the kernel of this alternate form has codimension 2 (and is defined over the ground field) and contains $[L,L]$ because $L$ is nilpotent of step 2. Every hyperplane $A'$ containing this kernel is an abelian ideal; we can pick it to be defined over the ground field.

If the derived subalgebra of $L$ is at least 2-dimensional, there exist two linear forms $f_1,f_2$ on $L$ such that the alternate bilinear forms $(x,y)\mapsto b_i(x,y)=f_i([x,y])$, $i=1,2$ are not proportional. They can be chosen to be defined over the ground field. Let $K_i$ be the kernel of $b_i$. Then $K_i$ is contained in $A$ (otherwise $b_i$ would be zero). Besides, $K_1$ and $K_2$ have codimension 2 (because $A$ is an isotropic subspace for $b_i$) and are not equal, because otherwise $b_1$ and $b_2$ would be alternate forms on the plane $L/K_1$ and would thus be proportional as the set of antisymmetric matrices of size 2 is 1-dimensional. So the codimension of $K_1+K_2$ is at most 1. Since it's contained in $A$, we deduce that $A=K_1+K_2$. So $A$ is defined over the ground field and the proof is complete.

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Very nice! Even better, it seems that this argument works in the infinite dimensional case, as well! –  Salvatore Siciliano May 11 '12 at 22:43
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