Maximal dimension of abelian ideals of a Lie algebra and extensions of the ground field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:06:25Z http://mathoverflow.net/feeds/question/90831 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90831/maximal-dimension-of-abelian-ideals-of-a-lie-algebra-and-extensions-of-the-ground Maximal dimension of abelian ideals of a Lie algebra and extensions of the ground field Salvatore Siciliano 2012-03-10T16:42:52Z 2012-05-11T21:09:39Z <p>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 <a href="http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf" rel="nofollow">http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf</a>). Do you know any example in which $\beta(L)&lt;\beta(L\otimes_{\mathbb F} \bar{{\mathbb F}})=n-1$, 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?)</p> <p>I am mainly interested in the case where $L$ is a restricted Lie algebra over a field of characteristic $p>0$. </p> http://mathoverflow.net/questions/90831/maximal-dimension-of-abelian-ideals-of-a-lie-algebra-and-extensions-of-the-ground/96712#96712 Answer by Yves Cornulier for Maximal dimension of abelian ideals of a Lie algebra and extensions of the ground field Yves Cornulier 2012-05-11T21:09:39Z 2012-05-11T21:09:39Z <p>There's no such example.</p> <p>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 $K$.</p> <p>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).</p> <p>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.</p> <p>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.</p>