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I am very interested in the Nicola Ciccoli's answer to a question formulated in math*overflow*:67717

My questions are:

  1. Can a nilpotent Lie bracket be rigid in the scheme of Lie brackets on $\mathbb{C}^n$? Is it easy to prove a rigid nilpotent Lie algebra exists?

  2. Do there exist rigid Lie brackets in the scheme of nilpotent Lie brackets on $\mathbb{C}^n$ for $n \geq 7$?

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1  
I'm not sure what you mean by the scheme of Lie brackets. Certainly if $[, ]$ is a bracket then so is $c [, ]$, so you at least want to quotient out by scalar multiplication for the question to be nontrivial. Do you also want to quotient by isomorphism? But then why is the result a scheme...? –  Qiaochu Yuan Jun 15 '11 at 9:13
    
Yes you quotient by isomorphism (obviously $(\mathfrak{g},[\cdot,\cdot])$ is always isomorphic to $(\mathfrak{g},c[\cdot,\cdot])$) ). –  YCor Nov 21 at 22:52

3 Answers 3

Vergne's conjecture is still open. It says that there is no complex $n$-dimensional nilpotent Lie algebra which is rigid in the variety $\mathcal{L}_n(\mathbb{C})$ of all $n$-dimensional complex Lie algebras. The Grunewald and O’Halloran conjecture, which implies Vergne's conjecture, states that every complex nilpotent Lie algebra is the proper degeneration of another Lie algebra of the same dimension. The Heisenberg Lie algebra, with $[x,y]=z$ is rigid in $\mathcal{N}_3(\mathbb{C})$, but not in $\mathcal{L}_3(\mathbb{C})$, since it is a proper degeneration of $sl_2(\mathbb{C})$.

R. Carles gives a necessary condition for a nilpotent Lie algebra $L$ to be rigid in $\mathcal{L}_n(\mathbb{C})$: it must be characteristically nilpotent, such that every ideal of codimension $1$ in $L$ is again characteristically nilpotent. It is not clear, whether Carles condition is also sufficient. Certainly the condition to be characteristically nilpotent is not sufficient. It is easy to construct characteristically nilpotent Lie algebras, which are not rigid. But even for the stronger condition of Carles, I found filiform nilpotent Lie algebras of dimension $n\ge 13$, which are characteristically nilpotent, and every ideal of codimension $1$ also being characteristically nilpotent. Unfortunately I do not know whether these algebras are rigid or not (there is a claim in the literature that no filiform nilpotent Lie algebra can be rigid in $\mathcal{L}_n(\mathbb{C})$, but I could not verify this, and did not find a valid proof).

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The condition by Carles is on any lie algebra or on nilpotent ones? –  Mariano Suárez-Alvarez Nov 21 at 22:57
    
Carles condition for rigidity in $R_m$ (solvable Lie algebras) is for any Lie algebra; but I used a corollary for nilpotent Lie algebras (thank you, I forgot to say this). –  Dietrich Burde Nov 30 at 21:00

I don't want to disappoint you, but I'm not a big expert in the field :(

  1. Is it easy? No. It was proved already by Dixmier in the 50's all nilpotent Lie algebra (of dimension greater than 1) have nonvanishing 2-cohomology. They are therefore not cohomologically rigid. This, in principle, does not exclude rigidity but makes it much less easier to check. Proving rigidity without cohomological arguments means being able to describe in quite some detail a specific irreducible component of the variety of Lie algebras of a fixed dimension.

In the paper

Grunewald- O'Halloran Journ. Alg. 162, 210--224 (1993)

many conditions are listed under which a nilpotent Lie algebra is not rigid.

If I remember correctly a nilpotent Lie algebras is rigid if it is Characteristically Nilpotent (basically all derivations are nilpotent). But between the known examples of CN Lie algebras none is explicitely known to be rigid. Vèrgne conjectured that there are no nilpotent Lie algebras which are rigid in the schem of nilpotent Lie brackets but as far as I know the conjecture is still open.

Just in case someone is curious about the $n\ge 7$ bound it is appropriate to remark that from dimension 8 onward one has the appearance of continuous families of non isomorphic Lie algebras.

I cannot add much to your second question.

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It is obvious that the nilpotent Lie bracket on $\mathbf{C}^3$

$\mu(X,Y)=Z$

is rigid in the scheme of nilpotent Lie brackets on $\mathbf{C}^3$ but this one is not characteristically nilpotent; Nicola Ciccoli, is this wrong?

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Yes, sorry, I overlapped the two questions, my remarks on the 2nd questions are plainly wrong. I edit my answer –  Nicola Ciccoli Jun 15 '11 at 16:02

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