A distinguished example of filiform deformation

TL;DR

This paper presents a specific 13-dimensional filiform complex Lie algebra with all codimension-1 ideals being characteristically nilpotent and demonstrates a non-trivial deformation of this algebra.

Contribution

It provides a unique example of a high-dimensional filiform Lie algebra with special nilpotency properties and constructs its non-trivial deformation.

Findings

Example of a 13-dimensional filiform Lie algebra with all codimension-1 ideals characteristically nilpotent

Construction of a non-trivial filiform deformation of this algebra

Insight into the structure and deformation theory of complex Lie algebras

Abstract

We exhibit an example of a filiform (complex) Lie algebra of dimension 13 with all its ideals of codimension 1 being characteristically nilpotent, and we construct a non trivial filiform deformation of it.

Equations

$$[e_0,e_i]$$

$$[e_1,e_2]$$

$$[e_1,e_4]$$

$$[e_1,e_6]$$

$$[e_1,e_8]$$

$$[e_1,e_{10}]$$

$$[e_2,e_3]$$

$$[e_2,e_5]$$

$$[e_2,e_7]$$

$$[e_2,e_9]$$

$$[e_3,e_4]$$

$$[e_3,e_6]$$

$$[e_3,e_8]$$

$$[e_4,e_5]$$

$$[e_4,e_7]$$

$$[e_5,e_6]$$

$$f_0=e_0+b{e}_1,f_1=e_2,f_2=[f_0,f_1],f_3=[f_0,f_2],\dots ,f_{11}=[f_0,f_{10}].$$

$${\mathfrak{h}}_b^i=⟨f_{i+1},\dots ,f_{11}⟩.$$

$$\overline{{\mathfrak{h}}_b}={\mathfrak{h}}_b/{\mathfrak{h}}_b^7.$$

$$[f_0,f_i]$$

$$[f_1,f_2]$$

$$[f_2,f_3]$$

$$D=\left(\begin{array}{ccccc}{m}_{1,1}& m_{1,2}& 0& \cdots & 0\\ m_{2,1}& m_{2,2}& 0& \cdots & 0\\ m_{3,1}& m_{3,2}& m_{3,3}& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ m_{8,1}& m_{8,2}& m_{8,3}& \cdots & m_{8,8}\end{array}\right)$$

$$E_{i,j}^k=\left(D[f_i,f_j]-[D{f}_i,f_j]-[f_i,D{f}_j]\right)[k]$$

$$E_{0,i}^{i+1}=m_{i+2,i+2}-m_{1,1}-m_{i+1,i+1},$$

$$m_{i+2,i+2}=i{m}_{1,1}+m_{2,2},\text{for 1≤i≤6.}$$

$$m_{8,8}=3m_{1,1}+2m_{2,2}.$$

$$m_{i+1,i+1}=(i+2)m_{1,1},\text{for 1≤i≤7}.$$

$$E_{1,2}^6=\frac{1}{10}{m}_{7,6}-\frac{27}{100}b{m}_{7,7}+\frac{27}{100}b{m}_{2,2}+\frac{27}{100}b{m}_{3,3}-\frac{1}{10}{m}_{4,3},$$

$$\begin{array}{c}\mathfrak{z}=⟨e_{11},e_{12}⟩,{\mathfrak{z}}^2=⟨e_9,\dots ,e_{12}⟩,{\mathfrak{z}}^3=⟨e_7,\dots ,e_{12}⟩,\\ {\mathfrak{z}}^4=⟨e_5,\dots ,e_{12}⟩,{\mathfrak{z}}^5=⟨e_3,\dots ,e_{12}⟩,{\mathfrak{z}}^6=\mathfrak{h}.\end{array}$$

$$D=\left(\begin{array}{ccccccc}{m}_{1,1}& m_{1,2}& & & & & \\ & m_{2,2}& & & & & \\ & & m_{3,3}& m_{3,4}& & & \\ & & & m_{4,4}& & & \\ & \text{∗}& & & \ddots & & \\ & & & & & m_{11,11}& m_{11,12}\\ & & & & & & m_{12,12}\end{array}\right)$$

$$E_{i,j}^k=\left(D[e_i,e_j]-[D{e}_i,e_j]-[e_i,D{e}_j]\right)[k].$$

$$E_{1,2}^3=0,E_{1,4}^5=0,E_{1,6}^7=0,E_{1,8}^9=0,E_{1,10}^{11}=0,$$

$$m_{3,4}=0,m_{5,6}=0,m_{7,8}=0,m_{9,10}=0,m_{11,12}=0.$$

$$D=\left(\begin{array}{cccc}{m}_{1,1}& 0& \cdots & 0\\ m_{2,1}& m_{2,2}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ m_{12,1}& m_{12,2}& \cdots & m_{12,12}\end{array}\right)$$

$$m_{4,4}$$

$$m_{5,5}$$

$$m_{6,6}$$

$$m_{7,7}$$

$$m_{8,8}$$

$$m_{9,9}$$

Full text

1. Introduction

It is already well known that a nilpotent Lie algebra $\mathfrak{n}$ with a codimension 1 ideal $\mathfrak{h}$ of rank $\geq 1$ (with non trivial semisimple derivations) is not rigid [C, GH]. From a semisimple derivation of $\mathfrak{h}$ one constructs a non trivial linear deformation of $\mathfrak{n}$ [GH].

If $\mathfrak{n}$ is it self of positive rank, then it admits a codimension 1 ideal of positive rank. So that the class of nilpotent Lie algebras with all its ideals of codimension 1 being characteristically nilpotent (without any semisimple derivation) is contained in the class of characteristically nilpotent Lie algebras. Let us call them strong characteristically nilpotent. It is not clear to us how large or how small is this class. If there exists a rigid nilpotent Lie algebra, in opposition to Vergne’s conjecture, it must be of this class.

Some time ago, Dietrich Burde told us about a family of filiform Lie algebras $\{\mathfrak{f}_{n}\}$, that they suspect to be of this class [BEdG].

In this paper we pick the first algebra of their list, $\mathfrak{f}_{13}$, we prove that it is strong characteristically nilpotent and we construct a filiform non trivial deformation of it.

This example provides further strong evidence supporting Vergne’s conjecture.

2. The algebra $\mathfrak{f}_{13}$

In this paper we consider the filiform Lie algebra $\mathfrak{f}_{13}$ (see [BEdG, §5]) defined over $\mathbb{C}$, which in the basis $\{e_{0},e_{1},\dots,e_{12}\}$ is defined by: 111Notice that in [BEdG] the basis is $\{e_{1},e_{2},\dots,e_{13}\}$.

[TABLE]

Remark 2.1*.*

In this paper $\mathbb{C}$ might be replaced by $\mathbb{R}$ or even $\mathbb{Q}$ with no further changes. Notice that the algebras $\{\mathfrak{f}_{n}\}$ in [BEdG] are defined over $\mathbb{Q}$. Our main interest is Vergne’s conjecture over $\mathbb{C}$.

A generic codimension 1 ideal of $\mathfrak{f}_{13}$ is of the form $\mathfrak{h}_{b}=\langle e_{0}+be_{1},\dots,e_{12}\rangle$, for some $b\in\mathbb{C}$. More precisely there is only one codimension 1 ideal left, the ideal $\mathfrak{h}=\langle e_{1},\dots,e_{12}\rangle$. In this section we show that all these ideals are characteristically nilpotent.

Let us start with the generic 12-dimensional ideal $\mathfrak{h}_{b}=\langle e_{0}+be_{1},\dots,e_{12}\rangle$, for some $b\in\mathbb{C}$. Define a new basis $\langle f_{0},f_{1},\dots,f_{11}\rangle$ for $\mathfrak{h}_{b}$ as follows:

[TABLE]

Notice that $\mathfrak{h}_{b}$ is filiform for all $b$ and its central descending series is

[TABLE]

In order to show that $\mathfrak{h}_{b}$ is characteristically nilpotent, it is useful to consider the 8-dimensional filiform quotient

[TABLE]

By abuse of notation, the quotient may be described using the (same) basis $\{f_{0},\dots,f_{7}\}$:

[TABLE]

Lemma 2.2**.**

The quotient $\overline{\mathfrak{h}_{b}}$ is characteristically nilpotent.

Proof.

Let $D$ be a derivation of $\overline{\mathfrak{h}_{b}}$. Since $D$ preserves the central descending series of $\overline{\mathfrak{h}_{b}}$, the matrix of $D$ in the basis $\{f_{0},\dots,f_{7}\}$ has the following form:

[TABLE]

For $0\leq i<j\leq 7$ and $0\leq k\leq 7$ let

[TABLE]

be the coordinate respect to $f_{k}$ of $D[f_{i},f_{j}]-[Df_{i},f_{j}]-[f_{i},Df_{j}]$. Since $D$ is a derivation, $E_{i,j}^{k}=0$ for all $i,j,k$.

A direct evaluation yields $E_{1,2}^{3}=-m_{1,2}$, and hence $m_{1,2}=0$. Also one directly obtains that, for $1\leq i\leq 6$,

[TABLE]

and therefore recursively

[TABLE]

Now $E_{1,4}^{7}=\frac{3}{35}m_{8,8}-\frac{3}{35}m_{2,2}-\frac{3}{35}m_{5,5}$, so that $m_{8,8}=m_{2,2}+m_{5,5}$ and using (2.1)

[TABLE]

By combining (2.1) with (2.2) we get that $m_{2,2}=3m_{1,1}$ and recursively that

[TABLE]

Therefore $D$ is lower triangular with all its diagonal entries being integer multiples of $m_{1,1}$.

Finally, equations $E_{0,2}^{4}=0$, $E_{0,3}^{5}=0$ and $E_{0,4}^{6}=0$ combined, yield $m_{7,6}=m_{4,3}$. And since

[TABLE]

from (2.3) it follows that $m_{1,1}=0$ and then $D$ is nilpotent. Consequently, $\mathfrak{h}_{b}$ is characteristically nilpotent. ∎

Let us now come to consider the single 12-dimensional ideal $\mathfrak{h}=\langle e_{1},\dots,e_{12}\rangle$. The proof that it is in fact characteristically nilpotent is analogous to the proof of Lemma 2.2. Notice that $\mathfrak{h}$ is not filiform, it is 6-step nilpotent, and its central ascending series is given by:

[TABLE]

Lemma 2.3**.**

The ideal $\mathfrak{h}$ is characteristically nilpotent.

Proof.

Let $D$ be a derivation of $\mathfrak{h}$. Since it preserves the central ascending series of $\mathfrak{h}$, the matrix of $D$ in the basis $\{e_{1},\dots,e_{12}\}$ is of the following $(2\times 2)$-block lower triangular form

[TABLE]

For $1\leq i<j\leq 12$ and $1\leq k\leq 12$, let as in the proof of Lemma 2.2,

[TABLE]

Recall that $E_{i,j}^{k}=0$ for all $i,j,k$.

The equations

[TABLE]

considered in the given order, yield directly

[TABLE]

Then since $E_{2,4}^{6}=m_{1,2}-\tfrac{1}{9}m_{3,4}$, $m_{3,4}=0$ and thus $D$ is lower triangular.

[TABLE]

Now from the equations $E_{0,j}^{j+2}=0$, for $1\leq j\leq 9$, we get that

[TABLE]

And from the equations $E_{1,2}^{5}=0$ and $E_{1,3}^{6}=0$, we get that

[TABLE]

From (2.4) and (2.6), it follows that $m_{3,3}=2m_{1,1}$, and from (2.5) and (2.7), it follows that $m_{2,2}=\tfrac{3}{2}m_{1,1}$. Therefore

[TABLE]

and thus

[TABLE]

We show now that $m_{1,1}=0$. To this end we consider the following sequence of equations and its consequences.

First, from $E_{0,1}^{5}=0$ we get $m_{6,4}=-\tfrac{1}{10}m_{3,1}+\tfrac{9}{10}m_{4,2}$, and then from $9E_{1,2}^{7}-E_{0,3}^{7}=0$ we get that

[TABLE]

Second, from $E_{0,2}^{6}=0$ we get $m_{7,5}=\tfrac{4}{5}m_{5,3}$, and then from $8E_{1,3}^{8}-E_{0,4}^{8}=0$ we get that

[TABLE]

Now from $\tfrac{3}{2}E2+E1=0$ it follows that

[TABLE]

From $E_{0,3}^{7}=0$ we get $m_{8,6}=\tfrac{10}{9}m_{1,1}-\tfrac{4}{63}m_{3,1}+\tfrac{5}{7}m_{4,2}$, and then from $\tfrac{3}{25}E_{0,5}^{9}-E_{1,4}^{9}=0$ we get that

[TABLE]

Finally, from $\tfrac{35}{18}E2-21E4$ we get that

[TABLE]

and from $20E3-\tfrac{225}{17}E5=0$ it follows that $m_{1,1}=0$, and therefore $D$ is nilpotent. ∎

Lemma 2.4**.**

Let $\mathfrak{n}$ be a nilpotent Lie algebra and $\mathfrak{n}^{i}$ the $i$-th term of its central descending series. If the quotient $\mathfrak{n}/\mathfrak{n}^{i}$ is characteristically nilpotent, then $\mathfrak{n}$ is characteristically nilpotent.

Proof.

Any nilpotent Lie algebra $\mathfrak{g}$ is generated, as a Lie algebra, by the set $\mathfrak{g}-[\mathfrak{g},\mathfrak{g}]$. So that $\mathfrak{n}$ is generated by $S=\mathfrak{n}-[\mathfrak{n},\mathfrak{n}]$ and $\mathfrak{n}/\mathfrak{n}^{i}$ by the projection set $\overline{S}$.

A non trivial semisimple derivation $D$ of $\mathfrak{n}$ is non trivial in $S$, hence the induced derivation on $\mathfrak{n}/\mathfrak{n}^{i}$ is semisimple and non trivial in $\overline{S}$.

Therefore if the quotient algebra $\mathfrak{n}/\mathfrak{n}^{i}$ has no semisimple derivations, then the algebra $\mathfrak{n}$ has not semisimple derivations as well. ∎

Lemmas 2.3, 2.2 and 2.4 prove the following result.

Proposition 2.5**.**

The algebra $\mathfrak{f}_{13}$ is strong characteristically nilpotent.

3. A non-trivial deformation of $\mathfrak{f}_{13}$

In this section let $\mathfrak{f}=\mathfrak{f}_{13}$. For the construction of a non trivial deformation of $\mathfrak{f}$ we follow closely the construction given in [TV, §4]. In this case $\mathfrak{h}=[\mathfrak{f},\mathfrak{f}]=\langle e_{2},\dots,e_{12}\rangle$ and we choose $D\in\operatorname{Der}(\mathfrak{h})$ given by

[TABLE]

and $D(e_{i})=0$, for $i=6\dots 12$. So that, for $t\in\mathbb{C}$ the deformed algebra $\mathfrak{f}_{t}$ is given by the bracket $[\ ,\ ]_{t}$ described by:

[TABLE]

and $[h,h^{\prime}]_{t}=[h,h^{\prime}]$, for all $h,h^{\prime}\in\mathfrak{h}$.

It turns out that $\mathfrak{f}_{t}\simeq\mathfrak{f}$ if and only if $t=0$. To prove this, let $g$ be an isomorphism between $\mathfrak{f}_{t}$ and $\mathfrak{f}$, and let $[g]$ be its matrix with respect to the basis $\{e_{0},\dots,e_{12}\}$ of $\mathfrak{f}_{t}$ and $\mathfrak{f}$. Since $g$ preserves the central descending series, then $[g]$ is of the form:

[TABLE]

That $g$ is an isomorphism is equivalent to

[TABLE]

for all $0\leq i<j\leq 12$. If $E_{i,j}^{k}$ is the coefficient of $e_{k}$ in $E_{i,j}$, that is

[TABLE]

then $g$ is an isomorphism if and only if $E_{i,j}^{k}=0$, for all $0\leq i<j\leq 12$ and $0\leq k\leq 12$.

We show first that the matrix $[g]$ above is lower triangular and its diagonal entries are all equal to 1 or equal to $\pm 1$.

Lemma 3.1**.**

In the matrix $[g]$ in (3.1), $m_{1,2}=0$ and $m_{i,i}=m_{1,1}^{i}$, for $2\leq i\leq 13$. Moreover, $m_{2,1}=0$ and $m_{1,1}=1$ or $m_{1,1}=-1$.

Proof.

That $m_{1,2}=0$ follows directly from the identity

[TABLE]

which is easy to evaluate. So we set $m_{1,2}=0$.

By a direct computation, we get that

[TABLE]

for $1\leq i\leq 11$, and hence $m_{i,i}=m_{2,2}m_{1,1}^{i-2}$, for $2\leq i\leq 13$. So we have in addition that

[TABLE]

from which it follows that $m_{2,2}=m_{1,1}^{2}$ and therefore $m_{i,i}=m_{1,1}^{i}$, for $1\leq i\leq 13$.

For the second part we consider, in the given order, the following equations:

[TABLE]

from which we get that

[TABLE]

Now, from the identity

[TABLE]

we obtain that $m_{2,1}=0$.

We continue by considering the equations

[TABLE]

from where we get that

[TABLE]

Now, from the identities

[TABLE]

it follows that

[TABLE]

Finally,

[TABLE]

and therefore $m_{1,1}=1$ or $m_{1,1}=-1$. ∎

We continue with another technical result that will simplify all the arguments afterwards.

Lemma 3.2**.**

The entries $m_{i+1,i}$ for $2\leq i\leq 12$, of the second diagonal of the matrix $[g]$, are all equal to $m_{3,2}$ if $m_{1,1}=1$, or $m_{i+1,i}=(-1)^{i}m_{3,2}$ if $m_{1,1}=-1$.

Proof.

On the one hand, if $m_{i,i}=1$ for all $i$, we then have that for $1\leq i\leq 10$

[TABLE]

from which it follows that $m_{i+1,i}=m_{3,2}$, for $2\leq i\leq 12$.

On the other hand, if $m_{i,i}=(-1)^{i}$, we then have that for $1\leq i\leq 10$

[TABLE]

from which it follows that $m_{i+1,i}=(-1)^{i}m_{3,2}$, for $2\leq i\leq 12$. ∎

These lemmas all together yield two possible forms for the matrix $[g]$, depending on whether $m[1,1]=1$ or $m[1,1]=-1$:

• •

If $m_{1,1}=1$, then

[TABLE]

• •

If $m_{1,1}=-1$, then

[TABLE]

We are ready to prove the main result of this section.

Theorem 3.3**.**

The filiform Lie algebra $\mathfrak{f}_{t}$ is isomorphic to $\mathfrak{f}$ if and only if $t=0$.

Proof.

The proof follows by repeatedly considering particular sets of identities $\{E_{i,j}^{k}=0\}$ and solving some of the entries $\{m_{i,j}\}$ until one gets that $t=0$. From now on let $\alpha=m_{1,1}$ and recall that $\alpha=1$ or $\alpha=-1$.

• •

By considering the equations

[TABLE]

we get that

[TABLE]

And from

[TABLE]

it follows that

[TABLE]

• •

By considering the equations

[TABLE]

we get that

[TABLE]

And from $0=E_{1,2}^{7}$, it follows that

[TABLE]

• •

By considering the equations

[TABLE]

we get that

[TABLE]

And from $0=E_{1,2}^{8}$, it follows that it follows that

[TABLE]

• •

By considering the equations

[TABLE]

we get that

[TABLE]

And from $0=E_{1,2}^{9}$, it follows that

[TABLE]

• •

Finally, by considering the equations

[TABLE]

we get that

[TABLE]

And from $0=E_{2,3}^{11}=-\tfrac{5}{36}t$, it follows that $t=0$.

∎