On pseudo-Hermitian quadratic nilpotent lie algebras

Abstract

We study nilpotent Lie algebras endowed with a complex structure and a quadratic structure which is pseudo-Hermitian for the given complex structure. We propose several methods to construct such Lie algebras and describe a method of double extension by planes to get an inductive description of all of them. As an application, we give a complete classification of nilpotent quadratic Lie algebras where the metric is Lorentz-Hermitian and we fully classify all nilpotent pseudo-Hermitian quadratic Lie algebras up to dimension 8 and their inequivalent pseudo-Hermitian metrics.

1 Introduction

A pseudo-Hermitian quadratic Lie algebra is a triple $({{\mathfrak {g}}},J,\varphi )$ where ${{\mathfrak {g}}}$ is a Lie algebra, J is a complex structure on ${{\mathfrak {g}}}$ and $\varphi$ is a non-degenerate ad-invariant bilinear form on ${{\mathfrak {g}}}$ that is compatible with J in the sense that $\varphi (Jx,Jy)=\varphi (x,y)$ for all $x,y\in {{\mathfrak {g}}}$. Geometrically, ${{\mathfrak {g}}}$ is the Lie algebra of a Lie group G endowed with a left-invariant complex structure and a bi-invariant pseudo-Hermitian metric. Further, it was seen in Andrada et al. (2008) that the complex structure on such Lie algebras defines a classical r-matrix so that they give rise to Lie bialgebras of complex type and, as a consequence, to certain Poisson Lie groups whose duals are complex Lie groups.

Even though the complex structure on a pseudo-Hermitian quadratic Lie algebra cannot be neither bi-invariant nor abelian, one may see that the class of such Lie algebras is a large one and includes nilpotent, solvable and non-solvable examples; one may even find examples of pseudo-Hermitian quadratic Lie algebras that are perfect. In this paper, we will focus our attention on the case of nilpotent ones.

Medina and Revoy in Medina and Revoy (1985) gave an inductive method to construct all solvable quadratic Lie algebras. The method is commonly known as double extension by a line and consists, essentially, in describing each solvable quadratic Lie algebra of dimension n as the Lie algebra obtained from a $n-2$-dimensional quadratic Lie algebra after a central extension by a 1-dimensional ideal and a suitable semidirect product by another 1-dimensional Lie algebra. When working with Lie algebras endowed with complex structures one wants the Lie algebra of dimension $n-2$ to be endowed with a complex structure too and, in most cases, this situation cannot be described by a single double extension by a line. Thus, in order to guarantee the existence of a complex structure in the Lie algebra of lower dimension, one needs to consider a couple of successive double extensions. In the nilpotent case this can be formulated as a certain generalized double extension in which the central extension is done by a 2-dimensional J-invariant subspace and one applies two semidirect products. As a rule, the two derivations with which one does the semidirect products do not span a Lie subalgebra of the large pseudo-Hermitian quadratic Lie algebra since they may have a nonzero bracket lying in the small Lie algebra. In our Proposition 2.2 we give a detailed description of this modified double extension method and in Theorem 2.4 we show that every indecomposable nilpotent pseudo-Hermitian quadratic Lie algebra can be constructed in that way. A simple application of such a method lets us describe all Lie algebras that can be endowed with Lorentz-Hermitian quadratic metrics.

Using our double extension method, sometimes called pHQ-double extension within the text, we give a complete classification of nilpotent pseudo-Hermitian quadratic Lie algebras up to dimension 8. It should be remarked that, although quadratic nilpotent Lie algebras up to dimension 10 which are indecomposable (as quadratic Lie algebras) have been classified in Kath (2007), one cannot easily deduce our classification from the results therein. On the one hand, indecomposability as a quadratic Lie algebra is not the same as indecomposability as a pseudo-Hermitian quadratic Lie algebra for one may find examples of pseudo-Hermitian quadratic Lie algebras that cannot be decomposed as a sum of two pseudo-Hermitian quadratic Lie algebras but can be split as the sum of two quadratic ones; this is the case, for instance, of the Lie algebra ${{\mathcal {W}}}_3\oplus \mathbb {R}$, where ${{\mathcal {W}}}_3$ is as defined in Favre and Santharoubane (1987), which is obviously decomposable as quadratic Lie algebra but indecomposable as pseudo-Hermitian quadratic Lie algebra. On the second hand, a quadratic Lie algebra might admit non-equivalent complex structures or pseudo-Hermitian quadratic metrics. In Theorem 3.6 we give the complete list of inequivalent pseudo-Hermitian quadratic structures on nilpotent Lie algebras up to dimension 8. Precisely, we show that there are only six inequivalent indecomposable pseudo-Hermitian quadratic nilpotent Lie algebras up to dimension 8.

2 Definitions, examples and first results

From now on, unless otherwise stated, all the Lie algebras considered in the paper are real Lie algebras.

Definition 1.1

A complex structure on a Lie algebra $\mathfrak {g}$ is a map $J \in \mathfrak {gl(g)}$ such that $J^2 = -I$, where I stands for the identity map, and equality

$$\begin{aligned} N_{J}(x,y)=[x, y] + J[Jx, y] + J[x, Jy] - [Jx, Jy] = 0 \end{aligned}$$

(1)

holds for all $x, y \in \mathfrak {g}$.

The complex structure J is said to be abelian if $[Jx,Jy]=[x,y]$ for every $x,y\in {{\mathfrak {g}}}$ whereas it is said to be bi-invariant when $[Jx,y]=J[x,y]$ holds for every $x,y\in {{\mathfrak {g}}}$.

Two Lie algebras endowed with complex structures $({{\mathfrak {g}}}_1,J_1)$, $({{\mathfrak {g}}}_2,J_2)$ are said to be holomorphically isomorphic if there exists an isomorphism of Lie algebras $\psi :{{\mathfrak {g}}}_1\rightarrow {{\mathfrak {g}}}_2$ such that $J_2\circ \psi =\psi \circ J_1$.

Remark 1.1

Let us point out a couple of interesting facts.

• (1) The so-called Nijenhuis tensor $N_{J}(x,y)=[x, y] + J[Jx, y] + J[x, Jy] - [Jx, Jy]$ is clearly antisymmetric and verifies the following identities:

  $$\begin{aligned} N_J(Jx,Jy)=-N_J(x,y),\quad N_J(Jx,y)=N_J(x,Jy)=-JN_J(x,y), \end{aligned}$$

  (2)

  for all $x,y\in {{\mathfrak {g}}}$.

• (2) Denote by ${{\mathfrak {g}}}^{1,0}=\text{ Ker }(J-iI)$ and ${{\mathfrak {g}}}^{0,1}=\text{ Ker }(J+iI)$ the corresponding eigenspaces of eigenvalues i and $-i$ in the complexification ${{\mathfrak {g}}}^{\mathbb {C}}$ of ${{\mathfrak {g}}}$, that is to say:

  $$\begin{aligned} {{\mathfrak {g}}}^{1,0}=\{x-iJx\,\ x\in {{\mathfrak {g}}}\},\quad {{\mathfrak {g}}}^{0,1}=\{x+iJx\,\ x\in {{\mathfrak {g}}}\}. \end{aligned}$$

  One has that equality (1) is equivalent to ${{\mathfrak {g}}}^{1,0}$ (or ${{\mathfrak {g}}}^{0,1}$) being a complex Lie subalgebra of ${{\mathfrak {g}}}^{\mathbb {C}}$. It is well known that J is abelian if and only if ${{\mathfrak {g}}}^{1,0}$ is an abelian Lie algebra and that J is bi-invariant if and only if $({{\mathfrak {g}}}, J)$ and $({{\mathfrak {g}}}^{1,0}, i)$ are holomorphically isomorphic (and, hence, ${{\mathfrak {g}}}$ can be considered as a complex Lie algebra).

  Moreover, for an arbitrary complex structure J on a Lie algebra ${{\mathfrak {g}}}$, denote by ${{\mathfrak {g}}}_J$ the vector space ${{\mathfrak {g}}}$ endowed with a new bracket defined by

  $$\begin{aligned}{}[x,y]_J=[Jx,y]+[x,Jy],\quad \ x,y\in {{\mathfrak {g}}}. \end{aligned}$$

  It can be seen that $[\cdot ,\cdot ]_J$ is actually a Lie bracket and that there is a natural holomorphic isomorphism $\psi :{{\mathfrak {g}}}_J\rightarrow {{\mathfrak {g}}}^{1,0}$ defined by $\psi (x)=Jx+ix$ (see, for instance, Andrada et al. (2008)).

  It should be remarked that the map $\eta :{{\mathfrak {g}}}_J\rightarrow {{\mathfrak {g}}}^{0,1}$ defined by $\eta (x)=Jx-ix$ is also an isomorphism of real Lie algebras. However, $\eta$ is not holomorphic but anti-holomorphic since

  $$\begin{aligned} \eta (Jx)=-x-iJx=-i(Jx-ix)=-i\eta (x),\quad \ x\in {{\mathfrak {g}}}. \end{aligned}$$

• (3) Notice that every Lie algebra with a complex structure $({{\mathfrak {g}}},J)$ has the structure of complex vector space defined by the multiplication by scalars given by:

  $$\begin{aligned} (\alpha +i\beta )\cdot x=\alpha x+\beta Jx,\quad \alpha ,\beta \in \mathbb {R},\ x\in {{\mathfrak {g}}}. \end{aligned}$$

  As a consequence, every Lie algebra admitting a complex structure must have even dimension. Actually, this is also true for every Lie algebra which admits a linear endomorphism J such that $J^2=-I$ (even if the Nijenhuis tensor is not identically zero).

Definition 1.2

Let $({{\mathfrak {g}}}, J)$ be a Lie algebra with a complex structure. If $\mathfrak {g}$ is also endowed with a non-degenerate symmetric bilinear map $\varphi : \mathfrak {g}\times \mathfrak {g} \rightarrow \mathbb {R}$ verifying the condition $\varphi ( Jx, Jy) = \varphi ( x, y),$ for all $x, y \in \mathfrak {g}$, then $(\mathfrak {g}, J, \varphi )$ is said to be a pseudo-Hermitian Lie algebra and the pair $(J, \varphi )$ a pseudo-Hermitian structure on $\mathfrak {g}$.

For a pseudo-Hermitian Lie algebra $(\mathfrak {g}, J, \varphi )$ one defines the fundamental 2-form or Kähler form by

$$\begin{aligned} \omega (x,y)=\varphi (x,Jy),\quad \ x,y\in {{\mathfrak {g}}}. \end{aligned}$$

It is an easy exercise to see that the map $(\cdot |\cdot ):{{\mathfrak {g}}}\times {{\mathfrak {g}}}\rightarrow {\mathbb {C}}$ defined by

$$\begin{aligned} (x|y)=\varphi ( x,y)+i\omega (x,y),\quad \ x,y\in {{\mathfrak {g}}}\end{aligned}$$

is actually a Hermitian form on the complex vector space ${{\mathfrak {g}}}$ defined as in Remark 1.1 (3), that is to say that $(\cdot |\cdot )$ is a sesquilinear form such that $(y|x)=\overline{(x|y)}$ for every $x,y\in {{\mathfrak {g}}}$.

Remark 1.2

It should be remarked that the signature of a pseudo-Hermitian metric $\varphi$ on a 2n-dimensional Lie algebra is always of the form $\text{ sig }(\varphi )=(2r,2s)$. We will use the term Hermitian metric exclusively for the positive definite case $\text{ sig }(\varphi )=(0,2n)$ and we will call a Lorentz-Hermitian metric the case of index 2, that is, $\text{ sig }(\varphi )=(2,2n-2)$.

Also, notice that, since $J^2=-I$, the condition $\varphi ( Jx, Jy) = \varphi ( x, y),$ is equivalent to $\varphi ( Jx, y) = -\varphi ( x, Jy),$ for all $x,y\in {{\mathfrak {g}}}$.

Definition 1.3

Let $(\mathfrak {g}_1, J_1, \varphi _1 )$, $(\mathfrak {g}_2, J_2, \varphi _2 )$ be two pseudo-Hermitian Lie algebras. We will say that they are equivalent if there exists a holomorphic isomorphism $\psi :{{\mathfrak {g}}}_1\rightarrow {{\mathfrak {g}}}_2$ which is also an isometry, that is to say, $\varphi _2(\psi x,\psi y)=\varphi _1(x,y)$ holds for all $x,y\in {{\mathfrak {g}}}_1.$

Definition 1.4

Let ${{\mathfrak {g}}}$ be a Lie algebra and $\varphi :{{\mathfrak {g}}}\times {{\mathfrak {g}}}\rightarrow \mathbb {R}$ a non-degenerate symmetric bilinear form. If for every $x,y,z\in {{\mathfrak {g}}}$ it holds that

$$\begin{aligned} \varphi ([x,y],z)+\varphi (y,[x,z])=0, \end{aligned}$$

then we say that $\varphi$ is a quadratic structure or a quadratic metric on ${{\mathfrak {g}}}$. In such a case, we also say that $({{\mathfrak {g}}},\varphi )$ is a quadratic Lie algebra. In many references, quadratic Lie algebras are also called metric Lie algebras as in Kath (2007).

If a quadratic Lie algebra $({{\mathfrak {g}}},\varphi )$ also admits a complex structure J such that $(\mathfrak {g}, J, \varphi )$ is pseudo-Hermitian, we will say that $(\mathfrak {g}, J, \varphi )$ is a pseudo-Hermitian quadratic Lie algebra.

Our next result shows that abelian complex structures do not exist in quadratic Lie algebras unless the Lie algebra itself is abelian (even if the metric is not pseudo-Hermitian).

Proposition 1.1

If $(\mathfrak {g},\varphi )$ is a non-abelian quadratic Lie algebra, then ${{\mathfrak {g}}}$ cannot admit an abelian complex structure.

Proof

Suppose that $(\mathfrak {g},\varphi )$ is a quadratic Lie algebra and J an abelian complex structure on ${{\mathfrak {g}}}$. Then we have

$$\begin{aligned} \varphi ( [Jx,Jy],z)= & {} -\varphi ( Jy,[Jx,z]) = \varphi (Jy,[x,Jz] ) \\= & {} -\varphi ( Jy,[Jz,x] )= \varphi ( [Jz,Jy],x )\\= & {} \varphi ( [z,y],x )=-\varphi ( [y,z],x )\\= & {} \varphi ( z,[y,x ])=\varphi ( z,[Jy,Jx ])\\= & {} -\varphi ( z,[Jx,Jy ])= - \varphi ([Jx,Jy], z ), \end{aligned}$$

for all $x,y,z\in {{\mathfrak {g}}}$, which clearly shows that ${{\mathfrak {g}}}$ must be abelian because $\varphi$ is non-degenerate. $\square$

Remark 1.3

The result above was already known in the pseudo-Hermitian case (see, for instance, (Andrada et al. 2008, Prop. 9)) but, as we have seen, it is also valid regardless of the compatibility of the complex structure and the metric, which has interesting implications. For example, bearing in mind that cotangent Lie algebras are always quadratic (see Bordemann (1997) and references therein), one deduces that a cotangent Lie algebra can never admit an abelian complex structure.

In contrast with Proposition 1.1, non-abelian quadratic Lie algebras may admit bi-invariant complex structures. For instance, if ${{\mathfrak {g}}}={{\mathfrak {s}}}_\mathbb {R}$ is the underlying real Lie algebra of a complex semisimple Lie algebra ${{\mathfrak {s}}}$ and $\varphi$ denotes the real part of the Killing form ${{\mathcal {K}}}$ on ${{\mathfrak {s}}}$, it is easy to prove that $({{\mathfrak {g}}},\varphi )$ is quadratic and that the map $J:{{\mathfrak {g}}}\rightarrow {{\mathfrak {g}}}$ defined by $Jx=ix$ for all $x\in {{\mathfrak {s}}}$ is a bi-invariant complex structure on ${{\mathfrak {g}}}$. Notice, however, that the triple $({{\mathfrak {g}}}, J,\varphi )$ is not pseudo-Hermitian since

$$\begin{aligned} \varphi (Jx,Jy)=\text{ Re }({{\mathcal {K}}}(ix,iy))=- \text{ Re }({{\mathcal {K}}}(x,y))=-\varphi (x,y), \end{aligned}$$

for every $x,y\in {{\mathfrak {s}}}$. Those metrics verifying $\varphi (Jx,Jy)=-\varphi (x,y)$ are usually called Norden metrics (see Bajo (1997) and references therein).

In fact, the following result is well known (see, for instance, (Andrada et al. 2008, Prop. 9)):

Proposition 1.2

Let $({{\mathfrak {g}}}, J,\varphi )$ be a pseudo-Hermitian quadratic Lie algebra. The complex structure J is bi-invariant if and only if ${{\mathfrak {g}}}$ is an abelian Lie algebra.

Remark 1.4

Although there are no non-abelian examples of pseudo-Hermitian quadratic Lie algebras with abelian or bi-invariant complex structures, the class of such Lie algebras is a wide one.

In (Andrada et al. 2008, Th. 24) it is seen that, up to dimension 6, every quadratic Lie algebra $({{\mathfrak {g}}},\varphi )$ with $\text{ sig }(\varphi )=(2r,2s)$ admits a $\varphi$-skewsymmetric complex structure J and, therefore, $({{\mathfrak {g}}}, J,\varphi )$ is pseudo-Hermitian quadratic.

We will next give some constructions of pseudo-Hermitian quadratic Lie algebras.

Our first construction uses the concept of $\hbox {T}^*$-extension defined by Bordemann in Bordemann (1997).

Proposition 1.3

Let $({{\mathfrak {g}}},J)$ be a Lie algebra with a complex structure and suppose that there exists a 2-cocycle $\theta \in Z^2({{\mathfrak {g}}},{{\mathfrak {g}}}^*)$ for the Chevalley-Eilenberg complex defined by the coadjoint representation of ${{\mathfrak {g}}}$ such that for all $x,y,z\in {{\mathfrak {g}}}$ the following equality holds:

$$\begin{aligned} \theta (x,y)z=\theta (Jx,Jy)z+\theta (Jy,Jz)x+\theta (Jz,Jx)y. \end{aligned}$$

(3)

Define on the vector space $T^*_\theta {{\mathfrak {g}}}={{\mathfrak {g}}}\oplus {{\mathfrak {g}}}^*$ the bilinear form $\varphi (x+f,y+g)=f(y)+g(x)$ and the bracket

$$\begin{aligned} {[}x+f,y+g]_T=[x,y]+\theta (x,y)+f\circ \,\textrm{ad}(y)-g\circ \,\textrm{ad}(x), \end{aligned}$$

for all $x,y\in {{\mathfrak {g}}}$, $f,g\in {{\mathfrak {g}}}^*$, where $\,\textrm{ad}(x)y=[x,y]$, and let $J_T$ be defined by $J_T(x+f)=Jx-f\circ J.$

The triple $(T^*_\theta {{\mathfrak {g}}},J_T,\varphi )$ is a pseudo-Hermitian quadratic Lie algebra.

Proof

First, notice that for every $x,y,z\in {{\mathfrak {g}}}$ we have

$$\begin{aligned} \theta (y,z)x=\theta (Jy,Jz)x+\theta (Jz,Jx)y+\theta (Jx,Jy)z=\theta (x,y)z, \end{aligned}$$

which proves the cyclic condition of (Bordemann 1997, Lemma 3.1). This implies that $(T^*_\theta {{\mathfrak {g}}},\varphi )$ is a quadratic Lie algebra by the results in Bordemann (1997). So, we only have to prove that $J_T$ is a complex structure on $T^*_\theta {{\mathfrak {g}}}$ and that it is skewsymmetric with respect to $\varphi$.

This last fact is clear since

$$\begin{aligned} \varphi (J_T(x+f),y+g)= & {} \varphi (Jx-f\circ J,y+g)=g(Jx)-f(Jy)\\= & {} -\varphi (x+f,Jy-g\circ J)\\= & {} -\varphi (x+f,J_T(y+g)), \end{aligned}$$

for every $x,y\in {{\mathfrak {g}}}$, $f,g\in {{\mathfrak {g}}}^*$.

To see that $J_T$ is a complex structure, first notice that $J_T^2(x+f)=J^2x+f\circ J^2=-x-f$, for all $x\in {{\mathfrak {g}}},f\in {{\mathfrak {g}}}^*$. Let us see that $N_{J_T}$ is identically zero. For $f,g\in {{\mathfrak {g}}}^*$ we clearly have

$$\begin{aligned} N_{J_T}(f,g)=[f,g]_T-J_T[f\circ J,g]_T-J_T[f,g\circ J]_T-[f\circ J,g\circ J]_T=0. \end{aligned}$$

Now, choosing $x\in {{\mathfrak {g}}},g\in {{\mathfrak {g}}}^*$ we get

$$\begin{aligned} N_{J_T}(x,g)= & {} [x,g]_T+J_T[Jx,g]_T-J_T[x,g\circ J]_T+[Jx,g\circ J]_T \\= & {} -g\circ \,\textrm{ad}(x)+g\circ \,\textrm{ad}(Jx)\circ J-g\circ J\circ \,\textrm{ad}(x)\circ J\\{} & {} -g\circ J\circ \,\textrm{ad}(Jx)\in {{\mathfrak {g}}}^*, \end{aligned}$$

but then $N_{J_T}(x,g)=0$ because

$$\begin{aligned} N_{J_T}(x,g)(y)= & {} -g([x,y]-[Jx,Jy]+J[x,Jy]+J[Jx,y])\\= & {} -g(N_J(x,y))=0. \end{aligned}$$

Finally, for $x,y\in {{\mathfrak {g}}}$ one gets

$$\begin{aligned} N_{J_T}(x,y)= & {} [x,y]_T+J_T[Jx,y]_T+J_T[x, Jy]_T-[Jx,Jy]_T\\= & {} N_J(x,y)+\theta (x,y)-\theta (Jx,y)\circ J\\{} & {} -\theta (x,Jy)\circ J-\theta (Jx,Jy). \end{aligned}$$

But, using the hypothesis on $\theta$ and its cyclicity, we have for all $z\in {{\mathfrak {g}}}$ that

$$\begin{aligned} 0= & {} \theta (x,y)z-\theta (Jz,Jx) y-\theta (Jy, Jz)x-\theta (Jx,Jy)z\\= & {} \theta (x,y)z-\theta (Jx,y) Jz-\theta (x,Jy) Jz-\theta (Jx,Jy)z, \end{aligned}$$

so that $N_{J_T}(x,y)=N_J(x,y)=0,$ which completes the proof. $\square$

Example 1.1

Using Proposition 1.3 we can construct a lot of nontrivial examples of pseudo-Hermitian quadratic Lie algebras. Let us mention some interesting cases:

• (1) If one considers $\theta =0$ then the Lie algebra $T^*_0{{\mathfrak {g}}}$ is actually the cotangent algebra $T^*{{\mathfrak {g}}}$ (Andrada et al. 2008, Example 11). Notice that this method let us construct a pseudo-Hermitian quadratic Lie algebra starting from any pair $({{\mathfrak {g}}},J)$ of a Lie algebra with a complex structure.

• (2) An interesting particular case of the construction above is when one considers ${{\mathfrak {g}}}$ the underlying real Lie algebra of a semisimple complex Lie algebra with the natural bi-invariant complex structure defined by $Jx=ix$, for $x\in {{\mathfrak {g}}}$. In this case the cotangent Lie algebra $T^*{{\mathfrak {g}}}$ provides an example of pseudo-Hermitian quadratic Lie algebra which is perfect (that is to say, $[T^*{{\mathfrak {g}}},T^*{{\mathfrak {g}}}]_T=T^*{{\mathfrak {g}}}$) and, hence, centerless. Obviously, the same construction can be done for a rank 2 compact semisimple Lie algebra endowed with a complex structure (see Samelson (1953)) to get that its cotangent Lie algebra is a perfect pseudo-Hermitian quadratic Lie algebra.

Example 1.2

One may find examples in which $\theta \ne 0$. For instance, let ${{\mathfrak {k}}}$ be the direct sum of the 3-dimensional Heisenberg algebra and $\mathbb {R}$, that is to say ${{\mathfrak {k}}}=\mathbb {R}\text{-span }\{x_1,x_2,x_3,x_4\}$ with the only nontrivial bracket $[x_1,x_2]=x_3$. Such a Lie algebra is the well-known algebra underlying the Kodaira-Thurston manifold (see Kodaira (1964) and Thurston (1976)) and it is well known that the linear map J defined by $Jx_1=x_2$, $Jx_2=-x_1$, $Jx_3=x_4$, $Jx_4=-x_3$ is an abelian complex structure on ${{\mathfrak {g}}}$.

It is quite straightforward to prove the following:

• (1) Every cyclic 2-cocycle $\theta \in Z^2({{\mathfrak {k}}},{{\mathfrak {k}}}^*)$ is a linear combination of the the cyclic cocycles $\theta _i$, $1\le i\le 4$ defined, up to skewsymmetry, by the following nonzero values:

  $$\begin{aligned}{} & {} \theta _1(x_1,x_2)=x_3^*,\quad \theta _1(x_1,x_3)=-x_2^*,\quad \theta _1(x_2,x_3)=x_1^*,\\{} & {} \theta _2(x_1,x_2)=x_4^*,\quad \theta _2(x_1,x_4)=-x_2^*,\quad \theta _2(x_2,x_4)=x_1^*,\\{} & {} \theta _3(x_1,x_3)=x_4^*,\quad \theta _3(x_1,x_4)=-x_3^*,\quad \theta _3(x_3,x_4)=x_1^*,\\{} & {} \theta _4(x_2,x_3)=x_4^*,\quad \theta _4(x_2,x_4)=-x_3^*,\quad \theta _4(x_3,x_4)=x_2^*. \end{aligned}$$

  Further, any such a 2-cocycle $\theta$ verifies the compatibility condition of eq. (3).

• (2) The pseudo-Hermitian $T^*$-extension of $({{\mathfrak {k}}}, J)$ by means of the cocycle $\theta =\sum _{i=1}^4\alpha _i\theta _i$ is the Lie algebra spanned by a basis $\{x_1,x_2,x_3,x_4,x_1^*,x_2^*,x_3^*,x_4^*\}$ with nontrivial brackets

  $$\begin{aligned}{} & {} [x_1,x_2]=x_3+\alpha _1x_3^*+\alpha _2x^*_4,\quad [x_1,x_3]=-\alpha _1x^*_2+\alpha _3x_4^*,\\{} & {} [x_1,x_4]=-\alpha _2x_2^*-\alpha _3x^*_3,\quad [x_2,x_3]=\alpha _1x^*_1+\alpha _4x_4^*,\\{} & {} [x_2,x_4]=\alpha _2x_1^*-\alpha _4x^*_3,\quad [x_3,x_4]=\alpha _3x^*_1+\alpha _4x_2^*,\\{} & {} [x_3^*,x_1]=x_2^*,\quad [x_3^*,x_2]=-x^*_1 \end{aligned}$$

  and the metric and complex structure defined by

  $$\begin{aligned} \varphi (x_i,x_j^*)=\delta _{ij},\quad \varphi (x_i,x_j)=\varphi (x_i^*,x_j^*)=0,\\ Jx_1 = x_2,\quad Jx_3=x_4,\quad Jx_1^*=x_2^*,\quad Jx_3^*=x_4^*. \end{aligned}$$

Notice that linearly independent 2-cocycles may give rise to isometric (further, holomorphically isometric) pseudo-Hermitian Lie algebras. A complete classification of non-equivalent cases will be detailed in Sect. 3, where the algebras $T^*_\theta {{\mathfrak {k}}}$ will play a relevant role.

The knowledge of a pseudo-Hermitian quadratic Lie algebra let us also construct many other pseudo-Hermitian quadratic Lie algebras by the following method:

Proposition 1.4

Let ${{\mathcal {A}}}$ be a real associative commutative algebra admitting a non-degenerate symmetric bilinear form $B:{\mathcal {A}}\times {{\mathcal {A}}}\rightarrow \mathbb {R}$ such that $B(ab,c)=B(b,ac)$ for all $a,b,c\in {{\mathcal {A}}}$ and let $({{\mathfrak {g}}},J,\varphi )$ be a pseudo-Hermitian quadratic Lie algebra.

The tensor product ${{\mathfrak {G}}}={{\mathfrak {g}}}\otimes _\mathbb {R}{{\mathcal {A}}}$ with the bracket, complex structure $J'$ and bilinear form $\phi$ defined by

$$\begin{aligned}{} & {} {[}x\otimes a,y\otimes b]_{{\mathfrak {G}}}=[x,y]\otimes ab,\quad J'(x\otimes a)=Jx\otimes a,\\{} & {} \phi (x\otimes a,y\otimes b)=\varphi (x,y)B(a,b), \end{aligned}$$

for every $x,y\in {{\mathfrak {g}}}$, $a,b\in {{\mathcal {A}}}$, is a pseudo-Hermitian quadratic Lie algebra.

Proof

It was already seen in (Hofmann and Keith 1986, Prop. A) that $({{\mathfrak {G}}},\phi )$ is a quadratic Lie algebra. Let us prove that $J'$ is a complex structure on ${{\mathfrak {G}}}$ and that it is $\phi$-skewsymmetric.

We clearly have $J'(Jx\otimes a)=J^2x\otimes a=-x\otimes a$, for all $x\in {{\mathfrak {g}}},a\in {{\mathcal {A}}}$, which shows that $(J')^2=-I$. Further, for every $x,y\in {{\mathfrak {g}}}$, $a,b\in {{\mathcal {A}}}$ we easily prove that $N_{J'}(x\otimes a,y\otimes b)=N_j(x,y)\otimes ab=0$, so that $J'$ is a complex structure and we also have

$$\begin{aligned} \phi (J'(x\otimes a),y\otimes b){} & {} =\varphi (Jx,y)B(a,b)=-\varphi (x,Jy)B(a,b)\\{} & {} =-\phi (x\otimes a, J'(y\otimes b)). \end{aligned}$$

Therefore, $({{\mathfrak {G}}},J',\phi )$ is a pseudo-Hermitian quadratic Lie algebra as claimed. $\square$

Corollary 1.5

If $({{\mathfrak {g}}},J,\varphi )$ is a pseudo-Hermitian quadratic Lie algebra, then the underlying real Lie algebra of its complexification ${{\mathfrak {G}}}=({{\mathfrak {g}}}^{\mathbb {C}})_\mathbb {R}$ is also a pseudo-Hermitian quadratic Lie algebra with the complex structure $J'$ and quadratic form $\phi$ defined by

$$\begin{aligned} J'(x+iy)=Jx+iJy,\quad \phi (x+iy,z+it)=\varphi (x,z)-\varphi (y,t), \end{aligned}$$

for all $x,y,z,t\in {{\mathfrak {g}}}$.

Proof

Notice that we can naturally identify ${{\mathfrak {G}}}={{\mathfrak {g}}}\otimes _\mathbb {R}{\mathbb {C}}$ by considering $x+iy=x\otimes 1+y\otimes i$ for all $x,y\in {{\mathfrak {g}}}$. Put $B(a,b)=\text{ Re }(ab)$ for $a,b\in {\mathbb {C}}$. It is easy to see that B is a non-degenerate, symmetric, bilinear form on ${\mathbb {C}}$ and that $B(ab,c)=B(b,ac)$ holds for all $a,b,c\in {\mathbb {C}}$. It thus suffices to apply Proposition 1.4 and see that, under the identification ${{\mathfrak {G}}}={{\mathfrak {g}}}\otimes _\mathbb {R}{\mathbb {C}}$, the map $J'$ and the bilinear form $\phi$ are as stated. $\square$

Example 1.3

When $({{\mathfrak {g}}},J,\varphi )$ is a pseudo-Hermitian quadratic Lie algebra, one can also use Proposition 1.4 to construct a family of nilpotent pseudo-Hermitian quadratic Lie algebras by appropriate tensorizations of ${{\mathfrak {g}}}$.

Let ${{\mathcal {A}}}=\mathbb {R}\text{-span }\{a,a^2,\dots ,a^k\}$ the nilpotent commutative associative algebra defined by the products $a^ia^j=a^{i+j}$ if $i+j\le k$ and $a^ia^j=0$ whenever $i+j> k$. It suffices to bear in mind that the form $B:{{\mathcal {A}}}\times {{\mathcal {A}}}\rightarrow \mathbb {R}$ defined by bilinearity from $B(a^i,a^j)=\delta _{i+j,k+1}$, where $\delta _{r,s}$ stands for the Kronecker symbol, is a non-degenerate symmetric bilinear form on ${{\mathcal {A}}}$ and it verifies

$$\begin{aligned} B(a^ia^j,a^\ell )=\delta _{i+j+\ell ,k+1}=B(a^j,a^ia^\ell ) \end{aligned}$$

for all $1\le i,j,\ell \le k$.

According to Proposition 1.4 the (nilpotent) Lie algebra ${{\mathfrak {G}}}={{\mathfrak {g}}}\otimes _\mathbb {R}{{\mathcal {A}}}$ admits a pseudo-Hermitian quadratic structure.

3 Double extension of pseudo-Hermitian quadratic Lie algebras

In Medina and Revoy (1985), the authors propose a method to construct $n+2$-dimensional quadratic Lie algebras from a given n-dimensional one by performing an appropriate central extension and a semidirect product defined by a certain skewsymmetric derivation. Such a method is known as double extension by a line. Explicitly, if $({{\mathfrak {g}}}_0,[\cdot ,\cdot ]_0,\varphi _0)$ is a quadratic Lie algebra and D is a $\varphi _0$-skewsymmetric derivation of ${{\mathfrak {g}}}_0$, then one considers the vector space ${{\mathfrak {g}}}=\mathbb {R}z\oplus {{\mathfrak {g}}}_0\oplus \mathbb {R}v$ with the bracket $[\cdot ,\cdot ]$ and the bilinear map $\varphi :{{\mathfrak {g}}}\times {{\mathfrak {g}}}\rightarrow \mathbb {R}$ defined by

$$\begin{aligned}{} & {} [z,x]=[z,v]=0,\quad [x,y]=[x,y]_0+\varphi _0(Dx,y)z,\quad [v,x]=Dx\\{} & {} \varphi (z,v)=1, \quad \varphi (x,y)=\varphi _0(x,y),\quad \varphi (z,\mathbb {R}z\oplus {{\mathfrak {g}}}_0)=\varphi (v,{{\mathfrak {g}}}_0\oplus \mathbb {R}v)=\{0\} \end{aligned}$$

for all $x,y\in {{\mathfrak {g}}}_0$. It results that $({{\mathfrak {g}}},[\cdot ,\cdot ],\varphi )$ is a quadratic Lie algebra. Further, one has the following result:

Proposition 2.1

(Medina-Revoy) Every non-abelian n-dimensional solvable quadratic Lie algebra $\mathfrak {g}$ is the double extension of a solvable quadratic Lie algebra $\mathfrak {g}_0$ of dimension $n - 2$ by a line. As a consequence, every solvable quadratic Lie algebra is obtained by a sequence of double extensions starting from an abelian Lie algebra.

In this section we will use two successive double extensions of a pseudo-Hermitian quadratic Lie algebra in order to obtain a new one. Explicitly, we have the following result:

Proposition 2.2

Let $\left( \mathfrak {g}_0, \left[ \cdot , \cdot \right] _0, J_0, \varphi _0 \right)$ be a pseudo-Hermitian quadratic Lie algebra. Let D, F be two $\varphi _0$-skewsymmetric derivations of ${{\mathfrak {g}}}_0$ such that $\left[ F+J_0D,J_0\right] =0$ and $\left[ F, D \right] = \,\textrm{ad}_{\mathfrak {g}_0} (s_0)$ for a certain $s_0\in {{\mathfrak {g}}}_0$.

Define the vector space ${{\mathfrak {g}}}$ as the direct sum of ${{\mathfrak {g}}}_0$ and a 4-dimensional space with generators $\left\{ z, z', v, v' \right\}$, namely

$$\begin{aligned} \mathfrak {g} = \mathbb {R}z \oplus \mathbb {R}z' \oplus \mathfrak {g}_0 \oplus \mathbb {R}v' \oplus \mathbb {R}v. \end{aligned}$$

Let us define on ${{\mathfrak {g}}}$ a skewsymmetric product $[\cdot ,\cdot ]$ and a symmetric bilinear form $\varphi$ by the following nonzero pairings:

$$\begin{aligned} {[}v,v']= & {} s_0,\quad [v,x]=F x - \varphi _0 (s_0, x)z',\quad {[}v',x]=Dx + \varphi _0 (s_0, x)z,\\ {[}x,y]= & {} \left[ x, y \right] _0 + \varphi _0 (Dx, y)z' + \varphi _0 (F x, y)z\\ \varphi (z, v)= & {} \varphi (z', v' ) = 1,\quad \varphi (x, y) = \varphi _0 (x, y), \end{aligned}$$

for any $x,y\in {{\mathfrak {g}}}_0$. Let us also consider $J\in {\mathfrak {g}}{\mathfrak {l}}({{\mathfrak {g}}})$ defined by

$$\begin{aligned} Jz = z',\quad Jz' = -z,\quad Jv = v',\quad Jv' = -v,\quad Jx = J_0 x, \end{aligned}$$

for all $x\in {{\mathfrak {g}}}_0$.

The triple $({{\mathfrak {g}}}, [\cdot ,\cdot ],\varphi )$ is a quadratic Lie algebra, J is a complex structure on ${{\mathfrak {g}}}$ and $\varphi$ is actually pseudo-Hermitian.

Proof

Denote by $({{\mathfrak {g}}}_1,\varphi _1)$ the double extension of $({{\mathfrak {g}}}_0,\varphi _0)$ by means of D. One then has ${{\mathfrak {g}}}_1=\mathbb {R}z'\oplus {{\mathfrak {g}}}_0\oplus \mathbb {R}v'$ with the nonzero brackets and pairings of $\varphi _1$ given by

$$\begin{aligned} {[}x,y]_1= & {} [x,y]_0+\varphi _0(Dx,y)z',\quad [v',x]_1=Dx,\\ \varphi _1(x,y)= & {} \varphi _0(x,y),\quad \varphi _1(z',v')=1, \end{aligned}$$

where $x,y\in {{\mathfrak {g}}}_0$. Let $D_1:{{\mathfrak {g}}}_1\rightarrow {{\mathfrak {g}}}_1$ be the linear map defined by

$$\begin{aligned} D_1(z')=0,\quad D_1(v')=s_0,,\quad D_1(x)=Fx-\varphi _0(s_0,x)z', \end{aligned}$$

for all $x\in {{\mathfrak {g}}}_0$. As F is $\varphi _0$-skewsymmetric, for every $\alpha ,\alpha ',\beta ,\beta '\in \mathbb {R}$ and $x,y\in {{\mathfrak {g}}}_0$ one has

$$\begin{aligned}{} & {} \varphi _1(D_1( \alpha z'+x+\beta v'),\alpha ' z'+y+ \beta ' v')\\{} & {} \qquad +\varphi _1(\alpha z'+x+\beta v',D_1( \alpha ' z'+y+\beta ' v'))\\{} & {} \quad =\varphi _1(Fx-\varphi _0(s_0,x)z'+\beta s_0,y+\beta ' v')\\{} & {} \qquad +\varphi _1(x+\beta v',Fy-\varphi _0(s_0,y)z'+\beta ' s_0,)\\{} & {} \quad =\varphi _0(Fx,y)+\beta \varphi _0(s_0,y)-\beta '\varphi _0(s_0,x)+ \varphi _0(x,Fy)\\{} & {} \qquad +\beta '\varphi _0(x,s_0)-\beta \varphi _0(s_0,y)=0, \end{aligned}$$

which shows that $D_1$ is skewsymmetric with respect to $\varphi _1$. Let us see that $D_1$ is a derivation of ${{\mathfrak {g}}}_1$. One clearly has $D_1[z',X]=[D_1z',X]+[z',D_1X]$ for all $X\in {{\mathfrak {g}}}_1$ and for every $x\in {{\mathfrak {g}}}_0$ we have that

$$\begin{aligned} D_1[v',x]_1-[D_1v',x]_1-[v',D_1x]_1= & {} FDx- \varphi _0(s_0,Dx)z'-[s_0,x]_0\\{} & {} -\varphi _0(Ds_0,x)z'-DFx=0 \end{aligned}$$

because D is $\varphi _0$-skewsymmetric and $[F,D]=\,\textrm{ad}_{{{\mathfrak {g}}}_0}(s_0)$. Besides, using again this last equality and the facts that F is a skewsymmetric derivation of $({{\mathfrak {g}}}_0,\varphi _0)$ and $\varphi _0$ is quadratic, we have for all $x,y\in {{\mathfrak {g}}}_0$ that

$$\begin{aligned} D_1[x,y]_1-[D_1x,y]_1-[x,D_1y]_1= & {} F[x,y]_0- \varphi _0(s_0,[x,y])z'\\{} & {} -[Fx,y]_0-\varphi _0(DFx,y)z'\\{} & {} -[x,Fy]_0-\varphi _0(Dx,Fy)z'\\= & {} -\varphi _0([s_0,x],y)z'+\varphi _0([F,D]x,y)z'=0. \end{aligned}$$

Thus, $D_1$ is a skewsymmetric derivation of $({{\mathfrak {g}}}_1,\varphi _1)$. Now, it is a simple calculation to see that the double extension of $({{\mathfrak {g}}}_1,\varphi _1)$ by means of $D_1$ is nothing but the pair $({{\mathfrak {g}}},\varphi )$ of the statement, proving that $({{\mathfrak {g}}},\varphi )$ is a quadratic Lie algebra.

Therefore, it only remains to prove that J is a complex structure and that it is $\varphi$-skewsymmetric. It is clear from the definition of J that $J^2=-I$ and we have for $x_k\in {{\mathfrak {g}}}_0$ and $\alpha _k,\alpha '_k,\beta _k\beta '_k\in \mathbb {R}$ $(k=1,2)$ that the following holds

$$\begin{aligned}{} & {} \varphi (J( \alpha _1z+\alpha _1' z'+x_1+\beta '_1 v'+\beta _1 v),\alpha _2z+\alpha _2' z'+x_2+\beta '_2 v'+\beta _2 v)\\{} & {} \quad = \varphi ( \alpha _1z'-\alpha _1' z+J_0x_1-\beta '_1 v+\beta _1 v',\alpha _2z+\alpha _2' z'+x_2+\beta '_2 v'+\beta _2 v)\\{} & {} \quad = \alpha _1\beta '_2-\alpha '_1\beta _2-\beta '_1\alpha _2+\beta _1\alpha '_2+\varphi _0(J_0x_1,x_2). \end{aligned}$$

Similarly we get

$$\begin{aligned}{} & {} \varphi (\alpha _1z+\alpha _1' z'+x_1+\beta '_1 v'+ \beta _1 v,J( \alpha _2z+\alpha _2' z'+x_2+\beta '_2 v'+\beta _2 v))\\{} & {} \quad =\alpha _2\beta '_1-\alpha '_2\beta _1-\beta '_2\alpha _1+\beta _2\alpha '_1+\varphi _0(J_0x_2,x_1), \end{aligned}$$

and thus J results to be skewsymmetric with respect to $\varphi$ because $J_0$ is $\varphi _0$-skewsymmetric. In order to see that the $N_J=0$, first notice that one obviously has $N_J(z,\cdot )=N_J(z',\cdot )=0$ because $z,z'=Jz$ are in the center of ${{\mathfrak {g}}}$ and that also $N_J(v,v')=0$ because $v'=Jv$. If $x\in {{\mathfrak {g}}}_0$ we get

$$\begin{aligned} N_J(v,x)= & {} Fx-\varphi _0(s_0,x)z'+J_0Dx+ \varphi _0(s_0,x)z'\\{} & {} +J_0FJ_0x+\varphi _0(s_0,J_0x)z-DJ_0x-\varphi _0(s_0,J_0x)z\\= & {} (F+J_0D+J_0FJ_0-DJ_0)x=J_0[F+J_0D,J_0]x=0. \end{aligned}$$

because $F+J_0D$ commutes with $J_0$ by hypothesis. Notice that, using (2), this also implies $N_J(v',x)=N_J(Jv,x)=-JN_J(v,x)=0$. Further, for $x,y\in {{\mathfrak {g}}}_0$ we have

$$\begin{aligned}{}[x,y]= & {} [x,y]_0+\varphi _0(Dx,y)z'+\varphi _0(Fx,y)z\\ J[Jx,y]= & {} J_0[J_0x,y]_0-\varphi _0(DJ_0x,y)z+\varphi _0(FJ_0x,y)z'\\ J[x,Jy]= & {} J_0[x,J_0y]_0+\varphi _0(J_0Dx,y)z-\varphi _0(J_0Fx,y)z'\\ {[}Jx,Jy]= & {} [J_0x,J_0y]_0-\varphi _0(J_0DJ_0x,y)z'-\varphi _0(J_0FJ_0x,y)z, \end{aligned}$$

from where we immediately get

$$\begin{aligned} N_J(x,y)=N_{J_0}(x,y)+\varphi _0([F+J_0D,J_0]x,y)z'+\varphi _0 (J_0[F+J_0D,J_0]x,y)z=0. \end{aligned}$$

Thus, the Nijenhuis tensor vanishes identically and we get that $({{\mathfrak {g}}},\varphi ,J)$ is a pseudo-Hermitian quadratic Lie algebra. $\square$

Definition 2.1

The pseudo-Hermitian quadratic Lie algebra $({{\mathfrak {g}}},J,\varphi )$ constructed as in Proposition 2.2 will be called the pseudo-Hermitian quadratic double extension of $({{\mathfrak {g}}}_0,J_0,\varphi _0)$ by a plane by means of (D, F) and $s_0$. For short, we will say that ${{\mathfrak {g}}}$ is a pHQ-double extension of ${{\mathfrak {g}}}_0$.

Remark 2.1

Let us remark some interesting facts.

• (1) Notice that the concept of pHQ-double extension defined above is different from the construction that Andrada, Barberis and Ovando give in (Andrada et al. 2008, Th. 21). For a nilpotent Lie algebra ${{\mathfrak {g}}}$, their construction is the particular case of ours for $D=0$.

• (2) If ${{\mathfrak {g}}}$ is a pHQ-double extension of a pseudo-Hermitian quadratic Lie algebra ${{\mathfrak {g}}}_0$ and the signature of the metric in ${{\mathfrak {g}}}_0$ is $\text{ sig }(\varphi _0)=(r,s)$, then for the metric in ${{\mathfrak {g}}}$ we have $\text{ sig }(\varphi )=(r+2,s+2)$.

• (3) If F, D verify $\left[ F+J_0D,J_0\right] =0$ and $\left[ F, D \right] = \,\textrm{ad}_{\mathfrak {g}_0} (s_0)$ then for the maps $F_1=D$, $D_1=-F$ one also has $[F_1,D_1]=\,\textrm{ad}_{\mathfrak {g}_0} (s_0)$ and

  $$\begin{aligned}{}[F_1+J_0D_1,J_0]= & {} [D-J_0F,J_0]=DJ_0-J_0FJ_0-J_0D-F\\= & {} J_0[F+J_0D,J_0]=0. \end{aligned}$$

  If we denote $\mathfrak {g}_1 = \mathbb {R}z_1 \oplus \mathbb {R}z_1' \oplus \mathfrak {g}_0 \oplus \mathbb {R}v'_1 \oplus \mathbb {R}v_1$ the pHQ-double extension by means of $(D_1,F_1)$ and ${{\mathfrak {g}}}$ is as in Proposition 2.2, one easily sees that the map $\Psi :{{\mathfrak {g}}}_1\rightarrow {{\mathfrak {g}}}$ defined by $\Psi (z_1)=z',\psi (z_1')=-z,\Psi (v_1)=v',\psi (v_1')=-v,\Psi (x)=x$ for all $x\in {{\mathfrak {g}}}_0$ is a holomorphic isomorphism that preserves the corresponding metrics.

• (4) Notice that the condition $[ F, D] = \,\textrm{ad}_{\mathfrak {g}_0} (s_0)$ does not completely define $s_0$ since the same identity holds for every $s_0'=s_0+w$ if $w\in {{\mathfrak {z}}}({{\mathfrak {g}}})$. The election of $s_0'$ instead of $s_0$ in the definition of the pHQ-double extension may give rise to nonisomorphic Lie algebras. For instance, when ${{\mathfrak {g}}}_0=\mathbb {R}^2$ with a Hermitian metric and we take $D=F=0$, the pHQ-double extension is abelian whenever $s_0=0$ whereas for $s_0\ne 0$ one gets a non-abelian Lie algebra.

Proposition 2.3

Let $({{\mathfrak {g}}},J,\varphi )$ be a pseudo-Hermitian quadratic Lie algebra of $\dim ({{\mathfrak {g}}})=n>4$ and denote by ${{\mathfrak {z}}}({{\mathfrak {g}}})$ its center.

If there exists an isotropic element $z\in {{\mathfrak {z}}}({{\mathfrak {g}}})$ such that $Jz\in {{\mathfrak {z}}}({{\mathfrak {g}}})$, then there exists a $(n-4)$-dimensional pseudo-Hermitian quadratic Lie algebra $({{\mathfrak {g}}}_0,J_0,\varphi _0)$ such that ${{\mathfrak {g}}}$ is a pHQ-double extension of ${{\mathfrak {g}}}_0$.

Proof

Since z and Jz are linearly independent, we can find an element $v\in {{\mathfrak {g}}}$ such that $\varphi (z,v)=1$ and $\varphi (Jz,v)=0$. Now, as $z\in {{\mathfrak {z}}}({{\mathfrak {g}}})$ is isotropic, one can use Medina and Revoy’s results in Medina and Revoy (1985) to see that there exists a quadratic Lie algebra $({{\mathfrak {g}}}_1,\varphi _1)$ such that ${{\mathfrak {g}}}=\mathbb {R}z\oplus {{\mathfrak {g}}}_1\oplus \mathbb {R}v$ is the double extension of ${{\mathfrak {g}}}_1$ by means of a certain skewsymmetric derivation $D_1$ of $({{\mathfrak {g}}}_1,\varphi _1)$. Notice that, in that case, ${{\mathfrak {g}}}_1$ is the $\varphi$-orthogonal to $\mathbb {R}z\oplus \mathbb {R}v$ and, hence, as $\varphi (Jz,v)=0$ and $\varphi (Jz,z)=0$, because J is skewsymmetric with respect to $\varphi$, we have that $Jz\in {{\mathfrak {g}}}_1$. Actually, Jz is an isotropic central element of ${{\mathfrak {g}}}_1$ because $\varphi (Jz,Jz)=\varphi (z,z)=0$ and $Jz\in {{\mathfrak {z}}}({{\mathfrak {g}}})$. Since we also have

$$\begin{aligned} \varphi (Jv,z)= & {} -\varphi (v,Jz)=0,\\ \varphi (Jv,v)= & {} -\varphi (v,Jv)=0,\\ \varphi (Jz,Jv)= & {} \varphi (z,v)=1, \end{aligned}$$

we get that $Jv\in {{\mathfrak {g}}}_1$ and, using again the results of Medina and Revoy (1985), we get that there exists a quadratic Lie algebra $({{\mathfrak {g}}}_0,\varphi _0)$ such that ${{\mathfrak {g}}}_1=\mathbb {R}Jz\oplus {{\mathfrak {g}}}_0\oplus \mathbb {R}Jv$ is the double extension of ${{\mathfrak {g}}}_0$ by means of a skewsymmetric derivation D of $({{\mathfrak {g}}}_0,\varphi _0)$. If we now denote $z'=Jz,v'=Jv$, we get that ${{\mathfrak {g}}}=\mathbb {R}z \oplus \mathbb {R}z' \oplus \mathfrak {g}_0 \oplus \mathbb {R}v' \oplus \mathbb {R}v$, where ${{\mathfrak {g}}}_0$ is the $\varphi$-orthogonal to $\mathbb {R}\text{-span }\{z,z',v,v'\}$. Besides, as $\mathbb {R}\text{-span }\{z,z',v,v'\}$ is clearly J-invariant and J is $\varphi$-skewsymmetric, we also have $J{{\mathfrak {g}}}_0={{\mathfrak {g}}}_0$.

From the construction of double extensions we have that the only a priori nonzero pairings for $\varphi$ are

$$\begin{aligned} \varphi (z,v)= & {} 1,\quad \varphi (z',v')=\varphi _1(z',v')=1,\\ \varphi (x,y)= & {} \varphi _1(x,y)=\varphi _0(x,y), \quad x,y\in {{\mathfrak {g}}}_0, \end{aligned}$$

and the bracket in ${{\mathfrak {g}}}$ is defined by $z,z'\in {{\mathfrak {z}}}({{\mathfrak {g}}})$ and

$$\begin{aligned} {[}x,y]_1= & {} [x,y]_0+\varphi _0(Dx,y)z',\quad [v',x]_1=Dx\\ {[}x,y]= & {} [x,y]_1+\varphi _1(D_1x,y)z,\quad [v',x]=[v',x]_1 +\varphi _1(D_1v',x)z,\\ {[}v,x]= & {} D_1x,\quad [v,v']=D_1v', \end{aligned}$$

for all $x,y\in {{\mathfrak {g}}}_0$. Notice that $D_1z'=0$ because, being $D_1$ skewsymmetric with respect to $\varphi _1$, we have $\varphi _1(D_1z',z')=0$ and for $x\in {{\mathfrak {g}}}_0$ and $\lambda \in \mathbb {R}$ we get

$$\begin{aligned} \varphi _1(x+\lambda v',D_1z')= & {} -\varphi _1(D_1x+\lambda D_1v',z')\\= & {} -\varphi ([v,x+\lambda v'],z')=\varphi (x+\lambda v',[v,z'])=0. \end{aligned}$$

This implies that $\varphi _1(D_1{{\mathfrak {g}}}_1,z')=\varphi _1({{\mathfrak {g}}}_1,D_1z')=\{0\}$, showing that $D_1{{\mathfrak {g}}}_1\subset \mathbb {R}z'\oplus {{\mathfrak {g}}}_0$. But $\varphi _1(D_1v',v')=0$ because $D_1$ is skewsymmetric and therefore we have:

$$\begin{aligned} D_1v'\in {{\mathfrak {g}}}_0,\quad D_1x=Fx+\alpha (x)z',\quad x\in {{\mathfrak {g}}}_0 \end{aligned}$$

for a certain $\alpha \in {{\mathfrak {g}}}_0^*$ and a certain linear endomorphism F of ${{\mathfrak {g}}}_0$. Define $s_0=D_1v'\in {{\mathfrak {g}}}_0$. The skewsymmetry of $D_1$ gives that F is $\varphi _0$-skewsymmetric and that

$$\begin{aligned} \alpha (x) =\varphi _1(D_1x,v')=-\varphi _1(x,D_1v')=-\varphi _0(x,s_0). \end{aligned}$$

The fact that $D_1$ is a derivation of ${{\mathfrak {g}}}_1$ gives

$$\begin{aligned} F[x,y]_0-\varphi _0([x,y]_0,s_0)z'= & {} [Fx,y]_0+\varphi _0(DFx,y)z'\\{} & {} + [x,Fy]_0+\varphi _0(Dx,Fy)z', \end{aligned}$$

from where we derive that F is a derivation of ${{\mathfrak {g}}}_0$ and that $[F,D]=\text{ ad}_{{{\mathfrak {g}}}_0}(s_0)$ because

$$\begin{aligned} \varphi _0([s_0,x],y)=\varphi (s_0,[x,y])=-\varphi _0(DFx,y)- \varphi _0(Dx,Fy)=\varphi ([F,D]x,y), \end{aligned}$$

for all $x,y\in {{\mathfrak {g}}}_0$. Summing up, we get that the bracket in ${{\mathfrak {g}}}$ is given by

$$\begin{aligned} {[}x,y]= & {} [x,y]_0+\varphi _0(Dx,y)z'+\varphi _0(Fx,y)z,\quad [v',x]=Dx+\varphi _1(s_0,x)z,\\ {[}v,x]= & {} Fx-\varphi _1(s_0,x)z',\quad [v,v']=s_0. \end{aligned}$$

Since $J{{\mathfrak {g}}}_0={{\mathfrak {g}}}_0$, it only remains to prove that the restriction $J_0=J_{|{{\mathfrak {g}}}_0}$ is a $\varphi _0$-skewsymmetric complex structure on ${{\mathfrak {g}}}_0$ and that $[F+J_0D,J_0]=0$. Using the brackets above and the fact that F is skewsymmetric for $\varphi _0$, we have for $x,y\in {{\mathfrak {g}}}_0$ that

$$\begin{aligned} {[}x,y]= & {} [x,y]_0+\varphi _0(Dx,y)z'+\varphi _0(Fx,y)z\\ J[Jx,y]= & {} J_0[J_0x,y]_0-\varphi _0(DJ_0x,y)z+\varphi _0(FJ_0x,y)z'\\ J[x,Jy]= & {} J_0[x,J_0y]_0+\varphi _0(J_0Dx,y)z-\varphi _0(J_0Fx,y)z'\\ {[}Jx,Jy]= & {} [J_0x,J_0y]_0-\varphi _0(J_0DJ_0x,y)z'-\varphi _0(J_0FJ_0x,y)z \end{aligned}$$

So, $N_J(x,y)=0$ is equivalent to $N_{J_0}(x,y)=0$ and

$$\begin{aligned} 0= & {} D+FJ_0-J_0F+J_0DJ_0=[F+J_0D,J_0],\\ 0= & {} F-DJ_0+J_0D+J_0FJ_0=J_0[F+J_0D,J_0]. \end{aligned}$$

Since one clearly has $J_0^2x=J^2x=-x$ and $\varphi _0(J_0x,y)=\varphi (Jx,y)=-\varphi (x,Jy)=-\varphi _0(x,J_0y),$ for every elements $x,y\in {{\mathfrak {g}}}_0$, we can conclude that $({{\mathfrak {g}}},J,\varphi )$ is the pHQ-double extension of $({{\mathfrak {g}}}_0,J_0,\varphi _0)$ by means of (D, F), $\square$

We can now state our main result. Its proof is heavily based on the result of Salamon (Salamon 2001, Corol. 1.4) (see also (Millionshchikov 2016, Prop. 3.2)) assuring that for a nontrivial nilpotent Lie algebra with a complex structure $({{\mathfrak {g}}}, J)$ one always has $[{{\mathfrak {g}}},{{\mathfrak {g}}}]+J[{{\mathfrak {g}}},{{\mathfrak {g}}}]\ne {{\mathfrak {g}}}.$

Theorem 2.4

Every pseudo-Hermitian quadratic nilpotent Lie algebra with indefinite metric is obtained from a series of pHQ-double extensions starting either from the zero Lie algebra or from an abelian Lie algebra with a definite metric.

Proof

Let $({{\mathfrak {g}}},J,\varphi )$ be a pseudo-Hermitian quadratic nilpotent Lie algebra with $\varphi$ indefinite. We can suppose that $\dim ({{\mathfrak {g}}})>4$ because in the 4-dimensional case one only has ${{\mathfrak {g}}}=\mathbb {R}^4$ endowed with an indefinite pseudo-Hermitian metric, which can be regarded as the pHQ-double extension of the zero Lie algebra by means of $D=F=0, s_0=0$.

As proved by Salamon in Salamon (2001), we have that $[{{\mathfrak {g}}},{{\mathfrak {g}}}]+J[{{\mathfrak {g}}},{{\mathfrak {g}}}]\ne {{\mathfrak {g}}}$ and therefore their orthogonal spaces with respect to $\varphi$ verify $[{{\mathfrak {g}}},{{\mathfrak {g}}}]^\bot \cap (J[{{\mathfrak {g}}},{{\mathfrak {g}}}])^\bot \ne \{0\}$. Since $[{{\mathfrak {g}}},{{\mathfrak {g}}}]^\bot ={{\mathfrak {z}}}({{\mathfrak {g}}})$ and $(J[{{\mathfrak {g}}},{{\mathfrak {g}}}])^\bot =J([{{\mathfrak {g}}},{{\mathfrak {g}}}]^\bot )=J{{\mathfrak {z}}}({{\mathfrak {g}}})$, we get that ${{\mathfrak {z}}}({{\mathfrak {g}}})\cap J{{\mathfrak {z}}}({{\mathfrak {g}}})\ne \{0\}.$ Let us prove that the restriction of $\varphi$ to $Z\times Z$, where $Z={{\mathfrak {z}}}({{\mathfrak {g}}})\cap J{{\mathfrak {z}}}({{\mathfrak {g}}})$ cannot be definite. Suppose, on the contrary, that $\varphi$ is definite on Z so that ${{\mathfrak {g}}}=Z\oplus Z^\bot$ is a $\varphi$-orthogonal direct sum of non-degenerate J-invariant ideals. Then one has that $Z^\bot =[{{\mathfrak {g}}},{{\mathfrak {g}}}]+J[{{\mathfrak {g}}},{{\mathfrak {g}}}]=[Z^\bot ,Z^\bot ]+J[Z^\bot ,Z^\bot ]$, which is impossible by Salamon’s result unless $Z^\bot =\{0\}$. But then ${{\mathfrak {g}}}=Z$ and we had that $\varphi$ was indefinite on ${{\mathfrak {g}}}$, a contradiction with our assumption of its definiteness on Z. Therefore, the restriction of $\varphi$ to $Z\times Z$ is not definite, as claimed. This implies that we can always find a nonzero isotropic vector $z\in Z={{\mathfrak {z}}}({{\mathfrak {g}}})\cap J{{\mathfrak {z}}}({{\mathfrak {g}}})$ and hence Proposition 2.3 guarantees that ${{\mathfrak {g}}}$ is a pHQ-double extension of another pseudo-Hermitian quadratic nilpotent Lie algebra $({{\mathfrak {g}}}_0,J_0,\varphi _0)$. Now, either $\varphi _0$ is definite and then ${{\mathfrak {g}}}_0$ must be abelian or $\varphi _0$ is again indefinite and we can apply the above reasoning to ${{\mathfrak {g}}}_0$. A successive application of that argument completes the proof. $\square$

Remark 2.2

Notice that, if ${{\mathfrak {g}}}$ is a pHQ-double extension of another pseudo-Hermitian quadratic Lie algebra ${{\mathfrak {g}}}_0$, and ${{\mathfrak {g}}}$ is nilpotent then ${{\mathfrak {g}}}_0$ must be nilpotent but also the derivations D, F defining the double extension should be nilpotent. Actually, if ${{\mathfrak {g}}}=\mathbb {R}z \oplus \mathbb {R}z' \oplus \mathfrak {g}_0 \oplus \mathbb {R}v' \oplus \mathbb {R}v$ is k-step nilpotent then $\,\textrm{ad}(v)^k=\,\textrm{ad}(v')^k=0$ and $\,\textrm{ad}(x)^k=0$ for all $x\in {{\mathfrak {g}}}$ but then, for each $y\in {{\mathfrak {g}}}_0$, we have

$$\begin{aligned} 0= & {} \,\textrm{ad}(v)^k(y)=F^ky-\varphi _0(s_0,F^{k-1}y)z'\\ 0= & {} \,\textrm{ad}(v')^k(y)=D^ky+\varphi _0(s_0,D^{k-1}y)z'\\ 0= & {} \,\textrm{ad}(x)^k(y)=\,\textrm{ad}_{{{\mathfrak {g}}}_0}(x)^ky+\varphi _0(Dx, \,\textrm{ad}_{{{\mathfrak {g}}}_0}(x)^{k-1}y)z'++\varphi _0(Fx,\,\textrm{ad}_{{{\mathfrak {g}}}_0}(x)^{k-1}y)z. \end{aligned}$$

The two first equalities show that F and D are nilpotent and the third one that ${{\mathfrak {g}}}_0$ is a nilpotent Lie algebra.

Using Theorem 2.4 and Remark 2.2 we can give a complete description of Lorentz-Hermitian quadratic Lie algebras.

Corollary 2.5

A non-abelian nilpotent pseudo-Hermitian quadratic Lie algebra $({{\mathfrak {g}}},J,\varphi )$ is Lorentz-Hermitian if and only if it is the orthogonal sum of an abelian Hermitian Lie algebra and the 6-dimensional Lorentz-Hermitian Lie algebra ${{\mathcal {L}}}=\mathbb {R}\text{-span }\{x_1,Jx_1,x_2,Jx_2,x_3,Jx_3\}$ with nonzero brackets and $\varphi$-pairings given by

$$\begin{aligned} {[}x_1,Jx_1]= & {} x_2, \quad [x_1,x_2]=- Jx_3,\quad [Jx_1,x_2]=x_3\\ \varphi (x_1,x_3)= & {} \varphi (Jx_1,Jx_3)=1,\quad \varphi (x_2,x_2)=\varphi (Jx_2,Jx_2)=1. \end{aligned}$$

Proof

Since up to dimension 4 every nilpotent quadratic Lie algebra is abelian (see Favre and Santharoubane (1987); Kath (2007)), we can consider that $\dim ({{\mathfrak {g}}})=n\ge 6$. As seen in the proof of Theorem 2.4, ${{\mathfrak {g}}}$ is a pHQ-double extension of a $(n-4)$-dimensional pseudo-Hermitian quadratic Lie algebra.

Recalling Remark 2.1 (2), the corresponding Lie algebra ${{\mathfrak {g}}}_0$ must have signature $(0,n-4)$. But the only nilpotent quadratic Lie algebras with positive definite metric are the abelian ones, so that ${{\mathfrak {g}}}_0=\mathbb {R}^{n-4}$ with an arbitrary complex structure and a Hermitian metric. It is well known that a nilpotent linear map which is skew-symmetric with respect to a definite scalar product must be identically zero. So, the derivations D and F defining the pHQ-double extension are both null. This means that the only nonzero brackets in ${{\mathfrak {g}}}=\mathbb {R}z \oplus \mathbb {R}z' \oplus \mathfrak {g}_0 \oplus \mathbb {R}v' \oplus \mathbb {R}v$ are

$$\begin{aligned} {[}v,v']=s_0,\quad [v,x]=-\varphi (s_0,x)z',\quad [v',x]=\varphi (s_0,x)z. \end{aligned}$$

Put $n=2k$, $\alpha =\varphi (s_0,s_0)$ and choose a basis $\{Js_0,s_1,Js_1,\dots ,s_{k-3},Js_{k-3}\}$ of $s_0^\bot$ in ${{\mathfrak {g}}}_0$. Then, the nonzero brackets in ${{\mathfrak {g}}}$ are

$$\begin{aligned} {[}v,v']=s_0,\quad [v,s_0]=-\alpha z',\quad [v',s_0]=\alpha z. \end{aligned}$$

Notice that $\alpha >0$ because $\varphi _0$ is positive definite. Put $x_1=\alpha ^{-1/4}v$, $x_2=\alpha ^{-1/2}s_0$ and $x_3=\alpha ^{1/4}z$. Since $v'=Jv$ and $z'=Jz$, we get that the nonnull brackets are

$$\begin{aligned} {[}x_1,Jx_1]= & {} \alpha ^{-1/2}s_0=x_2,\,\, [x_1,x_2]= -\alpha ^{-1/4}\alpha ^{-1/2}\alpha Jz=-Jx_3,\\ {[}Jx_1,x_2]= & {} \alpha ^{-1/4}\alpha ^{-1/2}\alpha z=x_3. \end{aligned}$$

Further, in ${{\mathcal {L}}}=\mathbb {R}\text{-span }\{x_1,Jx_1,x_2,Jx_2,x_3,Jx_3\}$ the metric is given by

$$\begin{aligned} \varphi (Jx_1,Jx_3)= & {} \varphi (x_1,x_3)=\alpha ^{-1/4}\alpha ^{1/4}\varphi (v,z)=1,\\ \varphi (Jx_2,Jx_2)= & {} \varphi (x_2,x_2)=\alpha ^{-1}\varphi (s_0,s_0)=1. \end{aligned}$$

Now, putting ${{\mathcal {A}}}= \mathbb {R}\text{-span }\{s_1,Js_1,\dots ,s_{k-3},Js_{k-3}\}$, one immediately gets that ${{\mathfrak {g}}}$ is the orthogonal sum ${{\mathfrak {g}}}={{\mathcal {L}}}\oplus {{\mathcal {A}}}$ where ${{\mathcal {A}}}$ is an abelian Lie algebra with Hermitian metric. $\square$

Remark 2.3

The Lie algebra ${{\mathcal {L}}}$ is isomorphic to ${{\mathcal {W}}}_3\oplus \mathbb {R}$ with the notation of Favre and Santharoubane (1987) and it is the Lie algebra denoted (0, 0, 0, 12, 14, 24) in the list of 6-dimensional nilpotent Lie algebras admitting complex structures given by Salamon in Salamon (2001). In Andrada et al. (2008), it is denoted by $L_3(1,2)\times \mathbb {R}.$

4 Classification of pseudo-Hermitian quadratic nilpotent Lie algebras up to dimension 8

We will now give a complete classification of the nilpotent Lie algebras of dimension less than or equal to 8 which admit a pseudo-Hermitian quadratic metric. Notice that our classification is actually done up to holomorphic isometric isomorphism but in the cases of abelian Lie algebras we will not list all the obvious distinct signatures.

The classification up to dimension 6 is now almost obvious. With the same notation as in Corollary 2.5, one has

Proposition 3.1

A nilpotent Lie algebra ${{\mathfrak {g}}}$ with $\dim ({{\mathfrak {g}}})\le 6$ admits a pseudo-Hermitian quadratic structure if and only if it is one of the abelian Lie algebras $\mathbb {R}^2$, $\mathbb {R}^4$, $\mathbb {R}^6$ with a metric of signature (2r, 2p) or ${{\mathcal {L}}}$ with either the metric of Corollary 2.5 or its opposite.

Proof

Recall that up to dimension 4 all the nilpotent quadratic Lie algebras are abelian (see Favre and Santharoubane (1987)) and a pseudo-Hermitian quadratic Lie algebra has even dimension. So, up to dimension 4 we only have the Lie algebras $\mathbb {R}^2$, $\mathbb {R}^4$. In the 6-dimensional case, by Theorem 2.4 we have that a pseudo-Hermitian quadratic Lie algebra ${{\mathfrak {g}}}$ is either $\mathbb {R}^6$ with a definite metric or a pHQ-double extension of $\mathbb {R}^2$ with a definite pseudo-Hermitian metric. In this second case one gets (changing the sign to the metric on ${{\mathfrak {g}}}$ if necessary) the Lie algebra ${{\mathcal {L}}}$ as in the proof of Corollary 2.5 if one chooses $s_0\ne 0$ and $\mathbb {R}^6$ with an indefinite metric when $s_0=0$. $\square$

Let us then study the case $\dim ({{\mathfrak {g}}})=8$. Using Theorem 2.4 we get that if ${{\mathfrak {g}}}$ is not a pHQ-double extension of $\mathbb {R}^4$ then it must be equivalent to $\mathbb {R}^8$ endowed with a definite metric. So, we can consider that ${{\mathfrak {g}}}$ is a pHQ-double extension of $\mathbb {R}^4$. Notice that for a definite metric on $\mathbb {R}^4$ we get either $\mathbb {R}^8$ with a metric of signature (2, 6) or (6, 2) or the orthogonal sum ${{\mathcal {L}}}\oplus \mathbb {R}^2$ (with the metric given in Corollary 2.5 or its opposite) so that we can consider that ${{\mathfrak {g}}}$ is a pHQ-double extension of $\mathbb {R}^4$ with a neutral pseudo-Hermitian metric (denoted, as usual, $\mathbb {R}^{2,2}$).

Let us start with the case in which the pHQ-double extension is done with two null derivations.

Proposition 3.2

Let $({{\mathfrak {g}}},J,\varphi )$ be the pHQ-doble extension of $\mathbb {R}^{2,2}$ by means of $D=F=0$ and $s_0\in \mathbb {R}^4$. Then it is equivalent to either $\mathbb {R}^{4,4}$ or $T^*_0{{\mathfrak {k}}}$ or else $T^*_{\pm \theta _1}{{\mathfrak {k}}}$, where ${{\mathfrak {k}}}$ and $\theta _1$ are as in Example 1.2.

Proof

From the definition of the pHQ-double extension we have that the nonzero brackets are

$$\begin{aligned} {[}v,Jv]=s_0,\quad [v,x]=-\varphi (s_0,x)Jz,\quad [Jv,x]=\varphi _0(s_0,x)z. \end{aligned}$$

If $s_0=0$ we obviously get the abelian Lie algebra $\mathbb {R}^8$ with a neutral metric so that we can consider $s_0\ne 0$.

Suppose that $\varphi _0(s_0,s_0)=0$. We may then find an isotropic vector $t_0\in \mathbb {R}^4$ such that $\varphi _0(s_0,t_0)=1,\varphi _0(Js_0,t_0)=0$. Our nontrivial brackets in the basis $\{v,Jv,z,Jz,s_0,Js_0,t_0,Jt_0\}$ are then

$$\begin{aligned} {[}v,Jv]=s_0,\quad [v,t_0]=-Jz,\quad [Jv,t_0]=z. \end{aligned}$$

If we define

$$\begin{aligned} x_1\!=\!v,\quad x_2\!=\!Jv,\quad x_3\!=\!s_0,\quad x_4\!=\!Js_0,\quad x_1^*\!=\!z,\quad x_2^*\!=\!Jz,\quad x_3^*\!=\!t_0,\quad x_4^*\!=\!Jt_0, \end{aligned}$$

we immediately get that ${{\mathfrak {g}}}$ is the Lie algebra $T^*_0{{\mathfrak {k}}}$ with the pseudo-Hermitian structure of Example 1.2.

On the other hand, if $\varphi (s_0,s_0)=\alpha \ne 0$, since the metric in $\mathbb {R}^4$ is neutral we can choose $t_0\in (\mathbb {R}s_0\oplus \mathbb {R}Js_0)^\bot$ such that $\varphi _0(t_0,t_0)=-\varphi _0(s_0,s_0)$ so that in the corresponding basis we get the nonzero brackets

$$\begin{aligned} {[}v,Jv]=s_0,\quad [v,s_0]=-\alpha Jz,\quad [Jv,s_0]=\alpha z. \end{aligned}$$

(4)

Choose $\lambda \in \mathbb {R}$ such that $\alpha =\epsilon 2\lambda ^4$ where $\epsilon =\pm 1$ is the sign of $\alpha$ and define

$$\begin{aligned} x_1= & {} \lambda ^{-1}v,\quad x_2=\lambda ^{-1}Jv,\quad x_3=(2\lambda ^2)^{-1}(s_0+t_0),\quad x_4=(2\lambda ^2)^{-1}(Js_0+Jt_0),\\ x_1^*= & {} \lambda z,\quad x_2^*=\lambda Jz,\quad x_3^*=\epsilon (2\lambda ^2)^{-1}(s_0-t_0),\quad x_4^*=\epsilon (2\lambda ^2)^{-1}(Js_0-Jt_0). \end{aligned}$$

We then have that $x_1^*,x_2^*,x_4,x_4^*\in {{\mathfrak {z}}}({{\mathfrak {g}}})$ and a simple calculation yields

$$\begin{aligned} {[}x_1,x_2]= & {} x_3+\epsilon x_3^*,\quad [x_2,x_3]=\epsilon x_1^*,\\ {[}x_3,x_1]= & {} \epsilon x_2^*,\quad [x_3^*,x_1]= x_2^*,\quad [x_3^*,x_2]= -x_1^*, \end{aligned}$$

which shows that the bracket of ${{\mathfrak {g}}}$ is that of $T^*_{\epsilon \theta _1}{{\mathfrak {k}}}$. Since the complex structure is obviously the same and we also have that the nonzero products for $\varphi$ are $\varphi (x_i,x_i^*)=1$, we are done. $\square$

Remark 3.1

Notice that from the identity (4) one easily sees that $T^*_{\pm \theta _1}{{\mathfrak {k}}}$ is holomorphically isomorphic to ${{\mathcal {L}}}\oplus \mathbb {R}^2$. Further, if one chooses on ${{\mathcal {L}}}$ the Lorentz-Hermitian metric of Corollary 2.5 one has that $T^*_{\theta _1}{{\mathfrak {k}}}$ is equivalent to ${{\mathcal {L}}}\oplus \mathbb {R}^{2,0}$ and $T^*_{-\theta _1}{{\mathfrak {k}}}$ is the same Lie algebra with the opposite metric. Besides, $T^*_{\theta _1}{{\mathfrak {k}}}$ and $T^*_{-\theta _1}{{\mathfrak {k}}}$ cannot be equivalent since the restriction of the metric to the derived ideal has signature (0, 1) for $T^*_{\theta _1}{{\mathfrak {k}}}$ and signature (1, 0) for $T^*_{-\theta _1}{{\mathfrak {k}}}$.

In order to classify the cases with $(D,F)\ne (0,0)$, we will always consider that $F\ne 0$. Notice that, according to Remark 2.1 (3), there is no loss of generality with that assumption. We first need the following technical lemma:

Lemma 3.3

In $\mathbb {R}^4$ with an arbitrary complex structure J and a neutral pseudo-Hermitian metric $\varphi$, let us consider $F,D\in {\mathfrak {gl}}(\mathbb {R}^4)$ two nilpotent $\varphi$-skewsymmetric maps such that $F\ne 0$ and $[F,D]=[F+JD,J]=0.$

If we denote by $\text{ Ker }(F), \text{ Im }(F)$ respectively the kernel and the image of F then we have

$$\begin{aligned} \text{ Ker }(F)=J(\text{ Ker }(F))=\text{ Im }(F). \end{aligned}$$

Proof

Notice that F is skewsymmetric with respect to $\varphi$ and hence must have even rank. Since $F\ne 0$ and F is singular (because it is nilpotent), we must have $\text{ rank }(F)=2$. This obviously implies that $\text{ Ker }(F)$ is 2-dimensional. Also recall that, if $F^*$ denotes the $\varphi$-adjoint for F, since F is $\varphi$-skewsymmetric, we have $\text{ Im }(F)=\text{ Ker }(F^*)^\bot =\text{ Ker }(F)^\bot$.

Let us first suppose that $\text{ Ker }(F)\cap J(\text{ Ker }(F))=\{0\}$. Then $\mathbb {R}^4=\text{ Ker }(F)\oplus J(\text{ Ker }(F))$. Since F is nilpotent, we may find a nonzero element $t_1\in \text{ Ker }(F)\cap \text{ Im }(F)$ and consider another $t_2\in \text{ Ker }(F)$ linearly independent with $t_1$. We then have that $\{t_1,t_2,Jt_1,Jt_2\}$ is a basis of $\mathbb {R}^4$. Let us denote by ${{\mathcal {F}}},{{\mathcal {D}}}, {{\mathcal {J}}}$ and ${{\mathcal {G}}}$ respectively the corresponding matrices of F, D, J and $\varphi$ with respect to such a basis. As $t_1\in \text{ Im }(F)$ it must be orthogonal to $\text{ Ker }(F)$ and, therefore, we have

$$\begin{aligned} \varphi (t_1,t_1)=\varphi (t_1,t_2)=\varphi (Jt_1,Jt_1)=\varphi (Jt_1,Jt_2)=\varphi (t_1,Jt_1)=\varphi (t_2,Jt_2)=0. \end{aligned}$$

Besides, there exist $\alpha ,\beta \in \mathbb {R}$ with $\beta \ne 0$ such that

$$\begin{aligned} \varphi (t_2,t_2)=\varphi (Jt_2,Jt_2)=\alpha ,\quad \varphi (t_1,Jt_2)=-\varphi (Jt_1,t_2)=\beta , \end{aligned}$$

so that the matrices ${{\mathcal {G}}}$ and ${{\mathcal {J}}}$ are given by

$$\begin{aligned} {{\mathcal {G}}}=\left( \begin{array}{rrrr}0 &{}0 &{}0&{}\beta \\ 0&{}\alpha &{}-\beta &{}0\\ 0&{}-\beta &{}0&{}0\\ \beta &{}0&{}0&{}\alpha \end{array}\right) ,\quad {{\mathcal {J}}}=\left( \begin{array}{rrrr}0 &{}0 &{}-1&{}0\\ 0&{}0&{}0&{}-1\\ 1&{}0&{}0&{}0\\ 0&{}1&{}0&{}0\end{array}\right) . \end{aligned}$$

On the other hand, there exists two $2\times 2$ matrices $S_1$, $S_2$ such that ${{\mathcal {F}}}$ has the following block decomposition

$$\begin{aligned} {{\mathcal {F}}}=\left( \begin{array}{c|c} 0&{}S_1\\ \hline 0&{}S_2\end{array}\right) , \end{aligned}$$

but one easily sees that the condition $F+F^*=0$ or, equivalently, ${{\mathcal {G}}}{{\mathcal {F}}}+{{\mathcal {F}}}^t{{\mathcal {G}}}=0$, implies that there exists $a\in \mathbb {R}$ such that

$$\begin{aligned} S_1=aI,\quad S_2=\left( \begin{array}{cc}0&{}a\alpha /\beta \\ 0&{}0\end{array}\right) . \end{aligned}$$

Now, since $[F,D]=0$ one has that $D(\text{ Ker }(F))\subset \text{ Ker }(F)$. So, the matrix of D with respect to the given basis must be of the form

$$\begin{aligned} {{\mathcal {D}}}=\left( \begin{array}{c|c} A_1&{}A_2\\ \hline 0&{}A_3\end{array}\right) \end{aligned}$$

for some $2\times 2$ matrices $A_1,A_2,A_3.$ The condition $[F+JD,J]=0$ leads to $A_1=A_3-aI$ and $A_2=-S_2.$ But F is nilpotent and this implies that both $A_3$ and $A_1=A_3-aI$ must be nilpotent, hence traceless, and this would imply $a=0$, a contradiction with the assumption $F\ne 0$.

As a consequence, we finally have $\text{ Ker }(F)\cap J(\text{ Ker }(F))\ne \{0\}$ so that we can find a vector $u\in \text{ Ker }(F)$ with $Ju\in \text{ Ker }(F)$ and this means $\text{ Ker }(F)=\mathbb {R}\text{-span }\{u,Ju\}=J(\text{ Ker }(F))$. If $\varphi (u,u)=\alpha \ne 0$, we would have that $\varphi$ is definite on $\text{ Ker }(F)$ but this is impossible since $\text{ Ker }(F)\cap \text{ Im }(F)\ne \{0\}$ by the nilpotency of F and, therefore, there exist an isotropic vector in $\text{ Ker }(F)$. Thus, $\varphi (u,u)=0$ and $\text{ Ker }(F)=\text{ Ker }(F)^\bot =\text{ Im }(F)$. $\square$

Remark 3.2

Recall that on a pseudo-Hermitian Lie algebra $({{\mathfrak {g}}}, J,\varphi )$ if $u_1\in {{\mathfrak {g}}}$ is isotropic then the subspace $\mathbb {R}u_1\oplus \mathbb {R}Ju_1$ is totally isotropic. Since $u_1^\bot \ne (Ju_1)^\bot$, we may find $v\in (Ju_1)^\bot$ such that $\varphi (u_1,v)=\lambda \ne 0$ so that $u_2=-\frac{\varphi (v,v)}{2\lambda ^2}u_1+\frac{1}{\lambda }v$ verifies $\varphi (u_2,u_2)=0$, $\varphi (u_1,u_2)=1$ and $\varphi (Ju_1,u_2)=0$.

Lemma 3.4

Let us consider $\mathbb {R}^4$ with an arbitrary complex structure J and a neutral pseudo-Hermitian metric $\varphi$. Let F, D be two nilpotent $\varphi$-skewsymmetric maps such that $F\ne 0$ and $[F,D]=[F+JD,J]=0$ and consider a nonzero vector $u_1\in \text{ Ker }(F)$ and a nonzero isotropic $u_2\in \mathbb {R}^4$ such that $\varphi (u_1,u_2)=1$, $\varphi (Ju_1,u_2)=0$.

The set $\{u_1,Ju_1,u_2,Ju_2\}$ is a basis of $\mathbb {R}^4$, the metric $\varphi$ is given by the only nonzero pairings $\varphi (u_1,u_2)=\varphi (Ju_1,Ju_2)=1,$ and there exist $a,b\in \mathbb {R}$ with $a\ne 0$ such that $F(u_2)=aJu_1$, $F(Ju_2)=-au_1$, $D(u_2)=bJu_1$, $D(Ju_2)=-bu_1$ and, besides, $D(u_1)=D(Ju_1)=0$.

Proof

The facts that $\{u_1,Ju_1,u_2,Ju_2\}$ is a basis of $\mathbb {R}^4$ and that $\varphi (u_1,u_2)=\varphi (Ju_1,Ju_2)=1$ are quite obvious so that we will only see the last part of the statement.

Notice that, according to Lemma 3.3, $\text{ Ker }(F)=\mathbb {R}\text{-span }\{u_1,Ju_1\}=\text{ Im }(F)$ so that $F(u_2)=\alpha _1u_1+\alpha _2Ju_1$, $F(Ju_2)=\beta _1u_1+\beta _2Ju_1$ but since F is skewsymmetric with respect to $\varphi$ we have $\alpha _1=\varphi (F(u_2),u_2)=0$, $\beta _2=\varphi (F(Ju_2),Ju_2)=0$ and

$$\begin{aligned} \alpha _2=\varphi (F(u_2), Ju_2)=-\varphi (u_2, F(Ju_2))=-\beta _1, \end{aligned}$$

so that putting $a=\alpha _2$ we get $F(u_2)=aJu_1$, $F(Ju_2)=-au_1$. Remark that we then have $[F,J]=0$.

The condition $[F,D]=0$ shows that $D(\text{ Ker }(F))\subset \text{ Ker }(F)$ so that the matrix of D with respect to the basis $\{u_1,Ju_1,u_2,Ju_2\}$ is of the form

$$\begin{aligned} {{\mathcal {D}}}=\left( \begin{array}{c|c} A_1&{}A_2\\ \hline 0&{}A_3\end{array}\right) \end{aligned}$$

for some $2\times 2$ matrices $A_i$. As we had obtained $[F,J]=0$, the condition $[F+JD,J]=0$ implies $[D,J]=0$ and a simple computation shows that each of the matrices $A_i$ must be of the form

$$\begin{aligned} A_i= \left( \begin{array}{rr}a_i&{}-b_i\\ b_i&{}a_i\end{array}\right) , \end{aligned}$$

for certain $a_i,b_i\in \mathbb {R}$. The nilpotency of D now implies that $a_1=b_1=a_3=b_3=0$, showing that $\text{ Ker }(F)\subset \text{ Ker }(D)$ and since D is $\varphi$-skewsymmetric we have $a_2=\varphi (F(u_2),u_2)=0$, so that the election $b=b_2$ gives the desired result. $\square$

Proposition 3.5

Let $({{\mathfrak {g}}},J,\varphi )$ be a nilpotent pHQ-double extension of $\mathbb {R}^{2,2}$ by means of a pair $(F,D)\ne (0,0)$ and a certain $s_0\in \mathbb {R}^4$ and let ${{\mathfrak {k}}}$ and $\theta _i$ $(1\le i\le 4)$ be as in Example 1.2. Then $({{\mathfrak {g}}},J,\varphi )$ is equivalent to either $T^*_0{{\mathfrak {k}}}$ or $T^*_{\theta _3}{{\mathfrak {k}}}$.

Proof

As seen in Remark 2.1 (3), we can consider $F\ne 0$. Lemma 3.3 clearly implies that $\text{ Ker }(F)$ is 2-dimensional and totally isotropic. We will now distinguish several cases.

If $s_0=0$, let us take an arbitrary $u_1\in \text{ Ker }(F)$ and choose an isotropic vector $u_2\in \mathbb {R}^4$ such that $\varphi _0(u_1,u_2)=1$ and $\varphi _0(Ju_1,u_2)=0$. Using Lemma 3.4 and the bracket given in Proposition 2.2 we get that there exist $a,b\in \mathbb {R}$, $a\ne 0$ such that the brackets in ${{\mathfrak {g}}}$ are given by

$$\begin{aligned} {[}v,Jv]= & {} 0,\quad [v,u_2]=aJu_1,\quad [v,Ju_2]=-au_1\\ {[}Jv,u_2]= & {} bJu_1,\quad [Jv,Ju_2]=-bu_1,\quad [u_2,Ju_2]=az+bJz. \end{aligned}$$

Recall that the quadratic form on ${{\mathfrak {g}}}$ is given by the nonzero products

$$\begin{aligned} \varphi (v,z)=\varphi (Jv,Jz)=\varphi (u_1,u_2)=\varphi (Ju_1,Ju_2)=1. \end{aligned}$$

Let us now define the vectors

$$\begin{aligned} x_1= & {} u_2,\quad x_3=az+bJz,\quad x_1^*=u_1,\quad x_3^*=(a^2+b^2)^{-1}(av+bJv),\\ x_2= & {} Jx_1,\quad x_4=Jx_3,\quad x_2^*=Jx_1^*,\quad x_4^*=Jx_3^*. \end{aligned}$$

It is obvious that $x_1^*,x_2^*,x_3,x_4$ are in the center of ${{\mathfrak {g}}}$ and an easy calculation shows that also $x_4^*$ is central and then the only nontrivial brackets are

$$\begin{aligned} {[}x_1,x_2]=x_3,\quad [x_3^*,x_1]=x_2^*,\quad [x_3^*,x_2]=-x_1^*. \end{aligned}$$

Since one easily verifies that $\varphi (x_i,x_j^*)=\delta _{ij}$ and $\varphi (x_i,x_j)=\varphi (x_i^*,x_j^*)=0$ for all $1\le i,j\le 4$, we see that ${{\mathfrak {g}}}=T^*_0{{\mathfrak {k}}}$ with the bracket and complex structure defined in Example 1.2.

If $s_0\ne 0$ and $s_0\in \text{ Ker }(F)$ we can choose $u_1=s_0\in \text{ Ker }(F)$ and find an isotropic $u_2\in \mathbb {R}^4$ verifying $\varphi _0(u_1,u_2)=1$ and $\varphi _0(Ju_1,u_2)=0$. Combining Lemma 3.4 and Proposition 2.2 one finds $a,b\in \mathbb {R}$, $a\ne 0$ such that the brackets are

$$\begin{aligned} {[}v,Jv]= & {} u_1,\quad [v,u_2]=aJu_1-Jz,\quad [v,Ju_2]=-au_1\\ {[}Jv,u_2]= & {} bJu_1+z,\quad [Jv,Ju_2]=-bu_1,\quad [u_2,Ju_2]=az+bJz. \end{aligned}$$

Consider the following vectors in $\mathbb {R}^4$:

$$\begin{aligned} x_1= & {} u_2-av-bJv,\quad x_3=2(a^2+b^2)u_1+2az+2bJz,\\ x_1^*= & {} \frac{1}{2a^2+2b^2}((a^2+b^2)u_1-az-bJz)\,\quad x_3^*=\frac{1}{4a^2+4b^2}(u_2+av+bJv)\\ x_2= & {} Jx_1,\quad x_4=Jx_3,\quad x_2^*=Jx_1^*,\quad x_4^*=Jx_3^*. \end{aligned}$$

As in the previous case, it is clear that $x_1^*,x_2^*,x_3,x_4\in {{\mathfrak {z}}}({{\mathfrak {g}}})$. A straightforward computation gives that $x_4^*\in {{\mathfrak {z}}}({{\mathfrak {g}}})$ and also

$$\begin{aligned}{}[x_1,x_2]= & {} x_3,\quad [x_3^*,x_1]=x_2^*,\quad [x_3^*,x_2]=-x_1^*\\ \varphi (x_i,x_j^*)= & {} \delta _{ij},\quad \varphi (x_i,x_j)=\varphi (x_i^*,x_j^*)=0,\quad 1\le i,j\le 4, \end{aligned}$$

which proves that $({{\mathfrak {g}}},J,\varphi )$ is again $T^*_0{{\mathfrak {k}}}$.

Let us now suppose that $s_0\not \in \text{ Ker }(F)$. Choose an element $u_1\in \text{ Ker }(F)$ such that $\varphi _0(u_1,s_0)=1$ and $\varphi _0(Ju_1,s_0)=0$. Take $\beta _1=\varphi _0(s_0,s_0)/2$ and consider $u_2=s_0-\beta _1u_1$. One immediately gets $\varphi (u_2,u_2)=\varphi (u_2,Ju_1)=0$ and $\varphi (u_1,u_2)=1$. Further, $s_0=\beta _1u_1+u_2$. From Lemma 3.4 and Proposition 2.2 we now get

$$\begin{aligned} {[}v,Jv]= & {} u_2+\beta _1u_1,\,\,\, [v,u_1]=-Jz,\,\,\, [v,u_2]=aJu_1-\beta _1Jz,\,\,\, [v,Ju_2]=-au_1\\ {[}Jv,u_1]= & {} z,\,\,\, [Jv,u_2]=bJu_1+\beta _1z,\,\,\, [Jv,Ju_2]=-bu_1,\,\,\, [u_2,Ju_2]=az+bJz. \end{aligned}$$

If we choose in this case

$$\begin{aligned} x_1= & {} {(a^2+b^2)^{-3/5}}(av+bJv),\quad x_3=(a^2+b^2)^{-1/5}u_2,\\ x_1^*= & {} {(a^2+b^2)^{-2/5}}(az+bJz)\,\quad x_3^*=(a^2+b^2)^{1/5}u_1\\ x_2= & {} Jx_1,\quad x_4=Jx_3,\quad x_2^*=Jx_1^*,\quad x_4^*=Jx_3^*, \end{aligned}$$

and we take $\alpha =(a^2+b^2)^{-2/5}\beta _1$, we get that the brackets in ${{\mathfrak {g}}}$ are given by

$$\begin{aligned} {[}x_1,x_2]= & {} x_3+\alpha x_3^*,\quad [x_1,x_3]=-\alpha x^*_2+x_4^*,\quad [x_1,x_4]=-x^*_3,\\ {[}x_2,x_3]= & {} \alpha x^*_1,\quad [x_3,x_4]=x^*_1,\quad [x_3^*,x_1]=x_2^*,\quad [x_3^*,x_2]=-x^*_1. \end{aligned}$$

Besides, one easily shows that we also have

$$\begin{aligned} \varphi (x_i,x_j^*)=\delta _{ij},\quad \varphi (x_i,x_j)=\varphi (x_i^*,x_j^*)=0,\quad 1\le i,j\le 4, \end{aligned}$$

and, hence, we get the pseudo-Hermitian structure of $T^*_\theta {{\mathfrak {k}}}$ for $\theta =\theta _3+\alpha \theta _1$.

Now, consider the vectors

$$\begin{aligned} y_1= & {} x_1+\alpha x_3, \quad y_2=x_2+\alpha x_4,\quad y_3=x_3,\quad y_4=x_4,\\ y_1^*= & {} x_1^*, \quad y_2^*=x_2^*,\quad y_3^*=x_3^*-\alpha x_1^*,\quad y_4^*=x_4^*-\alpha x_2^*. \end{aligned}$$

Notice that $Jy_1=y_2$, $Jy_3=y_4$, $Jy_1^*=y^*_2$, $Jy_3^*=y_4^*$ and a direct computation proves that

$$\begin{aligned} \varphi (y_i,y_j^*)=\delta _{ij},\quad \varphi (y_i,y_j)=\varphi (y_i^*,y_j^*)=0,\quad 1\le i,j\le 4. \end{aligned}$$

Moreover, $y_1^*,y_2^*,y_4^*$ are clearly central and we have

$$\begin{aligned} {[}y_1,y_2]= & {} [x_1,x_2]+\alpha [x_3,x_2]+\alpha [x_1,x_4]+\alpha ^2[x_3,x_4]\\= & {} x_3+\alpha x_3^*-\alpha ^2 x^*_1-\alpha x^*_3+\alpha ^2 x^*_1=y_3\\ {[}y_1,y_3]= & {} [x_1,x_3]=-\alpha x^*_2+x_4^*=y_4^*\\ {[}y_1,y_4]= & {} [x_1,x_4]+\alpha [x_3,x_4]=-x^*_3+\alpha x^*_1=-y^*_3\\ {[}y_2,y_3]= & {} [x_2,x_3]+\alpha [x_4,x_3]=\alpha x^*_1-\alpha x^*_1=0\\ {[}y_3,y_4]= & {} [x_3,x_4]=x^*_1=y_1^*\\ {[}y_3^*,y_1]= & {} [x_3^*,x_1]+\alpha [x_3^*,x_3]=x_2^*=y_2^*\\ {[}y_3^*,y_2]= & {} [x_3^*,x_2]+\alpha [x_3^*,x_4]=-x_1^*=-y_1^*, \end{aligned}$$

and the remaining brackets vanish. So, we actually have the Lie algebra $T^*_{\theta _3}{{\mathfrak {k}}}$. $\square$

We can now summarize all the results above. We will denote by ${{\mathcal {L}}}^{2,4}$ the Lie algebra with complex structure ${{\mathcal {L}}}$ of Corollary 2.5 endowed with the Lorentz-Hermitian metric and by ${{\mathcal {L}}}^{4,2}$ the same Lie algebra with the opposite metric. As before, the metrics and complex structures on $T^*_\theta {{\mathfrak {k}}}$ are the ones given in Example 1.2.

Since the direct sum of pseudo-Hermitian quadratic Lie algebras is clearly a pseudo-Hermitian quadratic Lie algebra, we will only give the classification of the indecomposable ones, that is to say, those pseudo-Hermitian quadratic Lie algebras that do not split as a sum of two pseudo-Hermitian quadratic ideals.

Theorem 3.6

For $n\le 8$ a n-dimensional indecomposable pseudo-Hermitian quadratic nilpotent Lie algebra is equivalent to one and only one of the following pseudo-Hermitian Lie algebras: $\mathbb {R}^{0,2}$, $\mathbb {R}^{2,0}$, ${{\mathcal {L}}}^{2,4}$, ${{\mathcal {L}}}^{4,2}$, $T^*_0{{\mathfrak {k}}}$, $T^*_{\theta _3}{{\mathfrak {k}}}$.

Proof

The result follows at once from the propositions above. It only remains to prove that those Lie algebras are inequivalent. Using the dimension of the Lie algebra and the signature of the metric, it suffices to see that the two $\hbox {T}^*$-extensions are inequivalent. But the center in $T^*_0{{\mathfrak {k}}}$ is 5-dimensional whereas that of $T^*_{\theta _3}{{\mathfrak {k}}}$ is 3-dimensional, so that they are not isomorphic (actually, one has that $T^*_0{{\mathfrak {k}}}$ is 2-step nilpotent while the other one is 3-step nilpotent). $\square$

It is clear that abelian pseudo-Hermitian quadratic Lie algebras are classified by the dimension of the Lie algebra and the signature of the metric. The following table gives a classification scheme for non-abelian pseudo-Hermitian quadratic Lie algebras $({{\mathfrak {g}}}, J,\varphi )$ with $\dim ({{\mathfrak {g}}})\le 8$ in terms of the dimensions of ${{\mathfrak {g}}}$ and $[{{\mathfrak {g}}},{{\mathfrak {g}}}]$ and the signatures of $\varphi$ and its restriction $\varphi _{[{{\mathfrak {g}}},{{\mathfrak {g}}}]}$ to $[{{\mathfrak {g}}},{{\mathfrak {g}}}]\times [{{\mathfrak {g}}},{{\mathfrak {g}}}]$. Although it is not necessary, we include a column with the nilpotency index of ${{\mathfrak {g}}}$ (Table 1).

Table 1 Non-abelian pseudo-Hermitian quadratic nilpotent Lie algebras up to dimension 8