The Betti Numbers for a Family of Solvable Lie Algebras

Abstract

We give a property of symplectic quadratic Lie algebras that their Lie algebra of inner derivations has an invertible derivation. A family of symplectic quadratic Lie algebras is introduced to illustrate this situation. Finally, we calculate explicitly the Betti numbers of a family of solvable Lie algebras in two ways: using the cohomology of quadratic Lie algebras and applying a Pouseele’s result on extensions of the one-dimensional Lie algebra by Heisenberg Lie algebras.

1 Introduction

Let $\mathfrak {g}$ be a complex Lie algebra endowed with a non-degenerate invariant symmetric bilinear form B, $\{X_1,\ldots ,X_n\}$ be a basis of $\mathfrak {g}$ and $\{\omega _1,\ldots ,\omega _n\}$ be its dual basis. Denote by $\{Y_1,\ldots ,Y_n\}$ the basis of $\mathfrak {g}$ defined by $B(Y_i,.)=\omega _i$, $1\le i\le n$. Pinczon and Ushirobira [5] discovered in that the differential $\partial$ on $\bigwedge (\mathfrak {g}^*)$, the space of antisymmetric forms on $\mathfrak {g}$, is given by $\partial :=-\{I,.\}$ where I is defined by:

$$\begin{aligned} I(X,Y,Z)=B([X,Y],Z),\quad \forall \ X,\ Y,\ Z\in \mathfrak {g}\end{aligned}$$

and $\{{,}\}$ is the super Poisson bracket on $\bigwedge (\mathfrak {g}^*)$ defined by

$$\begin{aligned} \{\Omega , \Omega ^{\prime }\} = (-1)^{k+1} \sum _{i,j} B(Y_i,Y_j){\iota }_{X_i}(\Omega ) \wedge {\iota }_{X_j}(\Omega ^{\prime }), \ \forall \ \Omega \in \bigwedge {}^k(\mathfrak {g}^*),\ \Omega ^{\prime } \in \bigwedge (\mathfrak {g}^*). \end{aligned}$$

In Sect. 1, by using this, we detail a result of Medina and Revoy [4] that there is an isomorphism between the second cohomology group $H^2(\mathfrak {g},\mathbb {C})$ and ${\text {Der}}_a(\mathfrak {g})/{\text {ad}}(\mathfrak {g})$ where ${\text {Der}}_a(\mathfrak {g})$ is the vector space of skew-symmetric derivations of $\mathfrak {g}$ and ${\text {ad}}(\mathfrak {g})$ is its subspace of inner ones.

A well-known theorem of Jacobson says that a Lie algebra over a field of characteristic zero is nilpotent if it admits an invertible derivation, but on the contrary, there exist nilpotent Lie algebras that have no invertible derivation. It is also well known that symplectic quadratic Lie algebras are nilpotent and hence so too are their algebras of inner derivations. We prove that the latter algebras have an invertible derivation. In particular, we have the following (Proposition 2.5).

Theorem 1

Let $(\mathfrak {g},B,\omega )$ be a symplectic quadratic Lie algebra. Consider the mapping ${\mathscr {D}}:{\text {ad}}(\mathfrak {g})\rightarrow {\text {ad}}(\mathfrak {g})$ defined by ${\mathscr {D}}({\text {ad}}(X))={\text {ad}}\left( \phi ^{-1}\left( \iota _X(\omega )\right) \right)$ with $\phi :\mathfrak {g}\rightarrow \mathfrak {g}^*$, $\phi (X)=B(X,.)$, then ${\mathscr {D}}$ is an invertible derivation of ${\text {ad}}(\mathfrak {g})$.

The reader is referred to [2] for further information about symplectic quadratic Lie algebras. A family of such algebras is given to illustrate this situation.

In Sect. 2, motivated by Corollary 4.4 in [4], we give the Betti numbers for a family of solvable quadratic Lie algebras defined as follows. For each $n\in \mathbb {N}$, let $\mathfrak {g}_{2n+2}$ denote the Lie algebra with basis $\{X_0,\ldots ,X_n,Y_0,\ldots ,Y_n\}$ and nonzero Lie brackets $[Y_0,X_i]=X_i$, $[Y_0,Y_i]=-Y_i$, $[X_i,Y_i]=X_0$, $1\le i\le n$. Denote by $B^k(\mathfrak {g}_{2n+2}) = B^k(\mathfrak {g}_{2n+2},\mathbb {C})$, $Z^k(\mathfrak {g}_{2n+2}) = Z^k(\mathfrak {g}_{2n+2},\mathbb {C})$, $H^k(\mathfrak {g}_{2n+2}) = H^k(\mathfrak {g}_{2n+2},\mathbb {C})$ and $b_{k}= b_{k}(\mathfrak {g}_{2n+2},\mathbb {C})$. By computing on super Poisson brackets, our second result is the following.

Theorem 2

The kth Betti numbers of $\mathfrak {g}_{2n+2}$ are given as follows:

• (1) If k is even, then one has

  $$\begin{aligned} b_{k}= \left| \begin{pmatrix} n \\ \frac{k}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k}{2}\end{pmatrix} -\begin{pmatrix} n \\ \frac{k-2}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-2}{2}\end{pmatrix}\right| . \end{aligned}$$

• (2) If k is odd, then one has

  • if $k<n+1$, then

    $$\begin{aligned} b_{k}= \begin{pmatrix} n \\ \frac{k-1}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-1}{2}\end{pmatrix} -\begin{pmatrix} n \\ \frac{k-3}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-3}{2}\end{pmatrix}, \end{aligned}$$

  • if $k=n+1$, then

    $$\begin{aligned} b_{n+1}= 2\begin{pmatrix} n \\ \frac{n}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{n}{2}\end{pmatrix} -2\begin{pmatrix} n \\ \frac{n+2}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{n+2}{2}\end{pmatrix}, \end{aligned}$$

  • if $k>n+1$, then

    $$\begin{aligned} b_{k}= \begin{pmatrix} n \\ \frac{k-1}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-1}{2}\end{pmatrix} - \begin{pmatrix} n \\ \frac{k+1}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k+1}{2}\end{pmatrix}. \end{aligned}$$

Our method is direct and different from Pouseele’s method given in [6] that we shall recall in Appendix 1. In Pouseele’s method, the Betti numbers of $\mathfrak {g}_{2n+2}$ are derived from the Betti numbers of the $2n+1$-dimensional Lie algebra $\mathfrak {f}$ defined by $[Y,X_i]=X_i$ and $[Y,Y_i]=-Y_i$ for all $1\le i\le n$.

For other results of Betti numbers for some families of nilpotent Lie algebras, we refer the reader to [1, 6] or [7].

2 A Characterization of Symplectic Quadratic Lie Algebras

Let $\mathfrak {g}$ be a complex Lie algebra endowed with a non-degenerate invariant symmetric bilinear form B. In this case, we call the pair $(\mathfrak {g},B)$ a quadratic Lie algebra. Denote by ${\text {Der}}_a(\mathfrak {g})$ the vector space of skew-symmetric derivations of $\mathfrak {g}$, that is, the vector space of derivations D satisfying $B(D(X),Y) = -B(X,D(Y))$ for all $X,\ Y\in \mathfrak {g}$, then ${\text {Der}}_a(\mathfrak {g})$ is a Lie subalgebra of ${\text {Der}}(\mathfrak {g})$.

Proposition 2.1

There exists a Lie algebra isomorphism T between ${\text {Der}}_a(\mathfrak {g})$ and the space $\{\Omega \in \bigwedge {}^2(\mathfrak {g}^*)\ |\ \{I,\Omega \}=0\}$. This isomorphism induces an isomorphism from ${\text {ad}}(\mathfrak {g})$ onto $\iota _\mathfrak {g}(I)=\{\iota _X(I)\in \bigwedge {}^2(\mathfrak {g}^*)\ |\ X\in \mathfrak {g}\}$.

Proof

Let $D\in {\text {Der}}_a(\mathfrak {g})$ and set $\Omega \in \bigwedge {}^2(\mathfrak {g}^*)$ by $\Omega (X,Y)=B(D(X),Y)$ for all $X,\ Y\in \mathfrak {g}$. Then D is a derivation of $\mathfrak {g}$ if and only if

$$\begin{aligned} \Omega ([X,Y],Z)+\Omega ([Y,Z],X)+\Omega ([Z,X],Y)=0 \end{aligned}$$

for all $X,\ Y,\ Z\in \mathfrak {g}$. It means $\{I,\Omega \}=0$. Define the map T from ${\text {Der}}_a(\mathfrak {g})$ onto $\{\Omega \in \bigwedge {}^2(\mathfrak {g}^*)\ |\ \{I,\Omega \}=0\}$ by $T(D)=\Omega$, then T is a one-to-one correspondence.

Now we shall show that $T([D,D^{\prime }])=\{T(D),T(D^{\prime })\}$ for all $D,\ D^{\prime }\in {\text {Der}}_a(\mathfrak {g})$. Indeed, set $\Omega =T(D)$, $\Omega ^{\prime }=T(D^{\prime })$ and fix an orthonormal basis $\{X_j\}_{j=1}^n$ of $\mathfrak {g}$. One has

$$\begin{aligned} \{\Omega ,\Omega ^{\prime }\}(X,Y)= & {} -\left( \sum _{j=1}^n {\iota }_{X_j}(\Omega ) \wedge {\iota }_{X_j}(\Omega ^{\prime })\right) (X,Y)\\= & {} -\sum _{j=1}^n \left( \Omega (X_j,X)\Omega ^{\prime }(X_j,Y)-\Omega (X_j,Y) \Omega ^{\prime }(X_j,X)\right) \\= & {} -\sum _{j=1}^n B\left( B(D(X_j),X)D^{\prime }(X_j)-B(D^{\prime }(X_j),X)D(X_j),Y)\right) \\= & {} -\sum _{j=1}^n B\left( D^{\prime }(D(X))-D(D^{\prime }(X)), Y)\right) = -B([D^{\prime },D](X), Y). \end{aligned}$$

That means, $T([D,D^{\prime }])=\{T(D),T(D^{\prime })\}$, and then, T is a Lie algebra isomorphism.

If $D={\text {ad}}(X_0)$, then $T(D)(X,Y)=B([X_0,X],Y)=I(X_0,Y,Z)=\iota _{X_0}(I)(X,Y)$. Therefore, $T(D)=\iota _{X_0}(I)$. $\square$

Corollary 2.2

$\{\iota _X(I),\iota _Y(I)\} =\iota _{[X,Y]}(I)$.

Corollary 2.3

[4] The cohomology group $H^2(\mathfrak {g},\mathbb {C})\simeq {\text {Der}}_a(\mathfrak {g},B)/{\text {ad}}(\mathfrak {g})$.

Definition 2.4

A non-degenerate skew-symmetric bilinear form $\omega :\mathfrak {g}\times \mathfrak {g}\rightarrow \mathbb {C}$ is called a symplectic structure on $\mathfrak {g}$ if it satisfies

$$\begin{aligned} \omega ([X,Y],Z)+\omega ([Y,Z],X)+\omega ([Z,X],Y)=0 \end{aligned}$$

for all $X,\ Y,\ Z\in \mathfrak {g}$.

A symplectic structure $\omega$ on a quadratic Lie algebra $(\mathfrak {g},B)$ is corresponding to a skew-symmetric invertible derivation D defined by $\omega (X,Y)=B(D(X),Y)$, for all $X,\ Y\in \mathfrak {g}$. As above, a symplectic structure is exactly a non-degenerate 2-form $\omega$ satisfying $\{I,\omega \}=0$. If $\mathfrak {g}$ has a such $\omega$, then we call $(\mathfrak {g},B,\omega )$ a symplectic quadratic Lie algebra.

For symplectic quadratic Lie algebras, the reader can refer to [2] for more details. Here we give a following property.

Proposition 2.5

Let $(\mathfrak {g},B,\omega )$ be a symplectic quadratic Lie algebra. Consider the mapping ${\mathscr {D}}:{\text {ad}}(\mathfrak {g})\rightarrow {\text {ad}}(\mathfrak {g})$ defined by ${\mathscr {D}}({\text {ad}}(X))={\text {ad}}\left( \phi ^{-1}\left( \iota _X(\omega )\right) \right)$ with $\phi :\mathfrak {g}\rightarrow \mathfrak {g}^*$, $\phi (X)=B(X,.)$, then ${\mathscr {D}}$ is an invertible derivation of ${\text {ad}}(\mathfrak {g})$.

Proof

As above, we have $\{I,\omega \}=0$ and then $\iota _X\left( \{I,\omega \}\right) =0$ for all $X\in \mathfrak {g}$. It implies $\{\iota _X(I),\omega \}=\{I,\iota _X(\omega )\}$ for all $X\in \mathfrak {g}$. Note that if X is nonzero, since $\omega$ is non-degenerate, then $\iota _X(\omega )$ is non-trivial. Set $Y=\phi ^{-1}\left( \iota _X(\omega )\right)$, then $\{I,\iota _X(\omega )\} = \iota _Y(I)$, and therefore, this defines an inner derivation. Let D be the derivation corresponding to $\omega$, then one has $[{\text {ad}}(X),D]={\text {ad}}(Y)$.

Let ${\text {ad}}(X)\in {\text {ad}}(\mathfrak {g})$. Set $\alpha =\phi (X)$. Since $\omega$ is non-degenerate, then there exists an element $Y\in \mathfrak {g}$ such that $\alpha =\iota _Y(\omega )$. In this case, ${\mathscr {D}}({\text {ad}}(Y))={\text {ad}}(X)$. That means, ${\mathscr {D}}$ onto and therefore it is bijective. $\square$

Next, we give a family of symplectic quadratic Lie algebras that has been defined in [3] as follows.

Example 2.6

Let $p \in \mathbb {N}\setminus \{ 0 \}$. We denote the Jordan block of size p by $J_1 := (0)$ and for $p \ge 2$,

$$\begin{aligned} J_p : = \begin{pmatrix} 0 &{} 1 &{} 0 &{} \dots &{} 0 \\ 0 &{} 0 &{} 1 &{} \dots &{} 0\\ \vdots &{} \vdots &{} \dots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \dots &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} \dots &{} 0 \end{pmatrix}. \end{aligned}$$

For $p \ge 2$, we consider $\mathfrak {q}= \mathbb {C}^{2p}$ with a basis $\{X_i,\ Y_i\}$, $1\le i\le p$, and equipped with a bilinear form B satisfying $B(X_i,X_j)=B(Y_i,Y_j)=0$ and $B(X_i,Y_j)=\delta _{ij}$. Let $C:\mathfrak {q}\rightarrow \mathfrak {q}$ with matrix

$$\begin{aligned} C=\begin{pmatrix} J_p &{} 0 \\ 0 &{} -{}^t J_p\end{pmatrix} \end{aligned}$$

in the given basis. Then $C \in \mathfrak {o}(2p)$.

Let $\mathfrak {h}=\mathbb {C}^2$ and $\{X_0,Y_0\}$ be a basis of $\mathfrak {h}$. Define on the vector space $\mathfrak {j}_{2p}=\mathfrak {q}\oplus \mathfrak {h}$ the Lie bracket $[Y_0,X] = C(X)$, $[X,Y] = B(C(X),Y)X_0$ and the bilinear form $\overline{B}(X_0,Y_0)=1$, $\overline{B}(X_0,X_0)=\overline{B}(Y_0,Y_0)=\overline{B}(X_0,X) = \overline{B}(Y_0,X)=0$ and $\overline{B}(X,Y)=B(X,Y)$ for all $X,\ Y\in \mathfrak {q}$. So $\mathfrak {j}_{2p}$ is a nilpotent Lie algebra, and it will be called a $2p+2$-dimensional nilpotent Jordan-type Lie algebra.

Denote by $\{\alpha ,\ \alpha _1,\ldots ,\ \alpha _p,\ \beta ,\ \beta _1,\ldots ,\ \beta _p\}$ the dual basis of $\{X_0,\ldots ,X_p,Y_0,\ldots ,Y_p\}$, then $I=\beta \wedge \sum _{i=1}^{p-1}\alpha _{i+1}\wedge \beta _i$. In this case, we choose $\omega = \alpha \wedge \beta + \sum _{i=1}^{p}i\alpha _{i}\wedge \beta _i$ then $\{I,\omega \} = 0$ and therefore $(\mathfrak {j}_{2p},B,\omega )$ is a symplectic quadratic Lie algebra. Notice that if we define ${\mathscr {D}}({\text {ad}}(Y_0))=-{\text {ad}}(Y_0)$, ${\mathscr {D}}({\text {ad}}(X_i))=i{\text {ad}}(X_i)$, and ${\mathscr {D}}({\text {ad}}(Y_i))=-i{\text {ad}}(Y_i)$, then ${\mathscr {D}}$ is an invertible derivation of ${\text {ad}}(\mathfrak {j}_{2p})$.

3 The Betti Numbers for a Family of Solvable Quadratic Lie Algebras

For each $n\in \mathbb {N}$, let $\mathfrak {g}_{2n+2}$ denote the Lie algebra with basis $\{X_0,\ldots ,X_n,Y_0,\ldots ,Y_n\}$ and nonzero Lie brackets $[Y_0,X_i]=X_i$, $[Y_0,Y_i]=-Y_i$, $[X_i,Y_i]=X_0$, $1\le i\le n$. Then $\mathfrak {g}$ is quadratic with invariant bilinear form B given by $B(X_i,Y_i)=1$, $0\le i\le n$, zero otherwise.

Let $\{\alpha ,\ \alpha _1,\ldots ,\ \alpha _n,\ \beta ,\ \beta _1,\ldots ,\ \beta _n\}$ be the dual basis of $\{X_0,\ldots ,X_n,Y_0,\ldots ,Y_n\}$ and set $V={\text {span}}\{\alpha _i\}$, $W={\text {span}}\{\beta _i\}$, $1\le i\le n$. It is easy to check that the associated 3-form of $\mathfrak {g}_{2n+2}$:

$$\begin{aligned} I=\beta \wedge \sum _{i=1}^n\alpha _i\wedge \beta _i. \end{aligned}$$

Denote by $\Omega _n:=\sum _{i=1}^n\alpha _i\wedge \beta _i$, then one has

$$\begin{aligned} B^2(\mathfrak {g}_{2n+2}) =\{\iota _X(I)|\ X\in \mathfrak {g}_{2n+2}\}={\text {span}}\left\{ \beta \wedge \alpha _i,\beta \wedge \beta _i, \Omega _n\ | \ 1\le i\le n\right\} . \end{aligned}$$

If $n=1$, then we can directly calculate that $H^2(\mathfrak {g}_4)=\{0\}$. If $n>1$, we have the nonzero super Poisson brackets:

• (i) $\{ I,\alpha \wedge \alpha _i\}=\alpha _i\wedge \Omega _n-\alpha \wedge \beta \wedge \alpha _i$ and $\{I,\alpha \wedge \beta _i\}=\beta _i\wedge \Omega _n+\alpha \wedge \beta \wedge \beta _i$,

• (ii) $\{I,\alpha \wedge \beta \}=I$,

• (iii) $\{ I,\alpha _i\wedge \alpha _j\}=2\beta \wedge \alpha _i\wedge \alpha _j$ and $\{ I,\beta _i\wedge \beta _j\}=-2\beta \wedge \beta _i\wedge \beta _j$.

It results that $Z^2(\mathfrak {g}_{2n+2}){=}{\text {span}}\left\{ \beta \wedge \alpha _i,\beta \wedge \beta _i, \alpha _i\wedge \beta _j\ |\ 1\le i,j\le n\right\}$ and then the second cohomology group $H^2(\mathfrak {g}_{2n+2}){=}{\text {span}}\left\{ [\alpha _i{\wedge }\beta _j]\right\} /{\text {span}}\left\{ \left[ \sum _{i=1}^n\alpha _i{\wedge }\beta _i\right] \right\}$, where $1\le i,j\le n$. So we recover the result of Medina and Revoy in [4] obtained by describing the space ${\text {Der}}_a(\mathfrak {g}_{2n+2})$ that $b_2=n^2-1$.

To get the Betti numbers $b_k$ for $k\ge 3$, we need the following lemma.

Lemma 3.1

The map $\{\Omega _n,.\}: \bigwedge ^k(V)\otimes \bigwedge ^m(W)\rightarrow \bigwedge ^k(V)\otimes \bigwedge ^m(W)$ with $k,m\ge 0$ is a vector space isomorphism if $k\ne m$ and $\{\Omega _n,\bigwedge ^k(V)\otimes \bigwedge ^k(W)\} =\{0\}$.

Proof

We have $\{\Omega _n,\alpha _{i_1}\wedge \ldots \wedge \alpha _{i_k}\}=k\alpha _{i_1}\wedge \ldots \wedge \alpha _{i_k}$, $\{ \Omega _n,\beta _{i_1}\wedge \ldots \wedge \beta _{i_m}\}=-m\wedge \beta _{i_1}\wedge \ldots \wedge \beta _{i_m}$ and $\{ \Omega _n,\alpha _{i_1}\wedge \ldots \wedge \alpha _{i_k}\wedge \beta _{j_1}\wedge \ldots \wedge \beta _{j_m}\}=(k-m)\alpha _{i_1}\wedge \ldots \wedge \alpha _{i_k}\wedge \beta _{j_1}\wedge \ldots \wedge \beta _{j_m}$, then the result follows. $\square$

By a straightforward computation on super Poisson brackets, we have the following corollary.

Corollary 3.2

The restrictions of the differential $\partial$ from $\alpha \wedge \bigwedge ^i(V)\otimes \bigwedge ^j(W)$ onto $\Omega _n\wedge \bigwedge ^i(V)\otimes \bigwedge ^j(W)\oplus \alpha \wedge \beta \wedge \bigwedge ^i(V)\otimes \bigwedge ^j(W)$ and from $\bigwedge ^i(V)\otimes \bigwedge ^j(W)$ onto $\beta \wedge \bigwedge ^i(V)\otimes \bigwedge ^j(W)$ with $i,j\ge 0$, $i\ne j$ are vector space isomorphisms.

Let us now give the cases for which $\ker (\partial )$ can be obtained. The following lemma is easy:

Lemma 3.3

We have $\partial \left( \bigwedge ^i(V)\otimes \bigwedge ^i(W) \right) {=}\partial \left( \beta \wedge \bigwedge ^i(V)\otimes \bigwedge ^j(W)\right) {=}\{0\}$ with $i,j\ge 0$. Moreover, $\partial \left( \alpha \wedge \beta \wedge \bigwedge ^i(V)\otimes \bigwedge ^j(W)\right) \subset \partial \left( \bigwedge ^{i+1}(V)\otimes \bigwedge ^{j+1}(W)\right)$ for all $i,\ j\ge 0$, $i\ne j$ and

• (i) $\partial \left( \alpha \wedge \beta \wedge \bigwedge ^i(V)\otimes \bigwedge ^i(W)\right) =\beta \wedge \Omega _n\wedge \bigwedge ^i(V)\otimes \bigwedge ^i(W)$,

• (ii) $\partial \left( \alpha \wedge \bigwedge ^i(V)\otimes \bigwedge ^i(W)\right) =\Omega _n\wedge \bigwedge ^i(V)\otimes \bigwedge ^i(W)$.

By the reason shown in (i) and (ii) of Lemma 3.3, we set the map

$$\begin{aligned} \phi _{k_1,k_2,n}:\bigwedge {}^{k_1}(\alpha _1,\ldots ,\alpha _n)&\otimes&\bigwedge {}^{k_2}(\beta _1,\ldots ,\beta _n)\rightarrow \bigwedge {}^{k_1+1}(\alpha _1,\ldots ,\alpha _n)\\&\otimes&\bigwedge {}^{k_2+1}(\beta _1,\ldots ,\beta _n) \end{aligned}$$

defined by $\phi _{k_1,k_2,n} (\omega )=\Omega _n\wedge \omega$, then we have the following result.

Proposition 3.4

• (i) If k is even, then

  $$\begin{aligned} \dim \ker (\partial _k)= & {} \begin{pmatrix} n \\ \frac{k}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k}{2}\end{pmatrix} + \sum _{i=0}^{k-1} \begin{pmatrix} n \\ i\end{pmatrix}\begin{pmatrix} n+1 \\ k-1-i\end{pmatrix}\\&+\,\dim \ker \phi _{\frac{k-2}{2},\frac{k-2}{2},n} -\begin{pmatrix} n \\ \frac{k-2}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-2}{2}\end{pmatrix}. \end{aligned}$$

• (ii) If k is odd, then

  $$\begin{aligned} \dim \ker (\partial _k)=\dim \ker \phi _{\frac{k-1}{2},\frac{k-1}{2},n} + \sum _{i=0}^{k-1} \begin{pmatrix} n \\ i\end{pmatrix}\begin{pmatrix} n+1 \\ k-1-i\end{pmatrix}. \end{aligned}$$

Using the formula $b_{k}(\mathfrak {g}_{2n+2}) = \dim \ker (\partial _k) + \dim \ker (\partial _{k-1}) - \begin{pmatrix} 2n+2 \\ k-1\end{pmatrix}$, the binomial identity

$$\begin{aligned} \begin{pmatrix} n \\ k-1\end{pmatrix} + \begin{pmatrix} n \\ k\end{pmatrix}=\begin{pmatrix} n+1 \\ k\end{pmatrix} \end{aligned}$$

and the formula

$$\begin{aligned} \sum _{i=0}^{k} \begin{pmatrix} n \\ i\end{pmatrix}\begin{pmatrix} n \\ k-i\end{pmatrix}=\begin{pmatrix} 2n \\ k\end{pmatrix} \end{aligned}$$

we obtain the following corollary.

Corollary 3.5

The kth Betti numbers of $\mathfrak {g}_{2n+2}$ are given as follows:

• (i) If k is even, then

  $$\begin{aligned} b_{k}(\mathfrak {g}_{2n+2})= \begin{pmatrix} n \\ \frac{k}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k}{2}\end{pmatrix} + 2\dim \ker \phi _{\frac{k-2}{2},\frac{k-2}{2},n} -\begin{pmatrix} n \\ \frac{k-2}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-2}{2}\end{pmatrix}. \end{aligned}$$

• (ii) If k is odd, then

  $$\begin{aligned} b_{k}(\mathfrak {g}_{2n+2})= & {} \begin{pmatrix} n \\ \frac{k-1}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-1}{2}\end{pmatrix} + \dim \ker \phi _{\frac{k-1}{2},\frac{k-1}{2},n}\\&+\dim \ker \phi _{\frac{k-3}{2},\frac{k-3}{2},n}-\begin{pmatrix} n \\ \frac{k-3}{2}\end{pmatrix}\begin{pmatrix} n \\ \frac{k-3}{2}\end{pmatrix}. \end{aligned}$$

Hence, it remains to compute $\dim \ker \left( \phi _{k,k,n}\right)$. Consider the power $\phi ^m_{k_1,k_2,n}$ of the map $\phi _{k_1,k_2,n}$, and let

$$\begin{aligned} K(m,k_1,k_2,n)=\dim \ker \left( \phi ^m_{k_1,k_2,n}\right) , \end{aligned}$$

then one has:

Lemma 3.6

• (i) The map

  $$\begin{aligned}&\theta ^m_{k_1,k_2,n+1}:\ker \left( \phi ^{m+1}_{k_1-1,k_2-1, n}\right) \oplus \ker \left( \phi ^m_{k_1-1,k_2,n}\right) \oplus \ker \left( \phi ^m_{k_1,k_2-1,n}\right) \\&\qquad \qquad \oplus \ker \left( \phi ^{m-1}_{k_1,k_2,n}\right) \rightarrow \ker \left( \phi ^m_{k_1,k_2,n+1}\right) \end{aligned}$$

defined by

$$\begin{aligned} \theta ^m_{k_1,k_2,n+1} (\omega _1,\omega _2,\omega _3,\omega _4)= & {} \alpha _{n+1}\wedge \beta _{n+1}\wedge \omega _1+\alpha _{n+1}\wedge \omega _2 + \beta _{n+1}\wedge \omega _3\\&+\,\omega _4-\frac{1}{m}\phi _{k_1-1,k_2-1,n}(\omega _1) \end{aligned}$$

is a vector space isomorphism.

• (ii) $K(m,k_1,k_2,n)=K(m+1,k_1-1,k_2-1,n-1) +K(m,k_1-1,k_2,n-1)+K(m,k_1,k_2-1,n-1)+K(m-1,k_1,k_2,n-1)$.

Proof

• (i) The map $\theta ^m_{k_1,k_2,n+1}$ is clearly injective. To prove $\theta ^m_{k_1,k_2,n+1}$ surjective, let us consider $\omega \in \bigwedge ^{k_1}(\alpha _1,\ldots ,\alpha _{n+1})\otimes \bigwedge ^{k_2}(\beta _1,\ldots ,\beta _{n+1})$ such that $\Omega ^m_{n+1}\wedge \omega =0$. Observe that $\omega$ can be written in the form $\omega =\alpha _{n+1}\wedge \beta _{n+1}\wedge \omega _1+\alpha _{n+1}\wedge \omega _2+\beta _{n+1}\wedge \omega _3+\omega _4$ where $\omega _1\in \bigwedge ^{k_1-1}(\alpha _1,\ldots ,\alpha _n)\otimes \bigwedge ^{k_2-1}(\beta _1,\ldots ,\beta _n)$, $\omega _2\in \bigwedge ^{k_1-1}(\alpha _1,\ldots ,\alpha _n)\otimes \bigwedge ^{k_2}(\beta _1,\ldots ,\beta _n)$, $\omega _3\in \bigwedge ^{k_1}(\alpha _1,\ldots ,\alpha _n)\otimes \bigwedge ^{k_2-1}(\beta _1,\ldots ,\beta _n)$ and $\omega _4\in \bigwedge ^{k_1}(\alpha _1,\ldots ,\alpha _n)\otimes \bigwedge ^{k_2}(\beta _1,\ldots ,\beta _n)$. By $\Omega ^m_{n+1}\wedge \omega =0$, we obtain $\Omega ^m_n\wedge \omega _2 = \Omega ^m_n\wedge \omega _3 = \Omega ^m_n\wedge \omega _4 = 0$, $\Omega ^m_n\wedge \omega _1 = - m\Omega ^{m-1}_n\wedge \omega _4$. It implies $\Omega ^{m+1}_n\wedge \omega _1=0$ and then $\omega _1\in \ker \left( \phi ^{m+1}_{k_1-1,k_2-1,n}\right)$. Moreover, $\Omega _n\wedge \omega _1 + m\omega _4\in \ker \left( \phi ^{m-1}_{k_1,k_2,n}\right)$ means

  $$\begin{aligned} \theta ^m_{k_1,k_2,n+1} \left( \omega _1,\omega _2,\omega _3,\omega _4+\frac{1}{m}\phi _{k_1-1,k_2-1,n}(\omega _1)\right) =\omega . \end{aligned}$$

• (ii) The assertion (2) follows (1).

$\square$

To calculate $K(m,k_1,k_2,n)$, we use the following boundary conditions from the definition of $\phi ^m_{k_1,k_2,n}$ in which we assume $\phi ^0_{k_1,k_2,n}$ is the identity map:

• (1) $K(0,k_1,k_2,n) = 0$ for all $k_1,\ k_2,\ n\ge 0$ .

• (2) $K(m,0,0,n) = {\left\{ \begin{array}{ll} 0, &{} \text{ if } m\le n, \\ 1, &{} \text{ if } m > n.\end{array}\right. }$

• (3) $K(m,0,1,n) = K(m,1,0,n)={\left\{ \begin{array}{ll} 0, &{} \text{ if } m=0 \text{ or } n > m, \\ n, &{} \text{ if } 1\le n\le m.\end{array}\right. }$

• (4) $K(m,k_1,k_2,0) = {\left\{ \begin{array}{ll} 1, &{} \text{ if } m\ge 1, k_1=k_2=0, \\ 0, &{} \text{ otherwise }.\end{array}\right. }$

By the condition (2), we extend $K(m,k_1,k_2,n)=0$ for negative $k_1$ or $k_2$, and by the condition (1), we set the condition (5) by $K(-m,k_1,k_2,n)=-K(m,k_1-m,k_2-m,n)$.

Lemma 3.7

$$\begin{aligned} K(m,k,k,n)=\sum _{p=0}^{n} \sum _{q=0}^{n}\begin{pmatrix} n \\ p\end{pmatrix}\begin{pmatrix} n \\ q\end{pmatrix} K(m+n-p-q,k-n+p,k-n+q,0). \end{aligned}$$

Proof

By induction on l, we prove that

$$\begin{aligned} K(m,k,k,n)=\sum _{p=0}^{l} \sum _{q=0}^{l}\begin{pmatrix} l \\ p\end{pmatrix}\begin{pmatrix} l \\ q\end{pmatrix} K(m+l-p-q,k-l+p,k-l+q,n-l). \end{aligned}$$

Let $l=n$ to get the lemma. $\square$

The Betti numbers of $\mathfrak {g}_{2n+2}$ are in the case $m=1$. By the conditions (4) and (5), we reduce the following.

Corollary 3.8

$$\begin{aligned} K(1,k,k,n) = {\left\{ \begin{array}{ll} 0,&{} \text{ if } k < \frac{1}{2}n, \\ \begin{pmatrix} n \\ k\end{pmatrix}\begin{pmatrix} n \\ k\end{pmatrix}-\begin{pmatrix} n \\ k+1\end{pmatrix}\begin{pmatrix} n \\ k+1\end{pmatrix},&\text{ if } k \ge \frac{1}{2}n.\end{array}\right. } \end{aligned}$$

Finally, by applying this formula, we obtain the Betti number of $\mathfrak {g}_{2n+2}$ according to Corollary 3.5.

4 Appendix 1: Another Way to Get the Betti Numbers of $\mathfrak {g}_{2n+2}$

In this part, we shall give another way to get the Betti numbers of $\mathfrak {g}_{2n+2}$. It is based on the following result.

Proposition 3.9

[6] Let $\mathfrak {g}$ be an extension of the one-dimensional Lie algebra $\left\langle Z\right\rangle$ by the Heisenberg Lie algebra $\mathfrak {h}_{2n+1}$, for some n,

$$\begin{aligned} 1\longrightarrow \mathfrak {h}_{2n+1}\longrightarrow \mathfrak {g}\longrightarrow \left\langle Z\right\rangle \longrightarrow 0 \end{aligned}$$

such that $\mathfrak {g}$ acts trivially on the center $\mathfrak {z}=\left\langle W\right\rangle$ of $\mathfrak {h}_{2n+1}$. Let $\mathfrak {f}=\mathfrak {g}/\mathfrak {z}$. Then

$$\begin{aligned} b_k(\mathfrak {g}) = \left\{ \begin{array}{ll} b_k(\mathfrak {f}) \ &{}\quad \text {for } k=0 \ \text {or }k=1, \\ b_k(\mathfrak {f})-b_{k-2}(\mathfrak {f}) \ &{}\quad \text {for } 2\le k \le n ,\\ 2\left[ b_{n+1}(\mathfrak {f})-b_{n-1}(\mathfrak {f})\right] \ \ \ &{}\quad \text {for } k=n+1, \\ b_{k-1}(\mathfrak {f})-b_{k+1}(\mathfrak {f}) \ &{}\quad \text {for } n+2\le k \le 2n,\\ b_{k-1}(\mathfrak {f}) \ &{}\quad \text {for } k=2n+1 \ \text {or }k=2n+2. \end{array}\right. \end{aligned}$$

It is easy to see that $\mathfrak {g}_{2n+2}$ is an extension of the one-dimensional Lie algebra $\left\langle Y_0\right\rangle$ by $\mathfrak {h}_{2n+1}$. To calculate the Betti numbers of $\mathfrak {g}_{2n+2}$, it needs to find the Betti numbers of the $2n+1$-dimensional Lie algebra $\mathfrak {f}$ with a basis $\{Y,X_1,\ldots ,X_n,Y_1,\ldots ,Y_n\}$ and the Lie bracket

$$\begin{aligned}{}[Y,X_i]=X_i,\ \ [Y,Y_i]=-Y_i \end{aligned}$$

for all $1\le i\le n$.

Let $\{Y^*,X_1^*,\ldots ,X_n^*,Y_1^*,\ldots ,Y_n^*\}$ be the dual basis of $\{Y,X_1,\ldots ,X_n,Y_1,\ldots ,Y_n\}$.

Proposition 3.10

• (1) One has

  $$\begin{aligned} \partial _k\left( Y^*\wedge \left( \bigwedge {}^{k-1}(X_1^*,\ldots ,X_n^*,Y_1^*,\ldots ,Y_n^*)\right) \right) =0. \end{aligned}$$

• (2) Assume $j+l=k$, then we have

  • if $j=l$, then

    $$\begin{aligned} \partial _k\left( \bigwedge {}^{j}(X_1^*,\ldots ,X_n^*)\otimes \bigwedge {}^{l}(Y_1^*,\ldots ,Y_n^*)\right) =0, \end{aligned}$$

  • if $j\ne l$, then

    $$\begin{aligned}&\partial _k\left( \bigwedge {}^{j}(X_1^*,\ldots ,X_n^*)\otimes \bigwedge {}^{l}(Y_1^*,\ldots ,Y_n^*)\right) =Y^*\wedge \left( \bigwedge {}^{j}(X_1^*,\ldots ,X_n^*)\right. \\&\left. \qquad \qquad \otimes \bigwedge {}^{l}(Y_1^*,\ldots ,Y_n^*)\right) . \end{aligned}$$

Proof

The assertion (1) is obvious. For (2), we use the following computation:

$$\begin{aligned} \partial _k\left( X_{i_1}^*\wedge \ldots \wedge X_{i_j}^*\wedge Y_{r_1}^*\wedge \ldots \wedge Y_{r_l}^*\right)= & {} (j-k) Y^*\\&\wedge X_{i_1}^*\wedge \ldots \wedge X_{i_j}^*\wedge Y_{r_1}^*\wedge \ldots \wedge Y_{r_l} \end{aligned}$$

for all $1\le i_1<\cdots <i_j\le n$ and $1\le r_1<\cdots <r_l\le n$.$\square$

The following corollary results.

Corollary 3.11

The Betti numbers of $\mathfrak {f}$ are given as follows:

$$\begin{aligned} b_k(\mathfrak {f}) = \begin{pmatrix} n \\ \left[ \frac{k}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{k}{2}\right] \end{pmatrix} \end{aligned}$$

where [x] denotes the integer part of x.

Applying this corollary, we have

$$\begin{aligned} b_k(\mathfrak {g}_{2n+2}) = \left\{ \begin{array}{ll} 1 \ &{}\quad \text {for } k=0 \ \text {or }k=1, \\ \begin{pmatrix} n \\ \left[ \frac{k}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{k}{2}\right] \end{pmatrix} - \begin{pmatrix} n \\ \left[ \frac{k-2}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{k-2}{2}\right] \end{pmatrix} \ &{}\quad \text {for } 2\le k \le n ,\\ 2\begin{pmatrix} n \\ \left[ \frac{n+1}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{n+1}{2}\right] \end{pmatrix} - 2\begin{pmatrix} n \\ \left[ \frac{n-1}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{n-1}{2}\right] \end{pmatrix} \ &{}\quad \text {for } k=n+1, \\ \begin{pmatrix} n \\ \left[ \frac{k-1}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{k-1}{2}\right] \end{pmatrix} - \begin{pmatrix} n \\ \left[ \frac{k+1}{2}\right] \end{pmatrix}\begin{pmatrix} n \\ \left[ \frac{k+1}{2}\right] \end{pmatrix} &{}\quad \text {for } n+2\le k \le 2n,\\ 1 \ &{}\quad \text {for } k=2n+1 \ \text {or }k=2n+2. \end{array}\right. \end{aligned}$$

and then Theorem 2 is obtained.

5 Appendix 2: The Second Cohomology Group of a Family of Nilpotent Lie Algebras

In this appendix, in the progress of our work, we give the second cohomology of a family of nilpotent Lie algebras that are double extensions of an Abelian Lie algebra (see [3] for more details about these Lie algebras).

Let us denote $\mathfrak {g}_{4n+2}$ a 2-nilpotent quadratic Lie algebra of dimension $4n+2$ spanned by $\{X,X_1,\ldots ,X_{2n},Y,Y_1,\ldots ,Y_{2n}\}$ where the Lie bracket is defined by $[Y,Y_{2i-1}]=X_{2i}$, $[Y,Y_{2i}]=-X_{2i-1}$, $[Y_{2i-1},Y_{2i}]=X$ and the bilinear form is given by $B(X,Y)=B(X_i,Y_i)=1$, zero otherwise. Let $\{\alpha ,\alpha _i,\beta ,\beta _i\}$ be the dual basis of $\{X,X_i,Y,Y_i\}$. We can check that the associated 3-form I of $\mathfrak {g}_{4n+2}$ is $I=\beta \wedge \Omega$ where $\Omega =\beta _1\wedge \beta _2+\beta _3\wedge \beta _4+\cdots +\beta _{2n-1}\wedge \beta _{2n}$. Therefore, it is easy to see that $\iota _{\mathfrak {g}_{4n+2}}(I)={\text {span}}\{\Omega ,\beta \wedge \beta _i\}$ for all $1\le i\le 2n$. We have the following proposition.

Proposition 3.12

$\dim (H^2(\mathfrak {g}_{4n+2},\mathbb {C}))=8$ if $n=1$ and $\dim (H^2(\mathfrak {g}_{4n+2},\mathbb {C}))=5n^2+n$ if $n>1$.

Proof

First, we need to describe $\ker (\partial _2)$. Let V be the space spanned by $\{\beta ,\beta _1,\ldots ,\beta _{2n}\}$, then $\{I,\omega \}=0$ for all $\omega \in V\wedge V$. By a straightforward computation, we have

• (1) $\{I,\beta \wedge \alpha _i\}=\{I,\alpha _{2i-1}\wedge \beta _{2i}\} = \{I,\alpha _{2i}\wedge \beta _{2i-1}\}=0$,

• (2) $\{ I,\alpha \wedge \beta \}=I$,

• (3) $\{I,\alpha \wedge \beta _{2i-1}\}=\beta _{2i-1}\wedge \Omega$, $\{I,\alpha \wedge \beta _{2i}\}=\beta _{2i}\wedge \Omega$,

• (4) $\{I,\alpha \wedge \alpha _{2i-1}\}=\alpha _{2i-1}\wedge \Omega +\beta \wedge \beta _{2i}\wedge \alpha$, $\{I,\alpha \wedge \alpha _{2i}\}=\alpha _{2i}\wedge \Omega -\beta \wedge \beta _{2i-1}\wedge \alpha$,

• (5) $\{I,\alpha _{2i-1}\wedge \alpha _{2j}\}=-\beta \wedge \beta _{2i}\wedge \alpha _{2j}-\beta \wedge \beta _{2j-1}\wedge \alpha _{2i-1}$, $\{I,\alpha _{2i}\wedge \alpha _{2j}\}=\beta \wedge \beta _{2i-1}\wedge \alpha _{2j}-\beta \wedge \beta _{2j-1}\wedge \alpha _{2i}$,

• (6) $\{I,\alpha _{2i-1}\wedge \beta _{2j}\}=-\{I,\alpha _{2j-1}\wedge \beta _{2i}\}=-\beta \wedge \beta _{2i}\wedge \beta _{2j}$, $i\ne j$,

• (7) $\{I,\alpha _{2i-1}\wedge \beta _{2j-1}\}=\{I,\alpha _{2j}\wedge \beta _{2i}\}=-\beta \wedge \beta _{2i}\wedge \beta _{2j-1}$,

• (8) $\{I,\alpha _{2i}\wedge \beta _{2j-1}\}=-\{I,\alpha _{2j}\wedge \beta _{2i-1}\}=\beta \wedge \beta _{2i-1}\wedge \beta _{2j-1}$, $i\ne j$.

As a consequence, if $n=1$, then it is direct that

$$\begin{aligned} \ker (\partial _2)= & {} V\wedge V\oplus {\text {span}}\{\beta \wedge \alpha _1,\beta \wedge \alpha _2,\alpha \wedge \beta -\alpha _1\wedge \beta _1,\alpha _1\wedge \beta _2,\alpha _1\\&\wedge \beta _1-\alpha _2\wedge \beta _2,\alpha _2\wedge \beta _1\}. \end{aligned}$$

Therefore, we obtain $\dim (H^2(\mathfrak {g}_{4n+2},\mathbb {C}))=8$.

In the case $n>1$, then $\Omega$ is indecomposable. Hence,

$$\begin{aligned}&\ker (\partial _2)=V\wedge V\oplus {\text {span}}\left\{ \beta \wedge \alpha _{2i-1},\beta \wedge \alpha _{2i},\alpha \wedge \beta -\sum _{i=1}^n\alpha _{2i-1}\wedge \beta _{2i-1},\right. \\&\quad \left. \alpha _{2i-1}\wedge \beta _{2j}+\alpha _{2j-1}\wedge \beta _{2i},\alpha _{2i-1}\wedge \beta _{2j-1}-\alpha _{2j}\wedge \beta _{2i},\alpha _{2i}\wedge \beta _{2j-1}+\alpha _{2j}\wedge \beta _{2i-1}\right\} \end{aligned}$$

with $1\le i,j\le n$ and it is easy to check that $\dim (H^2(\mathfrak {g}_{4n+2},\mathbb {C}))=5n^2+n$. $\square$