Killing-Yano 2-forms on 2-step nilpotent Lie groups

TL;DR

This paper characterizes 2-step nilpotent Lie groups that admit non-degenerate left invariant Killing-Yano 2-forms, showing they are precisely the complex Lie groups, with a focus on those from connected graphs.

Contribution

It identifies the specific class of 2-step nilpotent Lie groups supporting non-degenerate Killing-Yano 2-forms and describes the structure of these forms in graph-derived cases.

Findings

Only complex Lie groups among 2-step nilpotent groups admit such forms.

The space of invariant Killing-Yano 2-forms is one-dimensional for graph-derived groups.

Non-degenerate forms are exclusive to complex Lie groups in this setting.

Abstract

In this article we show that the only 2-step nilpotent Lie groups which carry a non-degenerate left invariant Killing-Yano 2-form are the complex Lie groups. In the case of 2-step nilpotent complex Lie groups arising from connected graphs, we prove that the space of left invariant Killing-Yano 2-forms is one-dimensional.

Equations

$$({\nabla }_X\omega )(Y,Z_1,\dots ,Z_{p-1})=-({\nabla }_Y\omega )(X,Z_1,\dots ,Z_{p-1}),$$

$$({\nabla }_X\omega )(Y,Z)=-({\nabla }_Y\omega )(X,Z),$$

$$({\nabla }_{X}T)Y=-({\nabla }_{Y}T)X$$

$$2⟨{\nabla }_{x}y,z⟩=⟨[x,y],z⟩-⟨[y,z],x⟩+⟨[z,x],y⟩,x,y,z\in \mathfrak{g},$$

$$⟨({\nabla }_x^{\mathfrak{u}}T{|}_{\mathfrak{u}})x,y⟩=⟨({\nabla }_x^{\mathfrak{g}}T)x,y⟩,x,y\in \mathfrak{u}.$$

$$T(\mathfrak{z})\subseteq \mathfrak{z},\text{hence also}T(\mathfrak{v})\subseteq \mathfrak{v};$$

$$T[x,y]=3[Tx,y]\text{for all }x,y\in \mathfrak{v}.$$

$$[x,y]=-J^2[x,y]=-3J[Jx,y]=-9[J^{2}x,y]=9[x,y]$$

$$T{|}_{\mathfrak{v}}=J{|}_{\mathfrak{v}},T{|}_{{\mathfrak{n}}^{\prime }}=3J{|}_{{\mathfrak{n}}^{\prime }},T{|}_{\mathfrak{a}}=T_0,$$

$$T{e}_i=a_i{f}_i,T{f}_i=-a_i{e}_i,i=1,\dots ,n,$$

$$T[e_i,e_j]=3a_i[f_i,e_j]=3a_j[e_i,f_j],$$

$$T[f_i,f_j]=-3a_i[e_i,f_j]=-3a_j[f_i,e_j].$$

$$[e_i,f_j]=\frac{a_i}{a_j}[f_i,e_j],[e_i,f_j]=\frac{a_j}{a_i}[f_i,e_j].$$

$$i_1=1,i_j=\min \{k\mid a_k\ne a_{i_{j-1}}\}$$

$$W_j=\underset{i:a_i=a_{i_j}}{⨁}{\mathfrak{v}}_i.$$

$${\mathfrak{n}}^{\prime }\oplus \mathfrak{v}={\oplus }_j{\mathfrak{n}}_j,\mathfrak{n}=\mathfrak{a}\oplus {\mathfrak{n}}_1\oplus \cdots \oplus {\mathfrak{n}}_r,$$

$$J[x,y]=\frac{1}{3a_{i_j}}T[x,y]=\frac{1}{3a_{i_j}}3[Tx,y]=\frac{1}{a_{i_j}}{a}_{i_j}[Jx,y]=[Jx,y].$$

$$T=\left(\begin{array}{cccccc}0& -3a{\mathrm{Id}}_p& & & & \\ 3a{\mathrm{Id}}_p& 0& & & & \\ & & 0& -a{\mathrm{Id}}_q& & \\ & & a{\mathrm{Id}}_q& 0& & \end{array}\right),J=\left(\begin{array}{cccccc}0& -{\mathrm{Id}}_p& & & & \\ {\mathrm{Id}}_p& 0& & & & \\ & & 0& -{\mathrm{Id}}_q& & \\ & & {\mathrm{Id}}_q& 0& & \end{array}\right),$$

$$\circ \text{for }i<j,[e_i,e_j]=\{\begin{array}{l}{z}_k,\text{ if }{z}_k\text{ is the edge joining }{e}_i\text{ and }{e}_j,\\ 0,\text{otherwise}.\end{array}$$

$$\circ z_r\text{ is central for any }r.$$

$$\text{if }[e_i,e_j]=[e_r,e_s]\ne 0\text{ then }i=r\text{ and }j=s.$$

$$[e_i,e_j]=-[f_i,f_j],[e_i,f_j]=[f_i,e_j],\text{ for any }i,j.$$

$$T{e}_i=\sum _{j=1}^n(⟨T{e}_i,e_j⟩e_j+⟨T{e}_i,f_j⟩f_j),$$

$$T{f}_i=\sum _{j=1}^n(⟨T{f}_i,e_j⟩e_j+⟨T{f}_i,f_j⟩f_j).$$

$$⟨T{e}_i,e_k⟩=⟨T{e}_i,f_k⟩=0\text{and}⟨T{f}_i,e_k⟩=⟨T{f}_i,f_k⟩=0.$$

$$\sum _j⟨T{e}_i,e_j⟩[e_j,e_i]+\sum _j⟨T{e}_i,f_j⟩[f_j,e_i]=0.$$

$$\sum _{j=1}^n⟨T{e}_i,e_j⟩[e_j,e_l]+\sum _{j=1}^n⟨T{e}_i,f_j⟩[f_j,e_l]=\sum _{j=1}^n⟨T{e}_l,e_j⟩[e_i,e_j]+\sum _{j=1}^n⟨T{e}_l,f_j⟩[e_i,f_j].$$

$$\sum _{j=1}^n⟨T{e}_i,e_j⟩[e_j,e_l]=\sum _{j=1}^n⟨T{e}_l,e_j⟩[e_i,e_j],$$

$$\sum _{j=1}^n⟨T{e}_i,f_j⟩[f_j,e_l]=\sum _{j=1}^n⟨T{e}_l,f_j⟩[e_i,f_j].$$

$$T{z}_k=3a_i{w}_k=3a_j{w}_k.$$

$$T{e}_i=a{f}_i,T{f}_i=-a{e}_i,T{z}_k=3a{w}_k,T{w}_k=-3a{z}_k,i=1,\dots ,n,k=1,\dots ,m.$$

Full text

Both authors were partially supported by CONICET, ANPCYT and SECYT-UNC (Argentina).

1. Introduction

A $p$-form $\omega$ on a Riemannian manifold $(M,g)$ is called Killing-Yano if it satisfies the Killing-Yano equation,

[TABLE]

where $\nabla$ is the Levi-Civita connection and $X,Y,Z_{1},\ldots,Z_{p-1}$ are arbitrary vector fields on $M$. These forms were introduced by K. Yano in [16] and they can be considered as generalizations of Killing vector fields. Indeed, a $1$-form is Killing-Yano if and only if its dual vector field is Killing.

Clearly, any parallel $p$-form is Killing-Yano. However, it is not easy to find examples of non-parallel Killing-Yano $p$-forms, as the following results show:

• •

If $(M^{7},g)$ is a compact manifold with holonomy $G_{2}$ then any Killing-Yano form is parallel [15].

• •

Every Killing-Yano $p$-form on a compact quaternion-Kähler manifold is parallel for any $p\geq 2$ [14].

• •

A compact simply connected symmetric space carries a non-parallel Killing-Yano $p$-form ($p\geq 2$) if and only if it is isometric to a Riemannian product $S^{k}\times N$, where $S^{k}$ is a round sphere and $k>p$ [6].

A natural source of examples in Riemannian geometry is provided by Lie groups equipped with left invariant Riemannian metrics. The study of left invariant Killing-Yano 2-forms on Lie groups with left invariant metrics started in [5] and was continued in [3, 4]. In particular, it was proved in [4] that there are no left invariant parallel non-degenerate 2-forms on nilpotent Lie groups. Therefore, if we are able to find non-degenerate Killing-Yano 2-forms on a nilpotent Lie group, then they will be automatically non-parallel. In this direction, the only known examples of left invariant Killing-Yano 2-forms on nilpotent Lie groups appear on 2-step nilpotent Lie groups, that is, when the commutator ideal is contained in the center of the group (see [5, Theorem 5.1]). In this article we refine this result by showing that non-degenerate Killing-Yano 2-forms appear only on complex 2-step nilpotent Lie groups endowed with a left invariant metric (Theorem 3.7) and, furthermore, any such Lie group admits a left invariant metric with non-degenerate Killing-Yano 2-forms (Proposition 3.5). In Section 4 we exhibit a family of examples where the dimension of the space of left invariant Killing-Yano 2-forms is 1.

2. Preliminaries

A $2$-form $\omega$ on a Riemannian manifold $(M,g)$ is called Killing-Yano if it satisfies the Killing-Yano equation,

[TABLE]

where $\nabla$ is the Levi-Civita connection and $X,Y,Z$ are arbitrary vector fields on $M$ (see [16]). The following result is clear.

Lemma 2.1**.**

Let $(M,g)$ be a Riemannian manifold, $\nabla$ the Levi-Civita connection and $\omega$ a $2$-form on $M$. The following conditions are equivalent:

1. ${\rm(i)}$

$(\nabla_{X}\omega)(Y,Z)+(\nabla_{Y}\omega)(X,Z)=0$; 2. ${\rm(ii)}$

$d\omega(X,Y,Z)=3(\nabla_{X}\omega)(Y,Z).$**

Let $(M,g)$ be a Riemannian manifold, $\nabla$ the Levi-Civita connection and $\omega$ a $2$-form on $M$ satisfying any of the conditions of Lemma 2.1. Then the skew-symmetric endomorphism $T$ of $TM$ defined by $\omega$ and $g$, that is $\omega(X,Y)=g(TX,Y)$ satisfies

[TABLE]

for all $X,Y$ vector fields on $M$. Conversely, if $T$ is a skew-symmetric endomorphism of $TM$ satisfying (3) then the 2-form $\omega(X,Y)=g(TX,Y)$ satisfies any of the conditions of Lemma 2.1. After this observation we will refer equivalently to a 2-form $\omega$ satisfying (2) or a skew-symmetric $(1,1)$-tensor $T$ satisfying or (3) as Killing-Yano (KY). Note that if $(J,g)$ is an almost Hermitian structure, then the fundamental $2$-form $\omega$ given by $\omega(X,Y)=g(JX,Y)$ is Killing-Yano if and only if $(J,g)$ is nearly Kähler.

Let $G$ be an $n$-dimensional Lie group and let $\mathfrak{g}$ be the associated Lie algebra of all left invariant vector fields on $G$. If $T_{e}G$ is the tangent space of $G$ at $e$, the identity of $G$, the correspondence $X\to X_{e}:=x$ from $\mathfrak{g}\to T_{e}G$ is a linear isomorphism. This isomorphism allows to define a Lie algebra structure on the tangent space $T_{e}G$ setting, for $x,y\in T_{e}G$, $[x,y]=[X,Y]_{e}$ where $X,Y$ are the left invariant vector fields defined by $x,y$, respectively.

A left invariant metric on $G$ is a Riemannian metric such that $L_{a}$, the left multiplication by $a\in G$, is an isometry for every $a\in G$. Conversely, every inner product on $T_{e}G$ gives rise, by left translations, to a left invariant metric. A Lie group equipped with a left invariant metric is therefore a homogeneous Riemannian manifold where many geometric invariants can be computed at the Lie algebra level. In particular, the Levi-Civita connection $\nabla$ associated to a left invariant metric $g$, when applied to left invariant vector fields, is given by:

[TABLE]

where $\langle\cdot,\cdot\rangle$ is the inner product induced by $g$ on $\mathfrak{g}$. Note that $\nabla g=0$ is equivalent to $\nabla_{x}\in\mathfrak{so}(\mathfrak{g},\langle\cdot,\cdot\rangle)$ for any $x\in\mathfrak{g}$.

A left invariant $p$-form $\omega$ on $G$ is a $p$-form such that $L_{a}^{*}\omega=\omega$ for all $a\in G$. We will consider left invariant $2$-forms $\omega$ on $(G,g)$ satisfying (2). Since $\nabla\omega$ and $d\omega$ are left invariant as well, we will study \omega\in\raise 1.0pt\hbox{\bigwedge}^{2}\mathfrak{g}^{*} satisfying (2) for $x,y,z\in\mathfrak{g}$. We will say then that $\omega$ is a Killing-Yano (KY) $2$-form on $\mathfrak{g}$. Clearly, the corresponding skew-symmetric $(1,1)$-tensor $T$ defined by $\omega$ and $g$ is left invariant, so it is determined by its value in $e\in G$, which will be denoted by $T:\mathfrak{g}\to\mathfrak{g}$. The endomorphism $T$ will be called a Killing-Yano (KY) tensor on $\mathfrak{g}$.

Lemma 2.2**.**

Let $T$ be a KY tensor on a Lie algebra $\mathfrak{g}$ with inner product $\langle\cdot,\cdot\rangle$. If $\mathfrak{u}$ is a $T$-invariant Lie subalgebra of $\mathfrak{g}$ then $T|_{\mathfrak{u}}$ is a KY tensor on $\mathfrak{u}$.

Proof.

Let $\nabla^{\mathfrak{u}}$ denote the Levi-Civita connection on $\mathfrak{u}$ associated to the restriction of $\langle\cdot,\cdot\rangle$ to $\mathfrak{u}$, and let $\nabla^{\mathfrak{g}}$ denote the Levi-Civita connection on $\mathfrak{g}$ associated to $\langle\cdot,\cdot\rangle$. Then it follows easily from (4) that

[TABLE]

Therefore if $T$ is KY on $\mathfrak{g}$ then $T|_{\mathfrak{u}}$ is KY on $\mathfrak{u}$. ∎

We recall also the following result proved in [4], which holds in particular for non abelian nilpotent Lie groups.

Theorem 2.3**.**

[4, Theorem 5.1]** If $\mathfrak{g}$ is a Lie algebra with an inner product $\langle\cdotp,\cdotp\rangle$ such that $\mathfrak{g}^{\prime}\cap\mathfrak{z}\neq\{0\}$, then there is no skew-symmetric invertible parallel tensor on $\mathfrak{g}$. In particular, this holds for any inner product on a non-abelian nilpotent Lie algebra.

3. Main results

We will consider next the case of a $2$-step nilpotent Lie algebra $\mathfrak{n}$ equipped with an inner product $\langle\cdotp,\cdotp\rangle$. We recall that a Lie algebra $\mathfrak{n}$ is called $2$-step nilpotent if it is not abelian and $[[\mathfrak{n},\mathfrak{n}],\mathfrak{n}]=0$, or equivalently, the commutator ideal $\mathfrak{n}^{\prime}:=[\mathfrak{n},\mathfrak{n}]$ is contained in the center $\mathfrak{z}$ of $\mathfrak{n}$. Such a Lie algebra is unimodular and its associated simply connected Lie group is diffeomorphic to a Euclidean space $\mathbb{R}^{n}$ via the exponential map.

If $\mathfrak{n}$ is a $2$-step nilpotent Lie algebra equipped with an inner product $\langle\cdotp,\cdotp\rangle$, then there is an orthogonal decomposition $\mathfrak{n}=\mathfrak{z}\oplus\mathfrak{v}$, where $\mathfrak{z}$ is the center of $\mathfrak{n}$ and $\mathfrak{v}$ is its orthogonal complement.

In [5] a characterization of $2$-step nilpotent Lie groups with left invariant metrics admitting a left invariant Killing-Yano tensor was obtained. Namely, the following result was proved:

Theorem 3.1**.**

[5, Theorem 3.1]** Let $T$ be a skew-symmetric endomorphism of $\mathfrak{n}=\mathfrak{z}\oplus\mathfrak{v}$. Then $T$ is a Killing-Yano tensor on $\mathfrak{n}$ if and only if

[TABLE]

and

[TABLE]

Note that (6) implies $[Tx,y]=[x,Ty]$ for any $x,y\in\mathfrak{v}$.

Corollary 3.2**.**

There are no nearly Kähler structures on 2-step nilpotent Lie algebras.

Proof.

If $(J,g)$ is a nearly Kähler structure on a 2-step nilpotent Lie algebra $\mathfrak{n}$, then $J$ is a KY tensor on $\mathfrak{n}$. Therefore, for $x,y\in\mathfrak{v}$,

[TABLE]

and therefore $[x,y]=0$, so that $\mathfrak{n}$ would be abelian. ∎

We show next that the existence of a KY tensor on a 2-step nilpotent Lie algebra $\mathfrak{n}$ imposes strong restrictions on $\mathfrak{n}$. Indeed,

Theorem 3.3**.**

Let $\mathfrak{n}$ be a $2$-step nilpotent Lie algebra with an inner product $\langle\cdotp,\cdotp\rangle$. If $T:\mathfrak{n}\to\mathfrak{n}$ is a KY tensor on $\mathfrak{n}$ then:

1. $(a)$

$\mathfrak{n}$* is isometrically isomorphic to a direct product of ideals $\mathfrak{n}=\mathfrak{n}_{1}\times\mathfrak{n}_{2}$, where $T|_{\mathfrak{n}_{1}}=0$, $\mathfrak{n}_{2}$ is $T$-invariant and $T|_{\mathfrak{n}_{2}}$ is an invertible KY tensor on $\mathfrak{n}_{2}$;* 2. $(b)$

$T$* is parallel if and only if $\mathfrak{n}_{2}$ is abelian. Moreover, if $\mathfrak{z}=\mathfrak{n}^{\prime}$ then $T=0$.*

Proof.

Since $T$ is skew-symmetric we have an orthogonal decomposition $\mathfrak{n}=\ker T\oplus\operatorname{Im}T$ into $T$-invariant subspaces. Recall that we also have another orthogonal decomposition $\mathfrak{n}=\mathfrak{z}\oplus\mathfrak{v}$, where $\mathfrak{z}$ is the center of $\mathfrak{n}$. According to Theorem 3.1 both $\mathfrak{z}$ and $\mathfrak{v}$ are $T$-invariant, therefore, if $x\in\ker T$ is decomposed as $x=x_{1}+x_{2}$ with $x_{1}\in\mathfrak{z}$ and $x_{2}\in\mathfrak{v}$, then $0=Tx=Tx_{1}+Tx_{2}$ with $Tx_{1}\in\mathfrak{z}$ and $Tx_{2}\in\mathfrak{v}$. Hence, $Tx_{1}=0$ and $Tx_{2}=0$, so that $\ker T=(\ker T\cap\mathfrak{z})\oplus(\ker T\cap\mathfrak{v})$. In the same way it is shown that $\operatorname{Im}T=(\operatorname{Im}T\cap\mathfrak{z})\oplus(\operatorname{Im}T\cap\mathfrak{v})$.

We will show next that both $\ker T$ and $\operatorname{Im}T$ are ideals of $\mathfrak{n}$. Indeed:

• •

if $x\in\ker T,\,y\in\mathfrak{n}$, then $x=x_{1}+x_{2}$, $y=y_{1}+y_{2}$ with $x_{1}\in\ker T\cap\mathfrak{z}$, $x_{2}\in\ker T\cap\mathfrak{v}$ and $y_{1}\in\mathfrak{z}$, $y_{2}\in\mathfrak{v}$. Hence, $T[x,y]=T[x_{2},y_{2}]=3[Tx_{2},y_{2}]=0$, using (6). Thus $\ker T$ is an ideal of $\mathfrak{n}$.

• •

if $Tx\in\operatorname{Im}T$ and $y\in\mathfrak{n}$, then $x=x_{1}+x_{2}$ and $y=y_{1}+y_{2}$ with $x_{1},y_{1}\in\mathfrak{z}$ and $x_{2},y_{2}\in\mathfrak{v}$. Then we have $[Tx,y]=[Tx_{2},y_{2}]=\frac{1}{3}T[x_{2},y_{2}]\in\operatorname{Im}T$, thus $\operatorname{Im}T$ is an ideal of $\mathfrak{n}$.

Therefore, setting $\mathfrak{n}_{1}:=\ker T$ and $\mathfrak{n}_{2}:=\operatorname{Im}T$ we obtain $\mathfrak{n}=\mathfrak{n}_{1}\times\mathfrak{n}_{2}$. Clearly, $T|_{\mathfrak{n}_{1}}=0$, $\mathfrak{n}_{2}$ is $T$-invariant and $T|_{\mathfrak{n}_{2}}$ is invertible. Moreover, it follows from Lemma 2.2 that $T|_{\mathfrak{n}_{2}}$ is a KY tensor on $\mathfrak{n}_{2}$. This proves $(a)$.

It is clear that $\mathfrak{n}_{2}$ is either abelian or $2$-step nilpotent. If $\mathfrak{n}_{2}$ is $2$-step nilpotent, it follows from Theorem 2.3 that $T|_{\mathfrak{n}_{2}}$ is not parallel, since it is invertible. Now, if $\mathfrak{n}_{2}=\operatorname{Im}T$ is abelian, then this means that $\operatorname{Im}T\subseteq\mathfrak{z}$. Hence $T|_{\mathfrak{v}}=0$ and it follows from (6) that $T|_{\mathfrak{n}^{\prime}}=0$. It is easy to verify that these conditions imply that $T$ is parallel, and $T=0$ if $\mathfrak{n}^{\prime}=\mathfrak{z}$. Therefore $(b)$ follows. ∎

Remark 3.4*.*

Note that when the KY tensor $T$ is invertible the KY 2-form $\omega$ associated to $T$, $\omega(X,Y)=g(TX,Y)$, is non-degenerate.

As a consequence of Theorem 3.3, we may restrict ourselves to the study of invertible KY tensors on 2-step nilpotent Lie groups. A large number of such examples appear in complex 2-step nilpotent Lie groups, as the following result shows:

Proposition 3.5**.**

Let $N$ be a complex 2-step nilpotent Lie group equipped with a left invariant Hermitian metric $g$. Then $(N,g)$ admits a non-parallel invertible KY tensor.

Proof.

Let $J$ denote the associated bi-invariant complex structure on $N$. Therefore the restriction of $J$ to $\mathfrak{n}$, the Lie algebra of $N$, satisfies: $J:\mathfrak{n}\to\mathfrak{n}$, $J^{2}=-\operatorname{Id}$ and $J\operatorname{ad}_{x}=\operatorname{ad}_{x}J$ for any $x\in\mathfrak{n}$, i.e., $J[x,y]=[x,Jy]=[Jx,y]$ for any $x,y\in\mathfrak{n}$.

Let $\langle\cdotp,\cdotp\rangle:=g|_{\mathfrak{n}}$ denote the induced inner product on $\mathfrak{n}$. If $\mathfrak{z}$ denotes the center of $\mathfrak{n}$, let $\mathfrak{v}:=\mathfrak{z}^{\perp}$, so that $\mathfrak{n}=\mathfrak{z}\oplus\mathfrak{v}$, orthogonal sum. Note that both $\mathfrak{z}$ and $\mathfrak{v}$ are $J$-invariant, as well as the commutator ideal $\mathfrak{n}^{\prime}=[\mathfrak{n},\mathfrak{n}]$.

Let us define a skew-symmetric isomorphism $T$ of $\mathfrak{n}$ in the following way: if we decompose $\mathfrak{z}=\mathfrak{n}^{\prime}\oplus\mathfrak{a}$ as an orthogonal sum, we set

[TABLE]

where $T_{0}:\mathfrak{a}\to\mathfrak{a}$ is any skew-symmetric isomorphism of $\mathfrak{a}$. It follows from Theorem 3.1 that $T$ is a left invariant Killing-Yano tensor on $N$. Moreover, according to Theorem 2.3, $T$ is not parallel. ∎

Remark 3.6*.*

We point out that the classification of complex 2-step nilpotent Lie algebras, up to isomorphism, is not known; furthermore, this is considered to be a wild problem in the literature (see for instance [12], where several partial results are collected).

The next theorem will show that the complex 2-step nilpotent Lie groups exhaust the family of 2-step nilpotent Lie groups with left invariant metric admitting invertible KY tensors.

Theorem 3.7**.**

If a 2-step nilpotent Lie group $N$ equipped with a left invariant metric $g$ admits an invertible left invariant KY tensor then $N$ is a complex Lie group. Moreover, $g$ is Hermitian with respect to this complex structure.

Proof.

Let $N$ be a $2$-step nilpotent Lie group equipped with a left invariant metric $g$ and a left invariant Killing-Yano tensor $T$. We denote by $\langle\cdotp,\cdotp\rangle:=g|_{\mathfrak{n}}$ the induced inner product on $\mathfrak{n}$, and we also denote by $T$ the restriction $T|_{\mathfrak{n}}$ of $T$ to $\mathfrak{n}$. Therefore, according to Theorem 3.1, $T$ preserves the center $\mathfrak{z}$ of $\mathfrak{n}$ and (6) holds.

Let us consider the orthogonal decomposition $\mathfrak{n}=\mathfrak{a}\oplus\mathfrak{n}^{\prime}\oplus\mathfrak{v}$, where $\mathfrak{z}=\mathfrak{a}\oplus\mathfrak{n}^{\prime}$ and $\mathfrak{v}=\mathfrak{z}^{\perp}$. Note that due to (6), $T$ preserves each of these subspaces, since $T$ is skew-symmetric. Therefore, there exists an orthonormal basis $\{e_{1},\ldots,e_{n},f_{1},\ldots,f_{n}\}$ of $\mathfrak{v}$ such that

[TABLE]

for some non-zero $a_{i}\in\mathbb{R}$. Moreover, interchanging $e_{i}$ with $f_{i}$ if necessary, we may assume that $a_{i}>0$ for all $i$ and, after a further reordering of the basis, we may assume also that $0<a_{1}\leq a_{2}\leq\cdots\leq a_{n}$. Let us denote $\mathfrak{v}_{i}=\text{span}\{e_{i},f_{i}\}$, so that $\mathfrak{v}=\oplus_{i}\mathfrak{v}_{i}$.

It follows from (6) that $[\mathfrak{v}_{i},\mathfrak{v}_{i}]=0$. Indeed, $T[e_{i},f_{i}]=3[a_{i}f_{i},f_{i}]=0$, hence $[e_{i},f_{i}]=0$. In particular, for each $i$ there exists $j$ such that $[\mathfrak{v}_{i},\mathfrak{v}_{j}]\neq 0$, since otherwise $\mathfrak{v}_{i}\subset\mathfrak{z}$. Let $(i,j)$ be a pair of indices such that $[\mathfrak{v}_{i},\mathfrak{v}_{j}]\neq 0$. Then at least one of $[e_{i},e_{j}]$, $[e_{i},f_{j}]$, $[f_{i},e_{j}]$ and $[f_{i},f_{j}]$ is non-zero. Then, according to (6), we have that

[TABLE]

Since $T$ is invertible, it follows that all four Lie brackets are non-zero. Furthermore, we have that

[TABLE]

As the constants are positive, we obtain $a_{i}=a_{j}$. Hence, $[e_{i},e_{j}]=-[f_{i},f_{j}]$, $[e_{i},f_{j}]=[f_{i},e_{j}]$ and $[\mathfrak{v}_{i},\mathfrak{v}_{j}]=\text{span}\{[e_{i},e_{j}],[e_{i},f_{j}]\}$, with $\dim[\mathfrak{v}_{i},\mathfrak{v}_{j}]=2$, and $[\mathfrak{v}_{i},\mathfrak{v}_{j}]$ is $T$-invariant.

Let us define a series of integers $\{i_{j}:j=1,\ldots,r\}$ in the following way:

[TABLE]

We define now the following subspaces of $\mathfrak{v}$:

[TABLE]

Clearly $\mathfrak{v}=\oplus_{j}W_{j}$ and, moreover, $[W_{j},W_{k}]=0$ if $j\neq k$. It then follows that the Lie subalgebras $\mathfrak{n}_{j}$ of $\mathfrak{n}$ defined by $\mathfrak{n}_{j}=W_{j}\oplus[W_{j},W_{j}]$ are in fact ideals of $\mathfrak{n}$.

Next note that $T^{2}|_{W_{j}}=-a_{i_{j}}^{2}\operatorname{Id}$, while $T^{2}|_{[W_{j},W_{j}]}=-9a_{i_{j}}^{2}\operatorname{Id}$, according to (6). As a consequence, since $\mathfrak{n}^{\prime}=[\mathfrak{v},\mathfrak{v}]$, we have that $\mathfrak{n}^{\prime}=\oplus_{j}[W_{j},W_{j}]$ and furthermore,

[TABLE]

a direct product of ideals. In order to show that $\mathfrak{n}$ carries a bi-invariant complex structure, set $J:\mathfrak{n}\to\mathfrak{n}$ as follows:

1. (i)

$J|_{\mathfrak{a}}$ is any almost complex structure $J_{0}$ on $\mathfrak{a}$ compatible with $\langle\cdotp,\cdotp\rangle$; 2. (ii)

$J|_{W_{j}}=\frac{1}{a_{i_{j}}}T|_{W_{j}}$; 3. (iii)

$J|_{[W_{j},W_{j}]}=\frac{1}{3a_{i_{j}}}T|_{[W_{j},W_{j}]}$.

Clearly, $J^{2}=-\operatorname{Id}$. In order to verify that it is bi-invariant, we need only check that $J[x,y]=[Jx,y]$ for any $x,y\in W_{j}$ for some $j$. Indeed,

[TABLE]

It is clear from the construction that $J$ is skew-symmetric with respect to $\langle\cdotp,\cdotp\rangle$. ∎

Remark 3.8*.*

It follows from the proof of Theorem 3.7 that $\mathfrak{n}$ can be decomposed as $\mathfrak{n}=\mathfrak{a}\times\mathfrak{n}_{1}\times\cdots\times\mathfrak{n}_{r}$, an orthogonal product of $T$-invariant ideals, with $\mathfrak{a}$ abelian and $\mathfrak{n}_{i}$ 2-step nilpotent. As a consequence, if $\mathfrak{n}$ is irreducible (i.e., it cannot be decomposed as an orthogonal direct product of ideals) then $\mathfrak{a}=0$, $r=1$, and the endomorphisms $T$ and $J$ of $\mathfrak{n}$ can be written, according to the orthogonal decomposition $\mathfrak{n}=\mathfrak{z}\oplus\mathfrak{v}$, as

[TABLE]

in some orthonormal basis of $\mathfrak{z}$ and $\mathfrak{v}$, for some $a\neq 0$. Here $\operatorname{Id}_{t}$ denotes the $t\times t$ identity matrix, $p=\dim\mathfrak{z}$ and $q=\dim\mathfrak{v}$.

Remark 3.9*.*

It was proved in [1] that any left invariant Hermitian metric on a unimodular complex Lie group is balanced, i.e. its fundamental 2-form is co-closed. It thus follows that the Hermitian metric $g$ in Theorem 3.7 is balanced with respect to the bi-invariant complex structure $J$.

Remark 3.10*.*

It is known that if $T$ is a KY tensor on a Riemannian manifold $(M,g)$ then $S:=-T^{2}$ is a Killing tensor on $M$ (see for instance [15, 9]). This means that $S$ is symmetric and $g((\nabla_{X}S)X,X)=0$ for any vector field $X$ on $M$. In contrast with the KY case, it was shown in [9] that many 2-step nilpotent Lie groups, not only the complex ones, admit Killing tensors for certain left invariant metrics.

Remark 3.11*.*

A symmetric $(1,1)$-tensor $B$ on a Riemannian manifold $(M,g)$ is called a Codazzi tensor if it satisfies $(\nabla_{X}B)Y=(\nabla_{Y}B)X$ for any vector fields $X,Y$ on $M$. In the case of a left invariant metric on a 2-step nilpotent Lie group, an analogous result to Theorem 3.1 holds. Indeed, a symmetric endomorphism of the Lie algebra $\mathfrak{n}=\mathfrak{v}\oplus\mathfrak{z}$ is a Codazzi tensor if and only if $B(\mathfrak{z})\subseteq\mathfrak{z}$ and $B[x,y]=[Bx,y]$ for all $x,y\in\mathfrak{n}$. However, in this case, it is easy to see that these conditions also imply that $B$ is parallel (compare [8, Proposition 2]).

4. Space of solutions

In this section we will show that for certain 2-step nilpotent complex Lie groups equipped with left invariant metrics, the space of Killing-Yano tensors is one-dimensional.

We will be concerned with 2-step nilpotent complex Lie groups arising from graphs. We recall briefly this construction, introduced in [7].

Let $\mathbb{G}=\mathbb{G}(S,E)$ be a simple, undirected graph with set of vertices $S=\{e_{1},\ldots,e_{n}\}$ and set of edges $E=\{z_{1},\ldots,z_{m}\}$, with $m\geq 1$. Let $\mathcal{B}=S\cup E$ and for a field $\mathbb{K}$ with characteristic different from 2, consider the Lie algebra $\mathfrak{n}_{\mathbb{G}}^{\mathbb{K}}$ whose underlying $\mathbb{K}$-vector space has $\mathcal{B}$ as a basis, and the Lie bracket is given by:

[TABLE]

Clearly, $\mathfrak{n}_{\mathbb{G}}^{\mathbb{K}}$ is a 2-step nilpotent Lie algebra, and it satisfies the following very strong condition:

[TABLE]

Many properties of $\mathfrak{n}_{\mathbb{G}}^{\mathbb{K}}$ can be deduced directly from $\mathbb{G}$. For instance, the center of $\mathfrak{n}_{\mathbb{G}}^{\mathbb{K}}$ is spanned by $E\cup\{\text{isolated vertices}\}$, and $\mathfrak{n}_{\mathbb{G}}^{\mathbb{K}}$ is indecomposable if and only if $\mathbb{G}$ is connected. Moreover, it was proved in [13] that given two graphs $\mathbb{G}_{1}$ and $\mathbb{G}_{2}$, the Lie algebras $\mathfrak{n}_{\mathbb{G}_{1}}^{\mathbb{K}}$ and $\mathfrak{n}_{\mathbb{G}_{2}}^{\mathbb{K}}$ are isomorphic if and only if $\mathbb{G}_{1}$ and $\mathbb{G}_{2}$ are isomorphic.

Example 1**.**

The 3-dimensional Heisenberg Lie algebra over $\mathbb{K}$ arises from the graph with two vertices joined by one edge. Moreover, it is clear from (8) that the $(2n+1)$-dimensional Heisenberg Lie algebra arises from a graph if and only if $n=1$.

We will study next KY tensors on 2-step nilpotent complex Lie algebras arising from graphs. So, let $\mathbb{G}=\mathbb{G}(S,E)$ be a graph as before, and let $\mathfrak{n}_{\mathbb{G}}^{\mathbb{C}}$ be the complex 2-step nilpotent Lie algebra arising from $\mathbb{G}$. When considered as a real Lie algebra, it will be denoted simply by $\mathfrak{n}_{\mathbb{G}}$. It has a real basis $\mathcal{B}^{\prime}=\{e_{1},f_{1},\ldots,e_{n},f_{n}\}\cup\{z_{1},w_{1},\ldots,z_{r},w_{r}\}$ such that its natural bi-invariant complex structure $J$ is given by $Je_{i}=f_{i},\,Jz_{i}=w_{i}$. The Lie bracket on $\mathfrak{n}_{\mathbb{G}}$ satisfies

[TABLE]

Note that $\mathfrak{n}_{\mathbb{G}}$ is not a real 2-step nilpotent Lie algebra arising from a graph, since it does not satisfy (8).

Let us define an inner product $\langle\cdotp,\cdotp\rangle$ on $\mathfrak{n}_{\mathbb{G}}$ by declaring the basis $\mathcal{B}^{\prime}$ to be orthonormal. If $\mathbb{G}$ has no isolated vertices then the center $\mathfrak{z}$ of $\mathfrak{n}_{\mathbb{G}}$ is $\text{span}\{z_{1},w_{1},\ldots,z_{r},w_{r}\}$, and its orthogonal complement $\mathfrak{v}$ is $\text{span}\{e_{1},f_{1},\ldots,e_{n},f_{n}\}$.

According to Proposition 3.3, $(\mathfrak{n}_{\mathbb{G}},\langle\cdotp,\cdotp\rangle)$ admits KY tensors. In the next result we determine the dimension of the space of KY tensors on $(\mathfrak{n}_{\mathbb{G}},\langle\cdotp,\cdotp\rangle)$ when $\mathbb{G}$ is connected.

Theorem 4.1**.**

When the graph $\mathbb{G}$ is connected, the space of KY tensors on $(\mathfrak{n}_{\mathbb{G}},\langle\cdotp,\cdotp\rangle)$ has dimension 1.

Proof.

Let $T$ be a KY tensor on $(\mathfrak{n}_{\mathbb{G}},\langle\cdotp,\cdotp\rangle)$. According to Theorem 3.1, $T$ preserves both $\mathfrak{z}$ and $\mathfrak{v}$ and, moreover, $T[x,y]=3[Tx,y]=3[x,Ty]$ for any $x,y\in\mathfrak{v}$. Since $\{e_{1},f_{1},\ldots,e_{n},f_{n}\}$ is an orthonormal basis of $\mathfrak{v}$, we have that

[TABLE]

Let us fix two indices $i,k\in\{1,\ldots,n\}$ with $i\neq k$. We will show that

[TABLE]

First, let us assume that $[e_{i},e_{k}]\neq 0$. It follows from $[Te_{i},e_{i}]=0$ that

[TABLE]

The first sum in the left-hand side lies in $\text{span}\{z_{1},\ldots,z_{r}\}$, whereas the second sum lies in $\text{span}\{w_{1},\ldots,w_{r}\}$. Therefore, each sum vanishes and, taking (8) into account, we have that $\langle Te_{i},e_{k}\rangle[e_{k},e_{i}]=0$ and $\langle Te_{i},f_{k}\rangle[f_{k},e_{i}]=0$, which imply $\langle Te_{i},e_{k}\rangle=\langle Te_{i},f_{k}\rangle=0$. Beginning with $[Tf_{i},f_{i}]=0$, the same computation gives $\langle Tf_{i},e_{k}\rangle=\langle Tf_{i},f_{k}\rangle=0$.

Let us now assume that $[e_{i},e_{k}]=0$. There exists $l\neq i$ such that $[e_{k},e_{l}]\neq 0$, since otherwise $e_{k}$ would be an isolated vertex of $\mathbb{G}$. Since $[Te_{i},e_{l}]=[e_{i},Te_{l}]$, and using (9), we obtain

[TABLE]

The first sum in each side lies in $\text{span}\{z_{1},\ldots,z_{r}\}$, whereas the second sum in each side lies in $\text{span}\{w_{1},\ldots,w_{r}\}$. Therefore we obtain that

[TABLE]

The non-zero element $[e_{k},e_{l}]$ appears in the left-hand side of (11), but if it appeared also in the right-hand side, according to (8) we would have that $\{k,l\}=\{i,j\}$ for some $j$, and this is impossible since both $k$ and $l$ are different from $i$. It follows that $\langle Te_{i},e_{k}\rangle=0$. The same reasoning can be applied in (12) (since also $[f_{k},e_{l}]\neq 0$), hence we obtain $\langle Te_{i},f_{k}\rangle=0$.

Analogously, beginning with $[Tf_{i},f_{l}]=[f_{i},Tf_{l}]$, we prove that $\langle Tf_{i},e_{k}\rangle=\langle Tf_{i},f_{k}\rangle=0$.

Thus, (10) holds and since $T$ is skew-symmetric we have that $Te_{i}=a_{i}f_{i}$ and $Tf_{i}=-a_{i}e_{i}$ for some $a_{i}\in\mathbb{R}$, $i=1,\ldots,n$. We will show next that $a_{i}=a_{j}$ for all $i,j$. Let us assume first that $[e_{i},e_{j}]=z_{k}$. It follows from (6) that $T[e_{i},e_{j}]=3[Te_{i},e_{j}]=3[e_{i},Te_{j}]$ and therefore we obtain that

[TABLE]

Therefore, $a_{i}=a_{j}$. If, on the other hand, $[e_{i},e_{j}]=0$, we can choose a sequence of indices $i=l_{0}<l_{1}<\cdots<l_{s-1}<l_{s}=j$ such that $[e_{l_{t}},e_{l_{t+1}}]\neq 0$, since $\mathbb{G}$ is connected. As a consequence, we have that $a_{i}=a_{l_{1}}=\cdots=a_{l_{s-1}}=a_{j}$. Setting $a:=a_{i}$ for any $i$, the KY tensor $T$ is given by

[TABLE]

Hence the space of KY tensors on $(\mathfrak{n}_{\mathbb{G}},\langle\cdotp,\cdotp\rangle)$ has dimension 1. ∎

Corollary 4.2**.**

When $\mathbb{G}$ is connected, any non-zero Killing-Yano tensor on $(\mathfrak{n}_{\mathbb{G}},\langle\cdotp,\cdotp\rangle)$ is invertible, thus non-parallel.

Note that in the proof of Theorem 4.1, the main properties used were the skew-symmetry of $T$ and equation (6). The constant appearing in this formula is not relevant for the proof, and as a consequence we obtain the following result (compare with [11, Section 5.1]).

Proposition 4.3**.**

If $I$ is a bi-invariant complex structure on $\mathfrak{n}_{\mathbb{G}}$ which is Hermitian with respect to $\langle\cdotp,\cdotp\rangle$, then $I=\pm J$.

We learnt recently that V. del Barco and A. Moroianu obtained in [10] the uniqueness, up to sign, of the orthogonal bi-invariant complex structure on a 2-step nilpotent metric Lie group which is de Rham irreducible.

Remark 4.4*.*

As mentioned in Example 1, the only complex Heisenberg Lie algebra $\mathfrak{h}_{2n+1}(\mathbb{C})$ which can be obtained from a graph is $\mathfrak{h}_{3}(\mathbb{C})$. However, it is easy to see that for $n\geq 2$, $\mathfrak{h}_{2n+1}(\mathbb{C})$ with its canonical $H$-type Hermitian metric has a space of KY tensors of dimension 1.