tcb@breakable

# Complex structures on stratified Lie algebras

Junze Zhang Electronic address<EMAIL_ADDRESS>School of Mathematics
and Statistics, University of New South Wales, Sydney

###### Abstract

This paper investigates some properties of complex structures on Lie algebras.
In particular, we focus on nilpotent complex structures that are characterized
by a suitable $J$-invariant ascending or descending central series
$\mathfrak{d}^{j}$ and $\mathfrak{d}_{j}$ respectively. In this article, we
introduce a new descending series $\mathfrak{p}_{j}$ and use it to give proof
of a new characterization of nilpotent complex structures. We examine also
whether nilpotent complex structures on stratified Lie algebras preserve the
strata. We find that there exists a $J$-invariant stratification on a step $2$
nilpotent Lie algebra with a complex structure.

##  1 Introduction

In recent years, complex structures on nilpotent Lie algebras have been shown
to be very useful for understanding some geometric and algebraic properties of
nilmanifolds. In [3] and [4], Cordero, Fernández, Gray and Ugarte introduced
nilpotent complex structures, studied $6$ dimensional nilpotent Lie algebras
with nilpotent complex structures, and provided a classification. Since the
ascending central series is not necessarily $J$-invariant, they introduced a
$J$-invariant ascending central series to characterize nilpotent complex
structures. More recently, Latorre, Ugarte and Villacampa defined the space of
nilpotent complex structures on nilpotent Lie algebras and further studied
complex structures on nilpotent Lie algebras with one dimensional center [10],
[11]. They also provided a theorem describing the ascending central series of
$8$ dimensional nilpotent Lie algebras with complex structures. In [7], Gao,
Zhao and Zheng studied the relation between the step of a nilpotent Lie
algebra and the smallest integer $j_{0}$ such that the $J$-invariant ascending
central series stops. Furthermore, they introduced a $J$-invariant descending
central series, which is another tool to characterize nilpotent complex
structures. These papers use the language of differential forms to
characterize nilpotent complex structures. Our proofs here in this paper are
purely Lie algebraic.

Let $G$ be a Lie group and $\mathfrak{g}\cong T_{e}G$ be its Lie algebra,
which we always assume to be real, unless otherwise stated. A linear
isomorphism $J:TG\rightarrow TG$ is an almost complex structure if $J^{2}=-I.$
By the Newlander–Nirenberg Theorem $\cite[cite]{[\@@bibref{}{MR88770}{}{}]},$
an almost complex structure $J$ corresponds to a left invariant complex
structure on $G$ if and only if

$[J_{e}X,J_{e}Y]-[X,Y]-J_{e}([J_{e}X,Y]+[X,J_{e}Y])=0,$ (1)

for all $X,Y\in\mathfrak{g}.$ Since we are interested only on Lie algebras in
this paper, from now on, we will write $J$ for $J_{e}.$ We will refer to
$\eqref{eq:Niv}$ as the $\textit{Newlander--Nirenberg condition}.$

##  2 Complex structures on nilpotent Lie algebras

In this section, we consider some properties of the central series of
nilpotent Lie algebras with complex structures $J$ and define $J$-invariant
central series. We define nilpotent complex structures, and relate their
properties to the dimension of the center $\mathfrak{z}$ of a nilpotent Lie
algebra.

######  Definition 2.1.

($\text{See},\text{ for instance}$, e.g., [9]) Let $\mathfrak{g}$ be a Lie
algebra. The descending central series and ascending central series of
$\mathfrak{g}$ are denoted $\mathfrak{c}_{j}(\mathfrak{g})$ and
$\mathfrak{c}^{j}(\mathfrak{g})$ respectively, for all $j\geq 0,$ and defined
inductively by

$\displaystyle\mathfrak{c}_{0}(\mathfrak{g})=\mathfrak{g},\text{
}\mathfrak{c}_{j}(\mathfrak{g})=[\mathfrak{g},\mathfrak{c}_{j-1}(\mathfrak{g})];$
(2) $\displaystyle\mathfrak{c}^{0}(\mathfrak{g})=\\{0\\},\text{ }\text{
}\mathfrak{c}^{j}(\mathfrak{g})=\\{X\in\mathfrak{g}:[X,\mathfrak{g}]\subseteq\mathfrak{c}^{j-1}(\mathfrak{g})\\}.$
(3)

######  Remark 2.2.

(i) Notice that $\mathfrak{c}^{1}(\mathfrak{g})=\mathfrak{Z}(\mathfrak{g})$,
$\mathfrak{c}_{1}(\mathfrak{g})=[\mathfrak{g},\mathfrak{g}],$ and

$\displaystyle\mathfrak{c}^{j}(\mathfrak{g})/\mathfrak{c}^{j-1}(\mathfrak{g})=\mathfrak{Z}\left(\mathfrak{g}/\mathfrak{c}^{j-1}(\mathfrak{g})\right)\text{
}\text{for all }j\geq 1,$

where $\mathfrak{Z}(\cdot)$ means the center of a Lie algebra. Furthermore,
$\mathfrak{c}_{j}(\mathfrak{g})/\mathfrak{c}_{j+1}(\mathfrak{g})$ $\subseteq$
$\mathfrak{Z}\left(\mathfrak{n}/\right.$
$\left.\mathfrak{c}_{j+1}(\mathfrak{g})\right)$ for all $j\geq 0.$ It is clear
that $\mathfrak{c}^{j}(\mathfrak{g})$ and $\mathfrak{c}_{j}(\mathfrak{g})$ are
ideals of $\mathfrak{g}$ for all $j\geq 0$.

(ii) A Lie algebra $\mathfrak{g}$ is called $\textit{nilpotent of step }k$,
for some $k\in\mathbb{N},$ if $\mathfrak{c}_{k}(\mathfrak{g})=\\{0\\}$ and
$\mathfrak{c}_{k-1}(\mathfrak{g})\neq\\{0\\}.$ We will denote nilpotent Lie
algebras by $\mathfrak{n}$ in this paper. See, e.g., [8, Section 5.2] or
$\cite[cite]{[\@@bibref{}{MR1920389}{}{}]}.$

###  2.1 $J$-invariant central series and nilpotent complex structures

Following [3, Definition 1], we define the J-invariant ascending central
series $\mathfrak{d}^{j}$ for nilpotent Lie algebras and introduce nilpotent
complex structures on nilpotent Lie algebras. Furthermore, we recall the
definition of the J-invariant descending central series $\mathfrak{d}_{j}$.
[7, Defintion 2.7]

######  Definition 2.3.

Let $\mathfrak{n}$ be a Lie algebra with a complex structure $J.$ Define a
sequence of $J$-invariant ideals of $\mathfrak{n}$ by
$\mathfrak{d}^{0}=\\{0\\}$ and

$\mathfrak{d}^{j}=\\{X\in\mathfrak{n}:[X,\mathfrak{n}]\subseteq\mathfrak{d}^{j-1},[JX,\mathfrak{n}]\subseteq\mathfrak{d}^{j-1}\\}$
(4)

for all $j\geq 1.$ We call the sequence $\mathfrak{d}^{j}$ the ascending
J-invariant $\textit{central series}.$ The complex structure $J$ is called
$\textit{nilpotent of step }j_{0}$ if there exists $j_{0}\in\mathbb{N}$ such
that $\mathfrak{d}^{j_{0}}=\mathfrak{n}$ and
$\mathfrak{d}^{j_{0}-1}\subset\mathfrak{n}$.

We define inductively the $J$-invariant descending central series by

$\mathfrak{d}_{0}=\mathfrak{n},\text{ }\text{
}\mathfrak{d}_{j}=[\mathfrak{d}_{j-1},\mathfrak{n}]+J[\mathfrak{d}_{j-1},\mathfrak{n}]\text{
}\text{ }$ (5)

all for $j\geq 1.$

######  Remark 2.4.

(i) For the ascending $J$-invariant central series $\mathfrak{d}^{j}$,

$\displaystyle\mathfrak{d}^{j}/\mathfrak{d}^{j-1}=\mathfrak{Z}(\mathfrak{n}/\mathfrak{d}^{j-1})\cap
J\mathfrak{Z}(\mathfrak{n}/\mathfrak{d}^{j-1})\qquad\text{ for all }j\geq 1.$

In particular, $\mathfrak{d}^{1}=\mathfrak{z}\cap J\mathfrak{z},$ which is the
largest $J$-invariant subspace of $\mathfrak{z}$ and, if $J$ is nilpotent,
then $\mathfrak{d}^{1}\neq\\{0\\}$. The nilpotency of $J$ implies that the
ascending $J$-invariant central series $\mathfrak{d}^{j}$ of $\mathfrak{n}$ is
strictly increasing until $\mathfrak{d}^{j_{0}}=\mathfrak{n}$. Furthermore, if
$\mathfrak{n}$ is a step $k$ nilpotent Lie algebra with a nilpotent complex
structure $J$ of step $j_{0}$, then $k\leq
j_{0}\leq\frac{1}{2}\dim\mathfrak{n}.$ See, e.g., [3] and
$\cite[cite]{[\@@bibref{}{gao2020maximal}{}{}]}.$

(ii) By definition, if $\mathfrak{n}$ admits a nilpotent complex structure,
then $\mathfrak{n}$ is nilpotent.

(iii) For all $j\geq 0,$ it is clear that
$\mathfrak{c}_{j}(\mathfrak{n})+J\mathfrak{c}_{j}(\mathfrak{n})\subseteq\mathfrak{d}_{j}$;
Furthermore, $\mathfrak{d}_{j}\unlhd\mathfrak{n}$ and
$\mathfrak{d}^{j}\unlhd\mathfrak{n}$ where $\unlhd$ is the notation of ideal.

(iv) Let $\mathfrak{n}$ be a Lie algebra with a complex structure $J.$ Then
$J$ preserves all terms of $\mathfrak{c}^{j}(\mathfrak{n})$ if and only if
$\mathfrak{d}^{j}=\mathfrak{c}^{j}(\mathfrak{n})$ for all $j\geq 0$. [3,
Corollary 5] Similarly, $J$ preserves all terms of
$\mathfrak{c}_{j}(\mathfrak{n})$ if and only if
$\mathfrak{d}_{j}=\mathfrak{c}_{j}(\mathfrak{n})$ for all $j.$

The following lemma provides a connection between $J$-invariant ascending and
descending central series.

######  Lemma 2.5.

Let $\mathfrak{n}$ be a Lie algebra with a complex structure $J$. Suppose that
$J$ is nilpotent of step $j_{0}.$

(i) Then $\mathfrak{n}/\mathfrak{d}^{j_{0}-1}$ is Abelian. Conversely, if
there exists $j_{0}\in\mathbb{N}$ such that
$\mathfrak{n}/\mathfrak{d}^{j_{0}-1}$ is Abelian, then $J$ is nilpotent of
step at most $j_{0}.$

(ii) Then $\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$ for all $j\geq 0$.
Conversely, if there exists $j_{0}\in\mathbb{N}$ such that
$\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$ for all $j\geq 0,$ then $J$
is nilpotent of step at most $j_{0}.$

###### Proof.

For part (i), suppose that $J$ is nilpotent of step $j_{0}$. By definition,
$\mathfrak{d}^{j_{0}}=\mathfrak{n}$ and
$\mathfrak{d}^{j_{0}-1}\subset\mathfrak{n}$. Then

$\displaystyle\mathfrak{Z}(\mathfrak{n}/\mathfrak{d}^{j_{0}-1})\cap
J\mathfrak{Z}(\mathfrak{n}/\mathfrak{d}^{j_{0}-1})=\mathfrak{n}/\mathfrak{d}^{j_{0}-1}.$

It is obvious that
$\mathfrak{Z}(\mathfrak{n}/\mathfrak{d}^{j_{0}-1})=\mathfrak{n}/\mathfrak{d}^{j_{0}-1}.$
Hence $\mathfrak{n}/\mathfrak{d}^{j_{0}-1}$ is Abelian.

Conversely, suppose that there exists $j_{0}\in\mathbb{N}$ such that
$\mathfrak{n}/\mathfrak{d}^{j_{0}-1}$ is Abelian. Then
$\\{0\\}\neq\mathfrak{c}_{1}(\mathfrak{n})\subseteq\mathfrak{d}^{j_{0}-1}$.
For all $X\in\mathfrak{n},$

$\displaystyle[X,\mathfrak{n}]\subseteq\mathfrak{d}^{j_{0}-1}\text{ and
}[JX,\mathfrak{n}]\subseteq\mathfrak{d}^{j_{0}-1}.$

We deduce that $\mathfrak{n}=\mathfrak{d}^{j_{0}}$ and therefore $J$ is
nilpotent of step at most $j_{0}$.

For part (ii), assume that $J$ is nilpotent of step $j_{0}.$ By definition,
$\mathfrak{d}_{0}=\mathfrak{n}=\mathfrak{d}^{j_{0}}.$ Next, assume that
$\mathfrak{d}_{s-1}\subseteq\mathfrak{d}^{j_{0}-s+1}$ for some
$s\in\mathbb{N}.$ Then

$\displaystyle\mathfrak{d}_{s}$
$\displaystyle=[\mathfrak{d}_{s-1},\mathfrak{n}]+J[\mathfrak{d}_{s-1},\mathfrak{n}]$
$\displaystyle\subseteq[\mathfrak{d}^{j_{0}-s+1},\mathfrak{n}]+J[\mathfrak{d}^{j_{0}-s+1},\mathfrak{n}]$
$\displaystyle\subseteq\mathfrak{d}^{j_{0}-s}+J\mathfrak{d}^{j_{0}-s}=\mathfrak{d}^{j_{0}-s}.$

Hence by induction, $\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$ for all
$j\geq 0.$

Conversely, suppose that there exists $j_{0}\in\mathbb{N}$ such that
$\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$ for all $j\geq 0$. In
particular, $\mathfrak{d}_{1}\subseteq\mathfrak{d}^{j_{0}-1}.$ By definition,
$\mathfrak{c}_{1}(\mathfrak{n})\subseteq\mathfrak{d}_{1}.$ It follows that

$[\mathfrak{n}/\mathfrak{d}^{j_{0}-1},\mathfrak{n}/\mathfrak{d}^{j_{0}-1}]\subseteq[\mathfrak{n},\mathfrak{n}]+\mathfrak{d}^{j_{0}-1}=\mathfrak{c}_{1}(\mathfrak{n})+\mathfrak{d}^{j_{0}-1}\subseteq\mathfrak{d}_{1}+\mathfrak{d}^{j_{0}-1}\subseteq\mathfrak{d}_{j_{0}-1},$

and thus $\mathfrak{n}/\mathfrak{d}^{j_{0}-1}$ is Abelian. From Lemma 2.5, $J$
is nilpotent of step at most $j_{0}.$ ∎

######  Remark 2.6.

Under the condition of Lemma $\ref{9},$ if $J$ is nilpotent of step $j_{0},$
$\mathfrak{d}_{j_{0}-1}\subseteq\mathfrak{d}^{1}\subseteq\mathfrak{z}.$ Then
$\mathfrak{d}_{j_{0}-1}$ is Abelian. Furthermore, there exists
$j_{0}\in\mathbb{N}$ such that $\mathfrak{n}/\mathfrak{d}^{j_{0}-1}$ is
Abelian if and only if $\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$ for
all $j\geq 0.$ This is proved by induction as in the proof of Lemma 2.5.

######  Corollary 2.7.

Let $\mathfrak{n}$ be a step $k$ nilpotent Lie algebra with a complex
structure $J.$ Then $J$ is nilpotent of step $k$ if and only if
$\mathfrak{d}_{j}\subseteq\mathfrak{d}^{k-j}$ for all $j\geq 0$.

###### Proof.

Suppose that $J$ is nilpotent of step $k.$ By $\text{Lemma }\ref{9},$
$\mathfrak{d}_{j}\subseteq\mathfrak{d}^{k-j}$ for all $j\geq 0.$ Conversely,
assume that $\mathfrak{d}_{j}\subseteq\mathfrak{d}^{k-j}$ for all $j.$ Again
by Lemma 2.5, $J$ is nilpotent of step at most $k.$ Furthermore, it follows
that $\\{0\\}\neq\mathfrak{c}_{k-1}(\mathfrak{n})\subseteq\mathfrak{d}_{k-1}.$
Therefore $\mathfrak{d}_{k-1}\neq\\{0\\}$ and $J$ is nilpotent of step $k.$ ∎

######  Remark 2.8.

From Remark $\ref{r41},$ $J$ is nilpotent of step $k$ if and only if
$\mathfrak{n}/\mathfrak{d}^{k-1}$ is Abelian.

We introduce a new descending central series whose descending ‘rate’ is slower
than that of $\mathfrak{c}_{j}(\mathfrak{n})$ but faster than that of
$\mathfrak{d}_{j}.$

######  Definition 2.9.

Let $J$ be a complex structure on a Lie algebra $\mathfrak{n}.$ We define the
sequence $\mathfrak{p}_{j}$ inductively by

$\mathfrak{p}_{0}=\mathfrak{n}\text{ and
}\mathfrak{p}_{j}=[\mathfrak{p}_{j-1},\mathfrak{n}]+[J\mathfrak{p}_{j-1},\mathfrak{n}]\text{
for all }j\geq 1.$ (6)

######  Remark 2.10.

It is clear that $\mathfrak{p}_{j+1}\subseteq\mathfrak{p}_{j}$ for all $j\geq
0.$ Furthermore, $\mathfrak{p}_{j}\unlhd\mathfrak{n}$ since
$[\mathfrak{p}_{j},\mathfrak{n}]\subseteq\mathfrak{p}_{j+1}\subseteq\mathfrak{p}_{j}$
for all $j\geq 0$.

######  Lemma 2.11.

Let $\mathfrak{n}$ be a Lie algebra with a complex structure $J$. Then
$\mathfrak{c}_{j}(\mathfrak{n})\subseteq\mathfrak{p}_{j}$ for all $j\geq 0$.
Furthermore, $\mathfrak{p}_{j}\subseteq\mathfrak{d}_{j}$ and
$J\mathfrak{p}_{j}\subseteq\mathfrak{d}_{j}$ for all $j\geq 0$.

###### Proof.

By definition, $\mathfrak{c}_{0}(\mathfrak{n})=\mathfrak{n}=\mathfrak{p}_{0}.$
It follows, by induction, that
$\mathfrak{c}_{j}(\mathfrak{n})\subseteq\mathfrak{p}_{j}$ for all $j\geq 0$.
Using $\eqref{eq:00},$
$[\mathfrak{d}_{j-1},\mathfrak{n}]\subseteq\mathfrak{d}_{j}$. By definition,
$\mathfrak{p}_{0}=\mathfrak{n}=\mathfrak{d}_{0}$ and
$J\mathfrak{p}_{0}=J\mathfrak{n}=\mathfrak{n}=\mathfrak{d}_{0}$. Next, suppose
that $\mathfrak{p}_{s}\subseteq\mathfrak{d}_{s}$ and
$J\mathfrak{p}_{s}\subseteq\mathfrak{d}_{s}$ for some $s\in\mathbb{N}.$ Then
by $\eqref{eq:p},$

$\displaystyle\mathfrak{p}_{s+1}=[\mathfrak{p}_{s},\mathfrak{n}]+[J\mathfrak{p}_{s},\mathfrak{n}]\subseteq[\mathfrak{d}_{s},\mathfrak{n}]\subseteq\mathfrak{d}_{s+1}\text{
and }J\mathfrak{p}_{s+1}\subseteq
J[\mathfrak{d}_{s},\mathfrak{n}]\subseteq\mathfrak{d}_{s+1}$

By induction, $\mathfrak{p}_{j}\subseteq\mathfrak{d}_{j}$ and
$J\mathfrak{p}_{j}\subseteq\mathfrak{d}_{j}$ for all $j\geq 0.$ ∎

######  Remark 2.12.

(i) Notice that
$\mathfrak{p}_{j}/\mathfrak{p}_{j+1}\subseteq\mathfrak{Z}\left(\mathfrak{n}/\mathfrak{p}_{j+1}\right)$
for all $j\geq 0.$ Indeed, for all $P\in\mathfrak{p}_{j}$ and
$Y\in\mathfrak{n},$ since $[P,Y]\subseteq\mathfrak{p}_{j+1},$ it is enough to
deduce

$[P+\mathfrak{p}_{j+1},Y+\mathfrak{p}_{j+1}]=[P,Y]+\mathfrak{p}_{j+1}\subseteq\mathfrak{p}_{j+1}.$

Hence
$\mathfrak{p}_{j}/\mathfrak{p}_{j+1}\subseteq\mathfrak{Z}\left(\mathfrak{n}/\mathfrak{p}_{j+1}\right)$.

(ii) By $\text{Lemma }\ref{p},$
$\mathfrak{p}_{j}+J\mathfrak{p}_{j}\subseteq\mathfrak{d}_{j}$ for all $j\geq
0$. We show that $\mathfrak{p}_{j}+J\mathfrak{p}_{j}\unlhd\mathfrak{n}$ for
all $j\geq 0.$ Indeed, for all $P,P^{\prime}\in\mathfrak{p}_{j},$

$\underbrace{[P+JP^{\prime},\mathfrak{n}]}_{\text{$\subseteq[\mathfrak{p}_{j}+J\mathfrak{p}_{j},\mathfrak{n}]$}}\subseteq\underbrace{[P,\mathfrak{n}]}_{\text{$\subseteq\mathfrak{p}_{j+1}$}}+\underbrace{[JP^{\prime},\mathfrak{n}]}_{\text{$\subseteq\mathfrak{p}_{j+1}$}}\subseteq\mathfrak{p}_{j+1}\subseteq\mathfrak{p}_{j+1}+J\mathfrak{p}_{j+1}\subseteq\mathfrak{p}_{j}+J\mathfrak{p}_{j}.$

Hence $\mathfrak{p}_{j}+J\mathfrak{p}_{j}\unlhd\mathfrak{n}$. From part (ii),
we can show that $\mathfrak{p}_{j}+J\mathfrak{p}_{j}$ is a $J$-invariant
descending central series. Indeed, for all
$T=P+JP^{\prime}\in\mathfrak{p}_{j}+J\mathfrak{p}_{j}$ and $Y\in\mathfrak{n},$

$\displaystyle[T+\mathfrak{p}_{j+1}+J\mathfrak{p}_{j+1},Y+\mathfrak{p}_{j+1}+J\mathfrak{p}_{j+1}]\subseteq[T,Y]+\mathfrak{p}_{j+1}+J\mathfrak{p}_{j+1}\subseteq\mathfrak{p}_{j+1}+J\mathfrak{p}_{j+1}.$

######  Theorem 2.13.

Let $\mathfrak{n}$ be a Lie algebra with a complex structure $J.$ The
following are equivalent:

(i) $J$ is nilpotent of step $j_{0}$;

(ii) $\mathfrak{p}_{j_{0}}=\\{0\\}$ and $\mathfrak{p}_{j_{0}-1}\neq\\{0\\};$

(iii) $\mathfrak{d}_{j_{0}}=\\{0\\}$ and $\mathfrak{d}_{j_{0}-1}\neq\\{0\\}$.

###### Proof.

We first show that (i) and (ii) are equivalent. Assume that $J$ is nilpotent
of step $j_{0}.$ From Lemma 2.5 part (ii),
$\mathfrak{d}_{j_{0}-1}\subseteq\mathfrak{d}^{1}.$ Hence by $\text{Lemma
}\ref{p},$

$\mathfrak{p}_{j_{0}}\subseteq[\mathfrak{d}_{j_{0}-1},\mathfrak{n}]\subseteq[\mathfrak{d}^{1},\mathfrak{n}]=\\{0\\}.$

Thus $\mathfrak{p}_{j_{0}}=\\{0\\}$. Assume, by contradiction, that
$\mathfrak{p}_{j_{0}-1}=\\{0\\}.$ We show that
$\mathfrak{p}_{j_{0}-j-1}+J\mathfrak{p}_{j_{0}-j-1}\subseteq\mathfrak{d}^{j}$
for all $j\geq 0$ by induction. By definition,
$\mathfrak{p}_{j_{0}-1}+J\mathfrak{p}_{j_{0}-1}=\\{0\\}=\mathfrak{d}^{0}.$
Next, suppose that
$\mathfrak{p}_{j_{0}-s-1}+J\mathfrak{p}_{j_{0}-s-1}\subseteq\mathfrak{d}^{s}$
for some $s\in\mathbb{N}.$ Then from Remark 2.12 part (ii),

$[\mathfrak{p}_{j_{0}-s-2}+J\mathfrak{p}_{j_{0}-s-2},\mathfrak{n}]\subseteq\mathfrak{p}_{j_{0}-s-1}+J\mathfrak{p}_{j_{0}-s-1}\subseteq\mathfrak{d}^{s}.$

This implies, using $\eqref{eq:1},$
$\mathfrak{p}_{j_{0}-s-2}+J\mathfrak{p}_{j_{0}-s-2}\subseteq\mathfrak{d}^{s+1}.$
By induction, $\mathfrak{p}_{j_{0}-j-1}+J\mathfrak{p}_{j_{0}-j-1}$
$\subseteq\mathfrak{d}^{j}$ for all $j\geq 0.$ In particular, let $j=j_{0}-1.$
Then $\mathfrak{n}\subseteq\mathfrak{d}^{j_{0}-1}$, which implies that $J$ is
nilpotent of step $j_{0}-1$ by definition. This is a contradiction. Therefore
$\mathfrak{p}_{j_{0}-1}\neq\\{0\\}.$

Conversely, suppose that $\mathfrak{p}_{j_{0}}=\\{0\\}$ and
$\mathfrak{p}_{j_{0}-1}\neq\\{0\\}$. We show that $J$ is nilpotent of step
$j_{0}.$ By definition,
$\mathfrak{p}_{j_{0}}+J\mathfrak{p}_{j_{0}}=\\{0\\}=\mathfrak{d}^{0}$. It
follows, by induction, that
$\mathfrak{p}_{j_{0}-j}+J\mathfrak{p}_{j_{0}-j}\subseteq\mathfrak{d}^{j}$ for
all $j\geq 0.$ Hence $\mathfrak{p}_{j_{0}-j}\subseteq\mathfrak{d}^{j}$. In
particular, let $j=j_{0}-1.$ Then

$\mathfrak{p}_{1}=[\mathfrak{n},\mathfrak{n}]\subseteq\mathfrak{d}^{j_{0}-1}\Rightarrow\mathfrak{n}/\mathfrak{d}^{j_{0}-1}\text{
is Abelian }.$

By $\text{Lemma }\ref{9},$ $J$ is nilpotent of step at most $j_{0}.$

We next show that $\mathfrak{d}^{j_{0}-1}\neq\mathfrak{n}$ by contradiction.
Assume, by contradiction, that $\mathfrak{n}=\mathfrak{d}^{j_{0}-1}.$ We show
that $\mathfrak{p}_{j-1}\subseteq\mathfrak{d}^{j_{0}-j}$ for all $j\geq 1$ by
induction. By definition,
$\mathfrak{p}_{0}=\mathfrak{n}=\mathfrak{d}^{j_{0}-1}.$ Next, suppose that
$\mathfrak{p}_{s-1}\subseteq\mathfrak{d}^{j_{0}-s}$ for some $s\in\mathbb{N}.$
Then

$\displaystyle\mathfrak{p}_{s}$
$\displaystyle=[\mathfrak{p}_{s-1},\mathfrak{n}]+[J\mathfrak{p}_{s-1},\mathfrak{n}]$
$\displaystyle\subseteq[\mathfrak{d}^{j_{0}-s},\mathfrak{n}]+[J\mathfrak{d}^{j_{0}-s},\mathfrak{n}]$
$\displaystyle\subseteq\mathfrak{d}^{j_{0}-s-1}.$

By induction, $\mathfrak{p}_{j-1}\subseteq\mathfrak{d}^{j_{0}-j}$ for all
$j\geq 1.$ In particular, let $j=j_{0}.$ We deduce that
$\mathfrak{p}_{j_{0}-1}\subseteq\mathfrak{d}^{0}=\\{0\\}.$ This implies that
$\mathfrak{p}_{j_{0}-1}=\\{0\\}$ which is a contradiction. Hence
$\mathfrak{d}^{j_{0}-1}\neq\mathfrak{n}.$ By definition, $J$ is nilpotent of
step $j_{0}.$

We now show (i) and (iii) are equivalent. Since $J$ is nilpotent of step
$j_{0},$ it follows, from Lemma 2.5 part (ii), that
$\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$ for all $j\geq 0.$ In
particular, let $j=j_{0}.$ By definition,
$\mathfrak{d}_{j_{0}}=\mathfrak{d}^{0}=\\{0\\}.$ We show that
$\mathfrak{d}_{j_{0}-1}\neq\\{0\\}.$ By Lemma 2.11,
$\\{0\\}\neq\mathfrak{p}_{j_{0}-1}+J\mathfrak{p}_{j_{0}-1}\subseteq\mathfrak{d}_{j_{0}-1}.$
Hence $\mathfrak{d}_{j_{0}-1}\neq\\{0\\}.$

Conversely, assume that $\mathfrak{d}_{j_{0}}=\\{0\\}$ and
$\mathfrak{d}_{j_{0}-1}\neq\\{0\\}$. By definition,
$[\mathfrak{d}_{j_{0}-1},\mathfrak{n}]\subseteq\mathfrak{d}_{j_{0}}=\\{0\\}.$
Hence, $\\{0\\}\neq\mathfrak{d}_{j_{0}-1}\subseteq\mathfrak{d}^{1}.$ Next,
assume that $\mathfrak{d}_{j_{0}-s}\subseteq\mathfrak{d}^{s}$ for some
$s\in\mathbb{N}.$ Then by definition,

$\displaystyle[\mathfrak{d}_{j_{0}-s-1},\mathfrak{n}]\subseteq\mathfrak{d}_{j_{0}-s}\subseteq\mathfrak{d}^{s}.$

By $\eqref{eq:1},$ $\mathfrak{d}_{j_{0}-s-1}\subseteq\mathfrak{d}^{s+1}.$ By
induction, $\mathfrak{d}_{j_{0}-j}\subseteq\mathfrak{d}^{j}$ for all $j\geq
0.$ Let $j=j_{0}.$ We find that
$\mathfrak{d}_{0}=\mathfrak{n}\subseteq\mathfrak{d}^{j_{0}}.$ Therefore
$\mathfrak{d}^{j_{0}}=\mathfrak{n}$ and $J$ is nilpotent of step at most
$j_{0}.$

We next show that $\mathfrak{d}^{j_{0}-1}\neq\mathfrak{n}.$ Suppose not, that
is, $\mathfrak{n}=\mathfrak{d}^{j_{0}-1}.$ By definition,
$\mathfrak{d}_{0}=\mathfrak{n}=\mathfrak{d}^{j_{0}-1}.$ It follows, by
induction, that $\mathfrak{d}_{j-1}\subseteq\mathfrak{d}^{j_{0}-j}$ for all
$j\geq 1$. Let $j=j_{0}.$ We find that
$\mathfrak{d}_{j_{0}-1}\subseteq\\{0\\}.$ This is a contradiction. Hence
$\mathfrak{d}^{j_{0}-1}\neq\mathfrak{n}$ and $J$ is nilpotent of step $j_{0}.$

Finally, since (i) is equivalent to both (ii) and (iii), we conclude that (ii)
and (iii) are equivalent. ∎

######  Remark 2.14.

Suppose that a Lie algebra $\mathfrak{n}$ admits a nilpotent complex structure
$J$ of step $j_{0}.$ Then

$\mathfrak{c}_{j}(\mathfrak{n})+J\mathfrak{c}_{j}(\mathfrak{n})\subseteq\mathfrak{p}_{j}+J\mathfrak{p}_{j}\subseteq\mathfrak{d}_{j}\subseteq\mathfrak{d}^{j_{0}-j}$
(7)

for all $j\geq 0$.

It is shown that, in $\cite[cite]{[\@@bibref{}{MR1665327}{}{}, \text{Corollary
}7]},$ if $\mathfrak{c}_{j}(\mathfrak{n})$ is $J$-invariant for all $j\geq 0,$
then $J$ is nilpotent. We will provide a different approach to this.

######  Corollary 2.15.

Let $\mathfrak{n}$ be a step $k$ nilpotent Lie algebra with a complex
structure $J.$ Suppose that all $\mathfrak{c}_{j}(\mathfrak{n})$ are
$J$-invariant. Then $\mathfrak{p}_{j}=\mathfrak{c}_{j}(\mathfrak{n})$ for all
$j\geq 0.$ Furthermore, $J$ is nilpotent of step $k$.

###### Proof.

Since all $\mathfrak{c}_{j}(\mathfrak{n})$ are $J$-invariant, by definition,
$\mathfrak{p}_{0}=\mathfrak{n}=\mathfrak{c}_{0}(\mathfrak{n}).$ We have, by
induction, that $\mathfrak{p}_{j}=\mathfrak{c}_{j}(\mathfrak{n})$ for all
$j\geq 0.$ Therefore $\mathfrak{p}_{k}=\mathfrak{c}_{k}(\mathfrak{n})=\\{0\\}$
and $\mathfrak{p}_{k-1}=\mathfrak{c}_{k-1}(\mathfrak{n})\neq\\{0\\}$. By
$\text{Theorem }\ref{14},$ $J$ is nilpotent of step $k.$ ∎

######  Corollary 2.16.

Let $\mathfrak{n}$ be a step $k$ nilpotent Lie algebra with a nilpotent
complex structure $J$ of step $k$. Suppose that
$\mathfrak{c}_{k-1}(\mathfrak{n})=\mathfrak{z}$. Then $\mathfrak{z}$ is
$J$-invariant.

###### Proof.

Since $J$ is nilpotent of step $k,$ by $\eqref{eq:ew}$,

$\mathfrak{z}+J\mathfrak{z}\subseteq\mathfrak{d}_{k-1}\subseteq\mathfrak{d}^{1}\subseteq\mathfrak{z}\Rightarrow[\mathfrak{z}+J\mathfrak{z},\mathfrak{n}]=\\{0\\}.$

Hence $J\mathfrak{z}=\mathfrak{z}.$ ∎

######  Corollary 2.17.

Let $\mathfrak{n}$ be a Lie algebra with a nilpotent complex structure $J$ of
step $j_{0}$. Then for all $j\geq 1,$ $\mathfrak{d}_{j_{0}-j}$ is not
contained in $\mathfrak{d}^{j-1}$.

###### Proof.

Since $J$ is nilpotent of step $j_{0},$ by $\text{Theorem }\ref{14},$
$\mathfrak{d}_{j_{0}-1}\neq\\{0\\}=\mathfrak{d}^{0}$. Hence
$\mathfrak{d}_{j_{0}-1}$ is not contained in $\mathfrak{d}^{0}.$ Next, suppose
that $\mathfrak{d}_{j_{0}-s+1}$ is not contained in $\mathfrak{d}^{s-2}$ for
some $\mathbb{N}\ni s\geq 2.$ We show that $\mathfrak{d}_{j_{0}-s}$ is not
contained in $\mathfrak{d}^{s-1}.$ Suppose not. That is,
$\mathfrak{d}_{j_{0}-s}\subseteq\mathfrak{d}^{s-1}.$ Then

$\displaystyle\mathfrak{d}_{j_{0}-s+1}$
$\displaystyle=[\mathfrak{d}_{j_{0}-s},\mathfrak{n}]+J[\mathfrak{d}_{j_{0}-s},\mathfrak{n}]$
$\displaystyle\subseteq[\mathfrak{d}^{s-1},\mathfrak{n}]+J[\mathfrak{d}^{s-1},\mathfrak{n}]\subseteq\mathfrak{d}^{s-2}.$

It follows that $\mathfrak{d}_{j_{0}-s+1}\subseteq\mathfrak{d}^{s-2}.$ This is
a contradiction. Hence $\mathfrak{d}_{j_{0}-s}$ is not contained in
$\mathfrak{d}^{s-1}.$ By induction, for all $j\geq 1,$
$\mathfrak{d}_{j_{0}-j}$ is not contain in $\mathfrak{d}^{j-1}$. ∎

We investigate the possible range of $\dim\mathfrak{z}$ for a Lie algebra
$\mathfrak{n}$ with a nilpotent complex structure $J.$

######  Proposition 2.18.

Let $\mathfrak{n}$ be a non-Abelian Lie algebra of dimension $2n$ with a
nilpotent complex structure $J.$ Then $2\leq\dim\mathfrak{z}\leq 2n-2.$

###### Proof.

Recall that $\mathfrak{d}^{1}=\mathfrak{z}\cap J\mathfrak{z}$, which is the
largest $J$-invariant subspace of $\mathfrak{z}$. Since $J$ is nilpotent, it
is clear that $\mathfrak{d}^{1}\neq\\{0\\}.$ Furthermore, since
$\mathfrak{d}^{1}$ is $J$-invariant, it follows that
$2\leq\dim\mathfrak{d}^{1}\leq\dim\mathfrak{z}.$ Then the lower bound of
$\dim\mathfrak{z}$ is $2.$

Next, we show that the upper bound of $\dim\mathfrak{z}$ is $2n-2.$ Since
$\mathfrak{n}$ is non-Abelian, it is possible to find $X,Y\in\mathfrak{n}$
such that $0\neq[X,Y]\in\mathfrak{c}_{1}(\mathfrak{n}).$ Then
$\mathrm{span}\\{X,Y\\}$ is $2$-dimensional and
$\mathrm{span}\\{X,Y\\}\cap\mathfrak{z}=\\{0\\}.$ Hence $\dim\mathfrak{z}\leq
2n-2.$

In conclusion, $2\leq\dim\mathfrak{z}\leq 2n-2.$ ∎

######  Remark 2.19.

From Proposition $\ref{16},$ we can further conclude that if
$\dim\mathfrak{z}=1,$ then the complex structure $J$ on $\mathfrak{n}$ is non-
nilpotent. In particular, the Lie algebra of $n\times n$ upper triangular
matrices does not admit a nilpotent complex structure.

##  3 Stratified Lie algebras with complex structures

In this section we consider a special type of nilpotent Lie algebras:
stratified Lie algebra. Recent results on nilpotent Lie algebras with a
stratification can be found in
$\cite[cite]{[\@@bibref{}{MR4127910}{}{}]},\cite[cite]{[\@@bibref{}{MR3521656}{}{}]},\cite[cite]{[\@@bibref{}{MR3742567}{}{}]}.$
We start with the definition of stratified Lie algebras.

######  Definition 3.1.

A nilpotent Lie algebra $\mathfrak{n}$ is said to admit a step k
stratification if it has a vector space decomposition of the form
$\mathfrak{n}_{1}\oplus\mathfrak{n}_{2}\oplus\ldots\oplus\mathfrak{n}_{k},$
where $\mathfrak{n}_{k}\neq\\{0\\},$ satisfying the bracket generating
property $[\mathfrak{n}_{1},\mathfrak{n}_{k}]=\\{0\\}$ and

$\displaystyle[\mathfrak{n}_{1},\mathfrak{n}_{j-1}]=\mathfrak{n}_{j}\qquad\text{
for all }j\in\\{2,\ldots,k\\}.$

A Lie algebra $\mathfrak{n}$ that admits a stratification is called a
$\textit{stratified Lie algebra}.$ A complex structure $J$ on a stratified Lie
algebra $\mathfrak{n}$ is said to be strata-preserving if it preserves each
layer of the stratification.

######  Remark 3.2.

Let $\mathfrak{n}$ be a step $k$ stratified Lie algebra. By induction,

$\mathfrak{c}_{j}(\mathfrak{n})=\bigoplus_{j+1\leq l\leq
k}\mathfrak{n}_{l}\qquad\text{ for all }j\geq 0.$ (8)

######  Proposition 3.3.

Let $\mathfrak{n}$ be a $2n$-dimensional step $n$ nilpotent Lie algebra for
some $n\in\mathbb{N}$. Suppose that $\dim\mathfrak{c}_{j}(\mathfrak{n})=2n-2j$
for all $1\leq j\leq n.$ Then $\mathfrak{n}$ does not admit a stratification.

###### Proof.

Assume, by contradiction, that $\mathfrak{n}$ admits a stratification. Since,
by $\eqref{eq:e}$, $\mathfrak{c}_{j}(\mathfrak{n})=\bigoplus_{j+1\leq l\leq
n}$ $\mathfrak{n}_{l}$ and $\dim\mathfrak{c}_{1}(\mathfrak{n})=2n-2$,
$\dim\mathfrak{n}_{1}=2.$ Since $\mathfrak{n}$ is a stratified Lie algebra,
$\mathfrak{n}_{2}=[\mathfrak{n}_{1},\mathfrak{n}_{1}].$ Thus
$\dim\mathfrak{n}_{2}=1$ and $\dim\mathfrak{c}_{2}(\mathfrak{n})=2n-3>2n-4.$
This is a contradiction. ∎

######  Proposition 3.4.

Let $\mathfrak{n}$ be a step $k$ stratified Lie algebra with a complex
structure $J$ and $k\geq 2$. Suppose that $\dim\mathfrak{n}_{1}=2.$ Then $J$
is not strata-preserving.

###### Proof.

Suppose, by contradiction, that there exists a strata-preserving complex
structure $J.$ Then $\dim\mathfrak{n}_{j}\in 2\mathbb{N}$ for all $j\geq 1.$
However, $\dim\mathfrak{n}_{1}=2$ implies that $\dim\mathfrak{n}_{2}=1,$ which
contradicts the assumption that $\dim\mathfrak{n}_{2}\in 2\mathbb{N}.$ Hence
$\mathfrak{n}$ does not have a strata-preserving complex structure. ∎

######  Remark 3.5.

Let $\mathfrak{n}$ be a step $3$ stratified Lie algebra with a strata-
preserving complex structure. Arguing in a similar way as in Proposition 3.4,
we conclude that $\dim\mathfrak{n}\neq 4$ or $6.$

We show that there always exists a stratification on step $2$ nilpotent Lie
algebra with a strata-preserving complex structure $J$.

######  Theorem 3.6.

Let $\mathfrak{n}$ be a step $2$ nilpotent Lie algebra with a complex
structure $J.$ Suppose that $\mathfrak{c}_{1}(\mathfrak{n})$ is $J$-invariant.
Then $\mathfrak{n}$ admits a $J$-invariant stratification.

###### Proof.

Define a $J$-invariant inner product $\psi$ by

$\displaystyle\psi(X,Y)=\phi(X,Y)+\phi(JX,JY),\text{ for all
}X,Y\in\mathfrak{n},$

where $\phi$ is an inner product on $\mathfrak{n}.$ We show that there exists
a stratification on $\mathfrak{n}$ such that $\mathfrak{n}_{1}$ and
$\mathfrak{n}_{2}$ are $J$-invariant. Define
$\mathfrak{n}_{2}=[\mathfrak{n},\mathfrak{n}]$ and
$\mathfrak{n}_{1}=\mathfrak{n}_{2}^{\perp}$, the orthogonal complement of
$\mathfrak{n}_{2}$ with respect to $\psi$. Then
$\mathfrak{n}_{2}=\mathfrak{c}_{1}(\mathfrak{n})$ is $J$-invariant and by
definition $\mathfrak{n}=\mathfrak{n}_{1}\oplus\mathfrak{n}_{2}$. Also note
that

$\mathfrak{n}_{2}=[\mathfrak{n}_{1}\oplus\mathfrak{n}_{2},\mathfrak{n}_{1}\oplus\mathfrak{n}_{2}]=[\mathfrak{n}_{1},\mathfrak{n}_{1}].$

This implies that $\mathfrak{n}_{1}$ generates $\mathfrak{n}.$ Thus $J$ is a
complex structure that preserves both $\mathfrak{n}_{1}$ and
$\mathfrak{n}_{2}$. ∎

######  Remark 3.7.

(i) Let $\mathfrak{g}$ be an arbitrary Lie algebra. A complex structure $J$ on
$\mathfrak{g}$ is called bi-invariant if $J[X,Y]=[JX,Y]$ for all
$X,Y\in\mathfrak{g}.$ That is, $J\circ\operatorname{ad}=\operatorname{ad}\circ
J.$ A complex structure $J$ is called Abelian if $[X,Y]=[JX,JY]$ for all
$X,Y\in\mathfrak{g}.$ See, e.g., [2], [14]. Notice that $J$ preserves all
terms of $\mathfrak{c}_{j}(\mathfrak{n})$ and $\mathfrak{c}^{j}(\mathfrak{n})$
if $J$ is bi-invariant, while if $J$ is Abelian, $J$ only preserves all terms
of $\mathfrak{c}^{j}(\mathfrak{n}).$

(ii) Suppose that $\mathfrak{n}$ is a step $k$ stratified Lie algebra with a
bi-invariant complex structure $J.$ From $\eqref{eq:e}$,
$\mathfrak{c}_{j}(\mathfrak{n})=\bigoplus_{j+1\leq l\leq k}\mathfrak{n}_{l}$,
it is clear that $\dim\mathfrak{n}_{j}\in 2\mathbb{N}$ for all
$j\in\\{1,\ldots,k\\}.$

######  Proposition 3.8.

Let $\mathfrak{n}$ be a step $k$ stratified Lie algebra with a strata-
preserving complex structure $J.$ Then
$J\mathfrak{c}_{j}(\mathfrak{n})=\mathfrak{c}_{j}(\mathfrak{n})$ for all
$j\geq 0$ and $J$ is nilpotent of step $k$.

###### Proof.

We first show that
$J\mathfrak{c}_{j}(\mathfrak{n})=\mathfrak{c}_{j}(\mathfrak{n})$ for all
$j\geq 0.$ Recall, from $\eqref{eq:1},$ that $\mathfrak{c}_{j}(\mathfrak{n})$
$=\bigoplus_{j+1\leq l\leq k}\mathfrak{n}_{l}$ and hence
$J\mathfrak{c}_{j}(\mathfrak{n})=\mathfrak{c}_{j}(\mathfrak{n})$ for all
$j\geq 0.$ By Corollary 2.15, $J$ is nilpotent of step $k.$ ∎

It is known that every step $2$ nilpotent Lie algebra maybe stratified (see,
e.g., [12]). We will provide another proof in Theorem 3.9, that every complex
structure on a step $2$ nilpotent Lie algebra is nilpotent of step $2$ or $3$.
See, e.g., [7, Theorem  1.3] and $\cite[cite]{[\@@bibref{}{MR2533671}{}{},
\text{Proposition }3.3]}.$ In what follows, we denote by
$\mathfrak{k}=\mathfrak{n}_{2}\cap J\mathfrak{n}_{2}$ the largest
$J$-invariant subspace contained in $\mathfrak{n}_{2}$ and we also remind the
reader that $\mathfrak{d}^{1}=\mathfrak{z}\cap J\mathfrak{z}$ is the largest
$J$-invariant subspace contained in $\mathfrak{z}.$

######  Theorem 3.9.

Let $\mathfrak{n}=\mathfrak{n}_{1}\oplus\mathfrak{n}_{2}$ be a step $2$
nilpotent Lie algebra with a complex structure $J$ and a $J$-invariant inner
product $\psi$.

(i) Suppose that $\mathfrak{k}=\\{0\\}$. Then $\mathfrak{d}_{1}$ is Abelian
and $J$ is nilpotent of step $2.$

(ii) Suppose that $\\{0\\}\neq\mathfrak{k}\subset\mathfrak{n}_{2}$. Then $J$
is nilpotent of step $3.$

(iii) Suppose that $\mathfrak{n}_{2}=\mathfrak{k}.$ Then $J$ is strata-
preserving and nilpotent of step $2.$

In conclusion, $J$ is nilpotent of either step $2$ or $3$.

###### Proof.

We start with parts (i) and (ii) together. Suppose that
$J\mathfrak{n}_{2}\neq\mathfrak{n}_{2}.$ Then,
$\mathfrak{p}_{2}=[J\mathfrak{n}_{2},\mathfrak{n}]\subseteq\mathfrak{n}_{2}.$
For all $Z_{2}\in\mathfrak{n}_{2},$ $X,JX\in\mathfrak{n},$ by the
Newlander–Nirenberg condition,

$[J\mathfrak{n}_{2},\mathfrak{n}]\ni[JZ_{2},JX]=J[JZ_{2},X]\in
J[J\mathfrak{n}_{2},\mathfrak{n}].$ (9)

This implies that $\mathfrak{p}_{2}$ is $J$-invariant in $\mathfrak{n}_{2}$.
We now consider the following two possibilities.

(i) If $\mathfrak{k}=\\{0\\},$ then from $\eqref{eq:iu},$ we get that
$\mathfrak{p}_{2}=\\{0\\}.$ By Theorem $\ref{14},$ $J$ is nilpotent of step
$2.$

(ii) If $\\{0\\}\neq\mathfrak{k}\subset\mathfrak{n}_{2},$ since
$\\{0\\}\neq\mathfrak{p}_{2}\subseteq\mathfrak{k}$ and
$J\mathfrak{p}_{2}\subset\mathfrak{n}_{2}$, then by definition,
$\mathfrak{p}_{3}=\\{0\\}.$ By Theorem $\ref{14},$ $J$ is nilpotent of step
$3.$

Finally, for part (iii), suppose that $\mathfrak{n}_{2}=\mathfrak{k}.$ We find
that $J$ preserves $\mathfrak{n}_{2}.$ By Theorem $\ref{5},$ $J$ is strata-
preserving. From $\text{Corollary }\ref{44},$ $J$ is nilpotent of step $2.$

In conclusion, $J$ is either nilpotent of step $2$ or $3.$ ∎

######  Remark 3.10.

(i) If $J$ is nilpotent of step $3,$ then there does not necessarily exist a
$J$-invariant stratification.

(ii) We recall, from $\cite[cite]{[\@@bibref{}{gao2020maximal}{}{},
\text{Theorem }1.3]},$ if $\mathfrak{z}$ is not $J$-invariant, then $J$ is
nilpotent of step $3.$ From Theorem $\ref{47},$ we have the following table:

$J$ | Strata-preserving | Non-strata-preserving
---|---|---
$J\mathfrak{z}=\mathfrak{z}$ |  $J$ nilpotent of step $2$ |  $J$ nilpotent of step $2$
$J\mathfrak{z}\neq\mathfrak{z}$ |  $J$ nilpotent of step $2$ |  $J$ nilpotent of step $3$
Table 1: nilpotency of $J$

From $\text{Table }\ref{2.1},$ if $J$ is nilpotent of step $2$, then $J$ is
either strata-preserving or center-preserving. More precisely, we conclude
that either $\mathfrak{k}=\mathfrak{n}_{2}\cap J\mathfrak{n}_{2}=\\{0\\}$ or
$J\mathfrak{n}_{2}=\mathfrak{n}_{2}.$ Indeed, if $\mathfrak{n}$ is step $2$
nilpotent Lie algebra with a nilpotent complex structure $J$ of step $2,$ $J$
may not be strata-preserving.

Notice that an even dimensional nilpotent Lie algebra with
$\dim\mathfrak{c}_{1}(\mathfrak{n})=1$ has step $2$. There does not exist a
$J$-invariant stratification for dimensional reasons. Suppose that
$\dim\mathfrak{c}_{1}(\mathfrak{n})\geq 2.$ We have the following theorem.

######  Theorem 3.11.

Let $\mathfrak{n}$ be a step $2$ stratified Lie algebra with a complex
structure $J$.

(i) Suppose that $\dim\mathfrak{n}_{2}=2$. Then

(a) $J$ is nilpotent of step $2;$ (b) if $\dim\mathfrak{d}^{1}=2,$ then
$J\mathfrak{n}_{2}=\mathfrak{n}_{2}$.

(ii) Suppose that $\dim\mathfrak{n}_{2}=2l$ for some $l\geq 2\in\mathbb{N}$.
Furthermore, assume that $\dim\mathfrak{d}^{1}\leq 4l-2$ and
$J\mathfrak{n}_{2}\neq\mathfrak{n}_{2}$. Then $J$ is nilpotent of step $3.$

###### Proof.

By Theorem $\ref{47},$ $J$ is nilpotent of either step $2$ or $3.$

Start with part (i). Assume that $\dim\mathfrak{n}_{2}=2.$ For part (a),
notice that $J$ could be either strata-preserving or not. If $J$ is strata-
preserving, by Theorem 3.9 part (iii), $J$ is nilpotent of step $2$.
Otherwise, $J$ is not strata-preserving. Since $\dim\mathfrak{n}_{2}=2$, it
follows that $\mathfrak{k}=\\{0\\}.$ Then by Theorem 3.9 part (i), $J$ is
Abelian and hence nilpotent of step $2.$

Next, for part (b), recall that $\mathfrak{d}^{1}=\mathfrak{z}\cap
J\mathfrak{z}$ is the largest $J$-invariant subspace of $\mathfrak{z}$.
Suppose that $\mathfrak{n}_{2}$ is not $J$-invariant. Then
$\mathfrak{k}=\\{0\\}.$ From part (i), $J$ is nilpotent of step $2.$ It
follows, from $\text{Theorem }\ref{14},$ that $\mathfrak{d}_{2}=\\{0\\}$ and
$\mathfrak{d}_{1}\subseteq\mathfrak{d}^{1}.$ However,
$\dim\mathfrak{d}_{1}=\dim\mathfrak{n}_{2}\oplus
J\mathfrak{n}_{2}=4>\dim\mathfrak{d}^{1}.$ This is a contradiction. Hence
$J\mathfrak{n}_{2}=\mathfrak{n}_{2}$.

We now show part (ii). Notice that $l\neq 1.$ Otherwise
$\dim\mathfrak{n}_{2}=\dim\mathfrak{d}^{1}=2.$ This implies that
$J\mathfrak{n}_{2}=\mathfrak{n}_{2}.$ Suppose, by contradiction, that $J$ is
not nilpotent of step $3.$ Hence $J$ is nilpotent of step $2.$ Then from
Remark 3.10 part (ii), $\mathfrak{k}=\\{0\\}$ and by definition,
$\mathfrak{d}_{1}=\mathfrak{n}_{2}\oplus
J\mathfrak{n}_{2}\subseteq\mathfrak{d}^{1}.$ However,
$\dim\mathfrak{d}_{1}=4l>\dim\mathfrak{d}^{1}.$ This is a contradiction. Hence
$\mathfrak{k}\neq\\{0\\}.$ By Theorem 3.9 part (ii), $J$ is nilpotent of step
$3.$ ∎

######  Remark 3.12.

We can extend the statement of part (i) into a higher step stratification as
follows:

Let $\mathfrak{n}$ be a step $k$ stratified Lie algebra with a nilpotent
complex structure $J$ of step $k$. Suppose that $\dim\mathfrak{n}_{k}=2$ and
$\dim\mathfrak{d}^{1}=2$. Then $J\mathfrak{n}_{k}=\mathfrak{n}_{k}.$

######  Corollary 3.13.

Let $\mathfrak{n}=\mathfrak{n}_{1}\oplus\mathfrak{n}_{2}$ be a step $2$
stratified Lie algebra with a complex structure $J$ such that
$\dim\mathfrak{n}_{2}=2.$ Then $J$ is center-preserving or strata-preserving
or both. Furthermore, suppose that $2\leq\dim\mathfrak{z}\leq 3$ or
$\dim\mathfrak{z}=4$ and $J\mathfrak{z}\neq\mathfrak{z}.$ Then there exists a
$J$-invariant stratification.

###### Proof.

By $\text{Theorem }\ref{49},$ $J$ is nilpotent of step $2$. Then by Table 1,
$J\mathfrak{n}_{2}=\mathfrak{n}_{2}$ or $J\mathfrak{z}=\mathfrak{z}$ or both
if $\mathfrak{n}_{2}=\mathfrak{z}$.

Furthermore, $\dim\mathfrak{d}^{1}=2$ since $2\leq\dim\mathfrak{z}\leq 3$ or
$\dim\mathfrak{z}=4$ and $J\mathfrak{z}\neq\mathfrak{z}.$ By part (ii) of
$\text{Theorem }\ref{49},$ $J\mathfrak{n}_{2}=\mathfrak{n}_{2}$. Furthermore,
by $\text{Theorem }\ref{5},$ there exists a $J$-invariant stratification. ∎

Suppose that $\mathfrak{n}$ is a $6$ dimensional step $2$ nilpotent Lie
algebra with a complex structure. In $\cite[cite]{[\@@bibref{}{MR1899353}{}{},
\text{Table } 1]},$ there is a complete classification of complex structures
on these algebras. However, no information is provided on whether or not $J$
preserves the strata.

######  Corollary 3.14 ([1, 4]).

Let $\mathfrak{n}$ be a $6$ dimensional step $2$ nilpotent Lie algebra with a
complex structure $J$ such that $\dim\mathfrak{c}_{1}(\mathfrak{n})=2.$ Then
$\mathfrak{n}$ admits a $J$-invariant stratification.

###### Proof.

By Theorem 3.9 and Proposition 2.18, $J$ is nilpotent and
$2\leq\dim\mathfrak{z}\leq 4$. If $\dim\mathfrak{z}=4,$
$\dim\mathfrak{c}_{1}(\mathfrak{n})=1$ and $J$ is not strata-preserving due to
dimensional reasons. We omit this case. Next, assume that
$\dim\mathfrak{z}\leq 3.$ This is a direct consequence of Corollary
$\ref{36}.$ ∎

In what follows, we focus on higher step stratified Lie algebras with complex
structures.

######  Proposition 3.15.

Let $\mathfrak{n}$ be a step $3$ stratified Lie algebra with a complex
structure $J$. Suppose that $J\mathfrak{n}_{3}=\mathfrak{n}_{3}$. Then $J$ is
nilpotent of step $3.$

###### Proof.

By the definition of the descending central series $\mathfrak{p}_{j}$ in
$\eqref{eq:p},$
$\\{0\\}\neq\mathfrak{p}_{2}=\mathfrak{n}_{3}+[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}].$
On the one hand, suppose that
$[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]=\\{0\\}.$ We deduce that
$\mathfrak{p}_{2}=\mathfrak{n}_{3}$ and hence $\mathfrak{p}_{3}=\\{0\\}$ by
definition. Using Theorem $\ref{14},$ $J$ is nilpotent of step $3.$ On the
other hand, suppose that
$[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]\neq\\{0\\}.$ Then by the
Newlander–Nirenberg condition, for all $U\in\mathfrak{c}_{1}(\mathfrak{n})$
and $X,JX\in\mathfrak{n}$

$\displaystyle
0\neq\underbrace{[JU,JX]-J[JU,X]}_{\text{$\in[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]+J[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]$}}=\underbrace{[U,X]+J[U,JX]}_{\text{$\in\mathfrak{n}_{3}$}}.$

Hence
$[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]\subseteq\mathfrak{n}_{3}.$
This implies that $\mathfrak{p}_{2}\subseteq\mathfrak{n}_{3}$ and therefore
$J\mathfrak{p}_{2}\subseteq\mathfrak{n}_{3}.$ Then
$\mathfrak{p}_{3}=[\mathfrak{p}_{2},\mathfrak{n}]+[J\mathfrak{p}_{2},\mathfrak{n}]=\\{0\\}.$
Again by Theorem 2.13, $J$ is nilpotent of step $3.$ ∎

######  Proposition 3.16.

Let $\mathfrak{n}$ be a $8$ dimensional step $3$ stratified Lie algebra with a
complex structure $J$ such that
$2\dim\mathfrak{n}_{3}=\dim\mathfrak{c}_{1}(\mathfrak{n})=4$. Suppose that
$J\mathfrak{n}_{3}\neq\mathfrak{n}_{3}$ and $\dim\mathfrak{z}\leq 3.$ Then $J$
is nilpotent of step $4.$ Furthermore,
$\mathfrak{d}_{2}=\mathfrak{n}_{3}\oplus J\mathfrak{n}_{3}$.

###### Proof.

Since $\mathfrak{n}_{3}\subseteq\mathfrak{z},$ $\dim\mathfrak{z}\geq 2.$ By
$\cite[cite]{[\@@bibref{}{MR4009385}{}{}, \text{Corollary }3.12]},$ $J$ is
nilpotent. Then using Remark 2.4 (i), $3\leq j_{0}\leq 4$, where $j_{0}$ is
the nilpotent step of $J.$ Suppose, by contradiction, that $J$ is nilpotent of
step $3.$ It follows, from the equation $\eqref{eq:ew}$, that
$\mathfrak{n}_{3}+J\mathfrak{n}_{3}\subseteq\mathfrak{d}_{2}\subseteq\mathfrak{d}^{1}\subseteq\mathfrak{z}.$
On the one hand, since $\dim\mathfrak{z}\leq 3,$ $\dim\mathfrak{d}^{1}=2.$ On
the other hand, since $J\mathfrak{n}_{3}\neq\mathfrak{n}_{3}$ and
$\dim\mathfrak{n}_{3}=2,$ $\mathfrak{n}_{3}\cap J\mathfrak{n}_{3}=\\{0\\}$ and
therefore $\dim\mathfrak{n}_{3}\oplus
J\mathfrak{n}_{3}=4>\dim\mathfrak{d}^{1}.$ This is a contradiction. So $J$ is
nilpotent of step $4.$

We now show that $\mathfrak{d}_{2}=\mathfrak{n}_{3}\oplus J\mathfrak{n}_{3}$.
It is sufficient to show that $\mathfrak{d}_{2}\subseteq\mathfrak{n}_{3}\oplus
J\mathfrak{n}_{3}.$ By definition,

$\displaystyle\mathfrak{d}_{2}$
$\displaystyle=[\mathfrak{d}_{1},\mathfrak{n}]+J[\mathfrak{d}_{1},\mathfrak{n}]$
$\displaystyle=\mathrm{span}\left\\{[T,X]+J[T^{\prime},X^{\prime}]:\text{
}\forall\text{ }T,T^{\prime}\in\mathfrak{d}_{1},\text{ }\forall\text{
}X,X^{\prime}\in\mathfrak{n}\right\\}.$

For all $T,T^{\prime}\in\mathfrak{d}_{1},$ we may write $T=U+JV$ and
$T^{\prime}=U^{\prime}+JV^{\prime}$ where
$U,V,U^{\prime},V^{\prime}\in\mathfrak{c}_{1}(\mathfrak{n}).$ Then

$0\neq[T,X]+J[T^{\prime},X^{\prime}]=\underbrace{[U,X]+J[U^{\prime},X^{\prime}]}_{\text{$\in\mathfrak{n}_{3}\oplus
J\mathfrak{n}_{3}$}}+[JV,X]+J[JV^{\prime},X^{\prime}].$ (10)

By the Newlander–Nirenberg condition,

$\displaystyle
0\neq\underbrace{[JV,X]+J[JV,JX]}_{\text{$\in[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]+J[J\mathfrak{c}_{1}(\mathfrak{n}),\mathfrak{n}]$}}=J[V,X]-[V,X]\in\mathfrak{n}_{3}\oplus
J\mathfrak{n}_{3}.$

Hence $[JV,X]+J[JV^{\prime},X^{\prime}]\in\mathfrak{n}_{3}\oplus
J\mathfrak{n}_{3}.$ From $\eqref{eq:90},$
$[T,X]+J[T^{\prime},X^{\prime}]\in\mathfrak{n}_{3}\oplus J\mathfrak{n}_{3}.$
Hence $\mathfrak{d}_{2}\subseteq\mathfrak{n}_{3}\oplus J\mathfrak{n}_{3}.$ In
conclusion, $\mathfrak{d}_{2}=\mathfrak{n}_{3}\oplus J\mathfrak{n}_{3}.$ ∎

## Acknowledgement

The results of this paper are contained in the author’s Master of Science
thesis at the University of New South Wales, prepared under the the
supervision of Michael. G. Cowling and Alessandro Ottazzi. I would like to
give deep thanks to both of them, for guiding and sharing their views on
mathematics. They also provided very useful comments and suggestions on the
project.

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# Conformation-dependent electron transport through a biphenyl molecule:
Circular current and related issues

Santanu K. Maiti<EMAIL_ADDRESS>Physics and Applied Mathematics
Unit, Indian Statistical Institute, 203 Barrackpore Trunk Road, Kolkata-700
108, India

###### Abstract

We investigate the conformation-dependent electron transfer in a biphenyl
molecule within a simple tight-binding framework. The overall junction current
and circular currents in two benzene rings driven by applied bias voltage are
calculated by using Green’s function formalism. Our analysis may provide the
possibilities of using organic molecules with loop substructures to design
molecular spintronic devices, indicating the emergence of molecular
spintronics.

###### pacs:

73.23.-b, 85.65.+h, 85.35.Ds

##  I Introduction

The study of electron transport through a molecular system attached to
metallic electrodes is regarded as one of the most promising research fields
in nanoscale technology and physics nitzan1 . With the discovery of advanced
molecular scale measurement methodologies like scanning electro-chemical
microscopy (SESM), scanning tunneling microscopy (STM), atomic force
microscopy (AFM), etc., it is now possible to measure current flow through
single molecules or cluster of molecules sandwiched between two electrodes
chen . The proposed idea of designing molecule-based diode by Aviram and
Ratner aviram  in 1974 first illustrates the possibility of using molecules as
active components of a device. Since then several ab-initio as well as model
calculations have been done to investigate molecular transport theoretically
ventra1 ; ventra2 ; sumit ; tagami ; orella1 ; orella2 ; arai ; walc ; san1 ;
san2 ; san3 ; san4 . But experimental realizations took a little longer time
to get feasible. In a pioneering experiment, Reed et al. investigated current-
voltage ($I$-$V$) characteristics in a single benzene molecule coupled to
metallic electrodes via thiol groups reed2 . Various other experiments using
molecules have also been reported in literature exploring many interesting
features e.g., ballistic transport, quantized conductance, negative
differential resistance (NDR), gate controlled transistor operation, memory
effects, bistable switching, conformational switching to name a few. Although
a lot of theoretical mag ; lau ; baer1 ; baer2 ; baer3 ; gold  and
experimental reed1 ; tali ; fish ; cui ; gim  studies have been made so far
using different molecules, yet several problems are to be solved for further
development of molecular electronics.

Most of the works associated with electronic conduction through molecular
bridge systems are mainly concerned on the overall conduction properties. But
a very few works are available where attention has been paid to the current
distribution within the molecule itself having single or multiple loop
substructures dist1 ; dist2 ; dist3 . Recently some interesting works have
been done by Nitzan et al. and few other groups where possible quantum
interference effects have been explored on current distribution through such
molecular geometries due to the existence of multiple pathways, yielding the
possibilities of voltage driven circular currents cir1 ; cir2 ; cir3 ; cir4 ;
cir5 ; cir6 . The appearance of circular currents in loop geometries have
already been reported in other contexts several years ago. This is commonly
known as persistent currents in mesoscopic conducting rings where the current
is induced by means of Aharonov-Bohm flux $\phi$ threaded by the ring butt2 ;
levy ; gefen ; skm1 ; skm2 ; skm3 ; skm4 ; skm5 ; skm6 . The reason behind
this current in isolated loop geometries is quite different from the previous
one where current is driven by an applied bias voltage. It has been verified
that the circular currents appear in molecular rings, driven by applied bias
voltage, produce considerable magnetic fields at the centers of these rings.
This phenomenon is somewhat interesting and can be used in different aspects
in the study of molecular transport. For example, in presence of a local spin
at the ring center one can regulate spin dependent transport through the
molecular wire by tuning the orientation of that local spin, and, also the
behavior of spin inelastic currents can be explained. In a recent work
Galperin et al. rai  have proposed some results towards this direction. One
can also utilize this circular current generated magnetic field in other way
to control spin dependent transport through a molecular wire without changing
the orientation of the local spin. In that case we can change the strength of
the magnetic field by some mechanisms. To test it, biphenyl molecule may the
best example where two benzene rings are connected by a single C-C bond. It
has been examined latha  that in the case of a biphenyl molecule, electronic
conductance changes significantly with the relative twist angle among two
benzene rings. For the planar conformation conductance becomes maximum, while
it gets decreased with increasing the twist angle. This phenomenon motivates
us to describe conformation-dependent circular currents in a biphenyl molecule
coupled to two metallic electrodes. We use a simple tight-binding (TB)
framework to describe the model quantum system and evaluate all the results
through Green’s function formalism. We believe that our present analysis will
certainly provide some important information that can be used to design
molecular spintronic devices in near future.

The structure of the paper is as follows. In section II, we describe the
molecular model and theoretical formulation for the calculations. The
essential results are presented in Section III which contains (a) transmission
probability as a function of injecting electron energy and junction current
through the molecular wire as a function of applied bias voltage for different
twist angles, and (b) conformation-dependent circular currents in two benzene
rings and associated magnetic fields at the ring centers. Finally, in section
IV, we summarize our main results and discuss their possible implications for
further study.

##  II Molecular Model and Theoretical Formulation

###  II.1 Tight-binding model

Figure 1 gives a schematic illustration of the molecular wire, where a
biphenyl molecule is coupled to two semi-infinite one-dimensional ($1$D)
metallic electrodes, commonly known as source and drain. Our analysis for the
present work is based on non-interacting electron picture, and, within this
framework, TB model is extremely suitable for analyzing electron transport
through a molecular bridge system.

Figure 1: (Color online). Schematic diagram of a biphenyl molecule attached to
two electrodes, namely, source and drain. The magenta arrow describes the
relative twist among the molecular rings.

The single particle Hamiltonian which captures the molecule and side-attached
electrodes gets the form:

$H=H_{mol}+H_{ele}+H_{tun}.$ (1)

The first term $H_{mol}$ corresponds to the Hamiltonian of the biphenyl
molecule sandwiched between two electrodes. Under nearest-neighbor hopping
approximation, the TB Hamiltonian of the molecule composed of $12$ ($N=12$)
atomic sites reads,

$\displaystyle H_{mol}$ $\displaystyle=$ $\displaystyle\sum_{i}\epsilon
c_{i}^{\dagger}c_{i}+\sum_{i}v\left[c_{i+1}^{\dagger}c_{i}+c_{i}^{\dagger}c_{i+1}\right]$
(2) $\displaystyle+$ $\displaystyle\sum_{j}\epsilon
c_{j}^{\dagger}c_{j}+\sum_{j}v\left[c_{j+1}^{\dagger}c_{j}+c_{j}^{\dagger}c_{j+1}\right]$
$\displaystyle+$ $\displaystyle
v_{4,7}\left[c_{4}^{\dagger}c_{7}+c_{7}^{\dagger}c_{4}\right]$

where the indices $i$ and $j$ are used for the left and right molecular rings,
respectively. $\epsilon$ denotes the on-site energy of an electron at
$i$-($j$-)th site and $v$ describes the isotropic nearest-neighbor coupling
between the molecular sites. $c_{i}^{\dagger}$($c_{j}^{\dagger}$) and
$c_{i}$($c_{j}$) are the creation and annihilation operators, respectively, of
an electron at the $i$-($j$-)th site. The last term in the right hand side of
Eq. 2 describes the coupling between two molecular rings. In terms of the
relative twist angle $\theta$ among these two rings, the coupling strength
$v_{4,7}$ gets the form: $v_{4,7}=v\cos\theta$.

The second and third terms of Eq. 1 describe the Hamiltonians for the $1$D
semi-infinite electrodes (source and drain) and molecule-to-electrode
coupling. In Wannier basis representation they are expressed as follows.

$\displaystyle H_{ele}$ $\displaystyle=$ $\displaystyle H_{S}+H_{D}$
$\displaystyle=$
$\displaystyle\sum_{\alpha=S,D}\left\\{\sum_{n}\epsilon_{0}d_{n}^{\dagger}d_{n}+\sum_{n}t_{0}\left[d_{n+1}^{\dagger}d_{n}+h.c.\right]\right\\},$

and,

$\displaystyle H_{tun}$ $\displaystyle=$ $\displaystyle H_{S,mol}+H_{D,mol}$
(4) $\displaystyle=$
$\displaystyle\tau_{S}[c_{p}^{{\dagger}}d_{0}+h.c.]+\tau_{D}[c_{q}^{{\dagger}}d_{N+1}+h.c.].$

Here, $\epsilon_{0}$ and $t_{0}$ correspond to the site energy and nearest-
neighbor hopping integral in the electrodes. $d_{n}^{{\dagger}}$ and $d_{n}$
are the creation and annihilation operators, respectively, of an electron at
the site $n$ of the electrodes. The coupling strength between the source and
the molecule is $\tau_{S}$, while it is $\tau_{D}$ between the molecule and
the drain. The source and drain are coupled to the molecule through $p$-th and
$q$-th atomic sites, respectively, those are variable.

###  II.2 Two-terminal transmission probability and junction current

To obtain transmission probability of an electron from source to drain
electrode through the molecule, we use Green’s function formalism. Within the
regime of coherent transport and in the absence of Coulomb interaction this
technique is well applied.

The single particle Green’s function operator representing the entire system
for an electron with energy $E$ is defined as,

$G=\left(E-H+i\eta\right)^{-1}$ (5)

where, $\eta\rightarrow 0^{+}$. Following the matrix forms of $H$ and $G$, the
problem of finding $G$ in the full Hilbert space of $H$ can be mapped exactly
to a Green’s function $\mathcal{G}$ corresponding to an effective Hamiltonian
in the reduced Hilbert space of the molecule itself and we have datta ,

$\mbox{\boldmath${\mathcal{G}}$}=\left(\mbox{\boldmath$E-H_{mol}-\Sigma_{S}-\Sigma_{D}$}\right)^{-1}.$
(6)

Here, $\Sigma_{S}$ and $\Sigma_{D}$ are the contact self-energies introduced
to incorporate the effect of coupling of the molecule to the source and drain,
respectively. In terms of this effective Green’s function ${\mathcal{G}}$,
two-terminal transmission probability $T$ through the molecular wire can be
written as datta ,

$T=\mbox{Tr}\mbox{\boldmath[$\Gamma_{S}\mathcal{G}^{r}\Gamma_{D}\mathcal{G}^{a}$]},$
(7)

where, $\Gamma_{S}$ and $\Gamma_{D}$ are the coupling matrices, and,
${\mathcal{G}}^{r}$ and ${\mathcal{G}}^{a}$ are the retarded and advanced
Green’s functions, respectively.

With the knowledge of the transmission probability we compute overall junction
current ($I_{T}$) as a function of bias voltage ($V$) using the standard
formalism based on quantum scattering theory.

$I_{T}(V)=\frac{2e}{h}\int\limits_{-\infty}^{\infty}T\,[f_{S}(E)-f_{D}(E)]\,dE.$
(8)

Here, $f_{S}$ and $f_{D}$ are the Fermi functions of the source and drain,
respectively. At absolute zero temperature the above equation boils down to
the following expression.

$I_{T}(V)=\frac{2e}{h}\int\limits_{E_{F}-\frac{eV}{2}}^{E_{F}+\frac{eV}{2}}T(E)\,dE,$
(9)

where, $E_{F}$ is the equilibrium Fermi energy.

###  II.3 Circular current and associated magnetic field

In order to calculate circular current in molecular rings of the biphenyl
molecule let us first concentrate on the current distribution in a simple loop
geometry illustrated in Fig. 2. A net current $I_{T}$

Figure 2: (Color online). Schematic view of current distribution through a
ring geometry coupled to two electrodes. The filled black circles correspond
to the positions of the atomic sites.

flows between two electrodes through a quantum ring, where $I_{1}$ and $I_{2}$
are the currents flowing through upper and lower arms of the ring,
respectively. We assign positive sign to the current propagating in the
counter-clockwise direction. With this current distribution, we define
circular current of the ring as cir1 ,

$I_{c}=\frac{1}{L}\left(I_{1}L_{1}+I_{2}L_{2}\right)$ (10)

where, $L_{1}$ and $L_{2}$ are the lengths of the upper and lower arms of the
ring, respectively, and $L=L_{1}+L_{2}$. Thus, in order to compute $I_{c}$,
following the above relation, we need to know the currents in different
branches of the loop geometry. We evaluate these currents by using Green’s
function formalism. At absolute zero temperature the current $I_{ij}$ flowing
from site $i$ to $j$ ($j=i\pm 1$) is given by the expression.

$I_{ij}=\int\limits_{E_{F}-\frac{eV}{2}}^{E_{F}+\frac{eV}{2}}J_{ij}(E)\,dE,$
(11)

where, $J_{ij}$ is the current density. In terms of the correlation function
$\mathcal{G}^{n}$ it can be written as dist3 ,

$J_{ij}=\frac{4e}{h}\mbox{Im}\mbox{\boldmath[$H_{ij}\mathcal{G}_{ij}^{n}$]},$
(12)

where, $\mathcal{G}^{n}$ $=$ $\mathcal{G}^{r}\Gamma_{S}\mathcal{G}^{a}$. This
correlation function is evaluated by setting the occupation function of the
source to unity and that of the drain to zero.

Finally, we determine the local magnetic field at any point $\vec{r}$ inside
the ring, associated with circular current $I_{c}$, from the Biot-Savart’s Law
cir1 ,

$\vec{B}(\vec{r})=\sum_{(i,j)}\int\frac{\mu_{0}}{4\pi}I_{ij}\frac{d\vec{r}\times(\vec{r}-\vec{r^{\prime}})}{|(\vec{r}-\vec{r^{\prime}})|^{3}},$
(13)

where, $\mu_{0}$ is the magnetic constant and $\vec{r^{\prime}}$ is the
position vector of an infinitesimal bond current element
$I_{ij}d\vec{r^{\prime}}$. Using the above expressions we evaluate circular
currents and associated magnetic fields in two molecular rings of the biphenyl
molecule.

Throughout this work, we assume that the entire voltage is dropped across the
molecule-to-electrode interfaces and this assumption is good enough for
molecules of smaller size. We also restrict ourselves at absolute zero
temperature and choose the units where $c=e=h=1$.

##  III Numerical Results and Discussion

In this section we present numerical results computed for transmission
probability, overall junction current and circular currents in a biphenyl
molecule under conventional biased conditions. Throughout our analysis we set
the on-site energies in the molecule as well as in the source and drain
electrodes to zero, $\epsilon=\epsilon_{0}=0$. The nearest-neighbor coupling
strength in the electrodes ($t_{0}$) is fixed at $2$eV, while in the molecule
($v$) it is set at $1$eV. The coupling strengths of the molecule to the source
and drain electrodes, characterized by the parameters $\tau_{S}$ and
$\tau_{D}$, are also set at $1$eV. We fix the equilibrium Fermi energy $E_{F}$
at zero and measure the energy scale in unit of $v$.

###  III.1 Transmission probability and junction current

In Fig. 3 we show the variation of transmission probability $T$ as a function
of injecting electron energy $E$ when the source and drain are coupled to the
biphenyl molecule at the sites $1$ ($p=1$) and $10$ ($q=10$), respectively.
The red curve corresponds to $\theta=0$, while the green and blue lines are
associated with $\theta=\pi/3$ and $\pi/2$, respectively.

Figure 3: (Color online). Transmission probability as a function of energy for
the biphenyl molecule when the electrodes are connected at the molecular sites
$1$ and $10$, as shown in Fig. 1. The red, green and blue lines correspond to
$\theta=0$, $\pi/3$ and $\pi/2$, respectively.

From the spectrum it is observed that the transmission probability exhibits
resonant peaks (red and green lines) for some particular energies and at these
resonances it ($T$) almost reaches to unity. All these resonant peaks are
associated with the energy eigenvalues of the molecule, and therefore, we can
say that the transmittance spectrum is a fingerprint of the electronic
structure of the molecule. The number of

Figure 4: (Color online). Total transport current as a function of bias
voltage when the electrodes are connected to the molecule following the same
configuration as prescribed in Fig. 3.

resonant peaks in $T$-$E$ spectrum and their corresponding widths for a
particular molecule-to-electrode configuration notably depend on the molecular
twist angle, which is clearly visible from the red and green curves. Depending
on the twist angle $\theta$, the resonating energy states are available at
different energies. For each of these energy eigenstates a resonant peak
appears in the transmission spectrum with a finite width associated with the
molecular coupling strength. Now if the resonating states with different
energies are very closely placed then the neighboring peaks can overlap with
each other which result a broader peak. For low enough molecular coupling, the
overlap between neighboring resonant peaks is no longer possible, and
therefore, separate peaks with identical broadening will be obtained in the
transmission spectrum. Obviously, for different states having identical energy
i.e., for degenerate states a single resonant peak is generated in the
transmission curve. The situation especially changes when $\theta=\pi/2$ i.e.,
one molecular ring becomes perpendicular with

Figure 5: (Color online). Total transport current for a specific voltage bias
as a function of twist angle for the biphenyl molecule. The source and drain
are coupled to the molecular sites $1$ and $10$, respectively.

respect to the other ring. In this particular case the transmission
probability completely disappears for the entire energy band spectrum. It is
shown by the blue curve in Fig. 3. With the increment

Figure 6: (Color online). Transmission probability as a function of energy for
the same parameter values used in Fig. 3, when the source and drain are
coupled to the molecular sites $6$ ($p=6$) and $9$ ($q=9$), respectively.

of the twist angle $\theta$ between these two molecular rings, the degree of
$\pi$-conjugation between them decreases, which results the reduction of the
junction conductance since the electronic transfer rate through the molecule
scales as the square of the $\pi$-overlap nitz . For $\theta=\pi/2$, the
$\pi$-conjugation between the molecular rings completely disappears, and
accordingly, vanishing transmission probability is obtained. Thus, rotating
one benzene molecule relative to the other one can regulate electronic
transmission through the biphenyl molecule and eventually one can get the
insulating phase, which leads to the possibility of getting a switching action
using this molecule.

The sharpness of the resonant peaks in $T$-$E$ spectrum strongly depends on
the molecular coupling strength to side-attached electrodes, and, it greatly
controls electron transfer through the bridge system. In the limit of weak-
coupling i.e., $\tau_{S}(\tau_{D})<<v$, sharp resonant peaks are observed in
transmission spectrum, whereas widths of these peaks get broadened in the
limit of strong-coupling, $\tau_{S}(\tau_{D})\sim v$.

Figure 7: (Color online). Total transport current as a function of bias
voltage when the electrodes are coupled to the molecule following the same
configuration as mentioned in Fig. 6.

The broadening of transmission peaks with the enhancement in coupling strength
is quantified by the imaginary parts of the self-energy matrices $\Sigma_{S}$
and $\Sigma_{D}$ which are incorporated in the transmittance expression via
the coupling matrices $\Gamma_{S}$ and $\Gamma_{D}$. This coupling effect on
electron transport has already been explored elaborately in literature sm1 ;
sm2 ; sm3 ; sm4 , and therefore, in the present work we do not illustrate it
further.

The fundamental features of electron transport will be more transparent from
our current-voltage characteristics. The overall junction current $I_{T}$

Figure 8: (Color online). Internal current distribution in the molecule for
its planar conformation when the bias voltage is fixed at $4$V. The source and
drain are attached to the atomic sites $1$ and $10$, respectively.

through the molecular wire following Landauer-Büttiker formalism (Eq. 9). In
Fig. 4 we display the variation of junction current $I_{T}$ as a function of
applied bias voltage $V$ for the biphenyl molecule where the electrodes are
connected at the molecular sites $1$ and $10$, same as in Fig. 1. The current
varies quite continuously with the voltage bias. Depending on the molecular
coupling to the source and drain electrodes, the current exhibits continuous-
like or step-like behavior since it is computed by integrating over the
transmission curve. For the weak molecular coupling step-like nature
associated with sharp resonant peaks in transmission spectrum will be
obtained, unlike to the continuous-like feature as observed in the limit of
strong-coupling. Therefore, for a fixed voltage bias one can regulate the
current amplitude by tuning molecular coupling strength, and, this phenomenon
provides an interesting behavior in designing molecular electronic devices.
Figure 4 reveals that the junction current decreases with increasing the
relative twist angle, following the $T$-$E$ characteristics. In addition to
this behavior, it is also important to note that the threshold bias voltage
$V_{th}$ of electron conduction firmly depends on the twist angle $\theta$,
which is clearly noticed by comparing the red and green curves in Fig. 4.

In order to explore the dependence of electron conduction through the biphenyl
molecule for any arbitrary angle of twist, in

Figure 9: (Color online). Internal current distribution in the molecule for
its planar conformation when the electrodes are coupled to the atomic sites
$6$ and $9$, for the same bias voltage taken in Fig. 8. The directions of the
corresponding magnetic fields at the ring centers are illustrated by the
encircled cross and dot representing downward (into page) and upward (out of
page) directions, respectively.

Fig. 5 we present the variation of junction current as a function of relative
twist angle $\theta$ for some typical bias voltages. The spectrum shows that
for the planar conformation total current amplitude becomes maximum and it
gradually decreases with the relative twist angle and eventually drops to zero
when $\theta=\pi/2$. Thus, at $\theta=\pi/2$ no electron conduction will take
place through this molecular bridge system, while for other choices of
$\theta$ electron can transfer through the molecule from the source to drain
electrode, which promotes a conformation-dependent switching action using the
biphenyl molecule.

A significant change in the transmission spectrum is realized when the
electrodes are coupled to the molecule in such a way that the upper and lower
arms of each molecular rings have unequal lengths. In Fig. 6 we present the
results for such a particular configuration where the source and drain are
attached to the molecular sites $6$ and $9$, respectively. The red, green and
blue curves correspond to the results for the identical parameter values
chosen in Fig. 3. From the transmission curves (red and green) we make out
that for a wide energy range across $E=0$, electron conduction does not take
place, and also the widths of some resonant peaks get reduced enormously
compared to the symmetric configuration where upper and lower arms in each of
the two molecular rings are identical (Fig. 3), even though the molecule-to-
electrode coupling strength is kept unchanged. This is solely due to the
effect of quantum interference among the electronic waves passing through
different arms of the molecular rings, and, it can be much clearly analyzed
through the following arguments. For a fixed molecular coupling, the
broadening of different resonant peaks which results from the overlap of
neighboring peaks depends on the location of energy levels, as discussed
earlier. The positions of these energy levels, on the other hand, are directly
associated with the molecule itself and the real parts of the self-energy
matrices $\Sigma_{S}$ and $\Sigma_{D}$ which correspond to the shift of the
energy eigenstates of the sample sandwiched between two electrodes. Thus for a
particular molecule-to-electrode configuration we get one set of resonating
energy levels, while for the other configuration a different set of energy
levels is obtained. These generate transmission peaks with different widths
associated with the level spacing. If the molecular coupling strength is low
enough, then a minor shift of molecular energy levels takes place, and
therefore, almost identical $T$-$E$ spectrum will be observed for different
molecule-to-electrode configurations. But, for moderate coupling strength one
can regulate electron conduction through the bridge system in a tunable way by
introducing more asymmetry among these two arms. This behavior is nicely
reflected in the current-voltage characteristics. The results are bestowed in
Fig. 7. It is clearly observed that for a particular bias voltage the current
amplitude decreases significantly compared to the symmetric configuration,
Fig. 4. The symmetry breaking among the molecular arms also tunes the
threshold voltage $V_{th}$ of electron conduction across the molecular wire
for a particular twist angle $\theta$, as found by comparing Figs. 4 and 7.

###  III.2 Circular current and magnetic field

Now we focus our attention on the behavior of circular currents and associated
magnetic fields at the ring centers of the molecule. These factors are highly
sensitive to the molecule itself as well as the molecule-to-electrode
interface geometry. To address these issues, we start with the current
distribution within the molecule when it is coupled to the electrodes in such
a way that the upper and lower arms of the molecular rings have identical
lengths. The current distribution is shown in Fig. 8, where the blue and green
arrows indicate the bond currents in upper and lower arms of the rings,
respectively. The arrow sizes represent the magnitudes of bond currents and
they are computed when the bias voltage is fixed at $4$V. Here $I_{T}$ is the
net junction current, shown by the red arrow, which is distributed among
different branches at the junction point. Due to the geometrical symmetry
reasons, the magnitudes of the bond currents in upper and lower arms of the
two rings are exactly identical and since they are propagating in opposite
directions, no net circulating current will appear which results vanishing
magnetic fields at the ring centers.

In order to establish circular currents in these two molecular rings, we

Figure 10: (Color online). Circular currents and associated magnetic fields at
the ring centers of the biphenyl molecule as a function of bias voltage when
the electrodes are coupled to the molecule following the configuration
prescribed in Fig. 9. The red and black curves describe the results for the
left and right rings, respectively. The twist angle $\theta$ is fixed at zero.

attach the electrodes asymmetrically such that the upper and lower arms of the
rings have unequal lengths. The internal current distribution for such a
particular configuration is illustrated in Fig. 9, where we set the same bias
voltage as taken in Fig. 8. In this situation the bond currents get unequal
magnitudes, and accordingly, circular

Figure 11: (Color online). Circular currents and corresponding magnetic fields
at the ring centers for a specific voltage bias ($V=1.5$V) as a function of
twist angle when the molecule is sandwiched between electrodes according to
the configuration taken in Fig. 9. The red and black lines correspond to the
identical meaning as in Fig. 10.

currents are established in the two molecular rings. The magnetic fields at
the ring centers associated to these circular currents are also shown. In the
left ring, the magnetic field is directed along the downward direction,
represented by encircled cross, while in the other ring it is directed along
the upward direction, illustrated by encircled dot. We calculate the magnetic
fields using Biot-Savart’s Law, Eq. 13, and scale them in unit of
$6\mu_{0}/4\pi R$, where $R$ is the perpendicular distance from the center to
any arm of the molecular ring and the factor $6$ appears due to the existence
of six bonds in each benzene ring.

Figure 10 demonstrates the magnitudes of the circular currents and the
associated magnetic fields at the ring centers as a function of voltage bias
for the planar conformation of the biphenyl molecule. From the spectrum we
notice that in one voltage regime ($\sim 0-1.3$V) no circulating current
appears, while other voltage regimes finite circular currents are available
and the sign of these currents also changes depending on the voltage region.
The associated magnetic fields also follow the same behavior, as illustrated
in the spectrum. This phenomenon can be explained as follows. The circular
current in a loop geometry is associated with energy eigenstates those can be
described by current carrying states with the current flowing in opposite
directions. Now, for a finite bias voltage whenever one of these resonant
states lies in the Fermi window, associated with the applied voltage bias and
the nature of voltage drops along the molecular wire, we will get the
corresponding circular current. When more than one resonant states come within
this Fermi window, all of them contribute to the current which provide a net
signal, and, in the particular case when they mutually cancel each other, the
net signal becomes zero. The sign of this net circular current or direction of
associated magnetic field depends on which resonant states dominate the
others, which again certainly depends on the applied bias voltage. From the
above analysis we can clearly understand the vanishing nature of circulating
current in the above mentioned voltage region ($\sim 0-1.3$V), since up to
this voltage window no resonant state appears which can contribute to the
circulating current. This, on the other hand, is also justified from the red
curve in $T$-$E$ spectrum, Fig. 6, which is drawn for the planar conformation
of the molecule. It indicates that within the energy window $-0.65$ to $+0.65$
i.e., when the bias voltage becomes $1.3$V, no electron transmission takes
place through the molecule, which results zero circulating current. For the
other voltage regimes finite circular currents are available depending on the
voltage region.

Finally, we focus on the variation of circular currents and corresponding
magnetic fields at the ring centers as a function of relative twist angle
$\theta$, when the voltage bias is kept constant. The results are shown in
Fig. 11, where two different colored curves represent the similar meaning as
in Fig. 10. Quite interestingly we see that the magnitudes of the magnetic
fields at the centers of two molecular rings decreases monotonically with
relative twist angle $\theta$, and for large enough $\theta$ they eventually
reduce to zero. Thus, for a particular bias voltage one can tune the strength
of magnetic field established at the ring center due to this circular current
simply by twisting one benzene molecule relative to the other, and hence, by
placing a local spin or a magnetic ion at the ring center, which will interact
to the magnetic field, spin selective transmission can be achieved through
this molecular system. This conformation-dependent spin selective transmission
will be investigated in a recent forthcoming paper.

##  IV Conclusion

In conclusion, we have investigated in detail the conformation-dependent two-
terminal electron transport through a biphenyl molecule within a simple tight-
binding framework using Green’s function formalism. Two principal results have
been obtained and analyzed. First, the dependence of molecular twist on
electronic transmission probability and the overall junction current have been
discussed. Our results lead to a possibility of getting the conformation-
dependent switching action using this molecule. Second, we have investigated
the variation of circular currents and associated magnetic fields developed at
the ring centers as a function of the relative twist angle. Tuning this angle,
one can tailor the strength of magnetic field at the ring centers, and, we
believe that the present analysis may provide the possibilities to design
molecular spintronic devices using organic molecules with loop substructures.

Throughout our work we have ignored the inter- and intra-site Coulomb
interactions as well as the effect of the electrodes, which we plan to
consider in our future works. Another important assumption is the zero
temperature approximation. Though all the results presented in this
communication are worked out at absolute zero temperature limit, the results
should remain valid even at finite temperatures ($\sim 300\,$K) since the
broadening of the energy levels of the biphenyl molecule due to its coupling
with the metal electrodes is much higher than that of the thermal broadening
datta .

##  V Acknowledgment

The author is thankful to Prof. Abraham Nitzan for many stimulating
discussions.

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