EFI-14-20

Comments on Quantum Higgs Vacua

Jeffrey A. Harvey, David Kutasov and Sungjay Lee

Enrico Fermi Institute and Department of Physics University of Chicago
5620 Ellis Av., Chicago Illinois 60637, USA

We study two-dimensional ${\cal N}=(4,4)$ supersymmetric gauge theories which
do not have classical Higgs branches, but are believed to have isolated
quantum Higgs vacua. We provide arguments for the existence of such vacua
based on brane dynamics in string theory and the supersymmetric partition sums
of the gauge theories.

##  1 Introduction

Many two dimensional supersymmetric gauge theories have a classical moduli
space of vacua consisting of Higgs and Coulomb branches connected at the
origin.#1#1#1In general there are also mixed branches; our discussion is easy
to generalize to include them. Due to the large infrared fluctuations
associated with scalar fields in two dimensions, one might expect the low
energy theory to be a sigma model on this moduli space. Instead, it often
splits into a direct sum of separate theories associated with the Coulomb and
Higgs branches [1].

A simple example of this phenomenon occurs in ${\cal N}=(4,4)$ supersymmetric
QED with $F$ charged hyper multiplets. Classically, this theory has a one
(quaternionic) dimensional Coulomb branch and an $F-1$ dimensional Higgs
branch, that intersect at the origin of both. Quantum mechanically, it splits
in the infrared into two decoupled SCFT’s, which correspond to the two
classical branches and in general have different central charges.

From the perspective of the Coulomb branch theory the classical origin is
pushed to infinity by a quantum correction to the metric. More precisely, the
one-loop exact metric on the Coulomb branch is [2]

$\displaystyle
ds^{2}=\left(\frac{1}{g^{2}}+\frac{F}{|\vec{\phi}|^{2}}\right)d\vec{\phi}\cdot
d\vec{\phi}\ ,$ (1)

where $\vec{\phi}$ are the scalars in the $(4,4)$ vector multiplet, and
$g^{2}$ is the gauge coupling constant. The origin of the Coulomb branch,
$\vec{\phi}=0$, which classically is a finite point, lies quantum mechanically
at the bottom of an infinite throat. This is regarded as a signal of the above
decoupling in this branch [1]. It is also useful to recall for future
reference that the Coulomb branch can be lifted by turning on an FI term $\xi$
for the $U(1)$ factor in the gauge group, which breaks the gauge symmetry and
leaves only the Higgs branch theory.

An interesting special case of the above theory is $F=1$, i.e., one charged
hyper multiplet. For $\xi\not=0$ this theory has an isolated Higgs vacuum in
which the gauge symmetry is broken. For $\xi=0$, the classical analysis
suggests that there is a Coulomb branch, but no Higgs vacuum. However, there
are reasons to believe that there is actually a separate isolated Higgs vacuum
which does not contain any light excitations and owes its existence to quantum
dynamics. This vacuum is important for describing fivebranes in matrix string
theory, and for understanding the D$1$-D$5$ system.

A similar quantum Higgs vacuum is expected to appear in theories with higher
rank gauge groups, such as ${\cal N}=(4,4)$ supersymmetric QCD with gauge
group $U(N)$ and $F=N$ hyper multiplets in the fundamental representation. One
can think of its existence as a consequence of the duality satisfied by the
Higgs branch theory, under which $(N,F)\to(F-N,F)$.

In this note we discuss these vacua from two points of view: brane dynamics in
string theory, and the supersymmetric index of the relevant gauge theory. In
section 2 we review the embedding of ${\cal N}=(4,4)$ SQCD in string theory as
the low energy theory on a system of intersecting NS$5$-branes and D-branes,
which makes the quantum vacuum structure more transparent, partly because it
makes the above Seiberg-like duality manifest. We use this description to
understand the isolated quantum Higgs vacua, and discuss briefly the closely
related problem of the behavior of the duality under mass deformations. In
section 3 we review the definition of the supersymmetric index of ${\cal
N}=(0,2)$ supersymmetric theories computed recently in [3, 4] using
localization (see also [5]). We then explain how the supersymmetric index of
these gauge theories can be used to deduce the existence of the isolated
quantum vacua.

##  2 Brane Perspective

Systems of NS$5$-branes and D-branes in string theory are useful for studying
the low energy behavior of many supersymmetric and non-supersymmetric vacua of
SYM theories in various dimensions (see e.g. [6] for a review), and this is
the case here as well. The brane system relevant for describing ${\cal
N}=(4,4)$ SQCD#2#2#2A dimensional reduction of that of [7], which was used to
study ${\cal N}=4$ supersymmetric gauge theories in $2+1$ dimensions. is
depicted in figure 1(a).

Figure 1: IIA brane realizations of ${\cal N}=(4,4)$ SQCD.

The ${\cal N}=(4,4)$ $U(N)$ gauge multiplet lives on $N$ D$2$-branes stretched
in the $x^{6}$ direction between two NS$5$-branes whose world volume fills the
$(2345)$ directions. Both types of branes are extended in the $1+1$
dimensional spacetime labeled by $(01)$. The hypermultiplets live at
intersections of the D$2$-branes and $F$ D$4$-branes stretched in $(01789)$.
To summarize, the worldvolume directions of the various branes are:

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---|---|---
NS5 | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ |  |  |  |
D2 | $\times$ | $\times$ |  |  |  |  | $\times$ |  |  |
D4 | $\times$ | $\times$ |  |  |  |  |  | $\times$ | $\times$ | $\times$

To study the isolated quantum Higgs vacua it is convenient to use an
alternative representation of this system, depicted in figure 1(b). The
representations of figures 1(a) and (b) are related by the Hanany-Witten
transition [6, 7]. Consider, for instance, the case $N=1$ (gauge group
$U(1)$). The classical Coulomb branch is parametrized by the location in the
$(2345)$ directions of the color D$2$-brane stretched between the fivebranes.
The Higgs branch is obtained in figure 1(a) by letting the D$2$-brane split
into segments on the D4-branes. The $F-1$ moduli correspond to the positions
of the segments stretched between adjacent D4-branes in the $(789)$
directions.#3#3#3The fourth modulus in the multiplet can be identified with
the $6$ component of the gauge field on the D$2$-branes. In figure 1(b) the
Higgs branch is described by reconnecting the color D$2$-brane to the longest
flavor D$2$-brane, and allowing the segments of the resulting two brane
connecting adjacent D$4$-branes to move in the $(789)$ directions [6]. The two
branches intersect at the origin of both, which is depicted in figure 1.

Turning on an FI term $\xi$ for the $U(1)$ corresponds in the brane
description to a relative displacement of the two NS$5$-branes in the
transverse directions $(789)$. As is clear from figure 1, this lifts the
classical Coulomb branch, and leaves behind only the Higgs branch.

For $N=F=1$ we recover the structure mentioned in section 1. For $\xi\not=0$
we find a vacuum, which from the brane perspective is described by a
configuration in which the color and flavor D$2$-branes in figure 1(b) combine
into a single D$2$-brane that connects the right NS$5$-brane to the
D$4$-brane, never intersecting the second NS$5$-brane. Since the low energy
theory on such a D$2$-brane is massive [7], this vacuum is isolated.

For $\xi=0$, the classical picture of figure 1 suggests that the above
isolated state becomes identical to the origin of the Coulomb branch, and in
particular there is no independent isolated vacuum. However, as we mentioned,
quantum mechanically we expect such a vacuum to exist. To understand this from
the brane point of view we recall that the classical configurations of figure
1 are corrected at finite $g_{s}$ by brane bending effects [8, 9]. A
D$2$-brane ending on an NS$5$-brane can be thought of as a charged particle in
the five dimensional gauge theory obtained by reducing the fivebrane theory on
$x^{1}$ (or, equivalently, a string charged under the self-dual $B_{\mu\nu}$
field in the six dimensional IIA fivebrane theory). Thus, the gauge field on
the fivebrane has a non-trivial profile $F_{0r}\sim 1/r^{3}$, where $r$ is the
radial direction in $(2345)$. Since this configuration is BPS, there is also a
non-trivial profile of the scalar field on the fivebrane $x^{6}(r)\sim
1/r^{2}$.

Figure 2: Classical (a) and quantum (b) descriptions of the Coulomb branch of
$U(1)$ ${\cal N}=(4,4)$ SQED with one charged hyper multiplet.

This has an important effect on the vacuum structure. A generic point on the
Coulomb branch, that classically is described by the configuration of figure
2(a), is replaced at finite $g_{s}$ by that of figure 2(b). In particular, the
kinetic term for the position of the color $D2$-brane in the $(2345)$
directions, $\vec{\phi}(x^{\mu})$, which classically is canonical, ${\cal
L}_{\text{kin}}\sim|\partial_{\mu}\vec{\phi}|^{2}$, develops a singularity at
$\vec{\phi}=0$, ${\cal
L}_{\text{kin}}\sim|\partial_{\mu}\vec{\phi}|^{2}/|\vec{\phi}|^{2}$. Thus, the
origin of the Coulomb branch is pushed to infinity, in agreement with the
discussion of [1, 2]. The supersymmetric brane configuration depicted in
figure 3 represents the isolated quantum Higgs vacuum of the theory.

The above discussion easily generalizes to the case of $U(N)$ gauge theory
with $F=N$ hypermultiplets (see figure 1). It is obvious in the brane
description that the classical theory (with $\xi=0$) has an $N$ dimensional
Coulomb branch, parametrized by the positions of the $N$ color D$2$-branes,
and no vacua with lower rank gauge group. Quantum mechanically, the vacuum
structure is much richer: each of the $N$ color D$2$-branes can combine with a
flavor D$2$-brane to form the configuration of figure 3, thereby reducing the
rank of the gauge group by one unit. When all $N$ color D$2$-branes do this,
one finds an isolated vacuum with no light excitations.

Figure 3: The quantum Higgs vacuum of the theory of figure 2.

One can take a broader viewpoint and consider the theory with general $N$ and
$F$ in figure 1. This theory has a rich vacuum structure that depends on $N$,
$F$. For $F<N$, a Higgs branch does not exist, and if we turn on a non-zero FI
parameter $\xi$ for the $U(1)$, there is no supersymmetric vacuum. For $F>N$
there is a Higgs branch of the classical moduli space in which the gauge
symmetry is fully broken. Its dimension is given by $N(F-N)$. This branch
gives rise in the quantum theory to a CFT with central charge $c=6N(F-N)$.

This CFT enjoys a duality $N\to F-N$, which is due to the fact that the Higgs
branches of the $U(N)$ and $U(F-N)$ theories with $F$ flavors coincide. In the
gauge theory this is familiar from studies of ${\cal N}=2$ SQCD in four
dimensions [10]. In two dimensions it was demonstrated that the two-sphere
partition functions [11, 12] and elliptic genera [4] of these two gauge
theories are the same, which supports the duality. In the brane language it is
a simple generalization of the results of [13].

For $F=N$ the electric theory is the Higgs branch of the $U(N)$ theory with
$N$ flavors, while the magnetic theory becomes a $U(0)$ theory, i.e. a
supersymmetric vacuum with no light excitations. This supersymmetric vacuum is
the isolated Higgs vacuum we discussed above.

We note in passing that the set of isolated vacua becomes richer when we
consider deformations. For example, consider turning on (equal) masses for
$F_{1}$ of the $F$ fundamental hypermultiplets.

Figure 4: Map of Higgs vacua of a $U(N)$ gauge theory with $F=N+1$ flavors
(top) and the dual $U(1)$ theory with $N+1$ flavors (bottom) when one of the
flavors has a non-zero mass.

In the brane language this corresponds to displacing $F_{1}$ of the
D$4$-branes from the origin of the $(2345)$ space to a particular point while
leaving the remaining $F_{2}=F-F_{1}$ D4-branes at the origin. In order for
the theory to have a Higgs branch, we need to divide the D$2$-branes into two
groups, of $N_{1}$ and $N_{2}$ with $N_{1}+N_{2}=N$, and displace them so that
they intersect the corresponding sets of D$4$-branes; see figure 4 for a
simple example. The existence of a Higgs vacuum implies that $0\leq N_{i}\leq
F_{i}$, which can be equivalently written as

$\displaystyle\max(F_{1}-\hat{N},0)\leq N_{1}\leq\min(F_{1},N)$ (2)

where $\hat{N}=F-N$. It is easy to see that this condition is invariant under
the duality map $N\to\hat{N}$, $N_{i}\to\hat{N}_{i}=F_{i}-N_{i}$, in agreement
with the fact that the low energy theory reduces in this case to a direct
product of CFT’s at the two intersections of $N_{i}$ and $F_{i}$ branes, with
central charges $c_{i}=6N_{i}(F_{i}-N_{i})$.

Whenever one of the $N_{i}$ coincides with the corresponding $F_{i}$, we get
an isolated vacuum in that theory, which corresponds under the duality map to
a theory with no light excitations but a supersymmetric vacuum. Thus, the
existence of the quantum Higgs vacua is necessary for the consistency of the
$(4,4)$ duality with deformations.

##  3 Supersymmetric Index

Two dimensional ${\cal N}=(4,4)$ $U(N)$ SQCD coupled to $F$ hypermultiplets in
the fundamental representation has a global $SU(F)$ flavor symmetry, as well
as an $SU(2)_{1}\times SU(2)_{2}\times SU(2)_{3}$ R-symmetry, under which the
vector multiplet scalars transform as $({\bf 1},{\bf 2},{\bf 2})$ and the
hyper multiplet scalars transform as $({\bf 2},{\bf 1},{\bf 1})$. The four
right-moving supercharges ${\cal Q}_{+}^{a\alpha}$ transform as $({\bf 2},{\bf
2},{\bf 1})$, while the four left-moving supercharges ${\cal
Q}_{-}^{a\dot{\alpha}}$ transform as $({\bf 2},{\bf 1},{\bf 2})$. The indices
$\alpha,\dot{\alpha}$ correspond to doublets under $SU(2)_{2}$ and
$SU(2)_{3}$, respectively. For $F>N$, when the Higgs branch theory is non-
trivial, $SU(2)_{2}$ and $SU(2)_{3}$ become parts of the infrared right and
left-moving ${\cal N}=4$ superconformal algebras, respectively.

Choosing particular supercharges ${\cal Q}_{+}^{++}$ and ${\cal Q}_{+}^{--}$
that generate an ${\cal N}=(0,2)$ subalgebra of the full ${\cal N}=(4,4)$ SUSY
algebra (i.e. $\\{{\cal Q}_{+}^{++},{\cal Q}_{+}^{--}\\}=\bar{L}_{0}$), we can
define a supersymmetric partition sum as follows

$\displaystyle{\cal E}(\tau;z,\zeta_{i})=\text{Tr}_{{\cal
H}_{\text{RR}}}\Big{[}(-1)^{F}q^{L_{0}}\bar{q}^{\bar{L}_{0}}e^{-2\pi
izJ_{R}}e^{-2\pi i\zeta_{1}(J_{1}-J_{2})}e^{-2\pi i\zeta_{2}\cdot
J_{f}}\Big{]}\ ,$ (3)

where $\\{J_{1},J_{2},J_{R},J_{f}\\}$ are Cartan subalgebra generators of the
R-symmetry group $SU(2)_{1}\times SU(2)_{2}\times SU(2)_{3}$ and the flavor
group $SU(F)$. Note that the global charges $J_{R},J_{1}-J_{2}$ and $J_{f}$
commute with the chosen supercharges ${\cal Q}_{+}^{++}$ and ${\cal
Q}_{+}^{--}$. Thus, states with $\bar{L}_{0}\not=0$ come in pairs whose
contributions to the partition sum (3) vanish. When the Higgs theory is non-
trivial (for $F>N$), one can think of this partition sum as the (equivariant)
elliptic genus of this theory, but it can be defined when the infrared SCFT is
trivial as well, and we will in fact be interested in this case below. Thus,
we will refer to (3) as a supersymmetric index.

The index (3) is independent of the FI parameter $\xi$, since the FI term is
${\cal Q}_{+}$-exact with ${\cal Q}_{+}={\cal Q}_{+}^{++}+{\cal Q}_{+}^{--}$
and crucially the asymptotic behavior of the potential on field space is
independent of this parameter. In particular, the superficially flat potential
on the Coulomb branch at $\xi=0$ is lifted by the nonzero background gauge
field that couples to $J_{1}-J_{2}$.

As mentioned above, for $\xi\not=0$ the Coulomb branch is lifted, so in that
case the index must come from the Higgs branch. In particular, for the case
$F=N$ where the vacuum is isolated (i.e. all field theoretic degrees of
freedom are massive), the index must be equal to $1$,

$\displaystyle{\cal E}(\tau;z,\zeta_{i})=1\ .$ (4)

This index was recently computed in [4, 5] using localization. Below we will
use their results to verify (4).

For $\xi=0$, we saw in the previous section that the infrared limit of the
gauge theory with $F=N$ is a direct sum of the $N$ dimensional Coulomb branch,
mixed branches in which the rank of the unbroken subgroup of $U(N)$ is lower
than $N$, and the isolated Higgs vacuum. The index is the sum of the
contributions of the various branches. The $\xi$ independence of the index (3)
implies that the result (4) must still be valid when $\xi=0$, and it is
natural to ask which branch(es) it is coming from. We will see later that the
Coulomb and mixed branches do not contribute to the index. Hence, the result
(4) must come from the conjectured isolated Higgs vacuum. We conclude that
this result requires the existence of the quantum Higgs vacuum.

The above statements must be true on general grounds, but it is useful to
verify them using the formalism of [4]. It is convenient to rewrite the index
(3) in the form

$\displaystyle{\cal E}(\tau;\vec{z})=\text{Tr}_{{\cal
H}_{\text{RR}}}\left[(-1)^{F}q^{L_{0}}\bar{q}^{\bar{L}_{0}}e^{2\pi
i\vec{z}\cdot\vec{J}}\right]\ ,$ (5)

where $\vec{J}=(J_{R},J_{1}-J_{2},J_{f})$. The authors of [4] showed that for
any ${\cal N}=(0,2)$ supersymmetric gauge theory one can write ${\cal E}$ as

$\displaystyle{\cal
E}(\tau;\vec{z})=\frac{1}{|W|}\sum_{u_{\ast}\in\mathfrak{M}_{\eta}}\operatorname*{JK-
Res}_{u_{\ast}}(Q_{\ast},\eta)Z_{\text{1-loop}}(u,\vec{z})\ ,$ (6)

where $u$ denotes the holomorphic coordinates parametrizing the moduli space
of flat gauge connection on the torus,
$u\in\left(\mathbb{C}/\mathbb{Z}+\tau\mathbb{Z}\right)^{r}$ for a rank $r$
gauge group, and $|W|$ is the order of the Weyl group. The one-loop
determinant $Z_{\text{1-loop}}$ is

$\displaystyle Z_{\text{1-loop}}(u)=Z_{\text{vec}}(u)\cdot
Z_{\text{chi}}(u)\cdot Z_{\text{fer}}(u)$ (7)

with the contributions from $(0,2)$ vector, chiral and Fermi multiplets given
by

$\displaystyle Z_{\text{vec}}(u)$
$\displaystyle=\left[\frac{2\pi\eta(\tau)^{2}}{i}\right]^{r}\prod_{a:\text{roots}}\frac{i\vartheta_{1}(\tau,\alpha\cdot
u)}{\eta(\tau)}\ ,$ $\displaystyle Z_{\text{chi}}(u)$
$\displaystyle=\prod_{\rho:\text{weights}}\frac{i\eta(\tau)}{\vartheta_{1}(\tau,\rho\cdot
u+\vec{z}\cdot\vec{J}_{c})}\ ,$ $\displaystyle Z_{\text{fer}}(u)$
$\displaystyle=\prod_{\rho:\text{weights}}\frac{i\vartheta_{1}(\tau,\rho\cdot
u+\vec{z}\cdot\vec{J}_{f})}{\eta(\tau)}\ ,$ (8)

where $\vec{J}_{c}$ and $\vec{J}_{f}$ are the $\vec{J}$ charges of chiral and
Fermi multiplets. Finally, ‘JK-Res’ is the Jeffrey-Kirwan residue whose
definition we review next.

Each $\vartheta_{1}$ factor in the denominator of $Z_{\text{1-loop}}$ (8)
defines a hyperplane in the $u$-plane,

$\displaystyle\rho\cdot u+\vec{z}\cdot\vec{J}_{c}=0\ ,$ (9)

where $Z_{\text{1-loop}}(u)$ becomes singular. Such a hyperplane is associated
with a charge vector $\rho$ in $\mathbb{R}^{r}$. Let us denote by
$\mathfrak{M}$ the set of singular points $u_{\ast}$ where $n\geq r$
hyperplanes intersect. We focus on the case $n=r$, which is the case of
interest for the theories discussed below. The Jeffrey-Kirwan residue depends
on a choice of an $r$-component vector $\eta$. For given $\eta$, one can
define a set $\mathfrak{M}_{\eta}$ of singular points in $\mathfrak{M}$ where
$r$ hyperplanes intersect and the corresponding $r$ charge vectors generate a
positive cone containing the vector $\eta$,

$\displaystyle\eta\in\text{Cone}[\rho_{1},\rho_{2},..,\rho_{r}]\ .$ (10)

In the case of $n=r$, the Jeffrey-Kirwan residue is defined as follows

$\operatorname*{JK-Res}_{u_{\ast}}\frac{1}{(\rho_{1}\cdot
u)\cdots(\rho_{r}\cdot
u)}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{|\text{det}(\rho_{1}\cdots\rho_{r})|}{}{}&\text{if}~{}u_{\ast}\in\mathfrak{M}_{\eta}\\\
0&\text{otherwise}.\end{array}\right.~{}$ (11)

For more details of the Jeffrey-Kirwan residue see [4, 14]. The index is of
course independent of the choice of $\eta$. For rank one gauge group with
$\eta>0$ ($\eta<0$), the index ${\cal E}(\tau,\vec{z})$ becomes the sum of
residues of $Z_{\text{1-loop}}(u)$ at the poles $u=u_{\ast}$ associated with
the fields of positive (negative) charge.

As a warm-up exercise, let us start with a $U(1)$ theory with a single charged
hyper multiplet of charge $+1$. From (8) one can show that the one-loop factor
corresponding to the $(4,4)$ vector multiplet is

$\displaystyle
Z_{vec}(u)=\frac{i\eta(q)^{3}}{\vartheta_{1}(\tau,\zeta_{1}-z)}\frac{\vartheta_{1}(\tau,2\zeta_{1})}{\vartheta_{1}(\tau,\zeta_{1}+z)}$
(12)

while the contribution from the charged hyper multiplet is

$\displaystyle Z_{hyper}(u)$
$\displaystyle=\frac{\vartheta_{1}(\tau,u-z)}{\vartheta_{1}(\tau,u-\zeta_{1})}\frac{\vartheta_{1}(\tau,-u-z)}{\vartheta_{1}(\tau,-u-\zeta_{1})}\
.$ (13)

Choosing $\eta>0$, the index thus becomes

$\displaystyle{\cal E}(\tau;\xi,z)$
$\displaystyle=\operatorname*{Res}_{u=\zeta_{1}}Z_{vec}(u)\cdot
Z_{hyper}(u)=1\ .$ (14)

One can further show that the contribution to the index from the Coulomb
branch vanishes. The low-energy theory on the Coulomb branch can be described
by two $(0,2)$ neutral chiral multiplets $\Sigma,\tilde{\Sigma}$ and two
neutral Fermi multiplets $\Upsilon,\tilde{\Upsilon}$. Here the Fermi multiplet
$\Upsilon$ contains the field strength $F_{01}$. The global charges are given
by $\vec{J}_{\Sigma}=(-1,1)$, $\vec{J}_{\tilde{\Sigma}}=(1,1)$,
$\vec{J}_{\Upsilon}=(0,0)$ and $\vec{J}_{\tilde{\Upsilon}}=(2,0)$. Since the
Fermi multiplet $\Upsilon$ is neutral under both $J_{R}$ and $J_{1}-J_{2}$,
one can show from (8) that the contribution from the Coulomb branch to the
index vanishes. This is essentially because the fermionic zero mode in
$\Upsilon$ will pair states of opposite fermion number and equal charges
$\vec{J}$.

We next turn to the case of general $N$, i.e., ${\cal N}=(4,4)$ $U(N)$ gauge
theory with $F=N$ fundamental hyper multiplets. We use the standard unit
vectors $e_{i}$ in $\mathbb{R}^{N}$, $i=1,2,..,N$, to parameterize the weight
vectors of the fundamental representation. The root vectors can be expressed
as $e_{i}-e_{j}$ with $i\neq j$. The one-loop determinant is in this case

$\displaystyle Z_{vec}(u)$
$\displaystyle=\left[\frac{i\eta(q)^{3}}{\vartheta_{1}(\tau,\zeta_{1}-z)}\frac{\vartheta_{1}(\tau,2\zeta_{1})}{\vartheta_{1}(\tau,\zeta_{1}+z)}\right]^{N}$
$\displaystyle\times\prod_{i\neq j}\frac{\vartheta_{1}(\tau,(e_{i}-e_{j})\cdot
u)}{\vartheta_{1}(\tau,(e_{i}-e_{j})\cdot
u+\zeta_{1}-z)}\frac{\vartheta_{1}(\tau,(e_{i}-e_{j})\cdot
u+2\zeta_{1})}{\vartheta_{1}(\tau,(e_{i}-e_{j})\cdot u+\zeta_{1}+z)}\ ,$ (15)

and

$\displaystyle
Z_{hyper}(u)=\prod_{i}\prod_{\mu:\text{weights}}\frac{\vartheta_{1}(\tau,e_{i}\cdot
u+\mu\cdot\zeta_{2}-z)}{\vartheta_{1}(\tau,e_{i}\cdot
u+\mu\cdot\zeta_{2}-\zeta_{1})}\frac{\vartheta_{1}(\tau,-e_{i}\cdot
u-\mu\cdot\zeta_{2}-z)}{\vartheta_{1}(\tau,-e_{i}\cdot
u-\mu\cdot\zeta_{2}-\zeta_{1})}\ ,$ (16)

where $\mu$ denote the weight vectors of the fundamental representations under
the flavor group $SU(N)$.

To compute the index#4#4#4Our analysis is parallel to that of [4]., we first
choose the $N$-component vector $\eta$ as follows

$\displaystyle\eta=e_{1}+e_{2}+..+e_{N}\ .$ (17)

One can show that any set of charge vectors generating a positive cone that
contains the vector $\eta$ should have either a pair of weight vector $e_{i}$
and root vector $e_{j}-e_{i}$ for certain $i$ and $j$ or weight vectors only.

In the case that the set of charge vectors have a pair of $e_{i}$ and
$e_{j}-e_{i}$, the Jeffrey-Kirwan residue becomes trivial. This is because the
corresponding singular points satisfy

$\displaystyle\left(e_{j}-e_{i}\right)\cdot u_{\ast}+\zeta_{1}\pm z$
$\displaystyle=0$ $\displaystyle e_{i}\cdot
u_{\ast}+\mu\cdot\zeta_{2}-\zeta_{1}$ $\displaystyle=0$ (18)

with a weight vector $\mu$ in the fundamental representation of $SU(N)$. This
implies that

$\displaystyle e_{j}\cdot u_{\ast}+\mu\cdot\zeta_{2}\pm z=0$ (19)

for which one of the factors in the numerator of $Z_{hyper}$ in (16) vanishes.

For the cone generated by the weight vectors $\\{e_{1},e_{2},..,e_{N}\\}$,
there are corresponding $N^{N}$ singular points satisfying

$\displaystyle e_{1}\cdot u_{\ast}+\mu_{1}\cdot\zeta_{2}-\zeta_{1}$
$\displaystyle=0\ ,$ $\displaystyle e_{2}\cdot
u_{\ast}+\mu_{2}\cdot\zeta_{2}-\zeta_{1}$ $\displaystyle=0\ ,$
$\displaystyle\vdots$ $\displaystyle e_{N}\cdot
u_{\ast}+\mu_{N}\cdot\zeta_{2}-\zeta_{1}$ $\displaystyle=0\ ,$ (20)

for arbitrary $N$ weight vectors $\mu_{i}$. However if two such weight vectors
coincide, for instance $\mu_{1}=\mu_{2}$, the Jeffrey-Kirwan residue vanishes
because

$\displaystyle\left(e_{1}-e_{2}\right)\cdot u_{\ast}=0$ (21)

at which the factor $\vartheta_{1}(\tau,(e_{1}-e_{2})\cdot u)$ in the
numerator of $Z_{vec}$ in (15) becomes zero. Thus we have $N!$ singular points
with $\mu_{i}\neq\mu_{j}$ for any $i,j$, at each of which the Jeffrey-Kirwan
residue can be shown to give one. Dividing by the order of the Weyl group (6)
one finds

$\displaystyle{\cal E}(\tau;z,\xi_{i})=1\ .$ (22)

As in the $U(1)$ case, one can verify that the contributions to the index from
the Coulomb and mixed branches vanish due to the existence of the neutral
Fermi multiplet. Therefore, one concludes that there must exist an isolated
quantum Higgs vacuum in the ${\cal N}=(4,4)$ $U(N)$ theory coupled to $N$
flavors.

## Acknowledgments

We thank Piljin Yi for helpful discussions. JH acknowledges the support of NSF
grant 1214409. DK is supported in part by DOE grant DE-FG02-13ER41958. The
work of SL is supported in part by the Ernest Rutherford fellowship of the
Science and Technology Facilities Council ST/J003549/1.

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# On the automorphism groups of some Leibniz algebras

L.A. Kurdachenko, A.A. Pypka
Oles Honchar Dnipro National University, Dnipro, Ukraine
e-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
I.Ya. Subbotin
National University, Los Angeles, USA
e-mail<EMAIL_ADDRESS>
###### Abstract

We study the automorphism groups of finite-dimensional cyclic Leibniz
algebras. In this connection, we consider the relationships between groups,
modules over associative rings and Leibniz algebras.

Key Words: automorphism group, (cyclic) Leibniz algebra, module over
associative ring.

2010 MSC: 20B27,17A32,17A36,17A99.

##  1 Introduction

Let $L$ be an algebra over a field $F$ with the binary operations $+$ and
$[,]$. Then $L$ is called a left Leibniz algebra if it satisfies the left
Leibniz identity

$[[a,b],c]=[a,[b,c]]-[b,[a,c]]$

for all $a,b,c\in L$.

Leibniz algebras first appeared in the paper of A. Blokh [5], while the term
“Leibniz algebra” appeared in the book of J.-L. Loday [20], and the article of
J.-L. Loday [21]. In [22], J.-L. Loday and T. Pirashvili began to study
properties of Leibniz algebras. The theory of Leibniz algebras has developed
very intensively in many different directions. Some of the results of this
theory were presented in the book [4]. Note that Lie algebras are a partial
case of Leibniz algebras. Conversely, if $L$ is a Leibniz algebra, in which
$[a,a]=0$ for every $a\in L$, then it is a Lie algebra. Thus, Lie algebras can
be characterized as anticommutative Leibniz algebras. The question about those
properties of Leibniz algebras that are absent in Lie algebras, and
accordingly about those types of Leibniz algebras that have essential
differences from Lie algebras, naturally arises. A lot has already been done
in this direction, including some results of the authors of this article. Many
new results can be found in the survey papers [7, 11, 15]. Other results
related to this topic can be found in [1, 2, 8, 9, 10, 13, 14, 16, 17].

In the study of Leibniz algebras the information about their endomorphisms and
derivations is very useful, as shown, for example, in [3, 19]. The
endomorphisms and derivations of infinite-dimensional cyclic Leibniz algebras
were investigated in [18].

Let $L$ be a Leibniz algebra. As usual, a linear transformation $f$ of $L$ is
called an endomorphism of $L$ if

$f([a,b])=[f(a),f(b)]$

for all $a,b\in L$. Clearly, a product of two endomorphisms of $L$ is also an
endomorphism of $L$, so that the set of all endomorphisms of $L$ is a
semigroup by a multiplication. In the same time, the sum of two endomorphisms
of $L$ is not necessary to be an endomorphism of $L$, so that we cannot speak
about the endomorphism ring of $L$.

Here we will use the term “semigroup” for the set that has an associative
binary operation. For a semigroup with an identity element, we will use the
term “monoid”. Clearly, an identical transformation is an endomorphism of $L$.
Therefore, the set $\mathbf{End}_{[,]}(L)$ of all endomorphisms of $L$ is a
monoid by a multiplication. As usual, a bijective endomorphism of $L$ is
called an automorphism of $L$.

Let $f$ be an automorphism of $L$. Then the mapping $f^{-1}$ is also an
automorphism of $L$. Indeed, let $x,y$ be arbitrary elements of $L$. Then
there are elements $u,v\in L$ such that $x=f(u)$, $y=f(v)$ and

$f^{-1}([x,y])=f^{-1}([f(u),f(v)])=f^{-1}(f[u,v])=[u,v]=[f^{-1}(x),f^{-1}(y)].$

Thus, the set $\mathbf{Aut}_{[,]}(L)$ of all automorphisms of $L$ is a group
by a multiplication.

It should be noted that endomorphisms of Leibniz algebras have hardly been
studied. It is also quite unusual that the structure of cyclic Leibniz
algebras is described relatively recently in [6]. In this paper, we began the
study of the structure of the automorphism groups of finite-dimensional cyclic
Leibniz algebras.

Let $L$ be a cyclic Leibniz algebra, $L=\langle a\rangle$, and suppose that
$L$ has finite dimension over a field $F$. Then there exists a positive
integer $n$ such that $L$ has a basis $a_{1},\ldots,a_{n}$ where

$a_{1}=a,a_{2}=[a_{1},a_{1}],\ldots,a_{n}=[a_{1},a_{n-1}],[a_{1},a_{n}]=\alpha_{2}a_{2}+\ldots+\alpha_{n}a_{n}.$

Moreover, $[L,L]=\mathbf{Leib}(L)=Fa_{2}+\ldots+Fa_{n}$ [6]. We fix these
designations.

The following types of cyclic Leibniz algebras appear here.

The first case: $[a_{1},a_{n}]=0$. In this case, $L$ is nilpotent, and we will
say that $L$ is a cyclic algebra of type (I). The structure of the
automorphism group of a cyclic Leibniz algebra of type (I) is described in
Section 1.

Now we need the following concepts. The left (respectively right) center
$\zeta^{\mathrm{left}}(L)$ (respectively $\zeta^{\mathrm{right}}(L)$) of a
Leibniz algebra $L$ is defined by the following rule:

$\zeta^{\mathrm{left}}(L)=\\{x\in L|\ [x,y]=0\ \mbox{for each }y\in L\\}$

(respectively,

$\zeta^{\mathrm{right}}(L)=\\{x\in L|\ [y,x]=0\ \mbox{for each }y\in L\\}).$

The left center of $L$ is an ideal of $L$, but it is not true for the right
center of $L$. Moreover, $\mathbf{Leib}(L)\leqslant\zeta^{\mathrm{left}}(L)$,
so that $L/\zeta^{\mathrm{left}}(L)$ is a Lie algebra. The right center of $L$
is a subalgebra of $L$, and in general the left and right centers are
different. They can even have different dimensions (see [12]).

The center $\zeta(L)$ of $L$ is defined by the following rule:

$\zeta(L)=\\{x\in L|\ [x,y]=0=[y,x]\ \mbox{for each }y\in L\\}.$

The center is an ideal of $L$.

Consider now the second type of cyclic Leibniz algebras. In this case,
$[a_{1},a_{n}]=\alpha_{2}a_{2}+\ldots+\alpha_{n}a_{n}$ and $\alpha_{2}\neq 0$.
Put $c=\alpha_{2}^{-1}(\alpha_{2}a_{1}+\ldots+\alpha_{n}a_{n-1}-a_{n})$. Then
$[c,c]=0$, $Fc=\zeta^{\mathrm{right}}(L)$, $L=[L,L]\oplus Fc$,
$[c,b]=[a_{1},b]$ for every $b\in[L,L]$ [6]. In particular,
$a_{3}=[c,a_{2}],\ldots,a_{n}=[c,a_{n-1}],[c,a_{n}]=\alpha_{2}a_{2}+\ldots+\alpha_{n}a_{n}$.
In this case, we will say that $L$ is a cyclic algebra of type (II).

For description of the automorphism group of a cyclic Leibniz algebra of type
(II) we consider the relationships between Leibniz algebras and modules over
associative ring. We will consider these relationships in Section 2. Using
these constructions, in Sections 3 we obtain the description of automorphism
group of a cyclic Leibniz algebra of type (II).

The third case: $[a_{1},a_{n}]=\alpha_{2}a_{2}+\ldots+\alpha_{n}a_{n}$ and
$\alpha_{2}=0$. Let $t$ be the first index such that $\alpha_{t}\neq 0$. In
other words, $[a_{1},a_{n}]=\alpha_{t}a_{t}+\ldots+\alpha_{n}a_{n}$. By our
condition, $t>2$. Then

$[a_{1},a_{n}]=\alpha_{t}[a_{1},a_{t-1}]+\ldots+\alpha_{n}[a_{1},a_{n-1}]=[a_{1},\alpha_{t}a_{t-1}+\ldots+\alpha_{n}a_{n-1}],$

which implies that
$\alpha_{t}a_{t-1}+\ldots+\alpha_{n}a_{n-1}-a_{n}\in\mathbf{Ann}_{L}^{\mathrm{right}}(a_{1})$.
Since $\alpha_{t}\neq 0$, then $\alpha_{t}^{-1}\neq 0$ and
$d_{t-1}=\alpha_{t}^{-1}(\alpha_{t}a_{t-1}+\ldots+\alpha_{n}a_{n-1}-a_{n})=a_{t-1}+\beta_{t}a_{t}+\ldots+\beta_{n}a_{n}\in\mathbf{Ann}_{L}^{\mathrm{right}}(a_{1})$.
Put

$\begin{split}d_{t-2}&=a_{t-2}+\beta_{t}a_{t-1}+\ldots+\beta_{n}a_{n-1},\\\
d_{t-3}&=a_{t-3}+\beta_{t}a_{t-2}+\ldots+\beta_{n}a_{n-2},\\\ &\ldots\\\
d_{1}&=a_{1}+\beta_{t}a_{2}+\ldots+\beta_{n}a_{n-t+1}.\end{split}$

Then

$\begin{split}[d_{1},d_{1}]&=[a_{1},d_{1}]=d_{2},\\\
[d_{1},d_{2}]&=[a_{1},d_{2}]=d_{3},\\\ &\ldots\\\
[d_{1},d_{t-2}]&=[a_{1},d_{t-2}]=d_{t-1},\\\
[d_{1},d_{t-1}]&=[a_{1},d_{t-1}]=0.\end{split}$

It follows that the subspace $U=Fd_{1}\oplus Fd_{2}\oplus\ldots\oplus
Fd_{t-1}$ is a nilpotent subalgebra. Moreover, a subspace
$[U,U]=Fd_{2}\oplus\ldots\oplus Fd_{t-1}$ is an ideal of $L$. Put further
$d_{t}=a_{t},d_{t+1}=a_{t+1},\ldots,d_{n}=a_{n}$.

The following matrix corresponds to this transaction:

$\left(\begin{array}[]{ccccccccccc}1&\beta_{t}&\beta_{t+1}&\ldots&\beta_{n}&0&0&\ldots&0&0&0\\\
0&1&\beta_{t}&\ldots&\beta_{n-1}&\beta_{n}&0&\ldots&0&0&0\\\
0&0&1&\ldots&\beta_{n-2}&\beta_{n-1}&\beta_{n}&\ldots&0&0&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
0&0&0&\ldots&1&\beta_{t}&\beta_{t+1}&\ldots&\beta_{n-1}&\beta_{n}&0\\\
0&0&0&\ldots&0&1&\beta_{t}&\ldots&\beta_{n-2}&\beta_{n-1}&\beta_{n}\\\
0&0&0&\ldots&0&0&1&\ldots&0&0&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
0&0&0&\ldots&0&0&0&\ldots&1&0&0\\\ 0&0&0&\ldots&0&0&0&\ldots&0&1&0\\\
0&0&0&\ldots&0&0&0&\ldots&0&0&1\end{array}\right)$

This matrix is non-singular, which shows that the elements
$\\{d_{1},\ldots,d_{n}\\}$ form a new basis. We note that a subspace
$V=Fd_{t}\oplus\ldots\oplus Fd_{n}$ is a subalgebra. Moreover, $V$ is an ideal
of $L$, because
$[a_{1},d_{t}]=d_{t+1},\ldots,[a_{1},d_{n-1}]=d_{n},[a_{1},d_{n}]=\alpha_{t}d_{t}+\ldots+\alpha_{n}d_{n}$.
Moreover, $[a_{1},d_{j}]=[d_{1},d_{j}]$ for all $j\geqslant t$ [6]. In this
case, we will say that $L$ is a cyclic algebra of type (III).

Thus, $L=A\oplus Fd_{1}$, $A=V\oplus[U,U]$ is a direct sum of two ideals,
$U=[U,U]\oplus Fd_{1}$ is a nilpotent cyclic subalgebra, i.e. is an algebra of
type (I), and $V\oplus Fd_{1}$ is a cyclic subalgebra of type (II). The
structure of automorphism group of a cyclic Leibniz algebra of type (III) is
described in Section 4.

##  2 The automorphism group of a cyclic Leibniz algebra of type (I).

######  Lemma 2.1.

Let $L$ be a Leibniz algebra over a field $F$, $f$ be an automorphism of $L$.
Then $f(\zeta^{\mathrm{left}}(L))=\zeta^{\mathrm{left}}(L)$,
$f(\zeta^{\mathrm{right}}(L))=\zeta^{\mathrm{right}}(L)$,
$f(\zeta(L))=\zeta(L)$, $f([L,L])=[L,L]$.

Proof. Let $x$ be an arbitrary element of $L$ and let
$z\in\zeta^{\mathrm{left}}(L)$. Since $f$ is an automorphism of $L$, there is
an element $y\in L$ such that $x=f(y)$. Then

$[f(z),x]=[f(z),f(y)]=f([z,y])=f(0)=0.$

It follows that $f(z)\in\zeta^{\mathrm{left}}(L)$.

On the other hand, there are the elements $u,v\in L$ such that $z=f(u)$,
$x=f^{-1}(v)$. Then

$[u,x]=[f^{-1}(z),f^{-1}(v)]=f^{-1}([z,v])=f^{-1}(0)=0.$

It follows that $u\in\zeta^{\mathrm{left}}(L)$, so that $z\in
f(\zeta^{\mathrm{left}}(L))$ and therefore
$f(\zeta^{\mathrm{left}}(L))=\zeta^{\mathrm{left}}(L)$. Using the similar
arguments, we obtain that
$f(\zeta^{\mathrm{right}}(L))=\zeta^{\mathrm{right}}(L)$ and
$f(\zeta(L))=\zeta(L)$.

If $x,y\in L$, then $f([x,y])=[f(x),f(y)]\in[L,L]$. It follows that
$f([L,L])\leqslant[L,L]$.

Conversely, let $w\in[L,L]$. Then
$w=\alpha_{1}[u_{1},v_{1}]+\ldots+\alpha_{t}[u_{t},v_{t}]$ for some
$u_{1},v_{1},\ldots,u_{t},v_{t}\in L$, $\alpha_{1},\ldots,\alpha_{t}\in F$.
Since $f$ is an automorphism of $L$, there are the elements
$a_{1},b_{1},\ldots,a_{t},b_{t}\in L$ such that $u_{j}=f(a_{j})$,
$v_{j}=f(b_{j})$, $1\leqslant j\leqslant t$. Then

$\begin{split}w&=\sum\limits_{1\leqslant j\leqslant
t}\alpha_{j}[u_{j},v_{j}]=\sum\limits_{1\leqslant j\leqslant
t}\alpha_{j}[f(a_{j}),f(b_{j})]\\\ &=\sum\limits_{1\leqslant j\leqslant
t}\alpha_{j}f([a_{j},b_{j}])=f\left(\sum\limits_{1\leqslant j\leqslant
t}\alpha_{j}[a_{j},b_{j}]\right)\in f([L,L]).\end{split}$

It follows that $[L,L]\leqslant f([L,L])$ and hence $[L,L]=f([L,L])$. $\Box$

Let $L$ be a Leibniz algebra. Define the lower central series of $L$

$L=\gamma_{1}(L)\geqslant\gamma_{2}(L)\geqslant\ldots\gamma_{\alpha}(L)\geqslant\gamma_{\alpha+1}(L)\geqslant\ldots\gamma_{\delta}(L)=\gamma_{\infty}(L)$

by the following rule: $\gamma_{1}(L)=L$, $\gamma_{2}(L)=[L,L]$ and
recursively $\gamma_{\alpha+1}(L)=[L,\gamma_{\alpha}(L)]$ for all ordinals
$\alpha$ and $\gamma_{\lambda}(L)=\bigcap_{\mu<\lambda}\gamma_{\mu}(L)$ for
the limit ordinals $\lambda$. As usually, we say that a Leibniz algebra $L$ is
called nilpotent, if there exists a positive integer $k$ such that
$\gamma_{k}(L)=\langle 0\rangle$. More precisely, $L$ is said to be nilpotent
of nilpotency class $c$ if $\gamma_{c+1}(L)=\langle 0\rangle$, but
$\gamma_{c}(L)\neq\langle 0\rangle$.

Define the upper central series

$\langle
0\rangle=\zeta_{0}(L)\leqslant\zeta_{1}(L)\leqslant\ldots\zeta_{\alpha}(L)\leqslant\zeta_{\alpha+1}(L)\leqslant\ldots\zeta_{\tau}(L)=\zeta_{\infty}(L)$

of a Leibniz algebra $L$ by the following rule: $\zeta_{1}(L)=\zeta(L)$ is the
center of $L$, and recursively
$\zeta_{\alpha+1}(L)/\zeta_{\alpha}(L)=\zeta(L/\zeta_{\alpha}(L))$ for all
ordinals $\alpha$, and
$\zeta_{\lambda}(L)=\bigcup_{\mu<\lambda}\zeta_{\mu}(L)$ for the limit
ordinals $\lambda$.

######  Corollary 2.2.

Let $L$ be a Leibniz algebra over a field $F$, $f$ be an automorphism of $L$.
Then $f(\zeta_{\alpha}(L))=\zeta_{\alpha}(L)$,
$f(\gamma_{\alpha}(L))=\gamma_{\alpha}(L)$ for all ordinals $\alpha$. In
particular, $f(\zeta_{\infty}(L))=\zeta_{\infty}(L)$,
$f(\gamma_{\infty}(L))=\gamma_{\infty}(L)$.

Using transfinite induction we can derive it from Lemma 2.1.

######  Lemma 2.3.

Let $L$ be a Leibniz algebra over a field $F$, $f$ be an endomorphism of $L$.
Then $f(\gamma_{\alpha}(L))\leqslant\gamma_{\alpha}(L)$ for all ordinals
$\alpha$. In particular, $f(\gamma_{\infty}(L))\leqslant\gamma_{\infty}(L)$.

Proof. If $x,y\in L$, then $f([x,y])=[f(x),f(y)]\in[L,L]$. It follows that
$f([L,L])\leqslant[L,L]$. Suppose that we have already proved that
$f(\gamma_{\beta}(L))\leqslant\gamma_{\beta}(L)$ for all ordinals
$\beta<\alpha$. If $\alpha$ is a limit ordinal, then
$\gamma_{\alpha}(L)=\bigcap\limits_{\beta<\alpha}\gamma_{\beta}(L)$. In this
case,

$f(\gamma_{\alpha}(L))=f\left(\bigcap\limits_{\beta<\alpha}\gamma_{\beta}(L)\right)\leqslant\bigcap\limits_{\beta<\alpha}f(\gamma_{\beta}(L))\leqslant\bigcap\limits_{\beta<\alpha}\gamma_{\beta}(L)=\gamma_{\alpha}(L).$

Suppose now that $\alpha$ is not a limit ordinal. Then $\alpha-1=\delta$
exists and $\gamma_{\alpha}(L)=[L,\gamma_{\delta}(L)]$. By induction
hypothesis, $f(\gamma_{\delta}(L))\leqslant\gamma_{\delta}(L)$. Let $w\in L$,
$v\in\gamma_{\delta}(L)$. Then

$f([w,v])=[f(w),f(v)]\in[L,\gamma_{\delta}(L)]=\gamma_{\alpha}(L).$

It follows that
$f([L,\gamma_{\delta}(L)])=f(\gamma_{\alpha}(L))\leqslant\gamma_{\alpha}(L)$.
$\Box$

######  Lemma 2.4.

Let $L$ be a cyclic finite-dimensional Leibniz algebra over a field $F$. Let
$S$ be a subset of all endomorphisms $f$ of $L$ such that $f(x)\in[L,L]$ for
each $x\in L$. Then $S=\\{f|\ f\in\mathbf{End}_{[,]}(L),f^{2}=0\\}$ and $S$ is
an ideal of $\mathbf{End}_{[,]}(L)$ with zero multiplication $f\circ g=0$ for
every $f,g\in S$.

Proof. Let $f$ be an endomorphism of $L$ such that $f(a_{1})\in[L,L]$. Then we
have $f(a_{2})=f([a_{1},a_{1}])=[f(a_{1}),f(a_{1})]$. An equality
$[L,L]=\mathbf{Leib}(L)$ and the fact that
$\mathbf{Leib}(L)\leqslant\zeta^{\mathrm{left}}(L)$ shows that $f(a_{2})=0$.
Similarly,

$\begin{split}f(a_{3})&=f([a_{1},a_{2}])=[f(a_{1}),f(a_{2})]=0,\\\ &\ldots\\\
f(a_{n})&=f([a_{1},a_{n-1}])=[f(a_{1}),f(a_{n-1})]=0.\\\ \end{split}$

It follows that $f(y)=0$ for all $y\in[L,L]$. We have
$f(a_{1})=\gamma_{2}a_{2}+\ldots+\gamma_{n}a_{n}$, so that

$f^{2}(a_{1})=f(f(a_{1}))=f(\gamma_{2}a_{2}+\ldots+\gamma_{n}a_{n})=\gamma_{2}f(a_{2})+\ldots+\gamma_{n}f(a_{n})=0,$

and similarly, $f^{2}(a_{j})=0$ for all $j$, $2\leqslant j\leqslant n$. It
follows that $f^{2}(x)=0$ for all $x\in L$. It means that $f^{2}$ is a zero
endomorphism.

Conversely, let $f$ be an endomorphism of $L$, $f^{2}=0$ and let
$f(a_{1})=\gamma_{1}a_{1}+\gamma_{2}a_{2}+\ldots+\gamma_{n}a_{n}$. Then

$\begin{split}0&=f^{2}(a_{1})=f(f(a_{1}))=f(\gamma_{1}a_{1}+\gamma_{2}a_{2}+\ldots+\gamma_{n}a_{n})\\\
&=\gamma_{1}f(a_{1})+(\gamma_{2}f(a_{2})+\ldots+\gamma_{n}f(a_{n}))\\\
&=\gamma_{1}^{2}a_{1}+((\gamma_{1}\gamma_{2}a_{2}+\ldots+\gamma_{1}\gamma_{n}a_{n})+(\gamma_{2}f(a_{2})+\ldots+\gamma_{n}f(a_{n})))\\\
&=\gamma_{1}^{2}a_{1}+v\end{split}$

where $v\in[L,L]$. Since $Fa_{1}\cap[L,L]=\langle 0\rangle$, $f^{2}=0$ implies
that $\gamma_{1}^{2}a_{1}=0$ and $v=0$. Thus, $\gamma_{1}^{2}=0$ and
$\gamma_{1}=0$. Hence, $S=\\{f|\ f\in\mathbf{End}_{[,]}(L),f^{2}=0\\}=\\{f|\
f\in\mathbf{End}_{[,]}(L),f(x)\in[L,L]\ \mbox{for each }x\in L\\}$.

Let $f\in S$ and $g$ be an arbitrary endomorphism of $L$. Then $(f\circ
g)(x)=f(g(x))\in[L,L]$. Using Lemma 2.3, we obtain that $(g\circ
f)(x)=g(f(x))\in[L,L]$. It follows that $S$ is an ideal of
$\mathbf{End}_{[,]}(L)$. Moreover, if $f,g\in S$, then $(f\circ
g)(x)=f(g(x))=0$, because $g(x)\in[L,L]$. $\Box$

As the first step, we consider the structure of the automorphism group of a
nilpotent finite-dimensional cyclic Leibniz algebra $L=\langle a\rangle$. In
this case, $[a_{1},a_{n}]=0$.

######  Lemma 2.5.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(I)}$ over a field $F$.
Then a linear mapping $f$ is an endomorphism of $L$ if and only if

$\begin{split}f(a_{1})&=\gamma_{1}a_{1}+\gamma_{2}a_{2}+\ldots+\gamma_{n}a_{n},\\\
f(a_{2})&=\gamma_{1}^{2}a_{2}+\gamma_{1}\gamma_{2}a_{3}+\ldots+\gamma_{1}\gamma_{n-1}a_{n},\\\
f(a_{3})&=\gamma_{1}^{3}a_{3}+\gamma_{1}^{2}\gamma_{2}a_{4}+\ldots+\gamma_{1}^{2}\gamma_{n-2}a_{n},\\\
f(a_{4})&=\gamma_{1}^{4}a_{4}+\gamma_{1}^{3}\gamma_{2}a_{5}+\ldots+\gamma_{1}^{3}\gamma_{n-3}a_{n},\\\
&\ldots\\\
f(a_{n-1})&=\gamma_{1}^{n-1}a_{n-1}+\gamma_{1}^{n-2}\gamma_{2}a_{n},\\\
f(a_{n})&=\gamma_{1}^{n}a_{n}.\end{split}$

Proof. Put $L_{1}=Fa_{1}\oplus\ldots\oplus Fa_{n}=L$,
$L_{2}=Fa_{2}\oplus\ldots\oplus Fa_{n}$, …, $L_{n-1}=Fa_{n-1}\oplus Fa_{n}$,
$L_{n}=Fa_{n}$. Then $\gamma_{1}(L)=L_{1}$, $\gamma_{2}(L)=L_{2}$, …,
$\gamma_{n}(L)=L_{n}$ and $\zeta_{1}(L)=L_{n}$, $\zeta_{2}(L)=L_{n-1}$, …,
$\zeta_{n}(L)=L_{1}$.

Lemma 2.3 shows that $f(L_{j})\leqslant L_{j}$ for all $j$, $2\leqslant
j\leqslant n$. Put $f(a_{1})=\sum\limits_{1\leqslant j\leqslant
n}\gamma_{j}a_{j}$.

$\begin{split}f(a_{2})&=f([a_{1},a_{1}])=[f(a_{1}),f(a_{1})]=\left[\sum\limits_{1\leqslant
j\leqslant n}\gamma_{j}a_{j},\sum\limits_{1\leqslant k\leqslant
n}\gamma_{k}a_{k}\right]\\\ &=\left[\gamma_{1}a_{1},\sum\limits_{1\leqslant
k\leqslant n}\gamma_{k}a_{k}\right]=\gamma_{1}\left(\sum\limits_{1\leqslant
k\leqslant n}\gamma_{k}[a_{1},a_{k}]\right)\\\
&=\gamma_{1}^{2}a_{2}+\gamma_{1}\gamma_{2}a_{3}+\ldots+\gamma_{1}\gamma_{n-1}a_{n};\\\
f(a_{3})&=f([a_{1},a_{2}])=[f(a_{1}),f(a_{2})]\\\
&=\left[\sum\limits_{1\leqslant j\leqslant
n}\gamma_{j}a_{j},\gamma_{1}^{2}a_{2}+\gamma_{1}\gamma_{2}a_{3}+\ldots+\gamma_{1}\gamma_{n-1}a_{n}\right]\\\
&=[\gamma_{1}a_{1},\gamma_{1}^{2}a_{2}+\gamma_{1}\gamma_{2}a_{3}+\ldots+\gamma_{1}\gamma_{n-1}a_{n}]\\\
&=\gamma_{1}^{3}a_{3}+\gamma_{1}^{2}\gamma_{2}a_{4}+\ldots+\gamma_{1}^{2}\gamma_{n-2}a_{n};\\\
f(a_{4})&=f([a_{1},a_{3}])=[f(a_{1}),f(a_{3})]\\\
&=\left[\sum\limits_{1\leqslant j\leqslant
n}\gamma_{j}a_{j},\gamma_{1}^{3}a_{3}+\gamma_{1}^{2}\gamma_{2}a_{4}+\ldots+\gamma_{1}^{2}\gamma_{n-2}a_{n}\right]\\\
&=[\gamma_{1}a_{1},\gamma_{1}^{3}a_{3}+\gamma_{1}^{2}\gamma_{2}a_{4}+\ldots+\gamma_{1}^{2}\gamma_{n-2}a_{n}]\\\
&=\gamma_{1}^{4}a_{4}+\gamma_{1}^{3}\gamma_{2}a_{5}+\ldots+\gamma_{1}^{3}\gamma_{n-3}a_{n};\\\
&\ldots\\\ f(a_{n-1})&=f([a_{1},a_{n-2}])=[f(a_{1}),f(a_{n-2})]\\\
&=\left[\sum\limits_{1\leqslant j\leqslant
n}\gamma_{j}a_{j},\gamma_{1}^{n-2}a_{n-2}+\gamma_{1}^{n-3}\gamma_{2}a_{n-1}+\gamma_{1}^{n-3}\gamma_{3}a_{n}\right]\\\
&=[\gamma_{1}a_{1},\gamma_{1}^{n-2}a_{n-2}+\gamma_{1}^{n-3}\gamma_{2}a_{n-1}+\gamma_{1}^{n-3}\gamma_{3}a_{n}]\\\
&=\gamma_{1}^{n-1}a_{n-1}+\gamma_{1}^{n-2}\gamma_{2}a_{n};\\\
f(a_{n})&=f([a_{1},a_{n-1}])=[f(a_{1}),f(a_{n-1})]\\\
&=\left[\sum\limits_{1\leqslant j\leqslant
n}\gamma_{j}a_{j},\gamma_{1}^{n-1}a_{n-1}+\gamma_{1}^{n-2}\gamma_{2}a_{n}\right]\\\
&=[\gamma_{1}a_{1},\gamma_{1}^{n-1}a_{n-1}+\gamma_{1}^{n-2}\gamma_{2}a_{n}]\\\
&=\gamma_{1}^{n}a_{n}.\end{split}$

Conversely, let $x=\lambda_{1}a_{1}+\lambda_{2}a_{2}+\ldots+\lambda_{n}a_{n}$
and $y=\mu_{1}a_{1}+\mu_{2}a_{2}+\ldots+\mu_{n}a_{n}$ be arbitrary elements of
$L$. Suppose that a linear mapping $f$ satisfies the above conditions. Then

$\begin{split}[x,y]&=[\lambda_{1}a_{1}+\lambda_{2}a_{2}+\ldots+\lambda_{n}a_{n},\mu_{1}a_{1}+\mu_{2}a_{2}+\ldots+\mu_{n}a_{n}]\\\
&=[\lambda_{1}a_{1},\mu_{1}a_{1}+\mu_{2}a_{2}+\ldots+\mu_{n}a_{n}]\\\
&=\lambda_{1}\mu_{1}[a_{1},a_{1}]+\lambda_{1}\mu_{2}[a_{1},a_{2}]+\ldots+\lambda_{1}\mu_{n}[a_{1},a_{n}];\\\
f([x,y])&=f(\lambda_{1}\mu_{1}[a_{1},a_{1}]+\lambda_{1}\mu_{2}[a_{1},a_{2}]+\ldots+\lambda_{1}\mu_{n}[a_{1},a_{n}])\\\
&=\lambda_{1}\mu_{1}f([a_{1},a_{1}])+\lambda_{1}\mu_{2}f([a_{1},a_{2}])+\ldots+\lambda_{1}\mu_{n}f([a_{1},a_{n}]);\\\
[f(x),f(y)]&=[f(\lambda_{1}a_{1}+\lambda_{2}a_{2}+\ldots+\lambda_{n}a_{n}),f(\mu_{1}a_{1}+\mu_{2}a_{2}+\ldots+\mu_{n}a_{n}]\\\
&=[\lambda_{1}f(a_{1})+(\lambda_{2}f(a_{2})+\ldots+\lambda_{n}f(a_{n})),\mu_{1}f(a_{1})+\ldots+\mu_{n}f(a_{n})]\\\
&=[\lambda_{1}f(a_{1}),\mu_{1}f(a_{1})+\mu_{2}f(a_{2})+\ldots+\mu_{n}f(a_{n})]\\\
&+[\lambda_{2}f(a_{2})+\ldots+\lambda_{n}f(a_{n}),\mu_{1}f(a_{1})+\mu_{2}f(a_{2})+\ldots+\mu_{n}f(a_{n})].\end{split}$

By our conditions, $f(a_{j})\in L_{2}$ for all $j\geqslant 2$, so that
$\lambda_{2}f(a_{2})+\ldots+\lambda_{n}f(a_{n})\in L_{2}$ and therefore

$[\lambda_{2}f(a_{2})+\ldots+\lambda_{n}f(a_{n}),\mu_{1}f(a_{1})+\mu_{2}f(a_{2})+\ldots+\mu_{n}f(a_{n})]=0.$

Thus,

$\begin{split}[f(x),f(y)])&=[\lambda_{1}f(a_{1}),\mu_{1}f(a_{1})+\mu_{2}f(a_{2})+\ldots+\mu_{n}f(a_{n})]\\\
&=\lambda_{1}\mu_{1}[f(a_{1}),f(a_{1})]+\lambda_{1}\mu_{2}[f(a_{1}),f(a_{2})]+\ldots+\lambda_{1}\mu_{n}[f(a_{1}),f(a_{n})].\end{split}$

By our conditions,

$\begin{split}[f(a_{1}),f(a_{1})]&=f([a_{1},a_{1}]),\\\
[f(a_{1}),f(a_{2})]&=f([a_{1},a_{2}]),\\\ &\ldots\\\
[f(a_{1}),f(a_{n})]&=f([a_{1},a_{n}]),\end{split}$

which implies that $f([x,y])=[f(x),f(y)]$. Hence, $f$ is an endomorphism of a
Leibniz algebra $L$. $\Box$

######  Corollary 2.6.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(I)}$ over a field $F$.
Then $\mathbf{End}_{[,]}(L)$ is isomorphic to a submonoid of
$\mathbf{M}_{n}(F)$ consisting of all matrices having the following form

$\left(\begin{array}[]{cccccccc}\gamma_{1}&0&0&0&\ldots&0&0&0\\\
\gamma_{2}&\gamma_{1}^{2}&0&0&\ldots&0&0&0\\\
\gamma_{3}&\gamma_{1}\gamma_{2}&\gamma_{1}^{3}&0&\ldots&0&0&0\\\
\gamma_{4}&\gamma_{1}\gamma_{3}&\gamma_{1}^{2}\gamma_{2}&\gamma_{1}^{4}&\ldots&0&0&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\gamma_{n-2}&\gamma_{1}\gamma_{n-3}&\gamma_{1}^{2}\gamma_{n-4}&\gamma_{1}^{3}\gamma_{n-5}&\ldots&\gamma_{1}^{n-2}&0&0\\\
\gamma_{n-1}&\gamma_{1}\gamma_{n-2}&\gamma_{1}^{2}\gamma_{n-3}&\gamma_{1}^{3}\gamma_{n-4}&\ldots&\gamma_{1}^{n-3}\gamma_{2}&\gamma_{1}^{n-1}&0\\\
\gamma_{n}&\gamma_{1}\gamma_{n-1}&\gamma_{1}^{2}\gamma_{n-2}&\gamma_{1}^{3}\gamma_{n-3}&\ldots&\gamma_{1}^{n-3}\gamma_{3}&\gamma_{1}^{n-2}\gamma_{2}&\gamma_{1}^{n}\end{array}\right).$

######  Corollary 2.7.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(I)}$ over a field $F$.
Then the automorphism group $\mathbf{Aut}_{[,]}(L)$ is isomorphic to a
subgroup $\mathbf{AC}(n)$ of $\mathbf{GL}_{n}(F)$ consisting of all matrices
having the following form

$\left(\begin{array}[]{cccccccc}\gamma_{1}&0&0&0&\ldots&0&0&0\\\
\gamma_{2}&\gamma_{1}^{2}&0&0&\ldots&0&0&0\\\
\gamma_{3}&\gamma_{1}\gamma_{2}&\gamma_{1}^{3}&0&\ldots&0&0&0\\\
\gamma_{4}&\gamma_{1}\gamma_{3}&\gamma_{1}^{2}\gamma_{2}&\gamma_{1}^{4}&\ldots&0&0&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\gamma_{n-2}&\gamma_{1}\gamma_{n-3}&\gamma_{1}^{2}\gamma_{n-4}&\gamma_{1}^{3}\gamma_{n-5}&\ldots&\gamma_{1}^{n-2}&0&0\\\
\gamma_{n-1}&\gamma_{1}\gamma_{n-2}&\gamma_{1}^{2}\gamma_{n-3}&\gamma_{1}^{3}\gamma_{n-4}&\ldots&\gamma_{1}^{n-3}\gamma_{2}&\gamma_{1}^{n-1}&0\\\
\gamma_{n}&\gamma_{1}\gamma_{n-1}&\gamma_{1}^{2}\gamma_{n-2}&\gamma_{1}^{3}\gamma_{n-3}&\ldots&\gamma_{1}^{n-3}\gamma_{3}&\gamma_{1}^{n-2}\gamma_{2}&\gamma_{1}^{n}\end{array}\right)$

where $\gamma_{1}\neq 0$.

######  Corollary 2.8.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(I)}$ over a field $F$.
Then a monoid of all endomorphisms of $L$ is a union of an ideal $S=\\{f|\
f\in\mathbf{End}_{[,]}(L),f^{2}=0\\}$ and an automorphism group
$\mathbf{Aut}_{[,]}(L)$. Moreover, $S$ is an ideal with zero multiplication
$f\circ g=0$ for every $f,g\in S$.

Proof. Indeed, consider the arbitrary endomorphism $f$ of $L$ and let

$f(a_{1})=\gamma_{1}a_{1}+\gamma_{2}a_{2}+\ldots+\gamma_{n}a_{n}.$

If $\gamma_{1}=0$, then Lemma 2.4 shows that $f\in S$. If $\gamma_{1}\neq 0$,
then Corollaries 2.6 and 2.7 shows that $f$ is an automorphism of $L$. $\Box$

######  Lemma 2.9.

Let $L$ be a cyclic finite-dimensional Leibniz algebra over a field $F$. Let
$G=\mathbf{Aut}_{[,]}(L)$ and $U$ be a subset of all automorphisms $f$ of $L$
such that $f(a_{1})=a_{1}+u$ for some $u\in[L,L]$. Then $U$ is a normal
subgroup of $G$ and $G/U$ is isomorphic to a subgroup of the multiplicative
group of a field $F$.

Proof. If $f$ is an arbitrary automorphism of $L$, then $f(a_{1})=\lambda
a_{1}+u$ for some $\lambda\in F$, $u\in[L,L]$. Lemma 2.4 shows that
$\lambda\neq 0$. We remark that a coefficient $\lambda$ is uniquely defined.
Indeed, suppose that $f(a_{1})=\lambda_{1}a_{1}+u_{1}$ for some
$\lambda_{1}\in F$, $u_{1}\in[L,L]$. Then $\lambda
a_{1}+u=\lambda_{1}a_{1}+u_{1}$. It follows that

$(\lambda-\lambda_{1})a_{1}=u_{1}-u.$

Since $Fa_{1}\cap[L,L]=\langle 0\rangle$, we obtain that
$\lambda-\lambda_{1}=0$ and $u_{1}-u=0$. Thus, $\lambda=\lambda_{1}$ and
$u_{1}=u$.

For the automorphism $f^{-1}$ we have

$f^{-1}(a_{1})=\sigma a_{1}+w$

for some $\sigma\in F$, $w\in[L,L]$. Then

$\begin{split}a_{1}&=(f^{-1}\circ f)(a_{1})=f^{-1}(f(a_{1}))=f^{-1}(\lambda
a_{1}+u)\\\ &=\lambda f^{-1}(a_{1})+f^{-1}(u)=\lambda(\sigma
a_{1}+w)+f^{-1}(u)\\\ &=\lambda\sigma a_{1}+\lambda w+f^{-1}(u).\end{split}$

Using Lemma 2.1, we obtain that $f^{-1}(u)\in[L,L]$, so that $\lambda
w+f^{-1}(u)\in[L,L]$. Since $Fa_{1}\cap[L,L]=\langle 0\rangle$, we obtain that
$\lambda\sigma a_{1}=a_{1}$ and $\lambda w+f^{-1}(u)=0$. Thus,
$\sigma=\lambda^{-1},w=-\lambda^{-1}f^{-1}(u)$.

If $f,g\in U$, then $f(a_{1})=a_{1}+u$, $g(a_{1})=a_{1}+v$ for some
$u,v\in[L,L]$. Then

$(f\circ g)(a_{1})=f(g(a_{1}))=f(a_{1}+v)=f(a_{1})+f(v)=a_{1}+u+f(v).$

Lemma 2.1 shows that $f(v)\in[L,L]$. It follows that $f\circ g\in U$.

Let now $f\in U$. As we have seen above, $f^{-1}(a_{1})=a_{1}-f^{-1}(u)$.
Using again Lemma 2.1, we obtain that $f^{-1}(u)\in[L,L]$, which means that
$f^{-1}\in U$. It follows that $U$ is a subgroup of a group $G$.

Let $h$ be an arbitrary element of $G$ and let again $f\in U$. Then
$h(a_{1})=\lambda a_{1}+y$ for some $y\in[L,L]$. By proved above,
$h^{-1}(a_{1})=\lambda^{-1}a_{1}-\lambda^{-1}h^{-1}(y)$. Thus,

$\begin{split}(h^{-1}\circ f\circ
h)(a_{1})&=h^{-1}(f(h(a_{1})))=h^{-1}(f(\lambda a_{1}+y))\\\ &=h^{-1}(\lambda
f(a_{1})+f(y))=\lambda h^{-1}(f(a_{1}))+h^{-1}(f(y))\\\ &=\lambda
h^{-1}(a_{1}+u)+h^{-1}(f(y))=\lambda h^{-1}(a_{1})+\lambda
h^{-1}(u)+h^{-1}(f(y))\\\
&=\lambda(\lambda^{-1}a_{1}-\lambda^{-1}h^{-1}(y))+\lambda
h^{-1}(u)+h^{-1}(f(y))\\\ &=a_{1}-h^{-1}(y)+\lambda
h^{-1}(u)+h^{-1}(f(y)).\end{split}$

Using Lemma 2.1, we obtain that $h^{-1}(y),h^{-1}(u),h^{-1}(f(y))\in[L,L]$. It
follows that $h^{-1}\circ f\circ h\in U$, so that $U$ is a normal subgroup of
$G$.

Finally, define the mapping $\vartheta:G\rightarrow\mathbf{U}(F)$ by the
following rule. Let $f$ be an arbitrary automorphism of $L$, $f(a_{1})=\lambda
a_{1}+u$ for some $\lambda\in F$, $u\in[L,L]$. Put $\vartheta(f)=\lambda$. If
$h$ is another automorphism of $L$, i.e. $h(a_{1})=\sigma a_{1}+y$ for some
$\sigma\in F$, $y\in[L,L]$, then

$\begin{split}(f\circ h)(a_{1})&=f(h(a_{1}))=f(\sigma a_{1}+y)\\\ &=\sigma
f(a_{1})+f(y)=\sigma(\lambda a_{1}+u)+f(y)\\\ &=(\sigma\lambda)a_{1}+\sigma
u+f(y)\\\ &=(\lambda\sigma)a_{1}+\sigma u+f(y).\end{split}$

Lemma 2.1 implies that

$\vartheta(f\circ h)=\lambda\sigma=\vartheta(f)\vartheta(h).$

Hence, $\vartheta$ is a homomorphism of a group $G$ in $\mathbf{U}(F)$.
Clearly, $\mathbf{Ker}(f)=U$. $\Box$

######  Corollary 2.10.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(I)}$ over a field $F$,
$G=\mathbf{Aut}_{[,]}(L)$. Then $G$ is a semidirect product of a normal
subgroup $U$, consisting of all automorphisms $f$ of $L$ such that
$f(a_{1})=a_{1}+u$ for some $u\in[L,L]$, and a subgroup $D=\\{f|\
f\in\mathbf{Aut}_{[,]}(L),f(a_{1})=\gamma a_{1},0\neq\gamma\in F\\}$.
Moreover, $D$ is isomorphic to the multiplicative group of a field $F$ and $U$
is isomorphic to a subgroup $\mathbf{UC}(n)$ of $\mathbf{AC}(n)$ consisting of
matrices having the following form

$E+\gamma_{2}\sum\limits_{1\leqslant k\leqslant
n-1}E_{k+1,k}+\gamma_{3}\sum\limits_{1\leqslant k\leqslant
n-2}E_{k+2,k}+\ldots+\gamma_{n}E_{n,1}.$

Proof. Let $f$ be a linear mapping of $L$, having in basis
$\\{a_{1},a_{2},\ldots,a_{n}\\}$ the following matrix

$\sum\limits_{1\leqslant k\leqslant n}\gamma^{k}E_{k,k}$

where $0\neq\gamma\in F$. Corollary 2.7 shows that $f$ is an automorphism of
$L$. Denote by $\mathbf{DmC}(n)$ the subset of $\mathbf{AC}(n)$, consisting of
matrices

$\mathbf{D}(\gamma)=\sum\limits_{1\leqslant k\leqslant n}\gamma^{k}E_{k,k}$

where $\gamma\neq 0$. It is not hard to see that $\mathbf{DmC}(n)$ is a
subgroup of $\mathbf{AC}(n)$ and $\mathbf{DmC}(n)\cong D$. Clearly, the
mapping $\theta:\mathbf{DmC}(n)\rightarrow\mathbf{U}(F)$ defined by the rule

$\theta\left(\sum\limits_{1\leqslant k\leqslant
n}\gamma^{k}E_{k,k}\right)=\gamma$

is an isomorphism. It shows that $\mathbf{DmC}(n)$, and hence $D$, is
isomorphic to a multiplicative group of a field $F$. In particular, it is
abelian.

Consider the set of matrices, having the following form

$\left(\begin{array}[]{ccccccc}1&0&0&0&\ldots&0&0\\\
\gamma_{2}&1&0&0&\ldots&0&0\\\ \gamma_{3}&\gamma_{2}&1&0&\ldots&0&0\\\
\gamma_{4}&\gamma_{3}&\gamma_{2}&1&\ldots&0&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\gamma_{n-1}&\gamma_{n-2}&\gamma_{n-3}&\gamma_{n-4}&\ldots&1&0\\\
\gamma_{n}&\gamma_{n-1}&\gamma_{n-2}&\gamma_{n-3}&\ldots&\gamma_{2}&1\end{array}\right)$

Each of these matrices is completely defined by its first column. Therefore,
we will denote this matrix by
$\mathbf{M}(\gamma_{2},\gamma_{3},\ldots,\gamma_{n})$. Denote the set of all
such matrices by $\mathbf{UC}(n)$. Clearly, we can write every matrix from
$\mathbf{UC}(n)$ in the following form

$E+\gamma_{2}\sum\limits_{1\leqslant k\leqslant
n-1}E_{k+1,k}+\gamma_{3}\sum\limits_{1\leqslant k\leqslant
n-2}E_{k+2,k}+\ldots+\gamma_{n}E_{n,1}.$

Using Corollary 2.7, we can obtain that the matrix of every automorphism from
a subgroup $U$ in basis $\\{a_{1},a_{2},\ldots,a_{n}\\}$ belong to
$\mathbf{UC}(n)$, and conversely. It is not difficult to show that
$U\cong\mathbf{UC}(n)$. Also it is not hard to prove that for every matrix
$M\in\mathbf{AC}(n)$ we have a decomposition

$M=\mathbf{M}(\gamma_{2},\gamma_{3},\ldots,\gamma_{n})\mathbf{D}(\gamma_{1}),$

and we obtain that

$\mathbf{AC}(n)=\mathbf{UC}(n)\mathbf{DmC}(n).$

An equality $U\cap D=\langle 1\rangle$ is obvious. $\Box$

Consider now a polynomial ring $F[X]$. Denote by $R(n)$ the ideal of $F[X]$,
generated by the polynomial $X^{n}$. Put $z=X+R(n)$. Then every element of a
factor-ring $F[X]/R(n)$ has a form

$\alpha_{0}+\alpha_{1}z+\alpha_{2}z^{2}+\ldots+\alpha_{n-1}z^{n-1},$

$\alpha_{0},\alpha_{1},\alpha_{2},\ldots,\alpha_{n-1}\in F$, and this
representation is unique. It is possible to show that

$\mathbf{U}(F[X]/R(n))=\\{\alpha_{0}+\alpha_{1}z+\alpha_{2}z^{2}+\ldots+\alpha_{n-1}z^{n-1}|\
\alpha_{0}\neq 0\\}.$

Put

$\mathbf{I}(F[X]/R(n))=\\{1+\alpha_{1}z+\alpha_{2}z^{2}+\ldots+\alpha_{n-1}z^{n-1}|\
\alpha_{0},\alpha_{1},\alpha_{2},\ldots,\alpha_{n-1}\in F\\}.$

Then it is not difficult to show that $\mathbf{I}(F[X]/R(n))$ is a subgroup of
$\mathbf{U}(F[X]/R(n))$.

######  Theorem 2.11.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(I)}$ over a field $F$.
Then $\mathbf{Aut}_{[,]}(L)$ is a semidirect product of a normal subgroup
$U\cong\mathbf{I}(F[X]/R(n))$ and a subgroup $D\cong\mathbf{U}(F)$.

Proof. Corollary 2.10 implies that $G=\mathbf{Aut}_{[,]}(L)$ is a semidirect
product of a normal subgroup $U\cong\mathbf{UC}(n)$ and a subgroup
$D\cong\mathbf{U}(F)$. Let $\Gamma,\Lambda\in\mathbf{UC}(n)$ where

$\begin{split}\Gamma&=E+\gamma_{1}\sum\limits_{1\leqslant k\leqslant
n-1}E_{k+1,k}+\gamma_{2}\sum\limits_{1\leqslant k\leqslant
n-2}E_{k+2,k}+\ldots+\gamma_{n-1}E_{n,1},\\\
\Lambda&=E+\lambda_{1}\sum\limits_{1\leqslant k\leqslant
n-1}E_{k+1,k}+\lambda_{2}\sum\limits_{1\leqslant k\leqslant
n-2}E_{k+2,k}+\ldots+\lambda_{n-1}E_{n,1}.\end{split}$

Put

$\Gamma\Lambda=E+\delta_{1}\sum\limits_{1\leqslant k\leqslant
n-1}E_{k+1,k}+\delta_{2}\sum\limits_{1\leqslant k\leqslant
n-2}E_{k+2,k}+\ldots+\delta_{n-1}E_{n,1}.$

Since $\Gamma\Lambda\in\mathbf{UC}(n)$, $\Gamma\Lambda$ is completely defined
by its first column. We have

$\begin{split}\delta_{1}&=\gamma_{1}+\lambda_{1},\\\
\delta_{2}&=\gamma_{2}+\gamma_{1}\lambda_{1}+\lambda_{2},\\\ &\ldots\\\
\delta_{j}&=\gamma_{j}+\gamma_{j-1}\lambda_{1}+\gamma_{j-2}\lambda_{2}+\ldots+\gamma_{1}\lambda_{j-1}+\lambda_{j},\\\
&\ldots\\\
\delta_{n-1}&=\gamma_{n-1}+\gamma_{n-2}\lambda_{1}+\gamma_{n-3}\lambda_{2}+\ldots+\gamma_{1}\lambda_{n-2}+\lambda_{n-1}.\end{split}$

Taking all this into account, we obtain the following isomorphism. Define a
mapping

$\phi:\mathbf{UC}(n)\rightarrow\mathbf{I}(F[X]/R(n))$

by the following rule: if $\Gamma\in\mathbf{UC}(n)$, i.e.

$\Gamma=E+\gamma_{1}\sum\limits_{1\leqslant k\leqslant
n-1}E_{k+1,k}+\gamma_{2}\sum\limits_{1\leqslant k\leqslant
n-2}E_{k+2,k}+\ldots+\gamma_{n-1}E_{n,1},$

then put
$\phi(\Gamma)=1+\gamma_{1}z+\gamma_{2}z^{2}+\ldots+\gamma_{n-1}z^{n-1}$. By
proved above, $\phi(\Gamma\Lambda)=\phi(\Gamma)\phi(\Lambda)$ for every
$\Gamma,\Lambda\in\mathbf{UC}(n)$. Clearly, the mapping $\phi$ is bijective,
so that $\phi$ is an isomorphism. $\Box$

##  3 On the relationships between Leibniz algebras and modules over
associative rings.

Let $L$ be a Leibniz algebra over a field $F$ and $A$ be an abelian ideal of
$L$. Denote by $\mathbf{End}_{F}(A)$ the set of all linear transformations of
$A$. Then $\mathbf{End}_{F}(A)$ is an associative algebra by the operations
$+$ and $\circ$. As usual, $\mathbf{End}_{F}(A)$ is a Lie algebra by the
operations $+$ and $[,]$ where $[f,g]=f\circ g-g\circ f$ for all
$f,g\in\mathbf{End}_{F}(A)$.

Let $u$ be an arbitrary element of $L$. Consider the mapping
$\mathbf{l}_{u}:A\rightarrow A$, defined by the rule
$\mathbf{l}_{u}(x)=[u,x]$, $x\in A$. For every $u,v\in L$ and $\lambda\in F$
we have

$\begin{split}\mathbf{l}_{u}(x+y)&=[u,x+y]=[u,x]+[u,y]=\mathbf{l}_{u}(x)+\mathbf{l}_{u}(y),\\\
\mathbf{l}_{u}(\lambda x)&=[u,\lambda
x]=\lambda[u,x]=\lambda\mathbf{l}_{u}(x).\end{split}$

Hence, $\mathbf{l}_{u}$ is a linear transformation of $A$. Furthermore,
$\beta\mathbf{l}_{u}(x)=\beta[u,x]=[\beta u,x]=\mathbf{l}_{\beta u}(x)$ for
every $x\in A$, which implies that $\beta\mathbf{l}_{u}=\mathbf{l}_{\beta u}$.
Moreover,

$(\mathbf{l}_{u}+\mathbf{l}_{v})(x)=\mathbf{l}_{u}(x)+\mathbf{l}_{v}(x)=[u,x]+[v,x]=[u+v,x]=\mathbf{l}_{u+v}(x),$

which follows that $\mathbf{l}_{u}+\mathbf{l}_{v}=\mathbf{l}_{u+v}$. Consider
the mapping $\vartheta:L\rightarrow\mathbf{End}_{F}(A)$, defined by the rule
$\vartheta(u)=\mathbf{l}_{u}$, $u\in L$. By above equalities, this mapping is
linear. A subspace $\mathbf{Im}(\vartheta)$ is a Lie subalgebra of Lie algebra
associated with $\mathbf{End}_{F}(A)$. Denote by $\mathbf{SL}(A)$ the
associative subalgebra of $\mathbf{End}_{F}(A)$ generated by
$\mathbf{Im}(\vartheta)$. Then the action of $L$ on $A$ can be extended in a
natural way to the action of $\mathbf{SL}(A)$. Then $A$ become to a module
over associative ring $\mathbf{SL}(A)$. This relationship we will use in a
following way.

Let $L$ be a cyclic Leibniz algebra of type (II). In this case,

$[a_{1},a_{n}]=\alpha_{2}a_{2}+\ldots+\alpha_{n}a_{n}$

and $\alpha_{2}\neq 0$. Put

$c=\alpha_{2}^{-1}(\alpha_{2}a_{1}+\ldots+\alpha_{n}a_{n-1}-a_{n}).$

Then $[c,c]=0$, $Fc=\zeta^{\mathrm{right}}(L)$, $L=[L,L]\oplus Fc$ and
$[c,b]=[a_{1},b]$ for every $b\in[L,L]$ [6]. In particular, $a_{3}=[c,a_{2}]$,
…, $a_{n}=[c,a_{n-1}]$, $[c,a_{n}]=\alpha_{2}a_{2}+\ldots+\alpha_{n}a_{n}$.

Put $A=[L,L]$. A linear transformation $\mathbf{l}_{c}:A\rightarrow A$ in
basis $\\{a_{2},\ldots,a_{n}\\}$ has the following matrix

$\left(\begin{array}[]{cccccc}0&0&0&\ldots&0&\alpha_{2}\\\
1&0&0&\ldots&0&\alpha_{3}\\\ 0&1&0&\ldots&0&\alpha_{4}\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\ 0&0&0&\ldots&0&\alpha_{n-1}\\\
0&0&0&\ldots&1&\alpha_{n}\end{array}\right).$

This matrix is non-degenerate. Hence, $\mathbf{l}_{c}$ is an $F$-automorphism
of a linear space $A$. We will consider $A$ as an $F\langle g\rangle$-module
where $\langle g\rangle$ is an infinite cyclic group and the action of $g$ on
$A$ defined by the following rule: $ga=\mathbf{l}_{c}(a)=[c,a]$ for each
element $a\in A$.

Consider now the dual situation. Let $A$ be a vector space over a field $F$
and let $R$ be a subalgebra of an associative algebra $\mathbf{End}_{F}(A)$ of
all $F$-endomorphisms of $A$. Then we can consider $R$ as a Lie algebra by the
operation

$[f,g]=f\circ g-g\circ f,$

$f,g\in R$. Choose in a Lie algebra $R$ some Lie-subalgebra $S$. Put
$L=A\oplus S$ and define an operation $[,]$ on $L$ by the following rule:

$\begin{split}[f,g]&=f\circ g-g\circ f\ \mbox{for all }f,g\in S,\\\ [a,b]&=0\
\mbox{for all }a,b\in A,\\\ [a,f]&=0\ \mbox{for all }a\in A,f\in S,\\\
[f,a]&=f(a)\ \mbox{for all }a\in A,f\in S,\\\
[a+f,b+g]&=[a,b]+[a,g]+[f,b]+[f,g]=f(b)+[f,g].\end{split}$

By such definition, the left center of $L$ includes $A$. Let $x,y,z$ be
arbitrary elements of $L$. Then $x=a+f$, $y=b+g$, $z=c+h$ for some $a,b,c\in
A$, $f,g,h\in S$. We have

$\begin{split}[x,y]&=[a+f,b+g]=[f,g]+f(b),\\\ [y,z]&=[b+g,c+h]=[g,h]+g(c),\\\
[x,z]&=[a+f,c+h]=[f,h]+f(c).\end{split}$

Then

$\begin{split}[[x,y],z]&=[[f,g]+f(b),c+h]=[f,g](c)+[[f,g],h],\\\
[x,[y,z]]&=[a+f,[g,h]+g(c)]=[f,[g,h]]+f(g(c)),\\\
[y,[x,z]]&=[b+g,[f,h]+f(c)]=[g,[f,h]]+g(f(c)).\end{split}$

Since $S$ is a Lie algebra, $[[f,g],h]=[f,[g,h]]-[g,[f,h]]$, and we obtain
that

$\begin{split}[x,[y,z]]-[y,[x,z]]&=[f,[g,h]]+f(g(c))-[g,[f,h]]-g(f(c))\\\
&=[f,[g,h]]-[g,[f,h]]+f(g(c))-g(f(c))\\\
&=[[f,g],h]+[f,g](c)=[[x,y],z].\end{split}$

This shows that $L$ is a Leibniz algebra. If the subalgebra $R$ is
commutative, then $S$ as a Lie algebra is abelian. In this case, the right
center of $L$ includes $S$.

Let now $A$ be a finite-dimensional vector space over a field $F$ and let $c$
be an $F$-automorphism of $A$. Let $R=F\langle c\rangle$ be an associative
subalgebra of $\mathbf{End}_{F}(A)$, generated by the automorphism $c$. This
subalgebra is commutative. Therefore, $R$ as a Lie algebra is abelian. Then a
subspace $Fc$ is a Lie subalgebra of $R$. Using the above construction, we can
construct a Leibniz algebra

$L=A\oplus Fc.$

By this way, we come to cyclic Leibniz algebra of type (II).

##  4 The automorphism group of a cyclic Leibniz algebra of type (II).

######  Lemma 4.1.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(II)}$ over a field $F$,
$D$ be a centralizer of a subspace $Fc$ in a monoid $\mathbf{End}_{[,]}(L)$.
Then $D$ is a submonoid of $\mathbf{End}_{[,]}(L)$. Moreover,
$C=D\cap\mathbf{Aut}_{[,]}(L)$ is a normal subgroup of
$\mathbf{Aut}_{[,]}(L)$.

Proof. Indeed, if $f,g$ are two endomorphisms of $L$ such that $f(c)=g(c)=c$,
then

$(f\circ g)(c)=f(g(c))=f(c)=c,$

so that $f\circ g\in D$. Since the identity mapping of $L$ belong to $D$, $D$
is a submonoid of $\mathbf{End}_{[,]}(L)$.

Let $f$ be an arbitrary element of $C$. Then

$c=(f^{-1}\circ f)(c)=f^{-1}(f(c))=f^{-1}(c),$

so that $f^{-1}\in C$.

Let $g$ be an arbitrary automorphism of $L$ and $f$ be an element of $C$.
Lemma 2.1 shows that $g(c)=\alpha c$ for some $0\neq\alpha\in F$. Then

$(g^{-1}\circ f\circ g)(c)=g^{-1}(f(g(c)))=g^{-1}(f(\alpha c))=g^{-1}(\alpha
c)=\alpha g^{-1}(c)=\alpha\alpha^{-1}c=c.$

Hence, $C$ is a normal subgroup of $\mathbf{Aut}_{[,]}(L)$. $\Box$

If $L$ is a cyclic Leibniz algebra of type (II), then every element of $L$ has
a form $a+\alpha c$ where $a\in[L,L]$, $\alpha\in F$, and its presentation in
such form is unique.

######  Lemma 4.2.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(II)}$ over a field $F$,
$D$ be a centralizer of a subspace $Fc$ in a monoid $\mathbf{End}_{[,]}(L)$.
Then $D$ is isomorphic to a multiplicative monoid of a factor-ring
$F[X]/\mathbf{a}(X)F[X]$ where
$\mathbf{a}(X)=\alpha_{2}+\alpha_{3}X+\ldots+\alpha_{n}X^{n-2}-X^{n-1}$.

Proof. Put $A=[L,L]$. We can consider $A$ as a module over a polynomial ring
$F[X]$, if we define the action of a polynomial
$\nu_{0}+\nu_{1}X+\ldots+\nu_{k}X^{k}$ on an arbitrary element $a\in A$ by the
following rule:

$(\nu_{0}+\nu_{1}X+\ldots+\nu_{k}X^{k})a=\nu_{0}a+\nu_{1}\mathbf{l}_{c}(a)+\ldots+\nu_{k}\mathbf{l}_{c}^{k}(a).$

Since

$\begin{split}a_{3}&=[c,a_{2}]=\mathbf{l}_{c}(a_{2}),\\\
a_{4}&=[c,a_{3}]=\mathbf{l}_{c}(a_{3})=\mathbf{l}_{c}(\mathbf{l}_{c}(a_{2}))=\mathbf{l}_{c}^{2}(a_{2}),\\\
&\ldots\\\
a_{n}&=[c,a_{n-1}]=\mathbf{l}_{c}(a_{n-1})=\mathbf{l}_{c}(\mathbf{l}_{c}^{n-3}(a_{2}))=\mathbf{l}_{c}^{n-2}(a_{2}),\end{split}$

and the fact that $\\{a_{2},a_{3},\ldots,a_{n}\\}$ is a basis of $A$, $A$
becomes a cyclic $F[X]$-module.

$\mathbf{l}_{c}^{n-1}(a_{2})=\mathbf{l}_{c}(\mathbf{l}_{c}^{n-2}(a_{2}))=\mathbf{l}_{c}(a_{n})=\alpha_{2}+\alpha_{3}\mathbf{l}_{c}(a_{2})+\ldots+\alpha_{n}\mathbf{l}_{c}^{n-2}(a_{2}),$

so that we can define $\mathbf{l}_{c}^{k}(a_{2})$ (and hence
$\mathbf{l}_{c}^{k}(a)$ for arbitrary $a\in A$) for each positive integer $k$.

If $f$ is an endomorphism of $L$, then Lemma 2.3 shows that $f(A)\leqslant A$.
Define now the mapping $f^{\downarrow}:A\rightarrow A$ by the rule:
$f^{\downarrow}(a)=f(a)$ for every $a\in A$. It is not hard to prove that
$f^{\downarrow}$ is a linear transformation of a vector space $A$. Suppose now
that $f\in D$. Then

$f^{\downarrow}([c,a])=f([c,a])=[f(c),f(a)]=[c,f(a)]=[c,f^{\downarrow}(a)].$

On the other hand,
$f^{\downarrow}([c,a])=f^{\downarrow}(\mathbf{l}_{c}(a))=f^{\downarrow}(Xa)$
and
$[c,f^{\downarrow}(a)]=\mathbf{l}_{c}(f^{\downarrow}(a))=Xf^{\downarrow}(a)$.
Thus, we obtain that

$f^{\downarrow}(Xa)=Xf^{\downarrow}(a).$

In other words, $f^{\downarrow}$ is an endomorphism of the $F[X]$-module $A$.

Further,
$f(a_{2})=\beta_{0}a_{2}+\beta_{1}a_{3}+\beta_{2}a_{4}+\ldots+\beta_{n-2}a_{n}$
for some $\beta_{0},\ldots,\beta_{n-2}\in F$. As we have seen above,

$\begin{split}a_{3}&=[c,a_{2}]=\mathbf{l}_{c}(a_{2})=Xa_{2},\\\
a_{4}&=\mathbf{l}_{c}^{2}(a_{2})=X^{2}a_{2},\\\ &\ldots\\\
a_{n}&=\mathbf{l}_{c}^{n-2}(a_{2})=X^{n-2}a_{2}.\end{split}$

Hence, we come to the following presentation

$\begin{split}f(a_{2})&=\beta_{0}a_{2}+\beta_{1}Xa_{2}+\beta_{2}X^{2}a_{2}+\ldots+\beta_{n-2}X^{n-2}a_{2}\\\
&=(\beta_{0}+\beta_{1}X+\beta_{2}X^{2}+\ldots+\beta_{n-2}X^{n-2})a_{2}.\end{split}$

Put

$\mathbf{d}_{f}(X)=\beta_{0}+\beta_{1}X+\beta_{2}X^{2}+\ldots+\beta_{n-2}X^{n-2}.$

Then $f(a_{2})=\mathbf{d}_{f}(X)a_{2}$.

If $a$ is an arbitrary element of $A$, then
$a=\sigma_{0}a_{2}+\sigma_{1}a_{3}+\sigma_{2}a_{4}+\ldots+\sigma_{n-2}a_{n}$
for some $\sigma_{0},\ldots,\sigma_{n-2}\in F$. As we have seen above,

$\begin{split}a&=\sigma_{0}a_{2}+\sigma_{1}Xa_{2}+\sigma_{2}X^{2}a_{2}+\ldots+\sigma_{n-2}X^{n-2}a_{2}\\\
&=(\sigma_{0}+\sigma_{1}X+\sigma_{2}X^{2}+\ldots+\sigma_{n-2}X^{n-2})a_{2}.\end{split}$

Then

$\begin{split}f(a)&=f(\sigma_{0}a_{2}+\sigma_{1}a_{3}+\sigma_{2}a_{4}+\ldots+\sigma_{n-2}a_{n})\\\
&=\sigma_{0}f(a_{2})+\sigma_{1}f(a_{3})+\sigma_{2}f(a_{4})+\ldots+\sigma_{n-2}f(a_{n})\\\
&=\sigma_{0}f(a_{2})+\sigma_{1}f(Xa_{2})+\sigma_{2}f(X^{2}a_{2})+\ldots+\sigma_{n-2}f(X^{n-2}a_{2})\\\
&=\sigma_{0}f(a_{2})+\sigma_{1}Xf(a_{2})+\sigma_{2}X^{2}f(a_{2})+\ldots+\sigma_{n-2}X^{n-2}f(a_{2})\\\
&=(\sigma_{0}+\sigma_{1}X+\sigma_{2}X^{2}+\ldots+\sigma_{n-2}X^{n-2})f(a_{2})\\\
&=(\sigma_{0}+\sigma_{1}X+\sigma_{2}X^{2}+\ldots+\sigma_{n-2}X^{n-2})\mathbf{d}_{f}(X)a_{2}\\\
&=\mathbf{d}_{f}(X)(\sigma_{0}+\sigma_{1}X+\sigma_{2}X^{2}+\ldots+\sigma_{n-2}X^{n-2})a_{2}=\mathbf{d}_{f}(X)a.\end{split}$

Thus, we can see that an endomorphism $f$ is defined by the polynomial
$\mathbf{d}_{f}(X)$.

Conversely, let $g(X)$ be an arbitrary polynomial.

We define the $F[X]$-endomorphism $\mathbf{s}(g)$ of $A$ by the rule:

$\mathbf{s}(g)(a)=g(X)a,a\in A.$

It is not hard to see that if $g(X),r(X)$ are two polynomials, then
$\mathbf{s}(g)\circ\mathbf{s}(r)=\mathbf{s}(gr)$.

For every $F[X]$-endomorphism $h$ of $F[X]$-module $A$ define the mapping
$h^{\uparrow}:L\rightarrow L$ by the following rule: for arbitrary element
$x=a+\gamma c$ of $L$ we put

$h^{\uparrow}(a+\gamma c)=h(a)+\gamma c.$

Let $y=b+\sigma c$ be another arbitrary element of $L$. Then

$\begin{split}h^{\uparrow}(x+y)&=h^{\uparrow}(a+\gamma c+b+\sigma
c)=h^{\uparrow}(a+b+(\gamma+\sigma)c)\\\
&=h(a+b)+(\gamma+\sigma)c=h(a)+h(b)+\gamma c+\sigma c\\\ &=h(a)+\gamma
c+h(b)+\sigma c=h^{\uparrow}(a+\gamma c)+h^{\uparrow}(b+\sigma c)\\\
&=h^{\uparrow}(x)+h^{\uparrow}(y).\end{split}$

Let $\mu\in F$, then

$\begin{split}h^{\uparrow}(\mu x)&=h^{\uparrow}(\mu(a+\gamma
c))=h^{\uparrow}(\mu a+\mu\gamma c)\\\ &=h(\mu a)+\mu\gamma c=\mu
h(a)+\mu\gamma c\\\ &=\mu(h(a)+\gamma c)=\mu h^{\uparrow}(a+\gamma c)\\\ &=\mu
h^{\uparrow}(x).\end{split}$

Hence, $h^{\uparrow}$ is a linear transformation of a vector space $L$.
Furthermore,

$\begin{split}h^{\uparrow}([x,y])&=h^{\uparrow}([a+\gamma c,b+\sigma
c])=h^{\uparrow}([\gamma c,b])\\\ &=h([\gamma c,b])=h(\gamma[c,b])=\gamma
h([c,b])=\gamma h(Xb)\\\ &=\gamma Xh(b);\\\
[h^{\uparrow}(x),h^{\uparrow}(y)]&=[h^{\uparrow}(a+\gamma
c),h^{\uparrow}(b+\sigma c)]\\\ &=[h(a)+\gamma c,h(b)+\sigma c]=[\gamma
c,h(b)]=\gamma[c,h(b)]\\\ &=\gamma Xh(b).\end{split}$

Thus, $h^{\uparrow}([x,y])=[h^{\uparrow}(x),h^{\uparrow}(y)]$, so that
$h^{\uparrow}$ is an endomorphism of a Leibniz algebra $L$. By definition,
$h^{\uparrow}(c)=c$, so that $h^{\uparrow}\in D$. Clearly, if $h_{1},h_{2}$
are different $F[X]$-endomorphism of $A$, then $h_{1}^{\uparrow}\neq
h_{2}^{\uparrow}$.

If $f\in D$, then

$\begin{split}f(a+\gamma c)&=f(a)+f(\gamma c)=f(a)+\gamma f(c)\\\
&=f(a)+\gamma c=f^{\downarrow}(a)+\gamma c\\\
&=(f^{\downarrow})^{\uparrow}(a+\gamma c).\end{split}$

In other words, $f=(f^{\downarrow})^{\uparrow}$.

Define now the mapping $\vartheta:F[X]\rightarrow D$ by the rule:
$\vartheta(g(X))=\mathbf{s}(g)^{\uparrow}$ for every polynomial $g(X)\in
F[X]$. If $f$ is arbitrary element of $D$, then $f^{\downarrow}$ is an
$F[X]$-endomorphism of $A$. As we have seen above, this mapping is defined by
polynomial $\mathbf{d}_{f^{\downarrow}}(X)$, more precisely,
$f^{\downarrow}=\mathbf{s}(\mathbf{d}_{f}(X))$. Then

$f=(f^{\downarrow})^{\uparrow}=(\mathbf{s}(\mathbf{d}_{f}(X)))^{\uparrow}=\vartheta(\mathbf{d}_{f}(X)),$

so that a mapping $\vartheta$ is surjective.

Let $g(X),r(X)$ be two polynomials. Recall that
$\mathbf{s}(gr)=\mathbf{s}(g)\circ\mathbf{s}(r)$. Let now $h_{1},h_{2}$ be two
$F[X]$-endomorphism of $A$. If $x=a+\gamma c$ is an arbitrary element of $L$,
then

$(h_{1}\circ h_{2})^{\uparrow}(a+\gamma c)=(h_{1}\circ h_{2})(a)+\gamma
c=h_{1}(h_{2}(a))+\gamma c$

and

$(h_{1}^{\uparrow}\circ h_{2}^{\uparrow})(a+\gamma
c)=h_{1}^{\uparrow}(h_{2}^{\uparrow}(a+\gamma
c))=h_{1}^{\uparrow}(h_{2}(a)+\gamma c)=h_{1}(h_{2}(a))+\gamma c.$

It proves an equality $(h_{1}\circ h_{2})^{\uparrow}=h_{1}^{\uparrow}\circ
h_{2}^{\uparrow}$. Now we have

$\vartheta(g(X)r(X))=(\mathbf{s}(gr))^{\uparrow}=(\mathbf{s}(g)\circ\mathbf{s}(r))^{\uparrow}=\mathbf{s}(g)^{\uparrow}\circ\mathbf{s}(r)^{\uparrow}=\vartheta(g(X))\vartheta(r(X)).$

Hence, $\vartheta$ is an epimorphism of a multiplicative monoid $F[X]$ on $D$.

If $g(X)\in\mathbf{Ker}(\vartheta)$, then $f=\vartheta(g(X))$ is an identity
automorphism of $L$, i.e. $f(x)=x$ for each $x\in L$. Then
$f(a)=f^{\downarrow}(a)=a$ for each $a\in A$. This means that $\mathbf{s}(g)$
is an identity automorphism of $A$, so that $g(X)a=a$ for each $a\in A$. In
particular, $g(X)a_{2}=a_{2}$. In other words, $(g(X)-1)a_{2}=0$ and then
$g(X)-1\in\mathbf{Ann}_{F[X]}(a_{2})$. In other words,
$\mathbf{Ker}(\vartheta)=\mathbf{Ann}_{F[X]}(a_{2})+1$. We note that
$\mathbf{Ann}_{F[X]}(a_{2})$ is an ideal of a ring $F[X]$. Now it is not hard
to prove that a multiplicative monoid $F[X]/\mathbf{Ker}(\vartheta)$ is
isomorphic to the multiplicative monoid of a factor-ring
$F[X]/\mathbf{Ann}_{F[X]}(a_{2})$. Finally,

$X^{n-1}a_{2}=[c,a_{n}]=\alpha_{2}a_{2}+\alpha_{3}a_{3}+\ldots+\alpha_{n}a_{n}=(\alpha_{2}+\alpha_{3}X+\ldots+\alpha_{n}X^{n-2})a_{2},$

so that $\mathbf{Ann}_{F[X]}(a_{2})=\mathbf{Ann}_{F[X]}(A)=\mathbf{a}(X)F[X]$
where $\mathbf{a}(X)=\alpha_{2}+\alpha_{3}X+\ldots+\alpha_{n}X^{n-2}-X^{n-1}$.
$\Box$

######  Theorem 4.3.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(II)}$ over a field $F$.
Then $\mathbf{Aut}_{[,]}(L)=G$ includes a normal subgroup $C$, which is
isomorphic to

$\mathbf{U}(F[X]/\mathbf{a}(X)F[X]),$

where $\mathbf{a}(X)=\alpha_{2}+\alpha_{3}X+\ldots+\alpha_{n}X^{n-2}-X^{n-1}$
such that $G/C$ is isomorphic to a subgroup of a multiplicative group of a
field $F$.

Proof. As in Lemma 4.1, denote by $D$ the centralizer of $Fc$ in
$\mathbf{End}_{[,]}(L)$ and let $C=D\cap\mathbf{Aut}_{[,]}(L)$ is a
centralizer of $Fc$ in $\mathbf{Aut}_{[,]}(L)$. By Lemma 2.1, $f(Fc)=Fc$ for
each $f\in C$, and it follows that $G/C$ is isomorphic to a subgroup of a
multiplicative group of a field $F$. An equality
$C=D\cap\mathbf{Aut}_{[,]}(L)$ and Lemma 4.2 imply that $C$ is isomorphic to
$\mathbf{U}(F[X]/\mathbf{a}(X)F[X])$ where

$\mathbf{a}(X)=\alpha_{2}+\alpha_{3}X+\ldots+\alpha_{n}X^{n-2}-X^{n-1}.$

$\Box$

##  5 The automorphism group of a cyclic Leibniz algebra of type (III).

######  Theorem 5.1.

Let $L$ be a cyclic Leibniz algebra of type $\mathrm{(III)}$ over a field $F$.
Then $\mathbf{Aut}_{[,]}(L)$ is a subdirect product of groups $G_{1}$ and
$G_{2}$ where $G_{1}$ is a group described in Theorem 2.11, $G_{2}$ is a group
described in Theorem 4.3.

Proof. We have $L=A\oplus Fd_{1}$, $A=V\oplus[U,U]$, $U=Fd_{1}\oplus
Fd_{2}\oplus\ldots\oplus Fd_{t-1}$ is a nilpotent cyclic subalgebra, i.e. is
an algebra of type (I). Moreover, a subspace $[U,U]=Fd_{2}\oplus\ldots\oplus
Fd_{t-1}$ is an ideal of $L$. Furthermore, $V=Fd_{t}\oplus\ldots\oplus Fd_{n}$
is an ideal of $L$, and $[a_{1},d_{j}]=[d_{1},d_{j}]$ for all $j\geqslant t$.
In other words, $V\oplus Fd_{1}$ is a cyclic subalgebra of type (II).

Let $G$ be an automorphism group of a Leibniz algebra $L$. Since $L/V\cong U$
is a cyclic nilpotent Leibniz algebra, $G_{1}=G/C_{G}(L/V)$ is a group, which
has been described in Theorem 2.11. Since $L/[U,U]\cong V\oplus Fd_{1}$ is a
cyclic Leibniz algebra of second type, $G_{2}=G/C_{G}(L/[U,U])$ is a group,
which has been described in Theorem 4.3.

Let $f\in C_{G}(L/V)\cap C_{G}(L/[U,U])$. Then for each $x\in L$ we have
$f(x)=x+a_{1}$ where $a_{1}\in V$ and $f(x)=x+a_{2}$ where $a_{2}\in[U,U]$. It
follows that $f(x)-x\in V\cap[U,U]=\langle 0\rangle$, so that $f(x)=x$. Thus,
$C_{G}(L/V)\cap C_{G}(L/[U,U])=\langle 1\rangle$ and Remak’s theorem yields
the embedding of $G$ into the direct product $G_{1}\times G_{2}$. $\Box$

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