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arXiv:2504.16759v1 [math.DG] 23 Apr 2025
Cyclic Riemannian Lie groups: description and curvatures
Fatima-Ezzahrae Abid1, Sa¨ıd Benayadib, Mohamed Boucetta1, Hamza El Ouali1, Hicham
Lebziouic,∗
aUniversit´e Cadi-Ayyad, Facult´e des sciences et techniques, BP 549 Marrakech Maroc
abid.fatimaezzahrae@gmail.com, m.boucetta@uca.ac.ma, eloualihamza11@gmail.com
bLaboratoire IECL, Universit´e de Lorraine, CNRS-UMR 7502, UFR MIM, Metz, France
said.benayadi@univ-lorraine.fr
cUniversit´e Sultan Moulay Slimane, ´
Ecole Sup´erieure de Technologie-Kh´enifra, B.P : 170, Kh´enifra, Maroc.
h.lebzioui@usms.ma
Abstract
A cyclic Riemannian Lie group is a Lie group Gequipped with a left-invariant Riemannian
metric hthat satisfies HX,Y,Zh([X,Y],Z)=0 for any left-invariant vector fields X,Y,Z. The initial
concept and exploration of these Lie groups were presented in Monatsh. Math. 176 (2015), 219-
239. This paper builds upon the results from the aforementioned study by providing a complete
description of cyclic Riemannian Lie groups and an in-depth analysis of their various curvatures.
Keywords: Riemannian Lie groups, Cyclic metrics, Lie algebras.
2020 Mathematics Subject Classification: 53C30, 53C25, 22E25, 22E46.
1. Introduction
A cyclic Riemannian Lie group is a Lie group Gequipped with a left-invariant Riemannian
metric hthat satisfies
h([X,Y],Z)+h([Y,Z],X)+h([Z,X],Y)=0
for any left-invariant vector fields X,Y,Z. The concept and initial exploration of these groups
were introduced in [3]. In this paper, we extend this initial study and provide a detailed descrip-
tion of connected and simply-connected cyclic Riemannian Lie groups. Specifically, we prove
the following theorem.
Theorem 1.1. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then (G,h)is isomorphic both as a Lie group and a Riemannian manifold to
Rr×(G(q,p,Ω),h0)×^
SL(2,R)×...×^
SL(2,R)
| {z }
m−copies
,r≥0,m≥0,p≥q≥0,
where:
1. Rris the abelian flat Lie group, ^
SL(2,R)is the universal cover of SL(2,R)endowed with
one of the left-invariant metrics described in Theorem 2.1,
∗Corresponding author
Preprint submitted to Elsevier April 24, 2025
2. (G(q,p,Ω),h0)is the Riemannian Lie group whose underlying manifold is Rq×Rpand
the group product is given by
(s,t).(s′,t′)=(s+s′,t+(ehs,Ω1i0t′
1,...,ehs,Ωpi0t′
p)),
Ω = ωi j1≤j≤p
1≤i≤qis a real (q,p)-matrix such that Ωthas rank q, (Ω1,...,Ωp)are the columns
of Ω,h,i0is the canonical Euclidean product of Rqand h0is the left-invariant metric
h0=
q
X
i=1
ds2
i+
p
X
i=1
e−2hs,Ωii0dt2
i.
This classification enables us to conduct a thorough investigation of the different curvatures
of cyclic Riemannian Lie groups.
The paper is organized as follows: In Section 2, we prove Theorem 1.1. In Section 3, we
investigate the properties of the different curvatures of cyclic Riemannian Lie groups. We prove
that the scalar curvature of a cyclic Riemannian Lie group is always negative. Additionally, we
determine cyclic Riemannian Lie groups with constant sectional curvature, negative sectional
curvature, negative Ricci curvature, constant Ricci curvature, parallel Ricci curvature or paral-
lel curvature (see Theorems 3.1-3.7). Finally, based on the results above, we give the list of
connected and simply-connected cyclic Riemannian Lie groups up to dimension 5 (see Table 1).
2. Characterization of cyclic Riemannian Lie groups: proof of Theorem 1.1
In this section, we give some fundamental properties of cyclic Riemannian Lie groups leading
to a proof of Theorem 1.1.
Let (G,h) be a connected Lie group endowed with a left-invariant Riemannian metric. It is a
well-known fact that the geometric properties of (G,h) can be read at the level of its Lie algebra g
identified to the vector space of left-invariant vector field. The metric hinduces a scalar product
h,ion gwhich determines hentirely.
We call (G,h) a cyclic Riemannian Lie group if, for any X,Y,Z∈g,
h[X,Y],Zi+h[Y,Z],Xi+h[Z,X],Yi=0.(2.1)
The Levi-Civita connection of hdefines a product (X,Y)7→ X⋆Yon gcalled Levi-Civita product
given by Koszul’s formula
2hX⋆Y,Zi=h[X,Y],Zi+h[Z,X],Yi+h[Z,Y],Xi.(2.2)
For any X∈g, we denote by LX:g−→ gand RX:g−→ g, respectively, the left multiplication
and the right multiplication by Xgiven by
LXY=X⋆Yand RXY=Y⋆X.
For any X∈g, LXis skew-symmetric with respect to h,iand adX=LX−RX, where adX:
g−→ gis given by adXY=[X,Y]. Let [g,g] be the derived ideal of gand Z(g) its center. Put
Nℓ(g)={X∈g,LX=0}and Nr(g)={X∈g,RX=0}.We have obviously
[g,g]⊥={X∈g,RX=R∗
X}.(2.3)
2
For any X∈g, let JXbe the skew-symmetric endomorphism given by JXY=adt
YXwhere adt
Yis
the adjoint of adYwith respect to h,i. Note that
[g,g]⊥={X∈g,JX=0}.(2.4)
The restriction K of the curvature to gand ric the Ricci curvature are given by
K(X,Y)=L[X,Y]−[LX,LY] and ric(X,Y)=tr(Z7→ K(X,Z)Y).
Proposition 2.1. Let (G,h)be a cyclic Riemannian Lie group. Then the following assertions
hold.
(i)For any X,Y,Z∈g,
hX⋆Y,Zi=−hX,[Y,Z]i.
In particular, LX=−JXand RX=−adt
X.
(ii) [g,g]⊥=Nℓ(g)={X∈g,adt
X=adX}. In particular, [g,g]⊥is an abelian subalgebra.
(iii)Z(g)=(g⋆g)⊥=Nℓ(g)∩Nr(g). In particular, Z(g)⊂[g,g]⊥.
Proof. (i) It is an immediate consequence of (2.2) and (2.1).
(ii) It is a consequence of (i), (2.4) and (2.3).
(iii) It is an immediate consequence of (2.1).
As an immediate consequence of Proposition 2.1, we recover [3, Proposition 5.5].
Proposition 2.2. Let (G,h)be a cyclic Riemannian Lie group. Then:
1. if G is nilpotent then G is abelian;
2. if G is solvable then it is 2-solvable.
Proof. According to Proposition 2.1 (iii), [g,g]∩Z(g)⊂[g,g]∩[g,g]⊥={0}and if gis nilpotent
non abelian then [g,g]∩Z(g),{0}which completes the proof of the first part. The second part
is a consequence of the first one since if Gis solvable then its derived ideal is nilpotent.
Let (G,k) be a connected Lie group endowed with a bi-invariant pseudo-Riemannian metric.
This means that, for any X,Y,Z∈g,
k([X,Y],Z)+k([X,Z],Y)=0.(2.5)
Note that if kis Riemannian then Gis isomorphic to the product of an abelian Lie group and a
compact semi-simple Lie group (see [6]).
Let hbe a left-invariant metric on Gdetermined by its restriction h,ito g. Then there exists
an invertible linear map δ:g−→ ggiven by the relation
hu,vi=k(δ(u),v).
Proposition 2.3. With the notations and hypothesis above the following assertions are equiva-
lent:
(i) (G,h)is a cyclic Riemannian Lie group.
(ii)The endomorphism δis a symmetric invertible anti-derivation of g, i.e., for any X,Y,Z∈g,
δ([X,Y]) =−[δ(X),Y]−[X, δ(Y)].
3
Proof. For any X,Y,Z∈g, by using (2.5), we get
IX,Y,Zh[X,Y],Zi=k(δ([X,Y]),Z)+k([δ(X),Y],Z)+k([X, δ(Y)],Z)
and the result follows.
We give now a different proof of [3, Theorem 4.2].
Proposition 2.4. Let (G,k)be a Lie group endowed with a bi-invariant Riemannian metric. If
G carries a cyclic left-invariant Riemannian metric then G is abelian. In particular, a compact
semi-simple Lie group cannot be a cyclic Riemannian Lie group.
Proof. Suppose that h,iis the restriction of a cyclic Riemannian metric on g. Then, according to
Proposition 2.3, δgiven by hu,vi=k(δ(u),v) is an invertible anti-derivation and its eigenvalues
are positive. Thus
g=E1⊕...⊕Er
where Eiis the eigenspace associated to λi. For any X∈Eiand Y∈Ej,
δ([X,Y]) =−(λi+λj)[X,Y].
Since −(λi+λj)<0 then [X,Y]=0. This shows that gis abelian.
In [3], it was proven that the universal covering group ^
SL(2,R) of SL(2,R) is the only con-
nected and simply-connected simple cyclic Riemannian Lie group. We give now another proof
of this result and complete it by classifying all cyclic Riemannian metrics on ^
S L(2,R) and ex-
hibiting their Ricci and scalar curvatures.
We consider sl(2,R) and B=(X1,X2,X3) the basis given by
X1= 0 1
1 0!,X2= 0 1
−1 0!and X3= 1 0
0−1!.
The Lie brackets are given by
[X1,X2]=2X3,[X2,X3]=−2X1and [X3,X2]=2X1.
Theorem 2.1. The universal covering group ^
SL(2,R)of SL(2,R)is the only connected and
simply-connected simple cyclic Riemannian Lie group. Let h,ibe the restriction of a cyclic
left-invariant Riemannian metric of ^
SL(2,R)to its Lie algebra sl(2,R),ric its Ricci curvature
and σits scalar curvature. Then, up to an automorphism of sl(2,R), the matrix of h,iin Bis
given by
µ+ν0 0
0µ0
0 0 ν
, µ > ν > 0,ric =8Diag(1,−1,−1) and σ=−8µ2−8µν −8ν2
(µ+ν)µν .
Proof. Suppose that Gis a simply-connected simple cyclic Riemannian Lie group. Then its Lie
algebra gis simple and its Killing form is nondegenerate and defines a pseudo-Riemannian bi-
invariant metric on G. By virtue of Proposition 2.3, ghas an invertible anti-derivation. According
4
to [2, Theorem 5.1], g⊗C=sl(2,C) and hence gis a real form of sl(2,C) and gis isomorphic
either to su(2) or sl(2,R). But SU(2) cannot be a cyclic Riemannian Lie group by virtue of
Proposition 2.4. On the other hand, left-invariant Riemannian metrics on ^
SL(2,R) were classified
in [4, Theorem 3.6] and it is an easy task to determine the cyclic ones and their Ricci or scalar
curvatures.
Theorem 2.2. Let (G,h)be a connected and simply-connected semi-simple cyclic Riemannian
Lie group. Then G is isomorphic both as a Lie group and a Riemannian manifold to the Rieman-
nian product ^
SL(2,R)×. . . ×^
SL(2,R)and each factor ^
SL(2,R)is endowed with one of the cyclic
metrics listed in Theorem 2.1.
Proof. Denote by Bthe Killing form of gand δthe symmetric isomorphism given by hu,vi=
B(δ(u),v). According to Proposition 2.3, δis an anti-derivation, i.e., for any u,v∈g,
δ([u,v]) =−[δ(u),v]−[u, δ(v)].
Let Ibe a semi-simple ideal of g. Since I=[I,I] the relation above shows that δ(I)=I. Let
g=g1⊕...⊕grbe the decomposition of gas simple ideals. For any i∈ {1, . . . , r}, the restriction
of h,ito giis cyclic and, by virtue of Theorem 2.1, giis isomorphic to sl(2,R). On the other
hand, for any i,j,u,v∈gi,w∈gj. We have
B([u,v],w)=B(u,[v,w]) =0
and since [gi,gi]=githis shows that B(gi,gj)=0 and hence hgi,gji=B(δ(gi),gj)=0. This
completes the proof.
Theorem 2.3. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then G is isomorphic both as a Lie group and a Riemannian manifold to the product G1×
^
SL(2,R)×...×^
SL(2,R)where G1is solvable and each factor ^
SL(2,R)is endowed with one of
the cyclic metrics listed in Theorem 2.1.
Proof. We denote by Rad(g) the radical of g(the largest solvable ideal). From the relation (2.1),
one can deduce easily that Rad(g)⊥is a Lie subalgebra and for any x∈Rad(g)⊥, the restric-
tion of adxto Rad(g) is symmetric. Now, for any x,y∈Rad⊥, ad[x,y]=[adx,ady] and hence
the restriction of ad[x,y]to Rad(g) is both symmetric and skew-symmetric and it vanishes. Since
g=Rad(g)⊕Rad(g)⊥, Rad(g)⊥is semi-simple and [Rad(g)⊥,Rad(g)⊥]=Rad(g)⊥and hence
[Rad(g)⊥,Rad(g)] =0 thus Rad(g)⊥is a semi-simple ideal and we can conclude by using Theo-
rem 2.2.
Theorems 2.2 and 2.3 reduce the determination of connected and simply-connected cyclic
Riemannian Lie groups to the determination of the solvable ones. Thanks to the following result,
the determination of connected and simply-connected cyclic Riemannian Lie groups is complete.
Before stating our result, let us introduce the model of solvable cyclic Riemannian Lie group.
Let p,q∈N∗and Ω = (ωi j)1≤j≤p
1≤i≤qa real (q,p)-matrix such that Ωthas rank q. We denote by
G(q,p,Ω) the Lie group whose underline manifold is Rq×Rpand the group product is given by
(s,t).(s′,t′)=(s+s′,t+(ehs,Ω1i0t′
1,...,ehs,Ωpi0t′
p)),
5
where (Ω1,...,Ωp) are the columns of Ωand h,i0is the canonical Euclidean product of Rq.
Let ( f1,..., fp) and (h1, . . . , hq) be the canonical bases of Rpand Rq, respectively. We have
hℓ
i=∂
∂si
and fℓ
i=ehs,Ωii0∂
∂ti
,
where uℓis the left invariant vector field associated to the vector u∈Rq×Rp.
We have obviously that the non vanishing brackets are
[hℓ
i,fℓ
j]=ωi j fℓ
j.(2.6)
Let h0be the left-invariant metric for which (hℓ
1,...,hℓ
q,fℓ
1,..., fℓ
p) is orthonormal. Then
h0=
q
X
i=1
ds2
i+
p
X
i=1
e−2hs,Ωii0dt2
i.
By using (2.6), it is easy to check that (G(q,p,Ω),h0) is a cyclic Riemannian Lie group.
Proposition 2.5. (G(q,p,Ω),h0)is isometric to (G(r,s,Θ),e
h0)if and only if q =r, p =s and
there exists Q orthogonal and P a permutation such that Ω = QΘP.
Proof. Denote by g1and g2the Lie algebras of G(q,p,Ω) and G(r,s,Θ), respectively. Then
there exists o rthonormal bases ( f1,..., fp), ( f′
1,..., f′
s) of [g1,g1] and [g2,g2], respectively and
orthonormal bases (h1, . . . , hq), (h′
1,...,h′
r) of [g1,g1]⊥and [g2,g2]⊥, respectively such that
[hi,fj]=ωi j fjand [h′
i,f′
j]=θi j f′
j.
Suppose that there exists an automorphism A:g1−→ g2which preserves the metrics. Then
A([g1,g1]) =[g2,g2] and A([g1,g1]⊥)=[g2,g2]⊥and hence p=sand q=r. For any u∈[g1,g1],
[A(u),A(fi)] =A[u,fi]=
q
X
j=1
ujωji
A(fi).
So A(fi) is a commune eigenvector for advfor any v∈[g2,g2] and hence A(fi)=f′
σ(i). Put
A(hi)=Pq
j=1qjih′j. Then
A[hi,fj]=ωi jA(fj)=ωi j f′
σ(j)
=[A(hi),f′
σ(j)]
=
q
X
l=1
qliθlσ(j)f′
σ(j).
So ωi j =(QtΘ)iσ(j)=(QtΘP)i j where Pis the permutation associated to σ. Since Ais an
isometry then Qis an orthogonal matrix.
Remark 1. Suppose p =q and Ωinvertible. Then G(q,p,Ω)is isomorphic to G(q,p,DP)where
Ω = OS is a the polar decomposition, S =PDPtand D is diagonal with positive entries.
Theorem 2.4. Let (G,h)be a connected and simply connected solvable cyclic Riemannian non
abelian Lie group. Then (G,h)is isometric to (G(q,p,Ω)×Rr,h0⊕ h ,i0)where Ωthas rank q
and p +q+r=dim G.
6
Proof. According to Proposition 2.1, [g,g]⊥is abelian, Z(g)⊂[g,g]⊥and since [g,g] is nilpotent
it is also abelian by virtue of Proposition 2.2. Moreover, according to Proposition 2.1, for any
u∈[g,g]⊥, aduis symmetric. Thus g=[g,g]⊕h⊕Z(g) where h⊕Z(g)=[g,g]⊥and hh,Z(g)i=0.
Since his abelian, we can diagonalize simultaneously adufor any u∈[g,g]⊥and hence there
exists an orthonormal basis ( f1,..., fp) of [g,g], an orthonormal basis (h1, . . . , hq) of hand Ω =
(ωi j)1≤j≤p
1≤i≤qsuch that the non vanishing brackets are
[hi,fj]=ωi j fj.
But h∩Z(g)={0}and hence the lines of Ωare linearly independent which means Ωthas rank q.
According to (2.6), [g,g]⊕his isomorphic to the Lie algebra of G(q,p,Ω) which completes the
proof.
Finally, the combination of Theorems 2.3 and 2.4 proves Theorem 1.1.
3. Curvatures of cyclic Riemannian Lie groups
In this section, we give some important properties of the different curvatures of cyclic Rie-
mannian Lie groups
Theorem 3.1. Let (G,h)be a cyclic Riemannian non-abelian Lie group. Then its scalar cur-
vature is negative. In particular, neither the sectional curvature or the Ricci curvature can be
positive.
Proof. It is well-known that any Riemannian metric on a non abelian solvable Lie group has
negative scalar curvature (see [6]) and, according to Proposition 2.1, any cyclic Riemannian
metric on sl(2,R) has a negative scalar curvature and the result follows from Theorem 1.1.
A Lie group (G,h) is called vectorial (see [3]) if there exists h∈gsuch that, for any u,v∈g,
[u,v]=hu,hiv− hv,hiu.(3.1)
These Riemannian Lie groups appeared first in [6].
Proposition 3.1. Let (G,h)be a vectorial Riemannian Lie group. Then (G,h)is cyclic and its
sectional curvature is constant, i.e., for any u,v∈g,
K(u,v) :=L[u,v]−[Lu,Lv]=−hh,hiu∧v,
where u ∧v(w)=hu,wiv− hv,wiu. Moreover, the Riemannian universal covering of (G,h)is
isometric to G(1,n−1,Ω)where n =dim G and Ω = (|h|2,...,|h|2).
Proof. A straightforward computation using (3.1) shows that (G,h) is cyclic and the Levi-Civita
product is given by
u⋆v=hu,vih− hh,viu,u,v∈g.
7
Not that Lh=0. Let us compute the curvature K(u,v)=L[u,v]−[Lu,Lv].
K(u,v)w=(u⋆v)⋆w−(v⋆u)⋆w−u⋆(v⋆w)+v⋆(u⋆w)
=−hv,hiu⋆w+hu,hiv⋆w− hv,wiu⋆h+hh,wiu⋆v
+hu,wiv⋆h− hh,wiv⋆u
=−hv,hihu,wih+hv,hihh,wiu+hu,hihv,wih− hu,hihh,wiv
− hv,wihu,hih+hv,wihh,hiu+hh,wihu,vih− hh,wihh,viu
+hu,wihv,hih− hu,wihh,hiv− hh,wihu,vih+hh,wihh,uiv
=|h|2(hv,wiu− hu,wiv)
=−hh,hiu∧v(w).
On the other hand, g=h⊥⊕Rh,h⊥is abelian and for any u∈h⊥[h,u]=|h|2u. By using (2.6),
we can see that the Lie algebra of Gis isomorphic to G(1,n−1,Ω) where Ω = (|h|2,...,|h|2).
This completes the proof.
Conversely, we have the following result.
Theorem 3.2. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group of
constant sectional curvature, i.e., there exists k ∈Rsuch that
K(u,v) :=L[u,v]−[Lu,Lv]=ku ∧v.(3.2)
Then k ≤0. If k =0then G is abelian and if k <0then (G,h)is vectorial and hence it is
isometric to (G(1,n−1,Ω),h0)where n =dim G and Ω = (λ, . . . , λ)with λ > 0.
Proof. According to Theorem 3.1, k≤0 and if k=0 then Gis abelian. Suppose now that
k<0. Since the sectional curvature is negative then gis solvable (see [6, Theorem 1.6]) and
hence, according to Proposition 2.1, [g,g]⊥=Nℓ(g),{0}. Let a∈Nℓ(g) with a,{0}. We have
ker ada=Ra. Indeed, since La=0 then, according to (3.2), for any b∈g,
L[a,b]=ka ∧b.
So if [a,b]=0 then a∧b=0 and hence aand bare colinear. Moreover, according to Proposition
2.1, adais symmetric and hence adaleaves a⊥invariant and its restriction is invertible. Then there
exists an orthonormal basis (e2,...,en), (λ2,...,λn) such that λi,0 for any iand
[a,ei]=λiei.
For any i∈ {2,...,n},
ei⋆ei=1
λi
L[a,ei]ei=−k
λi
a.
On the other hand, since ei⋆a=−[a,ei], from the relation
hei⋆ei,ai=−hei,ei⋆ai
we deduce that
λ2
i=|k||a|2.
8
For i,j, by using (3.2), we have
[ei,ej]=ei⋆ej−ej⋆ei=k
λi
a∧ei(ej)−k
λj
a∧ej(ei)=0
and
[Lei,Lej]=−kei∧ej.
Thus
k2
λiλj
[a∧ei,a∧ej]=−kei∧ej.
But for any skew-symmetric endomorphism, we have
[A,u∧v]=Au ∧v+u∧Av.
We deduce that
k2
λiλjha,aiei∧ej=−kei∧ej.
So, for any i,j,
λiλj=|k||a|2.
In conclusion, λ2=... =λn=±|a|√|k|. If we take h=±√|k|
|a|a, we get the desired result by
applying Proposition 3.1.
Remark 2. One can see that this result is still valid if we drop the hypothesis cyclic and suppose
only that there exists a ∈gsuch that La=0.
Theorem 3.3. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then the sectional curvature is negative if and only (G,h)is isometric to (G(1,n−1,Ω),h0)
where Ω = (λ1,...,λn−1)and the λiare non null and have the same sign.
Proof. Suppose that the sectional curvature is negative then gis solvable (see [6, Theorem 1.6])
and, by virtue of Theorem 2.4, [g,g] is abelian. Moreover, according to [5, Proposition 2], [g,g]
has codimension one and there exists u0∈[g,g]⊥such that the symmetric part of the restriction
adu0to [g,g] is positive definite. But, according to Proposition 2.1, adu0is symmetric. We deduce
that for any u∈[g,g]⊥\ {0}, the restriction of aduto [g,g] is invertible symmetric and all its
eigenvalues have the same sign.
Conversely, suppose that [g,g] is abelian and, for any u∈[g,g]⊥, the restriction of aduto
[g,g] is invertible symmetric and all its eigenvalues have the same sign. Take a unitary vector
h∈[g,g]⊥. Then ther e exists an orth onormal basis ( f2,..., fn) and (λ2, . . . , λn) non null and
having the same sign such that, for any i∈ {2,...,n},
[h,fi]=λifi.
Let ⋆be the Levi-Civita product. Then the non vanishing products are
fi⋆fi=λihand fi⋆h=−λifi.
9
A straightforward computation gives that, for i,j,kdifferent
K( fi,fj)fk=K( fi,fj)h=K( fi,h)fj=0,
hK( fi,fj)fi,fji=−λiλj,
hK( fi,h)fi,hi=−λ2
i.
This shows that the sectional curvature is negative and completes the proof.
The expression of the Ricci curvature of a cyclic Riemannian Lie groups is rather simple.
Proposition 3.2. Let (G,h)be a cyclic Riemannian Lie group. Then its Ricci curvature is given,
for any u,v∈g, by
ric(u,v)=−B(u,v)− h[H,u],vi,
where H is defined by the relation hH,ui=tr(adu)and B is the Killing form given by B(u,v)=
tr(adu◦adv). In particular, for any u ∈[g,g]⊥,ric(u,u)≤0and ric(u,u)=0if and only if
u∈Z(g).
Proof. It is well-known (see [1, pp. 4]) that the Ricci curvature is given by
ric(u,v)=−tr(Ru◦Rv)−1
2h[H,u],vi− 1
2h[H,v],ui.
According to Proposition 2.1, Ru=−adt
uand, since H∈[g,g]⊥, adH=adt
H. The last assertions
is a consequence of the fact that, for any u∈[g,g]⊥, adu=adt
uwhich completes the proof.
Let us compute now the Levi-Civita product, the Ricci curvature and the curvature of (G(q,p,Ω),h0).
Proposition 3.3. We denote by (h1,...,hq,f1,..., fp)the canonical basis of Rq×Rp. In the
following, we give only the non vanishing quantities.
1. The Levi-Civita products of (G(q,p,Ω),h0)are given by
fi⋆fi=
q
X
k=1
ωkihkand fi⋆hj=−ωji fi,1≤i≤p,1≤j≤q.
2. The curvatures of (G(q,p,Ω),h0)are given by
K( fi,fj)fi=−hΩi,Ωji0fj,i,j
K( fi,hj)fi=−ωji
q
X
k=1
ωkihk,
K( fi,hj)hk=ωjiωki fi.
3. The Ricci curvature of (G(q,p,Ω),h0)is given by
ric(hi,hj)=−hLi,Lji0and ric( fj,fj)=−
p
X
l=1hΩl,Ωji0.
4. The scalar curvature of (G(q,p,Ω),h0)is given by
σ=−
q
X
i=1
(|Li|2+tr(Li)2).
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5. The derivative of the Ricci curvature is given by
∇fi(ric)( fi,hk)=
p
X
l=1
(ωkl −ωki )hΩl,Ωii0.
6. The derivative of the curvature is given by
∇fi(K)( fi,fk,fk)=hΩi,Ωki0
q
X
m=1
(ωmi −ωmk)hm,
∇fi(K)( fj,hk,fi)=(ωki −ωk j)hΩj,Ωii0fj.
(L1,...,Lq)are the lines of Ω,(Ω1,...,Ωp)its columns and tr(( x1,...,xp)) =Pp
i=1xi.
Proof. The expressions of the Levi-Civita product follow immediately from Proposition 2.1 (i)
and the expressions of the Lie bracket given by 2.6. The mean curvature vector is given by
H=Pq
i=1tr(Li)hiand the Ricci curvature can be deduced easily by using its expression given
in Proposition 3.2. The expressions of the curvature and the derivative of the Ricci curvature
follows immediately.
Theorem 3.4. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then (G,h)has negative Ricci curvature if and only if (G,h)is isometric to (G(q,p,Ω),h0),Ωt
has rank q and, for any j =1,...,p,
p
X
l=1hΩl,Ωji0>0.
Proof. According to Theorems 2.3, 2.2 and 2.1, Gis the Riemannian product a cyclic solvable
group and r-copies of ^
SL(2,R). But, by virtue of Theorem 2.1, a cyclic metric on ^
SL(2,R) has
always a positive Ricci direction. This completes the proof.
Theorem 3.5. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then (G,h)is λ-Einstein if and only if (G,h)is isometric to (G(q,p,Ω),h0),(L1,...,Lq)form an
orthogonal family and, for any i =1,...,q and for any j =1,...,p,
hLi,Lii=−λand
p
X
l=1hΩl,Ωji0=−λ.
Theorem 3.6. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then (G,h)is Ricci parallel if and only if (G,h)is isometric to (Rr×G(q,p,Ω),h,i0⊕h0)and,
for any j ∈ {1,...,q}}, i ∈ {1,...,p}
p
X
l=1
(ωjl −ωji)hΩl,Ωii0=0.
Proof. The cyclic Riemannian metrics on ^
SL(2,R) listed in Theorem 2.1 are not Ricci parallel
and we can conclude by using Theorem 1.1 and Proposition 3.3.
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Recall that a connected and simply connected Riemannian manifold is symmetric if and only
if it is locally symmetric, i.e., its tensor curvature is parallel.
Theorem 3.7. Let (G,h)be a connected and simply-connected cyclic Riemannian Lie group.
Then (G,h)is symmetric if and only if (G,h)is isometric to (Rr×G(q,p,Ω),h,i0⊕h0)and
there exists an orthogonal family Fof Rqsuch, for any i ∈ {1,...,p},Ωi∈ F .
Proof. The cyclic Riemannian metrics on ^
SL(2,R) listed in Theorem 2.1 are not symmetric and
we can conclude by using Theorem 1.1 and Proposition 3.3.
Example 1. Let Ωbe (q,q)orthogonal matrix. By virtue of Theorems 3.5 and 3.7, (G(q,q,Ω),h0)
is a symmetric cyclic Einstein Riemannian Lie group.
We end this paper, by giving the list of connected and simply-connected cyclic Riemannian
Lie groups of dimension ≤5.
Dimension Riemannian Lie groups Conditions
2 (G(1,1,(λ)),h0)λ,0
(R×G(1,1,(λ)),h,i0⊕h0)λ,0
3 (G(1,2,(λ1, λ2)),h0) (λ1, λ2),(0,0)
(^
SL(2,R),h1)
(R2×G(1,1,(λ)),h,i0⊕h0)λ,0
R×(G(1,2,(λ1, λ2)),h,i0⊕h0) (λ1, λ2),(0,0)
4 (R×^
SL(2,R),h,i0⊕h1)
(G(1,3,(λ1, λ2, λ3)),h0) (λ1, λ2, λ3),(0,0,0)
(G(2,2,Ω),h0) det Ω,0
(R3×G(1,1,(λ)),h,i0⊕h0)λ,0
R2×(G(1,2,(λ1, λ2)),h,i0⊕h0) (λ1, λ2),(0,0)
(R2×^
SL(2,R),h,i0⊕h1)
5 (R×G(1,3,(λ1, λ2, λ3)),h,i0⊕h0) (λ1, λ2, λ3),(0,0,0)
(R×G(2,2,Ω),h,i0⊕h0) det Ω,0
(G(2,3, P
Q!,h0)P∧Q,0
(G(1,4,(λ1, λ2, λ3, λ4)),h0) (λ1, λ2, λ3, λ4)) ,0
Table 1: Connected and simply-connected cyclic Riemannian non-abelian Lie groups of dimension ≤5.
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