Ana Hini c Gali c La Trobe University, Australia Marcel ... · La Trobe University, Australia...

Nilpotent Lie algebrasCurvatures of a nilpotent Lie algebras

Intrinsic and extrinsic curvatures of subalgebras of nilpotent Lie algebras

On the curvatures of subalgebras of nilpotent Lie algebras

Ana Hinic GalicLa Trobe University, Australia

coauthors: Grant Cairns, Yury Nikolayevsky (La Trobe University, Australia)Marcel Nicolau (Universitat Autonoma de Barcelona, Spain)

PADGE2012, KULeuven, Belgium

August 29, 2012

Ana Hinic Galic La Trobe University, Australia On the curvatures of subalgebras of nilpotent Lie algebras



Table of contents

1 Nilpotent Lie algebras

2 Curvatures of a nilpotent Lie algebrasMetric Lie algebrasSectional curvatureRicci curvatureScalar curvature

3 Intrinsic and extrinsic curvatures of subalgebras of nilpotent Lie algebrasIntrinsic and extrinsic curvatures of subalgebras of nilpotent Lie algebrasIntrinsic and extrinsic sectional curvaturesIntrinsic and extrinsic Ricci curvaturesIntrinsic and extrinsic scalar curvatures




Nilpotent Lie algebras

Let g be an n-dimensional Lie algebra over R.

• Defined the following ideals:C0(g) = g ,C1(g) = [g, g],Ck+1(g) = [Ck(g), g], for all k ≥ 0.

Then we have the descending central series of g:

g = C0(g) ⊃ C1(g) ⊃ · · · ⊃ Ck(g) ⊃ . . . .

Definition

A Lie algebra g is called nilpotent if there is an integer k such that

Ck(g) = {0}.

The smallest integer k such that Ck(g) = {0} is called the nilindex of g.








g = C0(g) ⊃ C1(g) ⊃ · · · ⊃ Ck(g) ⊃ . . . .

Definition


Ck(g) = {0}.









g = C0(g) ⊃ C1(g) ⊃ · · · ⊃ Ck(g) ⊃ . . . .

Definition


Ck(g) = {0}.





Examples of nilpotent Lie algebras

1 Every abelian Lie algebra is nilpotent with the nilindex equal to 1.

2 The Heisenberg algebra h2k+1 defined in the basis {X1,X2, . . . ,X2k+1} by

[X2i−1,X2i ] = X2k+1 , i = 1, . . . , k.

The nilindex is equal to 2.

3 The n-dimensional algebra m0(n) defined in a basis {X1, . . . ,Xn} by thebrackets

[X1,Xi ] = Xi+1 for all 2 ≤ i ≤ n − 1.

The nilindex is equal to n − 1.




Metric Lie algebrasSectional curvatureRicci curvatureScalar curvature

Metric Lie algebras

• Let (G , g) be a simply-connected Lie group with left-invariant Riemannianmetric g .

• Then (g, 〈·, ·〉) is the corresponding Lie algebra of G equipped with an innerproduct (a metric Lie algebra).

• If g is a metric Lie algebra, with inner product 〈·, ·〉, the Levi-Civitaconnection on g is given by:

2〈∇XY ,Z〉 = 〈[X ,Y ],Z〉+ 〈[Z ,X ],Y 〉+ 〈[Z ,Y ],X 〉, ∀X ,Y ,Z ∈ g. (1)

• Decomposition:

∇XY =1

2[X ,Y ] + U(X ,Y ), where

〈U(X ,Y ),Z〉 =1

2(〈[Z ,X ],Y 〉+ 〈[Z ,Y ],X 〉) .





Metric Lie algebras

• Let (G , g) be a simply-connected Lie group with left-invariant Riemannianmetric g .

• Then (g, 〈·, ·〉) is the corresponding Lie algebra of G equipped with an innerproduct (a metric Lie algebra).

• If g is a metric Lie algebra, with inner product 〈·, ·〉, the Levi-Civitaconnection on g is given by:

2〈∇XY ,Z〉 = 〈[X ,Y ],Z〉+ 〈[Z ,X ],Y 〉+ 〈[Z ,Y ],X 〉, ∀X ,Y ,Z ∈ g. (1)

• Decomposition:

∇XY =1

2[X ,Y ] + U(X ,Y ), where

〈U(X ,Y ),Z〉 =1

2(〈[Z ,X ],Y 〉+ 〈[Z ,Y ],X 〉) .





Sectional curvature

• For a metric Lie algebra g, the sectional curvature for X ,Y ∈ g:

K(X ,Y ) = − R(X ,Y ,X ,Y )

〈X ,X 〉〈Y ,Y 〉 − 〈X ,Y 〉2 . (2)

• The numerator k of the curvature function K for X ,Y ∈ g is equal to

k(X ,Y ) =− R(X ,Y ,X ,Y )

=‖U(X ,Y )‖2 − 〈U(X ,X ),U(Y ,Y )〉 − 3

4‖[X ,Y ]‖2

− 1

2〈[X , [X ,Y ]],Y 〉 − 1

2〈[Y , [Y ,X ]],X 〉.

(3)

Theorem (Wolf, 1964)

Let (G , g) be a connected nonabelian nilpotent Lie group and let g be thecorresponding Lie algebra.Then there exist two-dimensional subspaces π1, π2, π3 ⊂ g such that thesectional curvatures satisfy

K(π1) < 0 < K(π3) and K(π2) = 0.





Sectional curvature

• For a metric Lie algebra g, the sectional curvature for X ,Y ∈ g:

K(X ,Y ) = − R(X ,Y ,X ,Y )

〈X ,X 〉〈Y ,Y 〉 − 〈X ,Y 〉2 . (2)

• The numerator k of the curvature function K for X ,Y ∈ g is equal to

k(X ,Y ) =− R(X ,Y ,X ,Y )

=‖U(X ,Y )‖2 − 〈U(X ,X ),U(Y ,Y )〉 − 3

4‖[X ,Y ]‖2

− 1

2〈[X , [X ,Y ]],Y 〉 − 1

2〈[Y , [Y ,X ]],X 〉.

(3)

Theorem (Wolf, 1964)

Let (G , g) be a connected nonabelian nilpotent Lie group and let g be thecorresponding Lie algebra.Then there exist two-dimensional subspaces π1, π2, π3 ⊂ g such that thesectional curvatures satisfy

K(π1) < 0 < K(π3) and K(π2) = 0.





Ricci curvature

• Ricci curvature tensor: Ric(X ,Y ) =∑n

i=1 R(Ei ,X ,Y ,Ei ) where{E1,E2, . . . ,En} is an orthonormal basis for g.

• Ricci curvature in the direction of X ∈ g (X 6= 0) is

Ric(X ) =Ric(X ,X )

‖X‖2 . (4)

• I. Dotti, 1982: For a metric Lie algebra g, the Ricci curvature function in adirection X ∈ g is given by

ric(X ) = Ric(X ,X ) =n∑

i=1

R(Ei ,X ,X ,Ei )

= −1

2

n∑i=1

‖[X ,Ei ]‖2 −1

2B(X ,X ) +

1

4

n∑i,j=1

〈X , [Ei ,Ej ]〉2 −n∑

i=1

〈[U(Ei ,Ei ),X ],X 〉,

(5)

where B(X ,Y ) = tr(ad(X ) ◦ ad(Y )) is the Killing form.





Ricci curvature

• Ricci curvature tensor: Ric(X ,Y ) =∑n

i=1 R(Ei ,X ,Y ,Ei ) where{E1,E2, . . . ,En} is an orthonormal basis for g.

• Ricci curvature in the direction of X ∈ g (X 6= 0) is

Ric(X ) =Ric(X ,X )

‖X‖2 . (4)

• I. Dotti, 1982: For a metric Lie algebra g, the Ricci curvature function in adirection X ∈ g is given by

ric(X ) = Ric(X ,X ) =n∑

i=1

R(Ei ,X ,X ,Ei )

= −1

2

n∑i=1

‖[X ,Ei ]‖2 −1

2B(X ,X ) +

1

4

n∑i,j=1

〈X , [Ei ,Ej ]〉2 −n∑

i=1

〈[U(Ei ,Ei ),X ],X 〉,

(5)

where B(X ,Y ) = tr(ad(X ) ◦ ad(Y )) is the Killing form.





Lemma

Let g be a nilpotent metric Lie algebra and {E1,E2, . . . ,En} an orthonormalbasis. Then for all X ,Y ∈ g

(a) ric(X ) = 14

∑ni,j=1〈X , [Ei ,Ej ]〉2 − 1

2

∑ni=1 ‖[X ,Ei ]‖2,

(b) Ric(X ,Y ) = 14

∑ni,j=1〈[Ei ,Ej ],X 〉〈[Ei ,Ej ],Y 〉 − 1

2

∑ni=1〈[X ,Ei ], [Y ,Ei ]〉.

Theorem (Milnor, 1976)

For any left-invariant metric on a nonabelian nilpotent Lie group there exists adirection of strictly negative Ricci curvature and a direction of strictly positiveRicci curvature.





Lemma

Let g be a nilpotent metric Lie algebra and {E1,E2, . . . ,En} an orthonormalbasis. Then for all X ,Y ∈ g

(a) ric(X ) = 14

∑ni,j=1〈X , [Ei ,Ej ]〉2 − 1

2

∑ni=1 ‖[X ,Ei ]‖2,

(b) Ric(X ,Y ) = 14

∑ni,j=1〈[Ei ,Ej ],X 〉〈[Ei ,Ej ],Y 〉 − 1

2

∑ni=1〈[X ,Ei ], [Y ,Ei ]〉.

Theorem (Milnor, 1976)

For any left-invariant metric on a nonabelian nilpotent Lie group there exists adirection of strictly negative Ricci curvature and a direction of strictly positiveRicci curvature.





Scalar curvature

The scalar curvature of a metric Lie algebra g is given by

s =n∑

i=1

Ric(Ei )

= −1

4

n∑i,j=1

‖[Ei ,Ej ]‖2 −1

2

n∑i=1

B(Ei ,Ei )− ‖n∑

i=1

U(Ei ,Ei )‖2.(6)

Lemma

Suppose that g is an n-dimensional metric Lie algebra with an orthonormalbasis {E1, . . . ,En} with respect to which the commutator coefficients are c k

i,j ;

that is, [Ei ,Ej ] =∑n

k=1 c ki,j Ek .

If g is nilpotent, then its scalar curvature is

s = −1

4

n∑i,j,k=1

(c ki,j

)2

.





Scalar curvature

The scalar curvature of a metric Lie algebra g is given by

s =n∑

i=1

Ric(Ei )

= −1

4

n∑i,j=1

‖[Ei ,Ej ]‖2 −1

2

n∑i=1

B(Ei ,Ei )− ‖n∑

i=1

U(Ei ,Ei )‖2.(6)

Lemma

Suppose that g is an n-dimensional metric Lie algebra with an orthonormalbasis {E1, . . . ,En} with respect to which the commutator coefficients are c k

i,j ;

that is, [Ei ,Ej ] =∑n

k=1 c ki,j Ek .

If g is nilpotent, then its scalar curvature is

s = −1

4

n∑i,j,k=1

(c ki,j

)2

.




Intrinsic and extrinsic curvatures of subalgebras of nilpotent Lie algebrasIntrinsic and extrinsic sectional curvaturesIntrinsic and extrinsic Ricci curvaturesIntrinsic and extrinsic scalar curvatures

Consider a Lie subgroup H of a simply-connected Lie group G and thecorresponding Lie algebras h and g respectively.

• ∇h: the Levi-Civita connection on H defined by the restriction of 〈·, ·〉g to h.

• The second fundamental form of g, defined by Gauss’ formula as

α(X ,Y ) = ∇X Y −∇hXY ,

has the explicit form

α(X ,Y ) =1

2

r∑j=1

(〈[fj ,X ],Y 〉+ 〈[fj ,Y ],X 〉) fj (7)

for an orthonormal basis {f1, . . . , fr} of the orthogonal complement h⊥ of h.

• Let {h1, . . . , hm} be an orthonormal basis for a subalgebra h of g.





Consider a Lie subgroup H of a simply-connected Lie group G and thecorresponding Lie algebras h and g respectively.

• ∇h: the Levi-Civita connection on H defined by the restriction of 〈·, ·〉g to h.

• The second fundamental form of g, defined by Gauss’ formula as

α(X ,Y ) = ∇X Y −∇hXY ,

has the explicit form

α(X ,Y ) =1

2

r∑j=1

(〈[fj ,X ],Y 〉+ 〈[fj ,Y ],X 〉) fj (7)

for an orthonormal basis {f1, . . . , fr} of the orthogonal complement h⊥ of h.

• Let {h1, . . . , hm} be an orthonormal basis for a subalgebra h of g.





Let h be an ideal of a nilpotent Lie algebra g and f ∈ h⊥.

• Then [f , h] ⊂ h and ad(f ) is nilpotent, so the restriction of ad(f ) to h isnilpotent and hence

tr(ad(f ) |h) = 0.

• Then (7) impliesm∑

i=1

α(hi , hi ) = 0

so the mean curvature ‖∑m

i=1 α(hi , hi )‖ of h is equal to zero.

Lemma

Let H be a connected Lie subgroup of a simply-connected nilpotent Lie groupG endowed with a left-invariant Riemannian metric and let the correspondingLie algebras be h and g, respectively.If h is an ideal of g, then H is a minimal submanifold of G .





Let h be an ideal of a nilpotent Lie algebra g and f ∈ h⊥.

• Then [f , h] ⊂ h and ad(f ) is nilpotent, so the restriction of ad(f ) to h isnilpotent and hence

tr(ad(f ) |h) = 0.

• Then (7) impliesm∑

i=1

α(hi , hi ) = 0

so the mean curvature ‖∑m

i=1 α(hi , hi )‖ of h is equal to zero.

Lemma

Let H be a connected Lie subgroup of a simply-connected nilpotent Lie groupG endowed with a left-invariant Riemannian metric and let the correspondingLie algebras be h and g, respectively.If h is an ideal of g, then H is a minimal submanifold of G .






• The intrinsic curvatures of h:Rh,Kh,Rich, s(h) the curvature operator, sectional curvature, Ricci curvatureand scalar curvature, respectively, defined by ∇h.

• The extrinsic curvatures of h:Rh

e ,Khe ,Rich

e , se(h) the curvature operator, sectional curvature, Ricci curvatureand scalar curvature by using the Levi-Civita connection ∇ of G .






• The intrinsic curvatures of h:Rh,Kh,Rich, s(h) the curvature operator, sectional curvature, Ricci curvatureand scalar curvature, respectively, defined by ∇h.

• The extrinsic curvatures of h:Rh

e ,Khe ,Rich

e , se(h) the curvature operator, sectional curvature, Ricci curvatureand scalar curvature by using the Levi-Civita connection ∇ of G .





Intrinsic and extrinsic sectional curvatures

• For X ,Y ∈ h, the extrinsic sectional curvature is just K(X ,Y ) and thecorresponding formula is given by

K(X ,Y ) = Kh(X ,Y )− 〈α(X ,X ), α(Y ,Y )〉 − ‖α(X ,Y )‖2

‖X‖2‖Y ‖2 − 〈X ,Y 〉2 ,

for linearly independent (not necessary orthonormal) vector fields X ,Y ∈ h.





• For X ,Y ∈ h, the extrinsic Ricci curvature tensor is

Riche (X ,Y ) =

m∑i=1

R(hi ,X ,Y , hi ).

• The extrinsic Ricci curvature in a direction of a unit vector field X ∈ h is

Riche (X ) = Rich

e (X ,X ). (8)

Theorem

Let h be an ideal of a nilpotent Lie algebra g and choose an orthonormal basis{h1, . . . , hm} for h. If X ∈ h, then the Ricci curvature function satisfies

riche (X ) = rich(X ) +

m∑i=1

‖α(X , hi )‖2,

while for X 6= 0, the extrinsic Ricci curvature satisfies

Riche (X ) = Rich(X ) +

m∑i=1

‖α(X , hi )‖2

‖X‖2 . (9)

In particular, Riche (X ) ≥ Rich(X ).





Corollary

If a is an abelian ideal of a nilpotent Lie algebra g, thenthe extrinsic Ricci curvature Rica

e (X ) is nonnegative for all X ∈ a.

You cannot just replace the Ricci curvature by the sectional curvature.It is not true that for an abelian ideal a, one has K(X ,Y ) ≥ 0 for all X ,Y ∈ a.Indeed, consider the five-dimensional algebra generated by the orthonormalbasis {X1, . . . ,X5} with relations

[X1,X2] = X3, [X1,X4] = X5.

X2, . . . ,X5 generate a four-dimensional abelian ideal. However, forX = (X2 + X3)/2 and Y = (X4 + X5)/2 we have

K(X ,Y ) = −1/4.





Corollary

If a is an abelian ideal of a nilpotent Lie algebra g, thenthe extrinsic Ricci curvature Rica

e (X ) is nonnegative for all X ∈ a.

You cannot just replace the Ricci curvature by the sectional curvature.It is not true that for an abelian ideal a, one has K(X ,Y ) ≥ 0 for all X ,Y ∈ a.Indeed, consider the five-dimensional algebra generated by the orthonormalbasis {X1, . . . ,X5} with relations

[X1,X2] = X3, [X1,X4] = X5.

X2, . . . ,X5 generate a four-dimensional abelian ideal. However, forX = (X2 + X3)/2 and Y = (X4 + X5)/2 we have

K(X ,Y ) = −1/4.





Theorem

Let h be an ideal of a nilpotent Lie algebra g and choose an orthonormal basis{h1, . . . , hm} for h. Then

se(h) = s(h) +m∑

i,j=1

‖α(hi , hj)‖2. (10)

In particular, se(h) ≥ s(h).

Corollary

If a is an abelian ideal of a nilpotent Lie algebra g, thenthe extrinsic scalar curvature se(a) of a is nonnegative.





Theorem

Let h be an ideal of a nilpotent Lie algebra g and choose an orthonormal basis{h1, . . . , hm} for h. Then

se(h) = s(h) +m∑

i,j=1

‖α(hi , hj)‖2. (10)

In particular, se(h) ≥ s(h).

Corollary

If a is an abelian ideal of a nilpotent Lie algebra g, thenthe extrinsic scalar curvature se(a) of a is nonnegative.





Corollary

If a1, a2 are abelian ideals of a nilpotent Lie algebra g, with a1 ⊂ a2, then

se(a1) ≤ se(a2).

An ideal with the maximal extrinsic scalar curvature may not be abelian.Consider the five-dimensional Lie algebra g generated by the orthonormal basis{X1, . . . ,X5} with the relations

[X1,X2] = 2X3, [X1,X3] =1

2X4, [X1,X4] = X5, [X2,X3] =

1

2X5.

Let hk be the ideal generated by Xk , . . . ,X5. So one has the seriesg = h1 ⊃ · · · ⊃ h5.

k 1 2 3 4 5

se(hk) −114

52

34

12

0

The maximum occurs at h2, which is not abelian.The algebra g in this example has the property that it has a unique maximalabelian ideal, h3.





Corollary

If X ,Y generate a two-dimensional ideal a of a nilpotent Lie algebra g, then

K(X ,Y ) ≥ 0.

You cannot just replace ideal by subalgebraConsider the five-dimensional algebra generated by the orthonormal basis{X1, . . . ,X5} with relations

[X1,X2] = X3, [X1,X4] = X5.

Then X = (X2 + X3)/2 and Y = (X4 + X5)/2 generate a two-dimensionalabelian subalgebra, while K(X ,Y ) = −1/4.





Corollary

If X ,Y generate a two-dimensional ideal a of a nilpotent Lie algebra g, then

K(X ,Y ) ≥ 0.

You cannot just replace ideal by subalgebraConsider the five-dimensional algebra generated by the orthonormal basis{X1, . . . ,X5} with relations

[X1,X2] = X3, [X1,X4] = X5.

Then X = (X2 + X3)/2 and Y = (X4 + X5)/2 generate a two-dimensionalabelian subalgebra, while K(X ,Y ) = −1/4.





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