On the geometry of biharmonic maps and biharmonic submanifolds · On the geometry of biharmonic...

On the geometry of biharmonicmaps and biharmonic

submanifolds

Adina Balmus, Stefano Montaldo and Cezar Oniciuc

Universita degli Studi di Cagliari

“Al.I. Cuza” University of Iasi

Constanta - August 2011

1 / 37

Chen definitionLet

i : M → Rn

be the canonical inclusion and H = (H1, . . . ,Hn) the mean cur-vature vector field.

Definition (B-Y. Chen) A submanifold M ⊂ Rn is biharmonic iff

∆H = (∆H1, . . . ,∆Hn) = 0

where ∆ is the Beltrami-Laplace operator on M w.r.t. the metricinduced by i.

2 / 37

Chen definitionLet

i : M → Rn



∆H = (∆H1, . . . ,∆Hn) = 0


• Why biharmonic?

m∆H = ∆(−∆i) = −∆2i

2 / 37

Chen definitionLet

i : M → Rn



∆H = (∆H1, . . . ,∆Hn) = 0


• Why biharmonic?

m∆H = ∆(−∆i) = −∆2i

• CMC submanifolds, |H| = constant, are not necessarily bi-harmonic.

2 / 37

Biharmonic submanifolds in En(c)

Leti : Mm → E

n(c)

be the canonical inclusion of a submanifold M in a constant sec-tional curvature c manifold.

3 / 37

Biharmonic submanifolds in En(c)

Leti : Mm → E

n(c)

be the canonical inclusion of a submanifold M in a constant sec-tional curvature c manifold.

Definition M is a biharmonic submanifold iff

∆iH = mcH

where• H ∈ C(i−1(TEn(c))) denotes the mean curvature vector fieldof M in E

n(c)

• ∆i is the rough Laplacian on i−1(TEn(c))

3 / 37

RemarkIf En(c) = S

n then one can consider Sn ⊂ Rn+1 and the inclusion

i : Mm → Sn ⊂ R

n+1

can be seen as a map into Rn+1.

4 / 37

RemarkIf En(c) = S


i : Mm → Sn ⊂ R

n+1


Alternative problem (Alias, Barros, Ferrandez)

∆H′ = (∆H1, . . . ,∆Hn+1) = λH′

where H′ is the mean curvature vector field of the inclusion as a

map into Rn+1.

4 / 37

RemarkIf En(c) = S


i : Mm → Sn ⊂ R

n+1


Alternative problem (Alias, Barros, Ferrandez)

∆H′ = (∆H1, . . . ,∆Hn+1) = λH′

where H′ is the mean curvature vector field of the inclusion as a

map into Rn+1.

This is NOT the biharmonic condition for submanifolds in Sn

∆H′ = mcH′

��HH⇔ ∆iH = mcH

4 / 37

Where does this definition come from?

To understand the origin of the biharmonic equation we need touse the theory of harmonic maps.

5 / 37

The energy Functional

Harmonic maps ϕ : (M,g) → (N,h) are critical points of theenergy

E (ϕ) =1

2

∫

M

|dϕ|2 vg

and they are solutions of the Euler-Lagrange equation

τ(ϕ) = traceg∇dϕ = 0

6 / 37

The energy Functional

Harmonic maps ϕ : (M,g) → (N,h) are critical points of theenergy

E (ϕ) =1

2

∫

M

|dϕ|2 vg

and they are solutions of the Euler-Lagrange equation

τ(ϕ) = traceg∇dϕ = 0

• If ϕ is an isometric immersion, with mean curvature vectorfield H, then:

τ(ϕ) = mH

6 / 37

The bienergy Functional

The bienergy functional (proposed by Eells–Lemaire) is

E2 (ϕ) =1

2

∫

M

|τ(ϕ)|2 vg

Critical points of E2 are called biharmonic maps and they aresolutions of the Euler-Lagrange equation (Jiang):

τ2(ϕ) = −∆ϕτ(ϕ) − traceg RN (dϕ, τ(ϕ))dϕ = 0

where RN is the curvature operator on N .

7 / 37

The bienergy Functional

The bienergy functional (proposed by Eells–Lemaire) is

E2 (ϕ) =1

2

∫

M

|τ(ϕ)|2 vg

Critical points of E2 are called biharmonic maps and they aresolutions of the Euler-Lagrange equation (Jiang):

τ2(ϕ) = −∆ϕτ(ϕ) − traceg RN (dϕ, τ(ϕ))dϕ = 0

where RN is the curvature operator on N .

• The biharmonic equation

τ2(ϕ) = 0

is a fourth-order non-linear elliptic equation (not easy to solve!).

7 / 37

Non existence of Biharmonic Maps

Harmonic maps are trivially biharmonicA non harmonic biharmonic map is called proper biharmonic

8 / 37



Proposition (Jiang) If M is compact and SecN ≤ 0 then bihar-monic ⇒ harmonic

8 / 37




Proof

8 / 37




Proof ϕ biharmonic ⇒

8 / 37




Proof ϕ biharmonic ⇒ ∆ϕτ = − traceRN (dϕ, τ)dϕ

8 / 37





∫

M

|∇τ |2 vg =∫

M

〈∆ϕτ, τ〉 vg = − trace

∫

M

〈RN (dϕ, τ)dϕ, τ〉 vg ≤ 0

8 / 37





∫

M

|∇τ |2 vg =∫

M


∫

M


⇒ ∇τ = 0.

8 / 37





∫

M

|∇τ |2 vg =∫

M


∫

M


⇒ ∇τ = 0. From

div〈τ, dϕ〉 = trace〈∇τ, dϕ〉 + |τ |2

8 / 37





∫

M

|∇τ |2 vg =∫

M


∫

M


⇒ ∇τ = 0. From

0 =

∫

M

div〈τ, dϕ〉 vg =

∫

M

|τ |2 vg

⇒ τ = 0. Q.E.D.

8 / 37

Remarks

• M compact + SecN ≤ 0 ⇒ there exists a harmonic map

ϕ : M → N

in each homotopy class (Eells–Sampson)

9 / 37

Remarks


ϕ : M → N


• Harmonic maps do not always exists. There exists no har-monic map from

T2 → S

2

in the homotopy class of Brower degree ±1 (Eells–Wood)

9 / 37

Remarks


ϕ : M → N



T2 → S

2


Problem Find biharmonic maps T2 → S

2 of degree ±1

9 / 37

Remarks


ϕ : M → N



T2 → S

2


Problem Find biharmonic maps T2 → S

2 of degree ±1

• So far we only know examples of biharmonic maps T2 → S

2

whose image is a curve.

9 / 37

Examples of proper biharmonic maps

10 / 37


• Any polynomial map of degree 3 between Euclidean spaces

10 / 37



Property (Almansi): let f : Rn → R be any harmonic functionthen

g(x) = |x|ℓf(x)is proper biharmonic for any harmonic f if and only if ℓ = 2.

10 / 37





• From the Hopf map H : C2 → R × C we get the proper

biharmonic map

C2 → R× C, (z, w) 7→ (|z|2 + |w|2)(|z|2 − |w|2, 2zw)

10 / 37





• From the Hopf map H : C2 → R × C we get the proper

biharmonic map

C2 → R× C, (z, w) 7→ (|z|2 + |w|2)(|z|2 − |w|2, 2zw)

• Let f(x1, ..., xn) =∑n

i=1aixi, ai ∈ R, then

g(x) = |x|2−nf(x)

is proper biharmonic (M–Impera)10 / 37


• The generalized Kelvin transformation

ϕ : Rm \ {0} → Rm \ {0}, ϕ(p) =

p

|p|ℓ

is proper biharmonic iff ℓ = m− 2 (B–M–O)

11 / 37


• The generalized Kelvin transformation

ϕ : Rm \ {0} → Rm \ {0}, ϕ(p) =

p

|p|ℓ

is proper biharmonic iff ℓ = m− 2 (B–M–O)

• The quaternionic multiplication

H → H, q 7→ qn

is biharmonic for any n ∈ N (Fueter, 1935)

11 / 37

Lets go back to biharmonic submanifolds

If ϕ : M → En(c) is an isometric immersion then

τ(ϕ) = mH, τ2(ϕ) = −m∆ϕH+ cm2

H

thus ϕ is biharmonic iff

∆ϕH = mcH

12 / 37

Lets go back to biharmonic submanifolds

If ϕ : M → En(c) is an isometric immersion then

τ(ϕ) = mH, τ2(ϕ) = −m∆ϕH+ cm2

H

thus ϕ is biharmonic iff

∆ϕH = mcH

Moreover, if ϕ : M → Rn is anisometric immersion, set ϕ =

(ϕ1, . . . , ϕn) and H = (H1, . . . ,Hn). Then

∆ϕH = (∆H1, . . . ,∆Hn)

and we recover Chen’s definition.

12 / 37

Biharmonic submanifolds of En(c)

13 / 37


Proposition [Chen (c = 0) - Oniciuc (c ≤ 0)] Let

ϕ : M → En(c)

be an isometric immersion with |H| = constant. If c ≤ 0, then ϕis biharmonic iff H = 0.

13 / 37



ϕ : M → En(c)


Proof Biharmonic ⇒ ∆ϕH = mcH.

13 / 37



ϕ : M → En(c)


Proof Biharmonic ⇒ ∆ϕH = mcH. Replacing in the Weitzen-

bock formula1

2∆ϕ|H|2 = 〈∆ϕ

H,H〉 − |∇H|2

we get, since c ≤ 0,

|∇H|2 = mc |H|2 ≤ 0

Thus we conclude that ∇H = 0.

13 / 37



ϕ : M → En(c)


Proof Biharmonic ⇒ ∆ϕH = mcH. Replacing in the Weitzen-

bock formula1

2∆ϕ|H|2 = 〈∆ϕ

H,H〉 − |∇H|2

we get, since c ≤ 0,

|∇H|2 = mc |H|2 ≤ 0

Thus we conclude that ∇H = 0.Next, for an isometric immersion we have:

|H|2 = − trace〈dϕ,∇H〉Q.E.D.

13 / 37

Geometric conditions for biharmonic submanifoldsBy decomposing the equation ∆ϕ

H = mcH in its normal andtangent components we find that an isometric immersion ϕ :Mm → E

n(c) is biharmonic iff

−∆⊥H− traceB(·, AH·) +mcH = 0 (normal)

2 traceA∇⊥

(·)H(·) + m

2grad(|H|2) = 0 (tangent)

A is the Weingarten operator - B the second fundamental form∇⊥ and ∆⊥ the connection and the Laplacian in the normal bun-dle.

14 / 37

Geometric conditions for biharmonic submanifoldsBy decomposing the equation ∆ϕ

H = mcH in its normal andtangent components we find that an isometric immersion ϕ :Mm → E

n(c) is biharmonic iff

−∆⊥H− traceB(·, AH·) +mcH = 0 (normal)

2 traceA∇⊥

(·)H(·) + m

2grad(|H|2) = 0 (tangent)

A is the Weingarten operator - B the second fundamental form∇⊥ and ∆⊥ the connection and the Laplacian in the normal bun-dle.

For hypersurfaces

∆⊥H− (mc− |A|2)H = 0

2A(

grad(|H|))

+m|H| grad(|H|) = 0

14 / 37

Non existence of biharmonic submanifolds

15 / 37


Proposition [Chen (c = 0), Caddeo–M–O (c ≤ 0)] If c ≤ 0, thereexists no proper biharmonic surfaces M2 ⊂ E

3(c).

15 / 37



3(c).

Proof

15 / 37



3(c).

Proof In this caseH = |H|η,

where η is a unit normal to M .

15 / 37



3(c).

Proof In this caseH = |H|η,

where η is a unit normal to M .Then, biharmonicity, Gauss and Codazzi equations imply that|H| is a solution of a polynomial equation with constant coeffi-cients, thus |H| is constant.

Q.E.D.

15 / 37

Chen’s Conjecture

Conjecture

Biharmonic submanifolds of En(c), n > 3, c ≤ 0, are minimal

16 / 37

Chen’s Conjecture

Conjecture

Biharmonic submanifolds of En(c), n > 3, c ≤ 0, are minimal

Partial solutions of the conjecture are known for:

• curves of Rn (Dimitric)• submanifolds of finite type in R

n (Dimitric)• hypersurfaces with at most two principal curvatures (B–M–O)• pseudo-umbilical submanifolds Mm ⊂ E

n(c), c ≤ 0, m 6= 4,(Caddeo–M–O, Dimitric)

• hypersurfaces of E4(c), c ≤ 0 (Hasanis–Vlachos, B–M–O)• spherical submanifolds of Rn (Chen)• submanifolds of bounded geometry (Ichiyama–Inoguchi–Urakawa)

16 / 37

Biharmonic submanifolds of Sn

All the non existence results described in the previous sectiondo not hold for submanifolds in the sphere.

Problem:Classify all biharmonic submanifolds of Sn

17 / 37

Main examples of biharmonic submanifolds in Sn

18 / 37


B1. The small hypersphere

Sm( 1√

2) S

m+1biharmonic

18 / 37


B1. The small hypersphere

Sm( 1√

2) S

m+1biharmonic

B2. The standard products of spheres

Sm1( 1√

2)× S

m2( 1√2) S

m+1biharmonic

m1 +m2 = m− 1 and m1 6= m2.

18 / 37


B3. Composition property

MmSn−1( 1√

2) S

nminimal biharmonic

proper biharmonic

19 / 37


B3. Composition property

MmSn−1( 1√

2) S

nminimal biharmonic

proper biharmonic

B4. Product composition property

Mm11

×Mm22

Sn1( 1√

2)× S

n2( 1√2) S

nminimal

proper biharmonic

n1 + n2 = n− 1, m1 6= m2

19 / 37

Biharmonic cuves in Sn

(Caddeo–M–O)

20 / 37


(Caddeo–M–O)

γ ⊂ Sn biharmonic curve

κ = curvature

20 / 37


(Caddeo–M–O)


κ = curvature

κ = 1

20 / 37


(Caddeo–M–O)


κ = curvature

κ = 1

S1( 1√

2) S

2

20 / 37


(Caddeo–M–O)


κ = curvature

κ = 1

S1( 1√

2) S

2

κ ∈ (0, 1)

20 / 37


(Caddeo–M–O)


κ = curvature

κ = 1

S1( 1√

2) S

2

κ ∈ (0, 1)

γ S1( 1√

2)× S

1( 1√2) S

3

geo

biharmonic

γ geodesic of slope 6= ±1, 0 and ∞

20 / 37

Biharmonic hypersurfaces in Sn

21 / 37


Mm ⊂ Sm+1 biharmonic

κ = number of distinct principal curvature

21 / 37




κ = 1

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

Sm1( 1√

2)× S

m2( 1√2) S

m+1

B2

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

Sm1( 1√

2)× S

m2( 1√2) S

m+1

B2

κ = 3

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

Sm1( 1√

2)× S

m2( 1√2) S

m+1

B2

κ = 3 Compact + CMC

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

Sm1( 1√

2)× S

m2( 1√2) S

m+1

B2


Non Existence

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

Sm1( 1√

2)× S

m2( 1√2) S

m+1

B2


Non Existence

Isoparametric

21 / 37




κ = 1

Sm( 1√

2) S

m+1

B1

κ ≤ 2

Sm1( 1√

2)× S

m2( 1√2) S

m+1

B2


Non Existence

Isoparametric

Ichiyama-Inoguchi-Urakawa

21 / 37

CMC Biharmonic submanifolds in Sn

Mm ⊂ Sn biharmonic + CMC

22 / 37


Mm ⊂ Sn biharmonic + CMCM3 ⊂ S

4

22 / 37



4

Compact

22 / 37



4

Compact

22 / 37



4

Compact |H| = 1 |H| ∈ (0, 1)

22 / 37



4

Compact |H| = 1 |H| ∈ (0, 1)

MmSn−1( 1√

2) S

nminimal biharmonic

B3

22 / 37



4

Compact |H| = 1 |H| ∈ (0, 1)

MmSn−1( 1√

2) S

nminimal biharmonic

B3

Compact

22 / 37



4

Compact |H| = 1 |H| ∈ (0, 1)

MmSn−1( 1√

2) S

nminimal biharmonic

B3

Compact

M is of 2-type

22 / 37



4

Compact |H| = 1 |H| ∈ (0, 1)

MmSn−1( 1√

2) S

nminimal biharmonic

B3

Compact

M is of 2-type

M is of 1-type22 / 37



4

Compact |H| = 1 |H| ∈ (0, 1)

MmSn−1( 1√

2) S

nminimal biharmonic

B3

Compact

M is of 2-type

M is of 1-type

?

22 / 37

Biharmonic submanifolds in Sn with ∇⊥H = 0

Mm ⊂ Sn biharmonic + PMC

23 / 37



M2 ⊂ Sn

23 / 37



M2 ⊂ Sn

M2Sn−1( 1√

2) S

nmin

B3

23 / 37



M2 ⊂ Sn

M2Sn−1( 1√

2) S

nmin

B3

∇⊥B = 0

23 / 37



M2 ⊂ Sn

M2Sn−1( 1√

2) S

nmin

B3

∇⊥B = 0

Mm11

×Mm22

Sn1( 1√

2)× S

n2( 1√2) S

nmin

B4

23 / 37

Pseudo-umbilical biharmonic submanifolds in Sn

CMC proper biharmonic submanifolds with |H| = 1 in Sn are B3

and they are pseudo-umbilical:

AH = |H|2Id

24 / 37




AH = |H|2Id

Question When a proper biharmonic pseudo-umbilical sub-manifold in S

n has |H| = 1, thus B3?

24 / 37




AH = |H|2Id

Question When a proper biharmonic pseudo-umbilical sub-manifold in S

n has |H| = 1, thus B3?

Theorem Let Mm be a compact pseudo-umbilical submani-fold in S

n, m 6= 4. Then M is proper biharmonic if and only if Mis B3.

24 / 37




25 / 37




Proof The tangent part of τ2 becomes

(m− 4) grad |H|2 = 0,

25 / 37





(m− 4) grad |H|2 = 0,

m 6= 4 ⇒ |H| = constant.

25 / 37





(m− 4) grad |H|2 = 0,


compact + CMC + pseudo-umbilical ⇒ PMC (H. Li)

25 / 37





(m− 4) grad |H|2 = 0,



⇒ M is minimal in Sn−1(a) ⊂ S

n, a ∈ (0, 1) (Chen)

25 / 37





(m− 4) grad |H|2 = 0,




n, a ∈ (0, 1) (Chen)

biharmonicity ⇒ a = 1/√2 (Caddeo–M–O)

25 / 37





(m− 4) grad |H|2 = 0,


��XXXXXcompact + CMC + pseudo-umbilical ⇒ PMC


n, a ∈ (0, 1) (Chen)

biharmonicity ⇒ a = 1/√2 (Caddeo–M–O)

25 / 37

The examples of Sasahara et alTheorem Let ϕ : M3 → S

5 be a proper biharmonic anti-invariantimmersion. Then the position vector field x0 = x0(u, v, w) in R

6

is given by

x0(u, v, w) = eiw(eiu, ie−iu sin√2v, ie−iu cos

√2v)

Moreover, |H| = 1/3.

26 / 37



6

is given by


√2v)


The immersion ϕ is PMC but NOT pseudo-umbilical

26 / 37



6

is given by


√2v)



Theorem Let φ : M2 → S5 be a proper biharmonic Legendre

immersion. Then the position vector field x0 = x0(u, v) of M inR6 is given by:

x0(u, v) =1√2

(

cos u, sin u sin(√2v),− sin u cos(

√2v),

sinu, cos u sin(√2v),− cos u cos(

√2v)

)

.

26 / 37



6

is given by


√2v)



Theorem Let φ : M2 → S5 be a proper biharmonic Legendre

immersion. Then the position vector field x0 = x0(u, v) of M inR6 is given by:

x0(u, v) =1√2

(

cos u, sin u sin(√2v),− sin u cos(

√2v),

sinu, cos u sin(√2v),− cos u cos(

√2v)

)

.

The immersion φ is NOT PMC26 / 37

Open Problems

Conjecture

The only proper biharmonic hypersurfaces in Sn are B1 or B2.

Conjecture

Any biharmonic submanifold in Sn has constant mean curvature.

27 / 37

Remark

An isometric immersion ϕ : Mm → En(c) is biharmonic iff

−∆⊥H− traceB(·, AH·) +mcH = 0 normal

2 traceA∇⊥

(·)H(·) + m

2grad(|H|2) = 0 tangent

Most of the classification results described depend only on thetangent part of τ2.

28 / 37

Remark



2 traceA∇⊥

(·)H(·) + m



Has the conditionτ2(ϕ)

⊤ = 0

a variational meaning?

28 / 37

Remark



2 traceA∇⊥

(·)H(·) + m



Has the conditionτ2(ϕ)

⊤ = 0

a variational meaning?

YES

28 / 37

The stress-energy tensorAs described by Hilbert, the stress-energy tensor associated toa variational problem is a symmetric 2-covariant tensor field Sconservative at critical points, i.e. divS = 0 at these points.

29 / 37


• In the context of harmonic maps, the stress-energy tensor is

S =1

2|dϕ|2g − ϕ∗h,

29 / 37



S =1

2|dϕ|2g − ϕ∗h, divS = −〈τ(ϕ), dϕ〉

(Baird–Eells)

29 / 37



S =1

2|dϕ|2g − ϕ∗h, divS = −〈τ(ϕ), dϕ〉

(Baird–Eells)

• For biharmonic maps the stress-energy tensor is

S2(X,Y ) =1

2|τ(ϕ)|2〈X,Y 〉+ 〈dϕ,∇τ(ϕ)〉〈X,Y 〉

−〈dϕ(X),∇Y τ(ϕ)〉 − 〈dϕ(Y ),∇Xτ(ϕ)〉with

divS2 = −〈τ2(ϕ), dϕ〉(Jiang, Loubeau–M–O)

29 / 37

The meaning of S2 = 0 (Loubeau–M–O)

30 / 37


A smooth map ϕ : (M,g) → (N,h) is biharmonic if it is a criticalpoints of the bienergy w.r.t. variations of the map.

30 / 37


A smooth map ϕ : (M,g) → (N,h) is biharmonic if it is a criticalpoints of the bienergy w.r.t. variations of the map.Opting for a different angle of attack, one can vary the metricinstead of the map and consider the functional

F : G → R, F (g) = E2(ϕ),

where G is the set of Riemannian metrics on M

30 / 37


A smooth map ϕ : (M,g) → (N,h) is biharmonic if it is a criticalpoints of the bienergy w.r.t. variations of the map.Opting for a different angle of attack, one can vary the metricinstead of the map and consider the functional

F : G → R, F (g) = E2(ϕ),

where G is the set of Riemannian metrics on M

Theorem

δ(F (gt)) = −1

2

∫

M

〈S2, ω〉 vg,

The tensor S2 vanishes precisely at critical points of the energy(bienergy) for variations of the domain metric, rather than forvariations of the map.

30 / 37

The condition S2 = 0 is rather strong, in fact

S2 = 0 ⇒ harmonic if:

• dim(M) = 2

• M is compact and orientable with dim(M) 6= 4

• ϕ is an isometric immersion and dim(M) 6= 4

• M is complete and ϕ has finite energy and bienergy

31 / 37

Isometric immersion

If ϕ : (M,g) → (N,h) is an isometric immersion from

divS2 = −〈τ2(ϕ), dϕ〉⇓

div S2 = − τ2(ϕ)⊤

32 / 37

Isometric immersion

If ϕ : (M,g) → (N,h) is an isometric immersion from

divS2 = −〈τ2(ϕ), dϕ〉⇓

div S2 = − τ2(ϕ)⊤

Problem

Study isometric immersions in space forms with divS2 = 0

32 / 37

Biharmonic submanifolds in a Riemannian manifold

An isometric immersion

ϕ : (M,g) → (N,h)

is biharmonic iff

∆⊥H+ traceB(·, AH·) + trace(RN (·,H)·)⊥ = 0

m2grad |H|2 + 2 traceA∇⊥

(·)H(·) + 2 trace(RN (·,H)·)⊤ = 0

33 / 37

Results for Bih. Sub. in non constant sec. curv.manifolds

• In three-dimensional homogeneous spaces (Thurston’s ge-ometries)

(Inoguchi, Ou–Wang, Caddeo–Piu–M–O)

• There exists examples of proper biharmonic hypersurfacesin a space with negative non constant sectional curvature

(Ou–Tang)

• It is initiated the study of biharmonic submanifolds in complexspace forms

(Ichiyama–Inoguchi–Urakawa, Fetcu–Loubeau–M–O, Sasahara)

• There are several works on biharmonic submanifolds in con-tact manifold and Sasakian space forms

(Inoguchi, Fetcu–O, Sasahara)

34 / 37

35 / 37

In a Sasakian manifold

(N,Φ, ξ, η, g)

a submanifold M ⊂ N tangent to ξ is called anti-invariant if Φmaps any tangent vector to M , which is normal to ξ, to a vectorwhich is normal to M .

36 / 37

Finite k-type submanifolds

An isometric immersion φ : M → Rn+1 (M compact) is called of

finite k-type ifφ = φ0 + φ1 + · · ·+ φk

where∆φi = λiφi, i = 1, . . . , k

and φ0 ∈ Rn+1 is the center of mass

A submanifold M ⊂ Sn ⊂ R

n+1 is said to be of finite type if it isof finite type as a submanifold of Rn+1.

A non null finite type submanifold in Sn is said to be mass-

symmetric if the constant vector φ0 of its spectral decompositionis the center of the hypersphere S

n, i.e. φ0 = 0.

37 / 37

Date post:	18-Jun-2020
Category:	Documents
Upload:	others
View:	7 times
Download:	0 times

On the geometry of biharmonic maps and biharmonic submanifolds · On the geometry of biharmonic...

Documents