Squares of positive ( p , p ) -forms

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Complex Analysis

Squares of positive ( p , p ) -forms

Carrés de ( p , p ) -formes positives

Zbigniew Błocki ^a , Szymon Pli´s ^{b , 1}

aInstytut Matematyki, Uniwersytet Jagiello´nski, Łojasiewicza 6, 30-348 Kraków, Poland bInstytut Matematyki, Politechnika Krakowska, Warszawska 24, 31-155 Kraków, Poland

a r t i c l e i n f o a b s t r a c t

Article history:

Received 30 November 2012

Accepted after revision 21 January 2013 Available online 7 February 2013 Presented by Jean-Pierre Demailly

We show that if

α

is a positive

(

2

,

2

)

-form, then so is

α

². We also prove that this is no longer true for forms of higher degree.

©

2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.

r é s u m é

Nous montrons que si

α

est une

(

2

,

2

)

-forme positive alors

α

²l’est aussi. Nous prouvons également que ceci n’est plus vrai pour les formes de degré supérieur.

©

2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.

1. Introduction

Recall that a ( p , p ) -form α in C

ⁿ

is called positive (we write α  ^{0) if for} ( 1 , 0 ) -forms γ

1

, . . . , γ

n−p

one has:

α _∧ i γ

1

∧ ¯ γ

1

∧ · · · ∧ ⁱ γ

n−^p

∧ ¯ γ

n−^p

 ⁰ .

This is a natural geometric condition, positive ( p , p ) -forms are for example characterized by the following property: for every p-dimensional subspace V and a test function ϕ _ 0, one has:



V

ϕα  ⁰ .

It is well known that positive forms are real (that is α ¯ = α ) and if β is a ( 1 , 1 ) -form then

α _ 0 , β  ⁰ ⇒ α _{∧ β } 0 . (1)

It was shown by Harvey and Knapp

[5]

(and independently by Bedford and Taylor

[1]) that (1) does not hold for all

( p , p ) and ( q , q ) -forms α and β , respectively. We refer to Demailly’s book

[2], pp. 129–132, for a nice and simple introduction to

positive forms.

Dinew

[3]

gave an explicit example of ( 2 , 2 ) -forms α , β in C

⁴

^{such that} α  ^0, β  ^{0 but} α ∧ β < 0. We will recall it in the next section. The aim of this note is to show the following, somewhat surprising result:

Theorem 1. Assume that α is a positive ( 2 , 2 ) -form. Then α

²

is also positive.

E-mail addresses:Zbigniew.Blocki@im.uj.edu.pl,umblocki@cyf-kr.edu.pl(Z. Błocki),splis@pk.edu.pl(S. Pli´s).

1 The second named author was partially supported by the NCN grant No. 2011/01/D/ST1/04192.

1631-073X/$ – see front matter

©

2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.

http://dx.doi.org/10.1016/j.crma.2013.01.009

It turns out that this phenomenon holds only for ( 2 , 2 ) -forms:

Theorem 2. For every p  3, there exists a ( p , p ) -form α in C

^2p

^{such that} α _ 0 but α

²

< 0.

We do not know if similar results hold for higher powers of positive forms.

The paper is organized as follows: in Section

2

we present Dinew’s criterion for positivity of ( 2 , 2 ) -forms in C

⁴

^{, which} reduces the problem to a certain property of 6 × 6 matrices. Further simplification reduces the problem to 4 × 4 matrices.

We then solve it in Section

3. This is the most technical part of the paper. Higher degree forms are analyzed in Section4,

where a counterpart of Dinew’s criterion is showed and Theorem

2

is proved.

2. Dinew’s criterion

Without loss of generality we may assume that n = ^{4. Let} ω

1

, . . . , ω

4

be a basis of (C

⁴

)

^∗

such that:

dV := ⁱ ω

1

∧ ¯ ω

1

∧ · · · ∧ ⁱ ω

4

∧ ¯ ω

4

= ω

1

∧ · · · ∧ ω

4

∧ ¯ ω

1

∧ · · · ∧ ¯ ω

4

> 0 .

Set

Ω

1

:= ω

1

∧ ω

2

, Ω

2

:= ω

1

∧ ω

3

, Ω

3

:= ω

1

∧ ω

4

, Ω

₄

:= ω

₂

_∧ ω

₃

, Ω

₅

:= − ω

₂

_∧ ω

₄

, Ω

₆

:= ω

₃

_∧ ω

₄

.

Then

Ω

_j

∧ Ω

k

=

 ω

₁

_{∧ · · · ∧} ω

₄

, if k = ⁷ − ^j ,

0 , otherwise .

With every ( 2 , 2 ) -form α we can associate a 6 × ^{6-matrix A} = ( ^a

jk

) by

α = 

j,k

a

_jk

Ω

j

∧ ¯ Ω

k

.

For

β = 

j,k

b

_jk

Ω

j

∧ ¯ Ω

k

we have:

α ∧ β = 

j,k

a

_jk

b

_7−j,7−k

dV . (2)

The key will be the following criterion from

[3]:

Theorem 3. α  ^{0 if} ¯ ^{z Az}

^T

 0 for all z ∈ C

⁶

^{with z}

1

z

6

+ ^z

2

z

5

+ ^z

3

z

4

= ^0.

Sketch of proof. For γ

1

= ^b

1

ω

1

+ · · · + ^b

4

ω

4

, γ

2

= ^c

1

ω

1

+ · · · + ^c

4

ω

4

, we have i γ

1

∧ ¯ γ

1

∧ i γ

2

∧ ¯ γ

2

=



4 j,k,l,m=1

b

j

_b ¯

_k

_c

_l

c ¯

m

ω

j

∧ ω

_l

∧ ¯ ω

_k

∧ ¯ ω

m

= 

j<l k<m

( b

_j

c

_l

− ^b

l

c

_j

)( b

_k

c

_m

− ^b

m

c

_k

) ω

j

∧ ω

l

∧ ¯ ω

k

∧ ¯ ω

m

.

It is now enough to show that the image of the mapping:

C

⁸

 ( b

1

, . . . , b

4

, c

1

, . . . , c

4

)

−→ ( ^b

1

c

₂

− ^b

2

c

₁

, b

₁

c

₃

− ^b

3

c

₁

, b

₁

c

₄

− ^b

4

c

₁

, b

₂

c

₃

− ^b

3

c

₂

, − ^b

2

c

₄

+ ^b

4

c

₂

, b

₃

c

₄

− ^b

4

c

₃

) ∈ C

⁶

is precisely { ^z ∈ C

⁶

^{: z}

1

z

6

+ ^z

2

z

5

+ ^z

3

z

4

= ⁰ } . Indeed, it is a well-known fact that the image of the Plücker embedding of the 4-dimensional Grassmannian G ( 2 , 4 ) in P (Λ

²

C

⁴

) P

⁵

is the quadric defined by the above equation. 2

Using Theorem

3

and an idea from

[3], we can show:

Proposition 4. The form

α

a

=



6 j=1

Ω

j

∧ ¯ Ω

j

+ ^a Ω

1

∧ ¯ Ω

6

+ ¯ ^a Ω

6

∧ ¯ Ω

1

is positive if and only if | ^a |  ^2.

Proof. We have:

¯

z Az

^T

= | z |

²

+ 2 Re ( a ¯ z

1

z

6

)  2 | z

1

z

6

| + 2 | z

2

z

5

+ z

3

z

4

| + 2 Re ( a ¯ z

1

z

6

).

If z

1

z

6

+ ^z

2

z

5

+ ^z

3

z

4

= ^{0 and} | ^a |  2 we clearly get ¯ z Az

^T

 0. If we take z

1

, z

6

with ¯ z

1

z

6

= −¯ ^a, | ^z

1

| = | ^z

6

| ^{and z}

2

, . . . , z

5

with z

2

z

5

+ ^z

3

z

4

= − ^z

1

z

6

then ¯ z Az

^T

= ² | ^a |( ² − | ^a |) ^. 2

By (2):

α

a

∧ α

b

= 2 

3 + Re ( a _b ¯ )  dV .

Therefore, α

2

, α

₋2

are positive, but α

2

∧ α

₋2

< 0.

In view of Theorem

3, we see that Theorem1

is equivalent to the following:

Theorem 5. Let A = ( ^a

jk

) ∈ C

^6×6

be hermitian and such that ^{z Az} ¯

^T

 ^{0 for z} ∈ C

⁶

^{with z}

1

z

₆

+ ^z

2

z

₅

+ ^z

3

z

₄

= ^{0. Then}



6 j,k=¹

a

_jk

a

_7−j,7−k

 0 .

We will need the following technical reduction:

Lemma 6. For every ( 2 , 2 ) -form α in C

⁴

, we can find a basis ω

1

, . . . , ω

4

of (C

⁴

)

^∗

such that:

α _∧ ω

1

∧ ω

2

∧ ¯ ω

1

∧ ¯ ω

_j

₌ α _∧ ω

1

∧ ω

2

∧ ¯ ω

2

∧ ¯ ω

_j

₌ 0 (3) for j = ³ , 4.

Proof. We may assume that α ₌ 0, then we can find ω

1

, ω

2

∈ (C

⁴

)

^∗

such that

α ∧ ω

1

∧ ω

2

∧ ¯ ω

1

∧ ¯ ω

2

= α ∧ ⁱ ω

1

∧ ¯ ω

1

∧ ⁱ ω

2

∧ ¯ ω

2

= ⁰ . (4) By V

1

denote the subspace spanned by ω

1

, ω

2

and by V

2

the subspace of all ω _{∈ (C}

⁴

)

^∗

satisfying (3) with ω

j

replaced by ω . Then dim V

1

= ^{2, dim V}

2

 2, and by (4) we infer V

1

∩ ^V

2

= { ⁰ } ^{, hence} ( C

⁴

)

^∗

= ^V

1

⊕ ^V

2

. 2

3. Proof of Theorem

5

By Lemma

6

we may assume that the matrix from Theorem

5

satisfies a

₂₆

= ^a

36

= ^a

46

= ^a

56

= ⁰

and

a

62

= ^a

63

= ^a

64

= ^a

65

= ⁰ .

Then



6 j,k=¹

a

_jk

a

_7−j,7−k

=



5 j,k=²

a

_jk

a

_7−j,7−k

+ ² 

a

11

a

66

+ | ^a

16

|

²

 .

Therefore Theorem

5

is in fact equivalent to the following result:

Theorem 7. Let A = ( ^a

jk

) ∈ C

^4×4

be hermitian and such that z Az ¯

^T

 ^{0 for z} ∈ C

⁴

^{with z}

1

z

4

+ ^z

2

z

3

= ^{0. Then}



4 j,k=¹

a

_jk

a

_5−j,5−k

 ⁰ . (5)

Proof. Write

A =

⎛

⎜ ⎝

λ

1

a b α

¯

a λ

2

β − c b ¯ ¯β λ

3

− ^d

¯

α _−¯ c −¯ ^d λ

4

⎞

⎟ ⎠ .

It satisfies the assumption of the theorem if and only if for every z ∈ C

⁴

of the form z = ( 1 , ζ, w , −ζ w ) one has ¯ z Az

^T

 0.

We can then compute

¯

z Az

^T

= λ

1

+ ^{2 Re} ( a ζ ) + λ

2

|ζ |

²

+ ^{2 Re} 

b − α ζ + β ¯ζ + ^c |ζ |

²

 w  + 

λ

₃

+ ^{2 Re} ( d ζ ) + λ

4

|ζ |

²



| ^w |

²

.

Therefore A satisfies the assumption if λ

j

 ^0,

| ^a |  

λ

1

λ

2

, | ^b |  

λ

1

λ

3

, | ^c |  

λ

2

λ

4

, | ^d |  

λ

3

λ

4

, (6)

and for every ζ ∈ C

 _b − α ζ + β ¯ζ + c |ζ |

²



²

 

λ

1

+ 2 Re ( a ζ ) + λ

2

|ζ |

²



λ

3

+ 2 Re ( d ζ ) + λ

4

|ζ |

²



. (7)

In our case (5) is equivalent to

4 Re ( a _d ¯ + ^b c ¯ )  ² (λ

₁

λ

₄

+ λ

2

λ

₃

) + ² 

| α _|

²

_{+ |β|}

²

^ .

We will in fact prove something more:

4 Re ( a _d ¯ + ^b c ¯ )  ( 

λ

1

λ

4

+ 

λ

2

λ

3

)

²

+ 

| α _{| + |β|} ^

²

. (8)

Without loss of generality, we may assume that:

Re ( a d ¯ ) > 0 , Re ( b ^c ¯ ) > 0 ,

for if for example Re ( a d ¯ )  0 then by (6) 4 Re ( a _d ¯ + ^b c ¯ )  ^{4 Re} ( b c ¯ )  ⁴ 

λ

1

λ

2

λ

3

λ

4

 ( 

λ

1

λ

4

+  λ

2

λ

3

)

²

.

Set u := ^Re ( a d ¯ ) and ζ := − ^r d ¯ / | ^d | ^{, where r} > 0 will be determined later. Then we can write the right-hand side of (7) as follows:



λ

1

− ^2ur

| ^d | + λ

2

r

²



λ

3

− ^2r | ^d | + λ

4

r

²



= 

λ

1

+ λ

2

r

²



λ

3

+ λ

4

r

²



+ ^4ur

²

− ^2r



λ

1

| ^d | + λ

3

u

| ^d | + ^r

²



λ

2

| ^d | + λ

4

u

| ^d |



 

λ

1

+ λ

2

r

²



λ

3

+ λ

4

r

²



+ 4ur

²

− 4r

²

( 

λ

1

λ

4

+  λ

2

λ

3

) √

u

= ( 

λ

1

λ

4

+ 

λ

2

λ

3

− ² √

u )

²

r

²

+ 

λ

1

λ

3

− 

λ

2

λ

4

r

²



2

.

For r = (

^λ_λ¹₂^λ_λ³₄

)

^1/4

from (7) we thus obtain:

  ^b _r + _d ¯

| ^d | α ₋ ^d

| ^d | β + ^cr 

  

λ

1

λ

4

+ 

λ

2

λ

3

− ² √ u .

We also have:

  ^b _r + _d ¯

| d | ( α − ¯β) + cr 

  

 ^b _r + cr 

 − | α | − |β|  2 

Re ( b c ¯ ) − | α | − |β|

and therefore:

2



Re ( a _d ¯ ) + 2 

Re ( b ¯ c )  

λ

1

λ

4

+ 

λ

2

λ

3

+ | α | + |β|.

To get (8), we can now use the following fact: if 0  ^a

1

 ^{x, 0}  ^a

2

 ^{x and a}

1

+ ^a

2

 ^x + ^{y then a}

²₁

+ ^a

²₂

 ^x

²

+ ^y

²

^{. This can} be easily verified: if a

1

+ a

2

 x then a

²₁

+ a

²₂

 x

²

and if a

1

+ a

2

 x then

x

²

+ ^y

²

 ^x

²

+ ( ^a

1

+ ^a

2

− ^x )

²

= ^a

²₁

+ ^a

²₂

+ ^2x ( x − ^a

1

)( x − ^a

2

). 2

4. (

p

,

p

) -Forms in C C C

^2p

In C

^2p

we choose the positive volume form:

dV := ^{i dz}

1

∧ ^d ¯ ^z

1

∧ · · · ∧ ^{i dz}

2p

∧ ^d ¯ ^z

2p

= ^dz

1

∧ · · · ∧ ^dz

2p

∧ ^{d z} ¯

1

∧ · · · ∧ ^d ¯ ^z

2p

.

By I we will denote the set of subscripts J = ( ^j

1

, . . . , j

p

) such that 1  ^j

1

< · · · < ^j

p

 2p. For every J ∈ I there exists unique J

^

∈ I such that J ∪ ^J

^

= { ¹ , . . . , 2p } . We also denote dz

J

= ^dz

j1

∧ · · · ∧ ^dz

jp

and ε

J

= ± 1 is defined in such a way that:

dz

J

∧ ^dz

J^

= ε

J

dz

1

∧ · · · ∧ ^dz

2p

.

Note that:

ε

J

ε

J^

= (− ¹ )

^p

. (9)

With every ( p , p ) -form α in C

^2p

we can associate an N × ^N-matrix ( a

_{J K}

) , where N =  I = ( 2p ) !

( p !)

²

,

by

α = 

J,K

a

_{J K}

i dz

_j1

∧ ^d ¯ z

_k1

∧ · · · ∧ ^{i dz}

jp

∧ ^d ¯ z

_kp

= ⁱ

^p2



J,K

a

_{J K}

dz

_J

∧ ^d ¯ z

_K

(10)

(note that ( − ¹ )

^p(p−1)/2

i

^p

= ⁱ

^p2

^{). Then} α

²

= 

J,K

ε

J

ε

K

a

_{J K}

a

_JK^

dV (11)

and for γ

1

, . . . , γ

p

∈ (C

^2p

)

^∗

α ∧ ⁱ γ

1

∧ ¯ γ

1

∧ · · · ∧ ⁱ γ

p

∧ ¯ γ

p

= 

J,K

a

_{J K}

γ

1

∧ · · · ∧ γ

p

∧ ^dz

J

∧ ( γ

1

∧ · · · ∧ γ

p

∧ ^dz

K

).

Therefore ( a

J K

) has to be positive semi-definite on the image of the Plücker embedding

 C

^2p



_∗



p

 ( γ

1

, . . . , γ

p

) −→

 γ

1

∧ · · · ∧ γ

p

∧ ^dz

J

dz

₁

∧ · · · ∧ ^dz

2p



J∈I

∈ C

^N

⁽¹²⁾

which is well known to be a variety in C

^N

(see, e.g.,

[4], p. 64).

We are now ready to prove Theorem

2:

Proof of Theorem

2. First note that it is enough to show it for p

= ^{3. For if} α is a ( 3 , 3 ) -form in C

⁶

^{such that} α  ^{0 and} α

²

< 0 then for p > 3 we set:

β := ^{i dz}

7

∧ ^d z ¯

₇

+ · · · + ^idz

2p

∧ ^d z ¯

_2p

.

We now have α _{∧ β}

^p−3

_ 0 but ( α _{∧ β}

^p−3

)

²

= α

²

_{∧ β}

^2p−6

< 0.

Set p = 3, so that N = 20, and order I = { ^J

1

, . . . , J

₂₀

} lexicographically. Then the image of the Plücker embedding (12) is in particular contained in the quadric:

z

₁

z

₂₀

− ^z

10

z

₁₁

+ ^z

5

z

₁₆

− ^z

2

z

₁₉

= ⁰ . (13)

For positive a , λ, μ to be determined later define:

α := ⁱ



λ( dz

_J1

∧ ^d z ¯

_J1

+ ^dz

J20

∧ ^d ¯ z

_J20

) + μ ^

k∈{²,5,10 11,16,19}

dz

_Jk

∧ ^d ¯ z

_Jk

+ ^a ( dz

_J1

∧ ^d ¯ z

_J20

+ ^dz

J20

∧ ^d ¯ z

_J1

)

 .

Then, similarly as in the proof of Proposition

4,

¯

z Az

^T

= λ 

| ^z

1

|

²

+ | ^z

20

|

²



+ μ ^ | ^z

2

|

²

+ | ^z

5

|

²

+ | ^z

10

|

²

+ | ^z

11

|

²

+ | ^z

16

|

²

+ | ^z

19

|

²



+ ^{2a Re} (¯ z

₁

z

₂₀

)

 ² (λ − ^a ) | ^z

1

z

₂₀

| + ² μ _{| −} z

₁₀

z

₁₁

+ ^z

5

z

₁₆

− ^z

2

z

₁₉

|

= ² (λ + μ ₋ a ) | ^z

1

z

20

|

if z satisfies (13). Therefore α  ^{0 if a}  λ + μ .

On the other hand, by (11) and (9):

α

²

= 2 

λ

²

+ 3 μ

²

− a

²

 dV .

We see that if we take a = λ + μ and λ > μ > 0, then α  ^{0 but} α

²

< 0. 2 References

[1] E. Bedford, B.A. Taylor, Simple and positive vectors in the exterior algebra ofCⁿ, preprint, 1974.

[2] J.-P. Demailly, Complex Analytic and Differential Geometry, monograph, 1997, available athttp://www-fourier.ujf-grenoble.fr/~demailly.

[3] S. Dinew, On positiveC(2,2)(C⁴)forms, preprint, 2006.

[4] J. Harris, Algebraic Geometry. A First Course, Grad. Texts in Math., vol. 133, Springer, 1995.

[5] R. Harvey, A.W. Knapp, Positive(p,p)forms, Wirtinger’s inequality, and currents, in: R.O. Kujala, A.L. Vitter III (Eds.), Value Distribution Theory, Part A, Dekker, 1974, pp. 43–62.