GÉOMÉTRIE ANALITIQUE COMPLEXE

UNIVERSITÉ PARIS-SUD

Devoir 1

Valerio Proietti

November 5, 2014

EXERCISE 1

Thanks to Corollary 3.4 of [1, p. 21] we know that 1/πz is a fundamental solution of the opera-tor ∂

∂z onC. As a consequence, if g is a smooth function (or even a distribution) with compactsupport in C, then the convolution f = 1/πz? g is a solution of the equation

∂

∂zf = g

To see the necessary condition, observe that

⟨ f ,∂zn

∂z⟩ =−⟨g , zn⟩ = 0

since z 7→ zn is holomorphic. It is also clear that our solution has the required regularity, sincewe can write

f = 1

πz? g = g ?

1

πz=

∫C

g (w − z)

πzdλ(z)

and take advantage of the smoothness of g .Conversely, if the condition on the powers of z is satisfied, by means of the series expansion

1

w − z= 1

w· 1

1− zw

= ∑n≥0

zn w−n−1 ∀z : |z| < |w |

we can write, for all |w | > R,

0 = f (w) =∫C

g (z)

π(w − z)dλ(z) =

∫suppg

g (z)

π(w − z)dλ(z) = ∑

n≥0

∫suppg

g (z)

πdλ(z)

where R is such that suppg ⊂ D(0,R) and the switch between integral and series is due touniform convergence of the geometric series.

EXERCISE 2

Since T is normal, we know that the coefficients of T and dT (when written in local coor-dinates) are bounded linear functionals on the space of continuous functions with compactsupport. As a consequence of Riesz-Markov representation theorem, we conclude that the

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generic coefficient of T (which will be denoted t in the sequel) is a (complex) Radon mea-sure. Recall that such measures are inner regular, id est

µt (E) = sup µ(F ) | F ⊂ E , F is compact

Since our manifolds have countable basis, we know that every open set admits an exhaustionby compact sets. We can safely assume that Y is closed.1

These preliminaries are useful for the following construction: given a compact set F ⊂ M àY , we build a bump functionφwhich equals 1 on F and has compact support in MàY . Then,thanks to the hypothesis on the support of T , we have

t (φ) = 0 =⇒ µt (F ) = 0 =⇒ suppµt ⊂ Y

We now choose a local chart for which Y = x1 = ·· · = xq = 0. Thus for every compactlysupported function f on M and 1 ≤ j ≤ q , we can write

(x j t )( f ) =∫

Mx j f dµt =

∫x j=0

x j f dµt +∫

Màx j=0x j f dµt = 0

and conclude x j T = 0. An analogous argoment shows x j dT = 0.For a (p −1)-form ω, we have

0 = (x j dT )(ω) = dT (x jω) = (−1)m−p+1[T (d x j ∧ω)+T (x j dω)] =(−1)m−p+1[T (d x j ∧ω)+ (x j T )(dω)] = (−1)m−p+1T (d x j ∧ω)

and for ω was arbitrarily chosen, we can infer that T is zero on all forms which have d x j intheir local expression. This is equivalent to say that we can factor out d x1 ∧ ·· ·∧d xq in thelocal expression for T . By means of partition of unity, we know that C ∞(X ) is a fine sheaf, andsince Dp (X ) is a C ∞(X )-module, it is soft. This basically means that Dp (X ) →Dp (Y ) : η 7→ η

is surjective.2

We are now in a position to define Θ. Taken η ∈ Dp (Y ), we pose Θη = Tη. This is welldefined because we proved suppµt ⊂ Y , and on this set any two extensions coincide.3

We have

ω=ωI ,J d x I ,J |I |+ |J | = p, I ⊂ 1, . . . , q, J ⊂ q +1, . . . ,m

Tω= (i∗Θ)ω=Θ(i∗ω) =

0 if I 6= ;Tω=Θω else

EXERCISE 3

We have v =∑|I |=q v I d zI and we choose k minimum such that no d zi occurs in this sum for

i > k. We proceed by induction on k, the case k = 1 being solved by application of Exercise 1.We write v = f ∧d zk +g so that only d zi for i ∈ 1, . . . ,k −1 is found in the expression of f , g .By assumption, we have 0 = ∂v = (∂ f )∧d zk +∂g , this clearly implies

∂v I

∂zi= 0 i > k

1Strictly speaking, this is not needed, but it’s psychologically helpful. Note that the support of a measure is thecomplement of the union of open sets of measure zero.

2This sheaf theoretical language breaks down to the fact that we can extend a smooth function on Y to a smoothfunction on X .

3Another possibility would be a local definition of Θ. In some sense, this would be even easier, since the coeffi-cients of T are measures, and we have no problems in taking the restriction (to Y ) of a measure.

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Therefore, the functions v I are holomorphic in the variables zi , i > k. We now take a bumpfunction φ which equals 1 on D(0,r )k and define the functions

FI = (φ(zk )v I )?1

πzk

where the convolution is considered only with respect to zk .We know that

∂FI

∂zk= v I

on D(0,r )k ⊂ C, in addition FI is holomorphic in zk+1, . . . , zm and differentiable in the othervariables. Having defined

F = (−1)q∑k∈I

FI d zIàk

we can calculate

∂i F =

0 i > k

− f ∧d zk i = k∂i = ∂

∂zid zi

Hence, v+∂F is ∂-closed and does not involve any d zi for i ≥ k. This terminates the inductivestep and the proof.

EXERCISE 4

We begin remarking that a simple computation yields

∇2(log

∣∣∣z − 1

j

∣∣∣)= 4∂

∂z

∂

∂z

(log

∣∣∣z − 1

j

∣∣∣)= 0

on the set Cà

1j

. The point left is where the function attains −∞, hence u j = log

∣∣∣z − 1j

∣∣∣ is

subharmonic. We claim u is harmonic on the set S =C0 à

1j | j ∈N+

, where C0 denotes the

punctured plane. To see this, choose z0 ∈ S and set

m = minj∈N+

∣∣∣z0 − 1

j

∣∣∣> 0

Note that m exist otherwise there would be a subsequence j (n) such that 0 < |z0 − j (n)−1| <n−1. For z ∈ D(z0,m/2), we have

m − m

2≤

∣∣∣z0 − 1

j

∣∣∣−|z − z0| ≤∣∣∣z − 1

j

∣∣∣≤ |z − z0|+∣∣∣z0 − 1

j

∣∣∣≤ m

2+|z0|+1 (1)

Due to continuity of |log (·)|, we conclude |u j | ≤ M for some M > 0 and for all j positive inte-gers. As a consequence,

u = ∑j≥1

2− j u j and∑j≥1

2− j ∂

∂zu j =

∑j≥1

2− j−1

|z − j−1|

converge uniformly, and this is sufficient to interchange derivative and series. We still needto check subharmonicity in 0, but this is easy, as the partial sums of −u(0) form an increasingsequence on which we can apply Beppo-Levi theorem on monotone convergence, to obtain

u(0) ≤µS(u;0,r ) ∀r ∈R

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because the inequality is valid for each term of the series (notation is as in [1, p. 33]). Beingthe exponential map increasing and convex, we see that eu is also subharmonic as well aslocally bounded, since (1) proves a local upper bound for u (valid for all complex numbersz0).

Lastly, using the sequence zn = 1n , we see eu(zn) = 0 whereas eu(0) > 0.

EXERCISE 5

If we denote the ellipsoid byΩ, then the function ρ(z) = |z1|2 +|z2|4 +|z3|6 −1, is the definingfunction ofΩ, i.e., ρ < 0 onΩ and ρ = 0,dρ 6= 0 on ∂Ω.

On the holomorphic tangent space Thz∂Ω, the Levi form is defined by

L∂Ω,z =1

|∇ρ(z)|∑

1≤i , j≤n

∂2ρ

∂zi∂z jd zi ⊗d z j

Therefore, carrying out simple calculations, we find

L∂Ω,z = (|z1|2 +4|z2|6 +9|z3|10)−12

1 0 00 4|z2|2 00 0 9|z3|4

Hence,

rankL∂Ω,z =

3 if z ∈ z2 6= 0∩ z3 6= 0

2 if z ∈ z2 = 04z3 = 0

1 if z ∈ D(0,1)× 0× 0

(Here, 4 is the simmetric difference of sets).As a sidenote, applying Theorem 7.12 in [1, p. 59], we see thatΩ is pseudoconvex.

EXERCISE 6

Let K be the holomorphic hull of K and set

K = (z1, z2) ∈C2 | |z1| ≤♣ r1, |z2| ≤♠ r2, |z1|1/ log r1 |z2|1 log r2 ≤♥ e

By definition, ∀ f ∈O (C) and z ∈ K , we have

| f (z)| ≤ supK

| f | (2)

We choose three functions: the projections, and (z1, z2) 7→ z1/ log r11 z1/ log r2

2 . Applying defini-tion (2) to the former yields conditions ♣ and ♠, whilst condition ♥ is obtained with thelatter. Hence K ⊂ K . For the reverse inclusion, consider Uα = Sα∩D(0, (r1,r2)), where Sα =(z1, z2) ∈C2 | z1/ log r1

1 z1/ log r22 =α and |α| ∈ (0,e]. For the implicit function theorem each Sα is

a Riemann surface.Now observe that ∂Uα = Sα∩∂D(0, (r1,r2)) ⊂ K , simply taking the modulus on the condi-

tion defining the surface. Then the inclusion map satisfies the hypothesis of Proposition 6.6din [1, p. 47], thus we get Uα ⊂ K .

To conclude, note that every point of K is in Uα for some α:

K ⊂e⋃

|α|=0Uα ⊂ K

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REFERENCES

[1] J.-P. Demailly. Complex analytic and differential geometry. 2012. URL: www- fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf (cit. on pp. 1, 4).

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