ScalingStacks

3.2. Explicit examples [028L]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context Β· Original author HTML

3.2. Explicit examples

In view of section 3.1, it is easy to construct examples of algebraic curves XβŠ‚β„‚2X\subset{\mathbb{C}}^{2} and functions in ℒ⁑(X){\mathcal{L}}(X) which do not admit an extension in ℒ⁑(β„‚2){\mathcal{L}}({\mathbb{C}}^{2}). We write z=(x,y)βˆˆβ„‚2z=(x,y)\in{\mathbb{C}}^{2}.

Example 3.2.

Let X={y=0}βˆͺ{y=1}βŠ‚β„‚2X=\{y=0\}\cup\{y=1\}\subset{\mathbb{C}}^{2} and Ξ·βˆˆβ„’β‘(X)\eta\in{\mathcal{L}}(X), where

η⁑(z)={ρ⁑(1,z),if​z=(x,0),ρ⁑(1,z)+1,if​z=(x,1).\eta(z)=\left\{\begin{array}[]{ll}\rho(1,z),\;\;\;\;\;\;\;{\rm if}\;z=(x,0),\\ \rho(1,z)+1,\;{\rm if}\;z=(x,1).\end{array}\right.

The function Ξ·~\widetilde{\eta} is not Ο‰\omega-psh on XΒ―={y=0}βˆͺ{y=t}\overline{X}=\{y=0\}\cup\{y=t\}, hence Ξ·\eta does not have an extension in ℒ⁑(β„‚2){\mathcal{L}}({\mathbb{C}}^{2}). Indeed, the maximum principle is violated along {y=0}\{y=0\} near the point a=[0:1:0]a=[0:1:0], since Ξ·~([t:1:0])=0\widetilde{\eta}([t:1:0])=0 for tβ‰ 0t\neq 0, while Ξ·~([t:1:t])=1\widetilde{\eta}([t:1:t])=1.

With a little more effort we can give an example as above where XX is an irreducible curve. Let ℂ⋆=β„‚βˆ–{0}{\mathbb{C}}^{\star}={\mathbb{C}}\setminus\{0\}.

Example 3.3.

Let XβŠ‚β„‚2X\subset{\mathbb{C}}^{2} be the irreducible cubic with equation x​y=x3+1xy=x^{3}+1. Then

XΒ―={[t:x:y]βˆˆβ„™2:xyt=x3+t3},XΒ―=Xβˆͺ{a},a=[0:0:1].\overline{X}=\{[t:x:y]\in{\mathbb{P}}^{2}:\,xyt=x^{3}+t^{3}\},\;\overline{X}=X\cup\{a\},\;a=[0:0:1].

The germ (XΒ―,a)(\overline{X},a) has two irreducible components X1,X2X_{1},\,X_{2}, both are smooth at aa, X1X_{1} being tangent to the line {x=0}\{x=0\}, and X2X_{2} to the line {t=0}\{t=0\}.

Note that in fact XβŠ‚β„‚β‹†Γ—β„‚X\subset{\mathbb{C}}^{\star}\times{\mathbb{C}} is the graph of the rational function y=x2+xβˆ’1y=x^{2}+x^{-1}, xβˆˆβ„‚β‹†x\in{\mathbb{C}}^{\star}. If (x,y)∈X(x,y)\in X and xβ†’0x\to 0 then (x,y)β†’a(x,y)\to a along X1X_{1}, while as xβ†’βˆžx\to\infty then (x,y)β†’a(x,y)\to a along X2X_{2}. The function

u⁑(x,y)=max⁑{βˆ’log⁑|x|,2​log⁑|x|+1}u(x,y)=\max\{-\log|x|,2\log|x|+1\}

is psh in ℂ⋆×ℂ{\mathbb{C}}^{\star}\times{\mathbb{C}}. It is easy to check that Ξ·:=u|Xβˆˆβ„’(X)\eta:=u\,|_{{}_{X}}\in{\mathcal{L}}(X) and

lim supX1βˆ‹[1:ΞΆ]β†’a(Ξ·(ΞΆ)βˆ’Ο(1,ΞΆ))=0,lim supX2βˆ‹[1:ΞΆ]β†’a(Ξ·(ΞΆ)βˆ’Ο(1,ΞΆ))=1.\limsup_{X_{1}\ni[1:\zeta]\to a}(\eta(\zeta)-\rho(1,\zeta))=0\;,\;\;\limsup_{X_{2}\ni[1:\zeta]\to a}(\eta(\zeta)-\rho(1,\zeta))=1.

Hence Ξ·\eta does not admit an extension in ℒ⁑(β„‚2){\mathcal{L}}({\mathbb{C}}^{2}).

We conclude this section with an example of a cubic XX in β„‚2{\mathbb{C}}^{2} and a psh function on XX of the form Ξ·=log⁑|P|\eta=\log|P|, where PP is a polynomial, so that Ξ·\eta admits a β€œtranscendental” extension with exactly the same growth, but small additional growth is necessary if we look for an β€œalgebraic” extension.

Proposition 3.4.

Let X={x=y3}X=\{x=y^{3}\} and η⁑(x,y)=log⁑|1+y|\eta(x,y)=\log|1+y|, so Ξ·|Xβˆˆβ„’1/3(X)\eta\,|_{{}_{X}}\in{\mathcal{L}}_{1/3}(X).

Given kβ‰₯1k\geq 1, there is a polynomial Qk​(x,y)Q_{k}(x,y) of degree k+1k+1 so that Qk​(y3,y)=(y+1)3​kQ_{k}(y^{3},y)=(y+1)^{3k}. In particular, ψk=13​k​log⁑|Qk|βˆˆβ„’(k+1)/3​k​(β„‚2)\psi_{k}=\frac{1}{3k}\log|Q_{k}|\in{\mathcal{L}}_{(k+1)/3k}({\mathbb{C}}^{2}) is an extension of Ξ·|X\eta\,|_{{}_{X}}.

There exists no polynomial Q⁑(x,y)Q(x,y) of degree kk so that Q⁑(y3,y)=(y+1)3​kQ(y^{3},y)=(y+1)^{3k}. However, Ξ·|X\eta\,|_{{}_{X}} has an extension in β„’1/3​(β„‚2){\mathcal{L}}_{1/3}({\mathbb{C}}^{2}).

Proof.

We construct QkQ_{k} by replacing y3y^{3} by xx in the polynomial

(y+1)3​k=βˆ‘j=03​k(3​kj)​yj.(y+1)^{3k}=\sum_{j=0}^{3k}{3k\choose j}y^{j}.

Since j=3​[j/3]+rjj=3[j/3]+r_{j}, rj∈{0,1,2}r_{j}\in\{0,1,2\}, it follows that

Qk​(x,y)=βˆ‘j=03​k(3​kj)​x[j/3]​yrj=3​k​xkβˆ’1​y2+l.d.t..Q_{k}(x,y)=\sum_{j=0}^{3k}{3k\choose j}x^{[j/3]}y^{r_{j}}=3kx^{k-1}y^{2}+l.d.t.\;.

We now check that there is no polynomial Q⁑(x,y)Q(x,y) of degree kk so that Q⁑(y3,y)=(y+1)3​kQ(y^{3},y)=(y+1)^{3k}. Indeed, if Q⁑(x,y)=βˆ‘j+l≀kcj​l​xj​ylQ(x,y)=\sum_{j+l\leq k}c_{jl}x^{j}y^{l} then

Q⁑(y3,y)=ck​0​y3​k+ckβˆ’1,1​y3​kβˆ’2+l.d.t.Q(y^{3},y)=c_{k0}y^{3k}+c_{k-1,1}y^{3k-2}+l.d.t.

does not contain the monomial y3​kβˆ’1y^{3k-1}.

Note that XΒ―={xt2=y3}=Xβˆͺ{a}\overline{X}=\{xt^{2}=y^{3}\}=X\cup\{a\}, where a=[0:1:0]a=[0:1:0], so the germ (XΒ―,a)(\overline{X},a) is irreducible. Proposition 3.1 implies that Ξ·|X\eta\,|_{{}_{X}} has an extension in β„’1/3​(β„‚2){\mathcal{L}}_{1/3}({\mathbb{C}}^{2}). ∎

We conclude with some remarks regarding our last example. If XX is an algebraic subvariety of β„‚n{\mathbb{C}}^{n} and ff is a holomorphic function on XX, ff is said to have polynomial growth if there is an integer N⁑(f)N(f) and a constant AA so that

|f⁑(z)|≀A​(1+β€–zβ€–)N⁑(f),βˆ€z∈X.|f(z)|\leq A(1+\|z\|)^{N(f)},\;\;\forall\,z\in X.

Then it is well known that there exists a polynomial PP of degree at most N⁑(f)+Ρ⁑(X)N(f)+\varepsilon(X) so that P|X=fP\,|_{{}_{X}}=f, where Ρ⁑(X)>0\varepsilon(X)>0 is a constant depending only on XX (see e.g. [Bj] and references therein). However, if XΒ―βŠ‚β„™N\overline{X}\subset{\mathbb{P}}^{N} is irreducible at each of its points at infinity then by Proposition 3.1 the psh function Ξ·=N​(f)βˆ’1​log⁑|f|βˆˆβ„’β‘(X)\eta=N(f)^{-1}\log|f|\in{\mathcal{L}}(X) has a psh extension in the Lelong class ℒ⁑(β„‚n){\mathcal{L}}({\mathbb{C}}^{n}).

On the other hand, Demailly [D1] has shown that in the case of the transcendental curve X={ex+ey=1}X=\{e^{x}+e^{y}=1\} any holomorphic function ff on XX, of polynomial growth, has a polynomial extension of the same degree to β„‚n{\mathbb{C}}^{n}. Hence it is natural to ask if for this curve one has that β„’(X)=β„’(β„‚n)|X{\mathcal{L}}(X)={\mathcal{L}}({\mathbb{C}}^{n})\,|_{{}_{X}}.

Original mathematics by the credited authors. Source-backed reader collection; mathematical self-containment is not assessed.