3. Algebraic subvarieties of ℂ n [028H]
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 · Original author HTML
3. Algebraic subvarieties of
If is an analytic subvariety of and is a positive number, we denote by
the Lelong class of psh functions on which verify
for all , where is a constant that depends on . We let . By Theorem A, functions admit a psh extension in each class , for every .
We assume in the sequel that is an algebraic subvariety of and address the question whether it is necessary to allow the arbitrarily small additional growth. More precisely, is it true that
|
|
|
i.e. is every psh function with logarithmic growth on the restriction of a
globally defined psh function with logarithmic growth? We will give a criterion for this to hold, but show that in general this is not the case.
3.1. Extension preserving the Lelong class
Consider the standard embedding
|
|
|
where denote the homogeneous coordinates on . Let be the Fubini-Study Kähler form and let
|
|
|
be its logarithmically homogeneous potential on .
We denote by the closure of in , so is an algebraic subvariety of . It is well known that the class is in one-to-one
correspondence with the Lelong class (see [GZ]). Let us look at the connection between -psh functions on and the class .
The mapping
|
|
|
is well defined and injective. However, it is in general not surjective, as shown by Examples 3.2 and 3.3 that follow.
Conversely, a function induces an upper semicontinuous function on defined in the obvious way:
|
|
|
The function is in general only weakly -psh on , i.e. it is bounded above on and it is -psh on the set of regular points of . This notion is in direct analogy to that of weakly psh function on an analytic variety (see [D2, section 1]). We do not pursue it any further here.
Note that if and only if . The following simple characterization is a consequence of Theorem B.
Proposition 3.1.
Let . The following are equivalent:
(i) There exists so that on .
(ii) .
(iii) For every point the following holds: if are the irreducible components of the germ then the value
|
|
|
is independent of .
In particular, if the germs are irreducible for all points then .
Proof.
Assume that holds. It follows that , where
|
|
|
is an -psh function on . Hence .
Conversely, if holds then by Theorem B there exists an -psh function on which extends . Hence is an extension of and .
The equivalence of and follows easily from [D2, Theorem 1.10].
∎
3.2. Explicit examples
In view of section 3.1, it is easy to construct examples of algebraic curves and functions in which do not admit an extension in . We write .
Example 3.2.
Let and , where
|
|
|
The function is not -psh on , hence does not have an extension in . Indeed, the maximum principle is violated along near the point , since for , while .
With a little more effort we can give an example as above where is an irreducible curve. Let .
Example 3.3.
Let be the irreducible cubic with equation . Then
|
|
|
The germ has two irreducible components , both are smooth at , being tangent to the line , and to the line .
Note that in fact is the graph of the rational function , . If and then along , while as then along . The function
|
|
|
is psh in . It is easy to check that and
|
|
|
Hence does not admit an extension in .
We conclude this section with an example of a cubic in and a psh function on of the form , where is a polynomial, so that admits a “transcendental” extension with exactly the same growth, but small additional growth is necessary if we look for an “algebraic” extension.
Proposition 3.4.
Let and , so .
Given , there is a polynomial of degree so that . In particular,
is an extension of .
There exists no polynomial of degree so that
. However, has an extension in .
Proof.
We construct by replacing by in the polynomial
|
|
|
Since , , it follows that
|
|
|
We now check that there is no polynomial of degree so that
. Indeed, if then
|
|
|
does not contain the monomial .
Note that , where , so the germ is irreducible. Proposition 3.1 implies that has an extension in .
∎
We conclude with some remarks regarding our last example. If is an algebraic subvariety of and is a holomorphic function on , is said to have polynomial growth if there is an integer and a constant so that
|
|
|
Then it is well known that there exists a polynomial of degree at most so that , where is a constant depending only on (see e.g. [Bj] and references therein). However, if is irreducible at each of its points at infinity then by Proposition 3.1 the psh function has a psh extension in the Lelong class .
On the other hand, Demailly [D1] has shown that in the case of the transcendental curve any holomorphic function on , of polynomial growth, has a polynomial extension of the same degree to . Hence it is natural to ask if for this curve one has that .