On the derivatives of the Lempert functions

0.15 MB·English

by  Nikolai Nikolov

ON THE DERIVATIVES OF THE LEMPERT
FUNCTIONS
8
0 NIKOLAI NIKOLOV AND PETER PFLUG
0
2
Abstract. We show that if the Kobayashi–Royden metric of a
n
complex manifold is continuous and positive at a given point and
a
J any non-zero tangent vector, then the ”derivatives” of the higher
order Lempert functions exist and equal the respective Kobayashi
8
1 metricsatthepoint. ItisageneralizationofaresultbyM.Kobaya-
shi for taut manifolds.

]
V
C
. 1. Introduction and results
h
t
a Let D ⊂ C be the unit disc. Let M be an n-dimensional complex
m
˜
manifold. Recall first the definitions of the Lempert function k and
M
[
the Kobayashi–Royden pseudometric κ of M:
M
1
v k˜∗ (z,w) = inf{|α| : ∃f ∈ O(D,M) : f(0) = z,f(α) = w},
M
2
9 k˜ = tanh−1k˜∗ ,
M M
8
2 κM(z;X) = inf{|α| : ∃f ∈ O(D,M) : f(0) = z,αf∗(d/dζ) = X},
.
1 where X is a complex tangent vector to M at z. Note that such an f
0
always exists (cf. [12]; according to [2], page49, this was already known
8
0 by J. Globevnik).
:
v TheKobayashipseudodistancek canbedefinedasthelargestpseu-
M
i ˜ (m)
X dodistanceboundedbyk . Notethatifk denotesthem-thLempert
M M
r function of M, m ∈ N, that is,
a
m
(m) ˜
k (z,w) = inf{ k (z ,z ) : z ,...,z ∈ M,z = z,z = w},
M X M j−1 j 0 m 0 m
j=1
then
(∞) (m)
k (z,w) = k := infk (z,w).
M M M
m
2000 Mathematics Subject Classification. 32F45.
Key words and phrases. Lempertfunctions,Kobayashipseudodistance,Kobaya-
shi–Royden pseudometric, Kobayashi–Busemanpseudometric.
ThisnotewaswrittenduringthestayofthefirstnamedauthorattheUniversit¨at
Oldenburg supported by a grant from the DFG, Az. PF 227/8-2 (November –
December 2006). He likes to thank both institutions for their support.
1
2 NIKOLAINIKOLOVAND PETER PFLUG
By a result of M.-Y. Pang (see [9]), the Kobayashi–Royden metric is
the ”derivative” of the Lempert function for taut domains in Cn; more
precisely, if D ⊂ Cn is a taut domain, then
˜
k (z,z +tX)
D
κ (z;X) = lim .
D
t90 t
In [6], S. Kobayashi introduces a new invariant pseudometric, called
the Kobayashi–Buseman pseudometric in [3]. One of the equivalent
ways to define the Kobayashi–Buseman pseudometric κˆ of M is just
M
to set κˆ (z;·) to be largest pseudonorm bounded by κ (z;·). Recall
M M
that
m m
κˆ (z;X) = inf{ κ (z;X ) : m ∈ N, X = X}.
M X M j X j
j=1 j=1
Thus it is natural to consider the new function κ(m), m ∈ N, namely,
M
m m
(m)
κ (z;X) = inf{ κ (z;X ) : X = X}.
M X M j X j
j=1 j=1
We call κ(m) the m-th Kobayashi pseudometric of D. It is clear that
M
(m) (m+1) (m) (m+1) (m)
κ ≥ κ and if κ (z;·) = κ (z;·) for some m, then κ (z;·)
M M M M M
(j) (2n−1) (∞)
= κ (z;·) for any j > m. It is shown in [8] that κ = κ := κˆ ,
D M M M
and 2n−1 is the optimal number, in general.
We point out that all the introduced objects are upper semicontin-
uous. Recall that this is true for κ (cf. [7]). It remains to check
M
˜
this for k . We shall use a standard reasoning. Fix r ∈ (0,1) and
M
z,w ∈ M. Let f ∈ O(D,M), f(0) = z and f(α) = w. Then
f˜= (f,id) : ∆ → M˜ = M×∆ is an embedding. Setting f˜(ζ) = f˜(rζ),
r
by [10], Lemma 3, we may find a Stein neighborhood S ⊂ M˜ of f˜(D).
r
Embed S as a closed complex manifold in some CN and denote by ψ
the respective embedding. Moreover, there is an open neighborhood
V ⊂ CN of ψ(S) and a holomorphic retraction θ : V → ψ(S). Then,
for z′ near z and w′ near w, we may find, as usual, g ∈ O(D,V) such
that g(0) = ψ(z′,0) and g(α/r) = ψ(w′,α). Denote by π the natu-
ral projection of M˜ onto M. Then h = π ◦ ψ−1 ◦ θ ◦ g ∈ O(D,M),
h(0) = z′ and h(α/r) = w′. So rk˜∗ (z′,w′) ≤ α, which implies that
M
limsup k˜ (z′,w′) ≤ k˜ (z,w).
M M
z′→z,w′→w
ToextendPang’sresultonmanifolds, wehavetodefinethe”derivati-
ves” of k(m), m ∈ N∗ = N ∪{∞}. Let (U,ϕ) be a holomorphic chart
M
ON THE DERIVATIVES OF THE LEMPERT FUNCTIONS 3
near z. Set
k(m)(w,ϕ−1(ϕ(w)+tY))
Dk(m)(z;X) = limsup M .
M t90,w→z,Y→ϕ∗X |t|
Note that this notion does not depend on the chart used in the defini-
tion and
Dk(m)(z;λX) = |λ|Dk(m)(z;X), λ ∈ C.
M M
(m)
Replacing limsup by liminf, we define Dk .
M
FromM.Kobayashi’spaper[5]itfollowsthat,ifM isatautmanifold,
then
κˆ (z;X) = Dk (z;X) = Dk (z;X),
M M M
thatis, theKobayashi–Buseman metricisthe”derivative” oftheKoba-
yashi distance. The proof there also leads to
(∗) κ(m)(z;X) = Dk(m)(z;X) = Dk(m)(z;X), m ∈ N∗.
M M M
We say that a complex manifold M is hyperbolic at z if k (z,w) >
M
0 for any w 6= z. We point out that the following conditions are
equivalent:
(i) M is hyperbolic at z;
(ii) liminf k˜ (z′,w) > 0 for any neighborhood U of z;
M
z′→z,w∈M\U
(iii) κ (z;X) := liminf κ (z′;X′) > 0 for any X 6= 0;
M z′→z,X′→X M
The implication (i)⇒(ii) ⇒(iii) are almost trivial (cf. [3]) and the
implication (iii)⇒(i) is a consequence of the fact that k is the integra-
M
ted form of κ .
M
In particular, if M is hyperbolic at z, then it is hyperbolic at any z′
near z.
Since if M is taut, then it is k-hyperbolic and κ is a continuous
M
function, the following theorem is a generalization of (∗).
Theorem 1. Let M be a complex manifold and z ∈ M.
(i) If M is hyperbolic at z and κ is continuous at (z,X), then
M
˜ ˜
κ (z;X) = Dk (z;X) = Dk (z;X).
M M M
(ii) If κ is continuous and positive at (z,X) for any X 6= 0, then
M
κ(m)(z;·) = Dk(m)(z;·) = Dk(m)(z;·), m ∈ N∗.
M M M
The first step in the proof of Theorem 1 is the following
Proposition 2. For any complex manifold M one has that
κ(m) ≥ Dk(m), m ∈ N∗.
M M
4 NIKOLAINIKOLOVAND PETER PFLUG
Note that when M is a domain, a weaker version of Proposition 2
can be found in [3], namely, κˆ ≥ Dk (the proof is based on the fact
M M
that Dk (z;·) is a pseudonorm).
M
2. Examples
The following examples show that the assumption on continuity in
Theorem 1 is essential.
• Let A be a countable dense subset of C . In [1] (see also [3]), a
∗
pseudoconvex domain D in C2 is constructed such that:
(a) (C×{0})∪(A×C) ⊂ D;
(b) if z = (0,t) ∈ D, t 6= 0, then κ (z ;X) ≥ C||X|| for some
0 D 0
˜
C = C > 0. (One can be shown that even Dk (z ;X) ≥ C||X||.)
t D 0
(3) (5)
Then it is easy to see that κ (·;e ) = Dk (·;e ) = k = 0 and
D 2 D 2 D
κˆ (z ;X) ≥ c||X||, where e = (0,1) and c > 0. Thus
D 0 2
(3) (5)
κˆ (z ;X) > κ (z ;e ) = Dk (z ;e ) = Dk (z ;X), X 6= 0.
D 0 D 0 2 D 0 2 D 0
This phenomena obviously extends to Cn, n > 2 (by considering
D×Dn−2). So the inequalities in Proposition 2 are strict in general.
• If D is a pseudoconvex balanced domain with Minkowski function
h , then (cf. [3])
D
˜
h = κ (0;·) = Dk (0;·).
D D D
˜ ˜
Therefore, Dk (0;X) > Dk (0;X) if κ (0;·) is not continuous at X.
D D D
On the other hand, if Dˆ denotes the convex hull of D, then
h = κˆ (0;·) = Dk (0;·) = Dk (0;·) = κˆ (0;·).
Dˆ D D D D
• Modifying the first example leads to a pseudoconvex domain D ⊂ C2
with
L (γ) > 0 = L (γ) = L (γ),
DkD kD DkD
where γ : [0,1] → C2, γ(t) := (ti/2,1/2), and L (γ) denotes the re-
•
spective length.
Indeed, choose a dense sequence (r ) in [0,i/2]. Put
j
∞ ∞
1 u(λ/2−r )
u(λ) = log |λ−1/k|, v(λ) = j , λ ∈ C,
X k2 4 X 2j2
k=1 j=1
and
D = {z ∈ C2 : ψ(z) = |z |ekzk2+v(z1) < 1}.
2
It is easy to see that v is a subharmonic function on C. Hence D is a
pseudoconvex domain with (C ×{0})∪( ∞ {r + 1/k}× C) ⊂ D.
Sj,k=1 j
ON THE DERIVATIVES OF THE LEMPERT FUNCTIONS 5
Observe that u|D < −1 and so D contains the unit ball B2. Note also
that
k (a,b) = 0, a,b ∈ γ([0,1]).
D
Set ψ(z) = kzk2/2−logψ(z). Fix z0 ∈ B withRez0 ≤ 0, Imz0 ≥ 1/e.
2 1 2
Sincbe u(λ) ≥ u(0) for Reλ ≤ 0, we have
||z0||/2 < ψ(z0) < 1−u(0) =: 8C.
b
Let ϕ ∈ O(D,D), ϕ(0) = z0. Following the estimates in the proof of
Example 3.5.10 in [3], we see that kϕ′(0)k < C. Hence, κ (z0;X) ≥
D
CkXk, X ∈ C2. Since k is the integrated form of κ , it follows that
D D
k (a,a−te ) ≥ Ct, a ∈ γ([0,1]), 0 ≤ t ≤ 1/2−1/e, e = (1,0).
D 1 1
Hence Dk (a;e ) ≥ C and therefore, L (γ) ≥ C/2 > 0, which
D 1 DkD
completes the proof of this example.
Note that it shows that, with respect to the lengths of curves, Dk
D
behaves different than the ”real” derivative of k (cf. [11] or [4], page
D
12). Moreover, it implies that, in general, Dk 6= Dk .
D D
Questions. It will be interesting to know examples showing that, in
˜
general,κ 6= Dk . ItremainsalsounclearwhetherDk isholomorphi-
D D D
cally contractible (see [3]). Recall that Dk = k ; but we do not
D D
R
know if Dk = k .
D D
R
3. Proofs
Proof of Proposition 2. First, we shall consider the case m = 1. The
key is the following
Theorem 3. ∗ [10] Let M be an n-dimensional complex manifold and
f ∈ O(D,M) regular at 0. Let r ∈ (0,1) and D = rD×Dn−1. Then
r
there exists F ∈ O(Dr,M), which is regular at 0 and F|rD×{0} = f.
˜
Since κ (z;0) = Dk (z;0) = 0, we may assume that X 6= 0. Let
M M
α > 0 and f ∈ O(D,M) be such that f(0) = z and αf (d/dζ) = X.
∗
Let r ∈ (0,1) and F as in Theorem 3. Since F is regular at 0, there
exist open neighborhoods U = U(z) ⊂ M and V = V(0) ⊂ D such
r
that F| : V → U is biholomorphic. Hence (U,ϕ) with ϕ = (F| )−1,
V V
is a chart near z. Note that ϕ (X) = αe , where e = (1,0,...,0).
∗ 1 1
If w and Y are sufficiently near z and αe , respectively, then
1
g(ζ) := F(ϕ(w)+ζY/α), ζ ∈ r2D,
∗We may replace Theorem 3 by the approach used in the proof of the upper
semicontinuity of k˜M.
6 NIKOLAINIKOLOVAND PETER PFLUG
belongs to O(r2D,M) with g(0) = w and g(tα) = ϕ−1(ϕ(w)+tY), t <
r2/α.Therefore, r2k∗ (w,ϕ−1(ϕ(w)+tY)) ≤ tα. Hencer2Dk˜ (z;X) ≤
M M
α. Letting r → e1 and α → κ (z;X) we get that Dk˜ (z;X) ≤
M M
κ (z;X).
M
Let now m ∈ N. By definition, κ(m)(z;·) is the largest function with
M
the following property:
For any X = m X one has that κ(m)(z;X) ≤ m κ (z;X ).
Pj=1 j M Pj=1 M j
(m) (m) (m)
To prove that κ ≥ Dk it suffices to check that Dk (z;·)
M M M
has the same property. Following the above notation and choosing
Y → ϕ X with m Y = Y, we set w = w and w = ϕ−1(ϕ(w)+
j ∗ j Pj=1 j 0 j
t j Y ). Since
Pk=1 j
m
(m) ˜
k (w,w ) ≤ k (w ,w ),
M q X M j−1 j
j=1
it follows by the case m = 1 that
m m
(m)
Dk (z;X) ≤ Dk (z;X ) ≤ κ (z;X ).
M X M j X M j
j=1 j=1
(2n−1)
Finally, let m = ∞ and n = dimM. Since κˆ = κ and k ≤
M M M
k(2n−1), we get that Dk ≤ κˆ using the case m = 2n−1. (cid:3)
M M M
Proof of Theorem 1. We may assume that X 6= 0. In virtue of Propo-
sition 2, we have to show that
(m) (m)
κ (z;X) ≤ Dk (z;X).
M M
For simplicity we assume that M is a domain in Cn.
(i) Fix a neighborhood U = U(z) ⋐ M. Applying the hyperbolicity
of M at z, there are a neighborhood V = V(z) ⊂ U and a δ ∈ (0,1)
such that, if h ∈ O(D,M) with h(0) ∈ V, then h(δD) ⊂ U. Hence, by
the Cauchy inequalities, ||h(k)(0)|| ≤ c/δk, k ∈ N.
Now choose sequences M ∋ w → z, C ∋ t → 0, and Cn ∋ Y → X
j ∗ j j
such that
k (w ,w +t Y )
M j j j j
→ Dk (z;X).
e |t | M
j e
There are holomorphic discs g ∈ O(D,M) and β ∈ (0,1) with g (0) =
j j j
w , g (β ) = w +t Y , and β ≤ k∗ (w ,w +t Y )+|t |/j. Note that
j j j j j j j M j j j j j
k∗ (w ,w +t Y ) ≤ c ||t Y || ≤ ce|t |.
M j j j j 1 j j 2 j
e Write
w +t Y = g (β ) = w +g′(0)β +h (β ).
j j j j j j j j j j
ON THE DERIVATIVES OF THE LEMPERT FUNCTIONS 7
Then
∞
k
||h (β )|| ≤ c βj ≤ c |β |2 ≤ c |t |2, j ≥ j .
j j X(cid:16) δ (cid:17) 3 j 4 j 0
k=2
Put Y = Y −h (β )/t . We have that g (0) = w and β g′(0)/t =
j j j j j j j j j j
Y → Xb. Therefore,
j
b
β k∗ (z ,w +t Y ) 1
κ (w ;Y ) ≤ j ≤ M j j j j + .
M j j |t | e |t | j
b j j
Hence with j → ∞, we get that κ (z;X) = κ (z;X) ≤ Dk (z;X).
M M M
(ii) The proof of the case m ∈ N is similar to the next oene and we
omit it. Now, we shall consider the case m = ∞.
Note first that our assumption implies thatM is hyperbolic atz and,
by the contrary,
∀ε > 0 ∃δ > 0 : ||w−z|| < δ,||Y −X|| < δ||X||
(1) ⇒ |κ (w;Y)−κ (z;X)| < εκ (z;X).
M M M
Moreover, the proof of (i) shows that
o(a,b)
˜
(2) k (a,b) ≥ κ (a;b−a+o(a,b)), where lim = 0.
M M
a,b→z ||a−b||
Choose now sequences M ∋ w → z, C ∋ t → 0, and Cn ∋ Y → X
j ∗ j j
such that
k (w ,w +t Y )
M j j j j
→ Dk (z;X).
M
|t |
j
There are points w = w ,...,w = w +t X in M such that
j,0 j j,mj j j j
mj 1
˜
(3) k (w ,w ) ≤ k (w ,w +t Y )+ .
X M j,k−1 j,k M j j j j j
k=1
Set w = w for k > m . Since
j,k j j
l
1 1
˜
k (w ,w ) ≤ k (w ,w ) ≤ k (w ,w +t Y )+ ≤ c |t |+ ,
M j j,l X M j,k−1 j,k M j j j j j 2 j j
j=1
then k (w ,w ) → 0 uniformly in l. Then the hyperbolicity of M at
M j j,l
z implies that w → z uniformly in l. Indeed, assuming the contrary
j,l
and passing to a subsequence, we may suppose that w 6∈ U for some
j,lj
U = U(z). Then
0 = lim k (w ,w ) ≥ liminf k˜ (z′,w) > 0,
M j j,l M
j→∞ z′→z,w∈M\U
a contradiction.
8 NIKOLAINIKOLOVAND PETER PFLUG
Fix now R > 1. Then (1) implies that
κ (z;w −w ) ≤ Rκ (w ;w −w +o(w ,w )), j ≥ j(R).
M j,k j,k−1 M j,k j,k j,k−1 j,k j,k−1
It follows by this inequality, (2) and (3) that
mj R
κ (z;w −w ) ≤ Rk (w ,w +t Yj)+ .
X M j,k j,k−1 M j j j j
k=1
Since κˆ (z;t Y ) is bounded by the first sum, we obtain that
M j j
k (w ,w +t Yj)+1/j
M j j j
κˆ (z;Y ) ≤ R .
M j
|t |
j
Note that κˆ (z;·) is a continuous function. Hence with j → ∞ and
M
R → 1, we get that κˆ (z;X) ≤ Dk (z;X). (cid:3)
M M
Remark. It follows by the above proofs and a standard diagonal
˜
process that κ (z;·) = Dk(z;·) if M is hyperbolic at z.
M
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Institute of Mathematics and Informatics, Bulgarian Academy of
Sciences, Acad. G. Bonchev 8, 1113 Sofia, Bulgaria
E-mail address: [email protected]
ON THE DERIVATIVES OF THE LEMPERT FUNCTIONS 9
Carl von Ossietzky Universita¨t Oldenburg, Institut fu¨r Mathe-
matik, Fakulta¨t V, Postfach 2503, D-26111 Oldenburg, Germany
E-mail address: [email protected]