Jucha, Piotr

Acta Mathematica

English

Jagiellonian Univeristy Repository

UNIVERSITATIS IAGELLONICAE ACTA MATHEMATICA, FASCICULUS XL2002THE WU METRIC IN ELEMENTARY REINHARDT DOMAINSby Piotr JuchaAbstract. In the paper we give the explicit formula for the Wu metric inelementary Reinhardt domains.Introduction. A new holomorphically invariant Hermitian pseudometricwas introduced by H. Wu in paper [4]. We will call it the Wu pseudometric.Theorem 0 gives a short description of the pseudometric.First let us introduce some notions concerning invariant metrics. Let Bnbe the unit ball in Cn and ∆ := B1. Denote by β0 the normalized Poincare´–Bergman metric of Bn at the point 0, i.e.β0 =n∑j,k=1dzj ⊗ dz¯k.Define the Kobayashi–Royden pseudometric on a domain D ⊂ Cn at apoint x ∈ D, for a vector v ∈ Cn,κM (x; v) :=inf{t > 0| ∃φ : U →M holomorphic,U ⊃ ∆, φ(0) = x, tφ′(0) = v};and the integrated form of κD, for x, y ∈ D,(∫ κD)(x, y) := inf{LκD(α)|α is a piecewise C1-curve in D from x to y},where LκD(α) :=∫ 10 κD(α(t);α′(t)) dt is the κD-length of a curve α.A domain D ⊂ Cn is called hyperbolic, if for any x0 ∈ D there exist aneighborhood U ⊂ D of x0 and a positive real number C such that κD(x; v) ≥C‖v‖ for all x ∈ U and v ∈ Cn. We say thatD is complete if any (∫ κD)-Cauchysequence (xν)∞ν=1 ⊂ D converges to a point x0 ∈ D in the natural topology.84Theorem 0. ([4]) The Wu pseudometric h, which is an (upper semicontin-uous) semi-definite Hermitian metric, can be defined on every domain D ⊂ Cnand has the following properties:(a) For 0 ∈ Bn, hBn,0 = β0.(b) If the Kobayashi–Royden pseudometric κD vanishes identically on D,then so does hD.(c) If D is hyperbolic, then hD is a Hermitian metric.(d) If D is complete hyperbolic, then hD is a continuous Hermitian metric.(e) If G ⊂ Cm is a domain and f : D → G is a holomorphic mapping, thenhG,f(x)(f ′(x)(u), f ′(x)(v)) ≤√nhD,x(u, v), for x ∈ D,u, v ∈ Cn. If κDis already a Hermitian pseudometric, then hG,f(x)(f ′(x)(u), f ′(x)(v)) ≤hD,x(u, v).Due to the relation between the Wu and Kobayashi–Royden pseudometricsit was possible to find the effective formulas for the Wu metric. It was studiedby C. K. Cheung and Kang–Tae Kim in [1] and [2] on Thullen domains. W.Zwonek computed the formulas for the Kobayashi–Royden pseudometric onelementary Reinhardt domains (in [5]). Using them, we give here the formulafor the Wu metric (Theorem 7).1. The Wu pseudometric. The idea and the sketch of the constructionof the Wu pseudometric on complex manifolds was first given in [4] (see also[1]). We recall here the details of the construction on domains in Cn as wellas simple properties for the sake of completeness.For a domain D ⊂ Cn let us introduce the family of sets (Fx)x∈D, whereFx := {f : Bn → D holomorphic |f(0) = x, f ′(0) : Cn → Cn is an isomorphism}.For any f ∈ Fx we can consider the Hermitian inner product on Cn inducedby the normalized Bergman metric:f!β0(u, v) := β0(f ′(0)−1(u), f ′(0)−1(v)), u, v ∈ Cn.Let Ψx := {f!β0| f ∈ Fx}. Then the family {Ψx}x∈D is a biholomorphicinvariant of D.Let Q be the space of all positive semi-definite Hermitian inner productson Cn with the natural topology of the vector space of all sesquilinear formson Cn. Define the partial ordering  in Q:α β if ∀ v ∈ Cn : α(v, v) ≤ β(v, v),and also the set of lower bounds of Ψx:l(Ψx) := {α ∈ Q| ∀β ∈ Ψx : α β}.85We are going to pick up an element of this set which will be the Wupseudometric of the domain D at the point x.First, let us make some remarks:Lemma 1. The set l(Ψx) is a compact subset of Q.Proof. Fix any β ∈ Ψx. Let us define the set l(β) of lower bounds of β:l(β) := {α ∈ Q|α β}.It suffices to prove that the set l(β) is compact since l(Ψx) =⋂β∈Ψx l(β).Choose a basis {e1, . . . , en} of Cn such that the matrix [β(ej , ek)]j,k=1,...,n isdiagonal.Then, for any α ∈ l(β), we have α(ej , ej) ≤ β(ej , ej) and also |α(ej , ek)|2 ≤α(ej , ej)α(ek, ek) ≤ β(ej , ej)β(ek, ek). This implies the boundedness of l(β).The closedness is obvious.Let Kx be the Kobayashi indicatrix at a point x:Kx := {v ∈ Cn|κD(x; v) < 1}and for any α ∈ QBα := {v ∈ Cn|α(v, v) < 1}be the “α-ball”. Notice that Bα ⊃ Bβ if α β. Moreover, the following holds:Lemma 2.Kx =⋃α∈ΨxBαProof. Notice that, for a holomorphic mapping f ∈ Fx such thatf ′(0)(w) = v, we have:κD(x; v)2 = κD(f(0); f ′(0)(w))2 ≤ κBn(0;w)2 = β0(w,w) = f!β0(v, v).In consequence, for any α ∈ Ψx we have Kx ⊃ Bα.To prove the other inclusion, fix v ∈ Cn \ {0} such that κD(x; v) = δ < 1.Choose φ : U → D from the definition of κD(x; v) and t ∈ (δ, 1) such thattφ′(0) = v. We may assume that φ′1(0) 6= 0. Define the mapping f : Bn → Cn,f(z1, . . . , zn) := φ(z1) + (0, z2, . . . , zn). For  > 0 small enough we obtain f ∈Fx. Moreover, we can check that f ′(0)−1(v) = (t, 0, . . . , 0). Thus f!β0(v, v) =t2 < 1, which finishes the proof.LetVx :=⋂γ∈l(Ψx){v ∈ Cn| γ(v, v) = 0}and take the orthogonal supplement Ix of the space Vx in Cn:Ix ⊕ Vx = Cn.86By the Cauchy–Schwarz inequality, any γ ∈ l(Ψx) is of the formγ = γ1 ⊕ γ2, where γ1 = γ|Ix×Ix , γ2 = γ|Vx×Vx ≡ 0,i.e. γ(u1 + v1, u2 + v2) = γ(u1, u2) for u1, u2 ∈ Ix, v1, v2 ∈ Vx.Lemma 3.(a) The set⋃α∈Ψx Bα ∩ Ix is bounded in Ix.(b) span{v ∈ Cn|κD(x; v) = 0} ⊂ Vx.Proof. Suppose that the set⋃α∈Ψx Bα ∩ Ix is unbounded. Then thereexist sequences (xν)∞ν=1 ⊂ ∂Bn ∩ Ix and (αν)∞ν=1 ⊂ Ψx such that limν→∞ xν =x0 ∈ ∂Bn ∩ Ix and αν(xν , xν) < 1ν .Hence we have 0 ≤ γ(x0, x0) = limν→∞ γ(xν , xν) ≤ limν→∞ αν(xν , xν) = 0,for any γ ∈ l(Ψx). But this implies that x0 ∈ Vx which cannot hold.To prove the other part, it suffices to show that {v|κD(x; v) = 0} ⊂{v| γ(v, v) = 0}, for any γ ∈ l(Ψx). Fix v 6= 0 satisfying κD(x; v) = 0. Rea-soning in the same way as in the proof of the previous lemma, for arbitrarilysmall t > 0, we can find an element α ∈ Ψx such that α(v, v) < t . But γ  αfor any α ∈ Ψx, hence γ(v, v) = 0.Lemma 4. There exists the unique element hx of the set l(Ψx) such thatfor any choice of a basis {e1, . . . , em} of the space Ixdet[γ(ej , ek)]j,k=1,...,m ≤ det[hx(ej , ek)]j,k=1,...,m, for any γ ∈ l(Ψx).Proof. It is easy to see the existence of such an element (by Lemma 1) andindependence of the choice of the basis. Suppose that h1 6= h2 are two elementssatisfying the above condition. Let h0 := 12h1 +12h2. We have h0 ∈ l(Ψx).Let us fix a basis {e1, . . . , em} of the space Ix. Notice that there must bedet[hl(ej , ek)]j,k=1,...,m 6= 0 for l = 1, 2. Indeed, define the Hermitian metrichr(ej , ek) := 0 for j 6= k, hr(ej , ej) := r > 0 and put hr|Vx×Vx ≡ 0. By Lemma3(a), we can find r > 0 so small that⋃β∈Ψx Bβ ∩ Ix ⊂ Bhr ∩ Ix. Of course,hr ∈ l(Ψx) and det[hr(ej , ek)]j,k=1,...,m = rm > 0.Let A and B be the matrices of h1 and h2, respectively, in this basis.However, since BA−1 is Hermitian, we could choose the basis {e1, . . . , em} insuch a way that the matrix BA−1 is diagonal with strictly positive coefficientsλ1, . . . , λm on the diagonal.Hencedet[h0(ej , ek)] =12mdet(A+B) =12mdet((I +BA−1)A)=12m(λ1 + 1) . . . (λm + 1) detA≥√λ1 . . . λm detA =√det(BA−1) detA = detA.87The equality holds only if λj = 1, j = 1, . . . ,m but that is impossible whenA 6= B (in our case). Thus, we obtain det[h1(ej , ek)] = det[h2(ej , ek)] <det[h0(ej , ek)] which cannot hold.We call this unique element hx the Wu pseudometric of the domain D atthe point x. (We will also write hx = hD,x.)Lemma 5. If the Kobayashi–Royden pseudometric κD(x; ·) is Hermitian atthe point x ∈ D, then hD,x(v, v) = κD(x; v)2. In particular, this is the casewhen D ⊂ C.Proof. We will show that Kx = Bhx , because each Hermitian metric isdetermined by its unit ball.Let κ˜x ∈ Q be such that κ˜x(v, v) = κD(x; v)2. We get κ˜x ∈ l(Ψx) directlyfrom the proof of Lemma 2. We have κ˜x(v, v) = hx(v, v) = 0 for v ∈ Vx by thedefinition of Vx. So, we only need to show that Kx = Bhx in Ix.Choose any basis {e1, . . . , em} of Ix. By the change of coordinates, weobtain the general property:det[α(ej , ek)]j,k=1,...,m dL2m(Bα ∩ Ix) = CdL2m(Bm)= det[β(ej , ek)]j,k=1,...,m dL2m(Bβ ∩ Ix), α, β ∈ l(Ψx),where the constant C > 0 depends only on the choice of the basis and dL2m is2m-dimensional Lebesgue measure.Lemma 2 implies Kx ∩ Ix ⊂ Bhx ∩ Ix. Suppose that these unit balls arenot equal, i.e. L2m(Kx ∩ Ix) < L2m(Bhx ∩ Ix). Then, due to the aboveproperty, there must be det[hx(ej , ek)]j,k=1,...,m < det[κ˜x(ej , ek)]j,k=1,...,m butthat contradicts Lemma 4.Geometrically speaking, “hx-ball” is the smallest posible ellipsoid contain-ing the Kobayashi indicatrix.2. Elementary Reinhardt domains. Let us introduce some notationsconcerning elementary Reinhardt domains. For µ = (µ1, . . . , µn) ∈ Rn \ {0}defineDµ := {z ∈ Cn : |z1|µ1 . . . |zn|µn < 1, if µj < 0 then zj 6= 0}.We say that µ is of rational type if there exist t > 0 and ν ∈ Zn+ such thatµ = tν; otherwise µ is said to be of irrational type.Without loss of generality we may assume that µ1, . . . , µl<0, µl+1, . . . , µn>0, where l ∈ {0, . . . , n}.88We remind the formulas for the Kobayashi–Royden pseudometric κDµ inelementary Reinhardt domain Dµ (see [5]). Below, γ denotesγ(x; v) =|v|1− |x|2 , for x ∈ ∆, v ∈ Cand κ∆∗ (where ∆∗ := ∆ \ {0}) is given by the formula (see [3])κ∆∗(x; v) = −|v|2|x| log |x| , for x ∈ ∆∗, v ∈ C.Theorem 6. Let (x, v) ∈ Dµ × Cn and J :={j ∈ {1, . . . , n} : xj = 0}={j1, . . . , jk}. Define µ˜l+1 := min{µl+1, . . . , µn}.(i) Assume that µ is of rational type and l < n, thenκDµ(x; v) =γ((xµ)1µ˜l+1 ; (xµ)1µ˜l+11µ˜l+1n∑j=1µjvjxj)if J = Ø(|x1|µ1 . . . |vj1 |µj1 . . . |vjk |µjk . . . |xn|µn) 1µj1+···+µjk if J 6= Ø.(ii) Assume that µ is of irrational type and l < n, thenκDµ(x; v) =γ(( n∏j=1|xj |µj) 1µ˜l+1 ;( n∏j=1|xj |µj) 1µ˜l+11µ˜l+1n∑j=1µjvjxj)if J = Ø(|x1|µ1 . . . |vj1 |µj1 . . . |vjk |µjk . . . |xn|µn) 1µj1+···+µjk if J 6= Ø.(iii) Assume that µ is of rational type and l = n, thenκDµ(x; v) = κ∆∗(xµ;xµn∑j=1µjvjxj).(iv) Assume that µ is of irrational type and l = n, thenκDµ(x; v) = κ∆∗(|x1|µ1 . . . |xn|µn ; |x1|µ1 . . . |xn|µnn∑j=1µjvjxj).Now, we state the main result:Theorem 7. Let (x, v) ∈ Dµ × Cn and J :={j ∈ {1, . . . , n} : xj = 0}.ThenhDµ,x(v, v) ={κDµ(x; v)2 if #J ≤ 10 if #J ≥ 2,where #J denotes the cardinality of J .89Proof. Notice that whenever #J ≤ 1, the Kobayashi–Royden pseudo-metric is Hermitian (both γ and κ∆∗ are Hermitian) and use Lemma 5. Other-wise, the linear span of the set of zeros of the Kobayashi–Royden pseudometriccoincides with Cn. Thus, by Lemma 3(b), the Wu pseudometric is zero.The Wu metric hD is not necessarily continuous with respect to the variablex ∈ D which is illustrated by the following example.Example. For the domain D(2,1) = {(x1, x2) ∈ C2| |x1|2|x2| < 1}, we havehD(2,1),(x1,x2)(1, 1) =|2x1x2 + x21|2(1− |x21x2|2)2 , if x1, x2 6= 0,while hD(2,1),(0,x2)(1, 1) = |x2|2 6= limx1→0 hD(2,1),(x1,x2)(1, 1) = 0, if x2 6= 0.However, the following is true:Theorem 8. The Wu metric (x, u, v) 7→ hDµ,x(u, v) is real analytic in thedomain D˜µ × C2n, where D˜µ := Dµ \ {(x1, . . . , xn) ∈ Cn|x1 . . . xn = 0}.Proof. The formulas in Theorem 6 and 7 imply the fact directly.References1. Cheung C. K., Kim Kang–Tae, Analysis of the Wu metric I: The case of convex Thullendomains, Trans. Amer. Math. Soc., 328 (1996), 1429–1457.2. Cheung C. K., Kim Kang–Tae, Analysis of the Wu metric II: The case of non-convexThullen domains, Proc. Amer. Math. Soc., 125 (1997), 1131–1142.3. Jarnicki M., Pflug P., Invariant Distances and Metrics in Complex Analysis, de Gruyter,1993.4. Wu H., Old and new invariant metrics, Several complex variables: Proc. of Mittag–LeﬄerInst. 1987–88, (J. E. Fornaess ed.) Math. Notes, Princeton Univ. Press, 38 (1993), 640–682.5. Zwonek W., Completeness, Reinhardt domains and the method of complex geodesics in thetheory of invariant functions, Dissertationes Math., 388 (CCCLXXXVIII), 2000.Received April 11, 2002Jagiellonian UniversityInstitute of MathematicsReymonta 430–059 Krako´w, Polande-mail : jucha@im.uj.edu.pl

The Wu metric in elementary Reinhardt domains

https://ruj.uj.edu.pl/xmlui/bitstream/handle/item/150020/jucha_the_wu_metric_in_elementary_reinhardt_2002.pdf?sequence=1&isAllowed=y

The Wu metric in elementary Reinhardt domains

Abstract

Similar works

Full text

Available Versions

Jagiellonian Univeristy Repository