the set of critical points of is discrete and so is the set of critical values of. Suppose that a 1 < b 1 < c < a 2 < b 2 and let c be the only critic

Transcription

1 1. The result DISCS IN STEIN MANIFOLDS Josip Globevnik Let be the open unit disc in C. In the paper we prove the following THEOREM Let M be a Stein manifold, dimm 2. Given a point p 2 M and a vector X tangent to M at p there is a proper holomorphic map f:! M such that f(0) = p and such that f 0 (0) = X for some > 0. This has been known in the special case when M is a bounded domain in C N with boundary of class C 2. It was proved in [FG] by using the fact that for such domains there are bounded strictly plurisubharmonic exhaustion functions without critical points near the boundary. 2. Outline of the proof We describe the idea of the proof in the special case when M is a pseudoconvex domain in C N. Let D C N be a pseudoconvex domain and let be a strictly plurisubharmonic exhaustion function for D. The idea is to start with an analytic disc f :! D such that f(0) = p; f 0 (0) = X for some > 0, and then to push f(); 2 b in directions that are approximately tangent to the level sets of while keeping f(0) and f 0 (0) xed, keeping f() essentially unchanged for all belonging to a xed compact subset of, and lowering (f()); 2, only a little. We perform this inductively. If we want to get a proper map in the limit we must, at each step, increase (f()); 2 b, for a certain amount and the sum of these amounts must equal +1. However, this amount depends on the lower bound of jgradj. In particular, if a < b and if has no critical value on [a; b] then there is a > 0 such that whenever a (f()) b ( 2 b), then, in the pushing process one can increase (f()) for at least for each 2 b. Thus, if has no critical values on (b; 1) one can go on and use the pushing process inductively to obtain a proper holomorphic map f:! D such that f(0) = p. In this way one can prove the theorem in the special case when M is a bounded pseudoconvex domain in C N with boundary of class C 2 [FG]. In the convex setting the idea of pushing boundaries of analytic discs in directions tangent to the level sets of was used earlier in [G1,G2,G3,G4]. In the present paper we consider the general case. Now we must show how to push f(); 2 b, through the critical levels of. We describe this when N = 2. First, we observe that in the pushing process described above the change is holomorphic on and for this one does not need to assume that f is holomorphic on. Perform such a pushing on f to get g. The change g? f is holomorphic on so if f is not holomorphic on but satises jf()? F ()j < ( 2 ) where F is continuous on and holomorphic on then g satises jg()? G()j < ( 2 ) where G (= F + (g? f)) is continuous on and holomorphic on. With no loss of generality assume that all critical points of are nondegenerate. Then 1

2 the set of critical points of is discrete and so is the set of critical values of. Suppose that a 1 < b 1 < c < a 2 < b 2 and let c be the only critical value of on [a 1 ; b 2 ]. Suppose that f:! D is continuous and holomorphic on and satises a 1 < (f()) < b 1 ( 2 b). Given, b 1 < < c, very close to c, one uses the pushing process above to get a continuous map f:! D, holomorphic on and such that < (f 1 ()) < c ( 2 b). Slightly changing f 1 we may assume that f 1 is smooth on b. Let a 2 < < < b 2. Let w 0 be a critical point of. It is known that if w? w 0 = (x 1 + iy 1 ; x 2 + iy 2 ) then, after a smooth change of coordinates near w 0, the function has the form (w 0 ) + x x 2 2 y 2 1 y 2 2, that is, the index of w 0 does not exceed 2. This, together with the fact that f 1 is smooth on b, enables us to push those f 1 (); 2 b, that are near the critical points of contained in fz 2 D: (z) = cg, through this level to get a continuous map f 2 : b! D with the property that either f 2 () is away from the critical points or (f 2 ()) > c. Away from the critical points we then push in the direction of grad to get a continuous map f 3 : b! D such that (f 3 ()) > c ( 2 b). In fact, we show that given > 0 there is a < c such that if < (f 1 ()) < c ( 2 b) then there is a continuous map f 3 :!D such that jf 3 ()? f 1 ()j < ( 2 ) and c < (f 3 ()) < a 2 ( 2 b). For our purpose we choose > 0 so small that if z 2 D then (z) and jz? wj < imply that w 2 D and (z) < b 2 and j(z)? (w)j <? a 2. The map f 3 is continuous on but is not holomorphic on. However, there is a map F (= f 1 ), continuous on, holomorphic on and such that jf 3 ()? F ()j < ( 2 ). We now perform the pushing process to get a continuous map f 4 :!D such that (f 4 ()) < ( 2 ), < (f 4 ()) < ( 2 b), and such that there is a continuous map g:!c 2, holomorphic on and such that jg()? f 4 ()j < ( 2 ). Our choice of now implies that g maps into D and satises a 2 < (g()) < b 2 ( 2 b). Thus, starting with a continuous map f:!d, holomorphic on and satisfying a 1 < (f()) < b 1 ( 2 b), after a detour through nonholomorphic maps we end up with a continuous map g:!d, holomorphic on and such that a 2 < (g()) < b 2 ( 2 b). Now one builds this into an inductive process to obtain a proper holomorphic map f:!d such that f(0) = p. The proof for general Stein manifolds is technically more complicated but uses the same idea. 3. Preliminaries By the embedding theorem for Stein manifolds [Ho] we may assume that M is a closed submanifold of C N for some N 2 IN. We rst show that it is enough to consider the case dimm = 2. Let p 2 M and let X 2 C N be tangent to M at p. If X = 0 then let V be a complex line through p. If X 6= 0 then let V be the complex line passing through p and p + X. One may assume that V is not contained in M. If it is, then, with no loss of generality, replace M with (M), where : C N!C N is a biholomorphic map such that (p) = p; (D)(p)(X) = X and such that (M) does not contain V. Let L(V ) be the family of all ane complex subspaces of C N of dimension N? 1 which contain V. By the transversality theorem, almost every 2 L(V ) meets M n V transversely. Further, since M does not contain V, M \ V is at most countable and so almost every 2 L(V ) is transverse to each of the tangent spaces 2

3 T p (M); p 2 M \ V. Thus, there is a 2 L(V ) that intersects M transversely. Obviously p 2 \ M and X is tangent to \ M at p. Thus, we may replace M by M \, a closed submanifold of of dimension dimm? 1. Repeating the process, we may, with no loss of generality, assume that dimm = 2. With no loss of generality assume that 0 62 M. The transversality theorem implies that after rotating M around p if necessary we may assume that the sphere fz 2 C N : jzj = jpjg intersects M transversely and so do all the spheres fz 2 C N : jz? aj = jp? ajg for suciently small a 2 C N. Sard's implies that for almost every a 2 C N the function z 7! jz? aj 2 is a Morse function on M. Thus, after translating M for a suitable small a we may assume that the function : M!IR dened by (z) = jzj 2 (z 2 M) is a Morse function and that (p) is not a critical value of. Since is a Morse function all its critical points are nondegenerate. In particular, the set of critical points of is discrete and so is the set of critical values of. By the theorem of Docquier and Grauert [GR, p. 257] there are an open neighbourhood of M in C N and a holomorphic map :!M such that (z) = z (z 2 M). 4. Small discs tangent to the level sets of Denote by IB the open unit ball in C N. For each q 2 C N n f0g let E(q) = fz 2 C N : hz? qjqi = 0g be the ane complex hyperplane passing through q and tangent to the sphere bjqjib, and for each q 2 M let T (q) be the ane complex subspace of dimension 2 passing through q and tangent to M at q. Assume that Q M is a compact set consisting of regular points of. For each q 2 Q, T (q) intersects E(q) transversely so E(q) \ T (q) = L(q) is a complex line and there is a > 0 such that (q + IB) \ E(q) \ M = (q) is a closed, one dimensional complex submanifold of q + IB, which is tangent to L(q) at q. Since Q is compact, a > 0 can be chosen that works for all q 2 Q. For each q 2 Q there is a ; 0 < <, such that for each r; 0 < r, (q) intersects b(q + rib) transversely, and the intersection (q + rib) \ (q) is biholomorphically equivalent to a disc. Again, by the compactness of Q, a > 0 can be chosen that works for all q 2 Q. Since E(q) is orthogonal to q it follows that the spheres in E(q) centered at q are the level sets of the function z 7! jzj 2 restricted to E(q). In particular, (z) = jqj 2 + r 2 = (q) + r 2 (z 2 (q) \ b(q + rib)): (4:1) By transversality everything varies smoothly if we vary q 2 M and r; 0 < r <, smoothly. Lemma 4.1 Given a compact set Q M of regular points of there is a 0 > 0 such that for every positive continuous function on b that satises () < 0 ( 2 b) there is a continuous map F : b!m such that (i) for each 2 b the function 7! F (; ) is holomorphic on (ii) F (; 0) = f() ( 2 b) (iii) (F (; ) > (f() ( 2 b; 2 n f0g) (iv) (F (; ) = (f()) + () ( 2 b; 2 b): Proof. Let be as in the preceding discussion and put 0 = 2. Let be a positive continuous function on b such that () < 0 ( 2 b) and let f: b!q be a continuous map. The preceding discussion shows that for each 2 b, b? f()+() 1=2 IB intersects (f()) 3

4 ? transversely and D() = (f()) \ f() + () 1=2 IB is biholomorphically? equivalent to a disc. If z belongs to the boundary of this disc, that is, if z 2 (f()) \ b f() + () 1=2 IB then (z) = (f()) + (). By the transversality and by the continuity of f and the discs D() change continuously with. Let a: b!bib be a continuous map such that for each 2 b, a() 2 T C (f()) = f()?f() + L(f()). For each 2 b, let be the biholomorphic map from to D() such that (0) = f() and that 0 (0) is a positive multiple of a(). The map is unique and by the well known properties of conformal maps [Po] extends continuously to. Obviously ( ()) = (f()) + () ( 2 b). Dene F (; ) = () ( 2 b; 2 ). By construction, F satises (i), (ii) and (iv) and (4.1) implies (iii). It remains to prove that F is continuous. To do this it is enough to prove that if 2 b then is arbitrarily close to, uniformly on, provided that is suciently close to. Passing to a smaller 0 at the beginning if necessary we may assume that there is an! > 0 such that for each 2 b, j? j <!, the orthogonal projection P onto L(f()) is one to one and regular on D(). Since D() vary continuously with it is enough to prove that the conformal map that maps P (D()) onto P (D()) that takes P (f()) to P (f()), and whose derivative takes P (a()) to a positive multiple of P (a()), is arbitrarily close to the identity, uniformly on, provided that is suciently close to. This follows from [Po, p. 26]. The proof is complete. 5. Pushing the boundaries of discs in tangent directions The following fact is essentially contained in [FG]. Lemma 5.1 Let G: b!c N be a continuous map such that for each 2 b, the map 7! G(; ) is holomorphic on and satises G(; 0) = 0. Given > 0 and r; 0 < r < 1, there is a polynomial P : C!C N such that (i) P () 2 G(; b) + IB ( 2 b) (ii) P (t) 2 G(; ) + IB ( 2 b; 0 t 1) (iii) jp ()j < (jj r) (iv) P (0) = 0 (v) P 0 (0) = 0. Proof. It suces to show that G can be approximated uniformly on b by maps of the form ~G(; ) = nx j=1 A j ()?m j where A j : C!C N are polynomials and m; n are positive integers. The poynomial P () = ~ G(; k ) = nx j=1 A j () kj?m then satises (i)-(iv) provided that the approximation of G by ~ G is suciently good and the integer k > m is chosen suciently large. 4

5 Approximate rst G(; ) on b uniformly by G(; r) with r < 1 suciently close to 1. We have G(; ) = 1 2i = 1 2i Z Z b b G(; )d? Z n G(; )d + n+1 1 n+1 2i b G(; )d n+1 (? ) : When jj r the last term is arbitrarily small provided that n is large enough. It follows that on b, G(; ) can be uniformly approximated by poynomials B 1 () + + B n () n with B j continuous on b. For each j; 1 j n, one can approximate, uniformly on b, B j () arbitrarily well by A j ()= m where A j is a polynomial and where m can be chosen the same for all j. This completes the proof. Lemma 5.2 Let t 1 < t 2 and assume that every t 2 [t 1 ; t 2 ] is a regular value of. There is a 0 > 0 with the following property: Suppose that f:!m is a continuous map, holomorphic in a neighbourhood of 0, such that t 1 (f()) t 2 ( 2 b) and such that jf()? H()j ( 2 ) where H:!C N is a continuous map, holomorphic on which satises H(0) = f(0) and H 0 (0) = f 0 (0). Given R; 0 < R < 1; " > 0 and a positive continuous function on b, () < 0 ( 2 b), there are r; R < r < 1, and a continuous map g:! M, holomorphic in a neighbourhood of 0, such that (i) j(g())? (f()) + () j < " ( 2 b) (ii) (g(t)) (f())? " (r t 1; 2 b) (iii) jg()? f()j < " (jj r) (iv) g(0) = f(0), (v) g 0 (0) = f 0 (0) and such that jg()? ()j < + " ( 2 ) where :!C N is a continuous map, holomorphic on which satises (0) = g(0) and 0 (0) = g 0 (0). If f is holomorphic on then g can be chosen to be holomorphic on. Proof. Let Q = fz 2 M: t 1 (z) t 2 g. The set Q is compact and consists of regular points of. Let 0 be as in Lemma 4.1. Suppose that f:!m is a continuous map, holomorphic in a neighbourhood of 0, such that f(b) Q and such that jf()? H()j ( 2 ) where H:!C N is a continuous map, holomorphic on and such that H(0) = f(0) and H 0 (0) = f 0 (0). Let " > 0 and 0 < R < 1 and let be a positive continuous function on b, () < 0 ( 2 b). By Lemma 4.1 there is a continuous map F : b!m such that (a) for each 2 b the map 7! F (; ) is holomorphic on (b) F (; 0) = f() ( 2 b) (c) (F (; )) (f()) ( 2 b; 2 ) (d) (F (; ) = (f()) + () ( 2 b; 2 b). Choose an open neighbourhood W M of F (b ) [ f() with compact closure. One can choose ; 0 < < ", so small that (z)? " (w) (z) + " whenever z 2 W; w 2 M; jw? zj < 2 (5:1) 5

6 Shrinking if necessary we may assume that j(w)? wj < (w 2 ): (5:2) Choose, 0 < <, so small that if z 2 F (b ) [ f(); w 2 C N ; jw? zj < 2 then w 2 ; (w) 2 W; j(w)? zj < : Choose r; R < r < 1, so close to 1 that ) (5:3) jf(t)? f()j < ( 2 b; r t 1): (5:4) Put G(; ) =?f() + F (; ) ( 2 b; 2 ). The map G satises the assumptions in Lemma 5.1 which provides a polynomial P : C!C N which satises (i)-(v) in Lemma 5.1. Put ~ f = f + P. We have (a') ~ f() 2 F (; b) + IB ( 2 b). By (5.4), 2 b and r t 1 imply that f(t) + P (t) 2 f() + P (t) + IB which, by (ii) in Lemma 5.1, implies that (b') ~ f(t) 2 F (; ) + 2IB ( 2 b; r t 1). Further, (iii)-(v) in Lemma 5.1 imply that (c') j f() ~? f()j < (jj r) (d') f(0) ~ = f(0) (e') f ~ 0 (0) = f 0 (0). Now, (a'), (b'), and (c') together with (5.3) imply that ~ f() and ( ~ f()) W. Dene g() = ( ~ f()) ( 2 ). It is clear that g is continuous on, holomorphic in a neighbourhood of 0 and maps to M. Let 2 b. By (a') there is an 2 b such that j ~ f()? F (; )j < so by (5.3), jg()? F (; ) <. Now (d) and (5.1) imply (i). Given t; r t 1, and 2 b, (b') implies that there is an s 2 such that j ~ f(t)? F (; s)j < 2 so by (5.3) it follows that jg(t)? F (; s)j <. Now (c) and (5.1) imply that (g(t)) (f())? " which proves (ii). Further, (c') and (5.3) imply that jg()? f()j < < " (jj r) which proves (iii). (iv) and (v) follow from (d'),(e') and the fact that jm = id. From the construction it is clear that g is holomorphic on provided that f is holomorphic on. Note that (5.2) implies that j? f() + P ()?? f() + P () j < ( 2 ) so jf()? H()j ( 2 ) implies that j? f() + P ()?? H() + P () j j? f() + P ()?? f() + P () j + jf()? H()j < + < " + ( 2 ) where = H + P is continuous on and holomorphic on and satises (0) = g(0) and 0 (0) = g 0 (0). This completes the proof. We will use the following consequence of Lemma 5.2. Lemma 5.3 Let t 1 < t 2 and assume that every t 2 [t 1 ; t 2 ] is a regular value of. Suppose that f:!m is a continuous map, holomorphic in a neighbourhood of 0, such that t 1 < (f() < t 2 ( 2 b) and such that jf()? H()j < ( 2 ) where H:!C N 6

7 is a continuous map, holomorphic on which satises H(0) = f(0) and H 0 (0) = f 0 (0). Given R; 0 < R < 1, and " > 0 there are r; R < r < 1, and a continuous map g:!m, holomorphic in a neighbourhood of 0, such that (i) t 2? " (g()) t 2 + " ( 2 b) (ii) (g(t) (f())? " ( 2 b; r t 1) (iii) jg()? f()j < " (jj r) (iv) g(0) = f(0) (v) g 0 (0) = f 0 (0), and such that jg()? G()j < + " ( 2 ) where G:!C N is a continuous map which is holomorphic on and satises G(0) = g(0) and G 0 (0) = g 0 (0). If f is holomorphic on then g can be chosen to be holomorphic on. Proof. Let 0 be the constant from Lemma 5.2. Choose n 2 IN so large that n 0 > t 2? t 1 and let () = [t 2? (f())]=n ( 2 ). Then is a positive continuous function on b which satises () < 0 ( 2 b). Choose > 0 so small that n < "=2 (5:5) and that j(x)? (z)j < "=2 whenever z 2 M; (z) < 2( + ") + (t 2 + ") 1=22 ; x 2 M; jx? zj < n; ) (5:6) that and that (f()) + n? 1 t2? (f() + (n? 1) < t 2 ( 2 b); (5:7) n () ( 2 b): (5:8) Let g 0 = f; G 0 = H and r 0 = r. Using Lemma 5.2 repeatedly with in place of " one constructs inductively r k ; 1 k n; R < r 1 < < r n < 1, continuous maps g k :!M, holomorphic in a neighbourhood of 0, and continous maps G k :!C N, holomorphic on, such that for each k; 1 k n, (gk ())? (g k?1 ()) + () < ( 2 b) (i') (ii') (g k (t)) (g k?1 ())? ( 2 b; r k t 1) (iii') jg k ()? g k?1 ()j < (jj r k ) (iv') g k (0) = g k?1 (0); g 0 k (0) = g0 k?1 (0) (v') jg k ()? G k ()j < + k ( 2 ) (vi') g k (0) = G k (0); g 0 k (0) = G0 k (0). Indeed, suppose that 0 m n? 1 and that we have already constructed r k ; g k ; G k for 0 k m. Now (i') for 1 k m implies that (gm ())? (f()) + m() < m ( 2 b), which, by (5.7), gives (g m ()) < (f()) + m() + m < t 2 ( 2 b) and which, by (5.8), gives (g m ()) > (f()) + m()? m > t 1 ( 2 b). Now we apply Lemma 5.2 to get r m+1 ; g m+1 and G m+1 with the required properties. 7

8 Put r = r 1 ; g = g n and G = G n. Notice that (i') implies that (gk ())? (f()) + k() < k ( 2 b; 1 k n): (5:9) In particular, for each k; 1 k n and for each 2 b we have jg k ()j 2 < jf()j 2 +n()+ n = t 2 +n which, by (v'), implies that jg k ()j < (t 2 +n) 1=2 + +n ( 2 b). Since G k is holomorphic on the maximum principle gives jg k ()j < (t 2 + n) 1=2 + + n ( 2 ) which, by (v') and by (5.5) implies that jg k ()j < 2( + n) + (t 2 + n) 1=2 < 2( + ") + (t 2 + ") 1=2 so (g k ()) < 2( + ") + (t 2 + ") 1=22 ( 2 ; 1 k n): (5:10) If 2 b then (5.9) implies that j(g())?t 2 j = (gn ())? (f())+n() < n < " which proves (i). (iv) and (v) follow from (iv'). If jj r then jj r k for all k; 1 k n, so (iii') implies that jg()? f()j < n < " which proves (iii). Further, (v') and (5.5) imply that jg()? G()j < + " ( 2 ). Moreover, (vi') implies that G(0) = g(0) and G 0 (0) = g 0 (0). To prove (ii), let 2 b and r t 1. If t r n then by (ii'), (g(t)) (f())? n > (f())? ". Let r k t r k+1 for some k; 1 k r k+1. Since t r k, (ii') implies that (g k (t)) (f())?k and by (5.5) it follows that (g k (t)) (f())?"=2. On the other hand, since t r k+1, (iii') implies that jg(t)? g k (t)j < (n? k) < n. By (5.10) and (5.6) it follows that j(g k (t))? (g(t))j < "=2 and the preceding discussion implies that (G(t)) > (f())? ". This proves (ii). By Lemma 5.2 it is clear that if f is holomorphic on then each g k ; 1 k n can be chosen holomorphic on. In particular, this holds for g = g n. This completes the proof. 6. Pushing to increase the values of near a critical point The following lemma shows how to push to higher levels of those points near a critical point that are not contained in a small exceptional set. Lemma 6.1 Let q be a critical point of. There are arbitrarily small neighbourhoods U V W of q in M and a set E V such that (A) Given " > 0 there are > 0 and a continuous map : W n E!W n E such that (i) j()? zj < " (z 2 W n E) (ii) j bw = id (iii) (z) 2 U implies that z 2 U (iv) ((z)) (z) (z 2 W n E) (v) ((z)) (z) + (z 2 U n E) (B) Given S, a closed subset of W of two dimensional Hausdor measure zero and " > 0 there is a continuous map : W!W such that (i') j(z)? zj < " (z 2 W ) (ii') j bw = id (iii') (S) misses E 8

9 Proof, Part 1. We look rst at the model case. Write points in IR 4 in the form (x 1 ; x 2 ; y 1 ; y 2 ) = (x; y) where x = (x 1 ; x 2 ); y = (y 1 ; y 2 ). Let D be the open unit disc in IR 2. Suppose that 0 < r < R < ~ R and let 0 < < R? r. Choose r 0 > r such that r 0 + < R and let ' be a smooth decreasing function on [0; 1) such that '(t) 1 (0 t r); '(t) 0 (t r 0 ) and dene h(x; y) = (x + '(jxj)'(jyj) x jxj ; y): Write A =? ~ RD 2 n? f0g RD. Then h: A!A is a continuous map such that (a) jh(x; y)? (x; y)j ((x; y) 2 A) (b) h j ( RD) ~ 2 n (RD) 2 = id (c) h(x; y) = (x + jxj x n [f0g RD]) (d) if h(x; y) 2 (rd) 2 then (x; y) 2 (rd) 2. Thus h keeps y xed and for jyj < r pushes a point (x; y); x 6= 0, away from the two dimensional subspace x 1 = x 2 = 0. The exceptional set where h is not continuous (and not even dened) is contained in f0g RD. Let a be a function of the form a(x; y) = x x 2 2 y 2 1 y 2 2. Applying h increases the values of a(x; y): (e) a? h(x; y) a(x; y) ((x; y) 2 A) (f) a? h(x; y) a(x; y) + 2 ((x; y) 2 (rd) 2 ; x 6= 0). Part 2. We now use the fact that the set f0g RD is small. Suppose that T is a closed subset of ( ~ RD)2 of two dimensional Hausdor measure zero. Then T \(RD) 2 is a compact set of two dimensional Hausdor measure zero whose image under the projection (x; y) 7! x is a compact subset of RD of two dimensional Hausdor measure zero which implies that arbitrarily close to 0 there is an x such that fxg RD misses T. Given > 0 choose such an x which satises jxj <. It is easy to construct a homeomorphism h:? ~ RD 2!? ~ RD 2 such that (a') jh(x; y)? (x; y)j < ((x; y) 2 ( ~ RD)2 ) (b') h j b( ~ RD) 2 = id and which maps fxg RD onto f0g RD. Clearly h(t ) misses f0g RD. Part 3. We now consider the general case. Let q be a critical point of. Since is strictly plurisubharmonic there are a neighbourhood W M of q, a neighbourhood P of 0 in C 2 and a biholomorphic map g: P!W; g(0) = q, such that ( g)(w)? (q) = (1+ 1 )x 2 1+(1+ 2 )x 2 2+(1? 1 )y 2 1 +(1? 2 )y 2 2 +o(jwj 2 ) (w 2 P ) where i 0; i = 1; 2, and where w = (x 1 + y 1 ; x 2 + y 2 ) [HL, p. 29]. Using the Morse lemma and passing to smaller P and W if necessary we may assume that g: P!W is a smooth change of coordinates such that g(0) = q and such that ( g)(w)? (q) = a(x; y) where a is as above. There is an ~ R > 0 such that ( ~ RD) 2 P. Replace P by ( ~ RD)2 and let W = g(p ). Let < 1 be a Lipschitz constant of g on ( ~ RD)2. Let " > 0. Choose r; R; 0 < r < R < ~ R, and choose, 0 < < R? r, so small that < ". Let U = g((rd) 2 ); V = g((rd) 2 ). Let E = g(f0g RD) and let h 9

10 and A be as in Part 1. Dene = g h g?1. The map is a continuous map from W n E to W n E. If z 2 W n E then g?1 (z) 2 A so by (a), jh(g?1 (z))? g?1 (z)j and consequently j(z)? zj = jg(h(g?1 (z)))? g(g?1 (z)j < ". So satises (i). Since g(bp ) = bw, (b) implies (ii). If (z) 2 U then h(g?1 (z)) 2 (rd) 2 so by (d), g?1 (z) 2 (rd) 2 and consequently z 2 g((rd) 2 ) = U which implies (iii). Let z 2 W n E. Then g?1 (z) 2 A so by (e), ((z))? (q) = (g(h(g?1 (z))))? (q) = a(h(g?1 (z))) a(g?1 (z)) = ( g)(g?1 (z))? (q) = (z)? (q) so satises (iv). Finally, if z 2 U then g?1 (z) 2 (rd) 2 so by (f), ((z))?(q) = a(h(g?1 (z))) a(g?1 (z))+ 2 = (z)?(q)+ 2 so satises (v) with = 2. This completes the proof of (A). Part 4. To prove (B), let g; and be as above. Let S be a closed subset of W of two dimensional Hausdor measure zero. Then T = g?1 (S) is a closed subset of ( RD)2 ~ of two dimensional Hausdor measure zero. By Part 2 there is a homeomorphism h: ( RD) ~ 2!( RD)2 ~ which satises (a') and (b') and is such that h(t ) misses f0g RD. Let = g h g?1. Then : W!W is a homeomorphism. Clearly (S) = g(h(t )) misses g(f0g RD) = E so satises (iii'). As in Part 3, (a') implies that satises (i'). By (b'), satises (ii'). This completes the proof. 7. Pushing to increase the values of away from critical points The following lemma shows how to push to higher levels of the points which are away from critical points. Lemma 7.1 Let a < c < b and assume that c is the only critical point of on [a; b]. Write P = fz 2 M: a (z) bg. Let z 1 ; z 2 ; ; z m be the critical points of contained in fz 2 M: (z) = cg. Let U j ; 1 j m; be open neighbourhoods of z j in M, with pairwise disjoint closures contained in fz 2 M: a < (z) < bg. For each j; 1 j m, let B j U j be an open neighbourhood of z j. Given " > 0 there are > 0 and a continuous map H: P!M such that (i) jh(z)? zj < " (z 2 P ) (ii) H j [ m j=1 B j = id (iii) (H(z)) (z) (z 2 P ) (iv) (H(z)) (z) + (z 2 P n [ m j=1 U j). Proof. The set Q = P n [ m j=1 B j is compact and consists of regular points of. Given q 2 M and r > 0 write D(q; r) = fz 2 T (q): jz? qj < rg. Further, let A(q) be the orthogonal complement of?q + T (q). A(q) is a complex subspace of C N of dimension N? 2. For each q 2 M there are an r > 0 and a holomorphic map ' q : D(q; r)!a(q) such that ' q (q) = (D' q )(q) = 0 and such that fz + ' q (z): z 2 D(q; r)g M. Since Q is compact there is an r > 0 that works for all q 2 Q and there is a constant < 1 such that for each q 2 Q, the function ' q and its rst and second derivatives are bounded by on D(q; r). For each q 2 Q consider the function s q : D(q; r)!ir dened by s q (z) = (z + ' q (z)). Passing to a smaller r and to a larger if necessary we may assume that 1= < j(grad s q )(z)j < (z 2 D(q; r)) and that the second derivative of s q is also bounded by on D(q; r); q 2 Q. Elementary analysis now shows that there is a constant t 0 > 0 10

11 such that for every q 2 Q and every t; 0 < t t 0, we have q + t(grads q )(q) 2 D(q; r) and s q? q + t(gradsq )(q)) s q (q) + t j(grad s q)(q)j 2 2 (7:1) Passing to a smaller t 0 if necessary we may assume that t 0 + < ". Let be a continuous real function on P such that 0 1, such that 0 on [ m j=1 B j and 1 on P n [ m j=1 U j. Dene the map H: P!M by H(q) = q + t 0 (q)(grad s q )(q) + ' q q + t0 (q)(grad s q )(q) (q 2 Q) and H(q) = q (q 2 [ m j=1 B j). Since all the quantities depend continuously on q and since 0 on [ m j=1 B j it follows that H: P!M is a continous map. Since j(grad s q )(q)j < and j' q j < and since t 0 + < ", (i) follows. By denition, H satises (ii). By (7.1), H satises (iii), and since 1 on P n [ m j=1 U j, (7.1) implies that (H(z)) > (z) (z 2 P n [ m j=1 U j) and since P n [ m j=1 U j is compact it follows that there is a > 0 such that (iv) holds. This completes the proof. 8. Pushing through a critical level in a nonholomorphic way Lemma 8.1 Let c > (p) be a critical value of and let " > 0. There is a < c with the following property: Let ':!M be a continuous map, holomorphic on and such that < ('()) < c ( 2 bd) and '(0) = p. There is a continuous map :!M, holomorphic in a neighbourhood of 0, such that j ()? '(z)j < " ( 2 ), such that (0) = p; 0 (0) = ' 0 (0) and such that ( ()) > c ( 2 b). Proof. We rst replace 'jb by a smooth map close to 'jb. Then we use Lemma 6.1 to push through the critical level those points '(); ( 2 b), that are close to the critical points. After that we use Lemma 7.1 to push through the critical level the points '(); ( 2 b), that are away from critical points. Finally, we extend the change continuously to all so that the extension vanishes identically in a neighbourhood of 0. We now pass to the details. Part 1. There are a; b such that a < c < b and such that c is the only critical point on [a; b]. Denote P = fz 2 M: a (z) bg. Let z 1 ; z 2 ; z m be the critical points of contained in fz 2 M: (z) = cg. By Lemma 6.1 there are open neighbourhoods W j of z j in M, with pairwise disjoint closures contained in fz 2 M: a < (z) < bg, and for each j; 1 j m, there are open neighbourhoods U j V j W j of z j and a set E j V j such that given " > 0 there are a j > 0 and a continuous map j : W j n E j!w j n E j such that (i) j j (z)? zj < " (z 2 W j n E j ) (ii) j j bw j = id (iii) j (z) 2 U j implies that z 2 U j (iv) ( j (z)) (z) (z 2 W j n E j ) (v) ( j (z)) (z) + j (z 2 U j n E j ). 11

12 Moreover, given a closed subset S j of W j of two dimensional Hausdor measure zero and > 0 there is a continuous map j W j!w j such that (i') j j(z)? zj < (z 2 W j ) (ii') j j bw j = id (iii') j(s j ) misses E j. We will show that given " > 0 there are > 0 and a continuous map F : P n [ m j=1e j!m such that jf (z)? zj < 2" (z 2 P n [ m j=1e j ) and such that (F (z)) (z) + (z 2 P n [ m j=1e j ): 9 >= >; (8:1) Let " > 0. By Lemma 7.1 there are 0 > 0 and a continuous map H: P!M such that (i") jh(z)? zj < " (z 2 P ) (ii") H j [ m j=1 B j = id (iii") (H(z)) (z) (z 2 P ) (iv") (H(z)) (z) + 0 (z 2 P n [ m j=1u j ). Let j and j ; 1 j m, be as above. Dene the map G: P n [ m j=1 E j!p by ( j (z) (z 2 W j n E j ; 1 j m) G(z) = z (z 2 P n [ m j=1w j ) By the properties of j the map G is continuous and satises jg(z)? zj < " (z 2 P n [ m j=1 E j). Consequently the map F = H G is continuous, and by (i") satises jf (z)? zj < 2" (z 2 P n [ m j=1 E j). Let = minf 0 ; 1 ; ; m g. Suppose that z 2 P n [ m j=1 E j. If z 2 U j n E j for some j; 1 j m, then by (v), (G(z)) (z) + so by (iii"), (F (z)) (G(z)) (z) +. If z 2 (W j n E j ) n U j then by (iv), (G(z)) (z), and by (iii), G(z) 2 W j n U j. Since G(z) 2 P n [ m j=1 U j, (iv") implies that (F (z) (G(z)) + (z) +. Finally, if z 2 P n [ m j=1 W j then G(z) = z. In particular, G(z) 2 P n [ m j=1 U j so by (iv 00 ); (F (z)) (z) +. This completes the proof of (8.1). Part 2. Let " > 0. Choose > 0 so small that if z 2 M; (z) c and w 2 C N ; jw? zj < 3 then w 2 and j(w)? zj < " (8:2) By (8.1) there are ; 0 < < c? b, and a continuous map F : P n [ m j=1 E j!m such that jf (z)? zj < (z 2 P n [ m j=1 E j) and such that (F (z)) (z) + 2 (z 2 P n [ m j=1 E j). Let = c? and suppose that ':!M is a continuous map, holomorphic on, '(0) = p, and such that < ('(z)) < c ( 2 b). Let ' = 'jb be the boundary map. If r < 1 is chosen suciently close to 1 then ' 1 : b!m dened by ' 1 () = '(r) ( 2 b) is a smooth map such that < (' 1 ()) < c ( 2 b) and such that j' 1 ()? ' ()j < ( 2 b). Now ' 1 (b) is a compact set of two dimensional Hausdor measure zero so for each j; 1 j m; S j = ' 1 (b) \ W j is a closed subset of W j of two dimensional 12

13 Hausdor measure zero. Given > 0 there are continuous maps j: W j!w j ; 1 j m) which satisfy (i'), (ii') and (iii'). Dene the map : M!M by j W j = j (1 j m); (z) = z (z 2 M n [ m j=1w j ): Provided that ; 0 < <, is small enough, ' 2 = ' 1 is a continuous map from b to M such that j' 2 ()? ' 1 ()j < ( 2 b), < (' 2 ()) < c ( 2 b) and ([ m j=1 E j) \ ' 2 (b) = ;. Since > b it follows that ' 2 (b) P n ([ m j=1 E j). Thus ' 3 = F ' 2 is a well dened continuous map from b to M such that j' 3 ()? ' 2 ()j < ( 2 b) and such that (' 3 ()) (' 2 ()) + 2 ( 2 b) > + 2 which implies that (' 3 ()) > c ( 2 b). Clearly j' 3 ()? ' ()j < 3 ( 2 b). Let!:!C N be a continuous extension of 7! ' 3 ()?' () from bd to which satises j!()j < 3 ( 2 ) and which vanishes identically in a neighbourhood of 0. Then ~' 3 = ' +!:!C N is a continuous map which on bd coincides with ' 3 and in a neighbourhood of 0 coincides with '. By construction, j ~' 3 ()? '()j < 3 ( 2 ) so by (8.2), ~' 3 (). Dene = ~' 3. Then is a continuous map from to M, holomorphic in a neighbourhood of 0, which satises (0) = p and 0 (0) = ' 0 (0). Further, (8.2) implies that j ()?'()j = j( ~' 3 ())? '()j < " ( 2 ). This completes the proof. 9. The induction step Lemma 9.1 Let (p) < a 1 < a 2 < b 1 < b 2 < 1. Assume that has at most one critical value on [a 1 ; b 2 ] and if c is such a value then a 2 < c < b 1. Suppose that f:!m is a continuous map, holomorphic on and such that a 1 < (f()) < a 2 ( 2 ). Given R; 0 < R < 1, and " > 0 there are r; R < r < 1, and a continuous map g:!m, holomorphic on and such that (i) b 1 < (g()) < b 2 ( 2 b) (ii) (g(t)) > (f())? " ( 2 b; r t 1) (iii) jg()? f()j < " (jj r) (iv) g(0) = f(0) (v) g 0 (0) = f 0 (0). Remark In particular, (ii) implies that (g()) > a 1? " (r jj 1). Proof. If has no critical value on [a 1 ; b 2 ] then the lemma follows from Lemma 5.3. Suppose that c; a 2 < c < b 1, is a critical value of. Let b 1 < < < < b 2. Let " > 0 and let 0 < R < 1. Choose > 0 so small that fz 2 M: (z) g + 2IB ; (9:1) and that ( 1=2 + 4) 2 < (9:2) x 2 M; (x) ; y 2 C N ; jy? xj < 2 implies that y 2 and j((y))? (x)j < minf? b 1 ; b 2? ; "=4g ) (9:3) 13

14 Passing to a smaller if necessary we may assume that < "=4. By Lemma 8.1 there is a ; a 2 < < c, such that if ':!M is a continuous map, holomorphic on and such that < ('()) < c ( 2 b) then there is a continuous map :!M; holomorphic in a neighbourhood of 0 such that j ()? '()j < ( 2 ); (0) = '(0); 0 (0) = ' 0 (0) and ( ()) > c ( 2 b) 9 >= >; (9:4) By Lemma 5.3 there are r; R < r < 1, and a continuous map f 1 :!M, holomorphic on, such that < (f 1 ()) < c ( 2 b); (9:5) (f 1 (t) (f())? ( 2 b; r t 1); (9:6) jf 1 ()? f()j < (jj r) (9:7) and f 1 (0) = f(0); f 0 1(0) = f 0 (0). By (9.4) there is a continuous map f 2 :!M, holomorphic in a neighbourhood of 0, such that (f 2 ()) > c; (9:8) jf 2 ()? f 1 ()j < ( 2 ) (9:9) and f 2 (0) = f 1 (0); f 0 2(0) = f 0 1(0). Note that (9.9), (9.5) and (9.3) imply that (f 2 ()) < ( 2 b). Further, (9.9) holds where f 1 is continuous on and holomorphic on. Now Lemma 5.3 gives an r 1 ; r < r 1 < 1, and a continuous map f 3 :!M, holomorphic in a neighbourhood of 0, such that < (f 3 ()) < ( 2 b) (9:10) (f 3 (t) (f 2 ())? ( 2 b; r 1 t 1) (9:11) jf 3 ()? f 2 ()j < (jj r 1 ) (9:12) and f 3 (0) = f 2 (0); f 0 3(0) = f 0 2(0), together with a continuous map F :!C N, holomorphic on, such that F (0) = f 3 (0); F 0 (0) = f 0 3(0) and such that jf ()? f 3 ()j < 2 ( 2 ): (9:13) By (9.10), jf 3 ()j < 1=2 ( 2 b) so by (9.13), jf ()j < 1=2 + 2 ( 2 b). Since F is holomorphic on the maximum principle implies that jf ()j < 1=2 + 2 ( 2 ) so by (9.13), jf 3 ()j < 1=2 + 4 ( 2 ). By (9.2) it follows that (f 3 ()) < ( 1=2 + 4) 2 < ( 2 ). Now (9.13) and (9.1) imply that F (). Dene g = F. The map g is a well dened continuous map from to M which is holomorphic on. Now (9.13) and (9.3) imply that j(g())? (f 3 ())j < min f? b 1 ; b 2? ; "=4g ( 2 ); (9:14) 14

15 which, by (9.10) implies (i). (iv) is obvious and (v) follows from the fact that F 0 (0) = f 0 (0) and that jm = id. Further, (9.7), (9.9) and (9.12) imply (iii) since < "=4. Since f 1 is holomorphic on, (9.5) implies that jf 1 ()j < c 1=2 ( 2 ) so (9.9) and (9.2) imply that jf 2 ()j 2 < (c 1=2 + ) 2 < ( 1=2 + 4) 2 < ( 2 ), that is, (f 2 ()) < ( 2 ): (9:15) Let 2 b and let r t 1. Assume rst that t r 1. By (9.14), (g(t)) (f 3 (t))? "=4. Since t r 1, (9.11) implies that (g(t)) (f 2 ())? 2"=4. By (9.15), (9.9) and (9.3), (f 2 ()) > (f 1 ())? "=4 so (g(t) (f 1 ())? 3:"=4 and (9.6) implies that (g(t)) (f())? ". Now let r t r 1. By (9.12) and (9.9), jf 3 (t))? f 1 ((t)j < 2. Since (f 1 ()) < ( 2 ), (9.6) and (9.3) imply that (f 3 (t)) (f())? "=2 and (9.14) implies that (g(t) (f())? 3"=4 > (f())? ". This proves (ii). The proof is complete. 10. The completion of the proof Let X be a vector tangent to M at p. Choose a j ; b j ; j 2 IN [ f0g, such that a 0 < (p) < b 0 < a 1 < b 1 < < a n < b n < a n+1 <, lim a n = +1, such that for each j 2 IN [ f0g there is no critical point of on [a j ; b j ], and such that for each j 2 N there is at most one critical point on (b j?1 ; a j ). Choose a decreasing sequence n > 0 such that for each n 2 IN z 2 M; (z) b n ; w 2 M; jz? wj < n imply that j(z)? (w)j < 1: (10:1) Choose > 0 so small that p+x 2 ( 2 ) and that a 0 < ((p+x)) < b 0 ( 2 ) and dene f 0 () = (p+x) ( 2 ). Then f 0 :!M is a continuous map, holomorphic on which satises f 0 (0) = p; f0(0) 0 = X. Using Lemma 9.1 one can construct inductively a sequence of continuous maps f n :!M, holomorphic on, and an increasing sequence r n of positive numbers, converging to 1, such that for each n 2 IN, (i) a n < (f n ()) < b n ( 2 b) (ii) (f n ()) > a n?1? 1 (r n jj 1) (iii) jf n ()? f n?1 ()j < n =2 n (jj r n ) (iv) f n (0) = p (v) f 0 n(0) = X. By (iii), f n converges uniformly on compacta in, so f = lim f n is holomorphic on. Since f n () M and since M is closed it follows that f() M. By (iv), f(0) = p and by (v), f 0 (0) = X. Suppose that n 2 IN and that r n jj r n+1. Since jj r n+1, (iii) gives jf()? f n () jf n+1 ()? f n ()j + jf n+2 ()? f n+1 ()j + n+1 =2 n+1 + n+2 =2 n+2 + n. Note that ( i) and the fact that f is holomorphic on imply that (f n ()) < b n ( 2 ) so the preceding discussion together with (10.1) implies that j(f()) a n?1? 2 (r n jj r n+1 ). Since lim n!1 = 1 and since lim n!1 a n = +1 it follows that f:!m is a proper map. This completes the proof. 15

16 Acknowledgement The author is indebted to Berit Stensones for a very stimulating discussion which was essential for obtaining the result presented here, and to Barbara Drinovec for pointing out an error in an earlier approach to construct discs in Stein manifolds. References [FG] F. Forstneric, J. Globevnik: Discs in pseudoconvex domains. Comment. Math. Helv. 67 (1992) [G1] J. Globevnik: Boundary interpolation and proper holomorphic maps from the disc to the ball. Math. Z. 198 (1988) [G2] J. Globevnik: Discs in the ball containing given discrete sets. Math. Ann. 281 (1988) [G3] J. Globevnik: Relative embeddings of discs into convex domains. Invent. Math. 98 (1989) [G4] J. Globevnik: On moduli of boundary values of holomorphic functions. Math. Z. 208 (1991) [GR] R. C. Gunning, H. Rossi: Analytic functions of several complex variables. Prentice-Hall, Englewood Clis, 1965 [Hi] M. W. Hirsch: Dierential topology. Springer-Verlag, New York - Heidelberg, 1976 [Ho] Lars Hormander: An introduction to complex analysis in several variables. Van Nostrand, Princeton, 1966 [HL] G. M. Henkin, J. Leiterer: Theory of functions on complex manifolds. Akademie-Verlag, Berlin 1984 [Mi] J. Milnor: Morse theory. Princeton University Press, Princeton 1963 [Po] Ch. Pommerenke: Boundary behaviour of conformal maps. Grindl. der Math. Wiss Springer-Verlag, Heidelberg-New York 1992 Institute of Mathematics, Physics and Mechanics University of Ljubljana, Ljubljana, Slovenia josip.globevnik@fmf.uni-lj.si 16