LINES ON BRIESKORN-PHAM SURFACES

Transcription

1 LIN ON BRIKORN-PHAM URFAC GUANGFNG JIANG, MUTUO OKA, DUC TAI PHO, AND DIRK IRMA Abstact By usng toc modfcatons and a esult of Gonzalez-pnbeg and Lejeune- Jalabet, we answe the followng questons completely On whch Beskon-Pham suface thee exst smooth cuves passng though the sngula pont? If thee exst, how many and what ae the defnng equatons? Intoducton Let (X, 0) be an analytc space gem embedded n (C n, 0), and (L, 0) be a smooth cuve gem n (C n, 0) nce L s locally bholomophc to a lne, we often say that L s a lne If fo a sngula pont O X sng, thee exsts a lne (smooth cuve) L n C n such that O L and L \ {O} X eg, we say that X contans (o has ) a lne passng though O On a sngula suface X n C 3 one can not always fnd a smooth cuve passng though (not contaned n) the sngula locus of X Gonzalez-pnbeg and Lejeune-Jalabet poved a cteon fo the exstence of smooth cuve on any (two dmensonal) suface We quote ths esult hee fo the convenence of the eade Let π : X (X, 0) be the mnmal esoluton of a sngula suface (X, 0) Let L be the set of lnes on (X, 0) Fo an exceptonal dvso, let L denote the set of lnes on (X, 0) wth the stct tansfom ntesectng tansvesally Let m be the maxmal deal of the local ng O X,0 The cycle Z X on X defned by the deal sheaf mo X s called the maxmal cycle of π Gonzalez-pnbeg and Lejeune-Jalabet Theoem ([, 2]) () Thee exsts lnes on (X, 0) f and only f the maxmal cycle Z X has at least one educed component; (2) The set of all the lnes on (X, 0) s a dsjont unon of the L s wth each a educed component n Z X And L s called the famly of lnes coespondng to In [3, 4], lnes on sufaces wth smple sngulates ae classfed by usng the classfcaton machney All the sufaces wth smple o smple ellptc sngulates passng though x-axs ae equvalent to (unde the coodnate tansfomaton pesevng the x-axs) some sufaces defned by explct equatons Remak that by the afoementoned theoem many sngula sufaces have lnes Among the examples ae mnmal sufaces, sandwched sufaces and so on In ths pape we ae nteested n the followng questons on the so-called Beskon-Pham type sufaces, denoted by G(p, q, ), and defned n C 3 by x p + y q + z = 0 99 Mathematcs ubject Classfcaton 4J7, 3225, 3245 Key wods and phases lne, Beskon-Pham suface, toc modfcaton, toson numbe The fst autho was suppoted by JP: P98028

2 2 GUANGFNG JIANG, MUTUO OKA, DUC TAI PHO, AND DIRK IRMA Fo what knd of (p, q, ), thee ae lnes on G(p, q, )? And fo what knd of (p, q, ), thee ae no lnes on G(p, q, )? Futhemoe, f thee ae lnes, how many? What ae the defnng equatons fo the lnes? We gve complete answes to these questons The method we use hee s the toc modfcatons of suface sngulates We efe the eade to the book [6] fo notons and notatons Thee s an nvaant, called the toson numbe [3], attached to each lne Ths toson numbe s defned as follows Fo a pa of analytc space gems (Σ, 0) (X, 0), embedded n (C n, 0), defned by the deals g h of O := O n,0 espectvely Denote O X := O/h, O Σ := O/g The mage deal of g n O X s denoted by ḡ The O Σ -module M := ḡ/ḡ 2 g/(g 2 + h) s called the conomal module of ḡ If the toson submodule T(M) of M has fnte length, then the length s called the toson numbe of (Σ, X) and s denoted by λ := λ(σ, X) Fo two nteges m > n, let m/n := [m, m 2,, m k ] be the contnued facton epesentaton of m/n Defne Φ(m/n) := max{ m = = m = 2, k} Denote the numbe of famles of lnes on G(p, q, ) by (p, q, ) := {L } By the above mentoned theoem, (p, q, ) = { educed components n Z X } 2 The Newton bounday and the dual Newton dagam Let s := gcd(p, q, ) and p = gcd(q, )/s, q = gcd(, p)/s, = gcd(p, q)/s Let p, q, be the nteges such that p/s =, q/s = p q, /s = p q Then p, q, ae pawsely copme The Newton bounday of G(p, q, ) s a tangle wth vetces A(p, 0, 0), B(0, q, 0), C(0, 0, ) on the plane x/p + y/q + z/ = The dual Newton dagam Γ of G(p, q, ) conssts of = T (, 0, 0), 2 = T (0,, 0), 3 = T (0, 0, ) and P = T (p q,, p q ), whee and n the followng we use T to ndcate that the vecto s a column vecto Accodng to [6, III (63)], n ode to obtan a canoncal toc esoluton of the suface G(p, q, ), we need to make a canoncal egula smplcal subdvson Σ of Γ, whch we descbe oughly as follows (see [6, 7] fo detals) Fo two vectos Q and Q 2 n Z 3, denote by det(q, Q 2 ) the geatest common dvso of the absolute values of the 2 2 mnos of the 3 2 matx (Q, Q 2 ) Obvously, det(, P) = p If p >, P s not egula and we need to subdvde t By [6, II(2)], thee exsts a unque ntege p > p > 0 such that the vecto R := (P + p )/ p s ntegal Then det(r, ) = and det(r, P) = p < p If p >, epeat the pocess fo R P, untl we obtan a egula subdvson of P In ths way, we obtan the canoncal egula smplcal subdvsons of P, 2 P, 3 P Let R = T (,, 2,, 3, ) ( γ) be the ponts beng added to P (see Fgue ) mlaly one adds j = T (s,j, s 2,j, s 3,j ) ( j σ) to 2 P and T k = T (t,k, t 2,k, t 3,k ) ( k τ) to 3 P Then Σ s a egula smplcal subdvson of Γ such that ts estcton to P ( =, 2, 3) s the canoncal egula smplcal subdvson of P descbed above The dual esoluton gaph G of G(p, q, ) s sta-shaped wth the cente P, the exceptonal dvso detemned by P In the followng we use the captal lettes to denote the ponts on Γ as well as the exceptonal dvsos coespondng to them by abusng the symbol The ams n G consst of sp = gcd(q, ) copes of PR, sq = gcd(, p) copes of P and s = gcd(p, q) copes of PT (loc ct)

3 LIN ON BRIKORN-PHAM URFAC 3 3 C T A B R R γ T τ P σ 2 Fgue Newton bounday and dual Newton dagam of Beskon-Pham suface Lemma Wth the notatons as above, we have l, l,+ θ l, s l,j s l,j+ θ l, t l,k t l,k+ θ l, l =, 2, 3, whee θ := p q, θ 2 :=, θ 3 := p q ae the coodnates of P Poof We pove the lemma fo the R s Let π be the toc modfcaton assocated wth the above egula smplcal cone subdvson Σ Let (π x) = +, (R )+p q (P)+D, whee D s the sum of the dvsos whch do not ntesect wth (R ) s Note that the exceptonal dvso (R ) s a dsjont unon of (R (α) ) ( α sp ) ach (R (α) ) possesses the followng popetes: ) (R (α) ) s somophc to P, 2) the self ntesecton numbe of (R (α) ) s not bgge than 2 by [7] and 3) (R (α) ) only ntesects wth (R (α) and (R (α) + ) (hee R(α) 0 := (α), and R (α) γ+ := P), and (R(α) ) (R (β) + ) = δ αβ ([6] P56) Hence fom (π x) (R (α) ) = +,(R (α) )2 +,2 = 0, we have (R (α) )2 = (+,2 )/, 2 whch mples,,2 Fom (π y) (R (α) ) = 0 + 2,(R (α) )2 + 2,2 = 0, we have 2,2 2, mlaly, we can pove the othe nequaltes 3 The exstence of Lnes Theoem 2 Let s = gcd(p, q, ) G(p, q, ) has lnes f and only f G(p/s, q/s, /s) has lnes Poof It follows fom the fact that the two sufaces have the same dual Newton dagam Theoem 3 Assume gcd(p, q, ) = and ( ) : p q G(p, q, ) has lnes f and only f at least one of the followng condtons holds I) Two of the thee nteges p, q, ae copme and the othe one s dvsble by at least one of the copme numbes II) The nequalty > /gcd(p, q) holds )

4 4 GUANGFNG JIANG, MUTUO OKA, DUC TAI PHO, AND DIRK IRMA Poof We contnue to use the notatons n the begnnng of 2 By the assumpton ( ), we have the nequaltes ( ) : p q p q We fst show the necessty (a) Assume that thee exsts n the coodnates of P Then by ( ) we have p q = whch mples that gcd(p, q) = and Ths coesponds to a specal case of (I) (b) Let det(p, ) = p > The coodnates of R ae, = (p q + p )/ p q + p / p > by ( ), 2, = q q = 3, We assume that 3, = Ths mples that = q = Then we have p =, q = p, = p, fom whch we have gcd(p, q) = and q Ths coesponds to a case of (I) (c) In case det(p, 2 ) = q >, the coodnates of ae s 2, = ( + q )/ q >, s, = p s 3, = p Assume that s 3, = p = Then p and gcd(p, q) = Ths s also a case of (I) (d) If det(p, 3 ) = >, the coodnates of T ae t, = p q t 2, = and t 3, = ( p q + )/ In ths case, we have only two possbltes (d-) Fstly, assume that t 2, = whch s equvalent to p q and gcd(p, ) = Ths s a case of (I) nce gcd(p, q, ) =, ths s also a case of (II) f p > (d-2) econdly, assume that t 3, = Ths mples = p q + fo some > > 0, whch mples that > p q Ths mples that > /gcd(p, q), whch coesponds to (II) Note that all the coodnates of R ( > n case p > ) and that of j (j > n case q > ) ae bgge than one Also the fst two coodnates of T k (k > n case > ) ae bgge than one By lemma, t 3,k = mples that t 3,α = fo all α =,, k Hence > / gcd(p, q) We show the suffcency Assume fst that (II): > /gcd(p, q) Then > p q and = p q and we get t 3, = Next, assume that (I) s satsfed Then by ( ), ethe () gcd(p, q) = and p o q o (2) gcd(p, ) = and p q () coesponds to a case of (a), (b), o (c) and (2) coesponds to a case of (a) o (d) Fo example assume that gcd(p, q) = and p Ths mples that = and p q Thus p = and ths coesponds to (c) (espectvely to (a)) f q > (esp f q = ) The othe cases ae teated smlaly Remak 4 The case that p = s a specal case of (I) and (II) 4 quatons, numbes and toson numbes of the lnes We contnue to use the notatons n the begnnng of 2 and to assume p q Gonzalez- pnbeg and Lejeune-Jalabet theoem also gves a way to fnd the defnng equatons of the lnes (as cuves n C 3 ) on a suface The computaton of the toson numbe can be done as follows Fo a lne L wth paametezaton x = x(t), y = y(t), z = z(t), one may assume that dx dt (0) 0 By the nvese functon theoem, we have t = t(x) Hence the lne L can be defned by y φ(x) = 0 and z ψ(x) = 0 We choose the followng local coodnate tansfomaton of (C 3, 0): x = x, y = y φ(x), z = z ψ(x) The defnng deal of L s thus g = (y, z ) The defnng equaton of the suface contanng L can be wtten as h ax y + bx z mod g 2 Then λ = dm O/(x (a, b) + g) (see [4])

5 LIN ON BRIKORN-PHAM URFAC 5 Lemma 5 (p, p, p) = and the lne can be paametezed by and the toson numbe λ = p x = bt, y = ct, z = t, wth b p + c p + = 0, Poof The poof s tval and s omtted Lemma 6 ) If p >, then (p,, ) = p( ) + Let T 0 := 3, T := P, and T,, T be the ponts on 3 P to make the canoncal egula smplcal subdvson of 3 P Then to each T k coespond p educed components T () k maxmal cycle of G(p,, ) The toson numbes fo the lnes n L () T k the same: λ = k(p )q; ( > k > 0) n the (k fxed) ae 2) If p > and q, then (p,, p) = p( + q ) Let T 0 := 3, and T,, T the q be ponts on 3 P to make the canoncal egula smplcal subdvson of 3 P Then to each T k = T (kq, k, ) coespond p educed components T () k (k > 0) n the maxmal cycle of G(p,, p) The toson numbes of the lnes n L () T fo each fxed k (k =,, q ) k ae the same λ = k(p )q Let be on P 2 (see Fgue ) wth as the last coodnate Thee ae p educed components () ( =,, p) n the maxmal cycle { coespondng to ( The )} toson ae the same: λ = mn (p ), ( ) + q numbes of the lnes n L () Poof We leave the easy poof to the eade In the followng, we assume that gcd(p, q, ) = and < p q The exact numbes of famles of lnes ae gven n Table The poof s n lemma 7 Lemma 7 Let < p < q = and gcd(p, q) = Thee ae q educed components R () n the maxmal cycle coespondng to R (see fgue ) And the q famles of lnes on G(p, q, q) can be paametezed by x = ct + q p, y = t, z = ω β ( + c p t p δ ) /q t, β =,, q, [ whee c 0 s a constant, ω β s a oot of ω q =, and δ = q p q p { ( )} of these lnes ae the same λ = mn q, (p ) + q p Poof We omt the easy poof ] The toson numbes Lemma 8 Let gcd(p, q) = and The maxmal cycle s educed and conssts of T k+ = T ((k +)q, (k +)p, ) (k = 0,, = / ) The lnes n L Tk+ can be paametezed by x = c kq u (+αq+k)/p t (k+)q, y = c kp u α+kp t (k+)p, z = cut, whee 0 α < p such that ( + αq + k)/p s an ntege, c 0 s a constant and u = u(t) s a unt satsfyng + u + u (α+pk)q v (k+) w k = 0 The toson numbe fo the lnes n L Tk+ s λ = (k + )(p )q

6 6 GUANGFNG JIANG, MUTUO OKA, DUC TAI PHO, AND DIRK IRMA Condtons (p, q, ) = q q gcd(p, q) = > q, = + Φ( ) < p < q > q, p, q + q = + Φ( ) + q > q, p, q [ ] + p = + Φ( ) + p > gcd(p,q) < p q gcd(p, q, ) = gcd(p, q) =, p, q o gcd(p, q) > gcd(p, q) gcd(p,q) { ( = gcd(p, q) + Φ /gcd(p,q) )} Table Numbe of lnes on Beskon-Pham sufaces Poof In ths case P = T ( q, p, ) and det(p, ) = det(p, 2 ) =, det(p, 3 ) = Let 0 α, β < p such that M k = p ( + αt k+ + βt k ) = T( ) + αq + k, α + pk, p s ntegal Fom the chat detemned by M k, T k+, T k, one can do all the computatons Lemma 9 Let gcd(p, q) = and p, q In the maxmal cycle thee ae q educed components R () ( q) coespondng to R = T (( + δ)/p, /q, ), whee 0 < δ < p such that ( + δ)/p s an ntege The lnes n L () R can be paametezed as x = ct (+δ)/p, y = ω ( + c q t δ) /q t /q, z = t, whee c 0 s a constant { and ω([ s] a oot ) of ω q =, } =,, q The toson numbes ae the same: λ = mn (p ) p +, (q )/q Moeove, f >, n the maxmal cycle thee ae also educed components coespondng to T k+ = T ((k + )q, (k + )p, ) (k = 0,, ) The paametezatons and the toson numbes of the lnes n L Tk+ ae the same as n lemma 8 Poof One only needs to pove the statement about L R (), whch can be done fom the chat detemned by, 2, R

7 LIN ON BRIKORN-PHAM URFAC 7 Lemma 0 Let gcd(p, q) = and p, q In the maxmal cycle, thee ae p educed components (j) ( j p) coespondng to = T (/p, ( + δ)/q, ), whee 0 < δ < q such that s ntegal The lnes n L (j) can be paametezed by x = ω j ( + c q t δ ) /p t /p, y = ct (+δ)/q, z = t, whee c 0 s a constant { and ([ ω j ] s a ) oot of ω p =, } j =,, p The toson numbes [ ae the same: λ = mn (q ) q +, (p )/p Moeove, f >, thee ae also ] educed components n the maxmal cycle coespondng to T k+ = T ((k + )q, (k + )p, ) (k = 0,, ) The paametezatons and the toson numbes of the lnes n L Tk+ ae the same as n lemma 8 Poof mla to that of lemma 9 Lemma Let gcd(p, q, ) =, < p q and > / gcd(p, q) Assume that ethe gcd(p, q) =, p, q o gcd(p, q) > Then to each T k+ = T( ) (k + )q (k + )p gcd(p, q), gcd(p, q) gcd(p, q),, (k = 0,, ) coespond gcd(p, q) educed components T (γ) k+ (γ =,, gcd(p, q)) n the maxmal cycle And they ae the only educed components n the maxmal cycle Fo fxed k, the lnes n L T (γ) k+ can be paametezed by x = c k q u (+(k p+α) q)/ p γ t (k+) q, y = c k p u k p+α γ t (k+) p, z = cu γ t, γ =,,, whee u γ s a unt satsfyng + u + c k u (k p+α)q t (k+) = 0, and = gcd(p, q), p = p/, q = q/, 0 α < p such that ( + (k p+α) q)/ p s an ntege And the toson numbes of the lnes n L (γ) T ae: λ = (k + )(p )q/ gcd(p, q) k+ Poof Notce that all the coodnates [ of R and ] j ae geate than and T k+ s as n the statement of the lemma (k = 0,, gcd(p,q) ) Denote T 0 := 3 Let 0 α, β < p be the nteges such that M k := p ( + αt k+ + βt k ) = T( ) + (k p + α) q, k p + α, p s ntegal Then the smplex detemned by M k, T k+, T k s egula Fom ths chat, one can obtan the gven paametezaton of the lnes n L (γ) T by a calculaton k+ Coollay 2 If any two of the p, q, ae copme and q = p+α, = q + β wth 0 < α < p, 0 < β < q, thee do not exst smooth cuves on G(p, q, ) Poof We gve an elementay poof Let C be a smooth cuve n (C 3, 0), paametezed by x = x(t) = a k t k, y = y(t) = b k t k, z = z(t) = c k t k, t < ǫ k= Then C G(p, q, ) f and only f k= k= [x(t)] p + [y(t)] q + [z(t)] = 0 fo all t < ǫ Fom ths and the assumpton on p, q,, one can deduce that a = b = c = 0

8 8 GUANGFNG JIANG, MUTUO OKA, DUC TAI PHO, AND DIRK IRMA xample 3 ) On 8 suface G(2, 3, 5) thee do not exst lnes 2) On the sufaces G(2, 3, 7), thee exsts one famly of lnes A egula subdvson of the dual Newton dagam s gven n Fgue 2 Whee P = T (2, 4, 6), Q = T (, 7, 3), P Q R T 2 Fgue 2 R = T (7, 5, 2), = T (3, 2, ), = T (,, ), 2 = T (2, 2, ), 3 = T (2,, 2), 4 = T (2,, ), T = T (4, 3, ) Only the exceptonal dvsos coespondng to P, Q, R, have non-empty ntesectons wth the stct tansfom of the suface G(2, 3, 7) On the chat coespondng to the tangle σ = QP, we have the modfcaton map epesented by x = u v 2 w 3, y = u 7 v 4 w 2, z = u 3 v 6 w, whee (u, v, w) s the coodnates on the chat coespondng to σ The stct tansfom of G(2, 3, 7) s defned by u + w + = 0 The exceptonal dvso coespondng to s defned by u + = 0, w = 0 The blowng down of the lne v = c 0, u + w + = 0 s the smooth cuve n (C 3, 0): x = c 2 ( + t) t 3, y = c 4 ( + t) 7 t 2, z = c 6 ( + t) 3 t, whee t C s the paamete Ths cuve s on G(2, 3, 7) wth toson numbe 3 Remak 4 If one apples the fomula n lemma to ths example, one gets a dffeent paametezaton Ths shows that the paametezatons of the lnes n the same famly mght be dffeent Ths s because the toc modfcatons of the suface and the chat used mght be dffeent Howeve, the toson numbes ae nvaant

9 LIN ON BRIKORN-PHAM URFAC 9 Refeences [] G Gonzalez-pnbeg, M Lejeune-Jalabet, Coubes lsses su les sngulatés de suface, C R Acad c Pas, t éte, I, 38 (994) [2] G Gonzalez-pnbeg, M Lejeune-Jalabet, Famles of smooth cuves on suface sngulates and wedges, Ann Polonc Math, LXVII2 (997) [3] G Jang, Functons wth non-solated sngulates on sngula spaces, Thess, Unvestet Utecht, 998 [4] G Jang, D esma, Local embeddngs of lnes n sngula hypesufaces, Ann Inst Foue (Genoble), 49 (999) no 4, [5] H B Laufe, Nomal two-dmensonal sngulates, Ann Math tudes, 7, Pnceton Unv Pess, Pnceton 97 [6] M Oka, Non-degeneate complete ntesecton sngulaty, Hemann, Édteus des cences et des Ats, 997 [7] M Oka, On the esoluton of the hypesuface sngulates, n: eds T uwa, P, Wagech, Complex analytc sngulates, Advanced tudes n Pue Math 8, Knokunya Company, Tokyo, 987, p Jang (tll Mach 3, 2000), Oka and Pho : Depatment of Mathematcs, Faculty of cence, Tokyo Metopoltan Unvesty, Mnam-Ohsawa -, Hachoj-sh,Tokyo, JAPAN -mal addess: jang@compmeto-uacjp -mal addess: pdta@compmeto-uacjp -mal addess: oka@compmeto-uacjp esma : Depatment of Mathematcs, Utecht Unvesty, Budapestlaan 6, POBox 8000, 3508 TA Utecht, The Nethelands -mal addess: sesma@mathuunl Jang (pemanent): Laonng 2000, People s Republc of Chna -mal addess: jzjgf@maljzpttlncn Depatment of Mathematcs, Jnzhou Nomal Unvesty, Jnzhou h,