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(1)On the Classification of Quartic Surfaces with a Triple Point Part II By ,. Tadashi. TAKAHASHI*, Kimio WATANABE"" and Teiichi HIGUCHI**". gl. Introduction. Let V be a quartic surface in P3 with a triple point singularity at Pci V. We may choose homogeneous coordinates [X, Y, Z) M7] with P at [O, O, O, 1]. Then, the defining equation of V is given by the following form:. f,(X, Y, Z)W+f,(X, Y, Z)=O, where fi denotes a homogeneous polynomial of degree i (i=3, 4), and the local affine coordinates at P is given by setting I7V==1.. The type of the cubic form f3 may be a classification invariant. In [9], we have classified those quartic surfaces with triple points whose cubic part f3 are. either NC, CC, QC or QT where NC (CC, QC, QT resp.) denotes a nodal curve (cuspidal curve, conic and chord, conic and tangent resp.).. The remaining cases for cubic form f3 are TG, TC., MS, TL and NS where TG (TC, MS, TL, NS resp.) denotes 3 general lines (3 concurrent lines, multiple and single lines, triple line, non-singular elliptic curve resp.), i.e., characterized L. only by lines, or positive genus curve.. [I]G TC TL. *. MS. NS. Graduate School of Education, University of Tsukuba,. ** Institute of Mathematics, University of Tsukuba.. *** Department of Mathematics, Faculty of Education, Yokohama. National University..

(2) 72 T. TAKAHAsHi, K. WATANABE and T. HiGucHi In this paper, we shall consider above 5 cases for f3, and classify those quartic surfaces with triple points whose cubic part f3 are of type either TG,. TC, MS, TL or NS. From now, assuming that V is normal (i, e., V has only isolated singularities),. we shall consider above 5 cases in turn. We shall use good tables of typical defiriing equations for minimally elliptic singularities given by Laufer [5] in the next section. In this paper and [9], the complete classification of quartic surfaces with triple points have been done.. ,. g2. Classification. : Case TG So we take f3(X,Y,Z)=XYZI Thus P is a T,,,,. singularity. And f3(X, Y, Z)=O is a line pair.. LEMMA 5. Let F==XYZW+f4(X, Y, Z). (a) Singularities of F=O other than that atP correspond to multiple intersections of X==O with f4(X, Y, Z)=O or Y=O with f4(X, Y, Z)==O or Z==O with f4(X, Y, Z)=O away from [1, O, O] and.[O, 1, O] and [O, O, 1].. . (b) A k-tuple intersection of f,,(X, Y, Z)=O with f4(X, Y, Z)=O away from. [1,O,O]and[O,1,O]and[O,O,1]correspondstoanAk-isingularity. ' ' .t' t. (c).Let '''' tt, ',. ' ttY==O at [1, O, O], ' K(Y):=multiplicity of f4(X, Y, Z)=O with '. tt t. ' with Z==O at [1, O, O], K(Z):=multiplicity of f4(X, Y, Z)==O ' L(X): ==multiplicity of f4(X, Y, Z)==O with X=O at [O, 1, O], L(Z):==multiplicity of f,(X, Y, Z)==O with Z==O at [O, 1, O],. M(X): ==multiplicity of A(X, Y, Z)=O with X=O at [O, O, 1], M(Y):=multiplicity of f4(X, Y, Z)==O with Y=O at [O, Q, 1],,. ,., 1. Then P is a T4+K,4+L,4+M singularity for. t. K==Max.(K(Y), K(Z)), L==Max.(L(X), L(Z)), M==Max,(M(X), M(Y)). where Min.(K(Y),K(Z))=Oor1,Min.(L(X),L(Z))tOor1, Min.(M(X), M(Y))=O or 1.. N. If K(Y), K(Z) are both at least 2 or Z('X), L(Z) are both at least 2 or M(X), M(Y) are both at least 2, then F==O has non-isolated singularity by P.. PRooF. (a) By a change of coordinates we may write. F=C,X4+C,X3Y+C,X3Z+C,X2Y2+C,X2Z2+C,XY3+C,XZ3 +C,,Y4+C,,Y8Z+C,,Y2Z2+C,,YZ3+C,,Y4+WXYZ..

(3) Classification of Quartic Surfaces with a Triple Point, II 73 If X=O has a multiple intersection with f4(X, Y, Z)==O, say at [O, 1, l], in P2, then we obtain the following conditions.. F(O, 1, 1)==C,,+C,,+C,,+C,,+C,,==O, ......................(1). (a/OZ)f,(O, 1, Zt+1).=,=C,,+2C,,+3C,,+4C,,=O,............(2) (O/aX)f,(O, 1, 1)...,==C,+C,=O. ............................(3) In this case, the conditions that F==O has a singular point at [O, 1, 1, O] is t. as follows.. F(O, 1, 1, O)=C,,+C,,+C,,+C,,+C,,=O, ................(4) (O/OX)F(O, 1, 1, O)=C,+C, = O,..........................(5). (O/OY)F(O, 1, 1, O)= 4C,,+3C,,+2C,,+C,,= O,........･.･･(6). (O/OZ)F(O, 1, 1, O)==C,,+2C,,+3C,,+4C,,=O,............(7). (0/O17V)F(O, 1, 1, O)=O. Now (1)=(4), (3)=='(5), (2)=:(7) and 4(1)::'=-(6)+(7). Therefore we obtain the. result. The converse is similar. Q.E.D. (b) With the multiple point as in (a) we set Y=1 and get gi:=F(X, 1, Z' +1, VV).. And let Coeff.(GJ, term):=as above proof. Then we submit the following transformations.. equation symbol. VV=W'-Coeff. (gi, XZ),........................g2 W'==l>V"TCoeff. (g2, XZ2),...........･･･････････g3 Then g3 is as follows.. ' ･]. g3==(Cio+Cii+Ci2+Ci3+Ci4)+(Cii+2Ci2+3Ci3+4Ci,)Z. +(C,,+3C,,+6C,,)Z2+(C,,+4C,,)Z3+C,,Z4+WX +(higher terms with respect to recognition principle).. }Ience we obtain the result. Q. E. D. (c) It holds as above lemma1and lemma 3. Q.E.D.. In this case; Max.(P+q+r:T.,,,.)==21. Then F=O has non-zero 3 parameters. But normal form of Arnold by analytic transformations has 1 parameter.. CiX7+C2Y7+C3Z7+XYZ==O becomes to X'7+Y'7tZ.'7+C,X¥Z=O by X' =(1/C,)(i17)X, Yi =(1/C,)(i/7)Y, Z/=(1/C,)(i/-7)Z. ･ Example. T7,7,7･. ThefollowingequationhasaT7,7,7singularityatP. -. t t.t.L tt ' '' F=R(X3Y+Y91+.Z3X)+XYZI?V.-･･.･･t '. '.

(4) T. TAKAHAsHI, K. WATANABE and T. HIGucHI. 74. Figure oE example T 7,7,7 x3y+y3z+xz3=o x=o. L(X)=3 L(Z)=1. O,1,O]. [1.0,O]. y=o. [O,O,1] M(X)=1 M(Y)=3. z =o. '. K(Y)=1 K(Z)=3. Dual graph of T. p,q,r. #Cvertices)=K #(vertices)=M. /f'.-t-i l,iii!]"E}'{?l-,INXh,i. ,O,,"o' o6 , L-- -- --." "". - - ts # (vertices) =L. O=-2. -. Fig. 13. Applying the above lemma to the various configuration of case TG one ob-. tains the following list. ･. *When K(Y)=K(Z)=L(X)=L(Z)=M(X)==M(Y)==O. (This is the list of. singularities other than that at P, where P is a T4,4,4 singularity,). list Intersection symbol. 14. 14. 14. Singularities 14. 14. 2, 2. 2A,. 14. 14. 12, 2. A, 14. 14. 4. A,. 14. 12, 2. 14. A,. 14. 1, 3. 14. A,. 14. 14. 1, 3. A, 14. 2, 2. 14. 2A,.

(5) Classification of Quartic Surfaces with a Triple Point, II 75 -------t-e---t--}-}----------}-----------t------ie----e---t-t----------------........................To be continued.. The notation 1`.1`.2,2 for example, means that A(i¥, Y, Z)==O has four simple intersections with X =O and four simple intersections with Y==O and two double intersections with Z==O away from [1, O, O], [O, 1, O] and [O, O, 1].. We can make the list.s of remaining cases as above.. Case TC So we take f3(X, Y, Z)=Y3+Z3. Thus P is a U-type singularity at P. (.. And f,(X, Y, Z)=O is a line pair.. LEMMA 6. Let F= (Y3+Z3)VV+f4(X, Y, Z). (a) Singularities of F==O other than that at P correspond to multiple inter-. sections of the each line of f,(X, Y, Z)=O with f,(X, Y, Z)=O away from [1, O, O] in p2.. (b) A le-tuple intersection of the each line of f3(X, Y, Z)==O with f4(X, Y,Z). =O away from [!, O, O] corresponds to an Ak"i singularity of F==O in P3. (c) Let Li, L2, L3 be the each lines of f,(X, Y, Z)=O, and K: =the multiplicity of Li with f4(X, Y, Z)==O at [1, O, O],. J: =the multiplicity of L2 with f4(X, Y, Z)=O at [1, O, O], M: ==the multiplicity of L3 with f4(X, Y, Z)=O at [1, O, O]. Then the singular point P is determined by the following condition list. Condition list. K. J. M. Type of singularity. o. o. o. U12. 1. 1. 1. U14. 2. 1. 1. U13+2. 3. 1. 1. U13+3. 4. 1. 1. U13+4. If two of K, J, M are both at least 2, then we have a non-isolated singularity. ' PRooF. (a) and (b) hold as above lemma 5. (c) We consider the case when f4(X, Y, Z)==O does not run through the point [1, O, O] (i.e., f4(1, O, O)SO). Then F=O is as follows.. F==(Y3+Z3)17V+CoX`+(the higher terms with respect to the. recognitionprinciple). (Co40) Hence we can take the weights for (X, Y, Z), it is (1/4, 1/3, 1/3). Therefore.

(6) T. TAKAHAsHI, K, WATANABE and T. HiGucHi. 76. we have a Ui2 singularity at P.. For the next step, we consider the case when f4(X, Y, Z)==O through the point [1, O, O] (i.e., Co==O). Then Ci is not zero or C2 is not zero. Because if. Co=Ci==C2==O, then F==Ohasanon-isolatedsingularity at P. And now if Ci=O, C2 7ZO or C40, C2=O, then F=O is a SQH., choose the weights, it is (2/9, 1/3, 1/3) for (X, Y, Z). Hence we apply the recognition principle, we have a Ui4 singularity at P.. Let Li, L2, L3 be the each lines of f3(X, Y, Z)=O, and K: =the multiplicity of Li with f4(X, Y, Z)==O at [1, O, O], J: ==the multiplicity of L2 with f4(X, Y, Z)==O at [1, O, O],. M: ==the multiplicity of L3 with f4(X, Y, Z)=O at [6, O, O].. If Co==O and Ci40 and C2 7EO, then F= O becomes one of the following equations list by a certain analytic transformations as above for K, L, M. Equations list. K. L. M. 2. 1. 1. Equations. (Y3+Z3)W+C,X3Y+C,X3Z+C,X2Y2+g(X, Y, Z)==O where all the terms of g(X, Y, Z)==O are outside of the Newton boundary (O, 3, O), (O, O, 3), (3, O, 1), (3, O, 1), (2, 2, O).. 3. 1. 1. (Y3+Z3)W+C,X3Y+C,X3Z+C,XY3+g(X, Y, Z)==O where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, 3, O), (O, O, 3), (3, 1, O), (3, O, 1), (1, 3, O).. 4. 1. 1. (y3+Z3)W+C,X3Y+C,X3Z+C,,Y4+g(X, Y, Z)==O where all the terms of g(X, Y, Z)==O are'outside of the Newton boundary (O, 3, O), (O, O, 3), (3, 1, O), (3, O, 1), (O, 4, O).. If two of K, 1, M are both at least 2, by (c) of the above lemma 5 we have a. non-isolated singularity at P. ' Q.E.D..

(7) Classification of Quartic Surfaces with a Triple Point, II 77. Dual graph of Ui3+n . ''. '. '. t' tt'. .t tt tt . r3. '. -----(n-2)-----. '. -3 , -3 '. tt O =-2. N. '. Fig. 14.. '. Applying the above lemma to the various configuration of case TC one ob-. tains the following ' list. ' '. *When K= J==M=O. '. (This is the list of singularities other thqn that at P, where P is a Ui2 singularity.). list Intersection symbol. 14. 14. 12, 2. 14. 14. 14. A,. Singularities. The 4. 14. 14. 1, 3. 14. 14. 2, 2. 14. 14. 4. 2A,. A,. 14. 12, 2. 14. A,. 14. 1, 3. 14. A,. 14. 2, 2. 14. 2A,. A,. ---t----e-tt---------------------------t----------,.......................To be continued. notation of this list is as same as Case TG.. "'. ---. -----e-----e-------. Iii2athapniepua5?,.t,h.e iists of remaining cases as above;. When F=O has a Ui3+4 singularity at P, the intersection figure of f3(X, Y, Z) ==o with f4(X, Y, Z)==O in P2 is as follows. '. f4(X,Y,Z)=O . R2 Jn L. [1,O,O]-. L2. L3. l.. ( K=4, J--1,. )t[=L ). Fig. 15.. Case MS Choose coordinates so that f3=YZ2. gularity.. Thus P. is a V-type or V*-type sin-.

(8) 78 T. TAKAHAsHi, K. WATANABE and T. HiGucHi LNMMA 7. Let F=YZ2VV+f,(X, Y, Z). (a) Singularities of F==O other then that at P correspond to multiple intersections of f3(X, Y, Z)=YZ2=O with f4(X, Y, Z)==O away from [1, O, O]. (b) A le-tuple intersection of f3(X, Y, Z)=O with f4(X, Y, Z) =O away from [1, O, O] corresponds to an Ale-i singularity.. (c) Let K: =the multiplicity of f4(X, Y, Z)==O with Z==O at [1, O, O],. J: =the multiplicity of f4(X, Y, Z)=O with Y='O at [1, O, O],. '. Par.: --the partition of intersections of f4(X, Y, Z)=O with. Z=O away from [1, O, O]. Then the singular pont P is determined by the following condition list Condition list. K. 1. Par.. o. o. 4. IV*ls. o. o. 1. 3. IV17. o. o. 2. 2. IVIs+1+1. o. o. 12. 2. IVIs+1. o. o. 14. Yls. 1. 1. 3. IV17+1. 1. 1. 1. 2. 2Vls+1+1. 1. 1. 13. 2Vls+1. 1. 2. 3. IV17+2. 1. 3. 3. IV17+3. 1. 4. 3. 17. 1. 2. 1. 2. 2Vls+2+1. 1. 3. 1. 2. ･2Vls+3+1. 1. 4. 1. 2. 2Vls+4+1. 1. 2. 13. 2Vls+2. 1. 3. 13. 2Vls+3. 1. 4. 13. 2Vls+4. 2. 1. 12. 2V17. Type of singularity. V17+4. '.

(9) Classification of Quartic Surfaces with a Triple Point, II. 2. 1. 2. 2V17+1. 3. 1. 1. 2V*ls. o. 3V*lg. 4. l. t. 1. 79. If K and J are both at least 2, then F=O has a non-isolated singularity at P.. PRooF. (a) and (b) hold as above lemma 5. (a) When f4(1, O, O)40 (i.e., K=J=O) and Par. =4. Then F==O is as follows.. F=C,X4+C.Y3Z+C,,Z4+C,X3Z+C,X2YZ+C,X2Z2. +C,XY2Z+C,XZ3+YZ2. Here by the following analytic transformation, F==O becomes as follows.. Rewrite Z' to Z.. Z == Z< - Y2,. F=CoX`-CnY5+CiiY3Z+Ci4Y8-4Ci4Y6Z+6Ci4Y4Z2"Ci4Y2Z3. C,,Z4-C,X3Y2+C,X3Z--C,X2Y3+C,X2YZ+C,X2Y4-2C,X2Y2Z. +C,X2Z2-C,XY4+C,XY2Z-C,XY6+3C,XY4Z-3C,XY2Z2 C,XZ3+Y5-2Y3Z+YZ2. (C,40,C,,40) Thus F=O is SQH., so we have a iV*is singularity at P･ (Now Co is not zero, since f4(1, O, O)==C!. And since we can move the point at which f4(X, Y, Z)=O has 4-tuple inte'rsection, we can take Ci=:C3=C6=Cio=O,. i.e., we suppose that f4(X, Y,Z)=O has 4-tuple intersection with Z=O at [1, 1, O]. And moreover Cii is not zero. Because if Cn==O, then by the above condition Cio=Cu=Ci2==O, F=O has a non-isolated singularity at P. And since. we can set VV=W'-CsX-Ci2Y-Ci3Z, Cs--Ci2==Ci3==O-). The case of K=J=O and Par. ==1.3 (i.e., f4(X, Y, Z)=O does not run through the peint [1,O, O] in P2 and f4(X, Y, Z)=O has 1-tuple intersection and 3-tuple intersection with Z=O away from [1, O, O] in P2.). Then F==O is as follows.. F=C,X4+C,X3Y+C,,Y3Z+C,,Z4+C,X3Z+C,X2YZ. +C,X2Z2+C,XY2Z+C,XZ2+YZ2. And by the following analytic transformation, F=O is as follows.. Rewrite Z' to Z.. Z=Zi-Y2,. F=C,X4+C,X3Y+C,,Y8-4C,,Y6Z+6C,,Y4Z2-4C,,Y2Z3+C,,Z4. --C,X3Y2+C,X3Z-C,X2Y3+C,X2YZ+C,X2Y4-2C,X2Y2Z.

(10) 80 T. TAKAHAsHi, K. WATANABE and T. HiGucHi +C,X2Z2-C,XY4+C,XY2Z-C,XY6+3C,XY4Z-3C,XY2Z2. +C,XZ3-Y5+YZ2. And we can set Cii=2 without loss bf generaiity. Then F==O is as follows.. F=YZ2+X4+X3Y+Y5+g(X, Y, Z), where all the terms of g=O are outside of the Newton boundary (O,1,2), (4, O, O), (3, 1, O), (O, 5, O).. We can also determine the type of V-type or V*-type singularity by its '. Newton boundary. Hence in this case, we have a iVi7 singularity at P.. Y 5. Figure of the Newton boundary (Ot1,2)r(4,OtO)r (3tLrO>r(O,5rO)･'. 1 '. iJ. 3t 4. z. x. 3. Fig. 16.. The case of K=J==O and Par. =2.2. Then we can set Co=1 as above. .And we can set C2=O by a linear transformation with respect to X. And moreover we can put Co=1, Ci=-2, C3==1, C6==O, Cio=O, i.e., F==O has a 2-tuple intersection at [1, O, O] and a 2-tuple intersection. at [1, -1, O] in P2. And we submit the following analytic transformations.. y.,.y'+X' Z==Zt--(1/2)(C,+C,+C,,)X2-(1/2)･3C,,XYt, 'Yi= Y'M-(1/2)C,,X2-(1/2)t3C,,XZi,. Z,=Zn-(1/2)C,,Y"2+(1/2)Ci,XY". And rewrite Y", Z" to Y, Z. Then F==O is as follows.. F=(X+Y)Z2+X2Y2+X5+Y5+g(X, Y, Z), where all the terms of g(X, Y, Z)==O are outside of the Newton boundary (1, O, 2), (O, 1, 2), (2, 2, O), (5, O, O), (O, 5, O).. Hence we have a iVis+i+i singularity at P..

(11) Classification of Quartic Surfaces with a Triple Point, II 81. The case of K=J=O and Par. =12.2. We set C2=O as above. And since we set that f4(1¥, Y, Z)==O has 2-tuple intersection with Z==O at [O, 1, O], we set Ci=O. And moreover we set Cs=Ci2 =Ci3. =O by a linear transformation 17V=l7V'-Cs¥-Ci2Y-Ci3Z. Then F=O is as follows.. F=C,.2¥4+C,,Y3Z+C,,Z4+C,X2Y2+C,X2YZ+C,X2Z2. +C,XY2Z+C,XZ3+YZ2. And we submit the following analytic transformation:. Z==Zt-(1/2)C,,Y2. Rewrite Z' to Z. Then F=O becomes as follows.. F==YZ2+X4+X2Y2+Y5+g(X, Y, Z) where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, 1, 2), (4, O, O), (2, 2, O), (O, 5, O).. Hence we have a iVis+i singularity at P.. The case of K =J==O and Par. =1`. In this case, we can set f4(X, Y, Z)==X(X+Y)(aX2+bXY+cY2) by a linear transformation with respect to X. (Now c is not zero. 'Because if c=O, then. A(X, Y, Z)= X(X+Y)(aX2+bXY)==O has a 2-tuple intersection with Z==O at [O, 1, O].) Hence C640. Then F==O is SQH., so weightS for (X, Y, Z) are (1/4, 1/4, 3/8). Therefore we have a Vis singularity at P. Similarly we consider the remaining cases.. The case of K=J=1 and Par. =3. We set Co=C6=C3=Cio=Cs=Ci2==Ci3==O, Ci==C2=1 from a certain situation as above. And we submit the following analytic transformation:. Z=Z'-(1/2)CnY2. Rewrite Z' to Z. Then F=O becomes as follows.. F= YZ2+X3Y+X3Z+Y5+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, 1, 2), (3, 1, O), (3, e, 1), (O, 5, O).. Hence we have a iVi7+i singularity at P.. The case of K==J==1 and Par. =1.2. We set Co=C6=Cio=Cs=Ci2=:Ci3==O, Ci==C2==C3==1 from a certain situation. And we submit the following analytic transformation:.

(12) 82 T. TAKAHAsHi, K. WATANABE and T. HiGucHi. Z = Zt- (1/2)C,,Y2, Rewrite Z' to Z. Then F==O becomes as follows.. F=YZ2+X3Y+X2Y2+Y5+X3Z+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, 1, 2), (3, 1, O), (2, 2, O), (O, 5, O), (3, O, 1), (3, O, 1).. Hence we have a 2Vis+i+i singularity at P.. The case of K=J=1 and Par. =13. We set Co==C3=C6==Cs=Ci2==Ci3= Ci4= O, Ci =C2=Cio ==1 from a certain situation.. And we submit the following analytic transformations:. Z=Zt-(1/2)C,X2-(1/2)C,XY, Y' =Yt-C,XZt-C,1!C'2. Rewrite Y' and Z' to Y and Z. Then F=O becomes as follows.. F==YZ2+X3Y+X3Z+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, 1, 2), (O, 4, O), (3, 1, O), (3, O, 1).. Hence we have a 2Vis+i singularity at P.. The case of K==1, f==2 and Par. ==3. We set Co=C2=C3= C6==Cio=Cs==Ci2 =Ci3=Ci4=O, Ci==1 from a certain situation. And we submit the following analytic transformation:. Z :Z'-(1/2)C,,Y2. Rewrite Z' to Z, Then F=O becomes as follows.. F=YZ2+X3Y+X2Z2+Y5+g(X, Y, Z). Hence we have a iVi7+2 singularity at P. t. The case of K==1, 1=3 and Par. ==3. We set Co--'C2==C3=C6= Cio=Cs==Ci2 =Ci3==Ck==Cs=O, Ci =1 from a certain situation. And we submit the following analytic transformations:. Z==Z/-(1/2)C,,Y2, Z/=Z"-(1/2)C,X2-(1/2)C,XY. Rewrite z" to z. Then F=o becomes as fol16ws.. F==YZ2+X3Y+YZ3+Y5+g(X, Y, Z),.

(13) Classification of Quartic Surfaces with a Triple Point, II 83 where all the terms of g(X, Y, Z) =O are outside of the Newton boundary. (O, 1, 2), (3, 1, O), (1, O, 3), (O, 5, O). ･ .. Hence we have a iVi7+3 singularity at P.. ' The case of K==1, J=4 and Par. =3. '. We set Co=C2--C3==C6==Cio==Cs=Ci2==Ci3==Cs=Cg=O, Ci==1 fromacertain situation. And we submit the following analytic transformations:. Z==Zi-(1/2)C,,Y2,. Z,=Z"-(1/2)C,X2-(1/2)C,XY. ,. Rewrite Z" to Z. Then F==O becomes as follows.. F=YZ2+X3Y+Z4+Y5+g(X, Y, Z),. '. where all the terms of g(X, Y,Z)=O are outside of the Newton boundary (O, 1, 2), (3, 1, O), (O, O, 4), (O, 5, O).. Hence we have a iVi7+4 singularity at P.. ' The case of K=1, J=2 and Par.=12. We set Co==C2=C6 =Cio==Cs==Ci2==Ci3--Ci4=O, Ci==C3=1 from a certain situation. And we submit the following analytic transformations:. z==z'-(1/2)c.y2,. zx=Z"-(1/2)C,X2-(1/2)C,XY. Rewrite Z" to Z, Then F==O becomes as follows.. F==YZ2+X3Y+X2Y2+Y5+X2Z2+g(X, Y, Z), where all the terms of g(X, Y,Z)=O are outside of the Newton boundary (O, 1, 2), (3, 1, O), (2, 2, O), (O, 5, O), (2, O, 2).. Hence we have a 2Vis+2+i singnlarity at P.. The case of K==1, J==3 and Par. ==1.2. We set Co== C2==C6= Cio= Cs=Ci2= Ci3 =Ci4==Cs=O, Ci=C3 ==1 from a certain situation. And we submit the following anlytic transformations:. Z=Zi-(1/2)C,,Y2, Z,=Z"-(1/2)C,X2-(1/2)C,XY. Rewrite Z" to Z. Then F==O becomes as follows.. F=YZ2+X3Y+X2Y2+Y5+XZ3+g(I¥, Y, Z), where all the terms of g(X, Y,Z)==O are outside of the Newton boi;!pdary.

(14) 84 T. TAKAHAsHi, K. WATANABE and T. HiGucHi (O, 1, 2), (3, 1, O), (2, 2, O), (O, 5, O), (1, O, 3).. Hence we have a 2Vis+3+i singularity at P.. The case of K=1, 1=4 and Par.=1.2. We set Co=C2==C6=Cio=Cs==Ci2=Ci3==Cs==Cg==O, Ci=C3=1 from acertain situation. And we submit the following analytic transformations:. Z=Zt-(1/2)C,,Y2, zt=Z"-(1/2)C,X2-(1/2)C,XY, Rewrite Z" to Z. Then F=O becomes as follows.. F==YZ2+X3Y+X2Y2+Y5+Z4+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, 1, 2), (3, 1, O), (2, 2, O), (O, 5, O), (O, O, 4).. Hence we have a 2Vis+4+i singularity at P.. The case of K==1, J==2 and Par. =13. We set Co=C2 =C3=C6==Cs==Ci2==Ci3=Ci4=O, Ci=:Cio=Cs= 1 from a certain situa-･ tion. And we submit the following analytic transformations:. Z= Zt-(1/2)C,,Y2,. zi=Z"-(1/2)C,X2-(1/2)C,XY,. Y=Yt-C,XZM. Rewrite Y' and Z" to Y and Z. Then F=O becomes as follows.. F=YZ2+X3Y+X2Z2+Y`+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary of (O, 1, 2), (3, 1, O), (2, O, 2), (O, 4, O).. Hence we have a 2Vis+2 singularity at P.. The case of K=1, J=3 and Par. ==12. We set Co==C3=C6==C2== Cs=Ci4=Cs=Ci2=Ci3=O, Ci=Cg=Cio=1 from a certain situation. And we submit the following analytic transformations:. Z==Z'-(1/2)C.Y2, zt=ZM.(1/2)C,X2-(1/2)C,XY. Rewrite Z" to Z. Then F=O becomes as follows.. F=YZ2+X3Y+Y4+XZ3+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary. i..

(15) ClassificationofQuarticSurfaqeswithaTriplePpint,II. 85. (O, 1, 2), (3, 1, O), (O, 4, O), (1, O, 3). ., .," Hencewehavea2Vis+3singularityatP, . ,'. ,.,.i''. ･ .1. The case of K==1, 1--4 and Par.=:13.. We set Co==C3=C6=C2=Cs= Cg=Cs=C12=Ci3=O, Cl=Clo==C14=1 from a certain situation. Andwesubmitthefollowinganalytictransformations: /., ..,"t'. Z=Zi-(1/2)C,,Y2, . ･ '. -'･L,･･･'i':･H'. '. Zt=Z.-(1/2)C,X2-(1/2)C,XY.. ' '. Rewrite Z' to Z. Then F=O becomes as follows.. ttt /t t.:tt tt. ''･ F==YZ2+X3Y+Y4+Z4+g(X,Y,Z),. ･. .t./..-...t.. where all the terms of g(X, Y,Z)=O are outside of the Newton boundary (O, 1, 2), (3, 1, O), (O, 4, O), (O, O, 4).. Hence we have a 2Vis+4 singularity at P.. ' '･･ '･ The･casebfK=2,J=1andPar.=12.･･･/･･-ri･''sf. ･ l. ,･･: ].:･･-l･--;･.. AWnedS&t,. '. g&=b=.Cfi=,Chi,`Ml,C/i,--.9.h2g=k/---/yO,'i,C,3,--..C,i;,'.,.C/1,i:.,fr,Om a ce"tai" s.ipy.-a.tion･. Z==Z'-(1/2)C,,Y2-, z,,.2"m(1/2)c,X2-(1/2)C,XY,. y=Yt-C,X2-C,XZ". Rewrite Y' and Z" to Y and Z. Then Fl- O bdcomes as'follows.. F=YZ2+Y4+X2Y2+X3Z+g(X, Y, Z), where all the terms of g(X, Y, Z)==O are outside of the Newton boundary. (O,1,2),(O,4,O),(2,2,O),(3,O,1). . Hence we have a 2Vi7 singularity at P.. The case of K=2, 1=1 and Par. ==2. We Set Co==Ci=C6==Cio =Ci4=Cs= Ci2 =Ci3=O, C2==C3--1 from a certain situation.. And we submit the following analytic transformations:. Z==Zt-(1/2)C,,Y2, Zt=Z"-(1/2)C,X2-(1/2)C,.X'Y,. Y=Yt-C,X2-C,XZ". Rewrite Y' and Z" to Y and Z. Then F=Q becomes as follows.. F=YZ2+X2Y2+Y5+X3Z+g(X, Y, Z),.

(16) 86 '. T. TAKAHAsHi, K. WATANABE and T. HiGucHr. where all the terms of g(X, Y,Z)=O are outside of the Newton boundary (O, 1, 2), (2, 2, O), (O, 5, O), (3, O, 1).. Hence we have a 2Vi7+i singularity at P.. The case of K=3, J==1 and Par. =1. We set Co=Ci=C2==C3=Cs=Ci2=Ci3==O, C6=C2=1 from a certain situation. Then F=O is as follows.. F==YZ2+XY3+C,,Y4+C,,Y3Z+C,,Z4+X3Z+C,X2YZ. 1. +C,X2Z2+C,XY2Z+C,XZ3. For this equation choose weights (4/19, 4/19, 7/19) for (X, Y, Z) and apply the recognition principle to obtain the result. Hence we have a 2V*is singularity. The case of K=4, J==1 and Par. =O. We set Co= Ci=C2==C3=Cs==Ci2 =Ci3==C6=O, Cio=C2=1 from a certain situation. Then F= O is as follows.. F=YZ2+Y4+X3Z+C,,Y3Z+C,,Z4+C,X2YZ+C,X2Z2+C,XY2Z+C,XZ3.. DualgraphofIVis+n cequation:yz2=x4+x2y2+yn+4.. (n)l) ) -3 -'----Cn-L)----3 N. Dual graph o-F 2Vls+n. yz2 = y4 + x3 Y + x ab Y.. ( eguation. ( n2 l' r 2a + 3b =n. -3 ------( n-1 )------3. O--, Fig. 17(1).. , 8 ). ).

(17) Classification of Quartic Surfaces with a Triple Point, II 87 For this equation choose weights (5/24, 1/4, 3/8) for (X, Y, Z) and apply the recognition, principle to obtain the result. Hence we have a 3V*ig singularity at P.. Finally note that if K 1 are both at least 2, then f4(X, Y, Z)= O must have a singularity at [1, O, O] and it follows that F==O is singular along the line. Y=Z=O. Q. E. D. '. '. '. '. ))ualgraphoEIVz7 cequation:yz2=x4+x3y+y5) , -3 ･-3. e. Dual graph o￡. 2VZ7. ( egualtion : yz2 = y4 + x2y2 + x3z ). :3. -3. Dual graph of IVi5+n+m /l. -3 --r--( n-1 )'---- -･3 -------(. m-1 )･-....--... ' 2 ,... c eguation : .(x+y)z2 = x2y + >(n+4 + ym+4.. n> Zr mk1 Fig. 17(2).. ).

(18) T. TAKAHAsHI, K. WATANABE and T. HIGucHI. 88. Dual graph Of 2Vls+n+m --3. (eguation:YZ2 = '. -3 ------C m-l )-----. n-: )-----. -------d <. x]ii}r + x2M2. ' tm?lrn):t. 2a,+ 3b =. +yn+4. + xazb.. m+8 )) ). Dual graphOf zVi7+n ,-. -----C n-z )------3. < eguation : yz2 = x3y + xay b+ y5.. +8. Cn2 1, 2a + 3b =n. -3. )). DuaZ graph Of 2VIL7+n. ----p--nyd (. n-1. C eguation : yz (. ) -----d-. =xy 22 n)Z) Fig.. ･-. 3. 2 + y4 + x3z.. ･-. 3. v. ) '. 17 (3).. Example.. '. in R2. f4(XtY,z) = o. 3-tuple. [1,O,O] ( K=lr J"4 ). y=o z=o. P is a "rJ7+4 singularitY. Fig. 18..

(19) Classification of Quartic Surfaces with a Triple Ppint, II 89 We can make the lists of singularities other than that at P as above.. Case TL Choose coordinates so that f3(X, Y, Z)=Z3. Thus P is a VLtype or V"-. typesingularity. ' ' LEMMA 8. Let F==Z3W+A(X, Y, Z).. (a) F=O has no singularity other than that at P. (b) Let Par. be the partition of intersections of ft(X, Y, Z)==O with Z=O. Then the only one singularity P is determined by the following condition list. L. Condition list. Type. Par.. of singularity. t'. 4. V'21. 2. 2. 2Y"2o. 1. 3. Vi2o. 12. 2. V"lg. 14. V'ls. PRooF. (a) For any point (?=[a,b,c,d](#P)a40orb40orcsO. Now let a be non-zero. We submit the following transformations:. X==a, Y=Y'-b, Z=Z'-c, W=w'-d. Then W' is free for the following conditions.. F(a, b, c, d)==O,. 0 F(a, b, c, d)==O, ox O F(a, b, c, d) =O, OY .. O F(a, b, c, d)==O, oz O. F(a, b, c, d)==O.. OVV ･ HenceF=OhasnosingularityotherthanthatatP. , Q.E.D. (b) Let Par. be the partition of intersections of f4(X, Y, Z)=O with Z=O.. The case of Par. =4. , In this case, we can set f4(X, Y,O)=Y4 (i.e., we suppose that f4(X, Y,Z)=O has a 4-tuple intersection with Z=O at [1,O, O].). Then if C2==O, F==O has a non-isolated singularity at [1, O, O, O]. Therefore C2 is not zero. Hence for non-.

(20) 90 T. TAKAHAsHi, K. WATANABE and T. HiGucHi zero C2, F=O is as follows.. F==C,,Y3Z+C,,Y2Z2+C,,YZ3+C,,Z4+C,X2YZ+C,X2Z2. +C,XY2Z+C,XYZ2+C,XZ3+C,X3Z+Y4+Z3. Thus F==O is semiquasihomogeneous. For this equation choose weights (2/9, 1/4, 1/3), for (X, Y, Z) and apply the recognition principle. Then we have a V'2i singularity at P.. The case of Par. =2.2.. i. As above, we set f4(X, Y, O)=X2Y2. Then since F==O has only isolated singularty,. from lemma 5 f4(X, Y, Z)=O has a 1-tuple intersection with X==O at [O, 1, O]. and a 1-tuple intersection with Y==O at [1,O,O]. Hence C2#O and Cii#O. Therefore. F=C,,Y2Z2+C,X2YZ+C,X2Z2+C,XY2Z+C,XYZ2. +C2X3Z+C.Y3Z+X2y2+z3. Thus F==O is as follows.. F.=Z3+X2Y2+C,X3Z+C,,Y3Z+g(X, Y, Z), where all the terms of g(X, Y, Z)=O are outside of the Newton boundary (O, O, 3), (2, 2, O), (3, O, 1), (O, 3, 1).. And we can also determine the type of Y'-type or V"-type singularity by the Newton boundary. Hence we have a 2V"2o singularity at P.. The case of Par. ==1.3. We set f4(X, Y, O)=XY3. Then as above C240, F=O is a semiquasihomogeneous. For F=:O choose weights (2/9, 7/27, 1/3) for (X, Y, Z) and apply the recogni-. ". tion principle. Then we have a V'2o singularity at P.. The case of Par. ==12.2. We set f4(X, Y, O)==Y`+X2Y2. Then as above C2tO, F=O is as follows.. F= C,X2YZ+C,X2Z2+C,XY2Z+C,XYZ2+C,,Y3Z+C,,Y2Z2. +C,X3Z+Y4+X2Y2+Z3. Thus F--O is as follows.. F=Z3+X2Y2+Y`+C,X3Z+g(X, Y, Z), where all the terms of g(X, Y, Z)==O are outside of the Newton boundary (O, O, 3), (2, 2, O), (O, 4, O), (3, O, 1).. Hence we have a V"ig'singularity at P.. '.

(21) Classification of Quartic Surfaces with a Triple Point, II. 91. The case of Par. ==1`.. We set f4(X, Y,O)= X`+Y`. Thus F=O is semiquasihomogeneous.. For F=O. choose weights (1/4, 1/4, 1/3) for (X, Y, Z) and apply the recognition principle. Q. E. D. Then we have a V'is singularity at P. DuaZ graph. of vrf. 19. ( equation , z3 = y4 + x2y2 + x3z ). ir. Dual graph oE alli'2o. ( eguation : z3 = x2y2 + x3z g･ y3z ). r. N. Fig. 19.. Example..

(22) l I. i i. 92 T. TAKAHAsHi, K. WATANABE and T. HiGucHi. x=o Y.=o st t. N .NNrNt. Nx/t fL(×,YtZ)=O. 1;'""N.. t 'N -N. tN -N tN. 1N. 1 ･N× .X [1,O,O] [O,1,O]. .N, z=O. /N. /2-tuple 2-tuple"× t. Par. = 2.2. -N tN. ,. P is a 2V"2o singularity.. 1. Fig, 20.. t. In this case, since F=O has no singularity other than that at P we have no list of singularities other than that at P. Case NS First note that for the singular point P itself to be a non-singular is equivalent to saying it has a type Ps singularitY. Choose coordinates sb that. f,(X, Y, Z)=X3+Y3+Z3+3RXYZ where 234--1.. LEMMA 9. Let. F..(X3+Y3+Z3+3RXYZ)W+f,(X, Y, Z). For any･ Point･Q (7EP) on V, F=O is non-singular or F==O has a -Ak (1:;l le .<.11). singularity at Q by the condition of coofcients of F=O.. PRooF. By a change of coordinates we consider at [1, -1, O, O]. Let sub. (--) be as above proof. And･ we submit the ･following setting and analytic. transformations of 12-steps.'i '. G,:=sub. (X=1, Y==Yt-1, F),. Rewrite Y' to Y, ,, 1-step.. T. D:-sup. (y-=o, z==o, pv-o, oiS5z,Gi), G2: ==sub. (Y=Y'-(1/3)･D･Z, Gi), Rewrite Y' to Y.. .4. lt. 2-step. ･. J: ==s.b. (y=o, z=o, w=o, oyOo2z G,), G,: =sub. (VV==VVt-(1/3)･1･ Z, G,),. 3-step. D,: ==sub. (y=o, z=o, mzt=o, aw9g,z G3),. '.

(23) Classification of Quartic Surfaces with a Triple Point, II. G,: =sub. (Y=Yi-(1/3)･(1/2)･D,･Z2, G,), Rewrite 17V' and Y' to J7V and Y.. 4-step. '. A:.=s.b.(y=o,z=o, m7==o, oyOo3,z G,),. G,: =sub. (M/ =I7Vt-(1/3)･(1/2)･k･Z2, G,),. 5-step. D,: =sub. (y==o, z=o, vv'=O, ovv9`o,z Gs), t. s. G,: ==sub. (Y =Yt-(1/3)･(1/6)･D,･Z3, G,), Rewrite 17V' and Y' to "r and Y.. 6-step. J,: =sub. (y =o, z=o, w==o, oyOo`,z G,), G,: =sub. (PV=M7i-(1/3)･(1/6)･J,･Z3, G,),. 7-step. D,: =sub. (y==o, z==o, wi==o, ol7v95o,z G,),. G,: ==sub. (Y=Yt-(1/3)･(1/24)･D,･Z4, G,), Rewrite W' and Y' to VV and Y.. 8-step. Ig:=sub. (y=o, z=o, vv=o, oyOo5,z G,), G,: ==sub. (J7V==W'-(1/3)･(1/24)･1,･Z4, G,),. 9-step. Dg: ==sub. (Y= O, Z=O, PV'== O, o17v96osz Gg),. Gio: =sub. (Y=Y'-(i/3)･(1/120)･Dg･Z5, G,), .. Rewrite M7' and Y' to J717 and Y.. 10-step. 1,,: ==sub. (y=o, z==o, m7=o, oyOo6,z G,,), G,,: ==sub. (I7I/= VVt-(1/3)･(1/120)･1,,･Z5, G,,),. 11-step. D,,:=sub.(Y=O, Z =O, 17Vi==O, owO;o,z Gii), G,, :, =sub. (Y =Yt-(1/3)･(1/720)･D,,･Z6, G,,), Rewr'ite 17V' and Y' to M7 and Y.. 12-step.. 93.

(24) 94 T. TAKAHAsHi, K. WATANABE and T. HiGucHi ,Il,:=sub.(y=o, z==o, mz =o, oyOo7,z G,,),. G,,: =sub. (W=Wt-(1/3)･(1/720)･A,･Z6, G,,), Rewrite M7' to VV.. Then ･. sub. (y==o, z=o,'w--o, oill51･z Gi3)==o' for i=i, 2, 3, 4, s, 6･. sub.(Y=O,Z=O,VV==O,-oya`o',lzG,,)==O fori=1,2,3,4,5,6. g. Therefore the condition of coeflicients of Za (O;:$a:S12) and Y in Gi3 =O deter-. minesaAk singularity of F==O. , ･ And moreover if the coeffients of' Za (OScrgll) and Y in Gi3=O are equal. r. '. to zero, then the coefficient of Zi2 in Gi,=O is equal to C,(23+1)'`. Since 23 7E -1,. if the ceflicient of Zi2 in Gi3==O, then C3 is equal to zero. But from the relation of cefficients for above condition, if C3 is equal to zero, then F=O has a. non-isolated singularity at [1, -1, O, O]. Q. E. D.. Example. , Following equation has a P, singularity at [O, O, O, 1] and a Aii singularity at [o, o, o, 1].. F=(6X3+6Y3+6Z3+18XYZ)VV-3X4-5X3Z+6X2Y2-24X2YZ +6X2Z2-24XY2Z+9XYZ2-8XZ3-3Y4-5Y3Z+6Y2Z2-8YZ3. We can make the lists of the singularities other than that at P. It is as same as that of Case NC or Case CC.. References [1] JEssop, C.M.: Quartic surfaces with singular points. Cambridge Univ., Press, (1916). [2] ARNoLD, V.I.: Normal forms of functions near' degenerate critical points, the Weyl groups Ak, Dk, Ele and Lagrange singularities. Func. Anal. App. 6, (1972), 254-272. [3] ARNoLD, V.I.: Normal forms of functions in the neighborhoods of degenerate critical points. Russ. Math. Surv. 29, (1974), 11-49.. [4] BRucE, J.W. and WALL, C.T.C.: On the classification of cubicsurfaces. J. London Math. Soc. 19, (1979), 245-256. [5] LAuFER, H.B.: On minimally elliptic singularities. Amer. J, Math. 99, (1977), 1257-1295.. [6] UMEzu, Y.: On Normal PropjectiveSurfaceswithTrivialDualizingSheaf. Tokyo J. Math., 4 (1981), 343-354.. [7] KARRAs, U.: Deformations of cusp singularjties. Proc. Symp. in Pure Math. A.M.S. 30, (1977), 37-44.. [8] HiGucHi, T., YosHiNAGA, E. and WATANABE, K.: Introduction to Complex Analysis of Several Variables (in Japanese). Morikita, '1980.. [9] TAKAHAsHi, T., WATANABE, K. and HiGucHi, T.: On the classification of quartic surfaces with a triple point, I. to appear in Sci. Rep. Yokohama National Univ.,. 1982.. 't. s. l.

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