On the Topological Types of Singularities of Brieskorn

On the Topological Types of Singularities of
Brieskorn-Pham Type
By
Etsuo YOSHINAGA" and Masahiko SUZUKI*"
(Received May 30, 1978)
gO. Introduction
The object of this paper is to prove the fo11owing
THEoREM. Let f,g be two polynomials of Brieskorn-Pham type of the
forms
f:zfi+･･･+z:n, 2SaiS.･･･Sa., aieN,
g=z9i+･･･+z9", 25biS.･･･Sb., biGIV.
If two hypersurfaces defined by f, g have the same topological type at O, then
we have ai == bi (i == 1,..., n).
Let f, g:C".C be analytic functions having an isolated singularity at O
with f(O)==g(O) :O. Then we say that two hypersurfaces Hi, H2 defined by f, g
have the same topological type at O if there exist two neighborhoods Ui, U2 of
O and a homeomorphism W from Ui to U2 such that
U(HinUi)=H2nU2, ut(O)==O.
gl. ProofofTheorem
l
Let f:C"-÷C be an analytic function having an isolated singularity at O
withf(O)=O, LetO<6<3(1, Wedefine
si = {t E C; ltl = 1},
SE = {z e C"; [zil2 + ･･･ + lz.l2 = e},
K=={zeCn;f(z)=O}nS,.
According to [2], the mapping ut(z): =f(z)11f(z)1: S,-K.Si is a locally trivial
smooth fibre bundle, Each fibre Fe=ut'i(eie)cS,-K has the homotopy type of
a bouquet of (n-1)-spheres. The generator of ni(Si) induces the monodromy
automorphism
* Department of Mathematics, Faculty of Education, Yokohama National University.
*" Department of Mathematics, University of Tsukuba,
38 E. YosHiNAGA and M. SuzuKi
h*: Hn-i(Fo; Z)' Hn-i(Fo; Z)'
We define the characteristic polynomial of the monodromy by
A(t): = det (tl. - h.).
Our proof bases on the following two theorems.
THEoREM 1 (D. T. Le [2]). Iftwo hypersurfaces with an isolated singularity
at O have the same topological type at O, then their monodromies are coajugate.
It follows from this theorem that if two hypersurfaces with an isolated singularity at O have the same topological type at O then their characteristic polynomials
coincide.
For simplification, we need the fo11owing definitions. To each monic polynomial (t-- cti)･･･(t- ctk) with cti,..., ctk E C", assign the divisor
div((t-cti)･･･(t---ctk))=<cti>+･･･+<ctk>
'
thought of as an element of the integral group ring ZC". Introduce the
notatlon
A.=div(tM-1)=<1>+<4>+･･･+<4m-i>
where 4=:exp (2nilm), and introduce the element
Em==M-IAm ,
which belongs to the rational group ring QC*. Ai =Ei=<1> will be written
briefiy as 1. The fo11owing theorem is the immediate result from J. Milnor and P.
Orlik [3].
THEoREM2, Let
f==zft+･･･+z#n, 2SaiS･･･Ea., aiEN･
Let A(t) be the characteristic polynomial ofthe hypersurface dcfined by f. Then
div A = 2(- 1)n-Sai,'''ai.E[aii,...,ai.]
= 2(-1)"-s [.",l'iiiia.'ii.] A[ai,,...,ai.]
to be summed over all the 2" subsets {ii,..., i,} of {1,..., n}, where denote by
[ai,..･, ak] their least common multiple.
THEoREM. Let f,g be two polynomials of Brieskorn-Pham type of the
forms
f==zfi+･･･+zSn, 2Sai5･･･5a., aiEN,
g=z?i+･･･+zB", 2;:Sbi;.S･･･:l;b., bi6IV･
If two hypersurfaces defined by f, g have the same topological type at O, then we
have ai= bi (i --- 1,..., n),
t
On the Topological Types of Singularities of Brieskorn-Pham Type 39
PRooF. Considerthecasethatniseven. Letusput:
Air･･i.: =[ait,.･., ai.]p
Bii'''is: =[bii,'':, bis],
Ci""is: = Elli;.iil.'III.';.'iE
and
Dii"'is: == bk'i19,?L"
The equation div Af= div A, and theorem 2 imply the following equation:
R (tAil'''is-1)Cil'''is H (tBit'''is-1)Dil'''is
s:even s:odd
stO
= n (tBir･･i.-1)Dil･･･i. n (tAil･･･i.ml)Cil･･･i.
s:even s:odd
s=si=o
We denote by P(t) the lefthand side of this equation and by 9(t) the righthand
side.
Now we show that the hypothesis ai<bi implies a contradiction. Since
9(t) has the factor t"i-1, P(t) has the factor tai-L fl (tBii･-･is-1)Dii･･･is
s:odd
does not have such a factor, so,,llJ.,.(tAii'''is-1)Dti''''s must have such factors.
sso
Namely, there exists Til.lr2 such that ai=a2==･･･=aT,<aT,+i････ CoMParing
the exponents of t"i-1 in bothhand sides, we have,
2) Cii-is= 2 Cii-is'
s:even
s=Si=O il,-.,s:odd
i. =< Tl
il,- " is 5 Tl
Since Ci,...i, == ZS', , this implies
O= ,¥.,(- 1)S(sTi)ai = -1+(1 -a,)Ti.
So ai =2 and Ti is even.
LetP,(t):=P(t)1 n (tAii-･･is-1)Cii-･･isand
g¥eoven
il,-"is5Tl
e,(t):=e(t)1 n (tAi,･･･i.-1)ci,...,..
s:odd
il"･"is5Tl
(1.1) In the case ofaT,<bi<aT,+i. Comparing the exponents of tbi-1 in the
bothhand sides of the equation Pi(t) =9i(t), we have bi==2 as same as above.
This is a contradiction for ai< bi･
(1.2) The case of aT+i<bi is impossible as same as (1.1).
(1.3) Inthecaseofbi=aT,+i･ Let
aTl+1=''' =aTt+T2<aTl+T2+1,
40 E. YosHiNAGA and M. SuzuKi
and
bi = ''' = bs < bs+i･
Comparing the exponents of tbi-1 in Pi(t)=9i(t), we have: in the case of
a lla Tl + 1,
2 C,,+2D,=
s+t:eyen
15I5 Ti s:odd
ISISS
TI+15I5Tl+T2
Au=aTl+t
2D,+ ]E) CiJ
s:even
ISI5S s+t:odd
15I5Ti
Ti+15JSTI+T2
Au=aTl+l
where I: ={ii,.･･,i,}, J:=={ji,･･･,J't}
namely
t
(1--ai)Ti(-1+(1-aT,+i)T2)=-1+(1-bi)S
soS=T2. Inthecaseofal*aT,+1,
2 C,+2D,=
t,e+ViegO=fOT,+T, Si:sOdids-s
2D,+ 2 C,,
s:even
t:odd,=O
ISISS Ti+15JETI+T2
namely
-1+(1 -aT,+ i)T2 = -1+(1 --- bi)S
so S=T2.
Now we may assume:
a1='''=aTl<aTl+1=''･=aTl+T2<'''<aTl+･･･+Tk+1=='''
=aTl+･･･+Tk+1<aTl+-･+Tk+1+1=='''=aTl+･･･+Tk+2<aTl+･･･+Tk+2+1l!S''･,
b1=･･･=bT,<･･･<bT,+..･+T.+1=･･･=bT,+･･･+Tk.1<bT2+･･･+Tk.,+1
=･･.=bT2+･･･+Tk+1+s<bT2+･･･+Tk.1+s+1'''
and aTi+1=b1,･･.,aT1+･･･+Tk+1==bT2+･･･+Tk+1･
Let us put:
?k+i(t):=Pi(t)1
n (tAi-1)Ci n (tBi-1)Di.
t,e+Vle-2r7IIOT,+...+T,., Sl:sOdld-tT,+･･･+T,.,
ek+i(t):--･-ei(t)1 n (tBi-1)Di n (tAi--1)cr
s:even,40Tl+ISI5Tl+･･･+Tk+1
s:odd
15IST2+-･+Tk+1
and a:==aTl+."+Tk.1+1, b:=bT2+･-Tk.1+1･
(k+1. 1) In the case of a>b, comparing the exponents of tb-1 in Pk+i(t)
9k+i(t), we have:
r
On the Topological Types of Singularities of Brieskorn-Pham Type
s+t:even
2
CIJ
s+t:odd
15I5T2+･-+Tk+1
T2+･･･+Tk+1+15J5T2+-･+Tk+1+s
sl1
1SI5 Ti
41
Dv =
BIJ=b
Tl+ISJ5Tl+-･+Tk+1
AIJ=b
Dv +
2
s+t:eyen
IEI5T2+-･+Tk+1
CIJ
>￡
s+t:odd
sl1
T2+･･･+Tk+1+15J5T2+･･･+Tk+1+S
1.<,ISTi
Bu=b
Tl+15JSTI+-･+Tk+1
ArJ=b
namely
2 (- 1)s
sl1
( Zi )ai( - 1)taj,･･･aj, =
2 (- 1)sb
slO
(
)bt.
･･･b is 2(m 1)t s
il
t
Since ,¥.,(-1)S(I`)ai=-1+(1-ai)'i==O, this implies b=2. This is a
contra-
diction.
(k+1. 2) In the case of a<b, comparing the exponents of ta-1 in
Pk + i(t)
= 2k+i(t), we have:
2
s+t:even
CIJ
CIJ
s+t:eyen
Tl+15I5Tl+-･+Tk+1
Tl+･-+Tk+1+ISJ5Tl+･･･+Tk+2
sl1
15IETi
Tl+15J5Tl+-･+Tk+2
AxJ=a
2
s+t:odd
Aw=a
CIJ
CIJ
s+t:odd
Tl+ISI5Tl+･･･+Tk+1
Tl+･-+Tk+1+5J5Tl+･･･+Tk+2
s51
15IS Ti
Aw=a
Tl+15J5Tl+-･+Tk+2
AiJ=a
namely
- 1)s (I')ai2(- 1)taj,･･･aj, + 2(- 1)Sai,
2(
sl1
ai.t;t1(-1)t( Tk+2 )at =O.
'''
t
Since the first term is zero, we have:
(1ma,,)R"･.(1-a,.)Rm{-1+(1-a)Tk+2}==O
where {a,,,...,a,.}:={ai;aila} and Rj:=#{i;ai=a,j}.
Hence a = 2, a contradiction.
(k+1. 3) In the case of a== b, comparing the exponents of ta-1 in
Pk+i(t) =
9k+i(t), we have:
s+t:even
s:O
15I5 Ti
2
CIJ +
s+t:eyen
Tl+IEJ5Tt+-･+Tk+1
AiJ=a
s+t:eyen
t40
2
IEIST2+･-+Tk+1
T2+･-+Tk+15JET2+
+Tk+1+s
BIJ=b
･･･
namely
Du +
2
s+t:odd
tto
ISI5Tl+･-+Tk+1
Tl+･･･+Tk+1+15J5Tl+
Z
stO
IEIS Ti
Tl+IEJ5Tl+-･+Tk+1
AiJ=a
DIJ =
tto
ISI5T2+･-+Tk+1
T2+･･･+Tk+t+15J5T2+
･･･ +Tk+1+S
BrJ=a
･- +Tk+2
AiJ=a
s+t:odd
2
CIJ +
CIJ +
s+t:odd
2
tto
IEI5Tl+･･･+Tk+1
Tt+･･･+Tk+1+15J5Tl+
+Tk+2
AiJ=a
--
CIJ
E. YosHiNAGA and M. SuzuKi
42
,1{.l)i ( - 1)S( [1i)aSi( - 1)'aj,''･aj, + 2( - 1)sai,･･･ai. ,].l)i ( - 1)t(tTk+2)at
= 2(- i)sbi,･･･b,. ,;, (- i)t(`9)bl.
Since the first term of the lefthand side is zero, so we have:
(1-art)Ri･･･(1-a,.)Rm{-1+(1-a)Tk+2}=
(1--bp,)P`'''(1-bp,)Ph{-1+(1mb)S}
where {a,....,a,.}:={ai;aila}, {bp,,･･･,bp,}:=={bi;bilb}
and Rj:=#{i;ai=a,,}, Pj:==#{i;bi=b,,}.
By the hypothesis of induction, {a,,,..･, a,.}=={b,,,･･･, bp,} or =={ai} U {bp,,･･･,
bp,} and if a,,=bpp then Ri=Pj. Hence we have (1-a,,)Ri･･･(1-a,m)R,n==
(1 - bp,)P`'''(1 - bp,)Ph = O,
)
so S=Tk+2･
Thus, in the case (1. 3), we have Ti+･･･+T.= T2+･･･+ T.=n for some m.
This contradicts Til2.
Therefore ai = bi.
We prove Theorem by induction. We assume ai---bi fori=1,..., k. Let us put:
Pt(t):=P(t)1 n (tAi-1)Cr n (tBi-1)Di,
s:even
ISISk s:odd
ISISk
e'(t):=e(t)1 n (tBi-1)Di fi (tAi-1)CJ.
s:even
s:odd
ISISk ISI-<k
Let ak;.:fak+i=･･･==ak+T<ak+T+i. If ak+i<bk+i, comparing the exponents
of tak+i-1 in bothhand sides of P'(t) = 9'(t), we have:
s+t:eleln CiJ+s+t:oliidl CiJ
ISISk ISISk
k+ISJSk+T
k+ISJSk+T
namely
Aw=ak+i Au=ak+i
2(-1)sai,･･･ai.,¥.,(-1)'(ll")aft.,==o.
Hence (1-a,,)Ri･･･(1-a,.)Rm{-1+(1--ak+i)T}=O, so ak+i=2 and T is even.
Where {a,,,..., a,.}: ={ai; ailak+i} and Rj: =#{i; ai=a,j}. But ak+i=2<bk+i
implies a contradiction as (1.3). Therefore we have ak+i=bk+i. This completes the proof of Theorem in the case that n is even.
The proof of Theorem in the case that n is odd is exactly same one in the case
thatnis even. Q. E. D.
References
[1] D.T.Lfi: Topologie des singularites des hypersurfaces complexes.
(1972), 171-182.
Asterisque 7 et 8
s
On the Topological Types of Singularities of Brieskorn-Pham Type
[2]
[3]
43
J. MiLNoR: Singular points ofcomplex hypersurfaces. Prinston University Press (1968).
J. MiLNoR and P. ORLiK: Isolated singularities defined by weighted homogeneous poly-
nomials. TopologyVol.9(1970),385-393.
Etsuo Yoshinaga
Department of Mathematics,
Faculty of Education,
Yokohama National University.
Masahiko Suzuki
Department of Mathematics,
;
University of Tsukuba.