Topology of polar weighted homogeneous hypersurfaces

0.22 MB·English

by  Mutsuo Oka

TOPOLOGY OF POLAR WEIGHTED HOMOGENEOUS
HYPERSURFACES
8
0 MUTSUO OKA
0
2
Abstract. Polar weighted homogeneous polynomials are the class of
n
special polynomials of real variable x ,y , i = 1,...,n with z = x +
a i i i i
√ 1y which enjoys a“ polar action”. In many aspects, theirbehavior
J − i
lookslikethatofcomplexweightedhomogeneouspolynomials. Westudy
4
basic properties of hypersurfaces which are defined by polar weighted
2
homogeneous polynomials.

]
G
A
h. 1. Introduction
t
a We consider a polynomial f(z,z¯) = c zν¯zµ where z= (z ,...,z ),
m ν,µ ν,µ 1 n
z¯ = (z¯ ,...,z¯ ), zν = zν1 zνn for ν = (ν ,...,ν ) (respectively ¯zµ =
[ z¯1µ1···z¯1nµn fornµ = (µ1,..1.,µ··n·) nas usuaPl. Here1z¯i is tnhe complex conjugate
1 of z . Writing z = x +√ 1y , it is easy to see that f is a polynomial of
i i i i
v −
2n-variables x ,y ,...,x ,y . Thus f can beunderstoodas an real analytic
8 1 1 n n
function f : Cn C. We call f a mixed polynomial of z ,...,z .
0 1 n
→
7 A mixed polynomial f(z,z¯) is called polar weighted homogeneous if there
3 exists integers q ,...,q and p ,...,p and non-zero integers m , m such
1 n 1 n r p
.
1 that
0
8 gcd(q ,...,q ) = 1, gcd(p ,...,p ) = 1,
1 n 1 n
0
n q (ν +µ )= m , n p (ν µ ) = m , if c = 0
v: i=1 j j j r i=1 j j − j p ν,µ 6
i We sayPf(z,z¯) is a polar weightedPhomogeneous of radial weight type
X
(q ,...,q ;m ) and of polar weight type (p ,...,p ;m ). We define vectors
r 1 n r 1 n p
a ofrationalnumbers(u ,...,u )and(v ,...,v )byu = q /m ,m vector =
1 n 1 n i i r i
−
p/m and we call them the normalized radial (respectively polar) weights.
i p
Using a polar coordinate (r,η) of C∗ where r > 0 and η S1 with S1 =
∈
η C η = 1 , we define a polar C∗-action on Cn by
{ ∈ || | }
(r,η) z= (rq1ηp1z ,...,rqnηpnz ), (r,η) R+ S1
1 n
◦ ∈ ×
(r,η) ¯z= (r,η) z = (rq1η−p1z¯ ,...,rqnη−pnz¯ ).
1 n
◦ ◦
Then f satisfies the functional equality
(1) f((r,η) (z,¯z)) = rmrηmpf(z,z¯).
◦
2000 Mathematics Subject Classification. 14J17, 32S25.
Key words and phrases. Polar weighted homogeneous, Polar action.
1
2 M.OKA
This notion is introduced by Ruas-Seade-Verjovsky [12] implicitly and then
by Cisneros-Molina [2].
It is easy to see that such a polynomial defines a global fibration
f : Cn f−1(0) C∗.
− →
The purpose of this paper is to study the topology of the hypersurface
F = f−1(1) for a given polar weighted homogeneous polynomial, which is a
fiber of the above fibration. Note that F has a canonical stratification
F = F∗I, F∗I = F C∗I
I⊂{1,2,...,n}
∐ ∩
Our main result is Theorem 10, which describes the topology of F∗I for a
simplicial polar weighted polynomial.
2. Polar weighted homogeneous hypersurface
This section is the preparation for the later sections. Proposition 2 and
Proposition 3 are added for consistency but they are essentially known in
the series of works by J. Seade and coauthors [12, 13, 10, 11, 14].
2.1. Smoothness of a mixed hypersurface. Letf(z,z¯)beamixedpoly-
nomial and we consider a hypersurface V = z Cn;f(z,z¯) = 0 . Put
{ ∈ }
z = x +iy . Then f(z,z¯) is a real analytic function of 2n variables (x,y)
j j j
with x = (x ,...,x ) and y= (y ,...,y ). Put f(z,z¯) = g(x,y)+ih(x,y)
1 n 1 n
where g, h are real analytic functions. Recall that
∂ = 1 ∂ i ∂ , ∂ = 1 ∂ +i ∂
∂zj 2 ∂xj − ∂yj ∂z¯j 2 ∂xj ∂yj
Thus (cid:16) (cid:17) (cid:16) (cid:17)
∂k = 1 ∂k i∂k , ∂k = 1 ∂h +i∂k
∂zj 2 ∂xj − ∂yj ∂z¯j 2 ∂xj ∂yj
(cid:16) (cid:17) (cid:16) (cid:17)
for any analytic function k(x,y). Thus for a complex valued function f, we
define
∂f = ∂g +i∂h, ∂f = ∂g +i∂g
∂zj ∂zj ∂zj ∂z¯j ∂z¯j ∂z¯j
We assume that g, h are non-constant polynomials. Then V is real codi-
mension two subvariety. Put
dRg(x,y) = ( ∂g ,..., ∂g , ∂g ,..., ∂g ) R2n
∂x1 ∂xn ∂y1 ∂yn ∈
dRh(x,y)= ( ∂h ,..., ∂h , ∂h,..., ∂h ) R2n
∂x1 ∂xn ∂y1 ∂yn ∈
For a complex valued mixed polynomial, we use the notation:
df(z,z¯) = (∂f ,..., ∂f ) Cn, d¯f(z,¯z) = (∂f ,..., ∂f ) Cn
∂z1 ∂zn ∈ ∂z¯1 ∂z¯n ∈
Recall that a point z V is a singular point of V if and only if two vec-
∈
tors dg(x,y), dh(x,y) are linearly dependent over R (see Milnor [4]). This
condition is not so easy to be checked, as the calculation of g(x,y), h(x,y)
from a given f(z,z¯) is not immediate. However we have
Proposition 1. The following two conditions are equivalent.
TOPOLOGY OF POLAR WEIGHTED HOMOGENEOUS HYPERSURFACES 3
(1) z V is a singular point of V and dimR(V,z)= 2n 2.
∈ −
(2) There exists a complex number α, α = 1 such that df(z,¯z) =
αd¯f(z,z¯). | |
Proof. First assume that dRg, dRh are linearly dependent at z. Suppose for
example that dg(x,y) = 0 and write dh(x,y) = tdg(x,y) for some t R.
6 ∈
This implies that
∂f = (1+ti) ∂g , ∂f =(1+ti) ∂g , thus
∂xj ∂xj ∂yj ∂yj
∂f = (1+ti) ∂g i ∂g , ∂f = (1+ti) ∂g +i ∂g .
∂zj ∂xj − ∂yj ∂z¯j ∂xj ∂yj
(cid:16) (cid:17) (cid:16) (cid:17)
Thus
df(z,z¯) = (1+ti) ∂g i ∂g ,..., ∂g i ∂g =2(1+ti)dzg(z,z¯)
∂x1 − ∂y1 ∂xn − ∂yn
(cid:16) (cid:17)
d¯f(z,z¯)= (1+ti) ∂∂xg1 +i∂∂yg1,...,∂∂xgn +i∂∂ygn = 2(1+ti)dz¯g(z,z¯)
(cid:16) (cid:17)
Here dzg = (∂∂zg1,..., ∂∂zgn) and dz¯g = (∂∂z¯g1,...,∂∂z¯gn). As g is a real valued
polynomial, using the equality dzg(x,y)= dz¯g(x,y) we get
1 ti
df(z,z¯)= − d¯f(z,z¯).
1+ti
Thus it is enough to take α = 1−ti.
1+ti
Conversely assume that df(z,z¯) = αd¯fz,z¯ for some α = a + bi with
a2+b2 = 1. Using the notations
∂g ∂g ∂g ∂g
d g = ( ,..., ), d g = ( ,..., ), etc,
x y
∂x ∂x ∂y ∂y
1 n 1 n
we get
(1 a)d g+bd g = bd h (1+a)d h
x y x y
− − −
bd g + (1 a)d g = (a+1)d h bd h.
x y x y
− − −
Solving these equations, we get
2b
dRg = (dxg,dyg) = (1 −a)2+b2dRh
−
which proves the assertion. (cid:3)
2.2. Polar weighted homogeneous hypersurfaces. Let f be a polar
weighted homogeneous polynomialof radial weight type(q ,...,q ;m )and
1 n r
of polar weight type (p ,...,p ;m ). By differentiating (1) in 1, we get
1 n p
§
(2) m f(z,¯z) = n q (∂f z + ∂f z¯)
r i=1 i ∂zi i ∂z¯i i
(3) mpf(z,z¯) = Pni=1pi(∂∂zfizi− ∂∂z¯fiz¯i).
We call these equalities Euler eqPualities. Recall that Cn has the canonical
hermitian inner product defined by
(z,w) = z w¯ + +z w¯ .
1 1 n n
···
4 M.OKA
Identifying Cn with R2n by z (x,y), the Euclidean inner product of
←→
R2n is given as (z,w)R = (z,w). Or we can also write as
ℜ
1
(z,w)R = ((z,w)+(z¯,w¯)).
2
Proposition 2. For any α = 0, the fiber F := f−1(α) is a smooth 2(n
α
6 −
1) real-dimensional manifold and it is canonically diffeomorphic to F =
1
f−1(1).
Proof. Take a point z F . We consider two particular vectors v , v
α r θ
∈ ∈
TzCn which are the tangent vectors of the respective orbits of R and S1:
v = d(r◦z) = (q z ,...,q z ),
r dr |r=1 1 1 n n
v = d(eiθ◦z) = (ip z ,...,ip z ).
θ dθ |θ=0 1 1 n n
Taking the differential of the equality
f((r,exp(iθ)) z)) = rmr exp(m θi)f(z,z¯),
p
◦
we see that dfz : TzCn TαC∗ satisfies
→
∂ ∂
df (v ) = m α , df (v ) = m
z r r z θ p
| |∂r ∂θ
where (r,θ) is the polar coordinate of C∗. This implies that f : Cn C
→
is a submersion at z. Thus F is a smooth codimension 2 submanifold. A
α
diffeomorphism ϕ :F F is simply given as ϕ(z) = (r1/mr,expiθ/mp) z
α 1 α
where α = rexp(iθ). → ◦(cid:3)
Theabove proofdoesnotworkfor α= 0. Recall thatthepolarR+ action
along the radial direction is written in real coordinates as
r (x,y)= (rq1x ,...,rqnx ,rq1y ,...,rqny ), r R+.
1 n 1 n
◦ ∈
Proposition 3. Let V = f−1(0). Assume that q > 0 for any j. Then V
j
is contractible to the origin O. If further O is an isolated singularity of V,
V O is smooth.
\{ }
Proof. A canonical deformation retract β :V V is given as β (z) = t z,
t t
→ ◦
0 t 1. (More precisely β (z) = lim β (z).) Then β = id and β is
0 t→0 t 1 V 0
≤ ≤
the contraction to O. Assume that z V O is a singular point. Consider
∈ \{ }
the decomposition into real analytic functions f(z) = g(x,y) + ih(x,y).
Using the radial R+-action, we see that
(4) g(r (x,y)) = rmrg(x,y), h(r (x,y)) = rmrh(x,y).
◦ ◦
This implies that g(x,y), h(x,y) are weighted homogeneous polynomials of
(x,y) and the Euler equality can be restated as
m g(x,y) = n p x ∂g (x,y)+y ∂g (x,y)
r j=1 j j∂xj j∂yj
(cid:16) (cid:17)
m h(x,y) = Pn p x ∂h (x,y)+y ∂h(x,y) .
r j=1 j j∂xj j∂yj
(cid:16) (cid:17)
P
TOPOLOGY OF POLAR WEIGHTED HOMOGENEOUS HYPERSURFACES 5
Differentiating the equalities (4) in r, we get
∂g ∂g ∂h ∂h
(r (x,y)) = rmr−qj (x,y), (r (x,y)) = rmr−qj (x,y).
∂x ◦ ∂x ∂x ◦ ∂x
j j j j
This implies that these differentials are also weighted homogeneous polyno-
mials of degree m q . Thus the jacobian matrix
r j
−
∂(g,h)
(r (x,y))
∂(x ,y ) ◦
(cid:18) i i (cid:19)
is the same with the jacobian matrix at z = (x,y) up to scalar multi-
plications in the column vectors by rmr−q1,...,rmr−qn,rmr−q1,...,rmr−qn
respectively. Thusanypoints oftheorbitr (x,y), r > 0aresingularpoints
◦
ofV. Thisisacontradiction totheassumptionthatO isanisolated singular
point of V, as lim r (x,y)= O. (cid:3)
r→0
◦
Proposition 4. (Transversality) Under the same assumption as in Propo-
sition 3, the sphere S = z Cn; z = τ intersects transversely with V
τ
{ ∈ | | }
for any τ > 0.
Proof. Letφ(x,y)= z 2 = n (x2+y2). Then S intersects transversely
k k j=1 j j τ
with V if and only if the gradient vectors dRg,dRh,dRφ are linearly inde-
P
pendent over R. Note that dRφ(x,y) = 2(x,y). Suppose that the sphere
Skzk is tangent to V at z = (x,y) V. Then we have for example, a linear
∈
relation dg(x,y) = αdh(x,y)+βdφ(x,y) with some α,β R. Note that
∈
the tangent vector v to the R+-oribit is tangent to V and it is written
r
v = (q x ,...,q x ,q y ,...,q y ) as a real vector. Then we have
r 1 1 n n 1 1 n n
0= dg(r◦(x,y) = n q x ∂g (x,y)+y ∂g (x,y)
dr |r=1 j=1 j j∂xj j∂yj
=P(vr(x,y(cid:16)),dg(x,y))R (cid:17)
= (vr(x,y),αdh(x,y))+(vr(x,y),βdφ(x,y))R
= 2β n q (x2+y2)
j=1 j j j
as (vr(x,y),dh(x,y))R = 0 by thPe same reason. This is the case only if
β = 0 which is impossible as V O is non-singular by Proposition 3. (cid:3)
\{ }
2.2.1. Remark. Let f(z,z¯) be a polar weighted homogeneous polynomial
with respective weights (q ,...,q ;m ) and (p ,...,p ;m ). Proposition 3
1 n r 1 n p
does not hold if the radial weights contain some negative q . Assume that
j
q 0 for any j and I := j q = 0 is not empty. Then it is easy to see
j 0 j
≥ { | }
thatf does nothave monomialwhich does notcontain anyz with i / I , as
i 0
∈
if such monomial exists, its radial degree is 0. This implies that V = f−1(0)
contains the coordinate subspace CI0 = z z = 0, i / I . We call CI0 the
i 0
{ | ∈ }
canonical retract coordinate subspace. Then Proposition 3 can be modified
as CI0 is a deformation retract of V. Of course, CI0 can be contracted
to O but this contraction is not through the action and not related to the
geometry of V.
6 M.OKA
2.2.2. Example. Consider the following examples.
g (z,¯z) = za1z¯ + +zanz¯ , a 1, and a 2( j)
1 1 2 ··· n 1 ∀ i ≥ j ≥ ∃
g (z,z¯) = za1z¯ + +zan−1z¯ +zan, a 1.
2 1 2 ··· n−1 n n ∀ i ≥
Proposition 5. (1) The radial weight vector (q ,...,q ) of g (z,z¯) is
1 n 1
semi- positive, i.e. q 0 for any j if a 1 for any i. ( j, a 2 is
j i j
≥ ≥ ∃ ≥
necessary for the existence of polar action.) It is not strictly positive
if and only ifn = 2m isevenand either (a) a = a = = a =
1 3 2m−1
···
1 or (b) a = a = = a = 0.
2 4 2m
···
In case (a) (respectively (b)), we have q = q = = q = 0
2 4 2m
···
and q 1, 0 j m 1 (resp. q = q = = q = 0 and
2j+1 1 3 2m−1
≥ ≤ ≤ − ···
q 1, 1 j m ).
2j
≥ ≤ ≤
(2) The radial weight vector (q ,...,q ) of g (z,z¯) is semi-positive. It
1 n 2
is not strictly positive if and only if a = 1. Let s be the integer
n
such that a = a = = a = 1 and a 2. Then
n n−2 n−2s n−2s−2
··· ≥
q = = q = 0 and q 1 otherwise.
n−1 n−2s+1 j
··· ≥
Proof. Wefirstconsiderg (z,z¯)= za1z¯ + +zanz¯ . Byaneasycalculation,
1 1 2 ··· n 1
using the notation a = a the normalized radial weigts (u ,...,u ) are
i+n i 1 n
given as
u = 1 m−1(a 1)a a , n= 2m
j a1···an−1 i=0 j+2i+1− j+2i+2··· j+n−1
u = 1 1+ Pm−1(a 1)a a , n = 2m+1
j a1···an+1 i=0 j+2i+1− j+2i+2··· j+n−1
(cid:16) (cid:17)
P
and the assertion follows immediately from this expression.
Next we consider g2(z,z¯) = z1a1z¯2 +···+znan−1z¯n +znan. Then the nor-
malized radial weigts (u ,...,u ) are given as
1 n
u = 1 1 + +( 1)n−j 1
j aj − ajaj+1 ··· − ajaj+1···an
aj+1−1 + + an−1 , n j : odd
ajaj+1 ··· ajaj+1···an −
= aj+1−1 + + an−1−1 + 1
 ajaj+1 ··· ajaj+1···an−1 ajaj+1···an

 n j :even
−
As ai 1, the assertion follows from the above expression. (cid:3)
≥
2.3. Simplicial mixed polynomial. Let f(z,z¯) = sj=1cjznjz¯mj be a
mixed polynomial. Here we assume that c ,...,c = 0. Put
1 s
6 P
s
fˆ(w) := c wnj−mj.
j
j=1
X
We call fˆthe the associated Laurent polynomial. This polynomial plays an
important role for the determination of the topology of the hypersurface
F = f−1(1). Note that
TOPOLOGY OF POLAR WEIGHTED HOMOGENEOUS HYPERSURFACES 7
Proposition 6. If f(z,z¯) is a polar weighted homogeneous polynomial of
polar weight type (p ,...,p ;m ), fˆ(w) is also a weighted homogeneous Lau-
1 n p
rent polynomial of type (p ,...,p ;m ) in the complex variables w ,...,w .
1 n p 1 n
A mixed polynomial f(z,z¯) is called simplicial if the exponent vectors
n m j = 1,...,a are linearly independent in Zn respectively. In
j j
{ ± | }
particular, simplicity implies that s n. When s = n, we say that f is full.
≤
Put n = (n ,...,n ), m = (m ,...,m ) in Nn. Assume that s n.
j j,1 j,n j j,1 j,n
≤
Consider two integral matrix N = (n ) and M = (m ) wherethe k-th row
i,j i,j
vectors are n , m respectively.
k k
Lemma 7. Let f(z,¯z) be a mixed polynomial as above. If f(z,z¯) is simpli-
cial, then f(z,¯z) is a polar weighted homogeneous polynomial. In the case
s= n, f(z,¯z) is simplicial if and only if det(N M) = 0.
± 6
Proof. First we assume that s = n and consider the system of linear equa-
tions
(n +m )u + +(n +m )u = 1
1,1 1,1 1 1,n 1,n n
···
(5)

···
 (nn,1+mn,1)u1+ +(nn,n+mn,n)un = 1
···
 (n1,1 m1,1)v1+ +(n1,n m1,n)vn =1
− ··· −
(6)

···
 (nn,1 mn,1)v1+ +(nn,n mn,n)vn = 1
− ··· −
It is easy to see that equations (5) and (6) have solutions if det N M = 0

± 6
which is equivalent for f to be simplicial by definition. Note that the solu-
tions (u ,...,u ) and (v ,...,v ) are rational numbers. We call them the
1 n 1 n
normalized radial (respectively polar) weights. Now let m , m be the least
r p
common multiple of the denominators of u ,...,u and v ,...,v respec-
1 n 1 n
tively. Then the weights are given as q = u m , p = v m , j = 1,...,n
j j r j j p
respectively.
Now suppose that s < n. It is easy to choose positive integral vec-
tors n , j = s + 1,...,n (and put m = 0, j = s + 1,...,n) such that
j j
det(N˜ M˜) = 0, where N˜ and M˜ are n n-matrices adding (n s) row
± 6 × −
vectors n ,...,n . Then the assertion follows from the case s = n. This
s+1 n
corresponds to considering the mixed polynomial:
s n
n m n
f(z,z¯)= c z j¯z j +0 z j.
j j ×
j=1 j=s+1
X X
(cid:3)
2.3.1. Example. Let
fa,b(z,z¯) = z1a1z¯2b1 +···+znanz¯1bn, ai,bi ≥ 1, ∀i
k(z,z¯)= zd(z¯ +z¯ )+ +zd(z¯ +z¯ ), d 2.
1 1 2 ··· n n 1 ≥
8 M.OKA
The associated Laurent polynomials are
fa,b(w)= w1a1w2−b1 +···+wnanw1−bn
kˆ(w) = wd(1/w +1/w )+ +wd(1/w +1/w ).
d1 1 2 ··· n n 1
Corollary 8. For the polynomial f , the following conditions are equiva-
a,b
lent.
(1) f is simplicial.
a,b
(2) f is a polar weighted homogeneous polynomial.
a,b
(3) (SC) a a = b b .
1 n 1 n
··· 6 ···
Proof. The assertion follows from the equality:
a 0 b
1 n
··· ±
b a 0
1 2
det(n±m)= det ±... ... ·.·.·. ... 
 
 0 bn−1 an 
 ··· ± 
 
a a ...a +( 1)n−1b b ...b forn+m
1 2 n 1 2 n
= −
( a1a2...an b1b2...bn for n m.
− −
The polynomial k(z,z¯) is a polar weighted homogeneous polynomial with
respective weight types (1,...,1;d+1) and (1,...,1;d 1). However it is
not simplicial. − (cid:3)
Now we consider an example which does not satisfy the simplicial con-
dition (SC) of Corollary 8: φ := zaz¯a + + zaz¯a. This does not have
a 1 1 ··· n n
any polar action as they are polynomials of z 2,..., z 2 and it takes
1 n
| | | |
only non-negative values. Note also that φ−1(1) is real codimension 1 as
a
φ (x,y)= n (x2+y2)a.
a j=1 j
As a typical simplicial polar weighted polynomial, we consider again the
P
following two polar weighted polynomials.
g (z,¯z) = za1z¯ + +zanz¯ , a 1, and a 2( j)
1 1 2 ··· n 1 ∀ i ≥ j ≥ ∃
g (z,z¯) = za1z¯ + +zan−1z¯ +zan, a 1.
2 1 2 ··· n−1 n n ∀ i ≥
The polynomial g (z,z¯) with a 2, ( i) is a special case of σ-twisted
1 i
≥ ∀
Brieskorn polynomial and has been studied intensively ([12]). In out case,
we only assume a 2 for some i. The existence of i with a 2 is the
i i
≥ ≥
condition for the existence of polar action. We consider two hypersurfaces
definedby V = g−1(0) for i = 1,2. Thecondition for a hypersurfacedefined
i i
by apolar weighted homogeneous polynomialto have an isolated singularity
ismorecomplicatedthanthatofthesingularitydefinedbyacomplexanaytic
hypersurface. For the above examples, we assert the following.
Proposition 9. For V , V , we have the following criterion.
1 2
(1) V C∗n, i = 1,2 are non-singular.
i
∩
(2) V = g−1(0) has no singularity outside of the origin if and only if
1 1
one of the following conditions is satisfied.
TOPOLOGY OF POLAR WEIGHTED HOMOGENEOUS HYPERSURFACES 9
(a) n is odd.
(b) n is even and there are (at least) two indices i,j(i < j) such
that a , a 2 and j i is odd.
i j
≥ −
(3) V = g−1(0) has no singularity outside of the origin if and only if
2 2
one of the following conditions is satisfied.
(a) a 2.
n
≥
(b) a = 1, n= 2m+1 is odd and a = 1 for any 1 j m+1.
n 2j−1
≤ ≤
Proof. We use Proposition 1. So assume that
(♯) : df(z,z¯) = αd¯f(z,z¯), α = 1.
| |
(1) We consider V . Suppose z V C∗n is a singular point. Note that
1 1
∈ ∩
dfz = (a1z1a1−1z¯2,··· ,anznan−1z¯1), d¯f(z,z¯)= (znan,z1a1,...,znan−−11)
(♯) implies that
(7) a z¯aj−1z¯ = αzaj−1, j = 1,...,n, α S1.
j j j+1 j−1 ∃ ∈
In this case, indices should be understood to be integers modulo n. So
z = z , and so on. If z C∗n, multiplying the absolute values of the
n+1 1
bonit=h1|szidi|eai.of the above equal∈ity, this give a contradiction: Qni=1ai|zi|ai =
Now we consider the smoothness on V O . Assume that z is a singular
1
Q \{ }
point of V O . For simplicity, we may assume that a 2 as g is
1 n 1
\{ } ≥
symmetric with the permutation i i+1.
Assume that z = 0. Then the →(ι+1)-th component of d¯f is non-zero.
ι
6
Thusby(♯), (ι+1)-th componentofdf isalsonon-zero. Thatis,zaι−1z¯ =
ι+1 ι+2 6
0. In particular, z = 0. We repeatthe same argumentand get a sequence
ι+2
6
of non-zero components z ,z ,.... Thus we arrive to the conclusion that
ι ι+2
either z = 0 (if n ι is odd ) or z = 0 ( if n ι is even ).
n−1 n
6 − 6 −
–If n ι is odd and z = 0, the last component of df is non-zero and we
n−1
− 6
have z ,z = 0 as we have assumed that a 2. This creates two non-zero
n 1 n
6 ≥
sequencez ,z ,z ,... andz ,z ,.... Thusweconcludethatz C∗n, which
n 2 4 1 3
∈
is impossible by the first argument.
–If n ι is even, z ,z ,...,z = 0. Thus we see that the first compo-
ι ι+2 n
nent of −d¯f is non-zero. By the sam6 e argument, we get non-zero sequence
z ,z ,....
2 4
Thus to show that z C∗n, it is enough to show that z = 0.
n−1
∈ 6
(a) Assumefirst n is odd. If ι is even, then we see that z ,z ,...,z = 0
ι ι+2 n−1
6
and we are done.
If ι is odd, we get z = 0, which implies the first component of d¯f(z,z¯) is
n
6
non-zero. Thus as the second round, we have non-zero sequence z ,z ,...
2 4
which contains z . Thus we are done.
n−1
(b) Now we assume that n is even but there is another integer 1 i < n
≤
such that a 2 and a 2 and i is odd. If ι is odd, we have shown that
i n
≥ ≥
z C∗n.
∈
10 M.OKA
If ι is even, we get z = 0 and thus z = 0. Then the sequence z ,z ,...
n 2 2 4
6 6
contains z . As a 2, looking at the i-th component of df, we get
i−1 i
≥
z z = 0. Thus we get a non-zero sequence z ,z ,... which contains
i i+1 i i+2
· 6
z , and we are done.
n−1
Now to show that one of the conditions (a) or (b) is necessary, we assume
that n is even and a = 1 for any odd ν and a 2. Thus putting n = 2m,
ν n
≥
f = (z z¯ +za2z¯ )+ +(z z¯ +za2mz¯ ).
1 2 2 3 ··· 2m−1 2m 2m 1
Consider the subvariety z = z = = z = 0. Then
1 3 n−1
···
df(z,z¯) = (z¯ ,0,z¯ ,0,...,z¯ ,0), d¯f(z,¯z) = (zan,0,...,za2m−2,0)
2 4 2m n 2m−2
the condition (♯) is written as
(♯) z = αzan, z = αza2, ,z = αza2m−2
2 n 4 2 ··· 2m 2m−2
which has real one-dimensional solution
z2j = αβjuγj ( j = 1,...,m), αβmuγma2m−1 = 1
j−1
β = 1+ a a a , γ = a a a
j i=1 2(j−1) 2(j−2)··· 2(j−i) j 2 4··· 2(j−1)
(2) We consider the case V . We will see first V C∗n is non-singular. Take
P 2 2
∩
a singular point of V . Then we have some α S1 so that
2
∈
(♯) : df(z,z¯) = αd¯f(z,¯z).
As we have
df(z,z¯) = (a za1−1z¯ , ,a zan−1−1z¯ ,a zan−1),
1 1 2 ··· n−1 n−1 n n n
d¯f(z,z¯) =(0,za1,...,zan−1)
1 n−1
we see that (♯) implies that za1−1z¯ = 0. Thus there are no singularities
1 2
on V C∗n. Suppose that z = 0 for some ι. If ι < n 1, this implies
2 ι
(ι+1)-∩th component of d¯f(z,z¯)6is non-zero. Thus (♯) impli−es that (ι+1)-th
component of df is non-zero. In particular, z is non-zero. (Of course,
ι+2
z = 0 if a > 1.) Repeating this argument, we arrive to the conclusion:
ι+1 ι+1
6
either z or z is non zero.
n−1 n
First assume that a 2. Comparing the last components of df(z,z¯)
n
and d¯f(z,¯z), we observe≥that z and z are both non-zero. Now we go
n−1 n
in the reverse direction. As the (n 1)-th component of df is non-zero,
the corresponding (n 1)-th compone−nt zan−2 of d¯f(z,z¯) is non-zero. Then
− n−2
(n 2)-th component of df(z,z¯) is non-zero. Going downword, we see that
−
z C∗n. However this is impossible, as we have already seen above.
∈
Next we assume that a = 1 and n is odd and a = 1 for any j. Note
n 2j−1
that the last component of df(z,z¯) is 1. Thus z = 0. If z = 0, we
n−1 n
6 6
get a contradiction as above z C∗n. Thus we may assume that z = 0.
n
Comparing (2j)-components of ∈df(z,z¯) and αd¯f(z,z¯), we get
z = 0, z = αza2,...,z = zan−3
2 4 2 n−1 n−3
which has no solution with z = 0.
n−1
6
Now we show that the condition (a) or (b) in (3) is necessary.