10 S. BERHANU
In thew coordinates, the defining function of Misgiven by p(w', wn) = p(w', Wn
+
2J=l w;i). We will show that if Wn = Un
+
J=T Vn is sufficiently small, p(O', Wn) ::; 0
whenever Vn
2::
u;i. Indeed,
p(O', Wn)
p(O', Wn
+
2J=lw;)
q?(O',~(wn
+
2J=Tw;)) ';S(wn
+
2J=Tw;)
q?(O', Un 4unvn) Vn 2u;
+
2v;
4 (
1 4
)4
2
2
2
2
Un  Vn  Vn  Un
+
Vn.
Suppose now Vn
2::
u;i. Then for Wn small, if Vn
2::
0, then vn
+
2v;i 0 and
so p(O', wn)::; 0; if Vn ::; 0, since Vn
2::
u;;, we have v;; ::;
u~
and so again for Wn
small, p(O', wn)::;
0.
We now let
p
= p and we have new coordinates z satisfying:
Yn
2::
x;
=
p(O',
Zn)::;
0.
The biholomorphic map G also leaves the form off unchanged, therefore in the
new coordinates we may still write
J(z',
Zn) = b(z', Zn)(z;
+
a!(z')zn
+
ao(z')),
b(O)
f.
0 and aj (0') = 0. We may also assume that b(O) =
1.
Let h(z) = b(O, z)z
2
.
We claim that there is a holomorphic function a(z) such
that a(O) = 1 and h(z) = h(a(z)z). To see this, consider the holomorphic function
F(z, a)= h(az)
~
h(z)
z
which is defined near (0, 1). F(O, 1) = 0 and
~~
(0, 1) = 2. Therefore, by the
implicit function theorem, there is t 0 and a holomorphic function a(z) defined
on lzl t such that a(O) = 1 and F(z,a(z)) = 0. Hence h(z) = h(a(z)z). From
the equation F(z, a(z)) = 0, we have a'(O) =
~ab
(0).
UZn
Let
W,
= {(O',zn):
Yn
2::
x;}n{(O',zn): lznl t}.
We know that since f(W,)
~
f(rJ),
0,
the neighborhood of 0 in M, f is open at 0
whenever the following holds: if (0', z)
rf.
W,, lzl t, then a(z)z E W,.
Suppose lzl t and (0', z)
rf.
W,.
If
z = x
+
J=Ty, y x
2
.
Notice that
{jb
2 3
a(z)z
=
z (O)z
+
O(z ).
OZn
Letting
"'ab
(0) = s
+
J=Tt, we have:
UZn
';S(a(z)z) y 2sxy t(x
2

y2
)
+
O(z
3
)
_!!_
+ ~x 2 
2sxy t(x
2

y2
)
+
O(z
3
)
2 2
~
+
(ty 2sx)y
+
(~
t)x
2
+
O(z 3
).
Now
:::.JL
0 since y x
2
and the term (ty 2sx)y can be absorbed in
:::.jL
if tis
small enough. Hence if
{jb
1
It I=
P~(O)I
::; 
2
,
un
then ';S(a(z)z) 0, ie. a(z)z
E
W,, implying that
f
is open at 0.
In thew coordinates, the defining function of Misgiven by p(w', wn) = p(w', Wn
+
2J=l w;i). We will show that if Wn = Un
+
J=T Vn is sufficiently small, p(O', Wn) ::; 0
whenever Vn
2::
u;i. Indeed,
p(O', Wn)
p(O', Wn
+
2J=lw;)
q?(O',~(wn
+
2J=Tw;)) ';S(wn
+
2J=Tw;)
q?(O', Un 4unvn) Vn 2u;
+
2v;
4 (
1 4
)4
2
2
2
2
Un  Vn  Vn  Un
+
Vn.
Suppose now Vn
2::
u;i. Then for Wn small, if Vn
2::
0, then vn
+
2v;i 0 and
so p(O', wn)::; 0; if Vn ::; 0, since Vn
2::
u;;, we have v;; ::;
u~
and so again for Wn
small, p(O', wn)::;
0.
We now let
p
= p and we have new coordinates z satisfying:
Yn
2::
x;
=
p(O',
Zn)::;
0.
The biholomorphic map G also leaves the form off unchanged, therefore in the
new coordinates we may still write
J(z',
Zn) = b(z', Zn)(z;
+
a!(z')zn
+
ao(z')),
b(O)
f.
0 and aj (0') = 0. We may also assume that b(O) =
1.
Let h(z) = b(O, z)z
2
.
We claim that there is a holomorphic function a(z) such
that a(O) = 1 and h(z) = h(a(z)z). To see this, consider the holomorphic function
F(z, a)= h(az)
~
h(z)
z
which is defined near (0, 1). F(O, 1) = 0 and
~~
(0, 1) = 2. Therefore, by the
implicit function theorem, there is t 0 and a holomorphic function a(z) defined
on lzl t such that a(O) = 1 and F(z,a(z)) = 0. Hence h(z) = h(a(z)z). From
the equation F(z, a(z)) = 0, we have a'(O) =
~ab
(0).
UZn
Let
W,
= {(O',zn):
Yn
2::
x;}n{(O',zn): lznl t}.
We know that since f(W,)
~
f(rJ),
0,
the neighborhood of 0 in M, f is open at 0
whenever the following holds: if (0', z)
rf.
W,, lzl t, then a(z)z E W,.
Suppose lzl t and (0', z)
rf.
W,.
If
z = x
+
J=Ty, y x
2
.
Notice that
{jb
2 3
a(z)z
=
z (O)z
+
O(z ).
OZn
Letting
"'ab
(0) = s
+
J=Tt, we have:
UZn
';S(a(z)z) y 2sxy t(x
2

y2
)
+
O(z
3
)
_!!_
+ ~x 2 
2sxy t(x
2

y2
)
+
O(z
3
)
2 2
~
+
(ty 2sx)y
+
(~
t)x
2
+
O(z 3
).
Now
:::.JL
0 since y x
2
and the term (ty 2sx)y can be absorbed in
:::.jL
if tis
small enough. Hence if
{jb
1
It I=
P~(O)I
::; 
2
,
un
then ';S(a(z)z) 0, ie. a(z)z
E
W,, implying that
f
is open at 0.