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f that guarantees that V(f) will be a nice space? First of all, note that this is a
local
question that has to do with what neighborhoods of points look like.
8.3.4. Theorem.
(The Implicit Parameterization Theorem) Let f :
R
n
Æ
R
m
be a dif-
ferentiable map. If f(
x
) = (f
1
(
x
),...,f
m
(
x
)) and if the m ¥ n Jacobian matrix f¢=(∂f
i
/∂x
j
)
has constant rank k < n on an open set containing V(f), then V(f) is a differentiable
manifold in
R
n
of dimension n - k.
Proof.
We need to show that we can parameterize a neighborhood of an arbitrary
point
x
0
in V(f). Without loss of generality assume that
x
0
is the origin. By Theorem
4.4.5 we can get local diffeomorphisms g and h of neighborhoods of the origin in
R
n
and
R
m
, respectively, so that
(
(
(
)
)
)
=
(
)
hfgx
,...,
x
x
,...,
x
, ,..., .
0
0
1
n
1
k
The map g can now be used to define the parameterization of a neighborhood of
x
0
that we need.
To analyze the set V(f) for the function f(x,y,z) = z - x
2
.
8.3.5. Example.
Solution.
We have that
fxyz
¢
(
,,
)
=
(
201
x
,, .
)
Clearly, f¢ has rank 1 on the zero set V(f) and so by Theorem 8.3.4 V(f) is a smooth
surface (two-dimensional manifold). In fact, V(f) is the parabolic “trough” shown in
Figure 8.9(a).
To analyze the set V(f) for the function f (x,y,z) = (x
2
+ 3y
2
+ 2z
2
8.3.6. Example.
-
1,z).
z
z
y
x
2
+ 3y
2
+ 2z
2
- 1 = 0
z = 0
y
x
x
z- x
2
= 0
(a)
(b)
Figure 8.9.
The varieties in Examples 8.3.5 and 8.3.6.
