| So all we have to do
is to compute a Groebner basis G w.r.t. lex order where
every ti is greater than
every xi: the elements
of G not involving t1,...,
tm define the
smallest variety containing the parametrization.
Tips.
Since we only want to eliminate the first m variables
t1,...,
tm
and we do not care about the others, we can use the more efficient m-th elimination order.
The Groebner applet
now supports this ordering so you can redo the all the
computations here and in the other pages using this type
of order.
Examples.
- The twisted cubic. Consider the
subset of points in R3
defined by
x = t
y = t2
z = t3.
The reduced Groebner basis for
I = <x - t,
y - t2,
z - t3>
w.r.t. lex order with
t>z>y>x is
G = {t - x, y - x2,
z - x3}, hence
V = V(y - x2,
z - x3) is the smallest
variety containing the parametrization.
But does the parametrization fill up all of V?
Over C, we could use the Extension Theorem
to prove that the partial solutions
(x, y, z)
of V(y - x2,
z - x3)
extend to the complete solutions
(t, x, y, z)
since the coefficient of t in
t - x is 1.
Over R, we must prove in addition that t
is real. This is obvious: given a real x,
we can find a real t from the equation
t - x = 0. Hence the
parametrization fills up the entire V. Here
is its drawing:
- The tangent surface of the
twisted cubic. Consider the polynomial
parametrization in R3
x = t + u
y = t2 + 2tu
z = t3 + 3t2u
of the tangent surface to the twisted
cubic. The reduced Groebner basis for
I =
<x - t - u, y - t2
- 2tu, z - t3
- 3t2u>
w.r.t. lex order with
t>u>x>y>z is
g1
= t + u - x
g2
= u2
- x2 + y
g3
= ux2
- uy - x3
+ 3/2xy
- 1/2z
g4
= uxy - uz - x2y
- xz + 2y2
g5
= uxz - uy2
+ x2z - 1/2xy2
- 1/2yz
g6
= uy3
- uz2 -
2x2yz + 1/2xy3
- xz2 + 5/2y2z
g7
= x3z
- 3/4x2y2
- 3/2xyz
+ y3 + 1/4z2.
So V(g7)
is the smallest variety containing the parametrization. We
will prove that equality occurs by showing
that all partial solutions
(x, y, z) of
V(g7)
= V(I2)
extend to the complete solutions
(t, u, x, y, z) of
V(I).
Over C it is simply a matter of using
the Extension Theorem one coordinate at a
time. First we go from the second elimination
ideal I2
to the first elimination ideal
I1
= <g2,..., g7>:
since the leading coefficient of u in
g2
is 1, (x, y, z) of
V(I2)
always extends to
(u, x, y, z) of
V(I1); then, we can
extend (u, x, y, z) of
V(I1) to
(t, u, x, y, z) of V(I)
just as easy because the leading coefficient of t
in g1
is 1.
Over R, we must prove in addition that
t and u are real. From
g2
= 0 we get
u2
= x2 - y; then,
from g3
= 0 we have
u3
= u(x2 - y)
= x3 - 3/2xy
+ 1/2z,
which says that u is real. It
follows that t is real too, since
from g1
= 0 is
t = x - u.
Thus we have proved that the tangent surface
to the twisted cubic equals
x3z
- 3/4x2y2
- 3/2xyz
+ y3 + 1/4z2
= 0 both in C3
and in R3.
 |
| The
tangent surface of the
twisted cubic |
- Let S be the parametric surface defined by
x = tu
y = t2
z = u2.
The reduced Groebner basis for
I =
<x - tu, y - t2,
z - u2>
w.r.t. lex order with t>u>x>y>z is
g1
= x2
- zy
g2
= u2
- z
g3
= tz - xu
g4
= xt - uy
g5
= ut - x
g6
= t2
- y.
So V =
V(I2)
= V(g1)
is the smallest variety containing S. Using the Extension
Theorem, it is easy to check that V = S
over C. Over R, we can see
that S covers half of V
(where y and z are both
non negative). The other half is
covered by the parametrization
x = tu
y = -t2
z = -u2.
Here is a picture of V:
We have so far considered polynomial parametrizations.
Some problems arise when we have rational
parametrizations. To illustrate what happens, let us
consider the set of points S defined by the following
rational parametrization:
x = u2/v
y = v2/u
z = u.
The equation
z3
= x2y is a consequence of
the above equations, hence we have
S C V, where
V
= V(z3 - x2y).
If we clear denominators in the above equations and
compute the reduced Groebner basis for
I
= <vx - u2,
uy - v2,
z - u> w.r.t. lex order with
u>v>x>y>z we get
V(I2)
= V(z(z3
- x2y)): this is not the smallest
variety containing S since it is strictly greater than V (it splits
into V and V(z)).
To avoid this difficulty, we need an extra dimension to
control the denominators.
|
Theorem 2 (Rational
Implicitization)[3]. Let K
be an infinite field and let S be the subset of points in
Kn
defined by the rational parametrization
| |
f1(t1, t2,...,
tm) |
| x1 = |
--------------------- |
| |
h1(t1, t2,...,
tm) |
| . |
|
| . |
|
| . |
|
| |
fn(t1, t2,...,
tm) |
| xn
= |
---------------------. |
| |
hn(t1, t2,...,
tm) |
Let J
be the ideal <h1x1-f1,
h2x2-f2,
..., hnxn-fn,
1-hw> in
K[w,t1,...,tm,x1,...,xn]
where h = h1h2
··· hn and let
Jm+1
= J I
K[x1,...,xn]
be the (m+1)th elimination ideal. Then
V(Jm+1)
is the smallest variety containing S.
|
| Let us apply the algorithm
suggested by Theorem 2 to the previous rational
parametrization S. We compute the Groebner basis of
J = <vx - u2,
uy - v2, z - u,
1 - uvw> w.r.t. lex order and
w>u>v>x>y>z to get J3
= J I
K[x,y,z] = z3
- x2y. Hence, Theorem 2
assures that V(J3)
= V(z3 - x2y)
is the smallest variety
containing S.
An exception to the general rule is when
in the rational parametrization there is only one
parameter t.
In this case the ideal
<h1x1-f1,
h2x2-f2,
..., hnxn-fn>
obtained by clearing denominators gives the right answer. We need only to
check that hi(t) and
fi(t) have no common roots.
Example.
Consider the following parametric rapresentation
of a curve in R2
| |
t |
| x= |
------- |
| |
t+1 |
| |
|
| |
t2-1 |
| y= |
------- |
| |
t2 |
The reduced Groebner basis for
I = <x(t+1)
- t, yt2 - t2
+ 1> w.r.t. lex order with t>x>y>z
is G = {yx2
- 2x + 1, t(y - 1) - xy + 1, t(x - 1) + x}, hence
V = V(yx2
- 2x + 1) is the smallest variety containing the
parametrization. Futhermore, the Extension Theorem
tells us that the parametric rapresentation fills up
all of the variety V
except for the point (1, 1).
You will see more examples following the links on the
left to Jeometry & Groebner applet.
|