Date: Wed, 21 Aug 96 00:50:22 CDT
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: algebraic geometry question
Newsgroups: sci.math.research
In article <3207EBD1.1A94@seidel.ece.ucsb.edu> you write:
>Consider the map
>
> f : R^m -> R^n
> z |-> [z'*F1*z z'*F2*z ... z'*Fn*z]'
>
>where {F1, ..., Fn} are fixed m x m symmetric (but not sign definite)
>matrix parameters. My question is, what does the image of a subspace
>of R^m look like in R^n? Numerical experiments seem to reveal that it
>looks like the surface of a cone, but I'd like to show this formally.
>It is trivial to show that it must be a cone; that it is a surface
>I guess follows when dim(subspace) < n. What I really need, though,
>is a simple algebraic characterization of the image in terms of a
>basis for the subspace of R^m and some properties of the matrices
>{F1, ..., Fn}, say their eigendecompositions or numerical ranges.
>
I'm not sure what you mean by 'cone'. Clearly it's true that if
v is in the image, then so is r.v for any non-negative scalar r --
is that what you meant?
I'm also not really sure what you hope to say, even in the case where
the subspace of the domain is all of R^m. In that case,
I interpret your question to mean, what is the projection
to {0} x R^n of the variety in R^m x R^n determined by the equations
y1 = Q1(x1, ..., xm)
...
yn = Qn(x1, ..., xm)
where the Q_i are n arbitrary quadratic forms in m variables.
I don;t know how to make use of the quadratic nature of these Q_i, but
in general there is a tool for looking at such projections: for a
specific set of n equations in n+m unknowns, one need only compute
a suitable Groebner basis of the ideal generated by the equations, and
truncate. I won't try to clarify what that means except to say that
there are reasonably elementary books on Groebner bases (which will
provide details) and packages for computer algebra systems which will
perform the computations.
I ran a couple of samples through maple. For example, if I give the input
> eq1:=x1^2+x1*x2-x3^2 - y1;
> eq2:=x1*x3 + 2*x2*x3 + 5*x3^2 - y2;
> eq3:= x1^2+7*x2^2-9*x3^2 - y3;
> eq4:=x1*x2 - 8*x2*x3 + 11*x3*x1 - y4;
and then the commands
> with(grobner);
> gbasis({eq1, eq2, eq3, eq4}, [x1, x2, x3, y1, y2, y3, y4]);
then I get some long output which includes one polynomial involving y alone:
4 3 3
190592 y4 + 5715978 y2 y4 + 6113912 y2 y3
3 2 2
- 41455100 y2 y1 + 114668 y1 y3 y4 - 16740674 y1 y4 y2
2 2 2
+ 5529592 y1 y3 y2 + 15081806 y1 y4 y2 - 42520656 y1 y3 y2
2 2 2 2
- 6398884 y1 y2 y4 - 1310658 y1 y3 y2 - 4886696 y2 y4
2 2 3
- 303182 y1 y3 y4 - 42472 y1 y3 y4 + 2311724 y1 y2 y3 y4 + 876928 y1 y4
2 2 4 4 2 2
+ 59071 y1 y3 + 1543537 y2 + 142029 y1 + 103026370 y1 y2
2 2 3 3 3
+ 928912 y1 y4 - 3226404 y1 y2 - 196656 y1 y3 - 168486 y1 y4
3 2 2 2 2
- 236208 y2 y4 + 1023532 y2 y3 y4 + 3698667 y3 y2 - 2034030 y2 y3 y4
2 2 2 3 3
+ 3429528 y2 y3 y4 + 14509 y3 y4 - 4444 y1 y3 + 4444 y3 y4
3 3
- 144520 y3 y4 + 17776 y3 y2
The zero set of this polynomial includes the image of the four quadratic
forms in the three variables. If you had intended to use polynomials of
_complex_ variables, this would be precisely (the Zariski closure of) the
image. Clearly this variety is too big in the real case, though, since
for example it includes whole lines through the origin (rather than
simple rays) as it is a homogeneous polynomial.
I don't know what else you want to know about this image. It's irreducible,
a hypersurface (degree 3 in 4-dimensional affine space), it's homogeneous,
and thus essentially a surface in 3-dimensional projective space.
dave