From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Translation (Re: invarianten
Date: 28 Sep 2001 04:46:34 GMT
Newsgroups: sci.math
Summary: Invariants of one-variable quartic polynomials under GL_2

Someone wrote, and someone else translated to English, this question:
> Do any of you know what the invariants of an equation are?
> My problem is this:
> A polynomial of degree 4 is given:
> ax^4 + 4bx^3 + 6cx^2+ 4dx + a
[Clearly the constant terms is meant to be  e ]
> and the invariants are
> g = ae - 4bd + 3c^2
> h = ace + 2bcd - c^3 - ad^2 - b^2 e
> How does one get that?

In article <600799ac.0109241726.7e6dbf3b@posting.google.com>,
Midianight <neil.fitzgerald@ic.ac.uk> wrote:

>Anyway, it looks like those expressions "g" and "h" are invariant
>under the transformation x -> x + k.

More generally you might consider the action of  GL(2,R), in which we
replace  x  by  (p x + q)/(r x + s).  You'll get a denominator of
(rx+s)^4, of course, but the corresponding numerator is another 
quartic, as the previous poster described (for the special case
p=s=1,r=0 ). Guess what: the two things called "invariants" are
simply multipled by powers of  p s - q r,  the determinant of the
corresponding matrix. So if you're willing to ignore those pesky
denominators,  g  and  h   (more commonly called  I  and  J ) are
invariants. (Actually what  h  is _really_ called is the
"catalecticant", which is worth knowing just for effect.)

And indeed ignoring them is a perfectly reasonable thing to do, in
some cases. For example, if you're interested in the set of roots of
f,  you see that these are moved under the corresponding Moebius 
transformation of the complex plane to the roots of the numerator
quartic -- the denominator won't affect the roots. If you're working
with the elliptic curve  y^2 = f(x), the additional factor in the
denominator can be swallowed up by a corresponding rational change
in  y. 

>Notice that those two expressions are clearly not the only ones with
>that property, because you can (for instance) add or multiply them
>together to get a third expression that is also "invariant".  Also,
>"a" on its own is an "invariant" in this sense.

Yes, working within the polynomial ring  Z[a,b,c,d,e]  we now have
two elements  g  and  h  which are invariant under the action of
PGL(2,R), but of course the subring they generate will also be
(element-wise) invariant. That is in fact the key defining property
of  g  and  h : they are the unique (up to scalar multiples) generators
for the ring of invariants.

The setting for this is, naturally, "Invariant Theory". Clearly it
ties in with Galois theory, but GL(2,R) is such a key group that this
ties in with many other disciplines too. [In fact, GL(2,R) is so
important, someone ought to write a book about it.] 

dave