At least if the input is a polynomial that question doesn't arise. Of
course if you want to manually specify the coefficients then the
conventions can be tricky. So far I only allow an actual polynomial as
input:
sage: R.<x,y,z,a30,a21,a12,a03,a20,a11,a02,a10,a01,a00> = QQ[]
sage: p = ( a30*x^3 + a21*x^2*y + a12*x*y^2 + a03*y^3 +
a20*x^2*z +
... a11*x*y*z + a02*y^2*z + a10*x*z^2 + a01*y*z^2 +
a00*z^3 )
sage: invariant_theory.ternary_cubic(p, x,y,z).coeffs()
(a30, a03, a00, a21, a20, a12, a02, a10, a01, a11)
sage: invariant_theory.ternary_cubic(p, x,y,z).scaled_coeffs()
(a30, a03, a00, 1/3*a21, 1/3*a20, 1/3*a12, 1/3*a02, 1/3*a10,
1/3*a01, 1/6*a11)
On Tuesday, September 11, 2012 1:21:29 PM UTC+1, John Cremona wrote:
>
> On 11 September 2012 13:13, Volker Braun <[email protected] <javascript:>>
> wrote:
> > On Tuesday, September 11, 2012 1:03:43 PM UTC+1, John Cremona wrote:
> >>
> >> Volker, will you also include what I call seminvariants?
> >
> >
> > Whats a semi-invariant to you? The leading term of a covariant?
> >
>
> Precisely.
>
> > sage: R.<a0, a1, a2, a3, a4, x0, x1> = QQ[]
> > sage: quadric =
> >
> invariant_theory.binary_quartic(a0*x1^4+4*a1*x0*x1^3+6*a2*x0^2*x1^2+4*a3*x0^3*x1+a4*x0^4,
>
>
> > x0, x1)
> > sage: quadric.g_covariant().lt()
> > a3^2*x0^4
> >
> > I don't think they should be methods of polynomials; Is x a linear
> > homogeneous or quadratic inhomogeneous term?
> >
>
> OK, as long as the covariants are there I can extract them.
>
> Your example above reveals another little issue: you put in the
> binomial coeffients, while sometimes one prefers not to. I think the
> fancy way to say this is that there is more than one integral
> representation for the same rational representation (OK, so there are
> fancier ways to say it than that). So it would be nice if that could
> be speicified by the user (and I will not argue as to which should be
> the default).
>
> John
>
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