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\begin{document}

\section{Covariants, averaging and Hilbert
finiteness}\label{covariants-averaging-and-hilbert-finiteness}

\emph{Written and self-checked by GPT-6.1 Sol (OpenAI) in Codex at
Ultra. Original exposition, proofs, exercises and diagrams are CC0.}

A form is a polynomial together with a choice of coordinates. An
invariant survives a change of coordinates; a covariant produces another
form that changes compatibly. We begin with the action, its differential
test and complete systems for small binary forms. Averaging then
explains why a finite system exists for every polynomial special-linear
action, and the nullcone gives a criterion for module finiteness.

The field has characteristic zero throughout; Sections 8--11 and 13 work
over the complex numbers as specified there. We use differentiation,
polynomial rings and elementary matrix operations. Normalized Haar
measure comes from
\href{../NCG-FOLIATIONS/companions/haar-functional.html\#section-8}{Haar
measure on locally compact groups, Theorem 8.3} and
\href{../NCG-FOLIATIONS/companions/haar-functional.html\#section-9}{Theorem
9.2} (Noncommutative geometry, foliations and index theory). The Hilbert
basis theorem, polynomial normality and the strong Nullstellensatz come
from
\href{../AG-CA/AG-CA-03.html\#2-polynomial-rings-and-finite-geometric-descriptions}{Noetherian
and Artinian rings},
\href{../AG-CA/AG-CA-05.html\#1-a-monic-equation-is-finite-control}{Integral
extensions, lying over, going up and going down}, and
\href{../AG-CA/AG-CA-06.html\#2-closed-points-detect-radical-equations}{The
Nullstellensatz and Jacobson rings} (Commutative algebra for geometry).

\subsection{1. Coordinates, substitution and determinant
weight}\label{section-1}

Write a binary form of degree \(d\) as \[
f(x,y)=\sum_{i=0}^d\binom di a_i x^{d-i}y^i.
\] The binomial factors are convenient over a characteristic-zero field.
They require separate treatment over the integers in the final lesson.

For \(g\in\operatorname{GL}_2(k)\), fix the left action \[
(g\cdot f)(v)=f(g^{-1}v),\qquad v=(x,y)^{\mathsf t}.
\] A polynomial \(C(a,x,y)\), homogeneous of coefficient degree
\(\delta\) and variable order \(\omega\), is a \textbf{covariant of
weight \(w\)} if \protect\phantomsection\label{equation-B1}{}

\[
C(g\cdot f,gv)=\det(g)^{-w}C(f,v).
\tag{B1}
\] An invariant is a covariant of order zero. The inverse determinant in
(B1) is part of our convention: it makes classical weights positive for
the inverse-substitution action. For the special linear group the
determinant factor disappears.

For \(n\) variables, use the same definition with
\(g\in\operatorname{GL}_n\). A form of degree \(d\) is
\(\sum_{|\alpha|=d}(d!/\alpha!)a_\alpha x^\alpha\). Systems of forms can
be graded separately in the coefficients of each member. A
\textbf{contravariant} has auxiliary variables transforming in the dual
representation; it must not be silently treated as a covariant with the
same variable action.

\protect\phantomsection\label{proposition-1-1}{}

\textbf{Proposition 1.1 (degree, order and weight).} A nonzero
bihomogeneous binary special-linear covariant extends to (B1), with \[
w=\frac{d\delta-\omega}{2}\in\mathbb Z_{\geq0}.
\]

\textbf{Proof.} The scalar matrix \(\lambda I\) multiplies the
coefficients by \(\lambda^{-d}\) and the variables by \(\lambda\). Hence
the left side of (B1) scales by \(\lambda^{-d\delta+\omega}\). A
determinant weight must therefore satisfy \(2w=d\delta-\omega\).

To prove integrality and nonnegativity without assuming them, use
\(\operatorname{diag}(t,t^{-1})\in\operatorname{SL}_2\). Its coefficient
action is \(a_i\mapsto t^{2i-d}a_i\). Let \(c(a)\) be the coefficient of
\(x^\omega\) in \(C\). It is nonzero: if \(C(a,1,0)\) vanished for every
coefficient vector, transitivity of \(\operatorname{SL}_2\) on nonzero
vectors, together with covariance, would make \(C\) vanish everywhere.
Since \(k\) is infinite, that would make the polynomial zero.

Every coefficient monomial in \(c\) has torus weight \(-\omega\). Its
weight is a sum of \(\delta\) integers \(2i-d\), so lies between
\(-d\delta\) and \(d\delta\), and is congruent to \(d\delta\) modulo
two. Thus \(\omega\leq d\delta\) and \(d\delta-\omega\) is even.

Over an algebraic closure choose \(\lambda\) with \(\lambda^2=\det g\)
and write \(g=\lambda s\), \(s\in\operatorname{SL}_2\). Scalar scaling
and special-linear covariance now give (B1). The resulting equality of
polynomials descends to \(k\). \(\square\)

For a multidegree \((\delta_j)\) in forms of degrees \(d_j\), the same
scalar calculation gives \(nw=\sum_jd_j\delta_j-\omega\) in \(n\)
variables whenever a determinant-weight covariant exists.

\subsection{2. A differential test for covariance}\label{section-2}

Let \(E_0=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)\)
and \(F_0=\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right)\).
The simultaneous action on coefficients and variables induces
\protect\phantomsection\label{equation-B2}{}

\[
E=y\frac{\partial}{\partial x}
-\sum_{i=1}^d i a_{i-1}\frac{\partial}{\partial a_i},
\qquad
F=x\frac{\partial}{\partial y}
-\sum_{i=0}^{d-1}(d-i)a_{i+1}\frac{\partial}{\partial a_i}.
\tag{B2}
\] For a system, sum the coefficient parts over all its forms. The
variable parts appear only once.

To check the first formula, substitute \(x-ty\) into \(f\). The
coefficient derivative is \[
-(d-i+1)\frac{\binom d{i-1}}{\binom di}a_{i-1}=-i a_{i-1}.
\] Meanwhile the simultaneous variable action is \(x\mapsto x+ty\). The
second formula follows by exchanging \(x,y\) and reversing the
coefficient indices. This derivation fixes the signs; a criterion
written for another action convention can have different signs.

\protect\phantomsection\label{theorem-2-1}{}

\textbf{Theorem 2.1 (infinitesimal criterion).} A polynomial
\(C(a,x,y)\) is special-linear invariant under the simultaneous action
if and only if \(EC=FC=0\).

\textbf{Proof.} Invariance under the upper and lower unipotent
one-parameter subgroups implies these two derivative conditions at
\(t=0\).

Conversely, the action of each subgroup on a polynomial is a polynomial
in \(t\). Its derivative is the translate of \(EC\), or of \(FC\), by
the chain rule and the subgroup law. If the corresponding derivation
kills \(C\), that derivative vanishes identically. In characteristic
zero a polynomial with zero derivative is constant. Its value at \(t=0\)
is \(C\), so \(C\) is invariant under both subgroups.

They generate \(\operatorname{SL}_2(k)\). Gaussian row operations reduce
any determinant-one matrix to a diagonal matrix, using these elementary
matrices. If \[
U(t)=\begin{pmatrix}1&t\\0&1\end{pmatrix},
\quad L(t)=\begin{pmatrix}1&0\\t&1\end{pmatrix},
\] then
\(W(t)=U(t)L(-t^{-1})U(t)=\left(\begin{smallmatrix}0&t\\-t^{-1}&0\end{smallmatrix}\right)\)
for \(t\ne0\), and \(W(t)W(1)^{-1}=\operatorname{diag}(t,t^{-1})\). This
includes the diagonal matrices in the generated group, proving the
criterion. \(\square\)

The proof uses unipotent subgroups and polynomial differentiation,
rather than an unstated assertion about connected Lie groups. It works
over every characteristic-zero field.

\subsection{3. Transvectants: contracting two variable
pairs}\label{section-3}

Use independent binary pairs \(v=(x,y)\) and \(u=(X,Y)\). Define \[
\Omega_{v,u}=\partial_x\partial_Y-\partial_y\partial_X.
\] For forms \(P,Q\) of orders \(m,n\), fix the normalized transvectant
\protect\phantomsection\label{equation-B3}{}

\[
(P,Q)_r=
\frac{(m-r)!(n-r)!}{m!\,n!}
\left.\Omega_{v,u}^{\,r}\bigl(P(v)Q(u)\bigr)\right|_{u=v},
\qquad 0\leq r\leq\min(m,n).
\tag{B3}
\] All constants in later computations refer to (B3); rescaled classical
invariants will be identified explicitly.

Expanding the operator gives the usable formula
\protect\phantomsection\label{equation-B4}{}

\[
(P,Q)_r=\frac{(m-r)!(n-r)!}{m!\,n!}
\sum_{j=0}^r(-1)^j\binom rj
\frac{\partial^rP}{\partial x^{r-j}\partial y^j}
\frac{\partial^rQ}{\partial x^j\partial y^{r-j}}.
\tag{B4}
\]

\protect\phantomsection\label{theorem-3-1}{}

\textbf{Theorem 3.1 (covariance of transvection).} If \(P,Q\) are
covariants of coefficient degrees \(\delta_1,\delta_2\), orders \(m,n\)
and weights \(w_1,w_2\), then \((P,Q)_r\) is a covariant of degree
\(\delta_1+\delta_2\), order \(m+n-2r\), and weight \(w_1+w_2+r\),
unless it is zero.

\textbf{Proof.} The chain rule shows that contracting two derivatives by
their alternating determinant picks up a factor \(\det g\) when both
variable pairs are replaced by \(gv,gu\): \[
\Omega_{v,u}\bigl(P(gv)Q(gu)\bigr)
=\det(g)\,(\Omega_{v,u}(P(v)Q(u)))\big|_{v\mapsto gv,\ u\mapsto gu}.
\] This identity is checked by multiplying the two Jacobian matrices;
their alternating product is the determinant. The derivatives commute,
so iteration gives \(\det(g)^r\). Apply it to the covariance identity
for the product of \(P,Q\), then divide by that factor. The diagonal
restriction \(u=v\) commutes with simultaneous substitution. This gives
determinant exponent \(-(w_1+w_2+r)\). Differentiation removes \(r\)
variables from each factor, and coefficient multiplication adds the
degrees. The scalar normalization changes none of these assertions.
\(\square\)

Exchanging the factors reverses the alternating contraction: \[
(Q,P)_r=(-1)^r(P,Q)_r.
\] Thus every odd self-transvectant vanishes. In particular, computing a
nonzero \((f,f)_1\) would expose a sign or implementation error.

\subsection{4. Reading a covariant from its leading
coefficient}\label{section-4}

The computations to follow will give \textbf{complete systems}: every
polynomial covariant is a polynomial in the displayed ones. To prove
this, we first turn covariance into a problem about a derivation on the
coefficient ring.

For a covariant \(C\) of order \(\omega\), write \[
C=c_0x^\omega+c_1x^{\omega-1}y+\cdots+c_\omega y^\omega.
\] The coefficient of \(x^\omega\) in \(EC=0\) is
\protect\phantomsection\label{equation-B5}{}

\[
\mathcal D c_0=0,\qquad
\mathcal D=\sum_{i=1}^d i a_{i-1}\partial_{a_i}.
\tag{B5}
\] The leading-coefficient map \(C\mapsto C(a,1,0)\) is injective on the
algebra of special-linear covariants, including sums of different
orders. Indeed, if that evaluation is zero, covariance and the
transitivity of \(\operatorname{SL}_2(k)\) on nonzero vectors make
\(C(a,v)=0\) for every \(v\ne0\) and every coefficient vector. A
polynomial vanishing there is zero because \(k\) is infinite.

For degrees two, three and four, abbreviate the coefficients by
\(a,b,c,d,e\), omitting those not present. Set \[
h=ac-b^2,\qquad t=a^2d-3abc+2b^3.
\] Direct differentiation gives
\(\mathcal Da=\mathcal Dh=\mathcal Dt=0\). For a cubic also put \[
D=a^2d^2-6abcd+4ac^3+4b^3d-3b^2c^2;
\] for a quartic put \[
i=ae-4bd+3c^2,\qquad
j=ace+2bcd-ad^2-b^2e-c^3.
\] These too are killed by \(\mathcal D\).

\protect\phantomsection\label{theorem-4-1}{}

\textbf{Theorem 4.1 (three coefficient kernels).} In characteristic
zero, \protect\phantomsection\label{equation-B6}{}

\[
\begin{aligned}
\ker(\mathcal D:k [a,b,c]\to k [a,b,c])&=k [a,h],\\
\ker(\mathcal D:k [a,b,c,d]\to k [a,b,c,d])
&\cong k [a,h,t,D]/(a^2D-t^2-4h^3),\\
\ker(\mathcal D:k [a,b,c,d,e]\to k [a,b,c,d,e])
&\cong k [a,h,t,i,j]/(t^2+4h^3-ia^2h+ja^3).
\end{aligned}
\tag{B6}
\] The isomorphisms send the formal generators to the polynomials just
defined.

\textbf{Proof.} Invert \(a\), and use \(s=b/a\). Then \(\mathcal Ds=1\).
The inverse coordinate substitutions are
\protect\phantomsection\label{equation-B7}{}

\[
b=as,\quad c=h/a+as^2,\quad
d=t/a^2+3hs/a+as^3,
\tag{B7}
\] and, in degree four, \protect\phantomsection\label{equation-B8}{}

\[
e=i/a+4st/a^2+6hs^2/a+as^4-3h^2/a^3.
\tag{B8}
\] Use only the available equations in each degree. These substitutions
show that the localized polynomial rings have coordinates \[
(a^{\pm1},h,s),\quad (a^{\pm1},h,t,s),\quad
(a^{\pm1},h,t,i,s),
\] respectively. In these coordinates \(\mathcal D=\partial_s\). Its
kernel consists of the polynomials independent of \(s\); this uses
characteristic zero.

Expanding the definitions gives
\protect\phantomsection\label{equation-B9}{}

\[
a^2D=t^2+4h^3,\qquad
t^2+4h^3-ia^2h+ja^3=0.
\tag{B9}
\] Here is also a way to verify the second expansion with fewer terms:
substitute (B7)--(B8) into the definition of \(j\); it becomes
\(j=(ia^2h-t^2-4h^3)/a^3\). An equality after inverting \(a\) is an
equality in the original polynomial ring.

Let \(S\) denote the corresponding proposed ring on the right of (B6).
Each of the two relations is primitive and linear in its last generator,
\(D\) or \(j\). Its leading coefficient, \(a^2\) or \(a^3\), has no
common factor with its constant coefficient, since the latter remains a
nonzero polynomial modulo \(a\). Gauss's lemma makes the relation
irreducible in the polynomial ring, so \(S\) is a domain. Gauss's lemma
is proved in the algebra lesson
\href{../AG-CA/AG-CA-05.html\#2-integral-closure-survives-localization}{Integral
extensions, lying over, going up and going down, Proposition 2.3}. After
inverting \(a\), equations (B7)--(B9) identify \(S [a^{-1}]\) with the
localized kernel. Because \(S\) is a domain, this also proves that \(S\)
embeds in the original coefficient ring \(R\).

We must check that no denominator in \(a\) creates a new polynomial
invariant. The necessary check is the injectivity of \(S/(a)\to R/(a)\).

For the quadratic it is \(k [h]\to k [b,c]\), \(h\mapsto-b^2\), which is
injective. For the cubic, \[
S/(a)=k [h,t,D]/(t^2+4h^3),
\quad h\mapsto-b^2,\quad t\mapsto2b^3,
\quad D\mapsto4b^3d-3b^2c^2.
\] Every element has a unique representative \(P(h,D)+tQ(h,D)\), by
division by the monic relation in \(t\). The elements \(b,D\) are
algebraically independent: over \(k(b)\), the expression for \(D\)
solves for \(d\), with \(c\) still free. After substitution, the first
summand uses even powers of \(b\), the second odd powers. Their sum is
zero only if both polynomials are zero. Thus this map is injective.

For the quartic the same normal form is \(P(h,i,j)+tQ(h,i,j)\), with \[
h=-b^2,\quad t=2b^3,\quad
i=-4bd+3c^2,\quad j=2bcd-b^2e-c^3
\] modulo \(a\). Now \(b,i,j\) are algebraically independent. Over
\(k(b)\), the equation for \(i\) solves for \(d\), and the equation for
\(j\) solves for \(e\), while \(c\) remains free. The even/odd argument
in \(b\) again proves injectivity.

Consequently, if \(u\in S\) is divisible by \(a\) in \(R\), its class in
\(S/(a)\) is zero, so \(u\in aS\). If \(u/a^N\in S [a^{-1}]\) belongs to
\(R\), repeat this observation \(N\) times to obtain \(u/a^N\in S\).
Thus \(S [a^{-1}]\cap R=S\). Every element of \(\ker\mathcal D\) lies in
that intersection, proving (B6). \(\square\)

The last part of the proof is often called \textbf{saturation}:
divisibility seen in the larger ring was already divisibility in the
proposed kernel. It is what upgrades a computation on the open set
\(a\ne0\) to a global presentation.

\subsection{5. Quadratics and cubics}\label{section-5}

\subsubsection{The quadratic}\label{the-quadratic}

Let \(f=ax^2+2bxy+cy^2\). Its basic invariant is \(h=ac-b^2\). The
complete covariant algebra is
\protect\phantomsection\label{equation-B10}{}

\[
k [f,h].
\tag{B10}
\] To see generation, take a covariant of coefficient degree \(\delta\)
and order \(\omega\). Its leading coefficient belongs to \(k [a,h]\) by
Theorem 4.1. Select the component of that expression having coefficient
degree \(\delta\) and torus weight \(-\omega\). The weights of \(a,h\)
are \(-2,0\); their coefficient degrees are \(1,2\). Replacing \(a\) by
\(f\) and \(h\) by \(h\) therefore gives a covariant of the same degree
and order with the same leading coefficient. Injectivity proves
equality. Algebraic independence of \(a,h\) proves that (B10) has no
relation.

The matrix of the quadratic is \[
Q=\begin{pmatrix}a&b\\b&c\end{pmatrix}.
\] Inverse substitution sends \(Q\) to \(g^{-\mathsf t}Qg^{-1}\); its
determinant \(h\) scales by \(\det(g)^{-2}\), as required by weight two.
If \(f=(u_1x+v_1y)(u_2x+v_2y)\), coefficient comparison gives
\protect\phantomsection\label{equation-B11}{}

\[
h=-\tfrac14(u_1v_2-v_1u_2)^2.
\tag{B11}
\] Thus \(h=0\) is exactly the repeated-root locus over an algebraic
closure. For the usual univariate coefficients the discriminant is
\(-4h\).

\subsubsection{The cubic}\label{the-cubic}

Write \[
f=ax^3+3bx^2y+3cxy^2+dy^3,\quad
H=(f,f)_2,\quad T=(f,H)_1,\quad \Delta=(H,H)_2.
\] Formula (B4) gives \protect\phantomsection\label{equation-B12}{}

\[
H=2\bigl((ac-b^2)x^2+(ad-bc)xy+(bd-c^2)y^2\bigr),
\qquad \Delta=-2D.
\tag{B12}
\] The leading coefficients of \(f,H/2,T,\Delta\) are \(a,h,t,-2D\).
Their orders are \(3,2,3,0\), and their coefficient degrees are
\(1,2,3,4\).

\protect\phantomsection\label{theorem-5-1}{}

\textbf{Theorem 5.1 (the cubic system).} The covariant algebra is
generated by \(f,H,T,\Delta\), with exactly one relation:
\protect\phantomsection\label{equation-B13}{}

\[
2T^2+H^3+\Delta f^2=0.
\tag{B13}
\]

\textbf{Proof.} The left side is a covariant, and its leading
coefficient is \(2t^2+8h^3-2Da^2=0\) by (B9). Leading-coefficient
injectivity proves (B13), including every coefficient of the forms.

For generation, Theorem 4.1 expresses the leading coefficient of any
covariant in \(a,h,t,D\). The torus weights are minus the orders just
listed. As in the quadratic case, select its coefficient-degree and
torus-weight component, then replace these four generators by
\(f,H/2,T,-\Delta/2\). Every monomial has the desired order and degree,
so injectivity identifies the resulting covariant with the original one.
Finally, the kernel presentation in (B6) proves that every relation is a
multiple of (B13), after the indicated rescaling. \(\square\)

For \(f=x^3+y^3\), the normalization gives \[
H=2xy,\quad T=x^3-y^3,\quad \Delta=-2.
\] Relation (B13) becomes \(2(x^3-y^3)^2+8x^3y^3-2(x^3+y^3)^2=0\). For
\(f=x^2y\), remembering that \(b=1/3\), it gives \[
H=-\tfrac29x^2,\quad T=\tfrac2{27}x^3,\quad\Delta=0.
\]

If \(f=\ell_1\ell_2\ell_3\), with \(\ell_r=u_rx+v_ry\) and
\([rs]=u_rv_s-v_ru_s\), then
\protect\phantomsection\label{equation-B14}{}

\[
\Delta=\tfrac2{27}[12]^2 [13]^2 [23]^2.
\tag{B14}
\] Here is a proof that fixes both the root meaning and the constant.
For distinct root lines, a projective linear transformation sends their
ordered triple to the root lines of \(xy(x-y)\). This follows by sending
the first two lines to the coordinate axes, then using a diagonal matrix
to send the third to \(x-y\). On replacing \(\ell_r\) by
\(\lambda_r\ell_r\), both sides of (B14) scale by
\((\lambda_1\lambda_2\lambda_3)^4\); under inverse substitution both
have determinant weight six. It is therefore enough to check
\(xy(x-y)\). Its coefficients are \(a=d=0,b=1/3,c=-1/3\), so \(D=-1/27\)
and \(\Delta=2/27\); the bracket product squared is one. This proves
(B14) on the dense open set of triples with distinct root lines, and
therefore as a polynomial identity everywhere. The ordinary discriminant
is \((27/2)\Delta\). In particular, its vanishing means precisely that
the cubic has a repeated root.

\subsection{6. Quartics and the cross-ratio of four
roots}\label{section-6}

Write \[
f=ax^4+4bx^3y+6cx^2y^2+4dxy^3+ey^4,\quad
H=(f,f)_2,\quad T=(f,H)_1.
\] Use the explicitly rescaled invariants
\protect\phantomsection\label{equation-B15}{}

\[
i=\tfrac12(f,f)_4=ae-4bd+3c^2,\qquad
j=\tfrac16(f,H)_4=ace+2bcd-ad^2-b^2e-c^3.
\tag{B15}
\] The factors \(1/2\) and \(1/6\) matter: applying (B3) without them
would give \(2i,6j\), and would change the usual discriminant formula.

The Hessian in this normalization is
\protect\phantomsection\label{equation-B16}{}

\[
\begin{split}
H=2\bigl(& (ac-b^2)x^4+2(ad-bc)x^3y\\
&+(ae+2bd-3c^2)x^2y^2
+2(be-cd)xy^3+(ce-d^2)y^4\bigr).
\end{split}
\tag{B16}
\] Its leading coefficient is \(2h\), and that of \(T\) is \(t\).

\protect\phantomsection\label{theorem-6-1}{}

\textbf{Theorem 6.1 (the quartic system).} The covariant algebra has
generators \(f,H,T,i,j\) and exactly one relation:
\protect\phantomsection\label{equation-B17}{}

\[
2T^2+H^3-if^2H+2jf^3=0.
\tag{B17}
\]

\textbf{Proof.} Its leading coefficient is twice the quartic relation in
(B9). All terms are covariants of coefficient degree six and order
twelve, so leading-coefficient injectivity proves the identity. For
generation and the absence of further relations, use the quartic kernel
in Theorem 4.1. Its generators have coefficient degrees \(1,2,3,2,3\)
and torus weights \(-4,-4,-6,0,0\). Select the required degree and
weight component, and lift by \(a,h,t,i,j\mapsto f,H/2,T,i,j\). This
repeats the already proved argument for the cubic. \(\square\)

In particular the quartic invariant ring is \(k [i,j]\): a covariant of
order zero can only use the generators of order zero. The kernel proof
also shows that \(i,j\) are algebraically independent.

The ordinary discriminant is
\protect\phantomsection\label{equation-B18}{}

\[
\operatorname{disc}(f)=256(i^3-27j^2).
\tag{B18}
\] To prove this, factor \(f=\prod_{r=1}^4\ell_r\). The root
discriminant \(\prod_{r<s}[rs]^2\) has coefficient degree six and
determinant weight twelve. On the open set of four distinct root lines,
projective substitution and rescaling of the factors reduce the
calculation to \[
f_\lambda=xy(x-y)(x-\lambda y),\qquad \lambda\ne0,1.
\] Formula (B15) gives \[
i=\frac{\lambda^2-\lambda+1}{12},\qquad
j=\frac{(\lambda+1)(2\lambda^2-5\lambda+2)}{432}.
\] Substitution yields \(256(i^3-27j^2)=\lambda^2(1-\lambda)^2\), which
is also its bracket product squared. Both sides transform with the same
weight and degree, so they agree on that open set, and polynomial
density proves (B18) for every quartic. Consequently the discriminant
vanishes exactly on the repeated-root locus.

For a quartic with four distinct roots, define the absolute invariant \[
J(f)=\frac{i^3}{i^3-27j^2}.
\] It is unchanged by \(\operatorname{GL}_2\) and by multiplying the
form by a nonzero scalar. For the displayed normal form,
\protect\phantomsection\label{equation-B19}{}

\[
J(\lambda)=\frac4{27}
\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}.
\tag{B19}
\] This equals the conventional elliptic \(j\)-invariant divided by
\(1728\). To check the normalization, the double cover of the root line
is \(y^2=x(x-1)(x-\lambda)\). Put \(x=X+(\lambda+1)/3\), \(Y=2y\).
Expansion gives \[
Y^2=4X^3-g_2X-g_3,\qquad
g_2=\frac43(\lambda^2-\lambda+1),\quad
g_3=\frac4{27}(\lambda+1)(2\lambda^2-5\lambda+2).
\] Here \(g_2^3-27g_3^2=16\lambda^2(\lambda-1)^2\). The conventional
definition \(j_{\rm ell}=1728g_2^3/(g_2^3-27g_3^2)\) therefore gives
exactly \(1728J(\lambda)\).

\protect\phantomsection\label{proposition-6-2}{}

\textbf{Proposition 6.2 (separation of the four-root configurations).}
Over an algebraic closure, two quartics with nonzero discriminant have
the same \(J\) if and only if their unordered root sets are projectively
equivalent. Equivalently, they differ by a linear change of variables
and a nonzero scalar.

\textbf{Proof.} The six possible values of the cross-ratio of an
unordered four-point configuration are
\protect\phantomsection\label{equation-B20}{}

\[
\lambda,\quad1-\lambda,\quad1/\lambda,\quad
1/(1-\lambda),\quad\lambda/(\lambda-1),\quad(\lambda-1)/\lambda.
\tag{B20}
\] For example, interchanging two entries in an ordered cross-ratio
yields \(1-\lambda\) or \(1/\lambda\); these two operations generate the
list. The other permutations leave one of these six values. Conversely,
choosing three of the points as \(0,1,\infty\) gives precisely these
possibilities for the fourth, by the unique projective transformation
taking one ordered triple to another.

Formula (B19) is unchanged by the two generating transformations, by
direct substitution. For a proof of the converse that includes
exceptional configurations with extra symmetries, use a variable \(z\).
Clearing denominators in \(J(z)=J(\lambda)\) gives the degree-six
polynomial \[
P_\lambda(z)=
\lambda^2(\lambda-1)^2(z^2-z+1)^3
-(\lambda^2-\lambda+1)^3z^2(z-1)^2.
\] For a generic \(\lambda\) its six distinct roots are the entries of
(B20), and its leading coefficient is \(\lambda^2(\lambda-1)^2\). Hence
\protect\phantomsection\label{equation-B21}{}

\[
P_\lambda(z)=\lambda^2(\lambda-1)^2
\prod_{\rho\text{ in }(20)}(z-\rho)
\tag{B21}
\] as an identity in \(k(\lambda)[z]\). The generic argument proves
equality of rational functions, so it specializes for every
\(\lambda\ne0,1\). The repetitions in the product then give the correct
multiplicities at special values. Thus equality of \(J\) forces the
other cross-ratio to be in (B20). The description of ordered triples
proves projective equivalence. The reverse implication was invariance.
\(\square\)

Over a field that is not algebraically closed, \(J\) separates geometric
orbits; it does not assert that the required transformation has
coefficients in that field. This distinction already occurs for
quadratics.

For \(x^4+y^4\) one has \(i=1,j=0\), so \(J=1\); it is a harmonic
configuration. For \(x^3y\), \(b=1/4\) and the other coefficients
vanish, giving \(i=j=0\) and zero discriminant. For distinct roots,
\(j=0\) describes the harmonic configurations: in the normal form it
means \(\lambda\in\{-1,2,1/2\}\). Likewise \(i=0\) means
\(\lambda^2-\lambda+1=0\), the equianharmonic configurations. Formula
(B21) accounts for the smaller numbers of distinct cross-ratios in both
cases.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Six cross-ratios of the unordered root configuration zero, one, three and infinity: three, one third, minus two, minus one half, three halves and two thirds. Every value gives J equal to 343 over 243.}]{figures/cross-ratio-orbit.png}}
\caption{Six cross-ratios of the unordered root configuration zero, one,
three and infinity: three, one third, minus two, minus one half, three
halves and two thirds. Every value gives J equal to 343 over 243.}
\end{figure}

The six displayed values form one orbit under (B20), and (B19) has the
same value at each. This illustrates Proposition 6.2 at a configuration
without extra symmetry; the polynomial identity (B21) also covers the
exceptional harmonic and equianharmonic orbits. Editable diagram.

\subsection{7. Determinants and mixed variables}\label{section-7}

For vector columns \(v_1,\ldots,v_m\in\mathbb C^n\), write \[
[i_1\cdots i_n]=\det(v_{i_1},\ldots,v_{i_n}).
\] Multiplication of the matrix by \(g\) multiplies every bracket by
\(\det g\), so these are special-linear invariants. For exactly as many
vectors as coordinates, the following matrix-orbit argument determines
the invariant ring directly.

For a dual vector \(\alpha\), the pairing \(\alpha(v)\) is invariant
under \(v\mapsto gv,\ \alpha\mapsto\alpha g^{-1}\). Determinants of
\(n\) dual rows instead scale by \(\det(g)^{-1}\). In symbolic notation,
the letters of a form are dual rows and its ordinary variables are
vector columns. Auxiliary dual variables describe contravariants.
Polynomials depending on several sorts of auxiliary variables are called
concomitants. These transformation rules keep the variable sorts
distinct.

For exactly \(n\) vector columns, bracket generation has a short direct
proof. If \(F\) is invariant, define
\(h(t)=F(\operatorname{diag}(t,1,\ldots,1))\). On invertible matrices,
two matrices have the same determinant exactly when one is obtained from
the other by left multiplication by a special-linear matrix. Thus
\(F(X)=h(\det X)\) there. Since \(h\) is polynomial and invertible
matrices form a dense open set, the equality holds for all \(X\).
Consequently \protect\phantomsection\label{equation-H1}{}

\[
\mathbb C [\operatorname{Mat}_n]^{\operatorname{SL}_n}=\mathbb C [\det X].
\tag{H1}
\] For \(n=2\) this proves the invariant ring of a pair of vectors is
\(\mathbb C [[12]]\).

In Sections 8--11 put \(G=\operatorname{SL}_n(\mathbb C)\), let \(W\) be
a finite-dimensional polynomial representation, and set
\(A=\mathbb C [W]\), with \((gF)(w)=F(g^{-1}w)\). Each \(A_d\) is a
finite-dimensional polynomial representation. Matrix entries of the
representation are regular functions on \(G\); inverse matrix entries
are polynomial there because \(g^{-1}=\operatorname{adj}(g)\).

\subsection{8. Compact averaging and the unitary trick}\label{section-8}

\protect\phantomsection\label{theorem-8-1}{}

\textbf{Theorem 8.1 (compact complete reducibility).} Every
finite-dimensional continuous complex representation of a compact group
\(K\) has an invariant positive-definite Hermitian inner product. Every
invariant subspace has an invariant complement.

\textbf{Proof.} Start with any positive-definite Hermitian form
\(\langle\ ,\ \rangle_0\), and normalized Haar measure \(\mu\). On a
compact group this measure is also right invariant: the right translate
of a left Haar measure is left Haar, has the same total mass, and hence
equals it by uniqueness. Define \[
\langle v,w\rangle_K=\int_K
\langle kv,kw\rangle_0\,d\mu(k).
\] It is Hermitian and invariant, by right translation in the integral.
For \(v\ne0\), the continuous positive function
\(\langle kv,kv\rangle_0\) has a positive minimum on \(K\); its integral
is positive. If \(U\) is invariant, unitarity makes its orthogonal
complement invariant: for \(v\perp U\) and \(u\in U\),
\(\langle kv,u\rangle_K=\langle v,k^{-1}u\rangle_K=0\). Splitting
successively proves complete reducibility. \(\square\)

For \(G=\operatorname{SL}_n(\mathbb C)\), take \[
K=\operatorname{SU}(n)=\{g:g^*g=I,\ \det g=1\}.
\] It is closed and bounded in a finite-dimensional real vector space,
hence compact.

\protect\phantomsection\label{lemma-8-2}{}

\textbf{Lemma 8.2 (density of the unitary subgroup).} A regular function
on \(G\) that vanishes on \(K\) is zero. The same holds for a
holomorphic function.

\textbf{Proof.} The real Lie algebra
\(\mathfrak k=\{X:X^*=-X,\ \operatorname{tr}X=0\}\) has complex span
\(\mathfrak{sl}_n(\mathbb C)\). Indeed every trace-zero matrix is the
sum of a skew-Hermitian trace-zero matrix and \(i\) times another one,
by separating its Hermitian and skew-Hermitian parts. Choose a real
basis \(X_1,\ldots,X_{n^2-1}\) of \(\mathfrak k\); it is a complex basis
of that complex span.

The holomorphic map \[
\Phi(z_1,\ldots,z_{n^2-1})
=\exp(z_1X_1)\cdots\exp(z_{n^2-1}X_{n^2-1})
\] has invertible derivative at zero, so the complex inverse-function
theorem gives a complex open neighbourhood of the identity in its image.
For real \(z_j\), its image lies in \(K\). If \(P\) vanishes on \(K\),
\(P\circ\Phi\) vanishes on a real open box. Applying the one-variable
identity theorem to one coordinate at a time makes it zero on a complex
neighbourhood of zero. Hence \(P\) vanishes on a complex open
neighbourhood in \(G\).

The group \(G\) is connected: elementary row operations express its
matrices as products of elementary unipotent matrices and
determinant-one diagonal matrices; the latter are products of diagonal
\(2\times2\) blocks, themselves products of elementary matrices by the
explicit calculation in Section 2. Each elementary subgroup is the image
of the connected plane \(\mathbb C\), and products give paths to the
identity. Holomorphic continuation on this connected complex manifold
now proves \(P=0\). The case \(n=1\) is immediate. \(\square\)

\protect\phantomsection\label{theorem-8-3}{}

\textbf{Theorem 8.3 (the unitary trick).} Every finite-dimensional
polynomial representation of \(G\) is completely reducible.

\textbf{Proof.} First, a \(K\)-invariant subspace \(U\) is
\(G\)-invariant. For \(u\in U\) and a linear functional \(\ell\)
vanishing on \(U\), the regular function \(g\mapsto\ell(gu)\) vanishes
on \(K\), and hence on \(G\) by Lemma 8.2. Therefore \(gu\in U\).

Given a \(G\)-invariant subspace, Theorem 8.1 supplies a \(K\)-invariant
complement; the preceding argument makes that complement
\(G\)-invariant. Repeating in finite dimension gives a direct sum of
irreducibles. \(\square\)

The passage from \(K\) to \(G\) uses holomorphic matrix coefficients. It
does not apply to an arbitrary continuous representation of the
underlying real group \(G\).

\subsection{9. The Reynolds projection}\label{section-9}

Let \(B=A^G\). Degree by degree, define
\protect\phantomsection\label{equation-H2}{}

\[
\rho(F)=\int_K kF\,d\mu(k).
\tag{H2}
\] The integral is just integration of the finitely many coordinates in
\(A_d\), not integration over an infinite-dimensional space.

\protect\phantomsection\label{theorem-9-1}{}

\textbf{Theorem 9.1 (Reynolds).} The map \(\rho:A\to B\) is the unique
degree-preserving, \(G\)-equivariant projection onto \(B\). It satisfies
\protect\phantomsection\label{equation-H3}{}

\[
\rho(bF)=b\rho(F)\quad(b\in B),\qquad
\rho(b)=b.
\tag{H3}
\]

\textbf{Proof.} Translation in (H2) makes its image \(K\)-fixed, hence
\(G\)-fixed by density. Averaging a fixed vector gives that vector, so
it is a projection. It preserves degrees. Left and right invariance of
Haar measure make it commute with \(K\), and density makes this
commutation hold with \(G\) on every finite-dimensional degree piece.
Pulling the \(K\)-fixed factor \(b\) out of the integral proves (H3).

For uniqueness, use Theorem 8.3 on each \(A_d\). Any equivariant map to
the trivial representation kills every nontrivial irreducible: a nonzero
map from an irreducible to a trivial representation would be injective,
forcing that irreducible to be trivial. On the sum of trivial summands a
projection onto \(B_d\) is the identity. These two conditions determine
it uniquely. \(\square\)

Thus a Reynolds projection is more than a linear choice of a complement.
Its invariant-linearity is what permits the next degree induction.

\subsection{10. Hilbert finiteness and normality}\label{section-10}

\protect\phantomsection\label{theorem-10-1}{}

\textbf{Theorem 10.1 (Hilbert finiteness).} For every finite-dimensional
polynomial representation \(W\) of \(\operatorname{SL}_n(\mathbb C)\),
the invariant ring \(B=\mathbb C [W]^G\) is generated by finitely many
homogeneous invariants.

\textbf{Proof.} Let \(I=A\,B_+\) be the ideal generated by all
positive-degree homogeneous invariants. The polynomial ring \(A\) is
Noetherian by the Hilbert basis theorem. Hence \(I\) has finitely many
ideal generators. Each is a finite \(A\)-linear combination of elements
of \(B_+\), so collecting the finitely many invariants occurring in
those expressions gives homogeneous invariants \(f_1,\ldots,f_s\)
generating \(I\).

If \(f\in B\) is homogeneous of positive degree \(d\), write
\(f=\sum_i a_if_i\). Taking degree \(d\) components lets us choose
\(a_i\) homogeneous of degree \(d-\deg f_i\), omitting a term when this
number is negative. Apply (H3): \[
f=\rho(f)=\sum_i\rho(a_i)f_i.
\] Every nonzero \(\rho(a_i)\) has degree strictly less than \(d\).
Induction on \(d\), starting with \(B_0=\mathbb C\), expresses \(f\) as
a polynomial in the \(f_i\). Every invariant is a sum of homogeneous
invariants, because the action preserves degree. Thus
\(B=\mathbb C [f_1,\ldots,f_s]\). \(\square\)

This proof establishes existence; the basis theorem does not supply a
numerical bound on the degrees of the chosen \(f_i\).

\protect\phantomsection\label{corollary-10-2}{}

\textbf{Corollary 10.2 (normality).} \(B\) is a Noetherian normal
domain.

\textbf{Proof.} It is a finite-type algebra over a field, hence
Noetherian, and a subring of the domain \(A\). If
\(z\in\operatorname{Frac}B\) is integral over \(B\), the same monic
equation makes it integral over \(A\). Polynomial rings are normal, by
\href{../AG-CA/AG-CA-05.html\#2-integral-closure-survives-localization}{Integral
extensions, lying over, going up and going down, Proposition 2.3}, so
\(z\in A\). Since it is a fraction of invariant elements, \(z\) is fixed
by \(G\). Hence \(z\in B\). \(\square\)

For example, the quadratic calculation in Section 5 gives
\(\mathbb C [\operatorname{Sym}^2(\mathbb C^2)^*]^{\operatorname{SL}_2}
=\mathbb C [ac-b^2]\). Applying Theorem 10.1 to coefficient variables
together with auxiliary vector or dual-vector variables proves finite
generation for their covariant and concomitant algebras as well.

\subsection{11. Equations for the nullcone give module
finiteness}\label{section-11}

Define the \textbf{nullcone} \[
\mathcal N=\{w\in W:b(w)=0\text{ for every }b\in B_+\}.
\] Theorem 10.1 makes this an algebraic set defined by finitely many
equations.

\protect\phantomsection\label{theorem-11-1}{}

\textbf{Theorem 11.1 (Hilbert's nullcone finiteness theorem).} Suppose
homogeneous positive-degree invariants \(q_1,\ldots,q_r\) have common
zero set \(\mathcal N\) in \(W\). Then \(B\) is a finite module over
\(\mathbb C [q_1,\ldots,q_r]\).

\textbf{Proof.} Choose homogeneous algebra generators \(b_1,\ldots,b_s\)
for \(B\). Each \(b_i\) vanishes on the common zero set of the \(q_j\).
The Nullstellensatz gives an integer \(N_i>0\) and an expression \[
b_i^{N_i}=\sum_j a_{ij}q_j\quad(a_{ij}\in A).
\] Applying \(\rho\) makes the coefficients invariant, so
\(b_i^{N_i}\in(q_1,\ldots,q_r)B\). Thus \(B/(q_1,\ldots,q_r)B\) is
spanned by the finitely many monomials \(\prod_i b_i^{e_i}\),
\(0\leq e_i<N_i\). Choose homogeneous lifts \(u_1,\ldots,u_t\) of a
homogeneous vector-space basis of that quotient.

For a homogeneous \(b\in B_d\), subtract a complex linear combination of
the \(u_i\) of degree \(d\) to obtain an element of
\((q_1,\ldots,q_r)B\). Write that remainder as \(\sum_jq_jc_j\),
choosing \(c_j\) of degree \(d-\deg q_j<d\). Induction on degree
expresses every \(c_j\), and hence \(b\), in the
\(\mathbb C [q_1,\ldots,q_r]\)-module generated by the \(u_i\). This
proves module finiteness. \(\square\)

This is stronger than the assertion that some finite generating set
exists: any homogeneous invariant equations cutting out the nullcone
give an integral, finite subalgebra. It does not assert that the \(q_j\)
are algebraically independent.

\subsection{12. Finite generation and a finite set of integer
solutions}\label{section-12}

\protect\phantomsection\label{lemma-12-1}{}

\textbf{Lemma 12.1 (Dickson).} Every subset of \(\mathbb N^m\), ordered
componentwise, has finitely many minimal elements. More strongly, every
infinite sequence in \(\mathbb N^m\) contains indices \(r<s\) with
\(v_r\leq v_s\).

\textbf{Proof.} A sequence of nonnegative integers has an infinite
nondecreasing subsequence. If some value occurs infinitely often, take
that constant subsequence. Otherwise every bounded set of values occurs
at only finitely many indices, so choose successively larger values at
larger indices. For the assertion in \(\mathbb N^m\), argue by induction
on \(m\). Take an infinite subsequence nondecreasing in the last
coordinate, and apply induction to the first \(m-1\) coordinates. The
resulting pair is comparable in every coordinate. An infinite set of
distinct minimal elements would give an infinite sequence without such a
pair. \(\square\)

\protect\phantomsection\label{theorem-12-2}{}

\textbf{Theorem 12.2 (Gordan's lemma).} For an integer matrix \(A\), the
additive monoid \[
M=\{v\in\mathbb N^m:Av=0\}
\] is generated by its finitely many minimal nonzero elements.

\textbf{Proof.} They form an antichain, so Lemma 12.1 makes them finite.
For \(v\in M\setminus\{0\}\), the finite box \(\{u:0\leq u\leq v\}\)
contains a minimal nonzero solution \(b\). It is minimal in all of
\(M\), since any smaller solution would still belong to the box.
Linearity gives \(A(v-b)=0\), and \(v-b\) is nonnegative. Repeating
subtraction strictly decreases the sum of the coordinates and therefore
terminates at zero. The subtracted elements express \(v\) as the
required sum. \(\square\)

The integer solutions record possible exponent patterns in products of
determinants. It does not by itself prove finiteness when the number of
symbolic letters is allowed to grow; that additional reduction must be
supplied.

\protect\phantomsection\label{theorem-12-3}{}

\textbf{Theorem 12.3 (Gordan's finiteness theorem).} The covariant
algebra of a finite system of binary forms over a characteristic-zero
field is finitely generated.

\textbf{Proof.} Over \(\mathbb C\), apply Theorem 10.1 to the polynomial
ring in all coefficient variables and \(x,y\). The special linear group
acts linearly on this finite-dimensional space of variables. Its
invariant ring is exactly the covariant algebra. The auxiliary vector
variables are part of the representation, so the conclusion concerns the
full covariant algebra.

To pass to any characteristic-zero field, first work over \(\mathbb Q\).
In each total degree the covariant subspace is the simultaneous kernel
of the two rational matrices for \(E,F\) in (B2). Extension of scalars
commutes with that kernel, since it is the solution space of finitely
many linear equations. Finite generation over \(\mathbb C\) gives a
bound \(N\) on the total degrees of a set of generators. Choose a
rational basis of the covariant spaces through degree \(N\), and let
\(B_0\) be the algebra they generate. In every degree, the quotient of
the rational covariant space by the corresponding component of \(B_0\)
becomes zero after tensoring with \(\mathbb C\); hence it was zero. Thus
\(B_0\) is the rational covariant algebra. Tensoring with \(k\), and
applying the same kernel argument in each degree, proves the theorem
over \(k\). \(\square\)

This is the Reynolds-and-Hilbert argument of Sections 9--10. Gordan's
original proof instead reduces iterated transvectants and uses
finiteness of their admissible exponent patterns. Kung and Rota's
\emph{The invariant theory of binary forms}, Section 6, gives two
constructive symbolic proofs; their Section 6.4, Proposition 6.3 and
Lemma 6.6, treats the integer-solution lemma. The kernel argument in
Sections 4--6 supplies an independent completeness proof for the small
forms.

\subsection{13. The nullcone of binary forms}\label{section-13}

For a binary \(d\)-ic, the root formulation has a complete elementary
proof. We need first a fact about symmetric root expressions.

\protect\phantomsection\label{lemma-13-1}{}

\textbf{Lemma 13.1 (descending a root expression).} Let \[
f(x,y)=\prod_{\nu=1}^d(u_\nu x+v_\nu y)
=\sum_{r=0}^d c_rx^{d-r}y^r.
\] A polynomial in the \(u_\nu,v_\nu\) that is symmetric in the \(d\)
pairs and homogeneous of degree \(q\) in each pair is a unique
polynomial, homogeneous of degree \(q\), in \(c_0,\ldots,c_d\). This
assertion and its proof hold with coefficients in any commutative ring,
including \(\mathbb Z\).

\textbf{Proof.} Set all \(u_\nu=1\). The resulting polynomial
\(P(v_1,\ldots,v_d)\) is symmetric and has degree at most \(q\) in each
\(v_\nu\). Use the usual leading-monomial proof of the fundamental
theorem of symmetric polynomials, with lexicographic order
\(v_1>\cdots>v_d\). A leading monomial has exponents
\(\alpha_1\geq\cdots\geq\alpha_d\), since exchanging a larger later
exponent with an earlier smaller one would give a larger monomial.
Subtract its coefficient times \[
e_1^{\alpha_1-\alpha_2}e_2^{\alpha_2-\alpha_3}
\cdots e_d^{\alpha_d}.
\] This product has the chosen leading monomial with coefficient one. It
uses \(\alpha_1\leq q\) elementary-symmetric factors, and has degree at
most \(q\) in each variable. The subtraction therefore stays within the
finite set of possible monomials of individual degree at most \(q\), and
strictly lowers the leading monomial. It terminates without division by
any integer or coefficient.

For \(u_\nu\ne0\), put \(z_\nu=v_\nu/u_\nu\). Homogeneity gives the
original polynomial as \((\prod u_\nu)^qP(z_1,\ldots,z_d)\). Here
\(c_0=\prod u_\nu\) and \(c_r=c_0e_r(z)\). A term using \(h\leq q\)
elementary-symmetric factors becomes their corresponding product of
\(c_r\)'s times \(c_0^{q-h}\). This is a homogeneous coefficient
polynomial of degree \(q\). The equality in the localization where the
\(u_\nu\) are inverted is a polynomial equality before localization:
multiplication by a monomial in polynomial variables is injective over
any coefficient ring.

For uniqueness, distinct products of the elementary symmetric functions
have distinct leading monomials, given by their successive cumulative
exponents, and leading coefficient one. They are therefore linearly
independent over any coefficient ring. A homogeneous relation of degree
\(q\) in the \(c_r\)'s, evaluated at every \(u_\nu=1\), becomes a
relation between these products in \(c_1=e_1,\ldots,c_d=e_d\); the power
of \(c_0\) in each term was fixed by total degree. Every coefficient is
zero. Different coefficient degrees give different degrees in each root
pair, so the entire coefficient-to-root substitution is injective as
well. \(\square\)

\protect\phantomsection\label{theorem-13-2}{}

\textbf{Theorem 13.2 (binary nullcone).} A binary form belongs to the
nullcone for \(\operatorname{SL}_2(\mathbb C)\) if and only if it has a
root of multiplicity strictly greater than \(d/2\). The zero form
belongs to the nullcone.

\textbf{Proof.} If a nonzero form has such a root, change coordinates so
its repeated factor is \(x^m\), \(m>d/2\). Under inverse substitution by
\(\operatorname{diag}(t^{-1},t)\), every monomial \(x^ry^{d-r}\)
occurring in it scales by \(t^{2r-d}\), with \(r\geq m\). Thus the
transformed coefficient vector tends to zero as \(t\to0\). Every
homogeneous positive-degree invariant is constant along this
transformation and continuous at zero, so its value on the form is zero.

Conversely, factor a nonzero form into \(d\) nonzero linear factors, and
assume no root line occurs more than \(d/2\) times. Make two copies of
each factor index, and arrange the \(2d\) copies so that indices with
the same root line form consecutive blocks. Every block has length at
most \(d\). Pair position \(r\) with position \(r+d\), for
\(1\leq r\leq d\). Two paired positions cannot belong to the same block,
so the bracket of their root covectors is nonzero.

The product \(M\) of these \(d\) brackets is therefore nonzero at this
tuple of root covectors. Each original factor index occurs twice, so
\(M\) has degree two in each root pair. Consider
\protect\phantomsection\label{equation-H4}{}

\[
\prod_{\sigma\in S_d}(T-\sigma M)
=T^{d!}+\sum_{j=1}^{d!}Q_jT^{d!-j}.
\tag{H4}
\] Each \(Q_j\) is symmetric in the root pairs, has degree \(2j\) in
each pair, and is special-linear invariant, since every bracket is.
Lemma 13.1 writes it as a coefficient polynomial \(I_j(f)\), homogeneous
of positive degree \(2j\). That coefficient polynomial is invariant: the
intertwining equality follows on every factored form, and every binary
form factors over \(\mathbb C\), so it holds on every coefficient
vector.

At least one \(Q_j\) is nonzero at the chosen tuple. Otherwise (H4)
would specialize to \(T^{d!}\), whose only root is zero, contradicting
the nonzero root \(M\). Hence some positive-degree invariant \(I_j\) is
nonzero at \(f\), and \(f\notin\mathcal N\). \(\square\)

For cubics the threshold is two: the nullcone is the double-root locus,
equivalently \(\Delta=0\). For quartics it is three. In Section 6 the
quartic invariant ring is \(\mathbb C [i,j]\), so the equations
\(i=j=0\) cut out precisely the triple-root locus. A double-root quartic
need not lie in the nullcone: \(x^2y^2\) has \(c=1/6\) and
\(i=1/12\ne0\).

The strict inequality matters for every even degree. A form with two
equally multiple roots is on the discriminant locus but has a nonzero
invariant.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A binary sextic with root multiplicities three, two and one. Two copies of each of its six factor slots are placed in twelve positions. Pairing each of the first six positions with the position six later joins different root lines and gives the nonzero product of brackets 14 squared, 25 squared and 36 squared.}]{figures/root-matching.png}}
\caption{A binary sextic with root multiplicities three, two and one.
Two copies of each of its six factor slots are placed in twelve
positions. Pairing each of the first six positions with the position six
later joins different root lines and gives the nonzero product of
brackets 14 squared, 25 squared and 36 squared.}
\end{figure}

This is the matching step of Theorem 13.2 for \(d=6\) and three distinct
root lines \(A,B,C\) of multiplicities \(3,2,1\). Each factor slot
occurs twice; every paired bracket joins different root lines. The
depicted nonzero product still needs the symmetric-polynomial step (H4)
to become a coefficient invariant. Editable diagram.

\subsection{Exercises}\label{exercises}

\protect\phantomsection\label{exercise-1}{}

\textbf{Exercise 1 --- A quadratic by two methods.} Prove that
\(h=ac-b^2\) is invariant under \(\operatorname{SL}_2\) and has
determinant weight two. Use both the differential criterion and a matrix
computation.

\hyperref[solution-1]{Solution}.

\protect\phantomsection\label{exercise-2}{}

\textbf{Exercise 2 --- Two cubic root types.} Compute \(H,T,\Delta\) for
\(x^3+y^3\) and \(x^2y\), and verify (B13).

\hyperref[solution-2]{Solution}.

\protect\phantomsection\label{exercise-3}{}

\textbf{Exercise 3 --- Balancing four coordinates.} Prove Dickson's
lemma and find the minimal generators of
\(\{v\in\mathbb N^4:v_1+v_2=v_3+v_4\}\).

\hyperref[solution-3]{Solution}.

\protect\phantomsection\label{exercise-4}{}

\textbf{Exercise 4 --- Signs and a quartic Hessian.} Prove the vanishing
of odd self-transvectants and compute the general quartic Hessian.

\hyperref[solution-4]{Solution}.

\subsection{Solutions}\label{solutions}

\subsubsection{Solution 1}\label{solution-1}

With no variable dependence, (B2) becomes \[
E=-a\partial_b-2b\partial_c,\qquad
F=-2b\partial_a-c\partial_b.
\] Consequently \(Eh=-a(-2b)-2ba=0\) and \(Fh=-2bc-c(-2b)=0\). Theorem
2.1 gives special-linear invariance. The coefficient degree is two and
order zero, so Proposition 1.1 gives weight two. Directly, inverse
substitution sends \(Q\) to \(g^{-\mathsf t}Qg^{-1}\), and taking its
determinant gives \(h(g\cdot f)=\det(g)^{-2}h(f)\).

\hyperref[exercise-1]{Exercise 1}.

\subsubsection{Solution 2}\label{solution-2}

The values from (B12) are \(2xy,x^3-y^3,-2\) in the first case. Their
substitution into (B13) gives the expansion displayed in Section 5. In
the second case \(a=c=d=0,b=1/3\), so \(H=-2x^2/9\). Formula (B4), with
orders three and two, gives \(T=2x^3/27\); since \(H\) is a square,
\(\Delta=0\). Thus \(2T^2+H^3=(8/729-8/729)x^6=0\). Formula (B14)
explains the discriminant difference.

\hyperref[exercise-2]{Exercise 2}.

\subsubsection{Solution 3}\label{solution-3}

The nondecreasing-subsequence induction in Lemma 12.1 is a proof of
Dickson's lemma for every \(m\). For this monoid the minimal nonzero
solutions are \[
(1,0,1,0),\quad(1,0,0,1),\quad
(0,1,1,0),\quad(0,1,0,1).
\] If a solution is nonzero, some coordinate on each side of the
equality is positive. Subtract the corresponding displayed vector,
preserving nonnegativity and the equality. Continue until all
coordinates vanish. Each displayed vector has coordinate sum two, the
smallest possible nonzero sum, so none can be decomposed or omitted.

\hyperref[exercise-3]{Exercise 3}.

\subsubsection{Solution 4}\label{solution-4}

Exchange the two variable pairs in \(\Omega^r\); each \(\Omega\) changes
sign, so \((Q,P)_r=(-1)^r(P,Q)_r\). For \(P=Q\) and odd \(r\), this says
that the transvectant equals its negative; characteristic zero forces
zero. For a quartic, \[
(f,f)_2=\frac{f_{xx}f_{yy}-f_{xy}^2}{72}.
\] Indeed (B4) has factor \(1/144\) and numerator
\(2f_{xx}f_{yy}-2f_{xy}^2\). Now \[
\begin{aligned}
f_{xx}&=12(ax^2+2bxy+cy^2),\\
f_{xy}&=12(bx^2+2cxy+dy^2),\\
f_{yy}&=12(cx^2+2dxy+ey^2).
\end{aligned}
\] Multiplying these quadratic polynomials gives exactly (B16).

\hyperref[exercise-4]{Exercise 4}.

\subsection{Further reading}\label{further-reading}

Kung and Rota,
\href{https://staff.math.su.se/shapiro/ProblemSolving/Rota.pdf}{The
invariant theory of binary forms}, Sections 2.2--2.3 and 6.4--6.6,
discusses symbolic evaluation, integer-solution finiteness and cubic
systems. Hilbert,
\href{https://gdz.sub.uni-goettingen.de/id/PPN235181684_0042}{Ueber die
vollen Invariantensysteme} (1893), Section 4, develops the nullcone
finiteness argument. The normalizations of binary covariants vary
between these accounts.

\end{document}