The degrees of a system of parameters of the ring of invariants of a binary form

. We consider the degrees of the elements of a homogeneous system of parameters for the ring of invariants of a binary form, give a divisibility condition, and a complete classi(cid:12)cation for forms of degree at most 8.


Hilbert's criterion
Hilbert's criterion gives a characterization of homogeneous systems of parameters as sets that define the nullcone.
Denote by V n the set of binary forms of degree n.The nullcone of V n , denoted N (V n ), is the set of binary forms of degree n on which all invariants vanish.By the Hilbert-Mumford numerical criterion (see [6] and [7,Chapter 2]) this is precisely the set of binary forms of degree n with a root of multiplicity > n 2 .Moreover, the binary forms with no root of multiplicity ≥ n 2 have closed SL(2, C)-orbits.The elements of N (V n ) are called nullforms.Another result from [6] that we will use is the following.Lemma 3.1.Fix integers j, t with t > 0. If an invariant of degree d is nonzero on a form a i x n−i y i with the property that all nonzero a i have i ≡ j (mod t), then d(n − 2j)/2 ≡ 0 (mod t).
Proof For an invariant of degree d with nonzero term a mi i we have m i = d and im i = nd/2.If i ≡ j (mod t) when a i = 0, then nd/2 = im i ≡ j m i = jd (mod t).
For odd n we recover the well-known fact that all degrees are even (take t = 1).Lemma 3.2.Fix integers j, t with t > 1 and 0 ≤ j ≤ n.Among the degrees d of a hsop, at least ⌊(n − j)/t⌋ satisfy d(n − 2j)/2 ≡ 0 (mod t).
Proof Subtracting a multiple of t from j results in a stronger statement, so it suffices to prove the lemma for 0 ≤ j < t.There are 1+⌊(n−j)/t⌋ =: 1+N coefficients a i with i ≡ j (mod t), so the subpace U of V n defined by a i = 0 for i ≡ j (mod t) has dimension 1 + N .If N = 0 there is nothing to prove, so we assume that N > 0. We claim that a general form f ∈ U has only zeroes of multiplicity strictly less than n/2.Indeed, write f = a j x n−j y j + a j+t x n−j−t y j+t + . . .+ a j+mt x n−j−mt y j+mt where j + (m + 1)t > n and m > 0. So f has a factor y of multiplicity j and a factor x of multiplicity n − j − mt.If j were at least n/2, then j + mt ≥ j + t > 2j ≥ n, a contradiction.If n − j − mt were at least n/2, then j + mt ≤ n/2 and hence t ≤ n/2 and hence j + (m + 1)t ≤ n, a contradiction.The remaining roots of f are roots of a j x mt + a j+t x (m−1)t y t + . . .+ a j+mt y mt , which is a general binary form of degree m in x t , y t and hence has mt distinct roots.
Let π : V n → V n //SL(2, C) be the quotient map; so the right-hand side is the spectrum of the invariant ring I. Set X := π(U ).We claim that X has dimension N .It certainly cannot have dimension larger than N , since acting with the one-dimensional torus of diagonal matrices on an element of U gives another element of U .To show that dim X = N we need to show that for general f ∈ U the fibre π −1 (π(f )) intersects U in a one-dimensional variety.By the above and the Hilbert-Mumford criterion, the SL(2, C)-orbit of f is closed.Moreover, its stabiliser is zero-dimensional.So by properties of the quotient map we have π −1 (π(f )) = SL(2, C) • f .Hence it suffices that the intersection of this orbit with U is one-dimensional.For this a Lie algebra argument suffices, in which we may ignore the Lie algebra of the torus: if (bx ∂ ∂y + cy ∂ ∂x )f lies in U , then we find that b = c = 0 if t > 2 (so that the contribution of one term from f cannot cancel the contribution from the next term); and b = 0 if j > 0 (look at the first term), and then also c = 0; and c = 0 if j + mt < n (look at the last term), and then also b = 0. Hence the only case that remains is t = 2, j = 0, and n ≥ 4 even.Then the equations ca 0 n + ba 2 2 = 0 and ca 2 (n − 2) + ba 4 4 = 0 are independent and force b = c = 0.
This concludes the proof that dim X = N .Intersecting X with the hypersurfaces corresponding to elements of an hsop reduces X to the single point in X representing the null-cone.In the process, dim X drops by N .But the only invariants that contribute to this dimension drop, i.e., the only invariants that do not vanish identically on X (hence on U ) are those considered in Lemma 3.1.Hence there must be at least N of these among the hsop.Lemma 3.3.Let t be an integer with t > 1.
(i) If n is odd, and j is minimal such that 0 ≤ j ≤ n and (n − 2j, t) = 1, then among the degrees of any hsop at least ⌊(n − j)/t⌋ are divisible by 2t.
(ii) If n is even, and j is minimal with 0 ≤ j ≤ 1 2 n and ( 1 2 n − j, t) = 1, then among the degrees of any hsop at least ⌊(n − j)/t⌋ are divisible by t.Theorem 3.4.Let t be an integer with t > 1.
(i) If n is odd, then among the degrees of any hsop at least ⌊(n − 1)/t⌋ are divisible by 2t (and all degrees are even).
(ii) If n is even, then among the degrees of any hsop at least ⌊(n − 1)/t⌋ are divisible by t, and if n ≡ 2 (mod 4) then at least n/2 by 2.
Proof (i) By part (i) of Lemma 3.3 we find a lower bound ⌊(n − j)/t⌋ for a j as described there.If that is smaller than ⌊(n − 1)/t⌋, then there is some multple at of t with n − j + 1 ≤ at ≤ n − 1.Put n = at + b, where 1 ≤ b ≤ j − 1.By definition of j we have (b − 2i, t) > 1 for i = 0, 1, ..., j − 1.If b is odd, say b = 2i + 1, we find a contradiction.If b is even, say b = 2i + 2, then t is even and n is even, contradiction.
(ii) By part (ii) of Lemma 3.3 we find a lower bound ⌊(n − j)/t⌋ for a j as described there.For t = 2 our claim follows.Now let t > 2. If ⌊(n − j)/t⌋ is smaller than ⌊(n − 1)/t⌋, then there is some multple at of t with n For example, it is known that there exist homogeneous systems of parameters with degree sequences 4 (n = 3); 2, 3 (n = 4); 4, 8, 12 (n = 5);

Poincaré series
If there exists a hsop with degrees d 1 , . . ., d n−2 , then the Poincaré series can be written as a quotient P (t) = a(t)/ (t di − 1) for some polynomial a(t) with nonnegative coefficients.If one does not have a hsop, but only a sequence of degrees, the conditions of Theorem 3.4 above are strong enough to guarantee that P (t) can be written in this way, but without the condition that the numerator has nonnegative coefficients.Proposition 4.1.Let d 1 , . . ., d n−2 be a sequence of positive integers satisfying the conditions of Theorem 3.4.Then P (t) (t di − 1) is a polynomial.
Proof Dixmier [4] proves that P (t)B(t) is a polynomial, where B(t) is defined by Consider a primitive t-th root of unity ζ.We have to show that if B(t) has root ζ with multiplicity m, then at least m of the d i are divisible by t, but this follows immediately from Theorem 3.4.Note that in case n ≡ 0 (mod 4) the factor ( We see that if n ≡ 0 (mod 4), n > 4, then P (t) can be written with a smaller denominator than corresponds to the degrees of a hsop.
We shall need the first few coefficients of P (t).Messy details arise for small n because there are too few invariants of certain small degrees.Let I be the ring of invariants of a binary form of degree (order) n, let I m be the graded part of I of degree m, and put h m = h n m = dim C I m , so that P (t) = m h m t m .The coefficients h n m can be computed by the Cayley-Sylvester formula: The dimension of the space of covariants of degree m and order a is zero when mn−a is odd, and equals N (n, m, t) − N (n, m, t − 1) if nm − a = 2t, where N (n, m, t) is the number of ways t can be written as sum of m integers in the range 0..n, that is, the number of Ferrers diagrams of size t that fit into a m × n rectangle.
We have Hermite reciprocity h n m = h m n , as follows immediately since reflection in the main diagonal shows N (n, m, t) = N (m, n, t).That means that Table 1 is symmetric.
Dixmier [4] gives the cases in which h m = 0. Since his statement is not precisely accurate, we repeat his proof.Proof (i) If n is odd, then all degrees are even.(ii) For n = 1 we have For n = 4 we have invariants of degrees 2, 3 and hence of all degrees m = 1.That means that h n 4 = 0.For n = 6 we have invariants of degrees 2, 15 and hence of all degrees m ≥ 14.That means that h n 6 = 0 for n ≥ 14.If n is odd this shows the presence of invariants of degrees 4, 6 and hence of all even degrees m > 2, provided n ≥ 15.For n = 5 we have invariants of degrees 4, 18 and hence of all even degrees m ≥ 16.That means that h n 5 = 0 for even n ≥ 16.If n is even this shows the presence of invariants of degrees 2, 5 and hence of all degrees m ≥ 4, provided n ≥ 16.It remains only to inspect the table for 4 ≤ m, n ≤ 14.

Dixmier's criterion
Dividing out the ideal spanned by p elements of a hsop diminishes the dimension by precisely (and hence at least) p.This means that the below gives a necessary and sufficient condition for a sequence of degrees to be the degree sequence of a hsop.Proposition 5.1.(Dixmier [4]) Let G be a reductive group over C, with a rational representation in a vector space R of finite dimension over ) be a sequence of positive integers.Assume that for each subsequence (j 1 , . . ., j p ) of (d 1 , . . ., d r ) the subset of points of V where all elements of all C[R] G j with j ∈ {j 1 , . . ., j p } vanish has codimension not less than p in V .Then This criterion is very convenient, it means that one can work with degrees only, without worrying about individual elements of a hsop.

Minimal degree sequences
If y 1 , ..., y r is a hsop, then also y e1 1 , ..., y er r for any sequence of positive integers e 1 , ..., e r , not all 1.This means that if the degree sequence d 1 , ..., d r occurs, also the sequence d 1 e 1 , ..., d r e r occurs.We would like to describe the minimal sequences, where such multiples are discarded.
There are further reasons for non-minimality.Proof We verify Dixmier's criterion.Consider a finite basis f 1 , ..., f s for the space of invariants of degree d ′ .Split the variety V in the s pieces defined by f i = 0 (1 ≤ i ≤ s) together with the single piece defined by f 1 = ... = f s = 0. Given p elements of the sequence d 1 , ..., d r−1 , d ′ + d ′′ we have to show that the codimension in V obtained by requiring all invariants of such degrees to vanish is at least p, that is, that the dimension is at most r − p.This is true by assumption if d ′ + d ′′ is not among these p elements.Otherwise, consider the s + 1 pieces separately.We wish to show that each has dimension at most r − p, then the same will hold for their union.For the last piece, where all invariants of degree d ′ vanish, this is true by assumption.But if some invariant of degree d ′ does not vanish, and all invariants of degree d ′ + d ′′ vanish, then all invariants of degree d ′′ vanish, and we are done.
Note that taking multiples is a special case of (repeated application of) this lemma, used with Let us call a sequence minimal if it occurs (as the degree sequence of the elements of a hsop), and its occurrence is not a consequence, via the above lemma or via taking multiples, of the occurrence of smaller sequences.We might try to classify all minimal sequences, at least in small cases.
Is it perhaps true that a hsop exists for any degree sequence that satisfies the conditions of Theorem 3.4 when there are sufficiently many invariants?E.g. when the coefficients of P (t) (1 − t di ) are nonnegative?
Example Some caution is required.For example, look at n = 6.The conditions of Theorem 3.4 are: at least three factors 2, at least one factor of each of 3, 4, 5.The sequence 6, 6, 6, 20 satisfies this restriction.Moreover, .Requiring all invariants of degree 6 to vanish is equivalent to the two conditions i 2 = i 6 = 0, and a hsop cannot contain three elements of degree 6.
Still, the above conditions almost suffice.And for n < 6 they actually do suffice.

n = 3
For n = 3 we only have simple multiples of the minimal degree.Proposition 6.2.A positive integer d is the degree of a hsop in case n = 3 if and only if it is divisible by 4.
If i 4 is an invariant of degree 4, then {i 4 } is a hsop.

n = 4
For n = 4 one has the sequence 2, 3, but for example also 5, 6.If i 2 and i 3 are invariants of degrees 2 and 3, then {i 2 , i 3 } is a hsop.Proposition 6.4.There is precisely one minimal degree sequence of hsops in case n = 4, namely 2, 3.

n = 5
Proposition 6.5.A sequence d 1 , d 2 , d 3 of three positive integers is the sequence of degrees of a hsop for the quintic if and only if all d i are even, and distinct from 2, 6, 10, 14, and no two are 4, 4 or 4, 22 and at least two are divisible by 4, at least one is divisible by 6, and at least one is divisible by 8.
The stated conditions suffice: We use (and verify below) that there are hsops with degrees 4, 8, 12 and with degrees 4, 8, 18.If all d i are divisible by 4, and we do not have a multiple of 4, 8, 12, then we have 4a, 4b, 24c where a and b have no factor 2 or 3, and not both are 1.It suffices to find 4, 4b, 24.Since 4, 8, 24 exists, we can decrease b by 2, and it suffices to find 4, 12, 24, which exists.
So, some d i , is not divisible by 4. We have one of the three cases 24a, 4b, 2c and 8a, 12b, 2c and 8a, 4b, 6c, where c is odd.In the middle case we have c ≥ 9 and it suffices to make 8, 12, 2c.Since 8, 12, 4 exists, we can reduce c by 2, and it suffices to make 8, 12, 18, which exists since 4, 8, 18 exists.
Finally in the last case we have c ≥ 3, and since 8, 4, 12 exists we can reduce c by 2. So it suffices to do 4, 8, 18, and that exists.Proposition 6.6.There are precisely two minimal degree sequences of hsops in case n = 5, namely 4, 8, 12 and 4, 8, 18.
Proof By the proof of the previous proposition, all we have to do is show the existence of hsops with the indicated degree sequences.It is well-known (see, e.g., Schur [9], p.86) that the quintic has four invariants i 4 , i 8 , i 12 , i 18 (with degrees as indicated by the index) that generate the ring of invariants, and every invariant of degree divisible by 4 (in particular i 2 18 ) is a polynomial in the first three.Thus, when i 4 , i 8 , i 12 vanish, all invariants vanish, and {i 4 , i 8 , i 12 } is a hsop.Knowing this, it is easy to see that also {i 4 , i 8 , i 18 } is a hsop: a simple Groebner computation shows that i 3 12 ∈ (i 4 , i 8 , i 18 ), hence N (V 5 ) = V(i 4 , i 8 , i 18 ).

n = 6
Similarly, we find for n = 6: Proposition 6.7.A sequence d 1 , d 2 , d 3 , d 4 of four positive integers is the sequence of degrees of a hsop for the sextic if and only if all d i are distinct from 1, 3, 5, 7, 9, 11, 13, and no two are in {2, 17}, and no three are in {2, 4, 8, 14, 17, 19, 23, 29}, and no three are in {2, 6, 17, 21}, and at least three are divisible by 2, at least one is divisible by 3, at least one by 4, and at least one by 5.
Proof For n = 6 the Poincaré series is Proof By the proof of the previous proposition, all we have to do is show the existence of hsops with the indicated degree sequences.It is well-known (see, e.g., Schur [9], p.90) that the sextic has five invariants i 2 , i 4 , i 6 , i 10 , i 15 (with degrees as indicated by the index) that generate the ring of invariants, where i 2 15 is a polynomial in the first four.This implies that N (V 6 ) = V(i 2 , i 4 , i 6 , i 10 ), so that {i 2 , i 4 , i 6 , i 10 } is a hsop.Now {i 2 , i 4 , i 6 , i 15 } and {i 2 , i 4 , i 10 , i 15 } are also hsops: we verified by computer that i 3 10 ∈ (i 2 , i 4 , i 6 , i 15 ) and i 5 6 ∈ (i 2 , i 4 , i 10 , i 15 ), so that N (V 6 ) = V(i 2 , i 4 , i 6 , i 15 ) = V(i 2 , i 4 , i 10 , i 15 ).

n = 7
For n = 7 we have to consider the invariants a bit more closely in order to decide which degree sequences are admissable for hsops.
We can save some work by observing that Dixmier [3] already showed the existence of hsops with degree sequences 4, 8, 8, 12, 30 and 4, 8, 12, 12, 20.It follows that [8]  Finally for p = 5 we have to show that each of these 23 sets determines the nullcone.But that follows immediately, since it is known already that [8,12,20 Consider the part of V defined by ψ = 0. Dixmier shows that if ψ = q 12 = p 20 = 0 (for certain invariants q 12 and p 20 of degrees 12 and 20, respectively), then f is a nullform.It follows that the subsets of V defined by ψ = q 12 = 0 or by ψ = p 20 = 0 have codimension at least 4 in V .
Proof We have We see that This shows that the given conditions are necessary.For sufficiency, use induction.The basis of the induction is provided by the 13 hsops constructed in the next proposition.Given a sequence of six numbers satisfying the conditions, order the numbers in such a way that the last is divisible by 7 and at least one of the last two is divisible by 5.All restrictions concern numbers at most 11, so if we split a number from the sequence into two parts each at least 12, such that the divisibility conditions remain true for the two resulting sequences, then by Lemma 6.1 and induction there exists a hsop with the given sequence as degree sequence.This means that one can reduce the first four numbers modulo 12, the fifth modulo 60, and the last modulo 420.It remains to check a 24 × 24 × 24 × 24 × 72 × 432 box, and this is done by a small computer program.Proposition 6.12.There are precisely 13 minimal degree sequences of hsops in case n = 8, namely We can save some work by observing that Shioda [10] already showed the existence of a hsop with degree sequence 2, 3, 4, 5, 6, 7. It follows that [d 1 , ..., d p ] ≥ p when (at least) p of the numbers 2, 3, 4, 5, 6, 7 divide some of the d i .
For p = 1, nothing remains to check.For p = 2, there only remains to show [9] ≥ 2, and this follows since there are two invariants of degree 9 without common factor, for example i 3 i 6 and i 4 i 5 .
Since adding a single condition diminishes the dimension by at most one, [3,8] ≥ 4 follows from [3,5,8] ≥ 5. (Given that i 2 vanishes since i 4  2 has degree 8, the condition that all invariants of degree 5 vanish is equivalent to the requirement that i 5 vanishes.)Similarly [5,8]  For example, we want dim V (i 2 , i 3 , i 4 , i 5 , i 8 ) = 1.Now i 2 , i 3 , i 4 , i 5 form part of a hsop, so V (i 2 , i 3 , i 4 , i 5 ) is irreducible and has dimension 2. Moreover i 8 does not vanish identically on V (i 2 , i 3 , i 4 , i 5 ) as we shall see, and it follows that dim V (i 2 , i 3 , i 4 , i 5 , i 8 ) = 1.
This argument works in all cases except that of V (i 2 , i 3 , i 4 , i 8 , i 9 ) and shows that each of the claimed sequences of degrees with the possible exception of 2, 3, 4, 8, 9, 210, is that of a hsop.In particular, e.g. 2, 3, 4, 5, 8, 42 is the sequences of degrees of a hsop.But now this argument also applies to V (i 2 , i 3 , i 4 , i 8 , i 9 ): V (i 2 , i 3 , i 4 , i 8 ) is irreducible of dimension 2 and i 9 does not vanish identically on it, and it follows that V (i 2 , i 3 , i 4 , i 8 , i 9 ) has dimension 1.

Proposition 6 . 3 .
A sequence d 1 , d 2 of two positive integers is the sequence of degrees of a hsop for the quartic if and only if neither of them equals 1, at least one is divisible by 2, and at least one is divisible by 3.Proof Clearly the conditions are necessary.In order to show that they suffice apply induction and the known existence of a hsop with degrees 2, 3.If d 2 > 7, then apply Lemma 6.1 to the two sequences d 1 , 6 and d 1 , d 2 − 6 to conclude the existence of a hsop with degrees d 1 , d 2 .If 2 ≤ d 1 , d 2 ≤ 7 and one is divisible by 2, the other by 3, then we have a multiple of the sequence 2, 3. Otherwise, one equals 6 and the other is 5 or 7.But 5, 6 is obtained from 2, 6 and 3, 6, and 7, 6 is obtained from 2, 6 and 5, 6.
Any sequence d 1 , ..., d n−2 of sufficiently large integers satisfying the divisibility conditions of Theorem 3.4 is the sequence of degrees of a hsop.
for A and B as in the table below.
Proof Minimality is immediately clear, so we only have to show existence.Apply Dixmier's criterion.As before we have to show that for all p and each subsequence d 1 , ..., d p of one of these 13 sequences the inequality [d 1 , ..., d p ] ≥ p holds.