The invariants of the binary nonic

This chapter is based on joint work with Brouwer [BP10a].

Let f ∈ V

9

,

f = a

0

x

9

+ 9a

1

x

8

y + . . . + 9a

8

xy

8

+ a

9

y

9

.

Consider the following covariants of V

9

:

c

1

= (f, f )

8

,

c

2

= (f, f )

6

,

c

3

= (f, f )

4

,

c

4

= (f, f )

2

,

c

5

= (f, c

1

)

2

,

c

6

= (f, c

2

)

6

,

c

7

= (c

2

, c

2

)

4

,

c

8

= (c

5

, c

5

)

6

,

c

9

= (c

5

, c

5

)

4

,

c

10

= (c

5

, c

8

)

2

,

c

11

= (c

9

, c

9

)

4

,

c

12

= (c

9

, c

11

)

4

,

c

13

= (c

2

, c

7

)

4

,

c

14

= (c

6

, c

6

)

2

,

c

15

= (c

2

, c

4

)

6

,

c

16

= (c

2

, c

4

)

4

,

c

17

= (c

2

, c

3

)

6

,

c

18

= (c

5

, c

6

)

3

,

c

19

= (c

1

, c

3

)

2

,

c

20

= (f, c

1

)

2

,

c

21

= (f, c

2

)

2

,

c

22

= (f, c

3

)

6

,

c

23

= (f, c

3

)

8

,

c

24

= (f, c

4

)

8

,

c

25

= (c

4

, c

4

)

10

,

c

26

= (c

3

, c

3

)

6

,

c

27

= (c

36

, c

6

)

3

,

c

28

= (c

3

, c

4

)

10

,

c

29

= (c

2

, c

4

)

2

,

c

30

= (c

3

, c

4

)

8

.

Lemma 4.8.1. If c

1

= 0, c

2

6= 0, and (c

2

, c

3

) ∈ N (V

3

⊕ V

6

), then f has a root

of multiplicity at least 6.

Proof. From (c

2

, c

3

) ∈ N (V

6

⊕ V

3

)it follows that both c

2

and c

3

are nullforms

and have a common root that has multiplicity 4 in c

2

and multiplicity 2 in c

3

.

Without loss of generality, we consider the following 3 cases: c

2

= x

6

, c

2

= x

5

y,

and c

2

= x

4

y(x + y).

Case 1: c

2

= x

6

. Then

and x

2

_{must divide c}

3

. We obtain a

9

= a

8

= 0, and substitute that in c

2

and

c

1

:

c

2

= (f, f )

6

= (−20a

62

+ 30a

5

a

7

)y

6

+ (−30a

5

a

6

+ 54a

4

a

7

)xy

5

+

(−90a

₅2

+ 114a

4

a

6

− 12a

3

a

7

)x

2

y

4

+

(−72a

4

a

5

+ 124a

3

a

6

− 60a

2

a

7

)x

3

y

3

+

(−90a

₄2

+ 114a

3

a

5

− 12a

2

a

6

− 18a

1

a

7

)x

4

y

2

+

(−30a

3

a

4

+ 54a

2

a

5

− 30a

1

a

6

+ 6a

0

a

7

)x

5

y+

(−20a

₃2

+ 30a

2

a

4

− 12a

1

a

5

+ 2a

0

a

6

)x

6

,

c

1

= (f, f )

8

= (70a

52

− 112a

4

a

6

+ 56a

3

a

7

)y

2

+

(28a

4

a

5

− 56a

3

a

6

+ 40a

2

a

7

)xy+

(70a

₄2

− 112a

3

a

5

+ 56a

2

a

6

− 16a

1

a

7

)x

2

.

Since we suppose c

2

= x

6

and c

1

= 0, the coecients of x

i

y

6−i

in c

2

, and of

x

j

_y

2−j

_{in c}

1

are 0 for 0 ≤ i ≤ 5 and 0 ≤ j ≤ 2.

If a

7

= 0, then it follows that a

6

= a

5

= a

4

= 0, and then x

6

| f, so f will have

a root of multiplicity 6. If a

7

6= 0, then

a

5

=

2a

2 6

3a

7

,

a

4

=

10a

3 6

27a

2 7

,

a

3

=

5a

4 6

27a

3 7

,

a

2

=

7a

5 6

81a

4 7

, a

1

=

28a

6 6

729a

5 7

, a

0

=

4a

7 6

243a

6 7

,

but then we have c

2

= 0, contrary to the assumption.

Case 2: c

2

= x

5

y. Then,

c

3

= (f, c

2

)

6

= − a

8

y

3

− 3a

7

xy

2

− 3a

6

x

2

y − a

5

x

3

,

and x

2

_{must divide c}

3

. We obtain a

8

= a

7

= 0, and substitute this in c

2

and c

1

:

c

2

= (f, f )

6

= (−20a

62

+ 2a

3

a

9

)y

6

+ (−30a

5

a

6

+ 6a

2

a

9

)xy

5

+

(−90a

₅2

+ 114a

4

a

6

+ 6a

1

a

9

)x

2

y

4

+ · · · +

(−90a

₄2

+ 114a

3

a

5

− 12a

2

a

6

)x

4

y

2

+

(−30a

3

a

4

+ 54a

2

a

5

− 30a

1

a

6

)x

5

y + · · ·

c

1

= (f, f )

8

= (70a

52

− 112a

4

a

6

+ 2a

1

a

9

)y

2

+ · · ·

Since we supposed c

2

= x

5

y

and c

1

= 0, the coecient c of y

2

in c

1

, and the

coecients d

i

of x

i

y

6−i

in c

2

vanish for 0 ≤ i ≤ 4, while d

5

6= 0. Now

5d

5

a

9

= −75a

4

d

0

+ 45a

5

d

1

− a

6

(9c + 22d

2

) = 0,

Case 3: c

2

= x

4

y(x + y). Then:

c

3

= (f, c

2

)

6

= (a

7

− a

8

)y

3

+ 3(a

6

− a

7

)xy

2

+ 3(a

5

− a

6

)x

2

y + (a

4

− a

5

)x

3

,

and x

2

_{must divide c}

3

. We obtain a

8

= a

7

= a

6

which we replace in c

2

and c

1

:

c

2

= (f, f )

6

= − 2(6a

4

a

6

− 15a

5

a

6

+ 10a

62

− a

3

a

9

)y

6

−

− 6(5a

3

a

6

− 9a

4

a

6

+ 5a

5

a

6

− a

2

a

9

)xy

5

−

− 6(15a

52

+ 3a

2

a

6

+ 2a

3

a

6

− 19a

4

a

6

− a

1

a

9

)x

2

y

4

−

− 2(36a

4

a

5

− 3a

1

a

6

+ 30a

2

a

6

− 62a

3

a

6

− a

0

a

9

)x

3

y

3

−

− 6(15a

2

4

− 19a

3

a

5

− a

0

a

6

+ 3a

1

a

6

+ 2a

2

a

6

)x

4

y

2

−

− 6(5a

3

a

4

− 9a

2

a

5

− a

0

a

6

+ 5a

1

a

6

)x

5

y−

− 2(10a

2

3

− 15a

2

a

4

+ 6a

1

a

5

− a

0

a

6

)x

6

,

c

1

= (f, f )

8

= 2(35a

52

− 8a

2

a

6

+ 28a

3

a

6

− 56a

4

a

6

+ a

1

a

9

)y

2

+

2(14a

4

a

5

− 7a

1

a

6

+ 20a

2

a

6

− 28a

3

a

6

+ a

0

a

9

)xy+

2(35a

42

− 56a

3

a

5

+ a

0

a

6

− 8a

1

a

6

+ 28a

2

a

6

)x

2

.

As we supposed c

2

= x

4

y(x + y)and c

1

= 0, the coecients of y

6

, xy

5

, x

2

y

4

,

x

3

_y

3

_{, x}

6

_{in c}

2

and all coecients of c

1

must vanish. We denote by J the ideal

generated by these coecients. Also, we denote by p

1

, p

2

the coecients of

x

4

_y

2

_{and x}

5

_y

_{in c}

2

:

p

1

= 15a

42

− 19a

3

a

5

− a

0

a

6

+ 3a

1

a

6

+ 2a

2

a

6

,

p

2

= 5a

3

a

4

− 9a

2

a

5

− a

0

a

6

+ 5a

1

a

6

.

A Gröbner basis computation shows that p

4

1

, p

22

∈ J

so that p

1

and p

2

vanish,

contradicting the assumption c

2

= x

4

y(x + y).

Consider now the following invariants of V

9

:

j

4,1

= (c

1

, c

1

)

2

,

j

8,1

= (c

2

, c

26

)

6

,

j

12,1

= ((c

7

, c

7

)

2

, c

7

)

4

,

j

12,2

= (c

9

, c

31

)

6

,

j

14,1

= (c

2

, c

27

)

3

)

6

,

j

16,1

= ((c

5

, c

5

)

2

, c

51

)

10

,

j

10,1

= (((c

25

, f )

6

, c

21

)

5

, c

2

)

6

.

Proposition 4.8.2. The seven invariants j

4

, j

8,1

, j

10,1

, j

12,1

, j

12,2

, j

14,1

, j

16,1

form a homogeneous system of parameters for the algebra O(V

9

)

SL2

of invariants

of the binary nonic.

Note that Dixmier (1985) proved the following.

Proposition 4.8.3. (Dixmier [Dix85]) The algebra O(V

9

)

SL2

has a homoge-

Proposition4.8.2gives an explicit system of parameters for O(V

9

)

SL2

with these

degrees. Brouwer & Popoviciu ([BP10a]) proved the existence of systems of

parameters for certain further sequences of degrees in the following.

Proposition 4.8.4. ([BP10a, Proposition 7.2]) The algebra of invariants of V

9

has systems of parameters with each of the ve sequences of degrees 4, 8, 10,

12, 12, 14, 16 and 4, 4, 10, 12, 14, 16, 24 and 4, 4, 8, 12, 14, 16, 30 and 4, 4,

8, 10, 12, 16, 42 and 4, 4, 8, 10, 12, 14, 48.

Before proving Proposition4.8.2, we give the following lemma.

Lemma 4.8.5. Let s ∈ V

6

, c ∈ V

3

be two non-zero binary forms. If both s and

care nullforms and if

0 = ((s

2

, s)

6

, c

2

)

6

= (((s, s)

2

, s)

1

, c

4

)

12

= (s, c

2

)

6

= (s, (c, c)

23

)

6

=

= (s, (c

3

, c)

3

)

6

,

then (s, c) ∈ N (V

6

⊕ V

3

).

Proof. Suppose that (s, c) /∈ N (V

6

⊕ V

3

). This means that s and c have no

common root which has multiplicity 4 in s and multiplicity 2 in c. Without loss

of generality we consider two cases:

(

s = x

4

(b

1

x

2

+ b

2

xy + b

3

y

2

),

c = y

3

_.

and

(

s = x

4

(b

1

x

2

+ b

2

xy + b

3

y

2

),

c = xy

2

_.

Case 1: c = y

3

_.

In this case we have (∼ is used for equalities up to a nonzero constant):

0 = ((s

2

, s)

6

, c

2

)

6

∼ b

33

,

0 = (((s, s)

2

, s)

1

, c

4

)

12

∼ b

2

(5b

22

− 18b

1

b

3

),

0 = (s, c

2

)

6

= b

1

,

and it follows that b

1

= b

2

= b

3

= 0, which implies s = 0, contradicting the

assumption s 6= 0.

Case 2: c = xy

2

_.

In this case we have (∼ is used for equalities up to a nonzero constant):

0 = (s, (c, c)

₂3

)

6

∼ b

1

,

0 = (s, c

2

)

6

∼ b

3

,

0 = (s, (c

3

, c)

3

)

6

∼ b

2

.

and it follows that b

1

= b

2

= b

3

= 0, which implies s = 0, contradicting the

Proof of Proposition

4.8.2. We introduce the invariants

j

4,2

= (c

2

, c

2

)

6

,

j

8,2

= (c

7

, c

7

)

4

,

j

8,3

= (c

8

, c

1

)

2

,

j

20,3

= (c

132

, (c

7

, c

7

)

2

)

4

,

j

20,4

= (c

25

, c

7 1

)

14

.

First we show that

N (V

9

) = V(j

4,1

, j

4,2

, j

8,2

, j

8,3

, j

12,1

, j

12,2

, j

14,1

, j

16,1

, j

20,3

, j

20,4

).

If j

4,1

= 0, then c

1

∈ V

2

is a nullform. Without loss of generality, we consider

the following two cases: c

1

= 0

and c

1

= x

2

.

Case 1: c

1

= 0.

If j

4,2

= j

8,2

= j

12,1

= j

20,3

= 0, then c

2

is a nullform.

If c

2

vanishes as well, then from Proposition

2.7.2

it follows that f has a

root of multiplicity 7, i.e. f is a nullform.

If c

2

6= 0, without loss of generality, we suppose x

4

| c

2

. Modulo the ideal

generated by the coecients of c

1

and the coecients of x

3

y

3

, x

2

y

4

, xy

5

, y

6

in

c

2

we have

j

8,1

= j

12,3

= j

12,4

= j

18,1

= j

20,1

= 0

where

j

12,3

= (c

14

, c

14

)

2

,

j

12,4

= ((c

22

, c

2

)

6

, c

26

)

6

,

j

18,1

= (((c

2

, c

2

)

2

, c

2

)

1

, c

64

)

12

,

j

20,1

= (c

2

, c

314

)

6

.

It follows, then, from Lemma4.8.5, that (c

2

, c

6

) ∈ N (V

6

⊕ V

3

), and then, from

Lemma4.8.1, that f has a root of multiplicity 6, i.e. f is a nullform.

Case 2: c

1

= x

2

.

Here we have (∼ is used for equalities up to a nonzero constant):

0 = j

20,4

= a

92

,

0 = j

16,1

∼ a

82

,

0 = j

12,2

∼ a

72

,

0 = j

8,3

∼ a

62

.

It follows that a

9

= a

8

= a

7

= a

6

= 0, and, if we combine this with c

1

= x

2

, we

get a

5

= 0

too, and then f is a nullform.

This shows that N (V

9

) = V(j

4,1

, j

4,2

, j

8,2

, j

8,3

, j

12,1

, j

12,2

, j

14,1

, j

16,1

, j

20,3

, j

20,4

).

We introduce now the following invariants:

j

8,4

= (c

7

, c

21

)

4

,

j

8,5

= (c

7

, c

17

)

4

,

j

10,2

= (c

18

, c

7

)

4

,

j

10,4

= ((c

6

, c

22

)

3

, (c

3

, c

3

)

8

)

4

,

The invariants j

8,1

, j

8,2

, . . . , j

8,5

are linearly independent and, together with

j

4,12

, j

4,22

, and j

4,1

j

4,2

, span the vector space of invariants of degree 8 which is of

dimension 8. (This can be seen, e.g., by a small computation in Mathematica.)

In a similar way it can be seen that the vector space of invariants of degree 10

is spanned by j

10,1

, j

10,2

, . . . j

10,5

. Using invariants of degree ≤ 16 we built a list

of 219 monomials of degree 20, each of them dividing one of the invariants j

4,1

,

j

4,2

, j

8,1

, . . . , j

8,5

, j

10,5

, and j

10,1

, to which we added

j

20,1

= (c

143

, c

2

)

6

,

j

20,2

= ((c

27

, c

2

)

4

, (c

24

, c

23

)

3

)

4

.

Denote I be the algebra O(V

9

)

SL2

of invariants, and I

i

its i-th graded part.

We evaluated the monomials at dim

C

I

20

= 217

random points in V

9

, giving as

result a matrix of (full) rank 217. Adding j

20,3

, j

20,4

, j

10,22

, j

10,32

, and j

10,42

to

the list of monomials and repeating the evaluation step gave (of course) again

matrices of rank 217. From the nullspaces of these matrices we obtained the

relations

j

20,3

, j

20,4

, j

10,22

, j

10,32

, j

10,42

∈ (j

4,1

, j

4,2

, j

8,1

, j

8,2

, . . . , j

8,5

, j

10,1

, j

10,5

)

(that is, j

20,1

and j

20,2

are not needed to span the elements mentioned). Using

invariants of degree ≤ 20 we built a list of 3561 monomials of degree 32, each of

them dividing one of the invariants j

4,1

, j

8,1

, j

8,5

, j

10,1

, j

10,5

, j

12,1

, j

12,2

, j

14,1

,

or j

16,1

. We evaluated the monomials at dim

C

I

32

= 2082random points in V

9

,

and this resulted in a matrix of rank 2082. The rank computations were made

modulo 32003, but as we obtained the maximal rank, these monomials must

span I

32

. It follows that

j

8,2

, j

8,3

, j

8,4

, j

4,2

∈

q

(j

4,1

, j

8,1

, j

8,5

, j

10,1

, j

10,5

, j

12,1

, j

12,2

, j

14,1

, j

16,1

),

and then we get

N (V

9

) = V(j

4,1

, j

8,1

, j

8,5

, j

10,1

, j

10,5

, j

12,1

, j

12,2

, j

14,1

, j

16,1

).

In the same way one can show that

N (V

9

) = V(j

4,2

, j

8,1

, j

8,5

, j

10,1

, j

10,5

, j

12,1

, j

12,2

, j

14,1

, j

16,1

).

It remains to remove two elements from one of these two sets of generators.

Let H = (j

4,1

, j

8,1

, j

10,1

, j

12,1

, j

12,2

, j

14,1

, j

16,1

). We computed dim

_C

I

i

∩ H

for i ≤ 60 and found dim

C

I

60

∩ H = 59107 = dim

C

I

60

, so that I

60

⊆ H. But

then H contains powers of all invariants of degrees 4, 10, 20, so that in particular

j

4,2

, j

10,5

∈

√

H. Now let H

0

= (j

4,1

, j

4,2

, j

8,1

, j

10,1

, j

12,1

, j

12,2

, j

14,1

, j

16,1

). We

computed dim

C

I

i

∩ H

0

for i ≤ 40 and found dim

C

I

40

∩ H

0

= 6612 = dim

C

I

40

,

so that I

40

⊆ H

0

. But then H

0

contains powers of all invariants of degree 8, so

that in particular j

8,5

∈

√

H

0

_{. But then}√_{H =}√_H

0

_{= I. Thus,}

Then, from Proposition2.7.1, it follows that the set

{j

4,1

, j

8,1

, j

10,1

, j

12,1

, j

12,2

, j

14,1

, j

16,1

}

is a system of parameters of O(V

9

)

SL2

.

Proposition 4.8.6. The algebra of invariants of the binary nonic is generated

by 92 invariants. The nonzero numbers d

i

of basic invariants of degree i are

i

4 8 10 12 14 16 18 20 22

d

i

2 5 5 14 17 21 25 2

1

Proof. The Poincaré series of O(V

9

)

SL2

is

P (t) =

a(t)

(1 − t

4

_{)(1 − t}

8

_{)(1 − t}

10

_{)(1 − t}

12

₎

2

_{(1 − t}

14

_{)(1 − t}

16

₎

with

a(t) = 1 + t

4

+ 5t

8

+ 4t

10

+ 17t

12

+ 20t

14

+ 47t

16

+ 61t

18

+ 97t

20

+

120t

22

+ 165t

24

+ 189t

26

+ 223t

28

+ 241t

30

+ 254t

32

+ 254t

34

+

241t

36

+ 223t

38

+ 189t

40

+ 165t

42

+ 120t

44

+ 97t

46

+ 61t

48

+

47t

50

+ 20t

52

+ 17t

54

+ 4t

56

+ 5t

58

+ t

62

+ t

66

,

so that

P (t) = 1 + 2t

4

+ 8t

8

+ 5t

10

+ 28t

12

+ 27t

14

+ 84t

16

+ 99t

18

+ 217t

20

+

273t

22

+ 506t

24

+ 647t

26

+ 1066t

28

+ 1367t

30

+ 2082t

32

+ 2649t

34

+

3811t

36

+ 4796t

38

+ 6612t

40

+ 8228t

42

+ 10960t

44

+ 13483t

46

+

17487t

48

+ 21274t

50

+ 26979t

52

+ 32490t

54

+ 40443t

56

+ 48242t

58

+

59107t

60

+ 69885t

62

+ 84470t

64

+ 99074t

66

+ ...

We apply the strategy described in Chap.

3.1. From the Poincaré series, the

maximal degree in which we have to look for generating invariants is 66. Up to

degree 22 we found the following generators:

deg generators 4 j4,1= (c1, c1)2, j4,2= (c2, c2)6, 8 j8,1= (c2, c26)6, j8,2= (c7, c7)4, j8,3= (c8, c1)2, j8,4= (c7, c21)4, j8,5= (c7, c17)4, 10 j10,1= (((c25, f )6, c21)5, c2)6, j10,2= (c18, c7)4, j10,3= (c18, c21)4, j10,4= ((c6, c22)3, (c3, c3)8)4, j10,5= (((c26, f )6, c20)3, c2)6, 12 j12,1= ((c7, c7)2, c7)4, j12,2= (c9, c31)6, j12,3= (c14, c14)2, j12,4= ((c22, c2)6, c26)6, j12,5= (c8, c8)2, j12,6= (c8, c13)2, j12,7= ((c7, c2)2, (c7, c3)4)6, j12,8= ((c7, c3)2, (c7, c4)4)10, j12,9= (c14, c8)2, j12,10= (c14, c13)2, j12,11= (c26, (c7, c3)4)6, j12,12= (c6c5, (c7, c3)2)10, j12,13= ((c6, c5)2, (c7, c2)2)6, j12,14= ((c8, c4)2, c22)12,

deg generators 14 j14,1= (c2, c27)3)6, j14,2= (c2, (c28c17, c1c2)5)6, j14,3= (c4, (c15c19, c1c4)9)14, j14,4= (c1, (c15c19, c1c4)15)2, j14,5= (c2, (c15c19, c1c4)13)6, j14,6= (c3, (c15c19, c1c4)11)10, j14,7= (c1, (c28c17, c1c2)7)2, j14,8= (c3, (c28c17, c1c2)3)10, j14,9= (c4, (c28c17, c1c2)1)14, j14,10= (((c16, c19)6, c2)5, c21)4, j14,11= (((c16, c19)4, c1)1, c22)12, j14,12= (((c16, c19)2, c2)1, c23)20, j14,13= ((c29c19, c2)1, c24)28, j14,14= ((c16, c19)7, c31)6, j14,15= ((c16, c19)1, c32)18, j14,16= ((c2c4, c1c3)1, c33)30, j14,17= ((c29, c3)1, c42)24, 16 j16,1= ((c5, c5)2, c51)10, j16,2= (c7c8, c26)6, j16,3= ((c1, c4)2, c3₇)12, j16,4= ((c2, c3)2, c3₇)12, j16,5= ((c2, c3)2, c46)12, j16,6= ((c1, c4)2, c46)12, j16,7= (c2c41, c25)14, j16,8= ((c2c3, c1c2)8, c41)8, j16,9= ((c2c3, c3c4)16, c4₁)8, j16,10= ((c2₄, c2c4)20, c4₁)8, j16,11= ((c24, c2c4)20, c27)8, j16,12= ((c3c4, c2c3)16, c27)8, j16,13= ((c3c2, c2c1)8, c27)8, j16,14= (((c1, c3)2, c16)6, c27)8, j16,15= (((c1, c3)2, c16)6, c4₁)8, j16,16= (((c2, c3)4, c30)4, c4₁)8, j16,17= (((c2, c3)4, c30)4, c27)8, j16,18= ((c29, c19)8, c27)8, j16,19= ((c29, c19)8, c41)8, j16,20= (c29c19, c42)24, j16,21= (c21c22, (c16, c19)2)16, 18 j18,1= (((c2, c2)2, c2)1, c64)12, j18,2= (c28, c18)4, j18,3= ((c7, c1c8)2, c18)4, j18,4= (c27, (c30, (c3, (f, (f, c2)3)6)4)4)8, j18,5= (c21, ((c29, c3)1, c42)22)4, j18,6= (c21, ((c2c4, c1c3)1, c33)28)4, j18,7= (c21, ((c16, c19)1, c32)16)4, j18,8= (c1c2, ((c16, c19)7, c31)2)8, 18 j18,9= (c21, ((c29c19, c2)1, c24)26)4, j18,10= (c21, (((c16, c19)2, c2)1, c23)18)4, j18,11= (c21, (((c16, c19)4, c1)1, c22)10)4, j18,12= (c1c2, ((c16, c19)6, c2)5· c21)8, j18,13= (c21, (c4, (c28c17, c1c2)1)12)4, j18,14= (c21, (c3, (c28c17, c1c2)3)8)4, j18,15= (c21, (c2, (c15c19, c1c4)13)4)4, j18,16= (c1c2, (c2, (c15c19, c1c4)13)2)8, j18,17= (c21, (c4, (c15c19, c1c4)9)12)4, j18,18= (c21, (c2, (c28c17, c1c2)5)4)4, j18,19= (c1c2, (c2, (c28c17, c1c2)5)2)8, j18,20= (c1c2, (c4, (c15c19, c1c4)9)10)8, j18,21= (c1c2, (c3, (c28c17, c1c2)3)6)8, j18,22= (c1c2, (c4, (c28c17, c1c2)1)10)8, j18,23= (c214, c18)4, j18,24= (c2c3, (((c16, c19)2, c2)1, c23)12)16, j18,25= (c2c3, ((c29c19, c2)1, c24)20)16, 20 j20,1= (c143, c2)6, j20,2= ((c27, c2)4, (c24, c23)3)4, 22 j22= (c21, (c2c3, ((c29c19, c2)1, c24)20)14)4.

Then we prove that no generators are needed in degrees 24, 26, . . . 58, 62, 66

by showing that for each i ∈ {24, 26, . . . 58, 62, 66} the vector space O(V

9

)

SLi 2

is spanned by monomials of degree i generated by the 92 invariants found in

degrees ≤ 22. The required computations for i ≤ 32 can be seen in Chap.

A.3.

The computations in degrees ≥ 34 were performed by A.E. Brouwer, with his

own software ([BP10a]).

In document Invariants of binary forms (Page 58-65)