Invariants for a single form, several forms.
degree | count |
---|---|
2 | (m choose 2) + (n+1 choose 2) |
3 | n(m+1 choose 2) + (n choose 3) |
4 | (m+1 choose 2)(n choose 2) |
total | (n choose 3) + (m+1 choose 2)(n+1 choose 2) + (m choose 2) + (n+1 choose 2) |
This is more or less classical. See also
A. E. Brouwer & M. Popoviciu, SL2-modules of small homological dimension, Transformation Groups 16 (2011) 599-617.
More in detail: for nV2 the numbers of basic covariants of degree d and order o are
d\o | 0 | 2 |
---|---|---|
1 | - | n |
2 | (n+1 choose 2) | (n choose 2) |
3 | (n choose 3) | - |
and the table for mV1+nV2 immediately follows, see below.
d\o | 0 | 1 | 2 |
---|---|---|---|
1 | - | m | n |
2 | (n+1 choose 2) + (m choose 2) | mn | (n choose 2) |
3 | (n choose 3) + n(m+1 choose 2) | m(n choose 2) | - |
4 | (m+1 choose 2)(n choose 2) | - | - |
type | deg | count | p=1 | p=2 | p=3 | p=4 | p=5 |
---|---|---|---|---|---|---|---|
type 1 | (1,3) | p | 1 | 2 | 3 | 4 | 5 |
type 2 | (2,4) | (p choose 2) | 0 | 1 | 3 | 6 | 10 |
type 3 | (2,2) | (p+1 choose 2) | 1 | 3 | 6 | 10 | 15 |
type 4 | (2,0) | (p choose 2) | 0 | 1 | 3 | 6 | 10 |
type 5 | (3,3) | (p+2 choose 3) | 1 | 4 | 10 | 20 | 35 |
type 6 | (3,1) | 2(p+1 choose 3) | 0 | 2 | 8 | 20 | 40 |
type 7 | (4,0) | (p+3 choose 4) | 1 | 5 | 15 | 35 | 70 |
type 8 | (4,2) | 3(p+2 choose 4) | 0 | 3 | 15 | 45 | 105 |
type 9 | (5,1) | 4(p+3 choose 5) | 0 | 4 | 24 | 84 | 224 |
type 10 | (6,0) | (p+2 choose 4)(p+1 choose 2)/3 | 0 | 1 | 10 | 50 | 175 |
total #inv | 1 | 7 | 28 | 91 | 255 | ||
total #cov | 4 | 26 | 97 | 280 | 689 |
The formulas for types 1-8 were given by Peano in
G. Peano, Sui sistemi di forme binarie di egual grado e sistema completo di quante si vogliano cubiche, Atti di Torino 17 (1882) 580-586.
The formulas for types 1-10 were given by Young in
A. Young, The irreducible concomitants of any number of binary quartics, Proc. London Math. Soc. 30 (1899) 290-307.but his formula for type 10 (namely, (p+4 choose 6)) is wrong. The correct formula is
(p+4 choose 6) + 3(p+3 choose 6) + (p+2 choose 6) = (p+2 choose 4)(p+1 choose 2)/3.
deg | count | p=1 | p=2 | p=3 | p=4 | p=5 |
---|---|---|---|---|---|---|
(2,0) | (p+1 choose 2) | 1 | 3 | 6 | 10 | 15 |
(3,0) | (p+2 choose 3) | 1 | 4 | 10 | 20 | 35 |
(4,0) | (p+1 choose 4)+(p+2 choose 4) | 0 | 1 | 6 | 20 | 50 |
(5,0) | (p choose 5)+2(p+1 choose 5)+3(p+2 choose 5) | 0 | 0 | 3 | 20 | 76 |
(6,0) | 10(p+2 choose 6) | 0 | 0 | 0 | 10 | 70 |
(2,2) | (p choose 2) | 0 | 1 | 3 | 6 | 10 |
(3,2) | (p choose 3)+2(p+1 choose 3) | 0 | 2 | 9 | 24 | 50 |
(4,2) | 3(p+1 choose 4)+3(p+2 choose 4) | 0 | 3 | 18 | 60 | 150 |
(5,2) | 3(p+2 choose 5)+2(p+3 choose 5) | 0 | 2 | 15 | 60 | 175 |
(1,4) | p | 1 | 2 | 3 | 4 | 5 |
(2,4) | (p+1 choose 2) | 1 | 3 | 6 | 10 | 15 |
(3,4) | 2(p+1 choose 3) | 0 | 2 | 8 | 20 | 40 |
(4,4) | 3(p+1 choose 4) | 0 | 0 | 3 | 15 | 45 |
(2,6) | (p choose 2) | 0 | 1 | 3 | 6 | 10 |
(3,6) | (p+2 choose 3) | 1 | 4 | 10 | 20 | 35 |
total #inv | 2 | 8 | 25 | 80 | 246 | |
total #cov | 5 | 28 | 103 | 305 | 781 |
This is from
A. Young, The irreducible concomitants of any number of binary quartics, Proc. London Math. Soc. 30 (1899) 290-307.