Discriminant of symmetric homogeneous polynomials

Laurent Bus´e

INRIA Sophia Antipolis, France [email protected]

Seminar of Algebraic Geometry, University of Barcelona October 3, 2014

Joint work with Anna Karasoulou (University of Athens) Outline

1 Tools from elimination theory • Discriminant of a hypersurface in a projective space • Resultant of homogeneous polynomials

2 The main formula • Homogeneous symmetric polynomials • Divided differences • Decomposition of the discriminant of a homogeneous symmetric polynomial

3 Sn-equivariant systems

• Sn-equivariant systems of homogeneous polynomials

• Resultant of a Sn-equivariant polynomial system The universal discriminant Disc belongs to An,d

n−1 I It is homogeneous of degree n(d − 1) ( usual grading of An,d ). I It is a prime element in An,d (irreducible in An,d ⊗ Q and not divisible by any integer > 1).

Discriminant of a hypersurface in a projective space

Notation:

I Fix integers n ≥ 1, d ≥ 2 and let k be a commutative ring.

I Pd (k) := Γ(O n−1 (d)) = k[x1,..., xn]d (Rk : Pd (k) = Pd ( ) ⊗ k). Pk Z Z I The universal homogeneous polynomial of degree d in n variables:

X α α α1 α2 αn Pn,d (x1,..., xn) = Uαx (x = x1 x2 ··· xn ). |α|=d

I The universal coefficient ring : An,d := Z[Uα : |α| = d]. Discriminant of a hypersurface in a projective space

Notation:

I Fix integers n ≥ 1, d ≥ 2 and let k be a commutative ring.

I Pd (k) := Γ(O n−1 (d)) = k[x1,..., xn]d (Rk : Pd (k) = Pd ( ) ⊗ k). Pk Z Z I The universal homogeneous polynomial of degree d in n variables:

X α α α1 α2 αn Pn,d (x1,..., xn) = Uαx (x = x1 x2 ··· xn ). |α|=d

I The universal coefficient ring : An,d := Z[Uα : |α| = d].

The universal discriminant Disc belongs to An,d

n−1 I It is homogeneous of degree n(d − 1) ( usual grading of An,d ). I It is a prime element in An,d (irreducible in An,d ⊗ Q and not divisible by any integer > 1). Smoothness criterion

For any algebraically closed field k and polynomial f ∈ Pd (k)

 ∂f ∂f ∂f  Disc(f ) 6= 0 ⇔ V n−1 f , , ,..., = ∅. Pk ∂x1 ∂x2 ∂xn

Resultant of the partial derivatives of F

Discriminant of a hypersurface in a projective space

Specialization : For any ring k and polynomial f ∈ Pd (k)

An,d → k (canonical ring map)

u(..., Uα,...) 7→ u(f )

Remark : Pn,d ∈ Pd (An,d ) and u(Pn,d ) = u

Definition (discriminant)

The discriminant of f ∈ Pd (k) is Disc(f ) ∈ k. Discriminant of a hypersurface in a projective space

Specialization : For any ring k and polynomial f ∈ Pd (k)

An,d → k (canonical ring map)

u(..., Uα,...) 7→ u(f )

Remark : Pn,d ∈ Pd (An,d ) and u(Pn,d ) = u

Definition (discriminant)

The discriminant of f ∈ Pd (k) is Disc(f ) ∈ k.

Smoothness criterion

For any algebraically closed field k and polynomial f ∈ Pd (k)

 ∂f ∂f ∂f  Disc(f ) 6= 0 ⇔ V n−1 f , , ,..., = ∅. Pk ∂x1 ∂x2 ∂xn

Resultant of the partial derivatives of F Resultant of homogeneous polynomials

Notation:

I Fix integers n ≥ 1, d1,..., dn ≥ 1.

I The universal homogeneous polynomial of degree di in n variables:

X α Pn,di (x1,..., xn) = Ui,αx (i = 1,..., n).

|α|=di

I The universal coefficient ring :

Ad1,...,dn := Z[Ui,α : i = 1,..., n |α| = di ].

The universal resultant Res belongs to Ad1,...,dn

I It is homogeneous w.r.t. Pn,di of degree d1 ... dn/di .

I It is a prime element in Ad1,...,dn . Resultant can be explicitly computed

I closed matrix-based formulas (Sylvester, Cayley, Macaulay, etc)

I many formal properties (multiplicativity, change of basis, etc)

Resultant of homogeneous polynomials

Specialization : For any ring k and polynomials fi ∈ Pdi (k), i = 1,..., n

Ad1,...,dn → k (canonical ring map)

u(..., Ui,α,...) 7→ u(f1,..., fn)

Definition (resultant)

The resultant of f1,..., fn is Res(f1,..., fn) ∈ k.

d1 d2 dn Remark : resultant is normalized by Res(x1 , x2 ,..., xn ) = 1. Common root detection

For any algebraically closed field k and polynomials fi ∈ Pdi (k)

Res(f ,..., f ) 6= 0 ⇔ V n (f ,..., f ) = ∅. 1 n Pk 1 n Resultant of homogeneous polynomials

Specialization : For any ring k and polynomials fi ∈ Pdi (k), i = 1,..., n

Ad1,...,dn → k (canonical ring map)

u(..., Ui,α,...) 7→ u(f1,..., fn)

Definition (resultant)

The resultant of f1,..., fn is Res(f1,..., fn) ∈ k.

d1 d2 dn Remark : resultant is normalized by Res(x1 , x2 ,..., xn ) = 1. Common root detection

For any algebraically closed field k and polynomials fi ∈ Pdi (k)

Res(f ,..., f ) 6= 0 ⇔ V n (f ,..., f ) = ∅. 1 n Pk 1 n

Resultant can be explicitly computed

I closed matrix-based formulas (Sylvester, Cayley, Macaulay, etc)

I many formal properties (multiplicativity, change of basis, etc) Example of the Clebsch surface : P4 3 P4 3 I f (x1,..., x4) = i=1 xi − ( i=1 xi ) 37 I Res(∂1f , . . . , ∂4f ) = −3 × 5

I a(4, 3) = 15/3 = 5 32 I Disc(f ) = −3 × 5 3 I But 3 divides f and Disc is homogeneous of degree 4 × 2 = 32 I Therefore Disc(f /3) = −5 smooth expect in characteristic 5.

Computation of the discriminant

Theorem/Definition (Demazure)

For any polynomial f ∈ Pd (k)

 ∂f ∂f ∂f  d a(n,d)Disc(f ) = Res , ,..., ∂x1 ∂x2 ∂xn where (d − 1)n − (−1)n a(n, d) := ∈ . d Z Computation of the discriminant

Theorem/Definition (Demazure)

For any polynomial f ∈ Pd (k)

 ∂f ∂f ∂f  d a(n,d)Disc(f ) = Res , ,..., ∂x1 ∂x2 ∂xn where (d − 1)n − (−1)n a(n, d) := ∈ . d Z

Example of the Clebsch surface : P4 3 P4 3 I f (x1,..., x4) = i=1 xi − ( i=1 xi ) 37 I Res(∂1f , . . . , ∂4f ) = −3 × 5

I a(4, 3) = 15/3 = 5 32 I Disc(f ) = −3 × 5 3 I But 3 divides f and Disc is homogeneous of degree 4 × 2 = 32 I Therefore Disc(f /3) = −5 smooth expect in characteristic 5. 2 I Example n = d = 2. P2,2 = U2x2 + U1xy + U0y ,

2 Disc(P2,2) = 4U2U0 − U1 ,

Disc(U0 = U2) = (2U0 − U1)(2U0 + U1). 3 2 2 3 I Example n = 2, d = 3. P2,3 = U3x + U2x y + U1xy + U0y ,

2 2 3 2 2 3 Disc(P2,2) = 27 U3 U0 +4 U3U1 −18 U3U2U0U1−U1 U2 +4 U2 U0,

3 Disc(U3 = U0, U2 = U1) = (U0 + U1)(3 U0 − U1) .

Main observation

I The discriminant of the generic homogeneous polynomial is irreducible in the universal ring of coefficients.

I However, it always factorizes with symmetric polynomials,

I and this leads to an easier computation (useful in various applications). Main observation

I The discriminant of the generic homogeneous polynomial is irreducible in the universal ring of coefficients.

I However, it always factorizes with symmetric polynomials,

I and this leads to an easier computation (useful in various applications).

2 I Example n = d = 2. P2,2 = U2x2 + U1xy + U0y ,

2 Disc(P2,2) = 4U2U0 − U1 ,

Disc(U0 = U2) = (2U0 − U1)(2U0 + U1). 3 2 2 3 I Example n = 2, d = 3. P2,3 = U3x + U2x y + U1xy + U0y ,

2 2 3 2 2 3 Disc(P2,2) = 27 U3 U0 +4 U3U1 −18 U3U2U0U1−U1 U2 +4 U2 U0,

3 Disc(U3 = U0, U2 = U1) = (U0 + U1)(3 U0 − U1) . Example : formula in the case n ≥ d = 2 The generic homogeneous symmetric polynomial is of the form

  n !2 Y X F = c(2)  xi xj  + c(1,1) xi 1≤i

and its discriminant is given by

( n−1
−c(2) (n − 1)c(2) + 2nc(1,1) if n is even, Disc(F ) = n−1 n−1
c(2) 2 c(2) + nc(1,1) if n is odd .

Goal of this talk

Our goal is to give a universal closed formula for the discriminant of a homogeneous symmetric polynomial

I First attempt by Perminov and Shakirov (arXiv:0910.5757)

I The main difficulty comes from the multiplicity of the factors Goal of this talk

Our goal is to give a universal closed formula for the discriminant of a homogeneous symmetric polynomial

I First attempt by Perminov and Shakirov (arXiv:0910.5757)

I The main difficulty comes from the multiplicity of the factors

Example : formula in the case n ≥ d = 2 The generic homogeneous symmetric polynomial is of the form

  n !2 Y X F = c(2)  xi xj  + c(1,1) xi 1≤i

and its discriminant is given by

( n−1
−c(2) (n − 1)c(2) + 2nc(1,1) if n is even, Disc(F ) = n−1 n−1
c(2) 2 c(2) + nc(1,1) if n is odd . The generic homogeneous symmetric polynomial of degree d X F (x1,..., xn) := cλeλ(x) ∈ Z[cλ : λ ` d][x1,..., xn]. λ`d

I Example for d = 2

2 F (x1,..., xn) = c(2)e2(x1,..., xn) + c(1,1)e1 (x1,..., xn).

Homogeneous symmetric polynomials

I Elementary symmetric polynomials in x = x1,..., xn

n X p Y ep(x)t = (1 + xi t) p≥0 i=1

I Given a partition λ = (λ1 ≥ · · · ≥ λk ), define

eλ(x) := eλ1 (x)eλ2 (x) ··· eλk (x) ∈ Z[x1,..., xn]. Pn I If i=1 λi = d we say that λ is a partition of d , denoted λ ` d. Homogeneous symmetric polynomials

I Elementary symmetric polynomials in x = x1,..., xn

n X p Y ep(x)t = (1 + xi t) p≥0 i=1

I Given a partition λ = (λ1 ≥ · · · ≥ λk ), define

eλ(x) := eλ1 (x)eλ2 (x) ··· eλk (x) ∈ Z[x1,..., xn]. Pn I If i=1 λi = d we say that λ is a partition of d , denoted λ ` d. The generic homogeneous symmetric polynomial of degree d X F (x1,..., xn) := cλeλ(x) ∈ Z[cλ : λ ` d][x1,..., xn]. λ`d

I Example for d = 2

2 F (x1,..., xn) = c(2)e2(x1,..., xn) + c(1,1)e1 (x1,..., xn). Discriminant of symmetric polynomials

The discriminant of the generic homogeneous symmetric polynomials is not irreducible

I Example n = d = 2:

Disc(ax 2 + bxy + ay 2) = 4a2 − b2 = (2a − b)(2a + b).

I Geometric intuition (via incidence variety)

 x e2 − x e ··· Pd−1(−1)i x i e   ∂F  1 2 1 1 i=0 1 d−1−i  S (x)  ∂x1 2 Pd−1 i i 1  x2 e2 − x2e1 ··· (−1) x2ed−1−i   .  =  i=0   .   .   . . .   .  ∂F  . . .  Sd (x) ∂xn 2 Pd−1 i i xn e2 − xne1 ··· i=0 (−1) xned−1−i

The rank of this Jacobian matrix drops if xi = xj for some i, j. ∂F ∂F Elementary observation : (xi − xj ) divides − ∂xi ∂xj

This naturally leads to divided differences. Definition (k − 1)th divided differences

Given {i1,..., ik } ⊂ {1,..., n}, there exists a unique homogeneous polynomial P{i1,...,ik } of degree d − k + 1 such that

{i1,...,ik } VandermondeMatrix(xi1 , xi2 ,..., xik ) · P (x1,..., xn) = 1 x ··· x k−2 P{i1}(x ,..., x ) i1 i1 1 n k−2 {i2} 1 xi2 ··· x P (x1,..., xn) i2 det ......

1 x ··· x k−2 P{ik }(x ,..., x ) ik ik 1 n

I For any two distinct integers p, q in {i1,..., ik },

{i1,i2,...,ik } {i1,i2,...,ik }\{ip } {i1,i2,...,ik }\{iq } (xiq − xip )P = P − P .

Divided differences

{1} {2} {n} I Let P (x), P (x),..., P (x) be hom. pol. of degree d ≥ 1 in R[x1,..., xn]. Assume that for all couple of integers (i, j)

{i} {j} P (x) − P (x) ∈ (xi − xj ) ⊂ R[x1,..., xn]. I For any two distinct integers p, q in {i1,..., ik },

{i1,i2,...,ik } {i1,i2,...,ik }\{ip } {i1,i2,...,ik }\{iq } (xiq − xip )P = P − P .

Divided differences

{1} {2} {n} I Let P (x), P (x),..., P (x) be hom. pol. of degree d ≥ 1 in R[x1,..., xn]. Assume that for all couple of integers (i, j)

{i} {j} P (x) − P (x) ∈ (xi − xj ) ⊂ R[x1,..., xn].

Definition (k − 1)th divided differences

Given {i1,..., ik } ⊂ {1,..., n}, there exists a unique homogeneous polynomial P{i1,...,ik } of degree d − k + 1 such that

{i1,...,ik } VandermondeMatrix(xi1 , xi2 ,..., xik ) · P (x1,..., xn) = 1 x ··· x k−2 P{i1}(x ,..., x ) i1 i1 1 n k−2 {i2} 1 xi2 ··· x P (x1,..., xn) i2 det ......

1 x ··· x k−2 P{ik }(x ,..., x ) ik ik 1 n Divided differences

{1} {2} {n} I Let P (x), P (x),..., P (x) be hom. pol. of degree d ≥ 1 in R[x1,..., xn]. Assume that for all couple of integers (i, j)

{i} {j} P (x) − P (x) ∈ (xi − xj ) ⊂ R[x1,..., xn].

Definition (k − 1)th divided differences

Given {i1,..., ik } ⊂ {1,..., n}, there exists a unique homogeneous polynomial P{i1,...,ik } of degree d − k + 1 such that

{i1,...,ik } VandermondeMatrix(xi1 , xi2 ,..., xik ) · P (x1,..., xn) = 1 x ··· x k−2 P{i1}(x ,..., x ) i1 i1 1 n k−2 {i2} 1 xi2 ··· x P (x1,..., xn) i2 det ......

1 x ··· x k−2 P{ik }(x ,..., x ) ik ik 1 n

I For any two distinct integers p, q in {i1,..., ik },

{i1,i2,...,ik } {i1,i2,...,ik }\{ip } {i1,i2,...,ik }\{iq } (xiq − xip )P = P − P . Given a partition λ = (λ1 ≥ · · · ≥ λk 6= 0) ` n, consider the morphism

ρλ : R[x1,..., xn] → R[y1,..., yk ]

P(x1,..., xn) 7→ P(y1,..., y1, y2,..., y2,..., yk ,..., yk ). | {z } | {z } | {z } λ1 λ2 λk

{i} {j} I define Fλ (y1,..., yk ) := ρλ(F (x1,..., xn)) ( j s.t. ρλ(xj ) = yi )

J I Divided differences : Fλ (y1,..., yk ), J ⊂ {1,..., k}

Notations for the main formula

Given F (x1,..., xn), an homogeneous symmetric polynomial

{i} I define F (x) := ∂F /∂xi for all i = 1,..., n

J I Divided differences : F (x1,..., xn), J ⊂ {1,..., n} Notations for the main formula

Given F (x1,..., xn), an homogeneous symmetric polynomial

{i} I define F (x) := ∂F /∂xi for all i = 1,..., n

J I Divided differences : F (x1,..., xn), J ⊂ {1,..., n}

Given a partition λ = (λ1 ≥ · · · ≥ λk 6= 0) ` n, consider the morphism

ρλ : R[x1,..., xn] → R[y1,..., yk ]

P(x1,..., xn) 7→ P(y1,..., y1, y2,..., y2,..., yk ,..., yk ). | {z } | {z } | {z } λ1 λ2 λk

{i} {j} I define Fλ (y1,..., yk ) := ρλ(F (x1,..., xn)) ( j s.t. ρλ(xj ) = yi )

J I Divided differences : Fλ (y1,..., yk ), J ⊂ {1,..., k} A combinatorial quantity Given a partition λ ` n, its multinomial coefficient is

 n  n! := . λ1, λ2, . . . , λk λ1!λ2! ··· λk !

I It counts the number of distributions of n distinct objects to k distinct recipients such that the recipient i receives exactly λi objects.

I The objects are not ordered inside the boxes, but the boxes are ordered.

I We do not want to count the permutations between the boxes having the same number of objects.

I Let sj be the number of boxes having exactly j objects, j ∈ [n], then Qn the number of these permutations is equal to j=1 sj !.

Finally, for any partition λ ` n we define the integer

1  n  mλ := Qn . j=1 sj ! λ1, λ2, . . . , λk The main formula

I Given a partition λ = (λ1 ≥ · · · ≥ λk 6= 0), we set l(λ) := k.

Theorem (n, d ≥ 2) P Let F (x1,..., xn) := λ`d cλeλ(x) (hom. sym. pol.) then

a(n,d) d−1
m0 d Disc (F ) = (−1) c(d) × m Y  {1} {1,2} {1,2,...,l(λ)−1} {1,2,...,l(λ)} λ Res Fλ , Fλ ,..., Fλ , Fλ λ`n where

( n−1 P Pl(λ) (d−1)(d−2)···(d−l(λ))  n(d − 1) − λ`n mλ j=1 (d−j) if d ≤ n m0 := 0 if d > n.

Computational advantages:

I Universal formula which is valid over any coefficient ring

I Rely on the computation of few resultants (min{n, d − 1}) Example n ≥ d = 3 – the formula

I The generic homogeneous polynomial of degree 3 :

3 F = c(3)e3 + c(2,1)e2e1 + c(1,1,1)e1 .

I The decomposition formula yields the equality

b n c n n 2 m 2 −(−1)  {1} Y  {1} {1,2}  (n−k,k) 3 m0 3 Disc(F ) = c(3) Res F(n) Res F(n−k,k), F(n−k,k) k=1

where the multiplicities are as follows:

n−1 I m0 = (n − 3)2 + 2 (straightforward computation), n
n n I m(n−k,k) = k for all k = 1,..., b 2 c except if n is even and k = 2 , 1 n
I If n is even then m( n , n ) = n . 2 2 2 2 I Let( n − k, k) ` n. We have {i,j} F = c(3) (xi + xj − e1) − c(2,1)e1 ∀i 6= j and a single Sylvester resultant computation gives

 {1} {1,2}  2 Res F(n−k,k), F(n−k,k) = c(3)× ! ! ! n − 1 n c + 3 c + 3n2c × 2 (3) 2 (2,1) (1,1,1) 1   − k(n−k) (n − 2) c3 + 24c + 3nc
c2 + (3n − 6) c2 c + nc3 . 2 (3) (1,1,1) (2,1) (3) (2,1) (3) (2,1)

Example n ≥ d = 3 – the factors

I For all i = 1,..., n {i} 2
2 F = c(3) e2 − xi e1 + xi + c(2,1) (e2 + e1(e1 − xi )) + 3c(1,1,1)e1 So it follows immediately that

  n − 1 n Res F {1} = c + 3 c + 3n2c . (n) 2 (3) 2 (2,1) (1,1,1) Example n ≥ d = 3 – the factors

I For all i = 1,..., n {i} 2
2 F = c(3) e2 − xi e1 + xi + c(2,1) (e2 + e1(e1 − xi )) + 3c(1,1,1)e1 So it follows immediately that

  n − 1 n Res F {1} = c + 3 c + 3n2c . (n) 2 (3) 2 (2,1) (1,1,1)

I Let( n − k, k) ` n. We have {i,j} F = c(3) (xi + xj − e1) − c(2,1)e1 ∀i 6= j and a single Sylvester resultant computation gives

 {1} {1,2}  2 Res F(n−k,k), F(n−k,k) = c(3)× ! ! ! n − 1 n c + 3 c + 3n2c × 2 (3) 2 (2,1) (1,1,1) 1   − k(n−k) (n − 2) c3 + 24c + 3nc
c2 + (3n − 6) c2 c + nc3 . 2 (3) (1,1,1) (2,1) (3) (2,1) (3) (2,1) I Consider the Clebsch surface which is given by the equation

3 3 3 3 3 h(x1, x2, x3, x4) = x1 +x2 +x3 +x4 −(x1+x2+x3+x4) = 3e3−3e2e1 = 0.

I Applying this formula, we get (c(3) = 1, c(2,1) = −1, c(1,1,1) = 0)

9 4 Disc(h/3) = Disc(e3 − e2e1) = −(−1) (−6 + 1)(−3 + 2) = −5.

Example n = 4 and d = 3

Formula 3 If n = 4, d = 3 and F = c(3)e3 + c(2,1)e2e1 + c(1,1,1)e1 then

10
9
Disc(F ) = −c(3) c(3) + 2 c(2,1) 6 c(2,1) + 16 c(1,1,1) + c(3) × 2 2 3
4 4 c(1,1,1)c(3) − 3 c(2,1) c(3) − 2 c(2,1) Example n = 4 and d = 3

Formula 3 If n = 4, d = 3 and F = c(3)e3 + c(2,1)e2e1 + c(1,1,1)e1 then

10
9
Disc(F ) = −c(3) c(3) + 2 c(2,1) 6 c(2,1) + 16 c(1,1,1) + c(3) × 2 2 3
4 4 c(1,1,1)c(3) − 3 c(2,1) c(3) − 2 c(2,1)

I Consider the Clebsch surface which is given by the equation

3 3 3 3 3 h(x1, x2, x3, x4) = x1 +x2 +x3 +x4 −(x1+x2+x3+x4) = 3e3−3e2e1 = 0.

I Applying this formula, we get (c(3) = 1, c(2,1) = −1, c(1,1,1) = 0)

9 4 Disc(h/3) = Disc(e3 − e2e1) = −(−1) (−6 + 1)(−3 + 2) = −5. d ! d !  {1} {2} {1} {1} 2 X X i 2 Res F , F = F (1, 1)F (1, −1)Kd = ai (−1) ai Kd . i=0 i=0

  Res F {1},..., F {n} = an−1(a + nb).

I Example n = 2: Universal S2-equivariant polynomial system:

{1} d d−1 d {2} {1} F (x1, x2) := a0x1 + a1x1 x2 + ··· + ad x2 , F (x1, x2) := F (x2, x1).

{i} I Example d = 1: F (x1,..., xn) = axi + be1(x), i = 1,..., n.

Equivariant polynomial systems

Definition A system of n homogeneous polynomials F {1},..., F {n} in n variables x = x1,..., xn is equivariant with respect to the canonical actions of the symmetric group Sn (Sn-equivariant) on the variables and polynomials if for all i ∈ {1, 2,..., n} and any permutation σ ∈ Sn

{i} {i} {σ(i)} σ(F ) := F (xσ(1), xσ(2),..., xσ(n)) = F (x1, x2,..., xn). d ! d !  {1} {2} {1} {1} 2 X X i 2 Res F , F = F (1, 1)F (1, −1)Kd = ai (−1) ai Kd . i=0 i=0

  Res F {1},..., F {n} = an−1(a + nb).

Equivariant polynomial systems

Definition A system of n homogeneous polynomials F {1},..., F {n} in n variables x = x1,..., xn is equivariant with respect to the canonical actions of the symmetric group Sn (Sn-equivariant) on the variables and polynomials if for all i ∈ {1, 2,..., n} and any permutation σ ∈ Sn

{i} {i} {σ(i)} σ(F ) := F (xσ(1), xσ(2),..., xσ(n)) = F (x1, x2,..., xn).

I Example n = 2: Universal S2-equivariant polynomial system:

{1} d d−1 d {2} {1} F (x1, x2) := a0x1 + a1x1 x2 + ··· + ad x2 , F (x1, x2) := F (x2, x1).

{i} I Example d = 1: F (x1,..., xn) = axi + be1(x), i = 1,..., n. Equivariant polynomial systems

Definition A system of n homogeneous polynomials F {1},..., F {n} in n variables x = x1,..., xn is equivariant with respect to the canonical actions of the symmetric group Sn (Sn-equivariant) on the variables and polynomials if for all i ∈ {1, 2,..., n} and any permutation σ ∈ Sn

{i} {i} {σ(i)} σ(F ) := F (xσ(1), xσ(2),..., xσ(n)) = F (x1, x2,..., xn).

I Example n = 2: Universal S2-equivariant polynomial system:

{1} d d−1 d {2} {1} F (x1, x2) := a0x1 + a1x1 x2 + ··· + ad x2 , F (x1, x2) := F (x2, x1).

d ! d !  {1} {2} {1} {1} 2 X X i 2 Res F , F = F (1, 1)F (1, −1)Kd = ai (−1) ai Kd . i=0 i=0

{i} I Example d = 1: F (x1,..., xn) = axi + be1(x), i = 1,..., n.   Res F {1},..., F {n} = an−1(a + nb). An important remark

Lemma The partial derivatives F {1}, F {2},..., F {n} of a symmetric homogeneous polynomial F (x1,..., xn) form a Sn-equivariant polynomial system.

Indeed, F is a polynomial in the elementary symmetric polynomials so for all i = 1,..., n

min{d,n} {i} ∂F X ∂ek F = = Sk (x1,..., xn). ∂xi ∂xi k=1

where Sk (x1,..., xn) are symmetric polynomials. Moreover, for any pair of integers i, j we have

j−1 ∂ej X = (−1)r x r e . ∂x i j−1−r i r=0

{i}
{σ(i)} Therefore, for any σ ∈ Sn, we have σ F = F as claimed. Theorem (n ≥ 2, d ≥ 1) {1} {n} Let F ,..., F be Sn-equivariant of degree d, then

   m0 Res F {1},..., F {n} = F {1,...,d+1} × m Y  {1} {1,2} {1,2,...,l(λ)−1} {1,2,...,l(λ)} λ Res Fλ , Fλ ,..., Fλ , Fλ λ`n where

( n−1 P Pl(λ) d(d−1)···(d−l(λ)+1)  nd − λ`n mλ j=1 (d−j+1) if d < n m0 := 0 if d ≥ n.

Resultant of a Sn-equivariant system

I Equivariant polynomial systems admit divided differences:

{i} {j} {i} {i} F (x) − F (x) = F (x) − σ(F (x)) ∈ (xi − xj ).

where σ ∈ Sn is such that σ(k) = k if k ∈/ {i, j} and σ(i) = j Resultant of a Sn-equivariant system

I Equivariant polynomial systems admit divided differences:

{i} {j} {i} {i} F (x) − F (x) = F (x) − σ(F (x)) ∈ (xi − xj ).

where σ ∈ Sn is such that σ(k) = k if k ∈/ {i, j} and σ(i) = j

Theorem (n ≥ 2, d ≥ 1) {1} {n} Let F ,..., F be Sn-equivariant of degree d, then

   m0 Res F {1},..., F {n} = F {1,...,d+1} × m Y  {1} {1,2} {1,2,...,l(λ)−1} {1,2,...,l(λ)} λ Res Fλ , Fλ ,..., Fλ , Fλ λ`n where

( n−1 P Pl(λ) d(d−1)···(d−l(λ)+1)  nd − λ`n mλ j=1 (d−j+1) if d < n m0 := 0 if d ≥ n. Question

I What are the geometrically irreducible factors in these decompositions ?

I In the case of the discriminant, it seems the decomposition we have given here provides these factors (this is not the case for the resultant of Sn-equivariant polynomial systems).

A word on the proof

I Basic strategy : to split the resultant iteratively by using • invariance of the resultant under elementary transformations • multiplicativity property of the resultant • divided differences

I Main difficulty : count the multiplicity of each distinct factor • Identical factors must be identified • These factors are not themselves irreducible, neither geometrically irreducible in general • Rely on a combinatorial counting

I The decomposition formula for the discriminant is obtained as a specialization of the resultant of a Sn-equivariant system A word on the proof

I Basic strategy : to split the resultant iteratively by using • invariance of the resultant under elementary transformations • multiplicativity property of the resultant • divided differences

I Main difficulty : count the multiplicity of each distinct factor • Identical factors must be identified • These factors are not themselves irreducible, neither geometrically irreducible in general • Rely on a combinatorial counting

I The decomposition formula for the discriminant is obtained as a specialization of the resultant of a Sn-equivariant system

Question

I What are the geometrically irreducible factors in these decompositions ?

I In the case of the discriminant, it seems the decomposition we have given here provides these factors (this is not the case for the resultant of Sn-equivariant polynomial systems).