$\mathfrak S_5$-equivariant syzygies for the Del Pezzo Surface of Degree 5

The Del Pezzo surface $Y$ of degree 5 is the blow up of the plane in 4 general points, embedded in $\mathbb{P}^5$ by the system of cubics passing through these points. It is the simplest example of the Buchsbaum-Eisenbud theorem on arithmetically-Gorenstein subvarieties of codimension 3 being Pfaffian. Its automorphism group is the symmetric group $\mathfrak S_5$. We give canonical explicit $\mathfrak S_5$-invariant Pfaffian equations through a $6 \times 6$ antisymmetric matrix. We give concrete geometric descriptions of the irreducible representations of $\mathfrak S_5$. Finally, we give $\mathfrak S_5$-invariant equations for the embedding of $Y$ inside $(\mathbb{P}^1)^5$, and show that they have the same Hilbert resolution as for the Del Pezzo of degree $4$.

forms (more precisely, the Pfaffians generate the ideal of polynomials vanishing on Y ). The simplest nondegenerate case is the case m = 2 of a generic 5 × 5 antisymmetric matrix A of linear forms: here r = −1 and we get the Del Pezzo surface Y 2 5 of degree 5 as Pfafffian locus of A, hence defined by the 5 quadratic equations which are the five 4 × 4-Pfaffians of A.
It suffices to choose a generic matrix, but an explicit matrix is better, as was for instance done in [GSHPBS12]. At any rate we are in this way far away from a normal form for the equations of Y 2 5 , which might be desirable for many purposes. What do we mean by a normal form? To explain this, recall that Y 2 5 is indeed isomorphic to the moduli space of ordered quintuples of points in P 1 , and this isomorphism follows after showing that its group of automorphisms is S 5 (Y 2 5 is indeed in bijection with the set of projective equivalence classes of quintuples where no point of P 1 occurs with multiplicity ≥ 3). Hence, we would like to have equations where this symmetry shows up, and which are 'invariant' for the symmetry group.
This requirement can be precisely specified as follows: we have the anticanonical (6-dimensional) vector space V := H 0 (ω −1 Y ) = H 0 (Y, O Y (−K Y )) and the 5dimensional vector space W of quadratic forms vanishing on Y (W ⊂ Sym 2 (V )). Both vector spaces are representations of the symmetric group S 5 and the 5×5 antisymmetric matrix A of linear forms is seen, by the Buchsbaum-Eisenbud theory, to be an invariant tensor A ∈ ((Λ 2 W ) ⊗ V ) S 5 .
The main result of the present paper is to explicitly and canonically determine it. Now, it looks like there is no unique such representation of the tensor A, because in the end we have still to choose a basis for both vector spaces V and W . However, if we take the natural irreducible 4-dimensional representation U 4 of S 5 , this is given by the invariant subspace x 1 + x 2 + · · · + x 5 = 0 in the 5-dimensional vector space U ′ 5 with coordinates (x 1 , . . . , x 5 ) which are naturally permuted by S 5 . So, there is a natural permutation representation U ′ 5 yielding a natural basis for U ′ 5 ⊃ U 4 . In our case, we first show that there is a natural basis s ij (1 ≤ i = j ≤ 3) of V , showing that V is the regular representation of S 3 tensored with the sign character, and then we show that the space W is a 5-dimensional invariant subspace in a natural 6-dimensional subspace W ′ of Sym 2 (V ), related to a natural permutation representation of S 5 1 , and with basis Q ij again related to the regular representation of S 3 (the space W is then generated by the differences between two such quadratic forms Q ij ).
This leads to a normal form: we produce in Theorem 3.3 (which we reproduce here immediately below) in an elegant numerical way an explicit antisymmetric 6 × 6 matrix A ′ with entries in V (i.e., with entries linear forms) whose 15 4×4-Pfaffians are exactly twice the differences between the quadratic forms Q ij .
Theorem 0.1. Let Y be the del Pezzo surface of degree 5, embedded anticanonically in P 5 . Then the ideal of Y is generated by the 4 × 4 -Pfaffians of the S 5equivariant anti-symmetric 6 × 6-matrix The action of the symmetric group S 5 on the entries of the matrix is given by the action on the vector space V , with basis s ij , i = j, 1 ≤ i, j ≤ 3, which is described in the proof of theorem 1.6, tensored with the sign character ǫ (we shall denote V ′ := V ⊗ ǫ and observe here that V , V ′ are isomorphic S 5 representations). The action on the matrix is determined by the action of S 5 on W ′ ⊗ W ′ , in turn determined by the action of S 5 on the vector space W ′ with basis the quadrics Q ij , i = j, 1 ≤ i, j ≤ 3. The latter action (which is explicitly described in remark 2.8) is a consequence of the former, since Q ij = s ij σ ij , where the σ ij 's are defined in section 1 and are as follows: σ ij = ǫ(s ik − s ki − s jk ).
In practice, the action of two generators of S 5 is explicitly given as follows. The transposition τ := (1, 2) acts as The cycle ϕ := (1, 2, 3, 4, 5) acts as s 12 → s 31 , s 13 → σ 12 = s 13 − s 31 − s 23 , s 21 → s 21 , s 23 → −σ 13 = s 12 −s 21 −s 32 , s 31 → −σ 32 = s 13 −s 31 +s 21 , s 32 → −σ 23 = s 12 −s 21 +s 31 , and Our main theorem is also interesting because, even if we work sometimes using character theory and over a field of characteristic 0, in the end we produce equations which define the Del Pezzo surface of degree 5 over any field of characteristic different from 2 (in this case symmetric and antisymmetric tensors coincide, and the situation should be treated separately).
We believe that our approach is also interesting from another point of view (as already announced in [Cat15] and [Cat17]): the symmetries of Y 2 5 bear some similarity to those of the icosahedron, as it is well known ( [Dol12], [CS16]), and the irreducible representations of S 5 can be explicitly described via the geometry of Y 2 5 . For instance, the representations V , W are irreducible, and together with the natural representation U 4 described above, they and their tensor product with the sign representation ǫ yield all irreducible representations of S 5 (V is the only one such that V ∼ = V ⊗ ǫ).
We also give explicit descriptions of the irreducible representations of S 5 via the geometry of Y 2 5 . While it is easy to describe U 4 and W ⊗ ǫ through natural permutation representations on S 5 /S 4 , respectively on S 5 / Aff(1, Z/5), we give another realization of the anticanonical vector space V , (1)), through an explicit permutation representation of dimension 24, the one on the set of oriented combinatorial pentagons (the cosets in S 5 /(Z/5)). This is interesting, because we have then a bijection between the set of the 12 (unoriented, nondegenerate) combinatorial pentagons and the set of the 12 geometric nondegenerate pentagons contained in Y , and formed with quintuples out of the 10 lines contained in Y (these geometric pentagons are hyperplane sections of Y ).
Finally, in a previous paper [BC18b] we have written explicit equations of Del Pezzo surfaces in products (P 1 ) h , and here, for the Del Pezzo of degree 5, we give in Theorem 4.3 S 5 -invariant equations inside (P 1 ) 5 .
An interesting phenomenon is that the minimal (multigraded) Hilbert resolution for this embedding of the Del Pezzo of degree 5 is exactly the same as for the natural embedding in (P 1 ) 5 of the Del Pezzo surfaces of degree 4, so that we have an example of a reducible Hilbert scheme.
1. Symmetries and the anticanonical system of the Del Pezzo surface of Degree 5 We recall some notation introduced in [Cat17], as well as some intermediate results established there.
The Del Pezzo surface Y := Y 2 5 of degree 5 is the blow-up of the plane P 2 in the 4 points p 1 , . . . , p 4 of a projective basis, that is, we choose Observe that p i corresponds to a vector e i , for i = 1, 2, 3, 4 , where e 1 , e 2 , e 3 is a basis of a vector space U ′ , and e 1 + e 2 + e 3 + e 4 = 0. In other words, P 2 = P(U ′ ), where U ′ is the natural irreducible representation of S 4 .
As already mentioned in the introduction, the Del Pezzo surface Y is indeed the moduli space of ordered quintuples of points in P 1 , and its automorphism group is isomorphic to S 5 .
The obvious action of the symmetric group S 4 permuting the 4 points extends in fact to an action of the symmetric group S 5 . This can be seen as follows. The six lines in the plane joining pairs p i , p j can be labelled as L i,j , with i, j ∈ {1, 2, 3, 4}, i = j.
Denote by E i,5 the exceptional curve lying over the point p i , and denote, for For each choice of 3 of the four points, {1, 2, 3, 4} \ {h}, consider the standard Cremona transformation σ h based on these three points. To it we associate the transposition (h, 5) ∈ S 5 , and the upshot is that σ h transforms the 10 (−1) curves E i,j via the action of (h, 5) on pairs of elements in {1, 2, 3, 4, 5}.
There are five geometric objects permuted by S 5 : namely, 5 fibrations ϕ i : Y → P 1 , induced, for 1 ≤ i ≤ 4, by the projection with centre p i , and, for i = 5, by the pencil of conics through the 4 points. Each fibration is a conic bundle, with exactly three singular fibres, corresponding to the possible partitions of type (2, 2) of the set {1, 2, 3, 4, 5} \ {i}.
The intersection pattern of the curves E i,j , which generate the Picard group of Y is dictated by the simple rule (recall that In this picture the three singular fibres of ϕ 1 are E 3,4 + E 2,5 , E 2,4 + E 3,5 , E 2,3 + E 4,5 .
An important observation is that Y contains exactly ten lines (i.e., ten irreducible curves E with E 2 = EK Y = −1), namely the lines E i,j . Their intersection pattern is described by the Petersen graph, whose vertices correspond to the curves E i,j , and whose edges correspond to intersection points (cf. figure 1).
Remark 1.1. 1) The action of Aut(Y ) ∼ = S 5 can be described as the action on P 2 by the following (birational) transformations: • the above vectorial action of S 4 on U ′ (by which S 4 ≤ S 5 acts on P 2 permuting p 1 , p 2 , p 3 , p 4 ), i.e., in coordinates, • for σ ∈ S 3 σ(x 1 : x 2 : x 3 ) = (x σ(1) : x σ(2) : x σ(3) ), and for σ = (3, 4): • the transposition (4, 5) is the standard Cremona transformation 2) The equation of the six lines of the complete quadrangle in P 2 (the lines L i,j joining p i and p j ) yields an eigenvector for the sign representation ǫ : In fact, the proper transform of div(Σ) is the sum of the 10 lines E i,j contained in Y , and it is a divisor in | − 2K Y | invariant for any automorphism. Since there is a natural action of which is an eigenvector for a character of S 5 . There are only two characters, the sign and the trivial one: restricting to the subgroup S 3 we see that it must be the sign.
We consider now the anticanonical vector space We know that V is, via adjunction, naturally isomorphic to the vector space of cubics passing through the 4 points, Since S 5 acts linearly on V we want to determine this action and translate this action on V ′ .
For this purpose we set, for 1 ≤ i = j ≤ 3, Proof. All 6 cubic forms vanish on p 1 , . . . , p 4 . s 12 , s 21 are the only ones not divisible by x 3 , and they are independent modulo (x 3 ), since their divisors on the line x 3 = 0 are, respectively 2p 1 + p 2 , p 1 + 2p 2 .
It suffices to show that the other 4 are independent after division by x 3 . Now, x 3 yield x 2 x 3 , respectively x 2 2 modulo (x 1 ), hence it suffices to show that s 13 x 3 x 1 , s 31 x 3 x 1 are independent: but these are respectively ( We set where ǫ := ǫ( 1 2 3 i j k ) is the sign of the permutation.
Then we have: (1) We observe as a parenthetical remark that the σ ij 's span a 4dimensional sub vector space of V . In fact, for all α, β, γ ∈ C such that α + β + γ = 0.
(2) Observe that for S 3 ≤ S 5 , S 3 acts on homogeneous polynomials by permuting x 1 , x 2 , x 3 , and that V ′ is the regular representation of S 3 .
Consider now the Euler sequence on P 2 := P(U ′ ), U ′ ∼ = C 3 : Therefore the isomorphism C[x 1 , x 2 , x 3 ] 3 ∼ = H 0 (P 2 , O P 2 (−K P 2 )) is given by the map: In this way we can read out on V ′ the given action of S 5 on V . We obtain immediately: Hence the representation V restricted to S 3 ≤ S 5 is the regular representation tensored by the sign character ǫ, while V restricted to S 4 ≤ S 5 is isomorphic to the subspace of S 3 ((U ′ ) ∨ ) ⊗ ǫ corresponding to the polynomials vanishing in p 1 , . . . , p 4 .
Proof. Direct computation for S 3 , for S 4 observe that the tranposition (3, 4) acts sending Remark 1.5. Since s ij σ ij = Σ and Σ is an eigenvector for the sign representation, the previous lemma implies immediately that for τ ∈ S 3 ≤ S 5 we have: We prove the following 2) The character of 6 V is the sign character ǫ = χ 2 .
Remark 1.7. Indeed, as we shall also see later, Recall that the transposition τ := (3, 4) acts as We use formula 1.1, obtaining that (3, 4) acts by: Next we consider the action of τ = (4, 5). This time we do not have an action on the space of polynomials, hence we work on the affine chart {x 3 = 1}.
For completeness we calculate explicitly the action of the other representatives of all the conjugacy classes (thereby giving a selfcontained proof that V is irreducible). The above explicit calculations show that the character vector for the representation V is χ V = (6, 0, −2, 0, 0, 1, 0). hence χ V , χ V = 1 and V is an irreducible representation.

Quadratic equations for the Del Pezzo surface of degree 5 and geometrical descriptions of S 5 -representations
We observe that by Riemann-Roch and Kodaira vanishing we have: The above formula can also be checked by direct calculation of the space of homogeneous polynomials F (x 1 , x 2 , x 3 ) of degree 6 vanishing of multiplicity 2 at the points p 1 , . . . , p 4 . This direct calculation also shows that the natural morphism of S 5 -representations: is surjective.
If we denote by W ′ the vector subspace of Sym 2 (V ) spanned by the Q ij , 1 ≤ i = j ≤ 3, then we have the following inclusions, where I is the sheaf of ideals of the functions vanishing on Y : 2.1. Geometrical and combinatorial pentagons. We want to give now some geometrical background to the choice of the sections s ij , σ ij , and the quadratic forms Q ij , which makes our calculations less mysterious.
Definition 2.1. Let S be a set. Then (1) an ordered combinatorial n-gon on S is a map p : Z/n → S; (2) the i-th side L i is the restriction of p to {i, i + 1}; (3) an ordered combinatorial n-gon on S is said to be nondegenerate if p is injective; (4) an oriented combinatorial n-gon on S is an equivalence class of ordered combinatorial n-gons on S for the action of Z/n given by composition on the source p(i) ∼ p(i + a); (5) a(n unoriented) combinatorial n-gon on S is an equivalence class of oriented combinatorial n-gons on S, where p(i) ∼ p(−i) (that is, an equivalence class of ordered n-gons for the action of the dihedral group D n , p(i) ∼ p(±i+a)); (6) the neighbouring n-gon of p(i) is the unoriented n-gon p(2i); it is nondegenerate, for n odd, iff p is nondegenerate, moreover in this case it has no sides in common with the n-gon p(i) once n ≥ 5; (7) a double combinatorial pentagon (n = 5) is the (unordered) pair of two neighbouring nondegenerate (unoriented) combinatorial pentagons. (8) If S is a linear space, then a geometrical n-gon is the union of n distinct lines L i associated to a combinatorial pentagon p(i) in such a way that L i is the line joining p(i) with p(i + 1).
Remark 2.2. If the set S has n elements, and we consider only nondegenerate n-gons, then the objects in item (1) are the elements of the set S n , those in item (4) are the cosets S n /(Z/n), for item (5) we get the cosets S n /D n , and finally for item (7) we have the cosets S 5 / Aff(1, Z/5). is an orbit for the action of S 5 on V , and the stabilizer of s 21 is the cyclic subgroup generated by (1, 2, 3, 4, 5). In particular S 4 acts simply transitively on M. Proof. We know that the only lines contained in Y are the 10 lines E ij , and it is straightforward to verify that their union is the divisor div(Σ). Let L 1 + L 2 · · · + L 5 be a geometrical pentagon; then L i intersects L i+1 , hence we get 5 distinct pairs A i of elements in {1, 2, 3, 4, 5} such that A i and A i+1 are disjoint, ∀i ∈ Z/5.
Hence A i+1 is disjoint from the union of A i and A i+2 , in particular A i and A i+2 have exactly one element in common.
We consider the 2 5-gon B i , where B i := A 2i . We denote by Hence B i = {p(i), p(i + 1)}, and we get a nondegenerate combinatorial pentagon.
Observe that we can recover the A i 's from the B j 's, simply letting A i := B 2i .
Similar formulae can be computed directly for div(s ij ), we may however observe that {div(s ij )} is the S 3 -orbit of div(s 12 ), hence div(s ij ) is a geometric pentagon.
Since div(Σ) is the union of the 10 lines of Y , follows that div(σ ij ) is associated to the neighbouring combinatorial pentagon of the one of div(s ij ).
Hence the quadrics Q ij correspond bijectively to double pentagons (these are 6).
We can summarize part of our discussion in the following corollary 3 . we get an order 4 transformation σ on M, such that σ 2 (m) = −m ∀m ∈ M, and inducing an involution on M/ ± 1, which exchanges ±s ij with ±σ ij , hence such that (M/ ± 1)/σ is in bijection with the set DP of double pentagons.
Proof. The bijection follows immediately from the fact that we have two transitive actions, and the stabilizer of s 21 is the cyclic subgroup generated by (1, 2, 3, 4, 5), which is also the stabilizer of the standard pentagon corresponding to the identity map.
We define then σ(m) by the property that it associates to an oriented pentagon p(i) the neighbouring oriented pentagon p(2i); from this definition follows that  Remark 2.5. Observe that σ corresponds to multiplication on the right by (1, 2, 4, 3) on the cosets gH, where H is any of the above three subgroups.
Next we show that all irreducible representations of S 5 of dimension different from 4 5 are contained in the permutation representation associated to the set M, i.e., to the set of oriented pentagons.
Theorem 2.6. Let M be the permutation representation associated to M, D the permutation representation associated to DP, P the permutation representation associated to P.
Then and together with the previous yields as direct sum the +1-eigenspace for −1 = σ 2 , which is clearly P , the representation associated to the pentagons. Observe that D ′ contains the sign representation ǫ.
Step 1: This follows immediately from Schur's Lemma: indeed M surjects onto V , hence V appears as a summand of M. But D and D ′ are reducible, hence P maps trivially to V , and V ′ is isomorphic to V . Therefore we have and we know that V ∼ = V ⊗ ǫ.
Step 2: We calculate the character of M.
In order to achieve this, we use the following obvious Lemma 2.7. Consider a transitive permutation representation of a finite group G on G/H. Then its character χ H is given by: In the case of the oriented pentagons, H is spanned by (1, 2, 3, 4, 5) and χ H (γ) = 0 unless γ has order dividing 5, in which case χ H (Id) = 24, while χ H ((1, 2, 3, 4, 5)) = 4, since the normalizer of H is the affine group Aff(1, Z/5).
Step 3: Subtracting twice the character of V , we obtain χ P = (12, 0, 4, 0, 0, 2, 0). Alternatively, and also in order to calculate the character of D, we again apply the above lemma.
If we consider pentagons and double pentagons (i.e. H = D 5 resp. H = Aff(1, Z/5)), observe that in both cases gHg −1 contains no elements of order 3 and in the case of pentagons no transpositions and no 4-cycles. Hence the character is zero, in the case of pentagons, unless we have 5-cycles, double transpositions, or the identity.
A 5-cycle fixes 2 pentagons, and exactly one double pentagon. A double transposition fixes 4 pentagons, and 2 double pentagons. A 4-cycle fixes 2 double pentagons: for instance, if i → 2i fixes { (0, a 1 , a 2 , a 3 , a 4 ), (0, a 2 , a 4 , a 1 , a 3 )}, then it must preserve the block decomposition {1, 2, 3, 4} = {a 1 , a 4 } ∪ {a 2 , a 3 } corresponding to the two sets of neighbours of 0; since it does not fix any pentagon, it must exchange these two sets, and then it must be either a 2 = 2a 1 or a 3 = 2a 1 . Up to choosing an appropriate representative, we may assume a 1 = 1; then it must be a 4 = −1, and the only two choices are a 2 = 2 or a 2 = −2.
Observe now that, for each τ ∈ S 3 , while, defining Hence W ′ and D ⊗ ǫ have the same character on S 3 . It follows that also W ′ is the direct sum of an irreducible five dimensional representation with a one dimensional one, and either W ∼ = D or W ∼ = D ⊗ ǫ. The first possibility is excluded since W ′ contains ǫ.
Remark 2.8. We have indeed precise formulae for the action on the space W ′ .
In particular: From the proof of Theorem 1.6 we see that other elements of S 5 act as follows: 2, 3, 4, 5): For future use we also show: Lemma 2.9. The determinant of W is the trivial representation, equivalently, 6 W ′ = ǫ.
3. S 5 -equivariant resolution of the sheaf of regular functions on the Del Pezzo surface Y ⊂ P 5 of degree 5 By Theorem 1.6, 2) we know that χ 6 V = χ 2 = ǫ.
Consider the Euler sequence on P 5 : , and this implies that where ǫ is the sign representation.
We take a Hilbert resolution of O Y (P := P 5 ): where W, W ′′ , U are S 5 -representations (a posteriori they shall be irreducible of respective dimensions 5, 5, 1, at this stage we only know that dim(W ) = 5).
Applying Hom(·, ω P ) to the above resolution, we obtain a resolution of which after twisting by O P (1) becomes: By the uniqueness of a minimal graded free resolution up to isomorphism, we get that (3.1) is isomorphic to (3.4), hence: Therefore we have proven the following: Proposition 3.1. The self dual Hilbert resolution of O Y is given by: Therefore we look for A ∈ (∧ 2 W ⊗ V ) S 5 , and we claim that A is uniquely determined up to scalars.
Lemma 3.2. The natural inclusion By Schur's lemma it follows that We want to find the (up to a constant) unique S 5 -equivariant map A, such that if A corresponds to a tensor α ∈ ( 2 W ⊗ V ) S 5 , then α ∧ α = B.
This proves the theorem.
4. S 5 -invariant equations of Y ⊂ (P 1 ) 5 As already mentioned in the first section there are five geometric objects permuted by the automorphism group S 5 of the del Pezzo surface Y of degree five: namely, 5 fibrations ϕ i : Y → P 1 , induced, for 1 ≤ i ≤ 4, by the projection with centre p i , and, for i = 5, by the pencil of conics through the 4 points (viewing the Del Pezzo surface as the moduli space of five ordered points on the projective line, these fibrations are just the maps to the moduli space of four ordered points on the projective line obtained forgetting the j-th of the five points). Each fibration is a conic bundle, with exactly three singular fibres, correponding to the possible partitions of type (2, 2) of the set {1, 2, 3, 4, 5} \ {i}.
Moreover, we showed 6 Theorem 4.2. Let Σ ⊂ (P 1 ) 4 =: Q, with coordinates , be the image of the Del Pezzo surface Y via ϕ 3 × ϕ 1 × ϕ 2 × ϕ 4 . Then the equations of Σ are given by the four 3 × 3-minors of the following Hilbert-Burch matrix: In particular, we have a Hilbert-Burch resolution: where H i is the pullback to Q of a point in P 1 under the i-th projection.
We give now the action of S 5 on the pencils ϕ i in order to determine the 10 (S 5invariant) equations of the image of Y under the map ϕ 3 × ϕ 1 × ϕ 2 × ϕ 4 × ϕ 5 , which correspond to the 10 possible projections of (P 1 ) 5 to (P 1 ) 3 .
In fact, S 5 acts by permuting the indices of ϕ i , but a permutation τ ∈ S 5 maps ϕ i to λ τ • ϕ τ (i) , where λ τ is a projectivity of P 1 .
A straightforward computation (using the formulae for the action of S 5 on P 2 by birational maps given in section 1) gives the following table.
But if this resolution were associated to a 5 × 3 matrix of the same type, then we would get a Del Pezzo surface of degree 4 and not of degree 5.
Hence (since K 2 is invariant for smooth deformations) we have a Hilbert scheme which is reducible (since the open set corresponding to smooth surfaces is disconnected).