Generalization of the addition and restriction theorems from free arrangements to the class of projective dimension one

We study a generalized version of Terao's famous addition theorem for free arrangements to the category of those with projective dimension one. Namely, we give a criterion to determine the algebraic structure of logarithmic derivation modules of the addition when the deletion and restrictions are free with a mild condition. Also, we introduce a class of divisionally SPOG arrangements whose SPOGness depends only on the intersection lattice like Terao's famous conjecture on combinatoriality of freeness.


Introduction
Let K be a field, V = K ℓ , and S = Sym * (V * ) = K[x 1 , . . ., x ℓ ] be the coordinate ring of V .For the derivation module Der S := ⊕ ℓ i=1 S∂ xi and a hyperplane arrangement A = {H 1 , . . ., H n }, where H i is defined as the zero locus of a non-zero linear form α Hi ∈ V * , the logarithmic derivation module D(A) of A is defined by The module D(A) is an S-graded reflexive module of rank ℓ, but not free in general.So we say that A is free with exponents exp(A) = (d 1 , . . ., d ℓ ) if D(A) ≃ ⊕ ℓ i=1 S[−d i ].In this article exp(A) is a multiset.Also in this article, we assume that all arrangements are essential unless otherwise specified, i.e., ∩ H∈A H = {0}.If A ̸ = ∅, then the submodule generated by the Euler derivation θ E ∈ D(A) forms a direct summand of D(A).Explicitly, D(A) = Sθ E ⊕D H (A), where D H (A) := {θ ∈ D(A) | θ(α H ) = 0}.So we may assume that d 1 = 1 = deg θ E ⩽ d i for i ⩾ 2 when an essential arrangement A is free.
Free arrangements have been a central topic in the research of hyperplane arrangements.Among them, the most important problem is so called Terao's conjecture asking whether the freeness of A depends only on the intersection lattice In other words, Terao's conjecture asks whether the freeness is combinatorial.This is completely open, but it was shown in [15] that the minimal free resolution of D(A) is T. Abe not combinatorial.To approach Terao's conjecture, one of the main tools is Terao's addition-deletion theorem.For the purpose of this article, we exhibit it in a slightly different way compared to its usual formulation: Theorem 1.1 (Addition and restriction theorems, [11]).Let H ∈ A, A ′ := A ∖ {H} and let A H := {H ∩L | L ∈ A ′ }.Assume that A ′ is free with exp(A ′ ) = (1, d 2 , . . ., d ℓ ).Then the following two conditions are equivalent: (1) A is free.
( In general the condition | exp(A ′ )∩exp(A H )| = ℓ−1 above is described as exp(A ′ ) ⊃ exp(A H ).However, note that | exp(A ′ ) ∩ exp(A H )| < ℓ − 1 often occurs even when both A ′ and A H are free.So we have the first question in this article: Problem 1.2.Assume that A ′ is free.Then which condition of D(A) makes A H free?More precisely, are there any explicit condition for D(A) to make A H free when A ′ is free in terms of freeness, projective dimension, free resolution and so on?Also, recent developments show the following projective dimensional version of the addition theorem.Note that A is free if and only if pd A = 0, where pd A denotes the projective dimension of the S-module D(A).Explicitly, we want to know the minimal free resolution of D(A) under the above conditions.Contrary to these problems, when A is free, we can describe D(A ′ ), which was proved in [4].To see this result, let us recall the definition of strictly plus-one generated (SPOG).Definition 1.5 ([4]).We say that A is strictly plus-one generated (SPOG) with POexp(A) = (1, d 2 , . . ., d ℓ ) and level d if there is a minimal free resolution of the following form:

. , θ ℓ is called the set of SPOG generators and θ the level element.
It was proved in [4] (see Theorem 2.4) that A ′ is SPOG if A is free and A ′ is not free.Interestingly, in this case the structure of D(A ′ ) is independent of that of D(A H ).However, in general A is neither free nor SPOG even if A ′ is free.The typical example is the case when A ′ : Then A is free with exponents (1, 1, 1, 1).If you add H : x 1 + x 2 + x 3 + x 4 = 0 to A ′ to get A, then it is well-known that A is neither free nor SPOG.In fact pd S D(A) = 2 in this case.When pd A ′ = pd A H = 0, then pd A ⩽ 1 by Theorem 1.3.Also Theorem 1.1 shows that one additional condition for exponents confirms that pd A = 0.So a weaker condition for exponents when pd A ′ = pd A H = 0 could determine the minimal free resolution of D(A).Namely, we can show the following, which answers Problems 1.2 and 1.4 partially: Theorem 1.6.Let A ′ be free with exp(A ′ ) = (1, d 2 , . . ., d ℓ ) ⩽ .Here for the set of integers (a 1 , . . ., a s ), the notation (a 1 , . . ., a s ) ⩽ means that a 1 ⩽ . . .⩽ a s .Let The organization of this article is as follows.In §2 we introduce several results and definitions for the proof of the main results in this article.In §3 we prove some useful results on the cardinality of the set of minimal generators.In §4 we prove the main results of this article.Several examples are also exhibited in §4.§5 is devoted to investigate the relation between Ziegler restriction of an arrangement and their SPOGness by using the methods introduced in the previous sections.

Preliminaries
In this section let us introduce several definitions and results for the proof of the main results in this article.First recall some combinatorics of arrangements.For the intersection lattice L(A), we can define the Möbius function µ : We call such B a polynomial B of (C, H).
Theorem 2.2 (Terao's factorization theorem, [12]).Assume that A is free with For H ∈ C and C ′ := C ∖ {H} we have the following Euler exact sequence Here for θ ∈ D(A) and the image f ∈ S/α H S of a polynomial f ∈ S by the canonical surjection S → S/α H S, ρ(θ) is defined by The Euler exact sequence is not right exact in general, but it is so when C ′ is free as follows.Theorem 2.5 (Division theorem, Theorem 1.1, [1]).Assume that A H is free and χ(A H ; t) | χ(A; t).Then A is free.Thus if we can show the freeness of A by using the division theorem several times, then the freeness of A is combinatorial, and such a free arrangement is called a divisionally free arrangement.Definition 2.6 (Definition 4.2, [5]).We say that A is stair-free if the freeness of A can be proved by using the addition and division theorems.

Theorem 2.7 (Theorem 4.3, [5]). If A is stair-free, then its freeness depends only on L(A).
Finally let us recall the fundamentals of the multiarrangement theory introduced by Ziegler in [16].A pair (A, m) is a multiarrangement if A is an arrangement and m : A → Z >0 .Multiarrangements were defined by Ziegler in [16] and used in several research of arrangements.We can define their logarithmic derivation module D(A, m) as follows: Then their freeness and exponents can be defined in the same manner as for D(A).For details, see [16].We have a canonical way to construct a multiarrangement from A.

Let H ∈ A. Then the Ziegler multiplicity m
The pair (A H , m H ) is called the Ziegler restriction of A onto H. Also recall that for H ∈ A, the submodule D H (A) of D(A) is defined by and the canonical inclusion as a section, we know that A is free if and only if D H (A) is free.Then D H (A) is closely related to D(A H , m H ) as we can see in the following several results:
Then the preimages of a set of generators for Im(π) by π generate D H (A).

2, [13]). A is free if and only if A is locally free along
For a multiarrangement we can introduce the concept of SPOG multiarrangements as follows.

Cardinality of minimal sets of generators
In this section we show some new results on the cardinality of a minimal set of generators for D(A), which will play a key role to prove our main theorem.

Definition 3.1. For an arrangement A, let g(A) denote the cardinality of a minimal set of generators for D(A). Clearly it is independent of the choice of the set of minimal generators.
Moreover for A = A ′ ∪ {H} we define an integer for a free arrangement A ′ that measures how far D(A ′ ) is from being tangent to H. Definition 3.2.Let A = A ′ ∪ {H} and assume that A ′ is free.Let F B(A ′ ) be the set of all homogeneous basis for D(A ′ ) and for each First we record the following easy facts.
On g(A) the following proposition is fundamental.
Proof.Let α H = x 1 .By the assumption on θ E , θ 2 , . . ., θ ℓ , it is clear that we can choose derivations are of positive degrees, and together with the derivations φ j (j = 1, . . ., k) form a minimal set of generators for D(A).Since their images by ρ generate the rank (ℓ − 1)-module D(A H ) due to Theorem 2.3, we can compute □ Next let us show a key result to prove Theorem 1.6.
Proof.The proof is essentially the same as that of [4, Theorem 1.9] (see [10, Proposition 3.6] too).For the completeness, let us give a sketch of the proof.We may assume that α H = x 1 and let (g i , g j ) = g ∈ S.
. By definition, we can choose g i , g j ∈ K[x 2 , . . ., x ℓ ] =: S ′ so we may assume that h i , h j , g ∈ S ′ .First let us prove that {θ k } k̸ =i,j ∪ {x Then we can express Taking the modulo x 1 = α H combined with Theorem 2.1 (see the proof of [4, Theorem 1.9] for details), we know that for some c, c ′ ∈ S. As a conclusion, it holds that θ ∈ Sθ i + Sθ j + Sφ, and By comparing the second Betti number of A calculated combinatorially and algebrogeometrically (see the proof of [4, Theorem 1.9] for details), we can see that h is a unit.Thus in fact A minimal free resolution of D(A) is easily obtained by the form of this minimal set of generators, which completes the proof.□ An immediate corollary of Proposition 3.4 is as follows: Corollary 3.6.Let A ′ be free and assume that g(A) = ℓ + 1.Then A is SPOG.
Proof.By Proposition 3.4, g(A) = ℓ + 1 only when s ⩽ 2. s = 0 cannot occur, and s = 1 implies that A is free by the addition theorem, thus g(A) = ℓ.So the rest case is when s = 2.In this case, Proposition 3.5 shows that A is SPOG.□ Since the freeness of A H and A ′ implies that pd A ⩽ 1 by Theorem 1.3, it is natural to study which condition on g(A) makes the arrangement A H free.
Proof.By the addition and restriction theorems, Theorem 2.3, Corollary 3.6 and the explicit form of the set of SPOG generators as in Proposition 3.5, the statement follows if g(A) ⩽ ℓ + 1. Assume that g(A) = ℓ + 2. Then by Proposition 3.4, we have a basis θ E , θ 2 , . . ., θ ℓ for D(A ′ ) such that θ i ̸ ∈ D(A) for i ⩾ ℓ − 2 and θ i ∈ D(A) for i ⩽ ℓ − 3. We may assume that α H = x 1 .Then clearly D(A) has a minimal set of generators with cardinality ℓ + 2 of the form by the same argument as in the proof of Proposition 3.4 or 3.5, where φ j is a linear combination of θ ℓ−2 , θ ℓ−1 , θ ℓ over K[x 2 , . . ., x ℓ ].By Theorem 2.3, the images of θ E , θ 2 , . . ., θ ℓ−3 , φ 1 , φ 2 by ρ have to generate D(A H ). Since rank S/α H S D(A H ) = ℓ−1, it holds that A H is free.□ By using results above, we can show the following proposition which is fundamental on the relation between free and SPOG arrangements.
Proof.By Proposition 3.4 and the assumption that g(A) = ℓ + 1, for the basis θ E , φ 2 , . . ., φ ℓ for D(A ′ ), we may assume that Then for constants c 1 , . . ., c s ∈ K.We may assume that c 1 ̸ = 0. Then we can choose θ 1 , φ 2 , . . ., φ n+t as a basis for M .Now assume that Let us prove that we may choose 0 for all i, then θ 1 , . . ., θ k are not independent over S.So we may assume that c k = 1, and we can choose θ 1 , . . ., θ k , φ k+1 , . . ., φ n+t as a basis for M .□ which satisfy the following conditions: (1) ψ t is a linear combination of {φ j } j∈[k−1]∖I over S for all t ∈ T , and

and (4) the image of G ∖ {ϕ j } j∈[k−1]∖I by the Euler restriction map ρ form a basis for D(A H ).
Then the S-module generated by for some g a ∈ S. Since we are interested in sets of generators, we may replace η u − i∈I g i φ i by η u to get an expression a=k g a η a by η u , we obtain Assume that b a=u g a φ a ̸ = 0, say g u = 1 by the reason of degrees in the conditions (1) and ( 2).Then replacing j∈J g j φ j + b a=u g a φ a by φ u , we can choose η u as φ u .So assume that all g a = 0 for u ⩽ a ⩽ b.Then Since η u ∈ D(A), the condition (3) shows that for some h j , h t ∈ S. Sending it by ρ, the Euler exact sequence shows that ρ(η u ) − t∈T h t ρ(ψ t ) = 0, contradicting the independency of the basis for D(A H ) in the condition (4), which completes the proof.□ Proof of Theorem 1.6.First we prove (2) ⇒ (1).It is easy to see that, for the basis θ E , φ 2 , . . ., φ ℓ for D(A ′ ) with deg Since rank S D(A H ) = ℓ − 1, it follows that A H is free with the given exponents.
Next we prove (1) ⇒ (2).Assume that A H is free with the given exponents above.In this assumption, Terao's addition theorem shows that A is not free since exp(A H ) ̸ ⊂ exp(A ′ ).By Theorem 2.

. , ρ(θ i−1 ). Hence
Now express θ in the following form by using the fact that d = deg θ ⩽ d j+1 : Here By the independency of images of θ k and θ by ρ, at least one of f i , f j is not zero.Assume that only one of them is not zero, say f i ̸ = 0 and f j = 0. Then (4.2) combined with the fact that Thus both f i and f j are not zero.Recall that φ i , φ j ̸ ∈ D(A).Thus the same proof as Proposition 3.5 shows that, letting Send φ by ρ, then we have for some b s ∈ S. Thus α H | h i and α H | b j , contradicting h i , h j ∈ S ′ .Thus we may assume that h = 1 and deg(g Second assume that a k ̸ = 0 for some k in the equation (4.1).Then we may assume that a j+1 = 1 and the equation (4.2) in this case is as follows: Replacing , we may assume that θ = φ j+1 .Continue this for θ j+1 , . . ., θ p , then we obtain either θ k = φ k+1 , or θ k = g i φ j − g j φ i modulo α H for j + 1 ⩽ k ⩽ p.So exchanging an appropriate θ k by θ, we obtain that θ = g j θ i − g i θ j modulo α H . Hence in both cases, Thus applying Lemma 4.2, we obtain that φ s = θ s for all 1 ⩽ s ⩽ ℓ with s ̸ = i, j and φ i , φ j ̸ ∈ D(A).Therefore, Proposition 3.5 shows that A is SPOG with POexp The following case is the most practical to apply Theorem 1.6.
Then the following two conditions are equivalent: and level d.
Proof.Clear by the proof of Theorem 1.6.□ Let us apply Theorem 1.6 to some examples.
Example 4.4.Let A be the Weyl arrangement of the type A 4 defined by A is well-known to be free with exp(A) = (1, 2, 3, 4).Let A ̸ ∋ H : It is easy to show that B H is free with exp(B H ) = (1,4,4).Note that In this setup, from exp(A) = (1, 2, 3, 4), the integers 2 and 3 are removed and d = 4 coincides with the remaining integer 4. Hence we can apply Theorem 1.6 to obtain that B is SPOG with POexp(B) = ( Note that Q(B H ) is Let L : x 2 = 0 and let C := B H∩L .Then it is easy to show that χ(C; t) = (t − 1)(t − 4) and χ(B H ; t) = (t − 1)(t − 4) 2 .Thus B H is divisionally free as in Theorem 2.5.Since A is divisionally free too, by Theorem 2.2, the freeness and exponents of A and B H are both combinatorial.Thus Theorem 1.6 shows that the SPOGness of B is combinatorially determined.
Proof of Theorem 1.8.Clear by Theorems 1.6 and 2.7.□ We can use Theorem 1.6 to show the combinatorial freeness of arrangements by using a non-free but SPOG arrangements.Let us check it by the following example: Example 4.5.Let A be the Weyl arrangement of the type B 4 defined by

T. Abe
Unfortunately, there are cases in which A ′ and A H are free, A is SPOG but Theorem 1.6 cannot be applied.
Example 4.9.Let Then A ′ is free with exp(A ′ ) = (1,5,7,9).Let H 1 : x 2 + x 3 + 7x 4 = 0, H 2 : x 1 + x 2 + x 3 = 0, and let A i := A ′ ∪ {H i }.Then A 1 is SPOG with POexp(A 1 ) = (1, 5, 8, 10) and level 15.So Theorem 1.6 shows that A H1 1 is free with exponents (1, 5, 15) and vice versa.On the other hand, A H2 2 is free with exponents (1,10,11), and D(A 2 ) has a minimal free resolution So in general, it can happen that A is of projective dimension one, is not SPOG, but A ′ and A H are both free.Note that the freeness of A H follows from g(A) ⩽ 6 and the freeness of A ′ by Theorem 3.7.

Ziegler restrictions and SPOG arrangements
Let us study a method to check whether A is SPOG or not by using Ziegler restrictions, i.e., a theory of multiarrangements.First recall the following two results which we will use later.
Theorem 5.1 (Theorem 2.3, [13]).Let E be a reflexive sheaf on P n (n ⩾ 3) and assume that E is locally free except for a finite number of points in P n .Then H 1 (E(e)) = 0 for all e << 0.

Generalization of the addition and restriction theorems
Proposition 5.2 (Proposition 2.5, [6] and the equation (1.5), [14]).Let A be an arrangement in V and m be a multiplicity on A. Then Next, we prove the characterization of SPOG arrangements in terms of that of the Ziegler restrictions as follows: for some g i ∈ S and c ∈ K. Since α ̸ = 0, we have a relation On the other hand, assume that there is a relation , we may assume that h 1 = 0. Since we have to determine the second syzygy, take a free module

T. Abe
Since H 1 ( D(A H , m H )(e)) = 0 for all e ∈ Z as above, the map H 1 ( D H (A)(e − 1)) is surjective.Note that there are at most finite number of non-local free points of D H (A). Assume not, then there is X ∈ L(A) such that A X is not free and dim X ⩾ 2.

Theorem 1 . 3 ([ 3 , 2 : 1 . 4 .
Theorem 1.11]).(1)Assume that pd A ′ = pd A H = 0. Then pd A ⩽ 1.(2) Assume that pd A ′ = 0 and pd A ⩽ 1.Then pd A H = 0. Now we have the second question in this article which is related to Problem 1.Problem Can we describe the algebraic structure of D(A) when A ′ and A H are both free, but | exp(A ′ ) ∩ exp(A H )| < ℓ − 1?

Problem 4 . 7 .
Generalize Theorem 1.6 to all cases when A ′ and A H are free,ℓ − 2 ⩽ | exp(A ′ ) ∩ exp(A H )| ⩽ ℓ − 1 and A is not free.In fact, to apply Theorem 1.6 the condition d ⩽ d j+1 is necessary.Let us see the following example.Example 4.8.Let A ′ be an arrangement in R 4 defined by

⊕
e∈Z H 0 ( D(A, m)(e)) = D(A, m), where D(A, m) is a sheaf on Proj(V ) obtained as the coherent sheaf associated to the module of D(A, m).
Se i such that by the map G : M → D H (A) defined by G(e i ) = θ i (i = 2, . . ., ℓ), G(e) = θ, M becomes the first syzygy of D H (A). Since θ 2 , . . ., θ ℓ , θ form a minimal set of generators for D H (A) and their images by π form a minimal set of generators for Algebraic Combinatorics, Vol. 7 #2 (2024) T. Abe D(A H , m H ), by the nine-lemma, we have an exact commutative diagram as follows: 0

1 (Example 5
D H (A)(e)) = 0 for all e ∈ Z.By using Proposition 5.2, it holds that π is surjective.Finally let us show that D(A H , m H ) is SPOG.By Yoshinaga's criterion (Theorem 2.10) and the surjectivity of π, it holds that (A H , m H ) is not free.Thus the first investigation of the generators forD(A H , m H ) shows that (A H , m H ) is SPOG.□ i − x j ) = 0.Then defineA := A 1 ∖ {x 1 = 0, x 5 = 0, x 2 = x 3 , x 1 = x 2 }.Let {x 1 = x 5 } = H ∈ A. Then by choosing appropriate coordinates x, y, z for H * , A H is isomorphic to xyz(x − w)(y − w)(z − w)(x − z)(y − z) = 0.Let {y = w} = X ∈ A H . Then A H ∖ {X} =: B is easily checked to be divisionally free by Theorem 2.5, with exponents (1, 2, 2, 2) and A X is free with exponents (1, 2, 3), which is also divisionally free.Thus Theorem 1.6 shows that A H is SPOG with POexp(A H ) = (1, 2, 3, 3) and level 3, which is combinatorial by Theorem 1.8.Now we can show by case-by-case argument that A is locally free along H and these local freeness depends only on L(A).Also, since b 2 (A) = 48 and b 2 (A H ) = 24, the b 2 -equality holds for (A, H).Thus the SPOGness of A H combined with local freeness along H and Theorem 5.7 shows that A is SPOG with POexp(A) = (1, 2, 3, 3, 3) and level 3, here 3 = |A| − |A H | = 12 − 9. Also the SPOGness of A is combinatorial by this argument.
Then the following two conditions are equivalent: (1) A H is free with exp(A H ) = (1, d 2 , . . ., di , . . ., dj , . . ., d ℓ ) ∪ (d) (2) A is SPOG with POexp(A) = (1, d 2 , . . ., d i + 1, . . ., d j + 1, . . ., d ℓ ) and level d.Theorem 1.6 can be regarded as an extension of the addition and restriction theorems (Theorem 1.1).Namely, Theorem 1.6 determines a minimal free resolution of D(A) as an SPOG-module when | exp(A ′ )∩exp(A H )| = ℓ−2.The condition d ⩽ d j+1 is necessary, see Example 4.8 for details.Now go back to Terao's conjecture.As we have seen, SPOG arrangements can be regarded as a close analogue of free arrangements.Thus to study Terao's conjecture by using an inductive approach, it is important to study combinatorial dependency of SPOG arrangements.For that purpose, let us introduce the following class of arrangements.{H} and A H are stair-free (see Definition 2.6 and Theorem 2.7), and exp(A ′ ), exp(A H ) and |A ′ | − |A H | satisfy the conditions in Theorem 1.6.Let S ℓ denote the set of stair-SPOG arrangements in an ℓ-dimensional vector space and let Definition 1.7.We say that A is stair-SPOG if there is H ∈ A such that both A ′ := A ∖ ℓ and B is Terao's polynomial, which give the required set of generators for D(A).□Proof of the main results In this section let us prove Theorems 1.6 and 1.8.For that, let us introduce the following two results.Lemma 4.1.Let N ⊂ M be S-graded free modules.Let θ 1 , . . ., θ n be a homogeneous basis for N with deg θ 1 ⩽ • • • ⩽ deg θ n and φ 1 , . . ., φ n+t be a homogeneous basis for M with deg φ 1 ⩽ • • • ⩽ deg φ n+t .If deg θ i = deg φ i for 1 ⩽ i ⩽ n, then we may choose θ 1 , . . ., θ n , φ n+1 , . . ., φ n+t as a basis for M .Proof.This is essentially the same as Theorem 4.42 in [9], but we give a proof for the completeness.Let d i := deg θ i = deg φ i .We prove by induction on 1 ⩽ i ⩽ n.Let θ 1 ∈ N ⊂ M .Let φ 1 , . . ., φ s be of degree d 1 and d s+1 > d 1 .
d 2 , ..., d ℓ ) and level d.Proof.Since π is surjective, there are θ 2 , ..., θ ℓ , θ ∈ D H (A) such that π(θ 2 ), ..., π(θ ℓ ) form a set of SPOG generators for D(A H , m H ) with a level element π(θ).For φ ∈ D H (A) let φ denote its image by the Ziegler restriction map π.Since π is surjective, Theorem 2.9 shows that θ 2 , ..., θ ℓ , θ together with θ E generate D(A).Let be the unique relation in the SPOG module D(A H , m H ), where α H ̸ = α ∈ V * and f i ∈ S. Then its preimages are of the formθ i ∈ α H D H (A)by the Ziegler exact sequence (2.2).Since D H (A) = ⟨θ 2 , . . ., θ ℓ , θ⟩ S which is a minimal set of generators because of the non-freeness of D H (A) and rank S D H i By (5.1), the left hand side of the above is in ker(G).So is the right hand side.Since α H ̸ = 0 in S and D H (A) is torsion free, we know that So we have a new relation among θ 2 , . . ., θ ℓ , θ but the degrees of this relation is lower than the original relation (5.2).Since the lowest degree relation in D(A H , m H ) is at degree d + 1 by the assumption, the lowest degree relation among θ 2 , . . ., θ ℓ , θ in D H (A) is (5.1), which is of degree d + 1. Hence applying the same argument to this new relation continuously, we can show that all the relations among θ 2 , . . ., θ ℓ i=2 h i e i + he ∈ ker(G) = K.Sending this by π, the commutativity shows that ℓ i=2 h i e i + he ∈ ker G. Since D(A H , m H ) is SPOG, ker G is generate by the unique relation αe − ℓ i=2 f i e i of degree d + 1.Thus the exactness of the middle column in the diagram shows that (5.3)ℓ i=2 h i e i + he = F (αe − ℓ i=2 f i e i ) + α H φfor some F ∈ S and φ ∈ M .Rewrite (5.3) into the following way:ℓ i=2 h i e i + he − F {(α − cα H )e − ℓ i=2 (f i + α H g i )e i } = α H (F ce + F ℓ i=2 g i e i + φ).ℓ i=2 g i e i + φ ∈ ker(G).ℓ , θ are generated by the unique relation (5.1), i.e., K ≃ S[−d − 1].Therefore, A is SPOG with the desired exponents and level.□ Algebraic Combinatorics, Vol. 7 #2 (2024)