PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 90(104) (2011), 23–46 DOI: 10.2298/PIM1104023P

GRÖBNER BASES FOR COMPLEX GRASSMANN MANIFOLDS

Branislav I. Prvulović

Communicated by Rade Živaljević

Abstract. By Borel’s description, integral cohomology of the complex Grassmann manifold G_{k,n} is a polynomial algebra modulo a well-known ideal. A strong Gröbner basis for this ideal is obtained when k = 2 and k = 3.

1. Introduction

Integral cohomology of complex Grassmannian G_{k,n} = U(n + k)/U(n) × U(k) is isomorphic to the polynomial algebra on the Chern classes c_{1}, c_{2}, …, c_{k} of the canonical complex vector bundle γ_{k} over G_{k,n} modulo the ideal I_{k,n} generated by the dual classes \overline{c}_{n+1} , \overline{c}_{n+2} , …, \overline{c}_{n+k} . Unfortunately, this description does not provide an efficient algorithm for determining whether a certain cohomology class is zero or not. But, if one has a Gröbner basis for I_{k,n} , this task is less demanding. In [5] and [6], the analogous problem for the mod 2 cohomology of real Grassmannians was considered and Gröbner bases for the corresponding ideals (in the cases k = 2 and k = 3) were presented. The theory of Gröbner bases over rings has complications that do not appear in the theory over fields. Nevertheless, for principal ideal domains, the generalization is good enough for our purposes.

In this paper, we show that calculations with ℤ coefficients, similar to those with ℤ_{2} coefficients in [5] and [6], provide strong Gröbner bases for the ideals I_{k,n} in ℤ[c_{1}, c_{2}, …, c_{k}] for k = 2, 3 and all n ⩾ k. These results are stated in Theorem 4.1 and Theorem 5.1. As a consequence of Theorem 4.1 (Corollary 4.2), we get the result of Hoggar (obtained in [3] by a calculation in terms of K-theory) concerning the structure of H^{∗}(G_{2,n}; ℤ) as an abelian group. In Corollary 5.1 we establish the analogous result for H^{∗}(G_{3,n}; ℤ).

2010 Mathematics Subject Classification: 13P10, 55R40.

Key words and phrases: Gröbner bases, complex Grassmannians, Chern classes.

Partially supported by Ministry of Science and Environmental Protection of Republic of Serbia, Project #174034.

2. Background on Gröbner bases

In this paper, we denote by ℕ_{0} the set of all nonnegative integers and the set of all positive integers is denoted by ℕ.

Let R be a principal ideal domain (PID) and R[x_{1}, x_{2}, …, x_{k}] the polynomial algebra over R on k variables. Different authors define monomials and terms in R[x_{1}, x_{2}, …, x_{k}] in various ways. Our terminology will be as follows. A monomial on variables x_{1}, x_{2}, …, x_{k} is a power product x_{1}^{a_{1}} x_{2}^{a_{2}} …x_{k}^{a_{k}} ∈ R[x_{1}, x_{2}, …, x_{k}], where a_{1}, a_{2}, …, a_{k} ∈ ℕ_{0}. The set of all monomials in R[x_{1}, x_{2}, …, x_{k}] will be denoted by M. A term in R[x_{1}, x_{2}, …, x_{k}] is a product αm of a coefficient α ∈ R and a monomial m ∈ M.

Let ⪯ be a fixed well ordering on M (a total ordering such that every nonempty subset of M has a least element) with the property that m_{1} ⪯ m_{2} implies m_{1}m_{3} ⪯ m_{2}m_{3} for all m_{1}, m_{2}, m_{3} ∈ M.

For a polynomial f = ∑ _{i=1}^{r} α_{i}m_{i} ∈ R[x_{1}, x_{2}, …, x_{k}], where m_{i} are pairwise different monomials and α_{i} ∈ R ∖ {0}, let M(f) := {m_{i} | 1 ⩽ i ⩽ r}. We define the leading monomial of f, denoted by LM(f), as maxM(f) with respect to ⪯. The leading coefficient of f, denoted by LC(f), is the coefficient of LM(f) and the leading term of f is LT(f) := LC(f) · LM(f).

The notion of a strong Gröbner basis (in [2], Becker and Weispfenning use the phrase D-Gröbner basis) for a given ideal I in R[x_{1}, x_{2}, …, x_{k}] can be defined in a number of equivalent ways. We have chosen the following one, which avoids the notion of reduction (see [2, p. 455] and [1, p. 251]).

Definition 2.1. Let G ⊂ R[x_{1}, x_{2}, …, x_{k}] be a finite set of nonzero polynomials and I_{G} = (G) the ideal in R[x_{1}, x_{2}, …, x_{k}] generated by G. We say that G is a strong Gröbner basis for I_{G} (with respect to ⪯) if for each f ∈ I_{G} ∖ {0} there exists g ∈ G such that LT(g) | LT(f) (meaning, as usual, that LT(f) = t · LT(g) for some term t).

Remark 2.1. If G is a strong Gröbner basis for I_{G} and f ∉ I_{G}, then there may still exist g ∈ G such that LT(g) divides LT(f), but one can see that f ≡ f_{1} modulo I_{G} for some polynomial f_{1} with the property that LT(f_{1}) is not divisible by any of LT(g), g ∈ G. Namely, if some LT(g) divides LT(f), say LT(f) = t · LT(g), then the polynomial f_{1} := f − t · g is ≡ f modulo I_{G} and LM(f_{1}) ≺ LM(f). If f_{1} does not have the desired property, we continue this process. Since ⪯ is a well ordering, the process must terminate.

Let G be an arbitrary finite subset of R[x_{1}, x_{2}, …, x_{k}] ∖ {0} and I_{G} the ideal generated by G. We now want to formulate a sufficient condition for G to be a strong Gröbner basis. If m ∈ M is a fixed monomial and if for f ∈ R[x_{1}, x_{2}, …, x_{k}] we have f = ∑ _{i=1}^{s} t_{i}g_{i}, where t_{i} are some terms and g_{i} some (not necessarily pairwise different) elements of G such that max_{1⩽i⩽s} LM(t_{i}g_{i}) ⪯ m, we say that ∑ _{i=1}^{s} t_{i}g_{i} is an m-representation of f with respect to G. An LM(f)-representation of f w.r.t. G is called a standard representation of f w.r.t. G.

We shall need the following lemma from [2]. We denote by lcm(a, b) and gcd(a, b) respectively the least common multiple and the greatest common divisor of a and b, where a and b are either monomials or elements of R.

Lemma 2.1. [2, p. 456] Let G be a finite set of nonzero polynomials from R[x_{1}, x_{2}, …, x_{k}] satisfying the following two conditions:

(i) For all g_{1}, g_{2} ∈ G there exists h ∈ G (which depends on g_{1} and g_{2}) such that LM(h) | lcm(LM(g_{1}), LM(g_{2})) and LC(h) | gcd(LC(g_{1}), LC(g_{2}));

(ii) Every nonzero f ∈ I_{G} has a standard representation w.r.t. G. Then G is a strong Gröbner basis.

Note that if LC(g) = 1 for all g ∈ G, then the condition (i) from Lemma 2.1 is certainly satisfied. Namely, one can take h to be g_{1}. Then, LM(h) = LM(g_{1}), so LM(h) | lcm(LM(g_{1}), LM(g_{2})). The other condition is clearly satisfied.

In order to formulate an important theorem, we need the following definition. Recall that we have a fixed ordering ⪯ on the monomials.

Definition 2.2. The S-polynomial of polynomials f, g ∈ R[x_{1}, x_{2}, …, x_{k}] is given by

S(f, g) := \frac{l}{LC(f)} · \frac{L}{LM(f)} · f − \frac{l}{LC(g)} · \frac{L}{LM(g)} · g,

where l = lcm(LC(f), LC(g)) and L = lcm(LM(f), LM(g)).

Let us note that since lcm(LC(f), LC(g)) is not uniquely determined in a PID, there is some indeterminacy in Definition 2.2, but any two least common multiples of the same pair of elements are associates and so, this indeterminacy makes no harm to the following theory. Nevertheless, we shall make the S-polynomial unique when R = ℤ, by requiring that lcm(LC(f), LC(g)) > 0. With this convention in mind, we see that (for R = ℤ), S-polynomial is antisymmetric, S(g, f) = −S(f, g).

We are now able to formulate the announced theorem.

Theorem 2.1. [2, p. 457] Let G be a finite subset of R[x_{1}, x_{2}, …, x_{k}], 0 ∉ G, and let I_{G} be the ideal in R[x_{1}, x_{2}, …, x_{k}] generated by G. If condition (i) from Lemma 2.1 holds and for all g_{1}, g_{2} ∈ G, S(g_{1}, g_{2}) either equals zero or has a standard representation with respect to G, then every nonzero f ∈ I_{G} has a standard representation w.r.t. G.

Remark 2.2. In the statement of this theorem in [2], Becker and Weispfenning reformulated the condition (i) from Lemma 2.1 in terms of G-polynomial of g_{1} and g_{2}, but we do not need this reformulation.

It is obvious from Definition 2.2 that LM(S(g_{1}, g_{2})) ≺ lcm(LM(g_{1}), LM(g_{2})) since lcm(LM(g_{1}), LM(g_{2})) cancels out in the upper expression. This means that if we have a standard representation of S(g_{1}, g_{2}) w.r.t. G, then we have an mrepresentation of S(g_{1}, g_{2}) w.r.t. G for a monomial m ≺ lcm(LM(g_{1}), LM(g_{2})). By a careful analysis of the proof of Theorem 2.1 in [2], one observes that the authors use only this weaker assumption (that S(g_{1}, g_{2}) has an m-representation w.r.t. G for some monomial m ≺ lcm(LM(g_{1}), LM(g_{2}))). Moreover, the corresponding theorem when R is a field (see [2, p. 219]) was given in this form.

By summarizing the preceding discussion, we obtain sufficient conditions for a set G ⊂ R[x_{1}, x_{2}, …, x_{k}] to be a strong Gröbner basis. These are stated in the following theorem which will be the crucial tool in proving our main results.

Theorem 2.2. Let G be a finite subset of R[x_{1}, x_{2}, …, x_{k}], 0 ∉ G, and I_{G} the ideal in R[x_{1}, x_{2}, …, x_{k}] generated by G. If for all g ∈ G, LC(g) = 1 and for all g_{1}, g_{2} ∈ G, S(g_{1}, g_{2}) either equals zero or has an m-representation with respect to G for some m ≺ lcm(LM(g_{1}), LM(g_{2})), then G is a strong Gröbner basis for I_{G}.

In the rest of the paper, we use the grlex ordering ⪯ on the monomials in R[x_{1}, x_{2}, …, x_{k}] with x_{1} > x_{2} > ⋯> x_{k}. It is defined as follows. The monomials are compared by the sum of the exponents and if these are equal for the two monomials, they are compared lexicographically from the left. That is, we shall write x_{1}^{a_{1}} x_{2}^{a_{2}} ⋯ x_{k}^{a_{k}} ≺ x_{1}^{b_{1}}x_{2}^{b_{2}} ⋯ x_{k}^{b_{k}} if either a_{1} +a_{2} +⋯ +a_{k} < b_{1} +b_{2} +⋯ +b_{k} or else a_{1} + a_{2} + ⋯+ a_{k} = b_{1} + b_{2} + ⋯+ b_{k} and a_{s} < b_{s}, where s = min{i | a_{i} ≠ b_{i}}.

3. Cohomology of G_{k,n}

Let G_{k,n} be the complex Grassmann manifold of k-dimensional complex vector subspaces in ℂ^{n+k} and let c_{1}, c_{2}, …, c_{k} be the Chern classes of the canonical bundle γ_{k} over G_{k,n} . It is known that the cohomology algebra H^{∗}(G_{k,n} ; ℤ) is isomorphic to the quotient ℤ[c_{1}, c_{2}, …, c_{k}]/I_{k,n} of the polynomial algebra ℤ[c_{1}, c_{2}, …, c_{k}] by the ideal I_{k,n} generated by polynomials \overline{c}_{n+1} , \overline{c}_{n+2} , …, \overline{c}_{n+k} . These are obtained from the equation (1 + c_{1} + c_{2} + ⋯+ c_{k})(1 + \overline{c}_{1} + \overline{c}_{2} + ⋯) = 1, that is

(3.1) 1 + \overline{c}_{1} + \overline{c}_{2} + ⋯ = \frac{1}{1 + c_{1} + c_{2} + ⋯+ c_{k}} = ∑_{t⩾0}(−1)^{t}(c_{1} + c_{2} + ⋯+ c_{k})^{t},

= ∑_{t⩾0} ∑_{a_{1} +···+a_{k} =t} (−1)^{t}[a_{1}, a_{2}, …, a_{k}]c_{1}^{a_{1}} c_{2}^{a_{2}} ⋯ c_{k}^{a_{k}}

= ∑_{a_{1} ,...,a_{k} ⩾0} (−1)^{a_{1} +···+a_{k}} [a_{1}, a_{2}, …, a_{k}]c_{1}^{a_{1}} c_{2}^{a_{2}} ⋯ c_{k}^{a_{k}}

where [a_{1}, a_{2}, …, a_{k}] (a_{j} ∈ ℕ_{0}) denotes the multinomial coefficient,

[a_{1}, a_{2}, …, a_{k}]= \frac{(a_{1} + a_{2} + ⋯+ a_{k})!}{a_{1}! · a_{2}! · · · ·· a_{k}!}

= ( {a_{1} + a_{2} + ⋯+ a_{k} \atop a_{1}} ) ( {a_{2} + ⋯+ a_{k} \atop a_{2}} ) …( {a_{k−1} + a_{k} \atop a_{k−1}} ) .

By identifying the homogenous parts of (cohomological) degree 2r in formula (3.1), we obtain the following proposition.

Proposition 3.1. For r ∈ ℕ,

\overline{c}_{r} = ∑_{a_{1} +2a_{2} +···+ka_{k} =r} (−1)^{a_{1} +···+a_{k}} [a_{1}, a_{2}, …, a_{k}]c_{1}^{a_{1}} c_{2}^{a_{2}} ⋯ c_{k}^{a_{k}} .

It is understood that a_{1}, a_{2}, …, a_{k} ∈ ℕ_{0}.

Let us add here that H^{∗}(G_{k,n} ; ℤ) ≅ ℤ[c_{1}, c_{2}, …, c_{k}]/I_{k,n} is a free (graduated) abelian group. Namely, the manifold G_{k,n} has a cell subdivision with no cells in odd dimensions (see [4, Problem 14-D]). Therefore, the (co)boundary operators in the cochain complex C^{∗}(G_{k,n} ; ℤ) are all trivial and H^{∗}(G_{k,n} ; ℤ) ≅ C^{∗}(G_{k,n} ; ℤ) is free. Furthermore, the number of 2i-cells in this CW -decomposition is p_{k,n} (i), where p_{k,n} (i) is the number of partitions of integer i into at most n nonnegative integers each of which is ⩽ k, and so the rank of the group H^{∗}(G_{k,n} ; ℤ) is ∑ _{i=0}^{nk} p_{k,n} (i).

4. Gröbner basis for I_{2,n}

The binomial coefficient for α ∈ ℤ, β ∈ ℕ is defined by \binom{α}{β} := \frac{α(α−1)···(α−β+1)}{β!} . Also, \binom{α}{0} := 1. If β is a negative integer, we define \binom{α}{β} to be equal to zero. Then it is easy to see that the well known formula

(4.1)  ( {α \atop β} ) = ( {α − 1 \atop β} ) + ( {α − 1 \atop β − 1} )

is valid for all α, β ∈ ℤ.

For k = 2, Proposition 3.1 gives us: \overline{c}_{r} = ∑ _{a+2b=r} (−1)^{a+b} ( {a + b \atop a} ) c_{1}^{a}c_{2}^{b}.

Let n ⩾ 2 be a fixed integer. In order to find a Gröbner basis for I_{2,n} = (\overline{c}_{n+1} , \overline{c}_{n+2} ), we define polynomials g_{m} (m ⩾ 0).

Definition 4.1. For m ∈ ℕ_{0}, let

g_{m} := ∑_{a+2b=n+1+m} (−1)^{n+1+a+b} ( {a + b − m \atop a} ) c_{1}^{a}c_{2}^{b}.

As before, it is understood that a, b ⩾ 0.

By comparing this definition with the above expression for \overline{c}_{r}, one observes that g_{0} = (−1)^{n+1} \overline{c}_{n+1} . Also,

c_{2} \overline{c}_{n} = ∑_{a+2b=n} (−1)^{a+b} ( {a + b \atop a} ) c_{1}^{a}c_{2}^{b+1} = ∑_{a+2b=n+2} (−1)^{a+b−1} ( {a + b − 1 \atop a} ) c_{1}^{a}c_{2}^{b}

= (−1)^{n} ∑_{a+2b=n+2} (−1)^{n+1+a+b} ( {a + b − 1 \atop a} ) c_{1}^{a}c_{2}^{b} = (−1)^{n}g_{1}.

The change of variable b ↦ b − 1 does not affect the requirement that b ⩾ 0 since for b = 0 the binomial coefficient \binom{a+b−1}{a} = \binom{n+1}{n+2} is equal to 0.

From the defining formula for g_{m}, one can see that if m ⩽ n + 2, then b must be such that m ⩽ b ⩽ \frac{n+1+m}{2} . Namely, a + b − m cannot be negative since a + b − m ⩽ −1 implies a + 2b ⩽ 2(a + b) ⩽ 2m − 2 ⩽ n + m contradicting the requirement that a + 2b = n + 1 + m. Now, a + b − m must be ⩾ a in order for \binom{a+b−m}{a} to be nonzero and we conclude that b ⩾ m. The second inequality comes from the condition a+2b = n+1+m. In particular, g_{n+2} = 0 and for 0 ⩽ m ⩽ n+1 we have

(4.2)  g_{m} = ∑_{b=m}^{[ \frac{n+1+m}{2} ]}(−1)^{b−m} ( {n + 1 − b \atop b − m} ) c_{1}^{n+1+m−2b} c_{2}^{b}.

Let G := {g_{0}, g_{1}, …, g_{n+1} }. We shall prove that, with respect to the grlex ordering, G is a strong Gröbner basis for I_{2,n}. It is obvious that the summand in (4.2) obtained for b = m provides the leading monomial LM(g_{m}) = LT(g_{m}) = c_{1}^{n+1−m} c_{2}^{m}. From this it will follow that an additive basis for H^{∗}(G_{2,n}; ℤ) is the set of all monomials c_{1}^{a}c_{2}^{b} such that a + b ⩽ n.

In order to show that G is a strong Gröbner basis for I_{2,n}, we define the ideal I_{G} := (G) = (g_{0}, g_{1}, …, g_{n+1} ) in ℤ[c_{1}, c_{2}]. As we have already noticed, \overline{c}_{n+1} = (−1)^{n+1} g_{0} ∈ I_{G}, \overline{c}_{n+2} = −c_{1} \overline{c}_{n+1} − c_{2} \overline{c}_{n} = (−1)^{n}c_{1}g_{0} + (−1)^{n+1} g_{1} ∈ I_{G}, so I_{2,n} ⊆ I_{G}.

It remains to prove that I_{G} ⊆ I_{2,n} and that G is a strong Gröbner basis. It turns out that the following proposition plays the crucial role in proving these facts.

Proposition 4.1. For each m ∈ ℕ_{0}, c_{2}g_{m} − c_{1}g_{m+1} = −g_{m+2} .

Proof. We proceed directly to the calculation.

c_{2}g_{m} − c_{1}g_{m+1} = ∑_{a+2b=n+1+m} (−1)^{n+1+a+b} ( {a + b − m \atop a} ) c_{1}^{a}c_{2}^{b+1}

− ∑_{a+2b=n+m+2} (−1)^{n+1+a+b} ( {a + b − m − 1 \atop a} ) c_{1}^{a+1} c_{2}^{b}

= ∑_{a+2b=n+m+3} (−1)^{n+a+b} ( {a + b − m − 1 \atop a} ) c_{1}^{a}c_{2}^{b}

− ∑_{a+2b=n+m+3} (−1)^{n+a+b} ( {a + b − m − 2 \atop a − 1} ) c_{1}^{a}c_{2}^{b}

= ∑_{a+2b=n+m+3} (−1)^{n+a+b} ( {a + b − m − 2 \atop a} ) c_{1}^{a}c_{2}^{b} = −g_{m+2} ,

by equality (4.1). We note that, for the similar reasons as above, the change of variable b ↦ b − 1 (a ↦ a − 1) does not affect the requirement that b ⩾ 0 (a ⩾ 0). The proposition follows. □

Corollary 4.1. I_{G} ⊆ I_{2,n}.

Proof. We already know that g_{0} = (−1)^{n+1} \overline{c}_{n+1} ∈ I_{2,n} and g_{1} = (−1)^{n}c_{2} \overline{c}_{n} = (−1)^{n+1} (c_{1} \overline{c}_{n+1} + \overline{c}_{n+2} ) ∈ I_{2,n}. Proposition 4.1 applies and by induction on m we have g_{m} ∈ I_{2,n} (0 ⩽ m ⩽ n + 1). The Corollary follows. □

Therefore G is a basis for I_{2,n} and we wish to prove that it is a strong Gröbner basis. We need the following lemma.

Lemma 4.1. For 0 ⩽ m < m + s ⩽ n + 1, we have

S(g_{m}, g_{m+s} ) = − ∑_{i=0}^{s−1} c_{1}^{i}c_{2}^{s−1−i} g_{m+2+i} .

Proof. We have that

lcm(LM(g_{m}), LM(g_{m+s} )) = lcm(c_{1}^{n+1−m} c_{2}^{m}, c_{1}^{n+1−m−s} c_{2}^{m+s} ) = c_{1}^{n+1−m} c_{2}^{m+s}

and so S(g_{m}, g_{m+s} ) = c_{2}^{s}g_{m} − c_{1}^{s}g_{m+s} .

We proceed by induction on s. For s = 1, we need to prove that S(g_{m}, g_{m+1} ) = −g_{m+2} . We calculate: S(g_{m}, g_{m+1} ) = c_{2}g_{m} −c_{1}g_{m+1} = −g_{m+2} , by Proposition 4.1. For the inductive step, we have

S(g_{m}, g_{m+s} ) = c_{2}^{s}g_{m} − c_{1}^{s}g_{m+s} = c_{2}^{s}g_{m} − c_{2}c_{1}^{s−1} g_{m+s−1} + c_{2}c_{1}^{s−1} g_{m+s−1} − c_{1}^{s}g_{m+s}

= c_{2}S(g_{m}, g_{m+s−1} ) + c_{1}^{s−1} S(g_{m+s−1} , g_{m+s} )

= −c_{2} ∑_{i=0}^{s−2} c_{1}^{i}c_{2}^{s−2−i} g_{m+2+i} − c_{1}^{s−1} g_{m+s+1}

= − ∑_{i=0}^{s−2} c_{1}^{i}c_{2}^{s−1−i} g_{m+2+i} − c_{1}^{s−1} g_{m+s+1} = − ∑_{i=0}^{s−1} c_{1}^{i}c_{2}^{s−1−i} g_{m+2+i} ,

again by Proposition 4.1 and the induction hypothesis. □

Theorem 4.1. Let n ⩾ 2. Then G = {g_{0}, g_{1}, …, g_{n+1} } defined above is a strong Gröbner basis for the ideal I_{2,n} in ℤ[c_{1}, c_{2}] with respect to the grlex ordering ⪯.

Proof. We have already shown that G is a basis for I_{2,n}. We want to apply Theorem 2.2. It is immediate from (4.2) that LC(g) = 1 for all g ∈ G. Let g_{m} and g_{m+s} (0 ⩽ m < m + s ⩽ n + 1) be two arbitrary elements of G. Since S-polynomial is antisymmetric, it suffices to show that S(g_{m}, g_{m+s} ) has required representation.

If m = n, then m + s must be n + 1 and, using Proposition 4.1, one obtains S(g_{m}, g_{m+s} ) = S(g_{n}, g_{n+1} ) = c_{2}g_{n} − c_{1}g_{n+1} = −g_{n+2} = 0.

If m ⩽ n−1, according to Lemma 4.1, S(g_{m}, g_{m+s} ) = − ∑ _{i=0}^{s−1} c_{1}^{i}c_{2}^{s−1−i} g_{m+2+i} . Note that for i ∈ {0, 1, …, s − 1}, m + 2 + i ⩽ m + s + 1 ⩽ n + 2 and so, either g_{m+2+i} ∈ G (if m + 2 + i ⩽ n + 1) or g_{m+2+i} = 0 (if m + 2 + i = n + 2). Define t = t(m, s) := c_{1}^{n−1−m} c_{2}^{m+s+1} . First of all, observe that

t ≺ c_{1}^{n+1−m} c_{2}^{m+s} = lcm(LM(g_{m}), LM(g_{m+s} )).

Now, for all i ∈ {0, 1, …, s − 1},

LM(c_{1}^{i}c_{2}^{s−1−i} g_{m+2+i} ) = c_{1}^{i}c_{2}^{s−1−i} LM(g_{m+2+i} ) = c_{1}^{i}c_{2}^{s−1−i} c_{1}^{n+1−m−2−i} c_{2}^{m+2+i}
= c_{1}^{n−1−m} c_{2}^{m+s+1} = t.

Theorem 2.2 applies and we conclude that G is a strong Gröbner basis for I_{2,n}. □

Corollary 4.2. Let n ⩾ 2. If c_{i} is the i-th Chern class of the canonical complex vector bundle γ_{2} over G_{2,n}, then the set {c_{1}^{a}c_{2}^{b} | a + b ⩽ n} is an additive basis for the free abelian group H^{∗}(G_{2,n}; ℤ).

Proof. A monomial c_{1}^{a}c_{2}^{b} corresponds to the partition of the (nonnegative) integer a + 2b:

\underbrace{1 + 1 + ⋯+ 1}_{a}+ \underbrace{2 + 2 + ⋯+ 2}_{b}.

Furthermore, it is obvious that this produces a one-to-one correspondence between the set {c_{1}^{a}c_{2}^{b} | a + b ⩽ n} and the set of all partitions of the nonnegative integers ⩽ 2n into at most n integers each of which is ⩽ 2. This means that the cardinality of the set {c_{1}^{a}c_{2}^{b} | a + b ⩽ n} is equal to rank(H^{∗}(G_{2,n}; ℤ)) = ∑ _{i=0}^{2n} p_{2,n} (i). Hence, it suffices to show that the set {c_{1}^{a}c_{2}^{b} | a + b ⩽ n} generates H^{∗}(G_{2,n}; ℤ).

Let σ ∈ H^{∗}(G_{2,n}; ℤ) ≅ ℤ[c_{1}, c_{2}]/I_{2,n} be a nonzero class and f ∈ ℤ[c_{1}, c_{2}] its representative. Since G is a strong Gröbner basis for I_{2,n} and f ∉ I_{2,n}, by Remark 2.1 we have that f ≡ f_{1} modulo I_{2,n} for some f_{1} ∈ ℤ[c_{1}, c_{2}] such that LT(f_{1}) is not divisible by any of LT(g) = LM(g), g ∈ G. Observe that {LM(g) | g ∈ G} is the set of all monomials c_{1}^{a}c_{2}^{b} such that a + b = n + 1. This means that the sum of the exponents in LM(f_{1}) and so, in every monomial from M(f_{1}), is ⩽ n. Since f_{1} also represents σ, this concludes the proof. □

Let us remark that our strong Gröbner basis G has some additional nice properties. It is minimal in the sense of Definition 4.5.6. from [1, p. 251] since no LT(g_{i}) divides LT(g_{j}) for i ≠ j. Moreover, it is reduced in the sense of the definition of this notion for Gröbner bases over fields, meaning that all leading coefficients in G are equal to 1 and no term of g_{i} is divisible by LT(g_{j}) for i ≠ j. This follows from formula (4.2) by observation that all monomials in M(g_{i}) except the leading one have the sum of the exponents < n + 1. Finally, one can verify that, since ℤ[c_{1}, c_{2}]/I_{2,n} is free, G produces unique normal forms (remainders).

Let us now calculate a few elements of the strong Gröbner basis G. By formula (4.2), g_{n+1} = c_{2}^{n+1} and g_{n} = c_{1}c_{2}^{n}. Using this and Proposition 4.1, we obtain c_{2}g_{n−1} = c_{1}g_{n} − g_{n+1} = c_{1}^{2}c_{2}^{n} − c_{2}^{n+1} = c_{2}(c_{1}^{2}c_{2}^{n−1} − c_{2}^{n}) and so we deduce that g_{n−1} = c_{1}^{2}c_{2}^{n−1} − c_{2}^{n}. Continuing in the same manner, one gets:

g_{n−2} = c_{1}^{3}c_{2}^{n−2} − 2c_{1}c_{2}^{n−1} ;
g_{n−3} = c_{1}^{4}c_{2}^{n−3} − 3c_{1}^{2}c_{2}^{n−2} + c_{2}^{n−1} ;
g_{n−4} = c_{1}^{5}c_{2}^{n−4} − 4c_{1}^{3}c_{2}^{n−3} + 3c_{1}c_{2}^{n−2} ;
g_{n−5} = c_{1}^{6}c_{2}^{n−5} − 5c_{1}^{4}c_{2}^{n−4} + 6c_{1}^{2}c_{2}^{n−3} − c_{2}^{n−2} , etc.

5. Gröbner basis for I_{3,n}

We now focus on the case k = 3 and our goal is to find a strong Gröbner basis for the ideal I_{3,n} in ℤ[c_{1}, c_{2}, c_{3}] (for all n ⩾ 3) and obtain some new information concerning the cohomology algebra H^{∗}(G_{3,n}; ℤ).

In the case k = 3, Proposition 3.1 gives us

\overline{c}_{r} = ∑_{a+2b+3c=r} (−1)^{a+b+c} ( {a + b + c \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}, r ∈ ℕ.

Let ⪯ be the grlex ordering on the monomials in ℤ[c_{1}, c_{2}, c_{3}] (with c_{1} > c_{2} > c_{3}) and let n ⩾ 3 be a fixed integer. In order to find a strong Gröbner basis for the ideal I_{3,n} = (\overline{c}_{n+1} , \overline{c}_{n+2} , \overline{c}_{n+3} ), we define the polynomials g_{m,l} ∈ ℤ[c_{1}, c_{2}, c_{3}], m, l ∈ ℕ_{0}.

Definition 5.1. For m, l ∈ ℕ_{0}, let
g_{m,l} := ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} ( {a + b + c − m − l \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}.

As before, it is understood that a, b, c ∈ ℕ_{0}.

Let us remark first that g_{0,0} = (−1)^{n+1} \overline{c}_{n+1} . Although in the expression for \overline{c}_{r} the product of binomial coefficients reduces to a trinomial coefficient, this is not the case for polynomials g_{m,l} for m > 0. Therefore, we are not able to use trinomial coefficients and their properties in the upcoming calculations with these polynomials.

In addition to that, we note that the coefficient \binom{a+b+c−m−l}{a}\binom{b+c−l}{b} may be nonzero when a + b + c − m − l < 0 (or b + c − l < 0). For example, if n = 4 we have

g_{5,0} = ∑_{a+2b+3c=10} (−1)^{1+a+b+c} ( {a + b + c − 5 \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

= −( {−1 \atop 0} ) ( {4 \atop 2}) c_{2}^{2}c_{3}^{2} + ( {0 \atop 0}) ( {5 \atop 5}) c_{2}^{5} − ( {−1 \atop 1} ) ( {3 \atop 0}) c_{1}c_{3}^{3} = c_{2}^{5} − 6c_{2}^{2}c_{3}^{2} + c_{1}c_{3}^{3}.

However, we can prove the following lemma.

Lemma 5.1. Let a, b, c, m, l be nonnegative integers. Then the following implication holds:

( {a + b + c − m − l \atop a} ) ( {b + c − l \atop b} ) ≠ 0

⟹ a + b + c < m + l or (b + c ⩾ m + l and c ⩾ l).

Proof. Assume that \binom{a+b+c−m−l}{a}\binom{b+c−l}{b} ≠ 0 and a + b + c ⩾ m + l. Then we have that \binom{a+b+c−m−l}{a} ≠ 0 and since both a + b + c − m − l and a are nonnegative we conclude that a + b + c − m − l ⩾ a, i.e., b + c ⩾ m + l.

If c < l, then b + c − l < b and since \binom{b+c−l}{b} ≠ 0 it must be b + c − l < 0. From this we have 0 ⩽ a+b+c−m−l < a−m ⩽ a, but this implies that \binom{a+b+c−m−l}{a} = 0 contradicting the assumption \binom{a+b+c−m−l}{a}\binom{b+c−l}{b} ≠ 0. This contradiction proves that c ⩾ l. □

Finally, we define the set G ⊆ ℤ[c_{1}, c_{2}, c_{3}], our candidate for the strong Gröbner basis.

Definition 5.2. G := {g_{m,l} | m + l ⩽ n + 1, m, l ∈ ℕ_{0}}.

We now prove an important property of G.

Proposition 5.1. For m, l ∈ ℕ_{0} such that m + l ⩽ n + 1, we have that the leading monomial LM(g_{m,l} ) = LT(g_{m,l} ) = c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l} and all other monomials in M(g_{m,l} ) have the sum of the exponents < n + 1.

Proof. Obviously, the (nonnegative) integers a := n + 1 − m − l, b := m, c := l satisfy the conditions a + 2b + 3c = n + 1 + m + 2l and the coefficient (−1)^{n+1+a+b+c} \binom{a+b+c−m−l}{a}\binom{b+c−l}{b} = \binom{a}{a}\binom{b}{b} = 1. So, the monomial c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l} does appear in g_{m,l} with coefficient 1.

Now, it suffices to prove the inequality a+b+c < n+1 for all other monomials c_{1}^{a}c_{2}^{b}c_{3}^{c} appearing in g_{m,l} with nonzero coefficient. If c_{1}^{a}c_{2}^{b}c_{3}^{c} is such a monomial, then a + 2b + 3c = n + 1 + m + 2l (i.e., a = n + 1 + m + 2l − 2b − 3c) and \binom{a+b+c−m−l}{a}\binom{b+c−l}{b} ≠ 0. According to Lemma 5.1, a+b+c < m+l or b+c ⩾ m+l and c ⩾ l.

In the first case a + b + c < m + l ⩽ n + 1 and we are done.

Otherwise, b+c ⩾ m+l and c ⩾ l give us that b+2c ⩾ m+2l, where the equality holds only if c = l and b = m. But then a = n + 1 + m + 2l − 2b − 3c = n + 1 − m − l and since c_{1}^{a}c_{2}^{b}c_{3}^{c} ≠ c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l}, we actually have b + 2c > m + 2l. This implies that a + b + c = n + 1 + m + 2l − b − 2c < n + 1. □

Let I_{G} be the ideal in ℤ[c_{1}, c_{2}, c_{3}] generated by G. Eventually, we shall prove that I_{G} = I_{3,n} = (\overline{c}_{n+1} , \overline{c}_{n+2} , \overline{c}_{n+3} ), but for the moment we prove that I_{G} contains I_{3,n}.

Proposition 5.2. I_{3,n} ⊆ I_{G}.

Proof. As we have already noticed, \overline{c}_{n+1} = (−1)^{n+1} g_{0,0} ∈ I_{G}. Since

−c_{1}g_{0,0} + g_{1,0} = −c_{1} ∑_{a+2b+3c=n+1} (−1)^{n+1+a+b+c} ( {a + b + c \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

+ ∑_{a+2b+3c=n+2} (−1)^{n+1+a+b+c} ( {a + b + c − 1 \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

= ∑_{a+2b+3c=n+1} (−1)^{n+2+a+b+c} ( {a + b + c \atop a} ) ( {b + c \atop b} ) c_{1}^{a+1} c_{2}^{b}c_{3}^{c}

+ ∑_{a+2b+3c=n+2} (−1)^{n+1+a+b+c} ( {a + b + c − 1 \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

= ∑_{a+2b+3c=n+2} (−1)^{n+1+a+b+c} ( {a + b + c − 1 \atop a − 1} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

+ ∑_{a+2b+3c=n+2} (−1)^{n+1+a+b+c} ( {a + b + c − 1 \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

= ∑_{a+2b+3c=n+2} (−1)^{n+1+a+b+c} ( {a + b + c \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c} = (−1)^{n+1} \overline{c}_{n+2} ,

we conclude that \overline{c}_{n+2} = (−1)^{n}c_{1}g_{0,0} + (−1)^{n+1} g_{1,0} ∈ I_{G}.

In order to show that \overline{c}_{n+3} ∈ I_{G} we calculate:

c_{1}^{2}g_{0,0} − 2c_{1}g_{1,0} + g_{2,0} = ∑_{a+2b+3c=n+1} (−1)^{n+1+a+b+c} ( {a + b + c \atop a} ) ( {b + c \atop b} ) c_{1}^{a+2} c_{2}^{b}c_{3}^{c}

−2 ∑_{a+2b+3c=n+2} (−1)^{n+1+a+b+c} ( {a + b + c − 1 \atop a} ) ( {b + c \atop b} ) c_{1}^{a+1} c_{2}^{b}c_{3}^{c}

+ ∑_{a+2b+3c=n+3} (−1)^{n+1+a+b+c} ( {a + b + c − 2 \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

= ∑_{a+2b+3c=n+3} (−1)^{n+1+a+b+c} ( {a + b + c − 2 \atop a − 2} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

+ ∑_{a+2b+3c=n+3} (−1)^{n+1+a+b+c} · 2 · ( {a + b + c − 2 \atop a − 1} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

+ ∑_{a+2b+3c=n+3} (−1)^{n+1+a+b+c} ( {a + b + c − 2 \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}.

First, we note that the change of variable a ↦ a − 2 in the first sum does not affect the requirement that a runs through ℕ_{0} since for a = 0 and a = 1 the binomial coefficient \binom{a+b+c−2}{a−2} is equal to zero. Similarly for the second sum. From formula (4.1) we deduce directly that \binom{α}{β} = \binom{α−2}{β} + 2\binom{α−2}{β−1} + \binom{α−2}{β−2} for all α, β ∈ ℤ, so we have

c_{1}^{2}g_{0,0} − 2c_{1}g_{1,0} + g_{2,0} = ∑_{a+2b+3c=n+3} (−1)^{n+1+a+b+c} ( {a + b + c \atop a} ) ( {b + c \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}
= (−1)^{n+1} \overline{c}_{n+3}

and the proposition is proved. □

In the subsequent calculations, the polynomials g_{m,l} with m + l = n + 2 will take part. We note that these polynomials are not necessarily elements of G, but, as Proposition 5.3 below states, they can be written as linear combinations of some elements of G.

In order to achieve this kind of presentation for g_{m,l} (m + l = n + 2), we prove the crucial fact which is stated in the following lemma. (We recall that the integer n ⩾ 3 is fixed.)

Lemma 5.2. Let m, l, a, b, c be nonnegative integers such that m + l = n + 2 and a + 2b + 3c = n + 1 + m + 2l. Then the following formula is true.

∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} ) = 0,

or, singling out the summand for j = 0,

( {a + b + c − n − 2 \atop a} ) ( {b + c − l \atop b} )

= ∑_{j=1}^{[ \frac{m}{2} ]}(−1)^{j−1} ( {m − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} ) .

Proof. We prove the lemma by induction on m. Let

S(m, l, a, b, c) := ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} )

The induction base will consist of three parts: m = 0, m = 1 and m = 2. Take m = 0 and nonnegative integers l, a, b, c such that l = n + 2 and a + 2b +

3c = n + 1 + 2l. The statement of the lemma in this case simplifies to:

S(0, l, a, b, c) = ( {a + b + c − n − 2 \atop a} ) ( {b + c − n − 2 \atop b} ) = 0.

Since a + 2b + 3c = n + 1 + 2l = 3n + 5, we have that 3c ⩽ a + 2b + 3c = 3n + 5,

so c ⩽ n + \frac{5}{3} < n + 2, i.e., b + c − n − 2 < b.

If b + c − n − 2 ⩾ 0, then \binom{b+c−n−2}{b} = 0 and we are done.

If b+c−n−2 < 0, then a+b+c−n−2 < a. Also, 3(a+b+c) ⩾ a+2b+3c = 3n+5

implying a + b + c ⩾ n + \frac{5}{3}. But since a + b + c is an integer, we actually have that a + b + c ⩾ n + 2. So, 0 ⩽ a + b + c − n − 2 < a, and we conclude that \binom{a+b+c−n−2}{a} = 0.

For m = 1, take l := n+1 and a, b, c ⩾ 0 such that a+2b+3c = n+1+1+2l = 3n + 4. In this case we need to prove

S(1, l, a, b, c) = ( {a + b + c − n − 2 \atop a} ) ( {b + c − n − 1 \atop b} ) = 0.

As in the case m = 0, we obtain that a+b+c ⩾ n+2 and c ⩽ n+1. If c < n+1, the proof is analogous to that of the first case. If c = n+1, then, since a+2b+3c = 3n+4, a must be 1 and b must be 0 and we obtain S(1, l, a, b, c) = \binom{0}{1}\binom{0}{0} = 0.

If m = 2, then l = n and let a, b, c be nonnegative integers such that a+2b+3c = n + 1 + 2 + 2l = 3n + 3. Here, the statement of the lemma reduces to

( {a + b + c − n − 2 \atop a} ) ( {b + c − n \atop b} ) − ( {a + b + c − n − 1 \atop a} ) ( {b + c − n − 1 \atop b} ) = 0,

since the left-hand side of this equality is S(2, l, a, b, c). In this case, from the condition a + 2b + 3c = 3n + 3 we can deduce that a + b + c ⩾ n + 1 and c ⩽ n + 1.

If c = n + 1, then necessarily a = b = 0, and we have

S(2, l, a, b, c) = ( {−1 \atop 0} ) ( {1 \atop 0}) − ( {0 \atop 0}) ( {0 \atop 0}) = 1 − 1 = 0.

If a + b + c = n + 1, since 0 ⩽ c ⩽ b + c ⩽ a + b + c and c + (b + c) + (a + b + c) =

3(n + 1), we conclude that c must be n + 1 and this case reduces to the previous one.

Suppose now that a + b + c ⩾ n + 2 and c ⩽ n. If c < n, then by the method of the case m = 0 one proves that both summands must be zero. If c = n, then there are two possibilities for the pair (a, b) such that the condition a + 2b + 3c = 3n + 3 is satisfied. First, if a = 3 and b = 0, we have

S(2, l, a, b, c) = ( {1 \atop 3}) ( {0 \atop 0}) − ( {2 \atop 3}) ( {−1 \atop 0} ) = 0 − 0 = 0.

Finally, if a = b = 1, we obtain

S(2, l, a, b, c) = ( {0 \atop 1}) ( {1 \atop 1}) − ( {1 \atop 1}) ( {0 \atop 1}) = 0 − 0 = 0

and the basis for the induction is completed.

For the induction step take m ⩾ 3, nonnegative integers l, a, b, c such that m + l = n + 2 and a + 2b + 3c = n + 1 + m + 2l and suppose that the statement of the lemma is true for all nonnegative integers < m. We need to prove that S(m, l, a, b, c) is zero. Since \binom{m−j}{j} = \binom{m−1−j}{j} + \binom{m−1−j}{j−1}, we have:

S(m, l, a, b, c) = ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 1 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} )

+ ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 1 − j \atop j − 1} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} ) .

Denoting these two sums by S_{1} and S_{2}, respectively, we have S(m, l, a, b, c) = S_{1}+S_{2}. Since \binom{b+c−l−j}{b} = \binom{b+c−l−j−1}{b} + \binom{b+c−l−j−1}{b−1}, we obtain:

S_{1} = ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 1 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b} )

+ ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 1 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b − 1} ) .

We now denote these two sums by S_{3} and S_{4} respectively and obtain S_{1} = S_{3} + S_{4} implying S(m, l, a, b, c) = S_{2} + S_{3} + S_{4}.

First, we consider the sum S_{4}. If m is odd, then [\frac{m}{2} ] = [\frac{m−1}{2} ] and if m is even, say m = 2r (r ⩾ 2), then the first binomial coefficient in the last summand of the sum S_{4} (for j = [\frac{m}{2} ] = r) is \binom{r−1}{r} = 0, so in either case

S_{4} = ∑_{j=0}^{[ \frac{m−1}{2} ]}(−1)^{j} ( {m − 1 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b − 1} )

= S(m − 1, l + 1, a, b − 1, c + 1) = 0,

by the induction hypothesis if b > 0 and if b = 0 it is obvious that S_{4} = 0.

Now, we have S(m, l, a, b, c) = S_{2} + S_{3} and we consider the sum S_{3}. Since \binom{m−1−j}{j} = \binom{m−2−j}{j} + \binom{m−2−j}{j−1}, this sum can be written as:

S_{3} = ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 2 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b} )

+ ∑_{j=0}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 2 − j \atop j − 1} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b} ) .

As before, we denote these two sums by S_{5} and S_{6} respectively and we have the equality S(m, l, a, b, c) = S_{2} + S_{5} + S_{6}.

Consider the sum S_{5} and its summand for j = [\frac{m}{2} ]. The first binomial coefficient in this summand is \binom{m−2−[m/2]}{[m/2]}. If m = 3, this binomial coefficient equals \binom{0}{1} = 0. If m ⩾ 4, we have that m − 2 − [\frac{m}{2} ] ⩾ [\frac{m}{2} ] − 2 ⩾ 0. Also, \frac{m}{2} − 1 < [\frac{m}{2} ] implying m − 2 − [\frac{m}{2} ] < [\frac{m}{2} ]. We conclude that \binom{m−2−[m/2]}{[m/2]} = 0, i.e., the summand obtained for j = [\frac{m}{2} ] is zero and so

S_{5} = ∑_{j=0}^{[ \frac{m}{2} ]−1} (−1)^{j} ( {m − 2 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b} )

= ∑_{j=0}^{[ \frac{m−2}{2} ]}(−1)^{j} ( {m − 2 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b} ) .

By looking at the sum S_{2} one easily sees that the first summand (for j = 0) equals zero (since \binom{m−1}{−1} = 0). This means that

S_{2} = ∑_{j=1}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 1 − j \atop j − 1} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} )

= ∑_{j=0}^{[ \frac{m}{2} ]−1} (−1)^{j+1} ( {m − 1 − j − 1 \atop j} ) ( {a + b + c − n − 2 + j + 1 \atop a} ) ( {b + c − l − j − 1 \atop b} )

= − ∑_{j=0}^{[ \frac{m−2}{2} ]}(−1)^{j} ( {m − 2 − j \atop j} ) ( {a + b + c − n − 1 + j \atop a} ) ( {b + c − l − j − 1 \atop b} ) .

Now the sums S_{2} and S_{5} are similar and since \binom{a+b+c−n−2+j}{a}−\binom{a+b+c−n−1+j}{a} = −\binom{a+b+c−n−2+j}{a−1}, we have that

S_{5} + S_{2} = − ∑_{j=0}^{[ \frac{m−2}{2} ]}(−1)^{j} ( {m − 2 − j \atop j} ) ( {a + b + c − n − 2 + j \atop a − 1} ) ( {b + c − l − j − 1 \atop b} )

= −S(m − 2, l + 2, a − 1, b, c + 1) = 0.

Again, we note that the upper sum is zero if a = 0 and if a > 0 we apply the induction hypothesis and obtain the latter equality.

We have reduced the sum S(m, l, a, b, c) to S_{6}. Finally, by considering the sum S_{6} we see that the summand for j = 0 is zero and so

S_{6} = ∑_{j=1}^{[ \frac{m}{2} ]}(−1)^{j} ( {m − 2 − j \atop j − 1} ) ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j − 1 \atop b} )

= − ∑_{j=0}^{[ \frac{m−2}{2} ]}(−1)^{j} ( {m − 3 − j \atop j} ) ( {a + b + c − n − 1 + j \atop a} ) ( {b + c − l − j − 2 \atop b} ) .

If m − 2 is odd, then [\frac{m−2}{2} ] = [\frac{m−3}{2} ]. If m − 2 is even, then [\frac{m−2}{2} ] = [\frac{m−3}{2} ] + 1, but, as in the case of the sum S_{4}, for m − 2 = 2r (r ⩾ 1 since m ⩾ 3) the first binomial coefficient in the summand obtained for j = [\frac{m−2}{2} ] = r equals \binom{r−1}{r} = 0. We conclude that S_{6} is equal to the sum

− ∑_{j=0}^{[ \frac{m−3}{2} ]}(−1)^{j} ( {m − 3 − j \atop j} ) ( {a + b + c − n − 1 + j \atop a} ) ( {b + c − l − j − 2 \atop b} )

= −S(m − 3, l + 3, a, b, c + 1) = 0,

by the induction hypothesis. Hence, S(m, l, a, b, c) = 0 and the proof of the Lemma 5.2 is completed. □

Proposition 5.3. Let m, l ∈ ℕ_{0} such that m + l = n + 2. Then

g_{m,l} = ∑_{j=1}^{[ \frac{m}{2} ]}(−1)^{j−1} ( {m − j \atop j} ) g_{m−2j,l+j} .

Proof. In a simplified notation, the product \binom{a+b+c−n−2+j}{a}\binom{b+c−l−j}{b} will be denoted by λ_{j} (a, b, c). Now, the previous lemma asserts that λ_{0}(a, b, c) = ∑ _{j=1}^{[ \frac{m}{2} ]}(−1)^{j−1} \binom{m−j}{j}λ_{j} (a, b, c) if a, b, c ⩾ 0 are such that a+2b+3c = n+1+m+2l.

Using the assumption m + l = n + 2 and Lemma 5.2, we have

g_{m,l} = ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} ( {a + b + c − m − l \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}
= ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} λ_{0}(a, b, c)c_{1}^{a}c_{2}^{b}c_{3}^{c}

= ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c}  ∑_{j=1}^{[ \frac{m}{2} ]}(−1)^{j−1} ( {m − j \atop j} ) λ_{j} (a, b, c)c_{1}^{a}c_{2}^{b}c_{3}^{c}

= ∑_{j=1}^{[ \frac{m}{2} ]}(−1)^{j−1} ( {m − j \atop j} ) ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} λ_{j} (a, b, c)c_{1}^{a}c_{2}^{b}c_{3}^{c}.

It remains to prove that ∑ _{a+2b+3c=n+1+m+2l} (−1)^{n+1+a+b+c} λ_{j} (a, b, c)c_{1}^{a}c_{2}^{b}c_{3}^{c} is equal to g_{m−2j,l+j} . But,

λ_{j} (a, b, c) = ( {a + b + c − n − 2 + j \atop a} ) ( {b + c − l − j \atop b} )

= ( {a + b + c − (m − 2j) − (l + j) \atop a} ) ( {b + c − (l + j) \atop b} )

and the proposition follows by Definition 5.1. □

In the following three propositions (5.4, 5.5 and 5.6) we give some convenient presentations for S-polynomials of elements of G. For the first one we need a few lemmas.

Lemma 5.3. For any integers α, β, γ, δ we have

( {α \atop β} ) ( {γ − 1 \atop δ − 1} ) − ( {α − 1 \atop β − 1} ) ( {γ \atop δ}) = ( {α − 1 \atop β} ) ( {γ \atop δ}) − ( {α \atop β} ) ( {γ − 1 \atop δ} ) .

38 PRVULOVIĆ

Proof. We calculate

( {α \atop β} ) ( {γ − 1 \atop δ − 1} ) − ( {α − 1 \atop β − 1} ) ( {γ \atop δ}) = ( {α \atop β} ) ( {γ − 1 \atop δ − 1} ) − ( {α \atop β} ) ( {γ \atop δ}) + ( {α \atop β} ) ( {γ \atop δ})
− ( {α − 1 \atop β − 1} ) ( {γ \atop δ}) = −( {α \atop β} ) ( {γ − 1 \atop δ} ) + ( {α − 1 \atop β} ) ( {γ \atop δ}) ,

by formula 4.1. □ Lemma 5.4. Let m, l ∈ ℕ_{0}, r ∈ ℕ and m + l < m + r + l ⩽ n + 1. Then

S(g_{m,l} , g_{m+r,l} ) = ∑_{i=0}^{r−1} c_{1}^{i}c_{2}^{r−1−i} (g_{m+i,l+1} − g_{m+2+i,l} ).

Proof. First, we observe that, according to Proposition 5.1, LM(g_{m,l} ) = c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l} and LM(g_{m+r,l} ) = c_{1}^{n+1−m−r−l} c_{2}^{m+r} c_{3}^{l}, so we have

lcm(LM(g_{m,l} ), LM(g_{m+r,l} )) = c_{1}^{n+1−m−l} c_{2}^{m+r} c_{3}^{l},

and since LC(g_{m,l} ) = LC(g_{m+r,l} ) = 1 (Proposition 5.1), we obtain that

S(g_{m,l} , g_{m+r,l} ) = c_{2}^{r}g_{m,l} − c_{1}^{r}g_{m+r,l} . We prove the lemma by induction on r. For r = 1, we need to verify the

equality S(g_{m,l} , g_{m+1,l} ) = g_{m,l+1} − g_{m+2,l} . We have S(g_{m,l} , g_{m+1,l} ) = c_{2}g_{m,l} − c_{1}g_{m+1,l}

= ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} ( {a + b + c − m − l \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b+1} c_{3}^{c}
− ∑_{a+2b+3c=n+1+m+1+2l}  (−1)^{n+1+a+b+c} ( {a + b + c − m − 1 − l \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a+1} c_{2}^{b}c_{3}^{c}
= ∑_{a+2b+3c=n+m+2l+3}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b − 1} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}
− ∑_{a+2b+3c=n+m+2l+3}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 2 \atop a − 1} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}.

By the previous lemma

( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b − 1} ) − ( {a + b + c − m − l − 2 \atop a − 1} ) ( {b + c − l \atop b} )

= ( {a + b + c − m − l − 2 \atop a} ) ( {b + c − l \atop b} ) −( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b} )

and we obtain:

S(g_{m,l} , g_{m+1,l} )

= ∑_{a+2b+3c=n+m+2l+3}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 2 \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}
− ∑_{a+2b+3c=n+m+2l+3}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

= −(g_{m+2,l} − g_{m,l+1} ) = g_{m,l+1} − g_{m+2,l} .

For the induction step we take r ⩾ 2 and calculate:

S(g_{m,l} , g_{m+r,l} ) = c_{2}^{r}g_{m,l} − c_{1}^{r}g_{m+r,l}
= c_{2}^{r}g_{m,l} − c_{1}^{r−1} c_{2}g_{m+r−1,l} + c_{1}^{r−1} c_{2}g_{m+r−1,l} − c_{1}^{r}g_{m+r,l}
= c_{2}S(g_{m,l} , g_{m+r−1,l} ) + c_{1}^{r−1} S(g_{m+r−1,l} , g_{m+r,l} )
= c_{2} ∑_{i=0}^{r−2} c_{1}^{i}c_{2}^{r−2−i} (g_{m+i,l+1} − g_{m+2+i,l} ) + c_{1}^{r−1} (g_{m+r−1,l+1} − g_{m+r+1,l} )

= ∑_{i=0}^{r−1} c_{1}^{i}c_{2}^{r−1−i} (g_{m+i,l+1} − g_{m+2+i,l} ),

by the induction hypothesis. □

Lemma 5.5. Let m, l ∈ ℕ_{0}, s ∈ ℕ and m + l < m + l + s ⩽ n + 1. Then

S(g_{m,l} , g_{m,l+s} ) = − ∑_{j=0}^{s−1} c_{1}^{j}c_{3}^{s−1−j} g_{m+1,l+1+j} .

Proof. Again using Proposition 5.1, we obtain

S(g_{m,l} , g_{m,l+s} ) = c_{3}^{s}g_{m,l} − c_{1}^{s}g_{m,l+s} .

We proceed by induction on s. For s = 1, the statement of the lemma reduces to S(g_{m,l} , g_{m,l+1} ) = −g_{m+1,l+1} . We have

S(g_{m,l} , g_{m,l+1} ) = c_{3}g_{m,l} − c_{1}g_{m,l+1}

= ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} ( {a + b + c − m − l \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c+1}

− ∑_{a+2b+3c=n+m+2l+3} (−1)^{n+1+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a+1} c_{2}^{b}c_{3}^{c}

= ∑_{a+2b+3c=n+m+2l+4}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

− ∑_{a+2b+3c=n+m+2l+4}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 2 \atop a − 1} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}.

As in some previous proofs, the change of variable a ↦ a − 1 in the second sum does not affect the requirement that a runs through ℕ_{0} since \binom{a+b+c−m−l−2}{a−1} = 0 for a = 0. In the first sum, the change of variable c ↦ c − 1 was made and we need to prove that for c = 0, the coefficient \binom{a+b−m−l−1}{a}\binom{b−l−1}{b} = 0, provided that a + 2b = n + m + 2l + 4.

If b ⩾ l+1, then 0 ⩽ b−l−1 < b and the second factor equals zero. If b ⩽ l, then a+b−m−l−1 ⩽ a−m−1 < a, so \binom{a+b−m−l−1}{a} ≠ 0 only if a+b−m−l−1 < 0, i.e., a+b < m+l+1. But then a+2b = a+b+b < m+l+1+l = m+2l+1 < n+m+2l+4 contradicting the fact that a+2b = n+m+2l+4. Hence, \binom{a+b−m−l−1}{a}\binom{b−l−1}{b} = 0.

Finally, since \binom{a+b+c−m−l−1}{a} − \binom{a+b+c−m−l−2}{a−1} = \binom{a+b+c−m−l−2}{a}, we get

S(g_{m,l} , g_{m,l+1} )

= ∑_{a+2b+3c=n+m+2l+4}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 2 \atop a} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}
= −g_{m+1,l+1}

and the induction base is completed.

Passing to the induction step, for s ⩾ 2 we have

S(g_{m,l} , g_{m,l+s} ) = c_{3}^{s}g_{m,l} − c_{1}^{s}g_{m,l+s}
= c_{3}^{s}g_{m,l} − c_{1}^{s−1} c_{3}g_{m,l+s−1} + c_{1}^{s−1} c_{3}g_{m,l+s−1} − c_{1}^{s}g_{m,l+s}
= c_{3}S(g_{m,l} , g_{m,l+s−1} ) + c_{1}^{s−1} S(g_{m,l+s−1} , g_{m,l+s} )
= −c_{3} ∑_{j=0}^{s−2} c_{1}^{j}c_{3}^{s−2−j} g_{m+1,l+1+j} − c_{1}^{s−1} g_{m+1,l+s} = − ∑_{j=0}^{s−1} c_{1}^{j}c_{3}^{s−1−j} g_{m+1,l+1+j}

and we are done. □

Note that lemmas 5.4 and 5.5 hold also when r = 0 (s = 0) since by definition S(f, f) = 0 and the sums on the right-hand side of the equalities are empty.

Now we generalize both Lemma 5.4 and Lemma 5.5.

Proposition 5.4. Let m, l, r, s ∈ ℕ_{0} such that m + l < m + l + r + s ⩽ n + 1. Then

S(g_{m,l} , g_{m+r,l+s} )

= ∑_{i=0}^{r−1} c_{1}^{s+i} c_{2}^{r−1−i} (g_{m+i,l+s+1} − g_{m+2+i,l+s} ) − ∑_{j=0}^{s−1} c_{1}^{j}c_{2}^{r}c_{3}^{s−1−j} g_{m+1,l+1+j} .

Proof. Using Proposition 5.1, we easily obtain that

lcm(LM(g_{m,l} ), LM(g_{m+r,l+s} )) = c_{1}^{n+1−m−l} c_{2}^{m+r} c_{3}^{l+s} ,

and so

S(g_{m,l} , g_{m+r,l+s} ) = c_{2}^{r}c_{3}^{s}g_{m,l} − c_{1}^{r+s} g_{m+r,l+s} .

Moving on, we have

S(g_{m,l} , g_{m+r,l+s} ) = c_{2}^{r}c_{3}^{s}g_{m,l} − c_{2}^{r}c_{1}^{s}g_{m,l+s} + c_{2}^{r}c_{1}^{s}g_{m,l+s} − c_{1}^{r+s} g_{m+r,l+s}
= c_{2}^{r}S(g_{m,l} , g_{m,l+s} ) + c_{1}^{s}S(g_{m,l+s} , g_{m+r,l+s} )
= − ∑_{j=0}^{s−1} c_{1}^{j}c_{2}^{r} c_{3}^{s−1−j} g_{m+1,l+1+j} + ∑_{i=0}^{r−1} c_{1}^{s+i} c_{2}^{r−1−i} (g_{m+i,l+s+1} − g_{m+2+i,l+s} ),

by lemmas 5.4 and 5.5. □

Lemma 5.6. Let m, l ∈ ℕ_{0}, s ∈ ℕ, m ⩾ s and m + l ⩽ n + 1. Then

S(g_{m,l} , g_{m−s,l+s} ) = − ∑_{j=0}^{s−1} c_{2}^{j}c_{3}^{s−1−j} g_{m−1−j,l+2+j} .

Proof. It is easy to see that by definition

S(g_{m,l} , g_{m−s,l+s} ) = c_{3}^{s}g_{m,l} − c_{2}^{s}g_{m−s,l+s} .

The proof is by induction on s. For the induction base, we want to show that S(g_{m,l} , g_{m−1,l+1} ) = −g_{m−1,l+2} . We have

S(g_{m,l} , g_{m−1,l+1} ) = c_{3}g_{m,l} − c_{2}g_{m−1,l+1}

= ∑_{a+2b+3c=n+1+m+2l}  (−1)^{n+1+a+b+c} ( {a + b + c − m − l \atop a} ) ( {b + c − l \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c+1}

− ∑_{a+2b+3c=n+m+2l+2}  (−1)^{n+1+a+b+c} ( {a + b + c − m − l \atop a} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a}c_{2}^{b+1} c_{3}^{c}

= ∑_{a+2b+3c=n+m+2l+4}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 1 \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}

− ∑_{a+2b+3c=n+m+2l+4}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 2 \atop b − 1} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}.

For the same reasons as in the proof of Lemma 5.5, we may assume that a, b and c still run through ℕ_{0}. Finally, since \binom{b+c−l−1}{b} − \binom{b+c−l−2}{b−1} = \binom{b+c−l−2}{b}, we obtain

S(g_{m,l} , g_{m−1,l+1} )

= ∑_{a+2b+3c=n+m+2l+4}  (−1)^{n+a+b+c} ( {a + b + c − m − l − 1 \atop a} ) ( {b + c − l − 2 \atop b} ) c_{1}^{a}c_{2}^{b}c_{3}^{c}
= −g_{m−1,l+2} .

Now let s ⩾ 2 and if the lemma holds for all positive integers < s, then

S(g_{m,l} , g_{m−s,l+s} ) = c_{3}^{s}g_{m,l} − c_{2}^{s}g_{m−s,l+s}

= c_{3}^{s}g_{m,l} − c_{2}^{s−1} c_{3}g_{m−s+1,l+s−1} + c_{2}^{s−1} c_{3}g_{m−s+1,l+s−1} − c_{2}^{s}g_{m−s,l+s}
= c_{3}S(g_{m,l} , g_{m−s+1,l+s−1} ) + c_{2}^{s−1} S(g_{m−s+1,l+s−1} , g_{m−s,l+s} )

= −c_{3} ∑_{j=0}^{s−2} c_{2}^{j}c_{3}^{s−2−j} g_{m−1−j,l+2+j} −c_{2}^{s−1} g_{m−s,l+s+1} = − ∑_{j=0}^{s−1} c_{2}^{j}c_{3}^{s−1−j} g_{m−1−j,l+2+j}

and the lemma follows. □

We finally use Lemma 5.6 to obtain two additional propositions concerning S-polynomials of the elements of G.

Proposition 5.5. Let m, l, ∈ ℕ_{0}, r, s ∈ ℕ, l ⩾ s, r ⩾ s and m+r+l−s ⩽ n+1. Then S(g_{m,l} , g_{m+r,l−s} )

= ∑_{i=0}^{r−s−1} c_{1}^{i}c_{2}^{r−1−i} (g_{m+i,l+1} − g_{m+2+i,l} ) − ∑_{j=0}^{s−1} c_{1}^{r−s} c_{2}^{j}c_{3}^{s−1−j} g_{m+r−1−j,l−s+2+j} .

Proof. By Proposition 5.1,

lcm(LM(g_{m,l} ), LM(g_{m+r,l−s} )) = c_{1}^{n+1−m−l} c_{2}^{m+r} c_{3}^{l},

implying

S(g_{m,l} , g_{m+r,l−s} ) = c_{2}^{r}g_{m,l} − c_{1}^{r−s} c_{3}^{s}g_{m+r,l−s}
= c_{2}^{r}g_{m,l} − c_{1}^{r−s} c_{2}^{s}g_{m+r−s,l} + c_{1}^{r−s} c_{2}^{s}g_{m+r−s,l} − c_{1}^{r−s} c_{3}^{s}g_{m+r,l−s}
= c_{2}^{s}S(g_{m,l} , g_{m+r−s,l} ) + c_{1}^{r−s} S(g_{m+r,l−s} , g_{m+r−s,l} )
= ∑_{i=0}^{r−s−1} c_{1}^{i}c_{2}^{r−1−i} (g_{m+i,l+1} − g_{m+2+i,l} ) − ∑_{j=0}^{s−1} c_{1}^{r−s} c_{2}^{j}c_{3}^{s−1−j} g_{m+r−1−j,l−s+2+j}  ,

by lemmas 5.4 and 5.6. □

Proposition 5.6. Let m, l, ∈ ℕ_{0}, r, s ∈ ℕ, l ⩾ s, r < s and m + l ⩽ n + 1.

Then

S(g_{m,l} , g_{m+r,l−s} )

= − ∑_{i=0}^{s−r−1} c_{1}^{i}c_{2}^{r}c_{3}^{s−r−1−i} g_{m+1,l−s+r+1+i}  − ∑_{j=0}^{r−1} c_{2}^{j}c_{3}^{s−1−j} g_{m+r−1−j,l−s+2+j}  .

Proof. In this case (r < s) Proposition 5.1 tells us that

lcm(LM(g_{m,l} ), LM(g_{m+r,l−s} )) = c_{1}^{n+1−m−l+s−r} c_{2}^{m+r} c_{3}^{l},

and we conclude

S(g_{m,l} , g_{m+r,l−s} ) = c_{1}^{s−r} c_{2}^{r}g_{m,l} − c_{3}^{s}g_{m+r,l−s}
= c_{1}^{s−r} c_{2}^{r}g_{m,l} − c_{2}^{r}c_{3}^{s−r} g_{m,l−s+r} + c_{2}^{r}c_{3}^{s−r} g_{m,l−s+r} − c_{3}^{s}g_{m+r,l−s}
= c_{2}^{r}S(g_{m,l−s+r} , g_{m,l} ) + c_{3}^{s−r} S(g_{m+r,l−s} , g_{m,l−s+r} )
= − ∑_{i=0}^{s−r−1} c_{1}^{i}c_{2}^{r}c_{3}^{s−r−1−i} g_{m+1,l−s+r+1+i}  − ∑_{j=0}^{r−1} c_{2}^{j}c_{3}^{s−1−j} g_{m+r−1−j,l−s+2+j}  ,

by lemmas 5.5 and 5.6. □

Observe that in the previous three propositions the S-polynomials of the elements of G are presented as some functions of polynomials g_{m,l} , where m+l ⩽ n+2. Those for which m+l ⩽ n+1 are the elements of G and those for which m+l = n+2 are either zero (for m = 0 and m = 1) or some linear combination of elements of G according to Proposition 5.3.

In order to prove that G is a basis for the ideal I_{3,n}, i.e., I_{G} = I_{3,n}, we list the following equalities:

(5.1)  g_{m+2,l} = g_{m,l+1} − c_{2}g_{m,l} + c_{1}g_{m+1,l} ,
(5.2)  g_{m+1,l+1} = c_{1}g_{m,l+1} − c_{3}g_{m,l} ,
(5.3)  g_{m−1,l+2} = c_{2}g_{m−1,l+1} − c_{3}g_{m,l} ,

which are obtained in the proofs of lemmas 5.4, 5.5 and 5.6 respectively as the induction bases.

Proposition 5.7. I_{G} = I_{3,n}.

Proof. According to Proposition 5.2, I_{3,n} ⊆ I_{G}, so it remains to prove that g ∈ I_{3,n} for all g ∈ G, i.e., g_{m,l} ∈ I_{3,n} for all m, l ∈ ℕ_{0} such that m + l ⩽ n + 1. The proof is by induction on m + l. We already have that g_{0,0} = (−1)^{n+1} \overline{c}_{n+1} ∈ I_{3,n}. Also, in the proof of Proposition 5.2 we established that

g_{1,0} = c_{1}g_{0,0} + (−1)^{n+1} \overline{c}_{n+2} = (−1)^{n+1} c_{1} \overline{c}_{n+1} + (−1)^{n+1} \overline{c}_{n+2} ∈ I_{3,n}

and that g_{2,0} = −c_{1}^{2}g_{0,0} + 2c_{1}g_{1,0} + (−1)^{n+1} \overline{c}_{n+3} ∈ I_{3,n}. By formula (5.1), g_{2,0} = g_{0,1} − c_{2}g_{0,0} + c_{1}g_{1,0} and so

g_{0,1} = g_{2,0} + c_{2}g_{0,0} − c_{1}g_{1,0} ∈ I_{3,n} .

Therefore, g_{m,l} ∈ I_{3,n} if m + l ⩽ 1.

Now, take g_{m,l} ∈ G such that m + l ⩾ 2 and assume that g_{_{m},_{l}} ∈ I_{3,n} if

m̃ + l̃< m + l. If l = 0, then m ⩾ 2 and by formula (5.1) we have

g_{m,0} = g_{m−2,1} − c_{2}g_{m−2,0} + c_{1}g_{m−1,0} ∈ I_{3,n} .

If l = 1, formula (5.2) gives us

g_{m,1} = c_{1}g_{m−1,1} − c_{3}g_{m−1,0} ∈ I_{3,n} .

Finally, if l ⩾ 2, we use formula (5.3) and obtain

g_{m,l} = c_{2}g_{m,l−1} − c_{3}g_{m+1,l−2} ∈ I_{3,n} ,

by the induction hypothesis. □

We are left to prove that G is a strong Gröbner basis for I_{3,n}. We are going to do that by showing that G satisfies sufficient conditions for being a strong Gröbner basis stated in Theorem 2.2.

Theorem 5.1. Let n ⩾ 3. The set G (see definitions 5.1 and 5.2) is a strong Gröbner basis for the ideal I_{3,n} in ℤ[c_{1}, c_{2}, c_{3}] with respect to the grlex ordering ⪯.

Proof. In order to apply Theorem 2.2, we first accord to Proposition 5.1 for the fact that LC(g) = 1 for all g ∈ G. Then, we take two arbitrary elements of G, say g_{m,l} and g_{_{m},_{l}} (g_{m,l} ≠ g_{_{m},_{l}}). Since S-polynomials are antisymmetric, without loss of generality we may assume that either (a) m < m̃ or else (b) m = m̃ and l < l̃. We distinguish three cases.

1^{∘} If condition (b) holds or if m < m̃ and l ⩽ l̃, writing m̃ = m + r, l̃= l + s,

r, s ∈ ℕ_{0}, we have m + l < m + l + r + s ⩽ n + 1, so the conditions of Proposition 5.4 are satisfied implying

S(g_{m,l} , g_{_{m},_{l}}) = S(g_{m,l} , g_{m+r,l+s} )

= ∑_{i=0}^{r−1} c_{1}^{s+i} c_{2}^{r−1−i} (g_{m+i,l+s+1} − g_{m+2+i,l+s} ) − ∑_{j=0}^{s−1} c_{1}^{j}c_{2}^{r}c_{3}^{s−1−j} g_{m+1,l+1+j} .

If m + l + r + s < n + 1, then all polynomials g_{m,l} appearing in this expression are elements of G. If m + l + r + s = n + 1, then g_{m+r+1,l+s} and eventually g_{m+1,l+s} (if r = 0) are not in G. But, according to Proposition 5.3, these two can be written as linear combinations of the elements of G and henceforth we consider these polynomials as the appropriate linear combinations.

By Proposition 5.1 the leading monomials of the elements of G all have the sum of the exponents equal to n+1. Therefore, the leading monomials of the summands in the first sum all have the sum of the exponents s+i+r −1−i+n+1 = n+r +s and in the second j +r +s−1−j +n+1 = n+r +s too. We define t = t(m, l, m̃, l̃) to be the maximum (with respect to ⪯) of all these leading monomials. Hence, the above expression is a t-representation of S(g_{m,l} , g_{_{m},_{l}}) w.r.t. G, t has the sum of the exponents equal to n + r + s and so

t ≺ c_{1}^{n+1−m−l} c_{2}^{m+r} c_{3}^{l+s} = lcm(LM(g_{m,l} ), LM(g_{_{m},_{l}})).

2^{∘} If m < m̃, l > l̃ and m̃ − m ⩾ l − l̃, writing m̃ = m + r, l̃ = l − s, r, s ∈ ℕ,

we have l ⩾ s, r ⩾ s and m + r + l − s ⩽ n + 1, i.e., the conditions of Proposition 5.5 are satisfied and by that proposition

S(g_{m,l} , g_{_{m},_{l}}) = S(g_{m,l} , g_{m+r,l−s} )

= ∑_{i=0}^{r−s−1} c_{1}^{i}c_{2}^{r−1−i} (g_{m+i,l+1} − g_{m+2+i,l} ) − ∑_{j=0}^{s−1} c_{1}^{r−s} c_{2}^{j}c_{3}^{s−1−j} g_{m+r−1−j,l−s+2+j} .

As in the previous case, for m + r + l − s = n + 1 the polynomials g_{m+r−s+1,l} and g_{m+r−1−j,l−s+2+j} (j = \overline{0, s − 1} ) are treated as linear combinations of elements of G (obtained in Proposition 5.3).

Again, we define t to be the maximum of all leading monomials in this expression and so we have a t-representation of S(g_{m,l} , g_{_{m},_{l}}) w.r.t. G. Since the sum of the exponents in the leading monomials is equal to i + r − 1 − i + n + 1 = n + r, i.e., r − s + j + s − 1 − j + n + 1 = n + r, we have

t ≺ c_{1}^{n+1−m−l} c_{2}^{m+r} c_{3}^{l} = lcm(LM(g_{m,l} ), LM(g_{_{m},_{l}})).

3^{∘} Finally, if m < m̃, l > l̃ and m̃ − m < l − l̃, again we put m̃ = m + r,

l̃ = l − s (r, s ∈ ℕ). In this case, l ⩾ s, r < s and m + l ⩽ n + 1, hence we may apply Proposition 5.6 and obtain

S(g_{m,l} , g_{_{m},_{l}}) = S(g_{m,l} , g_{m+r,l−s} )

= − ∑_{i=0}^{s−r−1} c_{1}^{i}c_{2}^{r}c_{3}^{s−r−1−i} g_{m+1,l−s+r+1+i}  − ∑_{j=0}^{r−1} c_{2}^{j}c_{3}^{s−1−j} g_{m+r−1−j,l−s+2+j}  .

Considering this case as the previous two, we observe that the sum of the exponents in the leading monomials is i+r+s−r−1−i+n+1 = n+s, i.e., j+s−1−j+n+1 = n + s. Defining t as before, we have

t ≺ c_{1}^{n+1−m−l+s−r} c_{2}^{m+r} c_{3}^{l} = lcm(LM(g_{m,l} ), LM(g_{_{m},_{l}})).

Therefore, by Theorem 2.2 we conclude that G is a strong Gröbner basis. □

Since LM(g_{m,l} ) = LT(g_{m,l} ) = c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l} (m, l ∈ ℕ_{0}, m + l ⩽ n + 1), we see that the set of all leading monomials in G is the set of all monomials with the sum of the exponents equal to n + 1. Therefore, a monomial c_{1}^{a}c_{2}^{b}c_{3}^{c} ∈ ℤ[c_{1}, c_{2}, c_{3}] is not divisible by any of these leading monomials if and only if a + b + c ⩽ n. Now, the proof of the following corollary is completely analogous to that of Corollary 4.2.

Corollary 5.1. Let n ⩾ 3. If c_{i} is the i-th Chern class of the canonical complex vector bundle γ_{3} over G_{3,n}, then the set {c_{1}^{a}c_{2}^{b}c_{3}^{c} | a + b + c ⩽ n} is a basis for the free abelian group H^{∗}(G_{3,n}; ℤ).

As in the case k = 2, one can verify that the strong Gröbner basis G from Theorem 5.1 is minimal, reduced and produces unique normal forms.

Let us now calculate a few elements of the strong Gröbner basis G. By Proposition 5.1, excluding the leading monomial LM(g_{m,l} ) = c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l}, the monomial c_{1}^{a}c_{2}^{b}c_{3}^{c} appears in g_{m,l} with nonzero coefficient only if a + b + c < n + 1, so then we have c ⩽ b + c ⩽ a + b + c ⩽ n and we conclude that a + 2b + 3c ⩽ 3n. Since a + 2b + 3c must be equal to n + 1 + m + 2l, we see that if n + 1 + m + 2l > 3n (i.e., m + 2l > 2n − 1) then g_{m,l} = LT(g_{m,l} ) = LM(g_{m,l} ) = c_{1}^{n+1−m−l} c_{2}^{m}c_{3}^{l}. In particular, we have the equalities:

g_{0,n+1} = c_{3}^{n+1} ;  g_{0,n} = c_{1}c_{3}^{n};  g_{1,n} = c_{2}c_{3}^{n}.

Starting from these three, we can calculate the polynomials g_{m,n−1} , g_{m,n−2} , g_{m,n−3} etc. in the following way. From formula (5.2), we have c_{3}g_{0,n−1} = c_{1}g_{0,n} − g_{1,n} = c_{1}^{2}c_{3}^{n} − c_{2}c_{3}^{n}, so

g_{0,n−1} = c_{1}^{2}c_{3}^{n−1} − c_{2}c_{3}^{n−1} .

Using formula (5.3), one obtains c_{3}g_{1,n−1} = c_{2}g_{0,n} − g_{0,n+1} = c_{1}c_{2}c_{3}^{n} − c_{3}^{n+1} , implying:

g_{1,n−1} = c_{1}c_{2}c_{3}^{n−1} − c_{3}^{n}.

Applying formula (5.1), we have

g_{2,n−1} = g_{0,n} − c_{2}g_{0,n−1} + c_{1}g_{1,n−1}
= c_{1}c_{3}^{n} − c_{1}^{2}c_{2}c_{3}^{n−1} + c_{2}^{2}c_{3}^{n−1} + c_{1}^{2}c_{2}c_{3}^{n−1} − c_{1}c_{3}^{n} = c_{2}^{2}c_{3}^{n−1} .

Obviously, continuing in the same manner, one can compute g_{m,l} ∈ G when l is close to n.

References

1. W. W. Adams and P. Loustaunau, An introduction to Gröbner Bases, Graduate Studies in Mathematics 3, American Mathematical Society, Providence, 1994.

2. T. Becker and V. Weispfenning, Gröbner Bases: a Computational Approach to Commutative Algebra, Graduate Texts in Mathematics, Springer-Verlag, New York, 1993.

3. S. G. Hoggar, On KO-theory of Grassmannians, Quart. J. Math. Oxford (2) 20 (1969), 447– 463.

4. J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton Univ. Press, 1974.

5. Z. Z. Petrović and B. I. Prvulović, On Groebner bases and immersions of Grassmann manifolds G_{2,n} , Homology Homotopy Appl. 13(2) (2011), 113–128.

6. Z. Z. Petrović and B. I. Prvulović, Groebner bases and some immersion theorems for Grassmann manifolds G_{3,n} , submitted.

Faculty of Mathematics (Received 06 07 2011) University of Belgrade Belgrade Serbia

    bane@matf.bg.ac.rs