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ARCHIVUM MATHEMATICUM (BRNO) Tomus 55 (2019), 319–331

A NOTE ON THE COHOMOLOGY RING OF THE ORIENTED GRASSMANN MANIFOLDS Ge_n,4

Tomáš Rusin

Abstract. We use known results on the characteristic rank of the canonical 4–plane bundle over the oriented Grassmann manifold^Gen,4 to compute the generators of theZ2–cohomology groupsH^j(^Gen,4) forn= 8,9,10,11. Drawing from the similarities of these examples with the general description of the cohomology rings ofGen,3 we conjecture some predictions.

1. Introduction

Let us denoteGn,k the Grassmann manifold of k–dimensional vector subspaces in Rⁿ, i.e. the space O(n)/(O(k)×O(n−k)). Next, denote Ge_n,k the oriented Grassmann manifold of orientedk-dimensional vector subspaces inRⁿ, the space SO(n)/(SO(k)×SO(n−k)). We may suppose that k≤n−kfor both of them.

The manifolds Gn,k and Gen,k come equipped with their canonical k-plane bundles, which we denoteγn,k andeγn,k respectively.

For the Grassmann manifoldGn,kthere is a concise description of itsZ2-cohomo- logy ring as a quotient ring of a polynomial ring (see [2])

(1.1) H^∗(Gn,k;Z2)∼=Z2[w1, w2, . . . , wk]/In,k,

where dim(wi) =iand the ideal In,k is generated bykhomogeneous polynomials

¯

w_n−k+1,w¯_n−k+2, . . . ,w¯n, where each ¯wi denotes thei-dimensional component of the formal power series

1 + (w1+w2+· · ·+wk) + (w1+w2+· · ·+wk)²+ (w1+w2+· · ·+wk)³+· · ·. Each indeterminatewi is a representative of theith Stiefel-Whitney classwi(γn,k) of the canonical k-plane bundleγn,k overGn,k.

However, the cohomology ring of the oriented Grassmann manifold Ge_n,k is not fully generated by the characteristic classes w_i(eγ_n,k) and is not known in general.

There are descriptions ofH^∗(Gen,k;Z2) for spheresGen,1∼=Sⁿ⁻¹, complex quadrics Gen,2, and in [1] forGen,3as well.

2010Mathematics Subject Classification: primary 57T15; secondary 57R20, 55R25.

Key words and phrases: oriented Grassmann manifold, characteristic rank, Stiefel-Whitney class.

Received May 30, 2019. Editor M. Čadek.

DOI: 10.5817/AM2019-5-319

In this paper we begin the study of theZ2-cohomology ring ofGe_n,4by considering the casesn= 8,9,10,11. We will abbreviate H^j(X;Z2) toH^j(X), denotewi = w_i(γ_n,k) andwe_i=w_i(eγ_n,k) as usual.

The paper is organized as follows. In the second section we review the general strategy on how to approach the study ofH^∗(Ge_n,k). It contains the tools which will be used later to perform the computations. The third section contains the main result of the paper, which is the complete description of all cohomology groups of Ge_n,4 forn= 8,9,10,11, along with partial information about the ring structure of H^∗(Gen,4). Some conjectures are also discussed based on these results.

2. Preliminaries

To obtain information aboutH^j(Gen,4), we first need to recall some general facts aboutH^j(Gen,k). We proceed similarly as in [4].

There is a covering projectionp:Ge_n,k →G_n,k, which is universal for (n, k)6=

(2,1). To this 2-fold covering, there is an associated line bundleξoverG_n,k, such thatw₁(ξ) =w₁(γ_n,k), to which we have Gysin exact sequence ([6, Corollary 12.3]) (2.1) −→^ψ H^j−1(G_n,k)−→^w1 H^j(G_n,k) ^p

∗

−→H^j(Ge_n,k)−→^ψ H^j(G_n,k)−→^w1 whereH^j−1(Gn,k)−→^w1 H^j(Gn,k) is the homomorphism given by the cup product with the first Stiefel-Whitney classw1=w1(γn,k).

Since the pullbackp^∗γn,kis isomorphic toeγn,k, the covering projectionp:Gen,k → G_n,k induces the ring homomorphism p^∗:H^∗(G_n,k)−→H^∗(Ge_n,k), which maps each Stiefel-Whitney class w_i to we_i.

Consequently, the image Im(p^∗:H^j(G_n,k) →H^j(Ge_n,k)) is a subspace of the Z2-vector space H^j(Gen,k) consisting only of cohomology classes, which can be expressed as polynomials in the Stiefel-Whitney characteristic classes ofeγn,k. We will call itthe characteristic subspaceand denote itC(j;n, k). Moreover (see [9]), the image Im(p^∗) of the ring homomorphism p^∗:H^∗(G_n,k) −→ H^∗(Ge_n,k) is a self-annihilating subspace of H^∗(Gen,k). That is, we have the following.

Lemma 2.1. For any xe ∈ C(j;n, k) and ey ∈ C(j⁰;n, k) we have xeye = 0 if j+j⁰=k(n−k) = dim(Gen,k).

From the exactness of the sequence (2.1), we have we1=p^∗(w1) = 0 and it is clear that a monomialwe^a₂²we^a₃³. . .we^a_k^k =p^∗(w^a₂²w^a₃³. . . w_k^ak) is zero inH^j(Gen,k) if and only ifw₂^a2w^a₃³. . . w_k^ak is aw1-multiple of some polynomial inH^∗(Gn,k). Let us therefore denotegi ∈Z2[w2, . . . , wk] the reduction of the polynomial ¯wi (see (1.1)) modulow₁ and by J_n,k the ideal in Z2[w₂, . . . , w_k] generated by g_n−k+1, . . . , g_n. The following lemma is a formal restatement of the previous observation.

Lemma 2.2. Monomialwe₂^a2we^a₃³. . .we_k^ak∈C(j;n, k)is equal to zero iffw^a₂²w^a₃³. . . w^a_k^k∈Jn,k.

The question whetherC(j;n, k) is equal to H^j(Ge_n,k) is related to the notion of the characteristic rank of a vector bundle, which was defined in [3], [7].

Definition 2.3. Let X be a connected, finite CW-complex andξ a real vector bundle overX. Thecharacteristic rankof the vector bundle ξ, charrank(ξ), is the greatest integer q, 0≤q≤dim(X), such that every cohomology class inH^j(X) for 0≤j≤qcan be expressed as a polynomial in the Stiefel-Whitney classesw_i(ξ) ofξ.

This implies that the characteristic rank ofeγn,k is equal to the greatest integer q, such that the homomorphismp^∗:H^j(Gn,k)→H^j(Gen,k) is surjective (that is C(j;n, k) = H^j(Ge_n,k)) for allj, 0 ≤ j ≤ q, or equivalently, by (2.1), that the homomorphismw₁:H^j(G_n,k)−→H^j+1(G_n,k) is injective for allj, 0≤j≤q.

Hence, in order to compute the characteristic rank ofeγ_n,k, it is necessary to study the kernel ofw₁:H^j(G_n,k)−→H^j+1(G_n,k). Let us denoteb_j(X) thejthZ2–Betti number of a manifoldX and then define αj(Gen,k) =bj(Gen,k)−dim(C(j;n, k)), the codimension of the subspaceC(j;n, k)⊆H^j(Gen,k).

There is a useful upper bound for this number described in the next proposition.

Proposition 2.4 ([5, Proposition 2.4. (3)]). For a non-negative integer x, we associate with H^n−k+x+1(G_n,k) (2≤k≤n−k) the set

N_x(G_n,k) :=

k−1

[

i=0

{w^b₂²· · ·w^b_k^kg_n−k+1+i; 2b₂+ 3b₃+· · ·+kb_k =x−i}. Ifx≤n−k−1and there aret linearly independent elements in the setNx(Gn,k), then

α_n−k+x(Ge_n,k)≤ |N_x(G_n,k)| −t , where|Nx(Gn,k)|is the cardinality of the setNx(Gn,k).

Whenj≤charrank(eγ_n,k) we have (see [4, (3)]) (2.2) b_j(Ge_n,k) =b_j(G_n,k)−b_j−1(G_n,k)

and the Betti numbers forGn,k are readily calculable from the Poincaré polyno- mial [2]

(2.3) Pt(Gn,k) =(1−t^n−k+1)· · ·(1−tⁿ) (1−t)· · ·(1−t^k) . 3. Computations

Recently, the number charrank(eγ_n,4) was completely determined [8] and adjusted to our notation we have the following.

Theorem 3.1([8, Theorem 6.6]). Letn≥8be an integer. Ift≥3is the unique integer such that2^t−1< n≤2^t, then

charrank(eγn,4) = min

4n−3·2^t−1−5,2^t−5 .

For better clarity of the forthcoming proofs we first list generators of the ideals J_n,4and derive some additional relations in cohomology implied by Lemma 2.2.

Lemma 3.2. We have

J_8,4=J_9,4= (g₆, g₇, g₈) = (w³₂+w²₃, w₂²w₃, w⁴₂+w₂w₃²+w²₂w₄+w²₄), J_10,4= (w²₂w₃, w⁴₂+w₂w₃²+w₂²w₄+w²₄, w₃³, w⁵₂+w²₃w₄+w₂w²₄),

J_11,4= (w⁴₂+w₂w₃²+w₂²w₄+w²₄, w₃³, w⁵₂+w²₃w₄+w₂w²₄, w₂⁴w₃+w₃w²₄).

Additionally, in H^∗(Ge8,4)andH^∗(Ge9,4)we have

we₄²=we₂²we₄, we³₃= 0, we⁵₂= 0, we₃we₄²= 0, we₂²we²₄=we⁴₂we₄.

In H^∗(Ge10,4)we have

we³₂we₄=we²₃we₄, we₃we₄²= 0, we⁶₂+we₂⁴we₄=we₂²we²₄, we³₄=we²₂we₄².

In H^∗(Ge_11,4)we have

we²₂we²₃+we³₂we₄=we²₃we₄, we²₂we₃we₄= 0, we³₂we²₄=we₃²we₄², we₂⁷=we⁴₂we²₃, we⁷₂=we⁵₂we₄, we₂we³₄= 0, we₂⁵we₃²=we⁴₂we²₄, we₂⁶we₄=we⁴₂we₄²,

Proof. Direct computation of the polynomials gi ∈ Z2[w2, w3, w4] shows that g5 = 0 and g6, . . . , g11 are as claimed. Since g5 = 0 we have (g5, g6, g7, g8) = (g6, g7, g8). However g9=w³₃=w2g7+w3g6, thus also (g6, g7, g8, g9) = (g6, g7, g8).

By definition, both J8,4 andJ9,4 are equal to (g6, g7, g8).

Now, sinceJ_8,4=J_9,4, by Lemma 2.2 the relations inH^∗(Ge_8,4) andH^∗(Ge_9,4) for the elements of the characteristic subspace will be the same. We will check them inH^∗(Ge8,4). The proof forH^∗(Ge9,4) is identical. Sincew2g6+g8∈J8,4, we have we²₂we4+we²₄= 0, which is equivalent to we₄²=we²₂we4. We have already shown that w³₃ ∈J8,4, thuswe₃³= 0. We have w⁵₂ =w₂²g6+w3g7 ∈J8,4, therefore we obtain we⁵₂= 0. Next we3we²₄=we²₂we3we4 by the first relation and the latter is zero because w4g7∈J8,4. Finally,we²₂we²₄=we₂⁴we4 by the same reason.

InH^∗(Ge10,4) we havewe³₂we₄+we²₃we₄= 0 sincew₂³w₄+w₃²w₄=w₃g₇+w₂g₈+g₁₀∈ J10,4. It is easy to check thatw3w₄² = (w₂²+w4)g7+w3g8+w2g9 ∈J10,4. Next we⁶₂+we₂⁴we4=we₂⁶+we2we²₃we4=we²₂we²₄by the first relation andw2g10∈J10,4. Finally we³₄=we⁴₂we4+we2we²₃we4+we²₂we²₄ sincew4g8∈J10,4and the first two summands are equal as they are we2-multiples of equal classes.

In H^∗(Ge11,4) we have we²₂we²₃+we₂³we4 =we₃²we4, sincew²₂w²₃+w₂³w4+w²₃w4 = w2g8+g10∈J11,4. Then we havew²₂w3w4=w3g8+w2g9+g11∈J11,4. Next, we have w³₂w²₄+w₃²w₄²= (w₃²+w2w4)g8+w2w3g9+w4g10+w3g11∈J11,4. Sincew₂²g10∈J11,4, we havewe⁷₂=we₂²we²₃we4+we³₂we₄², but the first summand is zero and the last is equal towe₃²we₄²by the third relation, which is equal towe₂⁴we²₃, becausew3g11∈J11,4. Since w³₂g8+w3g11 ∈J11,4, we havewe⁷₂+we⁵₂we4+we³₂we²₄+we₃²we²₄ = 0, but the last two summands are equal by the third relation. Sincew₂w₄g₈+w²₂g₁₀∈J_11,4, we have we⁷₂+we⁵₂we₄+we₂we₄³= 0, but the first two summands are equal by the previous relation.

Sincew₂w₃g₁₁ ∈J_11,4, we have we₂⁵we₃²=we₂we²₃we²₄ and the RHS is equal towe⁴₂we²₄ by the third relaton. Sincew₂w₄g₁₀∈J_11,4, we havewe⁶₂we₄+we₂we₃²we²₄+we²₂we³₄= 0, but the last summand iswe₂–multiple of zero and the middle one is equal towe⁴₂we²₄

and we obtain the desired result.

Theorem 3.3. We have the following generators ofH^j(Ge_8,4).

j gen. j gen.

0 we0 9 a4we2we3,we2we3we4

1 − 10 a4we³₂, a4we2we4,we₂³we4

2 we2 11 a4we3we4

3 we3 12 a4we₂⁴, a4we₂²we4,we₂⁴we4

4 a4,we₂²,we4 13 a4we2we3we4

5 we2we3 14 a4we₂³we4

6 a4we2,we³₂,we2we4 15 −

7 a4we3,we3we4 16 a4we₂⁴we4

8 a4we₂², a4we4,we⁴₂,we²₂we4

where a4 is an element in H⁴(Ge8,4)\C(4; 8,4).

Proof. We have charrank(eγ8,4) = 3, so forj ≤3 we haveC(j; 8,4) =H^j(Ge8,4), but C(4; 8,4) ⊂ H⁴(Ge_8,4) is a proper subspace and thus for the codimension we have α4(Ge8,4) = bj(Gen,k)−dim(C(j;n, k)) ≥ 1. On the other hand, from Proposition 2.4 we have α4(Ge8,4)≤1 sincex= 0 andN0(G8,4) ={g5}is a one element set. Let us denotea₄∈H⁴(Ge_8,4) an element outsideC(4; 8,4).

Now, let us first list all generators of C(j; 8,4) with the help of Lemma 3.2 before continuing further withH^j(Ge8,4). Note thatwe³₂=we₃² andwe₂²we₃= 0, since g6, g7 ∈J8,4. Also note that ifxe∈C(j; 8,4) is a nonzero element, it must be a wei–multiple of some nonzero element inC(j−i; 8,4) for somei∈ {2,3,4}.

InC(5; 8,4) there is only one nonzero element,we2we3. InC(6; 8,4) there are two,we₂³andwe2we4, because we₃²=we³₂.

InC(7; 8,4) we havewe₂²we3= 0 and thuswe3we4 is the only generator.

InC(8; 8,4) we havewe₂⁴=we2we₃²andwe²₂we4=we₄²as the two generators.

InC(9; 8,4) we havewe₂³we3=we₃³= 0, sowe2we3we4is the only nonzero element.

In C(10; 8,4) we have we⁵₂ = we²₂we²₃ = 0 and we₂³we4 = we²₃we4 = we2we₄² as the generator.

InC(11; 8,4) we havewe²₂we₃we₄= 0,we⁴₂we₃= 0, we₃we²₄= 0.

InC(12; 8,4) there is one generator we₂⁴we₄ equal to bothwe₂²we₄²andwe₂we²₃we₄. By Poincaré duality, to each nonzero element ex ∈ H^j(Ge_8,4) there exists a nonzero element ye∈ H^16−j(Ge8,4) such that xeey 6= 0. Thus Lemma 2.1 implies C(j; 8,4) = 0 for allj >12. Additionally, the dual to elementwe₂⁴we₄ must bea₄. Hence a₄we⁴₂we₄=a₄we₂we₃²we₄6= 0.

It remains to determineα_j(Ge_8,4) forj= 5,6,7,8. Forj≤7 Proposition 2.4 with x=j−4≤3 implies thatαj(Ge8,4)≤ |N_j−4(G8,4)|−t_j−4, wheret_j−4is the maximal number of linearly independent elements ofNj−4(G8,4). ForN1(G8,4) ={g6} we have t1 = 1. For N2(G8,4) = {w2g5, g7} we have t2 = 1, since g5 = 0. For N3(G8,4) ={w3g5, w2g6, g8}we havet3= 2 as allw2g6, g8, w2g6+g8 are nonzero.

Thusα5(Ge8,4)≤0,α6(Ge8,4)≤1 andα7(Ge8,4)≤1. On the other hand, we have showna₄we₂∈H⁶(Ge_8,4) anda₄we₃∈H⁷(Ge_8,4) to be nonzero and dual towe₂³we₄and we2we3we4 respectively. Therefore α6(Ge8,4), α7(Ge8,4)6= 0 and both must be equal to 1.

We determine α₈(Ge_8,4) with the help of the Euler characteristic. For the Grass- mann manifold G_8,4 we can compute its Euler characteristic from the Poin- caré polynomial (2.3) to obtain χ(G8,4) = 6. As Ge8,4 is a 2–fold cover, we have χ(Ge8,4) = 2·χ(G8,4) = 12. By this point we know the Betti numbers b₀(Ge_8,4), . . . , b₇(Ge_8,4). Poincaré duality and a simple calculation yieldsb₈(Ge_8,4) = 4.

Consequently,α8(Ge8,4) = 2 and since we already knowa4we²₂anda4we4are nonzero, we only need to show that they are distinct. That is done by considering their products withwe⁴₂ and realizing one is zero while the other is not.

By obvious adjustment of the last argument we also prove thata4we³₂6=a4we2we4

anda₄we⁴₂6=a₄we₂²we₄. All the remaining numbersα_j(Ge_8,4) are now determined by Poincaré duality combined with the knowledge of all C(j; 8,4) and the obvious

generators suffice to produce the required values.

Theorem 3.4. We have the following generators ofH^j(Ge_9,4).

j gen. j gen.

0 we₀ 11 a₈we₃

1 − 12 a₈we²₂, a₈we₄,we⁴₂we₄

2 we₂ 13 a₈we₂we₃

3 we₃ 14 a₈we₂³, a₈we₂we₄

4 we²₂,we₄ 15 a₈we₃we₄

5 we₂we₃ 16 a₈we⁴₂, a₈we²₂we₄ 6 we₂³,we₂we₄ 17 a₈we₂we₃we₄

7 we₃we₄ 18 a₈we³₂we₄

8 ea₈, w⁴₂,we₂²we₄ 19 −

9 we₂we₃we₄ 20 a₈we⁴₂we₄

10 a₈we₂,we₂³we₄

where a8 is an element in H⁸(Ge9,4)\C(8; 9,4).

Proof. First, asJ9,4=J8,4, we haveC(j; 9,4) =C(j; 8,4) for allj.

We have charrank(eγ9,4) = 7 soH^j(Ge9,4) =C(j; 9,4) =C(j; 8,4) forj≤7 and α₈(Ge_9,4) ≥ 1. To obtain an upper bound for α₈(Ge_9,4) we consider N₃(G_9,4) =

{w3g₆, w₂g₇, g₉}. Any two elements from this set are linearly independent, which meansα₈(Ge_9,4) = 1. Denotea₈an element in H⁸(Ge_9,4)\C(8; 9,4).

Clearly, the Poincaré dual towe⁴₂we₄isa₈and similarly as before we havea₈we⁴₂we₄= a₈we₂we²₃we₄6= 0.

Next, N4(G9,4) =

w²₂g6, w4g6, w3g7, w2g8 . We will show that these four ele- ments are linearly independent. Suppose that for some ci ∈ Z2, 1 ≤i ≤ 4 we have

c₁w²₂g₆+c₂w₄g₆+c₃w₃g₇+c₄w₂g₈= 0.

Since every element inZ2[w2, w3, w4] is of order 2, the equation implies (c₁+c₃+c₄)w₂²g₆+c₂w₄g₆+c₃(w₃g₇+w₂²g₆) +c₄(w₂g₈+w₂²g₆) = 0,

(c₁+c₃+c₄)(w⁵₂+w²₂w²₃) +c₂w₄g₆+c₃w₂⁵+c₄(w₂³w₄+w₂w₄²) = 0. Ifc1+c3+c4 orc3 or both are nonzero, the LHS is not divisible byw4, which is a contradiction. Thusc1+c3+c4=c3= 0 and the equation simplifies so much,we immediately deducec2=c4 = 0. Which in turn impliesc1= 0. We have proved independence and soα9(Ge9,4) = 0.

Now that Betti numbers b0(Ge9,4), . . . , b9(Ge9,4) are known, from calculating χ(G_9,4) = 6 and χ(Ge_9,4) = 12 we obtainb₁₀(Ge_9,4) = 2.

This gives enough information to quickly determine allαj(Ge9,4) and all are once again covered by the obvious generators derived froma8we⁴₂we4=a8we2we²₃we46= 0.

Theorem 3.5. We have the following generators ofH^j(Ge10,4).

j gen. j gen.

0 we0 13 −

1 − 14 a12we2, b12we2

2 we2 15 a12we3

3 we3 16 a12we²₂, a12we4, b12we²₂

4 we₂²,we4 17 a12we2we3

5 we2we3 18 a12we₂³, a12we2we4, b12we³₂ 6 we³₂,we²₃,we2we4 19 a12we3we4

7 we3we4 20 a12we₂⁴, b12we⁴₂ 8 we⁴₂,we2we²₃,we²₂we4 21 a12we2we3we4

9 we2we3we4 22 a12we²₃we4=b12we⁵₂

10 we⁵₂,we³₂we4 23 −

11 − 24 a12we₂⁴we4

12 a12, b12,we⁶₂,we⁴₂we4

wherea12, b12are linearly independent elements in H¹²(Ge9,4)\C(12; 9,4)such that a12we⁴₂we46= 0,a12we⁵₂= 0,b12we⁶₂6= 0 andb12we⁴₂we4= 0.

Proof. As before, we begin with determining the generators of C(j; 10,4). For j≤6 we haveC(j; 10,4) as stated since there are no relations.

InC(7; 10,4) we havewe²₂we₃= 0 and thuswe₃we₄is the only generator.

InC(8; 10,4) we havewe₂⁴+we₂we²₃+we₂²we₄=we₄²as the only relation, hence there are the three generators.

InC(9; 10,4) we havewe³₂we₃= 0 and we₃³= 0, sowe₂we₃we₄is the only generator.

InC(10; 10,4) we havewe₂²we²₃= 0,we³₂we4=we²₃we4, and amongwe⁵₂,we³₂we4, we2we²₄ either is equal to the sum of the other two sinceg10∈J10,4. Thus there are two generators.

In C(11; 10,4) we havewe²₂we3we4,we⁴₂we3 both multiples ofwe²₂we3= 0. Then we2we³₃ is a multiple ofwe³₃= 0 and we3we²₄= 0 by Lemma 3.2.

In C(12; 10,4) we have we⁴₂we4 = we2we²₃we4 and we⁶₂+we⁴₂we4 = we²₂we₄² = we³₄ by Lemma 3.2. Thus there are two generators, e.g.we⁶₂ andwe⁴₂we4.

We know that charrank(γe10,4) = 11, so by Poincaré duality and Lemma 2.1 we have C(j; 10,4) = 0 forj≥13.

Alsoα_j(Ge_10,4) = 0 forj≤11, so by now we have determined all Betti numbers except forb12(Ge10,4). We calculate it from the Euler characteristicχ(Ge10,4) = 20.

The result isb12(Ge10,4) = 4.

Soα₁₂(Ge_10,4) = 2 and there are two linearly independent elements inH¹²(Ge_9,4)\

C(12; 9,4). By Poincaré duality we can start by arbitrarily picking a pair (a⁰₁₂, b⁰₁₂) such thata⁰₁₂we⁴₂we₄6= 0 and b⁰₁₂we₂⁶6= 0. Then we adjustb⁰₁₂based on the fact that H²⁴(Ge10,4)∼=Z2. Ifb⁰₁₂we⁴₂we46= 0, we defineb12=a⁰₁₂+b⁰₁₂, so thatb12we₂⁴we4= 0.

Otherwise let b12=b⁰₁₂

Similarly, since H²²(Ge10,4)∼=H²(Ge10,4)∼=Z2, ifa⁰₁₂we₂⁵ 6= 0, we definea12 = a⁰₁₂+b12, elsea12=a⁰₁₂, so thata12we₂⁵= 0. We havea12, b12,we₂⁶,we⁴₂we4as generators ofH¹²(Ge_10,4).

Next, we haveH¹³(Ge10,4) =H¹¹(Ge10,4) = 0.

Sincea12we2 andb12we2 have different products withwe₂³we4 they are distinct and therefore linearly independent.

We havea12we36= 0, sincea12we2we₃²we4=a12we⁴₂we46= 0.

By considering the products of nonzero elements a₁₂we²₂, a₁₂we₄, b₁₂we₂² with we⁴₂,we₂we²₃,we₂²we₄ we see that they are independent. Indeed, suppose that for some c₁, c₂, c₃ ∈Z2 we havec₁a₁₂we²₂+c₂a₁₂we₄+c₃b₁₂we²₂= 0. Multiplying both sides bywe₂we₃²and recalling thatwe²₂we₃= 0, we obtainc₂= 0. Then multiplying bywe⁴₂ yieldsc₃= 0 and multiplying byw²₂we₄ givesc₁= 0.

We havea₁₂we₂we₃6= 0, sincea₁₂we₂we²₃we₄=a₁₂we₂⁴we₄6= 0.

Elements a12we³₂, a12we2we4, b12we₂³ are nonzero. They prove to be independent upon considering their products withwe³₂,we₃²,we2we4.

We havea12we3we46= 0, sincea12we2we²₃we4=a12we₂⁴we46= 0.

Elementsa12we₂⁴, b12we⁴₂ are nonzero. They prove to be independent upon consi- dering their products with we₂²,we4.

We havea12we2we3we46= 0, sincea12we2we²₃we4=a12we⁴₂we46= 0.

We havea12we²₃we46= 0 and b12we₂⁵6= 0, therefore they are equal.