AIMS Mathematics, 5(6): 7562–7581. DOI:10.3934/math.2020484 Received: 25 June 2020 Accepted: 23 September 2020 http://www.aimspress.com/journal/Math Published: 24 September 2020

Research article On Reidemeister torsion of flag of compact semisimple Lie groups

Cenap Ozel¨ 1,∗, Habib Basbaydar2, Yasar Sozen¨ 3, Erol Yilmaz2, Jung Rye Lee4 and Choonkil Park5,∗

1 Department of Mathematics, King Abdulaziz University, 21589 Jeddah, Makkah, Saudi Arabia 2 Department of Mathematics, AIBU Izzet Baysal University, Bolu, Turkey 3 Department of Mathematics, Hacettepe University, Ankara, Turkey 4 Department of Mathematics, Daejin University, Kyunggi 11159, Korea 5 Research Institute for Natural Sciences, Hanyang University, Seoul 04763, Korea

* Correspondence: Email: [email protected], [email protected]; Tel: +8222200892; Fax: +8222810019.

Abstract: In this paper we calculate Reidemeister torsion of flag K/T of compact semi- simple Lie group K = SUn+1 using Reidemeister torsion formula and Schubert calculus, where T is maximal torus of K. We find that this number is 1. Also we explicitly calculate ring structure of integral cohomology algebra of flag manifold K/T of compact semi-simple Lie group K = SUn+1 using root data, and Groebner basis techniques.

Keywords: Reidemeister torsion; flag manifolds; Weyl groups; Schubert calculus; Groebner-Shirshov bases; graded inverse lexicographic order Mathematics Subject Classification: 57Q10, 16Z05, 14M15, 14N15, 22E67

1. Introduction

Reidemeister torsion is a topological invariant and was introduced by Reidemeister in 1935. Up to PL equivalence, he classified the lens spaces S3/Γ, where Γ is a finite cyclic group of fixed point free orthogonal transformations [20]. In [11], Franz extended the Reidemeister torsion and classified the higher dimensional lens spaces S2n+1/Γ, where Γ is a cyclic group acting freely and isometrically on the sphere S2n+1. In 1964, the results of Reidemeister and Franz were extended by de Rham to spaces of constant curvature +1 [10]. Kirby and Siebenmann proved the topological invariance of the Reidemeister torsion for manifolds in 1969 [14]. Chapman proved for arbitrary simplicial complexes [7,8]. Hence, 7563 the classification of lens spaces of Reidemeister and Franz was actually topological (i.e., up to homeomorphism). Using the Reidemeister torsion, Milnor disproved in 1961. He constructed two homeomorphic but combinatorially distinct finite simplicial complexes. He identified in 1962 the Reidemeister torsion with Alexander polynomial which plays an important role in knot theory and links [16, 18]. In [21], Sozen¨ explained the claim mentioned in [27, p. 187] about the relation between a symplectic chain complex with ω−compatible bases and the Reidemeister torsion of it. Moreover, he applied the main theorem to the chain-complex

∂2⊗id ∂1⊗id 0 → C2(Σg; Ad%) → C1(Σg; Ad%) → C0(Σg; Ad%) → 0, where Σg is a compact Riemann surface of genus g > 1, where ∂ is the usual boundary operator, and where % : π1(Σg) → PSL2(R) is a discrete and faithful representation of the π1(Σg) of Σg [21]. Now we will give his description of Reidemesister torsion and explain why it is not unique by a result of Milnor in [17]. ∂n Let Hp(C∗) = Zp(C∗)/Bp(C∗) denote the homologies of the chain complex (C∗, ∂∗) = (Cn → ∂1 Cn−1 → · · · → C1 → C0 → 0) of finite dimensional vector spaces over field C or R, where Bp = Im{∂p+1 : Cp+1 → Cp}, Zp = ker{∂p : Cp → Cp−1}, respectively.

Consider the short-exact sequences:

0 → Zp ,→ Cp  Bp−1 → 0 (1.1)

0 → Bp ,→ Zp  Hp → 0, (1.2) where (1.1) is a result of 1st-Isomorphism Theorem and (1.2) follows simply from the definition of Hp. Note that if bp is a basis for Bp, hp is a basis for Hp, and `p : Hp → Zp and sp : Bp−1 → Cp are sections, then we obtain a basis for Cp. Namely, bp ⊕ `p(hp) ⊕ sp(bp−1).

If, for p = 0, ··· , n, cp, bp, and hp are bases for Cp, Bp and Hp, respectively, then the alternating product n − (p+1) n n Y h i( 1) Tor(C∗, {cp}p=0, {hp}p=0) = bp ⊕ `p(hp) ⊕ sp(bp−1), cp (1.3) p=0 n n is called the Reidemeister torsion of the complex C∗ with respect to bases {cp} , {hp} , h i p=0 p=0 where bp ⊕ `p(hp) ⊕ sp(bp−1), cp denotes the determinant of the change-base matrix from cp to bp ⊕ `p(hp) ⊕ sp(bp−1).

Milnor [17] proved that torsion does not depend on neither the bases bp, nor the sections sp, `p. 0 0 Moreover, if cp, hp are other bases respectively for Cp and Hp, then there is the change-base-formula:

n 0 !(−1)p Y [cp, cp] Tor(C , {c0 }n , {h0 }n ) = · Tor(C , {c }n , {h }n ). (1.4) ∗ p p=0 p p=0 [h0 , h ] ∗ p p=0 p p=0 p=0 p p

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Let M be a smooth n−manifold, K be a cell-decomposition of M with for each p = 0, ··· , n, c { p ··· p } − C p = e1 , , emp , called the geometric basis for the p cells p(K; Z). Hence, we have the chain- complex associated to M

∂n ∂1 0 → Cn(K) → Cn−1(K) → · · · → C1(K) → C0(K) → 0, (1.5)

n n where ∂p denotes the boundary operator. Then Tor(C∗(K), {cp}p=0, {hp}p=0) is called the Reidemeister torsion of M, where hp is a basis for Hp(K).

In [23], oriented closed connected 2m−manifolds (m ≥ 1) are considered and he proved the following formula for computing the Reidemeister torsion of them. Namely,

Theorem 1.1. Let M be an oriented closed connected 2m−manifold (m ≥ 1). For p = 0,..., 2m, let hp be a basis of Hp(M). Then the Reidemeister torsion of M satisfies the following formula:

m−1 Y p q 2m (−1) (−1)m T(M, {hp}0 ) = det Hp,2m−p(M) det Hm,m(M) , p=0 where det Hp,2m−p(M) is the determinant of the matrix of the intersection pairing (·, ·)p,2m−p : Hp(M) × H2m−p(M) → R in bases hp, h2m−p. It is well known that Riemann surfaces and Grasmannians have many applications in a wide range of mathematics such as topology, differential geometry, algebraic geometry, symplectic geometry, and theoretical physics (see [2,3,5,6, 12, 13, 22, 24–26] and the references therein). They also applied Theorem 1.1 to Riemann surfaces and Grasmannians. In this work we calculate Reidemeister torsion of compact flag manifold K/T for K = SUn+1, where K is a compact simply connected semi-simple Lie group and T is maximal torus [28]. The content of the paper is as follows. In Section 2 we give all details of cup product formula in the cohomology ring of flag manifolds which is called Schubert calculus [15, 19]. In the last section we calculate the Reidemesiter torsion of flag manifold SUn+1/T for n ≥ 3. The results of this paper were obtained during M.Sc studies of Habib Basbaydar at Abant Izzet Baysal University and are also contained in his thesis [1].

2. Schubert calculus and cohomology of flag manifold

Now, we will give the important formula equivalent to the cup product formula in the cohomology of G/B where G is a Kac-Moody˘ group. The fundamental references for this section are [15, 19]. To do this we will give a relation between the complex nil Hecke ring and H∗(K/T, C). Also we introduce a multiplication formula and the actions of reflections and Berstein-Gelfand-Gelfand type BGG operators Ai on the basis elements in the nil Hecke ring. Proposition 2.1. u v X w w ξ · ξ = pu,vξ , u,v6w w where pu,v is a homogeneous polynomial of degree `(u) + `(v) − `(w).

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Proposition 2.2.  ξw if r w > w,  i w  − X ∨ w0 r ξ = − 1 riw w − i  (w αi)ξ + ξ αi(γ )ξ otherwise.  γ  0  riw−→w −1 Theorem 2.3. Let u, v ∈ W. We write w = ri1 ··· rin as a reduced expression. X pw = A ◦ · · · ◦ Aˆ ◦ · · · ◦ Aˆ ◦ · · · ◦ A (ξu)(e) u,v i1 i j1 i jm in j1<···< jm ··· −1 r j1 r jm =v where m = `(v) and the notation Aˆi means that the operator Ai is replaced by the Weyl group action ri.

+ + Let C0 = S/S be the S -module where S is the augmentation ideal of S . It is 1-dimensional as C-vector space. Since Λ is a S -module, we can define C0 ⊗S Λ. It is an algebra and the action of R on w w Λ gives an action of R on C0 ⊗S Λ. The elements σ = 1 ⊗ ξ ∈ C0 ⊗S Λ is a C-basis form of C0 ⊗S Λ.

Proposition 2.4. C0 ⊗S Λ is a graded algebra associated with the filtration of length of the element of the Weyl group W.

Proposition 2.5. The complex linear map f : C0 ⊗S Λ → Gr C{W} is a graded algebra homomorphism. Theorem 2.6. Let K be the standard real form of the group G associated to a symmetrizable Ka˘c- Moody Lie algebra g and let T denote the maximal torus of K. Then the map

∗ θ : H (K/T, C) → C0 ⊗S Λ defined by θ(εw) = σw for any w ∈ W is a graded algebra isomorphism. Moreover, the action of w ∈ W w ∗ and A on H (K/T, C) corresponds respectively to that δw and xw ∈ R on C0 ⊗S Λ. Corollary 2.7. The operators Ai on H∗(K/T, C) generate the nil-Hecke algebra. Corollary 2.8. We can use Proposition 2.1 and Theorem 2.3 to determine the cup product εuεv in terms w ∗ of the Schubert basis {ε }w∈W of H (K/T, Z).

3. The Reidemeister torsion of compact flag manifold K/T for K = SUn+1

This section includes our calculations about Reidemeister torsion of flag manifolds using Theorem 1.1 and Proposition 2.1 because χ(SUn+1/T) = |W| = n! is always an even number. We know that the Weyl group W of K acts on the Lie algebra of the maximal torus T. lt is a finite group of isometries of the Lie algebra t of the maximal torus T. lt preserves the coweight lattice T v. For ((π/2)(eα+e−α)) each simple root α, the Weyl group W contains an element rα of order two represented by e in N(T). Since the roots α can be considered as the linear functionals on the Lie algebra t of the maximal torus T, the action of rα on t is given by

rα(ξ) = ξ − α(ξ)hα for ξ ∈ t, where hα is the coroot in t corresponding to simple root α.Also, we can give the action of rα on the roots by

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∗ rα(β) = β − α(hβ)α for α, β ∈ t , ∗ where t is the dual vector space of t. The element rα is the reflection in the hyperplane Hα of t whose equation is α(ξ) = 0. These reflections rα generate the Weyl group W. Set α1, α2, . . . , αn be roots of Weyl Group of SUn+1. Since the Cartan Matrix of Weyl Group of SU is n+1   2 i = j  M =  −1 |i − j| = 1 , i j   0 otherwise   −αi, i = j  rα (α j) = αi + α j, |i − j| = 1 i   α j, otherwise.

Proposition 3.1. The Weyl group W of S Un+1 is isomorphic to Coxeter Group An given by generators s1, s2,..., sn and relations

2 (i) si = 1 i = 1, 2,..., n;

(ii) si si+1 si = si+1 si si+1 i = 1, 2,..., n − 1;

(iii) si s j = s j si 1 ≤ i < j − 1 < n. Proof. (i)

rαi ◦ rαi (β) = rαi (β− < αi, β > αi)

= β− < αi, β > αi− < β− < αi, β > αi, αi > αi

= β− < αi, β > αi− < β, αi > αi+ < αi, β >< αi, αi > αi

= β− < αi, β > αi− < αi, β > αi + 2 < αi, β > αi = β.

(ii)

rαi ◦ rαi+1 ◦ rαi (β) = rαi ◦ rαi+1 (β− < αi, β > αi)

= rαi (β− < αi, β > αi− < αi+1, β− < αi, β > αi > αi+1)

= rαi (β− < αi, β > αi− < αi+1, β > αi+1

+ < αi+1, < αi, β > αi > αi+1)

= rαi (β− < αi, β > αi− < αi+1, β > αi+1

+ < αi, β >< αi+1, αi > αi+1)

= rαi (β− < αi, β > αi− < αi+1, β > αi+1− < αi, β > αi+1)

= β− < αi, β > αi− < αi+1, β > αi+1− < αi, β > αi+1

− < αi, β− < αi, β > αi− < αi+1, β > αi+1

− < αi, β > αi+1 > αi

= β− < αi, β > αi− < αi+1, β > αi+1− < αi, β > αi+1

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− < αi, β > αi+ < αi, β >< αi, αi > αi

+ < αi+1, β >< αi+1, αi > αi+ < αi, β >< αi+1, αi > αi

= β− < αi, β > αi− < αi+1, β > αi+1− < αi, β > αi+1

− < αi, β > αi + 2 < αi, β > αi− < αi+1, β > αi

− < αi, β > αi

= β− < αi, β > αi− < αi+1, β > αi− < αi+1, β > αi+1

− < αi, β > αi+1

= β − (< αi, β > + < αi+1, β >)(αi + αi+1).

rαi+1 ◦ rαi ◦ rαi+1 (β) = rαi+1 ◦ rαi (β− < αi+1, β > αi+1)

= rαi+1 (β− < αi+1, β > αi+1− < αi, β− < αi+1, β > αi+1 > αi)

= rαi+1 (β− < αi+1, β > αi+1− < αi, β > αi

+ < αi+1, β >< αi, αi+1 > αi)

= rαi+1 (β− < αi+1, β > αi+1− < αi, β > αi− < αi+1, β > αi)

= β− < αi+1, β > αi+1− < αi, β > αi− < αi+1, β > αi

− < αi+1, β− < αi+1, β > αi+1− < αi, β > αi

− < αi+1, β > αi > αi+1

= β− < αi+1, β > αi+1− < αi+1, β > αi− < αi, β > αi

− < αi+1, β > αi+1+ < αi, β >< αi+1, αi > αi+1

+ < αi+1, β >< αi+1, αi > αi+1

+ < αi+1, β >< αi+1, αi+1 > αi+1

= β− < αi+1, β > αi+1− < αi, β > αi− < αi+1, β > αi

− < αi+1, β > αi+1 + 2 < αi+1, β > αi+1− < αi, β > αi+1

− < αi+1, β > αi+1

= β− < αi+1, β > αi+1− < αi, β > αi− < αi+1, β > αi

− < αi, β > αi+1

= β − (< αi+1, β > + < αi, β >)(αi+1 + αi).

Hence rαi+1 ◦ rαi ◦ rαi+1 (β) = rαi+1 ◦ rαi ◦ rαi+1 (β). (iii)

rαi ◦ rα j (β) = rαi ◦ (β− < α j, β > α j)

= β− < α j, β > α j− < αi, β− < α j, β > α j > αi

= β− < α j, β > α j− < αi, β > αi+ < α j, β >< αi, α j > αi

= β− < α j, β > α j− < αi, β > αi.

rα j ◦ rαi (β) = rα j ◦ (β− < αi, β > αi)

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= β− < αi, β > αi− < α j, β− < αi, β > αi > α j

= β− < αi, β > αi− < α j, β > α j+ < αi, β >< α j, αi > α j

= β− < αi, β > αi− < α j, β > α j.

Hence rαi ◦ rα j (β) = rα j ◦ rαi (β). 

After this point si will represent rαi . Let us define the word  s s ··· s i < j  i i+1 j s =  s i = j i, j  i  1 i > j.

Theorem 3.2. [4, Theorem 3.1] The reduced Gr¨obner-Shirshov basis of the coxeter group An consists of relation

si, j si = si+ j si, j 1 ≤ i < j ≤ n together with defining relations of An. The following lemma is equivalent of [4, Lemma 3.2]. The only difference is the order of generators s1 > s2 > . . . sn in our setting.

Lemma 3.3. Using elimination of leading words of relations, the reduced elements of An are in the form

sn+1, jn+1 sn, jn sn−1, jn−1 ··· si, ji ··· s1, j1 1 ≤ i ≤ ji + 1 ≤ n + 1. Notice that jn+1 + 1 = n + 1 =⇒ jn+1 = n and sn+1,n = 1.

Algorithm 3.1. (Finding Inverse) Let w = sn, jn sn−1, jn−1 ··· s1, j1 . The inverse of w can be found using following algorithm. Invw = {}; Conw = Reverse(w); For k = 1 to k = n Find maximum sequence in Conw;

list = {sk, sk+1, sk+2,..., sk+ j}; Invw = list ∪ Invw; End For.

Example 3.4. Let s4,6 s3,5 s2,5 s1,3. The inverse of its is S 3 s2 s1 s5 s4 s3 s2 s5 s4 s3 s6 s5 s4. Invw = s1,4 S 3 s2 s5 s4 s3 s5 s4 s6 s5 Invw = s2,5 s1,4 S 3 s5 s4 s5 s6 Invw = s3,5 s2,5 s1,4 s5 s6 Invw = s5,6 s3,5 s2,5 s1,4.

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Lemma 3.5. Let w = (sn, jn )(sn−1, jn−1 ) ··· (si+1, ji+1 )(si, ji ) ··· (s1, j1 ) and

siw = (sn, j )(sn− , j ) ··· (si , j )(si, j ) ··· (s , j ), where  n 1 n−1 +1 i+1 i 1 1  ji+1 = ji + 1, ji = ji+1 i f ji < ji+1  siw =  ji+1 = ji, ji = ji+1 − 1 i f ji ≥ ji+1   jk = jk i f k , i, i + 1 Here if i = n, then we assume jn+1 = n.

Corollary 3.6. Let w = (sn, jn )(sn−1, jn−1 ) ··· (si+1, ji+1 )(si, ji ) ··· (s1, j1 ) and ··· ··· si−1(siw) = (sn, jˆn )(sn−1, jd ) (si+1,dj )(si,bj ) (s1,bj ), where  n−1 i+1 i 1  dji+1 = ji + 1, bji = ji−1 + 1, dji−1 = ji+1 i f ji < ji+1, ji−1 < ji+1   dji+1 = ji + 1, bji = ji−1, dji−1 = ji+1 − 1 i f ji < ji+1, ji−1 ≥ ji+1  s − (s w) = dj = j , bj = j + 1, dj = j − 1 i f j ≥ j , j < j − 1 i 1 i  i+1 i i i−1 i−1 i+1 i i+1 i−1 i+1   dji+1 = ji, bji = ji−1, dji−1 = ji+1 − 2 i f ji ≥ ji+1, ji−1 ≥ ji+1 − 1   bjk = jk i f k , i − 1, i, i + 1.

Proof. Let w = siw = (sn, j )(sn−1, j ) ··· (si+1, j )(si, j ) ··· (s1, j ). Then  n n−1 i+1 i 1  bji = ji−1 + 1, dji−1 = ji i f ji−1 < ji  si−1(w) =  bji = ji−1, dji−1 = ji − 1 i f ji−1 ≥ ji   bjk = jk i f k , i − 1, i.

(i) ji < ji+1 ⇒ ji+1 = ji + 1, ji = ji+1 So ji−1 < ji ⇒ ji−1 < ji+1, dji+1 = ji+1 = ji + 1, bji = ji−1 + 1 = ji−1 + 1, dji−1 = ji = ji+1.

(ii) ji < ji+1 ⇒ ji+1 = ji + 1, ji = ji+1 So ji−1 ≥ ji ⇒ ji−1 ≥ ji+1, dji+1 = ji+1 = ji + 1, bji = ji−1 = ji−1, dji−1 = ji − 1 = ji+1 − 1.

(iii) ji ≥ ji+1 ⇒ ji+1 = ji , ji = ji+1 − 1 So ji−1 < ji ⇒ ji−1 < ji+1, dji+1 = ji+1 = ji + 1, bji = ji−1 = ji−1, dji−1 = ji − 1 = ji+1 − 1.

(iv) ji ≥ ji+1 ⇒ ji+1 = ji, ji = ji+1 − 1 So ji−1 ≥ ji ⇒ ji−1 ≥ ji+1 − 1, dji+1 = ji+1 = ji , bji = ji−1 = ji−1, dji−1 = ji − 1 = ji+1 − 2.



Corollary 3.7. Let w = (sn, jn )(sn−1, jn−1 ) ··· (si+1, ji+1 )(si, ji ) ··· (s1, j1 ) and ··· ··· si+1(siw) = (sn, jˆn )(sn−1, jd ) (si+1,dj )(si,bj ) (s1,bj ). Then  n−1 i+1 i 1  dj = j + 2, dj = j , bj = j i f j < j , j < j  i+2 i i+1 i+2 i i+1 i i+1 i+1 i+2  dj = j + 1, dj = j − 1, bj = j i f j < j , j + 1 ≥ j  i+2 i i+1 i+2 i i+1 i i+1 i i+2 s (s w) =  dj = j + 1, dj = j , bj = j − 1 i f j ≥ j , j < j i+1 i  i+2 i i+1 i+2 i i+1 i i+1 i i+2   dji+2 = ji, dji+1 = ji+2 − 1, bji = ji+1 − 1 i f ji ≥ ji+1, ji ≥ ji+2   bjk = jk i f k , i, i + 1, i + 2.

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Proof. Let w = siw = (sn, j )(sn−1, j ) ··· (si+1, j )(si, j ) ··· (s1, j ). Then  n n−1 i+1 i 1  dji+2 = ji+1 + 1, dji+1 = ji+2 i f ji+1 < ji+2  si+1(w) =  dji+2 = ji+1, dji+1 = ji+2 − 1 i f ji+1 ≥ ji+2   bjk = jk i f k , i + 1, i + 2.

(i) ji < ji+1 ⇒ ji+1 = ji + 1, ji = ji+1 So ji+1 < ji+2 ⇒ ji + 1 < ji+2, dji+2 = ji+1 + 1 = ji + 2 , dji+1 = ji+2 = ji+2, bji = ji = ji+1.

(ii) ji < ji+1 ⇒ ji+1 = ji + 1, ji = ji+1 So ji+1 ≥ ji+2 ⇒ ji + 1 ≥ ji+2 , dji+2 = ji+1 = ji + 1, dji+1 = ji+2 − 1 = ji+2 − 1, bji = ji = ji+1.

(iii) ji ≥ ji+1 ⇒ ji+1 = ji, ji = ji+1 − 1 So ji+1 < ji+2 ⇒ ji < ji+2 , dji+2 = ji+1 + 1 = ji + 1, dji+1 = ji+2 = ji+2, bji = ji − 1 = ji+1 − 1.

(iv) ji ≥ ji+1 ⇒ ji+1 = ji, ji = ji+1 − 1 So ji+1 ≥ ji+2 ⇒ ji ≥ ji+2, dji+2 = ji+1 = ji, dji+1 = ji+2 − 1 = ji+2 − 1, bji = ji = ji+1 − 1.



i Using Lemma3 .3 and definitions of A and ri operators, we can obtain the followings.

Lemma 3.8. Let w = (s )(s − ) ··· (s )(s ) ··· (s ). Then ( n, jn n 1, jn−1 i+1, ji+1 i, ji 1, j1 εw1 i f j ≥ j Ai(εw) = i i+1 0 i f ji < ji+1, where w = (s )(s ) ··· (s )(s ) ··· (s ) with j = j , j = j − 1 and j = j 1 n, jn n−1, jn−1 i+1, ji+1 i, ji 1, j1 i+1 i i i+1 k k if k , i, i + 1. ( εsi−1 − εsi − εsi+1 i f i = j s j Lemma 3.9. ri(ε ) = εs j i f i , j.

The integral cohomology of SUn+1/T is generated by Schubert classes indexed

W = {sn jn sn−1, jn−1 ... s1 j1 : ji = 0 or i ≤ ji ≤ n}.

ri 2 Let xi = ε ∈ H (SUn+1/T, Z). We define an order between generators of the integral cohomology of sn j sn−1, j ...si j ...s1 j SUn+1/T. Since each element ε n n−1 i 1 can be represented by an n-tuple ( jn − n + 1, jn−1 − (n − 1) + 1,..., ji − i + 1,..., j1 − 1 + 1), we can define an order between n-tuples.

Definition 3.10. (Graded Inverse Lexicographic Order) Let α = (α1, α2, . . . , αn) and β = n (β1, β2, . . . , βn) ∈ Z≥0. We say α > β if |α| = α1 + α2 + . . . αn > |β| = β1 + β2 + . . . βn or |α| = |β| and in the vector difference α − β ∈ Zn, the right-most nonzero entry is positive. We will write sn j sn−1, j ...si j ...s1 j snk sn−1, j ...sik ...s1 j ε n n−1 i 1 > ε n k−1 i 1 if ( jn − n + 1, jn−1 − (n − 1) + 1,..., ji − i − 1,..., j1 − 1 + 1) > (kn − n + 1, kn−1 − (n − 1) + 1,..., ki − i − 1,..., k1 − 1 + 1).

Example 3.11. εs35 s23 s14 > εs35 s24 s13 since (3, 2, 4) > (3, 3, 3) in graded inverse lexicographic order.

∗ We will try to find a quotient ring Z[x1, x2,..., xn]/I which is isomorphic to H (SUn+1/T, Z). We also define an order between monomials as follows.

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α1 α2 αn β1 β2 βn Definition 3.12. We say x1 x2 ··· xn > x1 x2 ··· xn if |α| = α1 + α2 + ··· + αn > |β| = β1 + β2 + ··· + βn or |α| = |β| and in the vector difference α − β ∈ Zn the left-most non-zero entry is negative.

4 2 3 3 3 3 Example 3.13. x1 x2 x3 < x1 x2 x3, since (4, 2, 3) − (3, 3, 3) = (1, −1, 0). α α α 1 2 n snαn sn−1,αn−1 ...siαi ...s1α1 Lemma 3.14. x1 x2 ... xn = ε + lower terms.

si Proof. To prove this, we use induction on degree of the monomials. By definition xi = ε . Let us si s j compute xi x j = ε ε . Here we may assume that i ≤ j. If j − i > 1, the inverse of si s j is si s j. Hence

s j si i si Psi s j = r jA (ε ) = r j(1) = 1 in the cup product. If j = i + 1, the inverse of si+1 si is si si+1. In this case

i si i si {} Psi,si+1 = A ri+1(ε ) = A (ε ) = ε = 1.

If i = j, then we have to consider the word si,i+1. Its inverse si+1 si and

si,i+1 i si Psi si = ri+1A (ε ) = ri+1(1) = 1.

sk sl sk sl s j si sk sl Now we have to show that Psi s j = 0 if ε > ε . By definition of cup product the coefficient of ε is not zero only if si → sk sl and s j → sk sl. However, this is possible only if sk sl = s j si or sk sl = si,i+1 when j = i + 1. Clearly εsi si+1 < εsi+1 si . Hence εsi εsi+1 = εsi+1 si + lower terms and εsi εs j = εs j εsi if j − i > 1. In the case i = j, we have to look elements si sk and sk si. The inverse of sk si is equal to sk si itself if k − i > 1, hence sk si k si k si−1 − si si+1 Psi s j = A ri(ε ) = A (ε ε + ε ) = 0 since k − i > 1. Clearly εsi sk < εsi si+1 if k < i. Hence εsi εsi = εsi si+1 + lower terms. α α α 1 2 n snαn sn−1,αn−1 ...siαi ...s1α1 Assume that x1 x2 ... xn = ε + lower terms. α α α +1 α 1 2 i n snαn sn−1,αn−1 ...siαi+1...s1α1 We have to show x1 x2 ... xi ... xn = ε + lower terms by Bruhat ordering. 0 0 snαn sn−1,αn−1 ... siαi+1 ... s1α1 → w only if w = snαn sn−1,αn−1 ... siαi ... s1α1 where there exists an index j for which α j = α j + 1 and αk = αk if k , j. By given ordering

0 w = snαn sn−1,αn−1 ... siαi ... s1α1 > snαn sn−1,αn−1 ... siαi+1 ... s1α1 . 0−1 If j > i, then, by Algorithm3 .1, in w , we will not have a subsequence s j−1, s j−2 ... si after the j s j elements s j. Hence in the cup product before applying A we will not have the term ε . It means w0 Psi,w = 0. 0−1 If j = i, then, again by Algorithm3 .1, in w we will not have a subsequence s j−1, s j−2 ... si after j s j the elements s j. Hence in the cup product before applying A we will not have the term ε . It means w0 Psi,w = 1 if and only if j > i.  Example 3.15. Let l = 3, s3 s2 s1 x1 x2 x3 = ε + lower terms. 2 s3 s2 s12 x1 x2 x3 = ε + lower terms. s3 s23 s1 s3 s2 s12 s23 s12 s3 s13 s2 s13 Then we have ε > ε > ε > ε > ε . Since the inverse of s3 s23 s1 is s3 s13 and the s1 s1 s1 s2 inverse of s3 s2 s1 is s13,A3r1r2r3(ε ) = A3r1(ε ) = A3(−ε + ε ) = 0. s1 s1 s1 s2 Similarly, since the inverse of s3 s2 s12 is s2 s13,A2r1r2r3(ε ) = A2r1(ε ) = A2(−ε + ε ) = 1.

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Before finding the quotient ring Z[x1,..., xn]/I, we give some information about ring k[x1,..., xn]/I where k is a field. Fix a monomial ordering on k[x1,..., xn]. Let f ∈ k[x1,..., xn]. The leading monomial of f , denoted by LM( f ), is the highest degree monomial of f . The coefficient of LM( f ) is called leading coefficient of f and denoted by LC( f ). The leading term of f , LT( f ) = LC( f )LM( f ).

Let I ⊆ k[x1,..., xn] be an ideal. Define LT(I) = {LT( f ): f ∈ I}. Let < LT(I) > be an ideal generated by LT(I). Proposition 3.16. [9, Section 5.3, Propostions 1 and 4]

(i) Every f ∈ k[x1,..., xn] is congruent modulo I to a unique polynomial r which is a k-linear combination of the monomials in the complement of < LT(I) >.

(ii) The elements of {xα : xα << LT(I) >} are linearly independent modulo I.

(iii) k[x1,..., xn]/I is isomorphic as a k − vector space to

S = Span{xα : xα << LT(I) >}.

Theorem 3.17. [9, Section 5.3, Theorem 6] Let I ⊆ k[x1,..., xn] be an ideal.

(i) The k-vector space k[x1,..., xn]/I is finite dimensional.

mi (ii) For each i, 1 ≤ i ≤ n, there is a polynomial fi ∈ I such that LM( fi) = xi for some positive integer mi.

∗ Theorem 3.18. H (SUn+1/T, Z) isomorphic to Z[x1, x2,..., xn]/ < f1, f2,..., fn > where LT( fi) = n−i+2 xi with respect to monomial order given by Definition 3.12. ∗ Proof. Let I be the ideal such that H (SUn+1/T, R)  R[α1, α2, . . . , αn]/I. Since we found one to ∗ α1 α2 αn one correspondence between length l elements of H (SUn+1/T, Z) and monomials x1 x2 ··· xn , where α1 + α2 + ··· αn = l and for each i, 1 ≤ i ≤ n, αi ≤ n − i + 1, there should be a polynomial fi ∈ I such n−i+2 that LT( fi) = xi .  Example 3.19. Let n = 3. Then we have

αi ≤ n − i + 1, i = 1, 2, 3; α1 ≤ 3, α2 ≤ 2, α3 ≤ 1. For l = 1; x1, x2, x3; and 2 2 2 for l = 2; x1, x1 x2, x1 x3, x2 x3, x2. So we must have a polynomial f3 with LM( f3) = x3. 3 2 2 2 2 For l = 3; x1, x1 x2, x1 x3, x1 x2 x3, x1 x2, x2 x3, so 3 we must have a polynomial f2 with LM( f2) = x2. 3 3 2 2 2 2 For l = 4; x1 x2, x1 x3, x1 x2 x3, x1 x2, x1 x2 x3, so 4 we must have a polynomial f1 with LM( f1) = x1.

The complex of SUn+1/T is equal to (n + 1)n/2. So the highest element has length of (n + 1)n/2. n(n+1) Since the unique highest element has length of 2 , we now give the result about the multiplication n(n+1) − of elements of length k and of length 2 k.

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s s ···s s s ···s Theorem 3.20. Let A = ε n jn n−1, jn−1 1 j1 be an element of length k and B = ε npn n−1,pn−1 1p1 be an element n(n+1) − of length 2 k. The corresponding polynomials in Z[x1, x2,..., xn]/ < f1, f2,..., fn > has leading monomials x j1−1+1 x j2−2+1 ··· x ji−i+1 ··· x jn−n+1 and xp1−1+1 xp1−2+1 ··· xpn−n+1, respectively. Then 1 2 ( i 1 1 2 1 εsn,n sn−1,n,...,sin,...,s1n , i f j + p + 1 = n + i; A · B = i i 0, i f ji + pi + 1 , n + i.

Proof. The unique highest degree monomial in Z[x1, x2,..., xn]/ < f1, f2,..., fn > is n n−1 n−i+1 x1 x2 ··· xi ··· xn. The multiplication of leading monomials of corresponding monomials of A and B produce the monomial

j1+p1 j2+p2−2 ji+pi−2i+2 jn+pn−2n+2 x1 x2 ··· xi ··· xn .

If ji + pi − 2i + 2 = n − i + 1 → ji + pi + 1 = n + i for each i, i ≤ 1 ≤ n, then the multiplication gives n n−1 sn,n sn−1,n···sin···s1n sn,n sn−1,n···s1n the x1 x2 ... xn. Since this monomial correspondence the element ε , A · B = ε . If ji + pi + 1 , n + i, then the leading monomial and the monomials of lower degree must reduce to zero modulo < f1, f2,..., fn > in k[x1, x2,..., xn] when we apply the division algorithm. Hence A · B = 0. 

Now we can give the whole computation of the quotient ring Z[x1, x2, x3]/ < f1, f2, f3 >.

s1 s2 s3 Example 3.21. Let x1 = ε , x2 = ε , x3 = ε . For l = 2, we have

s3 s2 s2 s3 x2 x3 = ε + ε

2 s2 s3 s2 s1 x2 = ε + ε s3 s1 x1 x3 = ε

s2 s1 s1 s2 x1 x2 = ε + ε

2 s1 s2 x1 = ε , and         x x εs3 s2 εs3 s2 x x  2 3       2 3   2   s2 s3   s2 s3   2   x2   ε   ε   x2     s s   s s  −1    x x  = M  ε 3 1  and  ε 3 1  = M  x x , where  1 3       1 3   x x   εs2 s1   εs2 s1   x x   1 2       1 2   2   s1 s2   s1 s2   2  x1 ε ε x1  1 1 0 0 0   1 −1 0 1 −1       0 1 0 1 0   0 1 0 −1 1      M =  0 0 1 0 0  M−1 =  0 0 1 0 0  . Then we have      0 0 0 1 1   0 0 0 1 −1       0 0 0 0 1   0 0 0 0 1 

s3 s2 2 2 ε = x2 x3 − x2 + x1 x2 − x1 s2 s3 2 2 ε = x2 − x1 x2 + x2

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s3 s1 ε = x1 x3

s2 s1 2 ε = x1 x2 − x1 s1 s2 2 ε = x1.

2 Here we must have a relation involving x3 and we have it as

2 s3 s2 2 2 x3 = ε = x2 x3 − x2 + x1 x2 − x1. For l = 3;

2 s3 s2 s3 s3 s2 s1 s2 s3 s1 x2 x3 = ε + ε + ε s3 s2 s1 s2 s3 s1 s3 s1 s2 s1 s2 s3 x1 x2 x3 = ε + ε + ε + ε

2 s2 s3 s1 s2 s1 s2 s1 s2 s3 x1 x2 = ε + ε + ε 2 s3 s1 s2 s1 s2 s3 x1 x3 = ε + ε 2 s2 s1 s2 s1 s2 s3 x1 x2 = ε + ε 3 s1 s2 s3 x1 = ε and  x2 x   εs3 s2 s3   εs3 s2 s3   x2 x   2 3       2 3     s3 s2 s1   s3 s2 s1     x1 x2 x3   ε   ε   x1 x2 x3   2   s2 s3 s1   s2 s3 s1   2   x1 x   ε   ε   x1 x   2  = M   and   = M−1  2 , where  x2 x   εs3 s1 s2   εs3 s1 s2   x2 x   1 3       1 3   x2 x   εs2 s1 s2   εs2 s1 s2   x2 x   1 2       1 2   3   s1 s2 s3   s1 s2 s3   3  x1 ε ε x1  1 1 1 0 0 0   1 −1 0 1 0 0       0 1 1 1 0 1   0 1 −1 −1 1 0       0 0 1 0 1 1   0 0 1 0 −1 0  M =   M−1 =   .  0 0 0 1 0 1   0 0 0 1 0 −1       0 0 0 0 1 1   0 0 0 0 1 −1       0 0 0 0 0 1   0 0 0 0 0 1  Then we have

s3 s2 s3 2 2 ε = x2 x3 − x1 x2 x3 + x1 x3 s3 s2 s1 2 2 2 ε = x1 x2 x3 − x1 x2 − x1 x3 + x1 x2 s2 s3 s1 2 2 ε = x1 x2 − x1 x2 s3 s1 s2 2 3 ε = x1 x3 − x1 s2 s1 s2 2 3 ε = x1 x2 − x1 s1 s2 s3 3 ε = x1.

3 Here we must have a relation involving x2 and we now we have it as

3 s2 s3 s1 2 2 x2 = 2ε = 2(x1 x2 − x1 x2).

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For l = 4; we have

2 s3 s2 s3 s1 s3 s2 s1 s2 s2 s3 s1 s2 s3 s1 s2 s3 x1 x2 x3 = ε + ε + 2ε + 2ε 2 s3 s2 s1 s2 s2 s3 s1 s2 s3 s1 s2 s3 s2 s1 s2 s3 x1 x2 x3 = ε + ε + ε + ε 2 2 s2 s3 s1 s2 s2 s1 s2 s3 x1 x2 = ε + ε 3 s3 s1 s2 s3 x1 x3 = ε 3 s2 s1 s2 s3 x1 x2 = ε and         x x2 x εs3 s2 s3 s1 εs3 s2 s3 s1 x x2 x  1 2 3       1 2 3   2   s3 s2 s1 s2   s3 s2 s1 s2   2   x1 x2 x3   ε   ε   x1 x2 x3   2 2   s s s s   s s s s  −1  2 2   x x  = M  ε 2 3 1 2  and  ε 2 3 1 2  = M  x x , where  1 2       1 2   x3 x   εs3 s1 s2 s3   εs3 s1 s2 s3   x3 x   1 3       1 3   3   s2 s1 s2 s3   s2 s1 s2 s3   3  x1 x2 ε ε x1 x2  1 1 2 2 0   1 −1 −1 −1 2       0 1 1 1 1   0 1 −1 −1 0      M =  0 0 1 0 1  M−1 =  0 0 1 0 −1  .      0 0 0 1 0   0 0 0 1 0       0 0 0 0 1   0 0 0 0 1  Then

s3 s2 s3 s1 2 2 2 2 3 3 ε = x1 x2 x3 − x1 x2 x3 − x1 x2 − x1 x3 + 2x1 x2 s3 s2 s1 s2 2 2 2 3 ε = x1 x2 x3 − x1 x2 − x1 x3 s2 s3 s1 s2 2 2 3 ε = x1 x2 − x1 x2 s3 s1 s2 s3 3 ε = x1 x3 s2 s1 s2 s3 3 ε = x1 x2.

4 3 s1 s1 s2 s3 We must have a relation involving x1, which is x1 x1 = ε .ε = 0. For l = 5;

2 2 s3 s2 s3 s1 s2 s3 s2 s1 s2 s3 s2 s3 s1 s2 s3 x1 x2 x3 = ε + ε + ε 3 s3 s2 s1 s2 s3 s2 s3 s1 s2 s3 x1 x2 x3 = ε + ε 3 2 s2 s3 s1 s2 s3 x1 x2 = ε and         x2 x2 x εs3 s2 s3 s1 s2 εs3 s2 s3 s1 s2 x2 x2 x  1 2 3       1 2 3   x3 x x  = M  εs3 s2 s1 s2 s3  and  εs3 s2 s1 s2 s3  = M−1  x3 x x , where  1 2 3       1 2 3   3 2   s2 s3 s1 s2 s3   s2 s3 s1 s2 s3   3 2  x1 x2 ε ε x1 x2  1 1 1   1 −1 0      M =  0 1 1  M−1 =  0 1 −1  . So      0 0 1   0 0 1 

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s3 s2 s3 s1 s2 2 2 3 ε = x1 x2 x3 − x1 x2 x3 s3 s2 s1 s2 s3 3 3 2 ε = x1 x2 x3 − x1 x2 s2 s3 s1 s2 s3 3 2 ε = x1 x2.

Hence we don’t have any relation. For l = 6;

3 2 s3 s2 s3 s1 s2 s3 s3 s2 s3 s1 s2 s3 3 2 x1 x2 x3 = ε and ε = x1 x2 x3. Now let us multiple elements with lengths of k and 6 − k.

First M0 = 1 and | det(M0)| = 1.

Degree 1 ∗ Degree 5

Elements Leading Monomial in Polynomial Ring s1 ε x1 s2 ε x2 s3 ε x3 s3 s23 s12 2 2 ε x1 x2 x3 s3 s2 s13 3 ε x1 x2 x3 s23 s13 3 2 ε x1 x2

s3 s3 s23 s12 2 2 2 ε * ε x1 x2 x3 0 s3 s3 s2 s13 3 2 ε * ε x1 x2 x3 0 s3 s23 s13 3 2 ε * ε x1 x2 x3 1

s2 s3 s23 s12 2 3 ε * ε x1 x2 x3 0 s2 s3 s2 s13 3 2 ε * ε x1 x2 x3 1 s2 s23 s13 3 3 ε * ε x1 x2 0

s1 s3 s23 s12 3 2 ε * ε x1 x2 x3 1 s1 s3 s2 s13 4 ε * ε x1 x2 x3 0 s1 s23 s13 4 2 ε * ε x1 x2 0

Now we will calculate Reidemeister torsion of S U4/T by using above multiplication. From 0 0 1   multiplication of the second cohomology, we have M = 0 1 0 and | det(M )| = 1. 2   2 1 0 0

Degree 2 ∗ Degree 4

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Elements Leading Monomial in Polynomial Ring s3 s2 ε x2 x3 s23 2 ε x2 s3 s1 ε x1 x3 s2 s1 ε x1 x2 s12 2 ε x1 s3 s23 s1 2 ε x1 x2 x3 s3 s2 s12 2 ε x1 x2 x3 s23 s12 2 2 ε x1 x2 s3 s13 3 ε x1 x3 s2 s13 3 ε x1 x2 s3 s2 s3 s23 s1 3 2 ε * ε x1 x2 x3 0 s3 s2 s3 s2 s12 2 2 2 ε * ε x1 x2 x3 0 s3 s2 s23 s12 2 3 ε * ε x1 x2 x3 0 s3 s2 s3 s13 3 2 ε * ε x1 x2 x3 0 s3 s2 s2 s13 3 2 ε * ε x1 x2 x3 1 s23 s3 s23 s1 4 ε * ε x1 x2 x3 0 s23 s3 s2 s12 2 3 ε * ε x1 x2 x3 0 s23 s23 s12 2 4 ε * ε x1 x2 0 s23 s3 s13 3 2 ε * ε x1 x2 x3 1 s23 s2 s13 3 3 ε * ε x1 x2 0 s3 s1 s3 s23 s1 2 2 2 ε * ε x1 x2 x3 0 s3 s1 s3 s2 s12 3 2 ε * ε x1 x2 x3 0 s3 s1 s23 s12 3 2 ε * ε x1 x2 x3 1 s3 s1 s3 s13 4 2 ε * ε x1 x3 0 s3 s1 s2 s13 4 ε * ε x1 x2 x3 0

s2 s1 s3 s23 s1 2 3 ε * ε x1 x2 x3 0 s2 s1 s3 s2 s12 3 2 ε * ε x1 x2 x3 1 s2 s1 s23 s12 3 3 ε * ε x1 x2 0 s2 s1 s3 s13 4 ε * ε x1 x2 x3 0 s2 s1 s2 s13 4 2 ε * ε x1 x2 0

s12 s3 s23 s1 3 2 ε * ε x1 x2 x3 1 s12 s3 s2 s12 4 ε * ε x1 x2 x3 0 s12 s23 s12 4 2 ε * ε x1 x2 0 s12 s3 s13 5 ε * ε x1 x3 0 s12 s2 s13 5 ε * ε x1 x2 0 To calculate Reidemeister torsion of S U4/T we need multiplication of fourth cohomology bases 0 0 0 0 1   0 0 0 1 0   elements and then we have M = 0 0 1 0 0 and | det(M )| = 1. 4   4 0 1 0 0 0   1 0 0 0 0

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Degree 3 ∗ Degree 3

Elements Leading Monomial in Polynomial Ring s3 s23 2 ε x2 x3 s3 s2 s1 2 ε x2 x3 s23 s1 2 ε x1 x2 s3 s12 2 ε x1 x3 s2 s12 2 ε x1 x2 s13 3 ε x1 s3 s23 2 ε x2 x3 s3 s2 s1 ε x1 x2 x3 s23 s1 2 ε x1 x2 s3 s12 2 ε x1 x3 s2 s12 2 ε x1 x2 s13 3 ε x1

s3 s23 s3 s23 4 2 ε * ε x2 x3 0 s3 s23 s3 s2 s1 3 2 ε * ε x1 x2 x3 0 s3 s23 s23 s1 4 ε * ε x1 x2 x3 0 s3 s23 s3 s12 2 2 2 ε * ε x1 x2 x3 0 s3 s23 s2 s12 2 3 ε * ε x1 x2 x3 0 s3 s23 s13 3 2 ε * ε x1 x2 x3 1

s3 s2 s1 s3 s23 3 2 ε * ε x1 x2 x3 0 s3 s2 s1 s3 s2 s1 2 2 2 ε * ε x1 x2 x3 0 s3 s2 s1 s23 s1 2 3 ε * ε x1 x2 x3 0 s3 s2 s1 s3 s12 3 2 ε * ε x1 x2 x3 0 s3 s2 s1 s2 s12 3 2 ε * ε x1 x2 x3 1 s3 s2 s1 s13 4 ε * ε x1 x2 x3 0

s23 s1 s3 s23 4 ε * ε x1 x2 x3 0 s23 s1 s3 s2 s1 2 3 ε * ε x1 x2 x3 0 s23 s1 s23 s1 2 4 ε * ε x1 x2 0 s23 s1 s3 s12 3 2 ε * ε x1 x2 x3 1 s23 s1 s2 s12 3 3 ε * ε x1 x2 0 s23 s1 s13 4 2 ε * ε x1 x2 0

s3 s12 s3 s23 2 2 2 ε * ε x1 x2 x3 0 s3 s12 s3 s2 s1 3 2 ε * ε x1 x2 x3 0 s3 s12 s23 s1 3 2 ε * ε x1 x2 x3 1 s3 s12 s3 s12 4 2 ε * ε x1 x3 0 s3 s12 s2 s12 4 ε * ε x1 x2 x3 0 s3 s12 s13 5 ε * ε x1 x3 0

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s2 s12 s3 s23 2 3 ε * ε x1 x2 x3 0 s2 s12 s3 s2 s1 3 2 ε * ε x1 x2 x3 1 s2 s12 s23 s1 3 3 ε * ε x1 x2 0 s2 s12 s3 s12 4 ε * ε x1 x2 x3 0 s2 s12 s2 s12 4 2 ε * ε x1 x2 0 s2 s12 s13 5 ε * ε x1 x2 0 s13 s3 s23 3 2 ε * ε x1 x2 x3 1 s13 s3 s2 s1 4 ε * ε x1 x2 x3 0 s13 s23 s1 4 2 ε * ε x1 x2 0 s13 s3 s12 5 ε * ε x1 x3 0 s13 s2 s12 5 ε * ε x1 x2 0 s13 s13 6 ε * ε x1 0

To calculate Reidemeister torsion of SU4/T we need multiplication of sixth cohomology bases 0 0 0 0 0 1   0 0 0 0 1 0   0 0 0 1 0 0 elements and then we have M =   and | det(M )| = 1. 6 0 0 1 0 0 0 6   0 1 0 0 0 0   1 0 0 0 0 0 In general the matrix Mk represents the intersection pairing between the homology classes of degrees k and (n+1)n−k with real coefficient. So in general | det(M n(n+1) )| = 1. Hence the Reidemeister 2 torsion of SU4/T is 1 by the Reidmeister torsion formula for manifolds. By Theorems 1.1, 3.18 and 3.20, we obtain the following result.

Theorem 3.22. The Reidemeister torsion of S Un+1/T is always 1 for any positive integer n with n ≥ 3. Remark 3.23. We should note that we found this result by Schubert calculus. But, we choose any basis to define Reidemeister torsion. There are many bases for the Reidemeister torsion to be 1. Why we focus on this basis to compute the Reidemeister torsion is that we can use Schubert calculus and we have cup product formula in this algebra in terms of Schubert differential forms. Otherwise these computations are not easy. Also by Groebner techniques we can find the normal form of all elenents of Weyl group indexing our basis. So computations in this algebra is avaliable.

Remark 3.24. In our work, we consider flag manifold SUn+1/T for n ≥ 3. Then we consider the Schubert cells {cp} and the corresponding homology basis a {hp} associated to {cp} . We caculated that n n Tor(C∗(K), {cp}p=0, {hp}p=0) = 1.

0 If we consider the same cell-decomposition but other homology basis {hp} then by the change-base- formula (1.4), then we have n !(−1)p Y 1 Tor(C , {c }n , {h0 }n ) = · Tor(C , {c }n , {h }n ). ∗ p p=0 p p=0 [h0 , h ] ∗ p p=0 p p=0 p=0 p p

Remark 3.25. In the presented paper M = K/T is a flag manifold, where K = SUn+1 and T is the maximal torus of K. Clearly, M is a smooth orientable even dimensional(complex) closed manifold. So there is Poincare´ (or Hodge) duality. Therefore, we can apply Theorem 1.1 for M = K/T.

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Acknowledgments

We would like to express our sincere gratitude to the anonymous referee for his/her helpful comments that will help to improve the quality of the manuscript.

Conflict of interest

The authors declare that they have no competing interests.

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