Consensus and synchronization in discrete-time networks of multi-agents with stochastically switching topologies and time delays

1.   Introduction

Reidemeister torsion is a topological invariant and was introduced by Reidemeister in 1935. Up to PL equivalence, he classified the lens spaces $ \mathit{S}^{3}/\Gamma, $ where $ \Gamma $ 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 $ \mathit{S}^{2n+1}/\Gamma, $ where $ \Gamma $ is a cyclic group acting freely and isometrically on the sphere $ \mathit{S}^{2n+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, the classification of lens spaces of Reidemeister and Franz was actually topological (i.e., up to homeomorphism).

Using the Reidemeister torsion, Milnor disproved Hauptvermutung 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], Sözen explained the claim mentioned in [27,p. 187] about the relation between a symplectic chain complex with $ \omega- $compatible bases and the Reidemeister torsion of it. Moreover, he applied the main theorem to the chain-complex

where $ \Sigma_g $ is a compact Riemann surface of genus $ g > 1, $ where $ \partial $ is the usual boundary operator, and where $ \varrho:\pi_1(\Sigma_g)\to \mathrm{PSL}_2(\mathbb{R}) $ is a discrete and faithful representation of the fundamental group $ \pi_1(\Sigma_g) $ of $ \Sigma_g $ ^[21]. Now we will give his description of Reidemesister torsion and explain why it is not unique by a result of Milnor in ^[17].

Let $ \mathcal{H}_p(\mathcal{C}_\ast) = \mathcal{Z}_p(\mathcal{C}_\ast)/\mathcal{B}_p(\mathcal{C}_\ast) $ denote the homologies of the chain complex $ (\mathcal{C}_{\ast }, \partial_{\ast }) = (\mathcal{C}_n\stackrel{\partial_n}{\rightarrow} \mathcal{C}_{n-1}\rightarrow\cdots\rightarrow \mathcal{C}_1 \stackrel{\partial_1}{\rightarrow} \mathcal{C}_0\rightarrow 0) $ of finite dimensional vector spaces over field $ \mathbb{C} $ or $ \mathbb{R} $, where $ \mathcal{B}_p = \text{Im}\{\partial_{p+1}:\mathcal{C}_{p+1}\to \mathcal{C}_p\}, $ $ \mathcal{Z}_p = \text{ker}\{\partial_p:\mathcal{C}_p\to \mathcal{C}_{p-1}\}, $ respectively.

Consider the short-exact sequences:

where (1.1) is a result of $ 1^{\text{st}} $-Isomorphism Theorem and (1.2) follows simply from the definition of $ \mathcal{H}_p. $ Note that if $ {\mathfrak b}_p $ is a basis for $ \mathcal{B}_p, $ $ {\mathfrak h}_p $ is a basis for $ \mathcal{H}_p $, and $ \ell_p:\mathcal{H}_p\to \mathcal{Z}_p $ and $ s_p:\mathcal{B}_{p-1}\to \mathcal{C}_p $ are sections, then we obtain a basis for $ \mathcal{C}_p. $ Namely, $ \mathfrak{b}_p\oplus \ell_p(\mathfrak{h}_p)\oplus s_p(\frak{b}_{p-1}). $

If, for $ p = 0, \cdots, n $, $ \mathfrak{c}_p, $ $ \mathfrak{b}_p, $ and $ \mathfrak{h}_p $ are bases for $ \mathcal{C}_p, $ $ \mathcal{B}_p $ and $ \mathcal{H}_p, $ respectively, then the alternating product

is called the Reidemeister torsion of the complex $ \mathcal{C}_{\ast} $ with respect to bases $ \{\mathfrak{c}_p\}_{p = 0}^n, $ $ \{\mathfrak{h}_p\}_{p = 0}^{n}, $ where $ \left[\mathfrak{b}_p\oplus \ell_p(\mathfrak{h}_p)\oplus s_p(\frak{b}_{p-1}), \mathfrak{c}_p\right] $ denotes the determinant of the change-base matrix from $ \mathfrak{c}_p $ to $ \mathfrak{b}_p\oplus \ell_p(\mathfrak{h}_p)\oplus s_p(\frak{b}_{p-1}). $

Milnor ^[17] proved that torsion does not depend on neither the bases $ \frak{b}_p, $ nor the sections $ s_p, \ell_p. $ Moreover, if $ \frak{c}'_p, \frak{h}'_p $ are other bases respectively for $ \mathcal{C}_p $ and $ \mathcal{H}_p, $ then there is the change-base-formula:

Let $ M $ be a smooth $ n- $manifold, $ \mathbf{K} $ be a cell-decomposition of $ M $ with for each $ p = 0, \cdots, n, $ $ \mathfrak{c}_p = \{e^p_1, \cdots, e^p_{m_p}\}, $ called the geometric basis for the $ p- $cells $ \mathcal{C}_p(\mathbf{K}; \mathbb{Z}). $ Hence, we have the chain-complex associated to $ M $

where $ \partial_p $ denotes the boundary operator. Then $ \mathrm{Tor}(\mathcal{C}_{\ast}(\mathbf{K}), \{\mathfrak{c}_p\}_{p = 0}^n, \{\mathfrak{h}_p\}_{p = 0}^n) $ is called the Reidemeister torsion of $ M, $ where $ \mathfrak{h}_p $ is a basis for $ \mathcal{H}_p(\mathbf{K}). $

In ^[23], oriented closed connected $ 2m- $manifolds $ (m\geq 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\geq 1). $ For $ p = 0, \ldots, 2m, $ let $ \mathbf{h}_p $ be a basis of $ H_p(M). $ Then the Reidemeister torsion of $ M $ satisfies the following formula:

where $ \det H_{p, 2m-p}(M) $ is the determinant of the matrix of the intersection pairing $ (\cdot, \cdot)_{p, 2m-p}:H_p(M)\times H_{2m-p}(M) $ $ \to \mathbb{R} $ in bases $ \mathbf{h}_p, \mathbf{h}_{2m-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 ^{[2,3,5,6,12,13,22,24,25,26]} 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 = SU_{n+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 $ SU_{n+1}/T $ for $ n\geq 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 Kač-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, \mathbb C) $. Also we introduce a multiplication formula and the actions of reflections and Berstein-Gelfand-Gelfand type BGG operators $ A_i $ on the basis elements in the nil Hecke ring.

Proposition 2.1.

where $ p_{u, v}^w $ is a homogeneous polynomial of degree $ \ell (u) + \ell (v) - \ell (w) $.

Proposition 2.2.

Theorem 2.3. Let $ u, v \in W $. We write $ w^{-1} = r_{i_1} \cdots r_{i_n} $ as a reduced expression.

where $ m = \ell (v) $ and the notation $ \hat{A}_i $ means that the operator $ A_i $ is replaced by the Weyl group action $ r_i $.

Let $ \mathbb C_0 = S / S^+ $ be the $ S $-module where $ S^+ $ is the augmentation ideal of $ S $. It is 1-dimensional as $ \mathbb C $-vector space. Since $ \Lambda $ is a $ S $-module, we can define $ \mathbb C_0 \otimes_{S} \Lambda $. It is an algebra and the action of $ \mathcal R $ on $ \Lambda $ gives an action of $ \mathcal R $ on $ \mathbb C_0 \otimes_{S} \Lambda $. The elements $ \sigma^w = 1 \otimes \xi^w \in \mathbb C_0 \otimes_S \Lambda $ is a $ \mathbb C $-basis form of $ \mathbb C_0 \otimes_{S} \Lambda $.

Proposition 2.4. $ \mathbb C_0 \otimes_{S} \Lambda $ 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: \mathbb C_0 \otimes_{S} \Lambda \rightarrow {\operatorname{Gr}} \mathbb 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 Kač-Moody Lie algebra $ \mathbf g $ and let $ T $ denote the maximal torus of $ K $. Then the map

defined by $ \theta (\varepsilon^w) = \sigma^w $ for any $ w \in W $ is a graded algebra isomorphism. Moreover, the action of $ w \in W $ and $ A^w $ on $ H^*(K / T, \mathbb C) $ corresponds respectively to that $ \delta_w $ and $ x_w \in \mathcal R $ on $ \mathbb C_0 \otimes_S \Lambda $.

Corollary 2.7. The operators $ A^i $ on $ H^*(K / T, \mathbb C) $ generate the nil-Hecke algebra.

Corollary 2.8. We can use Proposition 2.1 and Theorem 2.3 to determine the cup product $ \varepsilon^u \varepsilon^v $ in terms of the Schubert basis $ \{\varepsilon^w\}_{w \in W} $ of $ H^*(K/ T, \mathbb Z) $.

3.   The Reidemeister torsion of compact flag manifold $ K/T $ for $ K= SU_{n+1} $

This section includes our calculations about Reidemeister torsion of flag manifolds using Theorem 1.1 and Proposition 2.1 because $ \chi(SU_{n+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 each simple root $ \alpha $, the Weyl group $ W $ contains an element $ r_\alpha $ of order two represented by $ e^{((\pi/2)(e_\alpha+e_{-\alpha}))} $ in $ N(T) $. Since the roots $ \alpha $ can be considered as the linear functionals on the Lie algebra t of the maximal torus $ T $, the action of $ r_\alpha $ on t is given by

where $ h_\alpha $ is the coroot in t corresponding to simple root $ \alpha $.Also, we can give the action of $ r_\alpha $ on the roots by

where $ \textbf{t}^* $ is the dual vector space of t. The element $ r_\alpha $ is the reflection in the hyperplane $ H_\alpha $ of t whose equation is $ \alpha(\xi) = 0 $. These reflections $ r_\alpha $ generate the Weyl group $ W $.

Set $ \alpha_1, \alpha_2, \ldots, \alpha_n $ be roots of Weyl Group of $ SU_{n+1} $. Since the Cartan Matrix of Weyl Group of $ SU_{n+1} $ is

$ M_{ij} = \left\{

$$\begin{array}{cc} 2 & i = j \\ -1 & |i-j| = 1 \\ 0 & \hbox{otherwise} \end{array}$$ \right. $,

$ r_{\alpha_i}(\alpha_j) = \left\{

$$\begin{array}{ll} -\alpha_i, & i = j \\ \alpha_i+\alpha_j, & |i-j| = 1 \\ \alpha_j, & \hbox{otherwise}. \end{array}$$ \right. $

Proposition 3.1. The Weyl group $ W $ of $ SU_{n+1} $ is isomorphic to Coxeter Group $ A_n $ given by generators $ s_1, s_2, \ldots, s_n $ and relations

(i) $ s_i^2 = 1 \qquad i = 1, 2, \ldots, n $;

(ii) $ s_is_{i+1}s_i = s_{i+1}s_is_{i+1} \qquad i = 1, 2, \ldots, n-1 $;

(iii) $ s_is_j = s_js_i \qquad 1\leq i < j-1 < n $.

Proof. (i)

(ii)

Hence $ r_{\alpha_{i+1}}\circ r_{\alpha_i}\circ r_{\alpha_{i+1}}(\beta) = r_{\alpha_{i+1}}\circ r_{\alpha_i}\circ r_{\alpha_{i+1}}(\beta) $.

(iii)

Hence $ r_{\alpha_i}\circ r_{\alpha_j}(\beta) = r_{\alpha_j}\circ r_{\alpha_i}(\beta) $.

After this point $ s_i $ will represent $ r_{\alpha_i} $.

Let us define the word

$ s_{i, j} = \left\{

$$\begin{array}{cc} s_is_{i+1}\cdots s_j & i < j \\ s_i & i = j \\ 1 & i > j. \end{array}$$ \right. $

Theorem 3.2. [4,Theorem 3.1] The reduced Gröbner-Shirshov basis of the coxeter group $ A_n $ consists of relation

together with defining relations of $ A_n $.

The following lemma is equivalent of [4,Lemma 3.2]. The only difference is the order of generators $ s_1 > s_2 > \ldots s_n $ in our setting.

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

$ s_{{n+1}, j_{n+1}}s_{n, j_n}s_{{n-1}, j_{n-1}}\cdots s_{i, j_i}\cdots s_{1, j_1} $ $ 1\leq i\leq j_i+1\leq n+1 $.

Notice that $ j_{n+1}+1 = n+1 \Longrightarrow j_{n+1} = n $ and $ s_{{n+1}, n} = 1 $.

Algorithm 3.1. (Finding Inverse) Let $ w = s_{n, j_n}s_{n-1, j_{n-1}}\cdots s_{1, j_1} $. 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 = \{s_k, s_{k+1}, s_{k+2}, \ldots, s_{k+j}\} $;

$ Invw = list\cup Invw $;

End For.

Example 3.4. Let $ s_{4, 6}s_{3, 5}s_{2, 5}s_{1, 3} $. The inverse of its is $ S_3s_2s_1s_5s_4s_3s_2s_5s_4s_3s_6s_5s_4 $.

$ Invw = s_{1, 4} $

$ S_3s_2s_5s_4s_3s_5s_4s_6s_5 $

$ Invw = s_{2, 5}s_{1, 4} $

$ S_3s_5s_4s_5s_6 $

$ Invw = s_{3, 5}s_{2, 5}s_{1, 4} $

$ s_5s_6 $

$ Invw = s_{5, 6}s_{3, 5}s_{2, 5}s_{1, 4} $.

Lemma 3.5. Let $ w = (s_{n, j_n})(s_{n-1, j_{n-1}})\cdots(s_{{i+1}, \; j_{i+1}})(s_{i, j_i})\cdots(s_{1, j_1}) $ and

$ s_{i}w $ = $ (s_{n, \overline{j_n}})(s_{n-1, \; \overline{j_{n-1}}})\cdots(s_{{i+1}, \; \overline{j_{i+1}}})(s_{i, \overline{j_i}})\cdots(s_{1, \overline{j_1}}) $, where

$ s_{i}w = \left\{

$$\begin{array}{ll} \overline{j_{i+1}} = j_i+1, \; \overline{j_i} = j_{i+1} & \; if \quad j_i < j_{i+1} \\ \overline{j_{i+1}} = j_i, \; \overline{j_i} = j_{i+1}-1 & \; if\quad j_i\geq j_{i+1} \\ \overline{j_k} = j_k & if \quad k\neq i, i+1 \end{array}$$ \right. $

Here if $ i = n $, then we assume $ j_{n+1} = n $.

Corollary 3.6. Let $ w = (s_{n, j_n})(s_{n-1, j_{n-1}})\cdots(s_{{i+1}, j_{i+1}})(s_{i, j_i})\cdots(s_{1, j_1}) $ and

$ s_{i-1}(s_{i}w) $ = $ (s_{n, \hat{j_n}})(s_{n-1, \widehat{j_{n-1}}})\cdots(s_{{i+1}, \widehat{j_{i+1}}})(s_{i, \widehat{j_i}})\cdots(s_{1, \widehat{j_1}}) $, where

$ s_{i-1}(s_{i}w) = \left\{

$$\begin{array}{ll} \widehat{j_{i+1}} = j_i+1, \; \widehat{j_i} = j_{i-1}+1, \; \widehat{j_{i-1}} = j_{i+1} & if \quad j_i < j_{i+1}, \; j_{i-1} < j_{i+1} \\ \widehat{j_{i+1}} = j_i+1, \; \widehat{j_i} = j_{i-1}, \; \widehat{j_{i-1}} = j_{i+1}-1 & if \quad j_i < j_{i+1}, \; j_{i-1}\geq j_{i+1} \\ \widehat{j_{i+1}} = j_i, \; \widehat{j_i} = j_{i-1}+1, \; \widehat{j_{i-1}} = j_{i+1}-1 & if \quad j_i\geq j_{i+1}, \; j_{i-1} < j_{i+1}-1 \\ \widehat{j_{i+1}} = j_i, \; \widehat{j_i} = j_{i-1}, \; \widehat{j_{i-1}} = j_{i+1}-2 & if \quad j_i\geq j_{i+1}, \; j_{i-1}\geq j_{i+1}-1 \\ \widehat{j_k} = j_k & if \quad k\neq i-1, \; i, \; i+1. \end{array}$$ \right. $

Proof. Let $ \overline{w} = s_iw = (s_{n, \overline{j_n}})(s_{n-1, \overline{j_{n-1}}})\cdots(s_{{i+1}, \overline{j_{i+1}}})(s_{i, \overline{j_i}})\cdots(s_{1, \overline{j_1}}) $. Then

$ s_{i-1}(\overline{w}) = \left\{

$$\begin{array}{ll} \widehat{j_i} = \overline{j_{i-1}}+1, \; \widehat{j_{i-1}} = \overline{j_{i}} & if \quad \overline{j_{i-1}} < \overline{j_i} \\ \widehat{j_i} = \overline{j_{i-1}}, \; \widehat{j_{i-1}} = \overline{j_{i}}-1 & if \quad \overline{j_{i-1}}\geq\overline{j_i} \\ \widehat{j_k} = \overline{j_k} & if \quad k\neq i-1, i. \end{array}$$ \right. $

(i) $ j_i < j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i+1 $, $ \overline{j_i} = j_{i+1} $ So $ \overline{j_{i-1}} < \overline{j_i} $ $ \Rightarrow $ $ j_{i-1} < j_{i+1} $, $ \widehat{j_{i+1}} = \overline{j_{i+1}} = j_i+1, $ $ \widehat{j_i} = \overline{j_{i-1}}+1 = j_{i-1}+1 $, $ \widehat{j_{i-1}} = \overline{j_i} = j_{i+1}. $

(ii) $ j_i < j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i+1 $, $ \overline{j_i} = j_{i+1} $ So $ \overline{j_{i-1}}\geq\overline{j_i} $ $ \Rightarrow $ $ j_{i-1}\geq j_{i+1} $, $ \widehat{j_{i+1}} = \overline{j_{i+1}} = j_i+1, $ $ \widehat{j_i} = \overline{j_{i-1}} = j_{i-1} $, $ \widehat{j_{i-1}} = \overline{j_i}-1 = j_{i+1}-1 $.

(iii) $ j_i\geq j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i $, $ \overline{j_i} = j_{i+1}-1 $ So $ \overline{j_{i-1}} < \overline{j_i} $ $ \Rightarrow $ $ j_{i-1} < j_{i+1} $, $ \widehat{j_{i+1}} = \overline{j_{i+1}} = j_i+1 $, $ \widehat{j_i} = \overline{j_{i-1}} = j_{i-1} $, $ \widehat{j_{i-1}} = \overline{j_i}-1 = j_{i+1}-1 $.

(iv) $ j_i\geq j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i $, $ \overline{j_i} = j_{i+1}-1 $ So $ \overline{j_{i-1}}\geq\overline{j_i} $ $ \Rightarrow $ $ j_{i-1}\geq j_{i+1}-1 $, $ \widehat{j_{i+1}} = \overline{j_{i+1}} = j_i $, $ \widehat{j_i} = \overline{j_{i-1}} = j_{i-1} $, $ \widehat{j_{i-1}} = \overline{j_i}-1 = j_{i+1}-2 $.

Corollary 3.7. Let $ w = (s_{n, j_n})(s_{n-1, j_{n-1}})\cdots(s_{{i+1}, j_{i+1}})(s_{i, j_i})\cdots(s_{1, j_1}) $ and

$ s_{i+1}(s_{i}w) $ = $ (s_{n, \hat{j_n}})(s_{n-1, \widehat{j_{n-1}}})\cdots(s_{{i+1}, \widehat{j_{i+1}}})(s_{i, \widehat{j_i}})\cdots(s_{1, \widehat{j_1}}) $. Then

$ s_{i+1}(s_{i}w) = \left\{

$$\begin{array}{ll} \widehat{j_{i+2}} = j_i+2, \; \widehat{j_{i+1}} = j_{i+2}, \; \widehat{j_{i}} = j_{i+1} & if \quad j_i < j_{i+1}, \; j_{i+1} < j_{i+2} \\ \widehat{j_{i+2}} = j_i+1, \; \widehat{j_{i+1}} = j_{i+2}-1, \; \widehat{j_{i}} = j_{i+1} & if \quad j_i < j_{i+1}, \; j_i+1\geq j_{i+2} \\ \widehat{j_{i+2}} = j_i+1, \; \widehat{j_{i+1}} = j_{i+2}, \; \widehat{j_{i}} = j_{i+1}-1 & if \quad j_i\geq j_{i+1}, \; j_{i} < j_{i+2} \\ \widehat{j_{i+2}} = j_i, \; \widehat{j_{i+1}} = j_{i+2}-1, \; \widehat{j_{i}} = j_{i+1}-1 & if \quad j_i\geq j_{i+1}, \; j_{i}\geq j_{i+2} \\ \widehat{j_k} = j_k & if \quad k\neq i, i+1, i+2. \end{array}$$ \right. $

Proof. Let $ \overline{w} = s_iw = (s_{n, \overline{j_n}})(s_{n-1, \overline{j_{n-1}}}) \cdots(s_{{i+1}, \overline{j_{i+1}}})(s_{i, \overline{j_i}})\cdots(s_{1, \overline{j_1}}) $. Then

$ s_{i+1}(\overline{w}) = \left\{

$$\begin{array}{ll} \widehat{j_{i+2}} = \overline{j_{i+1}}+1, \; \widehat{j_{i+1}} = \overline{j_{i+2}} & if \quad \overline{j_{i+1}} < \overline{j_{i+2}} \\ \widehat{j_{i+2}} = \overline{j_{i+1}}, \; \widehat{j_{i+1}} = \overline{j_{i+2}}-1 & if \quad \overline{j_{i+1}}\geq\overline{j_{i+2}} \\ \widehat{j_k} = \overline{j_k} & if \quad k\neq i+1, i+2. \end{array}$$ \right. $

(i) $ j_i < j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i+1 $, $ \overline{j_i} = j_{i+1} $ So $ \overline{j_{i+1}} < \overline{j_{i+2}} $ $ \Rightarrow $ $ j_{i}+1 < j_{i+2} $, $ \widehat{j_{i+2}} = \overline{j_{i+1}}+1 = j_i+2 $, $ \widehat{j_{i+1}} = \overline{j_{i+2}} = j_{i+2} $, $ \widehat{j_{i}} = \overline{j_i} = j_{i+1} $.

(ii) $ j_i < j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i+1 $, $ \overline{j_{i}} = j_{i+1} $ So $ \overline{j_{i+1}}\geq\overline{j_{i+2}} $ $ \Rightarrow $ $ j_{i}+1\geq j_{i+2} $, $ \widehat{j_{i+2}} = \overline{j_{i+1}} = j_i+1 $, $ \widehat{j_{i+1}} = \overline{j_{i+2}}-1 = j_{i+2}-1 $, $ \widehat{j_{i}} = \overline{j_i} = j_{i+1} $.

(iii) $ j_i\geq j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i $, $ \overline{j_i} = j_{i+1}-1 $ So $ \overline{j_{i+1}} < \overline{j_{i+2}} $ $ \Rightarrow $ $ j_{i} < j_{i+2} $, $ \widehat{j_{i+2}} = \overline{j_{i+1}}+1 = j_i+1 $, $ \widehat{j_{i+1}} = \overline{j_{i+2}} = j_{i+2} $, $ \widehat{j_{i}} = \overline{j_i}-1 = j_{i+1}-1 $.

(iv) $ j_i \geq j_{i+1} $ $ \Rightarrow $ $ \overline{j_{i+1}} = j_i $, $ \overline{j_i} = j_{i+1}-1 $ So $ \overline{j_{i+1}}\geq\overline{j_{i+2}} $ $ \Rightarrow $ $ j_{i}\geq j_{i+2} $, $ \widehat{j_{i+2}} = \overline{j_{i+1}} = j_i $, $ \widehat{j_{i+1}} = \overline{j_{i+2}}-1 = j_{i+2}-1 $, $ \widehat{j_{i}} = \overline{j_i} = j_{i+1}-1 $.

Using Lemma $ 3.3 $ and definitions of $ A^i $ and $ r_i $ operators, we can obtain the followings.

Lemma 3.8. Let $ w = (s_{n, j_n})(s_{n-1, j_{n-1}})\cdots(s_{{i+1}, \; j_{i+1}})(s_{i, j_i})\cdots(s_{1, j_1}) $. Then

$ A^{i}({\varepsilon}^w) = \left\{

$$\begin{array}{ll} {\varepsilon}^{w_1} \quad if \quad j_i\geq j_{i+1} \\ 0 \quad if \quad j_i < j_{i+1}, \\ \end{array}$$ \right. $

where $ w_1 = (s_{n, \overline{j_n}})(s_{n-1, \overline{j_{n-1}}})\cdots(s_{{i+1}, \overline{j_{i+1}}})(s_{i, \overline{j_i}})\cdots(s_{1, \overline{j_1}}) $ with $ \overline{j_{i+1}} = j_i $, \quad $ \overline{j_{i}} = j_{i+1}-1 $ and $ \overline{j_k} = j_k $ if $ \quad k\neq i, i+1 $.

Lemma 3.9. $ r_{i}({\varepsilon}^{s_j}) = \left\{

$$\begin{array}{ll} {\varepsilon}^{s_{i-1}}-{\varepsilon}^{s_i}-{\varepsilon}^{s_{i+1}} \quad if \quad i = j \\ {\varepsilon}^{s_j} \quad if \quad i\neq j. \\ \end{array}$$ \right. $

The integral cohomology of $ SU_{n+1}/T $ is generated by Schubert classes indexed

Let $ x_i = {\varepsilon}^{r_i}\in H^2(SU_{n+1}/T, \mathbb{Z}). $ We define an order between generators of the integral cohomology of $ SU_{n+1}/T $. Since each element $ {\varepsilon}^{s_{nj_n}s_{n-1, j_{n-1}}\ldots s_{ij_i}\ldots s_{1j_{1}}} $ can be represented by an $ n $-tuple $ (j_n-n+1, j_{n-1}-(n-1)+1, \ldots, j_i-i+1, \ldots, j_1-1+1), $ we can define an order between $ n $-tuples.

Definition 3.10. (Graded Inverse Lexicographic Order) Let $ \alpha = (\alpha_1, \alpha_2, \ldots, \alpha_n) $ and $ \beta = (\beta_1, \beta_2, \ldots, \beta_n)\in \mathbb{Z}^n_{\geq0} $. We say $ \alpha > \beta $ if $ |\alpha| = \alpha_1+\alpha_2+\ldots\alpha_n > |\beta| = \beta_1+\beta_2+\ldots\beta_n $ or $ |\alpha| = |\beta| $ and in the vector difference $ \alpha-\beta\in\mathbb{Z}^n $, the right-most nonzero entry is positive. We will write $ {\varepsilon}^{s_{nj_n}s_{n-1, j_{n-1}}\ldots s_{ij_i}\ldots s_{1j_{1}}} > {\varepsilon}^{s_{nk_n}s_{n-1, j_{k-1}}\ldots s_{ik_i}\ldots s_{1j_{1}}} $ if $ (j_n-n+1, j_{n-1}-(n-1)+1, \ldots, j_i-i-1, \ldots, j_1-1+1) > (k_n-n+1, k_{n-1}-(n-1)+1, \ldots, k_i-i-1, \ldots, k_1-1+1) $.

Example 3.11. $ {\varepsilon}^{s_{35}s_{23}s_{14}} > {\varepsilon}^{s_{35}s_{24}s_{13}} $ since $ (3, 2, 4) > (3, 3, 3) $ in graded inverse lexicographic order.

We will try to find a quotient ring $ \mathbb{Z}[x_1, x_2, \ldots, x_n]/I $ which is isomorphic to $ H^*(SU_{n+1}/T, \mathbb{Z}) $. We also define an order between monomials as follows.

Definition 3.12. We say $ x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n} > x_1^{\beta_1}x_2^{\beta_2}\cdots x_n^{\beta_n} $ if $ |\alpha| = \alpha_1+\alpha_2+\cdots+\alpha_n > |\beta| = \beta_1+\beta_2+\cdots+\beta_n $ or $ |\alpha| = |\beta| $ and in the vector difference $ \alpha-\beta\in\mathbb{Z}^n $ the left-most non-zero entry is negative.

Example 3.13. $ x_1^4x_2^2 x_3^3 < x_1^3 x_2^3 x_3^3 $, since $ (4, 2, 3)-(3, 3, 3) = (1, -1, 0) $.

Lemma 3.14. $ x_1^{\alpha_1}x_2^{\alpha_2}\ldots x_n^{\alpha_n} = {\varepsilon}^{s_{n\alpha_n}s_{n-1, \alpha_{n-1}}\ldots s_{i\alpha_i}\ldots s_{1\alpha_{1}}}+\mathrm{lower}\; \mathrm{terms}. $

Proof. To prove this, we use induction on degree of the monomials. By definition $ x_i = {\varepsilon}^{s_i} $. Let us compute $ x_ix_j = {\varepsilon}^{s_i}{\varepsilon}^{s_j} $. Here we may assume that $ i\leq j $. If $ j-i > 1 $, the inverse of $ s_is_j $ is $ s_is_j $. Hence

in the cup product. If $ j = i+1 $, the inverse of $ s_{i+1}s_i $ is $ s_is_{i+1} $. In this case

If $ i = j $, then we have to consider the word $ s_{i, {i+1}} $. Its inverse $ s_{i+1}s_i $ and

Now we have to show that $ P_{s_is_j}^{s_ks_l} = 0 $ if $ {\varepsilon}^{s_ks_l} > {\varepsilon}^{s_js_i} $. By definition of cup product the coefficient of $ {\varepsilon}^{s_ks_l} $ is not zero only if $ s_i\rightarrow s_ks_l $ and $ s_j\rightarrow s_ks_l $. However, this is possible only if $ s_ks_l = s_js_i $ or $ s_ks_l = s_{i, i+1} $ when $ j = i+1 $. Clearly $ {\varepsilon}^{s_is_{i+1}} < {\varepsilon}^{s_{i+1}s_i} $. Hence $ {\varepsilon}^{s_i}{\varepsilon}^{s_{i+1}} = {\varepsilon}^{s_{i+1}s_i}+\mathrm{lower}\; \mathrm{terms} $ and $ {\varepsilon}^{s_i}{\varepsilon} ^{s_j} = {\varepsilon}^{s_j}{\varepsilon}^{s_i} $ if $ j-i > 1 $. In the case $ i = j $, we have to look elements $ s_i s_k $ and $ s_k s_i $. The inverse of $ s_ks_i $ is equal to $ s_ks_i $ itself if $ k-i > 1 $, hence

since $ k-i > 1 $. Clearly $ {\varepsilon}^{s_is_k} < {\varepsilon}^{s_is_{i+1}} $ if $ k < i $. Hence $ {\varepsilon}^{s_i}{\varepsilon}^{s_i} = {\varepsilon}^{s_is_{i+1}}+\mathrm{lower}\; \mathrm{terms} $.

Assume that $ x_1^{\alpha_1}x_2^{\alpha_2}\ldots x_n^{\alpha_n} = {\varepsilon}^{s_{n\alpha_n}s_{n-1, \alpha_{n-1}}\ldots s_{i\alpha_i}\ldots s_{1\alpha_{1}}}+\mathrm{lower}\; \mathrm{terms} $.

We have to show $ x_1^{\alpha_1}x_2^{\alpha_2}\ldots x_i^{\alpha_i+1}\ldots x_n^{\alpha_n} = {\varepsilon}^{s_{n\alpha_n}s_{n-1, \alpha_{n-1}}\ldots s_{i\alpha_i+1}\ldots s_{1\alpha_{1}}}+\mathrm{lower}\; \mathrm{terms} $ by Bruhat ordering.

$ s_{n\alpha_n}s_{n-1, \alpha_{n-1}}\ldots s_{i\alpha_i+1}\ldots s_{1\alpha_{1}}\rightarrow w' $ only if $ w' = s_{n\overline{\alpha_n}}s_{n-1, \overline{\alpha_{n-1}}}\ldots s_{i\overline{\alpha_i}}\ldots s_{1\overline{\alpha_{1}}} $ where there exists an index $ j $ for which $ \overline{\alpha_j} = \alpha_j+1 $ and $ \overline{\alpha_k} = \alpha_k $ if $ k\neq j $.

By given ordering

If $ j > i $, then, by Algorithm $ 1 $, in $ w'^{-1} $, we will not have a subsequence $ s_{j-1}, s_{j-2}\ldots s_i $ after the elements $ s_j $. Hence in the cup product before applying $ A^j $ we will not have the term $ {\varepsilon}^{s_j} $. It means $ P_{s_i, w}^{w'} = 0. $

If $ j = i $, then, again by Algorithm $ 1 $, in $ w'^{-1} $ we will not have a subsequence $ s_{j-1}, s_{j-2}\ldots s_i $ after the elements $ s_j $. Hence in the cup product before applying $ A^j $ we will not have the term $ {\varepsilon}^{s_j} $. It means $ P_{s_i, w}^{w'} = 1 $ if and only if $ j > i $.

Example 3.15. Let $ l = 3 $,

$ x_1x_2x_3 = {\varepsilon}^{s_3s_2s_1}+\mathrm{lower}\; \mathrm{terms} $.

$ x_1^2x_2x_3 = {\varepsilon}^{s_3s_2s_{12}}+\mathrm{lower}\; \mathrm{terms} $.

Then we have $ {\varepsilon}^{s_3s_{23}s_1} > {\varepsilon}^{s_3s_2s_{12}} > {\varepsilon}^{s_{23}s_{12}} > {\varepsilon}^{s_3s_{13}} > {\varepsilon}^{s_2s_{13}} $. Since the inverse of $ s_3s_{23}s_1 $ is $ s_3s_{13} $ and the inverse of $ s_3s_2s_1 $ is $ s_{13} $, $ A_3r_1r_2r_3({\varepsilon}^{s_1}) = A_3r_1({\varepsilon}^{s_1}) = A_3(-{\varepsilon}^{s_1}+{\varepsilon}^{s_2}) = 0 $.

Similarly, since the inverse of $ s_3s_2s_{12} $ is $ s_2s_{13} $, $ A_2r_1r_2r_3({\varepsilon}^{s_1}) = A_2r_1({\varepsilon}^{s_1}) = A_2(-{\varepsilon}^{s_1}+{\varepsilon}^{s_2}) = 1 $.

Before finding the quotient ring $ \mathbb{Z}[x_1, \ldots, x_n]/I $, we give some information about ring $ \Bbbk[x_1, \ldots, x_n]/I $ where $ \Bbbk $ is a field. Fix a monomial ordering on $ \Bbbk[x_1, \ldots, x_n] $. Let $ f\in \Bbbk[x_1, \ldots, x_n] $. 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\subseteq k[x_1, \ldots, x_n] $ be an ideal. Define $ LT(I) = \{LT(f): f\in 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\in \Bbbk[x_1, \ldots, x_n] $ is congruent modulo $ I $ to a unique polynomial $ r $ which is a $ \Bbbk $-linear combination of the monomials in the complement of $ < LT(I) > $.

(ii) The elements of $ \{x^\alpha: x^\alpha\not\in < LT(I) > \} $ are linearly independent modulo $ I $.

(iii) $ \Bbbk[x_1, \ldots, x_n]/I $ is isomorphic as a $ \Bbbk-vector $ space to

Theorem 3.17. [9,Section 5.3,Theorem 6] Let $ I\subseteq \Bbbk[x_1, \ldots, x_n] $ be an ideal.

(i) The $ \Bbbk $-vector space $ \Bbbk[x_1, \ldots, x_n]/I $ is finite dimensional.

(ii) For each $ i $, $ 1\leq i\leq n $, there is a polynomial $ f_i\in I $ such that $ LM(f_i) = x_i^{m_i} $ for some positive integer $ m_i $.

Theorem 3.18. $ H^*(SU_{n+1}/T, \mathbb{Z}) $ isomorphic to $ \mathbb{Z}[x_1, x_2, \ldots, x_n]/ < f_1, f_2, \ldots, f_n > $ where $ LT(f_i) = x_{i}^{n-i+2} $ with respect to monomial order given by Definition $ 3.12 $.

Proof. Let $ I $ be the ideal such that $ H^*(SU_{n+1}/T, \mathbb R)\cong\mathbb R[\alpha_1, \alpha_2, \ldots, \alpha_n]/I $. Since we found one to one correspondence between length $ l $ elements of $ H^*(SU_{n+1}/T, \mathbb Z) $ and monomials $ x_{1}^{\alpha_1}x_{2}^{\alpha_2}\cdots x_{n}^{\alpha_n} $, where $ \alpha_1+\alpha_2+\cdots \alpha_n = l $ and for each $ i $, $ 1\leq i\leq n $, $ \alpha_i\leq n-i+1 $, there should be a polynomial $ f_i \in I $ such that $ LT(f_i) = x_{i}^{n-i+2} $.

Example 3.19. Let $ n = 3 $. Then we have

$ \alpha_i\leq n-i+1 $, $ i = 1, 2, 3 $;

$ \alpha_1\leq 3 $, $ \alpha_2\leq 2 $, $ \alpha_3\leq 1 $.

For $ l = 1;\; x_1, \; x_2, \; x_3 $; and

for $ l = 2;\; x_1^2, \; x_1x_2, \; x_1x_3, \; x_2x_3, \; x_2^2 $. So we must have a polynomial $ f_3 $ with $ LM(f_3) = x_3^2 $.

For $ l = 3;\; x_1^3, \; x_1^2x_2, \; x_1^2x_3, \; x_1x_2x_3, \; x_1x_2^2, \; x_2^2x_3 $, so

we must have a polynomial $ f_2 $ with $ LM(f_2) = x_2^3 $.

For $ l = 4;\; x_1^3x_2, \; x_1^3x_3, \; x_1^2x_2x_3, \; x_1^2x_2^2, \; x_1x_2^2x_3 $, so

we must have a polynomial $ f_1 $ with $ LM(f_1) = x_1^4 $.

The complex dimension of $ SU_{n+1} / T $ is equal to $ (n+1)n / 2 $. So the highest element has length of $ (n+1)n/2 $.

Since the unique highest element has length of $ \frac{n(n+1)}{2} $, we now give the result about the multiplication of elements of length $ k $ and of length $ \frac{n(n+1)}{2}-k $.

Theorem 3.20. Let $ A = {\varepsilon}^{s_{nj_{n}}s_{n-1, j_{n-1}}\cdots s_{1j_{1}}} $ be an element of length $ k $ and $ B = {\varepsilon}^{s_{np_{n}}s_{n-1, p_{n-1}}\cdots s_{1p_{1}}} $ be an element of length $ \frac{n(n+1)}{2}-k $. The corresponding polynomials in $ \mathbb{Z}[x_1, x_2, \ldots, x_n]/ < f_1, f_2, \ldots, f_n > $ has leading monomials

$ x_{1}^{j_1-1+1}x_{2}^{j_2-2+1}\cdots x_{i}^{j_i-i+1}\cdots x_{1}^{j_n-n+1} $ and $ x_{1}^{p_1-1+1}x_{2}^{p_1-2+1}\cdots x_{1}^{p_n-n+1} $, respectively. Then

$ A\cdot B = \left\{

$$\begin{array}{ll} {\varepsilon}^{s_{n, n}s_{n-1, n}, \ldots, s_{in}, \ldots, s_{1n}}, & if \; j_i+p_i+1 = n+i; \\ 0, & if \; j_i+p_i+1\neq n+i. \end{array}$$ \right. $

Proof. The unique highest degree monomial in $ \mathbb{Z}[x_1, x_2, \ldots, x_n]/ < f_1, f_2, \ldots, f_n > $ is $ x_{1}^{n}x_{2}^{n-1}\cdots x_{i}^{n-i+1}\cdots x_{n} $. The multiplication of leading monomials of corresponding monomials of $ A $ and $ B $ produce the monomial

If $ j_i+p_i-2i+2 = n-i+1\rightarrow j_i+ p_i+1 = n+i $ for each $ i $, $ i\leq 1\leq n $, then the multiplication gives the $ x_{1}^{n}x_{2}^{n-1}\ldots x_{n} $. Since this monomial correspondence the element $ {\varepsilon}^{s_{n, n}s_{n-1, n}\cdots s_{in}\cdots s_{1n}} $, $ A\cdot B = {\varepsilon}^{s_{n, n}s_{n-1, n}\cdots s_{1n}} $. If $ j_i+p_i+1\neq n+i, $ then the leading monomial and the monomials of lower degree must reduce to zero modulo $ < f_1, f_2, \ldots, f_n > $ in $ \Bbbk[x_1, x_2, \ldots, x_n] $ when we apply the division algorithm. Hence $ A\cdot B = 0 $.

Now we can give the whole computation of the quotient ring $ \mathbb{Z}[x_1, x_2, x_3]/ < f_1, f_2, f_3 > $.

Example 3.21. Let $ x_1 = {\varepsilon}^{s_1} $, $ x_2 = {\varepsilon}^{s_2} $, $ x_3 = {\varepsilon}^{s_3} $.

For $ l = 2 $, we have

and

$ \left(

$$\begin{array}{c} x_2x_3 \\ x_2^2 \\ x_1x_3 \\ x_1x_2 \\ x_1^2 \\ \end{array}$$ \right) = M \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2} \\ {\varepsilon}^{s_2s_3} \\ {\varepsilon}^{s_3s_1} \\ {\varepsilon}^{s_2s_1} \\ {\varepsilon}^{s_1s_2} \\ \end{array}$$ \right) $ $ \quad $ and $ \quad $ $ \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2} \\ {\varepsilon}^{s_2s_3} \\ {\varepsilon}^{s_3s_1} \\ {\varepsilon}^{s_2s_1} \\ {\varepsilon}^{s_1s_2} \\ \end{array}$$ \right) = M^{-1} \left( $$\begin{array}{c} x_2x_3 \\ x_2^2 \\ x_1x_3 \\ x_1x_2 \\ x_1^2 \\ \end{array}$$ \right) $, where

$ M = \left(

$$\begin{array}{ccccc} 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}$$ \right) \quad$ $ M^{-1} = \left( $$\begin{array}{ccccc} 1 & -1 & 0 & 1 & -1 \\ 0 & 1 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}$$ \right). $ Then we have

Here we must have a relation involving $ x_3^2 $ and we have it as

For $ l = 3 $;

and

$ \left(

$$\begin{array}{c} x_2^2x_3 \\ x_1x_2x_3 \\ x_1x_2^2 \\ x_1^2x_3 \\ x_1^2x_2 \\ x_1^3 \\ \end{array}$$ \right) = M \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2s_3} \\ {\varepsilon}^{s_3s_2s_1} \\ {\varepsilon}^{s_2s_3s_1} \\ {\varepsilon}^{s_3s_1s_2} \\ {\varepsilon}^{s_2s_1s_2} \\ {\varepsilon}^{s_1s_2s_3} \\ \end{array}$$ \right) $ $ \quad $ and $ \quad $ $ \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2s_3} \\ {\varepsilon}^{s_3s_2s_1} \\ {\varepsilon}^{s_2s_3s_1} \\ {\varepsilon}^{s_3s_1s_2} \\ {\varepsilon}^{s_2s_1s_2} \\ {\varepsilon}^{s_1s_2s_3} \\ \end{array}$$ \right) = M^{-1} \left( $$\begin{array}{c} x_2^2x_3 \\ x_1x_2x_3 \\ x_1x_2^2 \\ x_1^2x_3 \\ x_1^2x_2 \\ x_1^3 \\ \end{array}$$ \right) $, where

$ M = \left(

$$\begin{array}{cccccc} 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array}$$ \right) \quad $ $ M^{-1} = \left( $$\begin{array}{cccccc} 1 & -1 & 0 & 1 & 0 & 0 \\ 0 & 1 & -1 & -1 & 1 & 0 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array}$$ \right). $

Then we have

Here we must have a relation involving $ x_2^3 $ and we now we have it as

For $ l = 4 $; we have

and

$ \left(

$$\begin{array}{c} x_1x_2^2x_3 \\ x_1^2x_2x_3 \\ x_1^2x_2^2 \\ x_1^3x_3 \\ x_1^3x_2 \\ \end{array}$$ \right) = M \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2s_3s_1} \\ {\varepsilon}^{s_3s_2s_1s_2} \\ {\varepsilon}^{s_2s_3s_1s_2} \\ {\varepsilon}^{s_3s_1s_2s_3} \\ {\varepsilon}^{s_2s_1s_2s_3} \\ \end{array}$$ \right) $ $ \quad $ and $ \quad $ $ \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2s_3s_1} \\ {\varepsilon}^{s_3s_2s_1s_2} \\ {\varepsilon}^{s_2s_3s_1s_2} \\ {\varepsilon}^{s_3s_1s_2s_3} \\ {\varepsilon}^{s_2s_1s_2s_3} \\ \end{array}$$ \right) = M^{-1} \left( $$\begin{array}{c} x_1x_2^2x_3 \\ x_1^2x_2x_3 \\ x_1^2x_2^2 \\ x_1^3x_3 \\ x_1^3x_2 \\ \end{array}$$ \right) $, where

$ M = \left(

$$\begin{array}{ccccc} 1 & 1 & 2 & 2 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}$$ \right) \quad$ $ M^{-1} = \left( $$\begin{array}{ccccc} 1 & -1 & -1 & -1 & 2 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}$$ \right). $

Then

We must have a relation involving $ x_1^4 $, which is $ x_1x_1^3 = {\varepsilon}^{s_1}.{\varepsilon}^{s_1s_2s_3} = 0 $.

For $ l = 5 $;

and

$ \left(

$$\begin{array}{c} x_1^2x_2^2x_3 \\ x_1^3x_2x_3 \\ x_1^3x_2^2 \end{array}$$ \right) = M \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2s_3s_1s_2} \\ {\varepsilon}^{s_3s_2s_1s_2s_3} \\ {\varepsilon}^{s_2s_3s_1s_2s_3} \end{array}$$ \right) $ $ \quad $ and $ \quad $ $ \left( $$\begin{array}{c} {\varepsilon}^{s_3s_2s_3s_1s_2} \\ {\varepsilon}^{s_3s_2s_1s_2s_3} \\ {\varepsilon}^{s_2s_3s_1s_2s_3} \end{array}$$ \right) = M^{-1} \left( $$\begin{array}{c} x_1^2x_2^2x_3 \\ x_1^3x_2x_3 \\ x_1^3x_2^2 \end{array}$$ \right) $, where

$ M = \left(

$$\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array}$$ \right) $ $ \quad $ $ M^{-1} = \left( $$\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{array}$$ \right). $ So

Hence we don't have any relation.

For $ l = 6 $;

$ x_1^3x_2^2x_3 = {\varepsilon}^{s_3s_2s_3s_1s_2s_3} $ and $ {\varepsilon}^{s_3s_2s_3s_1s_2s_3} = x_1^3x_2^2x_3 $.

Now let us multiple elements with lengths of $ k $ and $ 6-k $.

First $ M_0 = 1 $ and $ |\det(M_0)| = 1 $.

Now we will calculate Reidemeister torsion of $ SU_4 /T $ by using above multiplication. From multiplication of the second cohomology, we have $ M_2 =

$$\begin{pmatrix}0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\end{pmatrix}$$ $ and $ |\det(M_2)| = 1. $

To calculate Reidemeister torsion of $ SU_4 /T $ we need multiplication of fourth cohomology bases elements and then we have $ M_4 =

$$\begin{pmatrix}0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\end{pmatrix}$$ $ and $ |\det(M_4)| = 1. $

To calculate Reidemeister torsion of $ SU_4 /T $ we need multiplication of sixth cohomology bases elements and then we have $ M_6 =

$$\begin{pmatrix}0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$$ $ and $ |\det(M_6)| = 1. $

In general the matrix $ M_k $ represents the intersection pairing between the homology classes of degrees $ k $ and $ (n+1)n - k $ with real coefficient. So in general $ |\det(M_{\frac{n(n+1)}{2}})| = 1. $ Hence the Reidemeister torsion of $ SU_4 /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 $ SU_{n+1} /T $ is always $ 1 $ for any positive integer $ n $ with $ n\geq 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 $ SU_{n + 1}/T $ for $ n \geq 3. $ Then we consider the Schubert cells $ \{\mathfrak{c}_p\} $ and the corresponding homology basis a $ \{ \mathfrak{h}_p\} $ associated to $ \{\mathfrak{c}_p\} $. We caculated that $ \mathrm{Tor}(\mathcal{C}_{\ast}(\mathbf{K}), \{\mathfrak{c}_p\}_{p = 0}^n, \{\mathfrak{h}_p\}_{p = 0}^n) = 1 $.

If we consider the same cell-decomposition but other homology basis $ \{\mathfrak{h}'_p\} $ then by the change-base-formula (1.4), then we have

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

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.