Local coordinate systems on quantum flag manifolds

This paper consist of 3 sections. In the first section, we will give a brief introduction to the"Feigin's homomorphisms"and will see how they will help us to prove our main and fundamental theorems related to quantum Serre relations and screening operators. In the second section, we will introduce Local integral of motions as the space of invariants of nilpotent part of quantum affine Lie algebras and will find two and three point invariants in the case of $U_q(\hat{sl_2}) $ by using Volkov's scheme. In the third section, we will introduce lattice Virasoro algebras as the space of invariants of Borel part $U_q(B_{+})$ of $U_q(g)$ for simple Lie algebra $g$ and will find the set of generators of Lattice Virasoro algebra connected to $sl_2$ and $U_q(sl_2)$. And as a new result, we found the set of some generators of lattice Virasoro algebra.

In the second section, we will introduce Local integral of motions as the space of invariants of nilpotent part of quantum affine Lie algebras and will find two and three point invariants in the case of Uq(ŝ l 2 ) by using Volkov's scheme. In the third section, we will introduce lattice Virasoro algebras as the space of invariants of Borel part Uq(B + ) of Uq(g) for simple Lie algebra g and will find the set of generators of Lattice Virasoro algebra connected to sl 2 and Uq(sl 2 ) And as a new result, we found the set of some generators of lattice Virasoro algebra.

Feigin's homomorphisms and quantum groups
In this section we will introduce Feigin's homomorphisms and will see that how they will help us to prove our main and fundamental theorems on screening operators. "Feigin's homomorphisms" was born in his new formulation on quantum Gelfand-Kirillov conjecture, which came on a public view at RIMS in 1992 for the nilpotent part U q (n), that are now known as "Feigin's Conjecture". In that mentioned talk, Feigin proposed the existence of a family of homomorphisms from a quantized enveloping algebra to rings of skew-polynomials. These "homomorphisms" are became very useful tools for to study the fraction field of quantized enveloping algebra. [6] Feigin's homomorphisms on U q (n). Here we will briefly try to show that what are Feigin's homomorphisms and how they will guide us to reach and to prove that the screening operators are satisfying in quantum Serre relations.
Set C as an arbitrary symmetrizable Cartan matrix of rank r, and n = n + the standard maximal nilpotent sub-algebra in the Kac-Moody algebra associated with C (thus, n is generated by the elements E 1 , ..., E r satisfying in the Serre relations). As always U q (n) is the quantized enveloping algebra of n. And A = (A ij ) = (d i c ij ) is the symmetric matrix corresponding to C for non-zero relatively prime integers d 1 , ..., d n such that d i a ij = d j a ji for all i, j. And set g as a Kac-Moody Lie algebra attached to A, on generators E i , F i , H i , 1 ≤ i ≤ n . [11] Let us to mention some of the structures related to g that we will use them here: the triangular decomposition g = n − ⊕ h ⊕ n + ; the dual space h * ; elements of h * will be referred to as weights; the root space decomposition n ± = ⊕ α∈∆± g α , g αi = CE i ; the root lattice Λ ∈ h * , {α 1 , · · · , α n } ⊂ ∆ + ⊂ h * being the set of simple roots; the invariant bilinear form Λ × Λ → Z defined by < α i , α j >= d i a ij . [11] Set A 1 and A 2 as a Λ− graded associative algebras and define a q− twisted tensor product as the algebra A 1⊗ A 2 isomorphic with A 1 ⊗ A 2 as a linear space with multiplication given by (a 1 ⊗ a 2 ) · (a ′ 1 ⊗ a ′ 2 ) := q <α ′ 1 ,α2> a 1 a ′ 1 ⊗ a 2 a ′ 2 , where α ′ 1 = deg(a ′ 1 ) and α 2 = deg(a 2 ). And by this definition A 1⊗ A 2 become a Λ− graded algebra. Proposition 1.1. Set g an arbitrary Kac-Moody algebra, then the map there is no such map as U ± q (g) → U ± q (g)⊗U ± q (g) in the case that g is an associative algebra. [9] And as always after defining a co-multiplication,∆, then we can extend it by a iteration as a sequence of maps [1] (1.4)∆ n : Now set C[X i ] as a ring of polynomials in one variable and by equipping it by grading structure degX i = α i for any simple root α i , we can regard it as a Λ− graded. By this grading there will be a morphism of Λ− graded associative algebras x i By following this construction for any sequence of simple roots β i1 , · · · , β i k , there will be a morphism of Λ− graded associative algebras (the cause of double indexation here is the appearance of i j s more than once in the sequence).
And finally, C[X 1i1 ]⊗ · · ·⊗C[X ki k ] is an algebra of skew polynomials C[X 1i1 , · · · , X ki k ], with Λ− grading X sis X tit = q <αi s ,αi t > X tit X sis , for s > t. But let us to simplify it as X i X j = q <degXi,degXj > X j X i ; the one that we will use it always.
So very briefly we constructed the already mentioned family of morphisms (Feigin's homomorphisms) from U − q (g) (the maximal nilpotent sub-algebra of a quantum group associated to an arbitrary Kac-Moody algebra) to the algebra of skew polynomials.
1.1. the contribution between Quantum Serre relations and screening operators. Theorem 1.7. Set Q = q 2 and points x 1 , · · · , x n such that x i x j = Qx i x j for i < j. And set Σ x = x 1 + · · · + x n . If Q N = 1 and x N i = 0 for some natural number N , then we claim that (Σ x ) N = 0 Proof. It's straightforward, just needs to use q-calculation.
Theorem 1.8. Suppose we have two different types of points x i , Namely, set (x 2i−1 ) i , that we will call them of type 1 and (x 2i ) i , that we will call them of type 2 for i ∈ I = {1, 2}, and the following q− commutative relations: Set Σ x 1 = Σ i∈I x 2i+1 and Σ x 2 = Σ i∈I x 2i . We will call them screening operators. Then we claim that Σ x 1 and Σ x 2 are satisfying on quantum Serre relations: Proof. It's straightforward, just needs to use q-calculation.
Theorem 1.10. Prove Theorem 1.2.1 in a general case, i.e. Set points X i ∈ {X 1 , · · · , X n } and Y i ∈ {Y 1 , · · · , Y n } with the following relations; We claim that Σ x 1 and Σ y 1 are satisfying in quantum Serre relations.
Proof. Proof by induction on k.
As we see in theorem 4.1.1, it's true for k = 2. Suppose that is true for k = n, we will prove that it's true for k = n + 1.
As we set it out, n is a nilpotent Lie algebra, so the Cartan sub-algebra of n is equal to n with Chevally generators Σ X 1 and Σ y 1 as they are satisfying in quantum Serre relations. So we can define U q (n) : be the quantum polynomial ring in one variable. We define: Here⊗ means quantum twisted tensor product. We define the embedding U q (n) → U q (n)⊗C q [X]: Σ X 1 → Σ x 1 + X ; Σ y 1 → Σ y 1 . Claim 1: (Σ x 1 + X) and Σ y 1 are satisfying on quantum Serre relations.
proof of claim 1: We will have the new operators Σ x 1 ′ = X 1 +· · ·+X n +X n+1 and Σ y 1 ′ = Y 1 +· · ·+Y n . Now define: And Define: proof of claim 2: lets do some part of this computation that maybe make confusion: And Σ And by substituting these, we have the result. So our definition is well defined. Now set Y = Y n+1 and we are done.
1.3. affinized Lie algebraŝl (2). As we know,M 2 = 2 −2 −2 2 is the generalized Cartan matrix forŝl (2). SetM q2 = q 2 q −2 q −2 q 2 and call it Cartan type matrix related toM 2 . sl(2) is satisfying in Theorems 1.2.1 and 1.2.2 as well; but what we need is just to change the quantum Serre relations in the following case: And to change the q− commutation relations also; according to our new Cartan type matrix But lets try to prove it in the case of Laurent skew q−polynomials C[X, X −1 ].
Proof. Proof by induction on k.
and as we checked out, it's straightforward to show that they are satisfying in quantum Serre relations (1.11). Suppose that it's true for k = n components x 1 , · · · , x n . Again as before we define:

Local integral of motions; Volkov's scheme
Set two screening operators satisfy.
In this section we will try to find a solution for this equation as Volkov planed. We call these kind of solutions as "Local integral of motions". In the sense of Feigin-Pugai [13], the main idea for to solve such kind of equations is to add "spectral parameter" β to k points screening operator and to define an analogue of R− matrix: Example U q (ŝl 2 ); two point invariants. Let us try to solve this equation for just two points x 1 and x 2 . In this case, our equations (2.3) will reduce to the following ones: There is a solution for these equation in [2], but we are interested on re finding them again here. For to do this, let us change the equations (2.4) to the following one, for simplicity. Set 2 ) The solutions of these equations are identical to the previous one, we did this change just because to find the solutions in these ones are easier than the previous one and less confusing. Then both of (2.5) will reduce to this linear difference equation: Let us do it for one of them (the first one) for to see the procedure (there is an identical approach for the second one): 1 and then distribute x 2 in it from the left and then bring it out to the right hand side, by using the q−commutation relations. The idea is to disappear x 2 from the both sides by multiplying the equation by x −1 2 from the right side. So we have Lets try to find R 1,2 (α 1 ; β): But we need some thing more, so lets continue; For to find its recursive sequence we have to pass the following steps: And then by comparing the coefficients in both side of the equation, we reach to the following key rule recursive sequence that we will use it for to find our final solution in the case of two points.
And now let us to set an general agreement for to simplify writing: Set (β) n := (1 − β)(1 − qβ)(1 − q 2 β) · · · (1 − q n−1 β) and let our summation be finite, i.e. set i ∈ {0, · · · , n} and R(α 1 ; β) = Σ n i=0 C i,0 α i 1 and in the next step we can extend our radius of convergence. Now lets try to find it: In infinity when i → +∞ we have q −i → 1; So we have: Example U q (ŝl 2 ); three point invariants. As what we had for previous example in two points; we will proceed the same steps for to find the solution of (2.2) for three points ; such that α i α j = qα j α i for i, j ∈ I as usual, and R(α 0 , α 1 ) = Σ n,m C n,m α n 0 α m 1 .
We are trying to solve the following difference equation subject to R; The process is exactly same as the previous one, so we will skip writing them here. For this equation We got the following recursive sequences, that will guide us to reach to our main solution for n, m = 1, ..., +∞ .
β−q −m C 0,m−1 And by considering the second sequence as our main key, and following it; we arrived to a nice and important sequence : That is compatible with (2.8) when m = 0. And this can show the correctness of our calculation.

Lattice Virasoro algebra
In this section we are interested on solutions Σ 1x of system of difference equation ] q = 0 that will be a generator of lattice Virasoro algebra. If we be able to find such kind of solution; then we can extend it to an another generator by a shift operator: Lattice Virasoro algebra connected to sl 2 . Here as always, we have the q−commutation relation X i X j = qX j X i , i < j between the points in sl 2 . Let us try to find three points invariants; this means to try to solve the following system of difference equation: One can find easily the trivial solutions of the second equation as follows: as we see, non of them have zero grading. So we should find another solution. By just keeping to look at them for a while, we can see that by multiplying these kind of solutions, one can find a zero grading expression, but it's not satisfying for these two ones. Again we note that for a solution; it's inverse is again a solution, so by this remark, we have two options here. We can inverse Σ 11x or Σ 12x and then multiply it with the other one. In both case we will have same set of generators except that in the first case (inverse of Σ 11x ), lattice Virasoro algebra is generated by elements of form Σ ix = X i X −1 i+1 X i+2 (X i + X i+1 + X i+2 ) −1 and in the second case (inverse of Σ 12x ), lattice Virasoro algebra is generated by elements of form . And by a simple fact that our space of working is closed under multiplication, so these new recently found objects can be a trivial solution for our system of difference equation. And then by shift operators (3.1), we will have the set of generators for our lattice Virasoro algebra connected to sl 2 .
Lattice Virasoro algebra connected to U q (sl 2 ). Set A = C[q]<xj ,xi> (xixj−qxj xi) the algebra of polynomials in variables q, x i over C for i ∈ I(our ordered index set), such that Our first project is to extend the usual binomial expansion to this algebra, for example we can see the shape of such expansion in a lower exponent 3: But for to prove it in a general case, we will use some techniques from combinatorics: Suppose x j and x i as above and set ω as a word formed by x j and x i . Then it's easy to see that any such kind of words can be permuted to x l j x k i along with a factor power of q, by using the q−commutation rule. For example, i , as the first x i should pass 5 x j 's and the second and third x i should pass 3 x j 's and forth and fifth x i should pass 1 x j and sixth will be stable. Now according to this fact , each word ω consist of k x i 's and n − k = l x j 's in (x i +x j ) n . That corresponds to a partition which lies inside an (n−k)×k rectangle. On the other hand each such partition corresponds to a unique word ω. Lets look at our example again; we have ω = x i x j x j x i x i x j x j x i x i x j x i , and the partition is 533110. If ω = q m x n−k j x k i , then as we see m is the sum of the parts of the partition. And the generating function for all partitions lie inside an (n − k) × k rectangle is the definition of the q− binomial coefficient n n−k q = n k q . So for a positive complex number n we will have But what will happen for the negative exponents? It's our next deal for to face. What we need to prove is to see what n k q will be when we replace n with −n? According to the definition q− binomial coefficient, we have . Now by replacing n with −n we will have: So as what we had for a positive exponent, we will have the result for negative exponent as follows: And it's identical to write the summation (3.2) from −∞ to 0 as follows: Formulation for to extend to four and more invariant points.
] will be zero and we will have to check the correctness of = 0. If it was true? then we will have a generator for lattice Virasoro algebra and by the shift operator, we will have all set of generators for it.
Let us define a Poisson bracket as Drinfeld defined and then compute some results by using that: {X j , X n i } := Lim q→1 adX j X n i 1−q and then we have: ad Xi X n i = 0, in both classical and quantum case.
ad Xi X n j = (1 − q n )X i X n j and ad Xj X n i = (1 − q n )X j X n i for i < j in quantum case. ad Xi X n j = 0 and ad Xj X n i = 0 for i < j in classical case. Now let us find the Poisson bracket for X 1 and X n i in classical case For example we have {X 1 , X 2 } = X 1 X 2 ∂ X2 By using this operators in the case (3.0.9) we see that this part when q → 1 will be zero and so after this time we just will deal with Σ Xi = Σ k i=1 + U + . And also we can find these relations in a more general case for a one variable function on X i as follows: And let us consider E i , F i , H i as the generators for U q (sl 2 ), then E i and H i will produce the Borel part B + ; One of the ways that we can act B + on the C[X i , X −1 i ] is as follows (3.9) π : is a simple root related to H i and P an arbitrary homogeneous element of C[X i , X −1 i ] Definition 3.11. Generators of lattice Virasoro algebra associated to simple Lie algebra g constitute the functional basis of space Inv Uq(B+ (C[X i , X −1 i ]). And for to find these generators we need to solve the following functional equations; [13] [Σ ix , Σ X ] = 0 and H i Σ ix = 0 ( * * ) Now the next question is that "how many variable X i is enough for to find a nontrivial solution for equations ( * * )?
The answer is if we deal with q in a general position, then one can expect that the dimension is dim(B + ) + 1. [13] So in the case of sl 2 it will be 3, the number of variables. [13] Now let us to go back to our example; So in general if we repeat the process for any X j<1 , we will have (3.12) {X According to the Poisson bracket and our early calculation, equations ( * * ) will have the following form As well as there is a same process with a minor differ for when we have j > 1 ; (3.14) {X And also , then by using the previous discussion we have ( . Now consider the following representation of (sl 2 ) q ; We need the highest weight vector of this representation that is the solution of equations (3.0.25). So we should have There is a solution for these equations in [13]. Here we use the same solution and procedure.
The Idea is to suppose existing of the local fields [13] (F x 1,k ) (0) , (F x 2,k+1 ) (0) , · · · (here (0) means that the degree is 0) and the modules created from these local fields as follows for degrees i and j. According to [13] we try to find the exchange algebra relations ]] (i − times) And then again will use the shift operator and will shift it once for to find another module as follows ]] (j − times) Let us to proceed as what we planned: as [13]. Then . Let us to call it (F x 1,2 ) ( 1 2 ) ; because it's degree is 1 2 and to find another module from it by using shift X 1 → X 3 and X 2 → X 4 as follows: And again in a same process we will have (3.20) Now let us multiply (3.19) and (3.20) (because we need zero degree) with each other for to see what will happen?
And again let us proceed as [13] and to find: And again by following [13], by adding (3.21) and (3.22) for to find an object with zero grading that can be a four point invariant generator of lattice Virasoro algebra But now let us follow precisely the notation from [13] for to not be confused that is a four point generator of lattice Virasoro algebra, but we are not looking for such kind of solutions, because to extend it to a general form is somehow difficult. So we are looking for a simple solution. Now let us to define another such kind of solutions as we experienced.
Let us calculate it; (3.28) Now let us calculate the other part We want to show that Σ X will commute with Σ = ( For to show it, one can easily check that the contributions of many entries vanishes. Namely, the elements X j with indexes j from minus infinity to 1 and from 5 to plus infinity definitely commute due to the rules mentioned above. And after that we can concentrate on the sum 2 )(x 2 + x 3 + x 4 ) for to do this job, let us divide the project to the following small projects. We must demonstrate that the following equations are satisfying.
The proof is identical to the proof for fractional exponent 1 2 . Just what you need is to set It is true for k = 3. Suppose that it is true for k − 1 component. Then for k component we have: ; So the rest will come from k = 3 and we are done.
And then by using the shift operators (3.1), we will have the set of all generators.
Results; Generators of lattice Virasoro algebra coming from 3 and 4dimensional representation of sl 2 .