(JointprojectwithSilviaBoumova)December17,2020 VesselinDrensky FromaDiophantinetransportproblemfrom2016anditspossiblesolutionfrom1903toclassicalproblemsinalgebra LieJorOnlineSeminar:Algebras,representations,andapplicationsUniversidadedeS˜aoPaulo/Universit

LieJor Online Seminar: Algebras, representations, and applications Universidade de S˜ao Paulo/University of Campinas

From a Diophantine transport problem from 2016 and its possible solution from 1903

to classical problems in algebra

Vesselin Drensky

Institute of Mathematics and Informatics Bulgarian Academy of Sciences

drensky@math.bas.bg

(Joint project with Silvia Boumova)

December 17, 2020

Partially supported by the Bulgarian National Science Fund under Grant KP-06 N 32/1 of 07.12.2019

“Groups and Rings – Theory and Applications”.

This project is a result of the cooperation of a big international team Bulgarians in Bulgaria and abroad:

Silvia Boumova (University of Sofia and Institute of Mathematics and Informatics)

Georgi Genov (Institute of Mathematics and Informatics)

Rumen Dangovski (High School of Mathematics, Sofia, now in MIT) Vesselin Drensky (Institute of Mathematics and Informatics)

Elitza Hristova (Institute of Mathematics and Informatics).

Plamen Koev (San Jose State University, California) Plamen Koshlukov (State University of Campinas, Brazil)

Boyan Kostadinov (ex Ph.D. student – Technical University of Sofia)

Foreighn mathematicians:

Francesca Benanti (Palermo, Italy) Lucio Centrone (Bari, Italy)

M´aty´as Domokos (Budapest, Hungary) S¸ehmus Fındık (Adana, Turkey) Antonio Giambruno (Palermo, Italy) Chander Kanta Gupta (Winnipeg, Canada)

Ralf Holtkamp (Bochum, now in Hamburg, Germany) Roberto La Scala (Bari, Italy)

Angela Valenti (Palermo, Italy)

The motivation for this talk

The motivation for this talk was a recent Diophantine transport problem about how to transport profitably a group of persons or objects. We start with some examples and then we survey classical facts about solving systems of linear Diophantine equations and inequalities in nonnegative integers. We emphasize on the method of Elliott from 1903 and its further development by MacMahon in his “Ω-Calculus” or Partition Analysis.

This is only the tip of the iceberg

The main results are for solving problems in classical and noncommutative invariant theory and theory of algebras with polynomial identities.

The Diophantine transport problem In 2016 the following paper was posted at arXiv:1611.02627 Aureliano M. Robles-P´erez, Jos´e Carlos Rosales, Aequat. Math. 92 (2018), 661-670.

The following problem was the starting point of the paper:

A transport company is dedicated to carrying cars from the factory to the authorized dealer. For that, the company uses small and large trucks with a capacity of three and six cars, respectively. Moreover, we have the following restrictions.

The use of those trucks represents, for the company, a cost of 1200 and 1500 euros, respectively.

The company charges the dealer 300 euros for each transported car.

An additional car is added to the final load without charge to the customer.

If the company considers that a transport order is cost-effective when it has a profit of at least 900euros, how many cars must be transported in order to achieve that purpose?

Mathematization of the problem

Ifx andy are the numbers of tracks with a capacity of 3 and 6 cars, respectively, and nis the number of cars ordered by the dealer, then the problem is equivalent to the solution in nonnegative integers of the system

300n ≥ 1200x+ 1500y+ 900 n+ 1 ≤ 3x+ 6y

or, after an obvious simplification, to the solution of the system

n ≥ 4x+ 5y+ 3 n+ 1 ≤ 3x+ 6y

The authors consider a more general problem.

Let

a1x1+· · ·+apxp+α ≤n≤b1+· · ·+bpx−β

be a system of inequalities with nonnegative integer coefficients and letT be the set of all n∈N0 ={0,1,2, . . .} such that the system has at least one solution in N^p₀. Then the authors show thatthe setT ∪ {0}is a submonoid of the additive monoid (N0,+)and give algorithmic processes to compute T.

In another paper the authors consider a similar problem and offer a solution in a similar way.

Certain travel agency, which specializes in city tours for organized groups, uses small and large buses in its services. The small bus seating capacity is 30 passengers and a large bus is for up to 50 passengers. Moreover, the hire of each bus costs 310 euros, for the small ones, and 480 euros, for the large ones. Finally, the price of the city tour is 10 euros per passenger, but free for the responsible leader of the group. At above situation, we

propose the following problem: If a group of n+ 1persons (including the leader) wants to use its services, is the travel agency going to get profits?

Aureliano M. Robles-P´erez, Jos´e Carlos Rosales, Numerical semigroups in a problem about cost-effective transport, Forum Math. 29 (2017), No. 2,

Our purpose

We want to find all solutions in nonnegative integers

S ={s= (s₁, . . . , s_k)∈N^k0 |sis a solution of the system}

of a linear system of equations and inequalities with integer coefficients a₁₁x₁ +· · ·+ a_1kx_k + a₁ = 0

· · ·

a_l1x1 +· · ·+ a_lkx_k + a_l = 0 a_l+1,1x₁ +· · ·+ a_l+1,kx_k + a_l+1 ≥0

· · ·

am1x1 +· · ·+ a_mkx_k + am ≥0.

The leitmotif of this part of the talk is to apply a slight modification of the method as presented in the original paper by Elliott from 1903

E. B. Elliott, On linear homogeneous diophantine equations, Quart. J.

Pure Appl. Math., 34 (1903), 348-377, to show how to calculate the function

χ_S(t₁, . . . , t_k) =X

s∈S

t^s₁¹· · ·t^s_k^k,

which describes the solutions of the system, and to derive the parametric form of the solutions. Later we shall give concrete calculations for the example in the paper by Robles-P´erez and Rosales in Aequat. Math. Using similar methods one can handle also the example in the other paper by

A simpler problem

Chicken McNuggets were introduced by McDonald’s in 1979. They consist of small pieces of reconstituted boneless chicken meat that have been battered and deep fried.

Initially Chicken McNuggets were sold in packs of 6, 9, and 20.

Today in the United States they are sold in packs of 4, 6, 10, 20, 40, and 50 pieces. In Bulgaria the packs contain 6 and 9 pieces, and also 5 pieces if you buy Chicken McNuggets Meal (with medium Coca-Cola and medium French fries).

The Chicken McNugget problem Very soon the following problem became popular.

What numbers of Chicken McNuggets can be ordered using only packs with 6, 9, or 20 pieces?

An introduction to this problem and to several related questions can be found in

Scott T. Chapman, Chris O’Neill, Factoring in the Chicken McNugget Monoid, Mathematics Magazine, 91 (2018), No. 5, 323-336.

Solution of the Chicken McNugget problem

Using only packs with 6 and 9 pieces we can obtain all 3npieces starting from 6.

Since 20≡2 (mod 3), using one pack with 20 pieces and several packs with 6 and 9 pieces we can obtain all 20 + 3n,n≥2, pieces, i.e. all integers ≡2 (mod 3) starting from 26.

Similarly, combining two packs with 20 pieces with packs with 6 and 9 pieces we can obtain all integers≡1(mod 3) starting from 46.

Hence we can obtain all integers starting from 44.

McNugget numbers

The positive integers can be obtained by adding together orders of McDonald’s Chicken McNuggets in boxes of 6, 9, and 20 (prior to consuming any) are called McNugget numbers. Obviously, McNugget numbers nare the solutions in nonnegative integers of the equation

6a+ 9b+ 20c=n, n≥1.

The only positive integers which are not McNugget numbers are 1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 25, 28, 31, 34, 37, and 43.

This is the sequence A065003 (https://oeis.org/A065003) in the On-Line Encyclopedia of Integer Sequences founded by Sloane in 1964.

Even there is a video clip posted in Youtube for a man who drives from a McDrive to a McDrive trying to order 43 Chicken McNuggets:

How to order 43 Chicken McNuggets – Numberphile (2,729,831 views, Oct 9, 2012)

https://www.youtube.com/watch?v=vNTSugyS038

Numerical semigroups

Ifa1, . . . , a_k∈N, then the setS of all integers which can be presented in the form

n=a1z1+· · ·+akzk, z1, . . . , zk∈N0,

is a submonoid of (N0,+). The submonoidS is called a numerical semigroup if|N0\S|<∞.

Theorem. Ifa₁, . . . , a_k∈N, then they generate a numerical semigroup if and only if(a1, . . . , ak) = 1.

The largest integer which does not belong to a numerical semigroup S is called the Frobenius number of S.

For example, the Frobenius number of McNuggets numbers is equal to 43.

The Frobenius coin problem

The problem to find the Frobenius number of a numerical semigroup is stated in an attractive form as the Coin problem:

What is the largest monetary amount that cannot be obtained using only coins of specified denominations?

With only 2 pence and 5 pence coins, one cannot make 1 or 3 pence, but one can make any other integral amount.

Frobenius never stated the problem explicitly in a written text but there is some information that he mentioned it in his lectures.

J.L. Ram´ırez Alfons´ın, The Diophantine Frobenius problem, Oxford Lecture Series in Mathematics and its Applications 30, Oxford: Oxford University Press, 2005.

The Frobenius number is known explicitly only in the case of coins with two coprime nominalsaandb. Then the Frobenius number isab−(a+b).

Usually this result is attributed to Sylvester because in 1884 he published a close entertaining problem. There are also some publications which show that he knew the result.

Math. Questions from the Educational Times, 41 (1884), 21.

Computational complexity of the problem to compute the Frobenius number

In the general case we do not know a formula for the Frobenius number of thek-generated numerical semigroup S. Butfor a fixed kthere is an algorithm which computes the Frobenius number for polynomial time.

The input data are the logarithms of the generators a1, . . . , ak of S.

R. Kannan, Lattice translates of a polytope and the Frobenius problem, Combinatorica 12 (1992), No. 2, 161-177.

The P versus NP problem

In computational complexity theory, NP (nondeterministic polynomial time) is a complexity class used to classify decision problems. NP is the set of decision problems for which the problem instances, where the answer is “yes”, have proofs verifiable in polynomial time. (An equivalent

definition of NP is the set of decision problems solvable in polynomial time by a non-deterministic Turing machine.)

The general class of questions for which some algorithm can provide an answer in polynomial time is called “class P” or just “P”.

The P versus NP problem is a major unsolved problem in computer science. It asks whether every problem whose solution can be quickly verified (technically, verified in polynomial time) can also be solved quickly (again, in polynomial time).

This is one of the seven Millennium Prize Problems of the Clay Mathematics Institute for a US$1,000,000 prize for the first correct solution.

The Frobenius coin problem is NP-hard

The problem H is NP-hardif it is at least as hard as the hardest problems in NP. This means that if we have a hypothetical solution for H, then every problem Lin NP can be reduced in polynomial time to H. In other words, assuming that the solution for H takes 1 unit time, it can be used to solve Lin polynomial time.

Theorem. The knapsack problem can be reduced to the Frobenius problem in polynomial time.

J.L. Ram´ırez-Alfons´ın, Complexity of the Frobenius problem, Combinatorica 16 (1996), No. 1, 143-147.

The knapsack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.

Back to linear Diophantine systems Most of the next considerations can be found in

S. Boumova, V. Drensky, B. Kostadinov, A Diophantine transport problem from 2016 and its possible solution from 1903, arXiv:2003.06766v1

[math.NT], Math. and Education in Math., Proc. of the Forty-ninth Spring Conf. of the Union of Bulgar. Mathematicians, 2020, 89-113.

http://www.math.bas.bg/smb/2020 PK/tom/pdf/089-113.pdf In particular, one can find there detailed references.

The ideas of Euler, Gordan, and Hilbert seen from nowadays The following fact was known already by Euler in 1748:

L. Euler, Introductio in Analysin Infinitorum, Vol. I, Bousquet, Lausanne, M.-M. 1748.

L. Euler, Vollstaendige Anleitung zur Algebra. 2. Theil, Von Aufl¨osung algebraischer Gleichungen und der unbestimmten Analytic, Kays. Acad.

der Wissenschaften, St. Petersburg, 1770.

L. Euler, De partitione numerorum in partes tam numero quam specie dates, Noui Commentarii Academiae Scientiarum Imperialis

Petropolitanae, 14 (1769): I (1770) 168-187.

Lemma. Let all coefficients aij andbj of the system a11x1 +· · ·+ a1kxk= b1

· · ·

a_m1x₁ +· · ·+ a_mkx_k= b_m,

be nonnegative integers such that for each j= 1, . . . , kat least one of the coefficientsaij,i= 1, . . . , m, is different from 0. Then the number of solutions in nonnegative integers of the system is equal to the coefficient of t^b₁¹· · ·t^b_m^m of the expansion as a power series of the product

k

Y

j=1

1 1−t^a₁^1j· · ·t^am^mj

.

If we replace in the lemma the product

k

Y

j=1

1 1−t^a₁^1j· · ·t^am^mj

by

k

Y

j=1

1

1−zjt^a₁^1j· · ·t^am^mj

,

then the coefficient oft^b₁¹· · ·t^b_k^k will be a polynomial χ_S(z1, . . . , z_k) =X

s∈S

z₁^s1· · ·z_k^sk

in z₁, . . . , z_k, whereS⊂N^k₀ is the set of the solutionss= (s₁, . . . , s_k) of the system considered in the lemma.

Example. Given the system

x₁ + x₂ + x₃ = 10

x₁ + 2x₂ + 3x₃ = 15 we expand the product

1

(1−z₁t₁t₂)(1−z₂t₁t²₂)(1−z₃t₁t³₂)

as a power series and find that the coefficient oft¹⁰₁ t¹⁵₂ is equal to z₁⁷z2z₃²+z⁶₁z₂³z3+z₁⁵z⁵₂. Hence the system has three solutions

(s₁, s₂, s₃) = (7,1,2),(6,3,1),(5,5,0).

Back to the McNugget problem We change the problem:

What is the minimal number of packs if we want to buyn pieces of Chicken MgNuggets?

Solution. We consider the formal power series

f(x, t) = 1

(1−x⁶t)(1−x⁹t)(1−x²⁰t) =X

fnpxⁿt^p.

The coefficient f_np is equal to the number of ways we can presentn in the form n= 6a+ 9b+ 20c such thata+b+c=p.

f(x, t) = 1 +x⁶t+x⁹t+x¹²t²+x¹⁵t²+x¹⁸(t²+t³) +x²⁰t +x²¹t³+x²⁴(t³+t⁴) +x²⁶t²+x²⁷(t³+t⁴)

+x²⁹t²+x³⁰(t⁴+t⁵) +x³²t³+x³³(t⁴+t⁵) +x³⁵t³+x³⁶(t⁴+t⁵+t⁶) +x³⁸(t³+t⁴) +x³⁹(t⁵+t⁶)

+x⁴⁰t²+x⁴¹t⁴+x⁴²(t⁵+t⁶+t⁷) +x⁴⁴(t⁴+t⁵) +x⁴⁵(t⁵+t⁶+t⁷) +x⁴⁶t³+x⁴⁷(t⁴+t⁵) +x⁴⁸(t⁶+t⁷+t⁸) +x⁴⁹t³+x⁵⁰(t⁵+t⁶) +· · · For example, we can buy 47 pieces in 4 packs or 5 packs.

If we want to find which are the solutions of the system n= 6a+ 9b+ 20c, a+b+c=p we consider the formal power series

f(x, t1, t2, t3) = 1

(1−x⁶t1)(1−x⁹t2)(1−x²⁰t3) =X

xⁿt^a₁t^b₂t^c₃.

The problem is of polynomial complexity

If we want to find the solutions of the equation 6a+ 9b+ 20c=nfor which p=a+b+cis minimal, it is sufficient to consider the product

X

6a≤n

x^6at^a₁· X

9b≤n

x^9bt^b₂· X

20c≤n

x^20ct^c₃.

In the same way we can prove

Theorem. IfS is a numerical semigroup generated bya1, . . . , a_k then the presentation of the elements of S as a sum of minimal number of

generators can be found in polynomial time.

Homogeneous linear Diophantine systems with arbitrary integer coefficients Now we shall consider the system

a11x1 +· · ·+ a1kxk = 0

· · ·

a_m1x₁ +· · ·+ a_mkx_k = 0,

where the coefficients aij are arbitrary integers. As before, we shall be interested in solutions in nonnegative integers. Every solution of the system is a sum of minimal(or fundamentalorirreducible) solutions. The lattest cannot be presented as a sum of other solutions.

In 1873 Gordan proved the following theorem.

Theorem. Every homogeneous linear Diophantine systems with arbitrary integer coefficients has a finite number of minimal solutions.

P. Gordan, Ueber die Aufl¨osung linearer Gleichungen mit reellen Coefficienten, Math. Ann., 6 (1873), No. 1, 23-28.

The proof of Gordan is effective. If q = (q1, . . . , q_k) is a minimal solution, the proof gives an upper bound for the integers q_j,j= 1, . . . , k.

Homogeneous Diophantine linear systems from computational complexity point of view By the Hilbert Basis theorem every ideal of the polynomial algebra

Q[x1, . . . , xk]is finitely generated. The following property is known as the Dickson lemma with easy proof by induction. As it is mentioned by Dickson, it is a direct consequence of the Hilbert Basis theorem applied to monomial ideals in Q[x1, . . . , xk].

L.E. Dickson, Finiteness of the odd perfect and primitive abundant numbers with ndistinct prime factors, Amer. J. Math., 35 (1913), No. 4, 413-422.

Lemma. Let J be a subset of N^k₀. Then J has a finite subset {q⁽ⁱ⁾= (q₁⁽ⁱ⁾, . . . , q_k⁽ⁱ⁾)|i= 1, . . . , n}

with the property that for any q= (q₁, . . . , q_k)∈J there exists a q⁽ⁱ⁾ such that q_j⁽ⁱ⁾≤q_j,j = 1, . . . , k.

Restated in the language of monomials we have

Let J be a subset of the set [X_k]of commuting monomials in kvariables.

Then J has a finite subset

{u⁽ⁱ⁾ =x^q₁ⁱ¹· · ·x^q_k^ik |i= 1, . . . , n}

with the property that for any u=x^q1· · ·x^qk ∈J there exists a u⁽ⁱ⁾ which

This lemma was used by Gordan in 1899 in his proof of the Hilbert Basis theorem.

P. Gordan, Neuer Beweis des Hilbert’schen Satzes ¨uber homogene Functionen, G¨ott. Nachr., 1899, 240-242.

Clearly, as a corollary we immediately obtain also a nonconstructive proof of the theorem of Gordan for the finite generation of numerical semigroups.

Behavior of the minimal solutions in terms of recursion theory Roughly speaking, from the point of view of computability theory, a primitive recursive function can be computed by a computer program such that for every loop in the program the number of iterations can be

bounded from above before entering the loop.

(The formal definition is that primitive recursive functions are obtained from the basic functions 0, successor and projection by finitely many applications of substitution and induction.)

If we add to the basic functions the functions f₁, . . . , f_m, then the obtained by substitution and induction functions are primitive recursive in f1, . . . , fm.

Recursive functions solve problems which depend on solutions to smaller

Theorem. (Seidenberg) Given a functionf :N0 →N0 let us consider ascending chains of idealsA0(A1 (· · ·(As in Q[x1, . . . , x_k], whereAi

is generated by polynomials of degree≤f(i). Then there is an integer g_f^(k) depending only onf and ksuch that the length of any such chain is

≤g_f^(k). Moreover one could explicitly write down a formula forg_f^(k) in terms off andk.

A. Seidenberg, On the length of a Hilbert ascending chain, Proc. Amer.

Math. Soc., 29 (1971), 443-450.

This inspires a problem for homogeneous Diophantine linear systems Problem. Given a functionf :N→N, what can one say about the functionpf :N→Nsuch that pf(k) is the maximal integer with the property: There exists a a homogeneous linear Diophantine system in k variables with arbitrary integer coefficients which has a system of minimal solutions

I_f ={q⁽ⁱ⁾= (q⁽ⁱ⁾₁ , . . . , q_k⁽ⁱ⁾)|i= 1, . . . , p_f(k)} ⊂N^k0

such that q₁⁽ⁱ⁾+· · ·+q_k⁽ⁱ⁾≤f(k).

Results of Moreno Soc´ıas

Theorem. In the notation of the previous problem for every kthere is a primitive recursive function p_f in f, but there is no bound p_f which is primitive recursive inf in general.

G. Moreno Soc´ıas, Length of polynomial ascending chains and primitive recursiveness, Math. Scand., 71 (1992), No. 2, 181-205.

Corollary from the proof of the theorem. For everyk there is a

homogeneous linear Diophantine system with integer coefficients and a set of minimal solutionsI_k={q⁽ⁱ⁾_k = (q⁽ⁱ⁾_1k, . . . , q⁽ⁱ⁾_kk)}such that

q⁽ⁱ⁾_1k +· · ·+q⁽ⁱ⁾_kk ≤iand the number of solutions|I_k|is not bounded by a primitive recursive function ink.

The Ackermann function a(k, n) :N²₀→N0 is defined by a(0, n) =n+ 1, a(k+ 1,0) =a(k,1),

a(k+ 1, n+ 1) =a(k, a(k+ 1, n)).

It is known that a(k, n) is recursive and grows faster than any primitive recursive function.

W. Ackermann, Zum Hilbertschen Aufbau der reellen Zahlen, Math. Ann., 99 (1928), 118-133.

Bad behaviour of the minimal solutions

Theorem. (Moreno Soc´ıas) Letd∈N. Then there exists a homogeneous linear Diophantine system with integer coefficients andk unknowns such that its set of minimal solutions

I_d={q⁽ⁱ⁾= (q₁⁽ⁱ⁾, . . . , q_k⁽ⁱ⁾)|i=d, d+ 1, . . . , p} ⊂N^k0

has the property q⁽ⁱ⁾₁ +· · ·+q⁽ⁱ⁾_k =i,i=d, d+ 1, . . . , p, and pis equal to a(k, d−1)−1, wherea(k, n) is the Ackermann function.

G. Moreno Soc´ıas, An Ackermannian polynomial ideal, Applied Algebra, Algebraic Algorithms and Error-Correcting Codes, Proc. 9th Int. Symp., AAECC-9, New Orleans/LA (USA) 1991, Lect. Notes Comput. Sci., 539, Springer-Verlag, Berlin, 1991, 269-280.

From the homogeneous case

to the solution of general linear Diophantine constraints Now we consider an arbitrary linear system of equations and inequalities with integer coefficients

a11x1 +· · ·+ a_1kx_k + a1 = 0

· · ·

a_l1x₁ +· · ·+ a_lkx_k + a_l = 0 a_l+1,1x1 +· · ·+ a_l+1,kx_k + a_l+1 ≥0

· · ·

a_m1x₁ +· · ·+ a_mkx_k + a_m ≥0

and want to reduce its solution to the case of homogeneous equations.

We introduce new unknowns y, y_l+1, . . . , y_m and replace the system with the system

a₁₁x₁ +· · ·+ a_1kx_k + a₁y = 0

· · ·

al1x1 +· · ·+ alkxk + aly = 0 a_l+1,1x₁ +· · ·+ a_l+1,kx_k + a_l+1y − y_l+1 = 0

· · ·

am1x1 +· · ·+ amkxk + amy − ym = 0.

The following easy theorem shows how to reduce the solution of a general system to a homogeneous one.

Theorem. Let

{q⁽ⁱ⁾ = (r⁽ⁱ⁾₁ , . . . , r⁽ⁱ⁾_k , s⁽ⁱ⁾, s⁽ⁱ⁾_l+1, . . . , s⁽ⁱ⁾_m)|i= 1, . . . , n} ⊂N^k+1+m−l₀ be the set of minimal solutions of the homogeneous system of equations from the previous slide and lets⁽ⁱ⁾ = 1for i= 1, . . . , c,s⁽ⁱ⁾ = 0for i=c+ 1, . . . , dand s⁽ⁱ⁾>1 fori=d+ 1, . . . , n. Then the set of all solutions of the original system are of the form

q=q⁽ⁱ⁾+tc+1q⁽ⁱ⁾+· · ·+t_dq^(d), i= 1, . . . , c, tc+1, . . . , t_d∈N0.

Remark. Let some of the inequalities in the system , e.g.

am1x1+· · ·+a_mkx_k+am >0

be strict. Then we replace it in the associated homogeneous system by a_m1x₁+· · ·+a_mkx_k+ (a_m+ 1)y−y_m= 0.

The method of Elliott

There are many methods based on different ideas for solving systems of linear Diophantine equations and inequalities. We shall explain in detail the method of Elliott from 1903. Originally, it was developed for systems of homogeneous linear Diophantine equations. But for our applications we shall restate the method for systems of linear Diophantine inequalities.

We fix the system of Diophantine inequalities a₁₁x₁+· · ·+a_1kx_k+a₁ ≥0

· · ·

am1x1+· · ·+amkxk+am ≥0,

and consider the set of solutions of the system in nonnegative integers

The first idea of Elliott

A usual way to describe a setA⊂R^k is in terms of its characteristic (or indicator) function ch_fA :R^k → {0,1} defined by

chf_A(p1, . . . , pk) =

(1,(p₁, . . . , p_k)∈A 0,(p1, . . . , pk)∈/ A.

Let P ⊂N^k₀. By analogy with the characteristic function ofP we call the formal power series

χ_P(t₁, . . . , t_k) =X

p∈P

t^p₁¹· · ·t^p_k^k, p= (p₁, . . . , p_k), the characteristic series of P.

WhenP ⊂N^k₀ is the set of solutions of a system of linear Diophantine equations Elliott suggests to call χP(t1, . . . , t_k) the generating function of the set of solutions.

The second idea of Elliott

To find the characteristic series of a set of solutions S ⊂N^k₀ of a system of homogeneous linear Diophantine equations Elliott involves Laurent series.

Let P ⊂N^k₀, let f(t₁, . . . , t_k) =X

p∈P

α_pt^p₁¹· · ·t^p_k^k ∈C[[t₁, . . . , t_k]], p= (p₁, . . . , p_k), α_p∈C, be a formal power series, and let S be a subset of P. We call the formal power series

χ_S(f;t₁, . . . , t_k) =X

p∈S

α_pt^p₁¹· · ·t^p_k^k,

The next lemma is one of the key moments in the approach of Elliott. Its proof is obvious.

Lemma. Let P ⊂N^k₀, let f(t1, . . . , t_k) =X

p∈P

αpt^p₁¹· · ·t^p_k^k ∈C[[t1, . . . , t_k]], p= (p1, . . . , p_k), αp∈C, be a formal power series, and let S ⊂P be the set of solutions in P of the Diophantine equation

a₁x₁+· · ·+a_kx_k = 0, a_i ∈Z.

If the Laurent series

ξ_S(t₁, . . . , t_k, z) =f(t₁z^a1, . . . , t_kz^ak) =

∞

X

n=−∞

X

p∈P

α_pt^p₁¹· · ·t^p_k^kzⁿ,

p= (p1, . . . , pk),n=a1p1+· · ·+akpk, has the form ξS(t1, . . . , tk, z) =

∞

X

n=−∞

fn(t1, . . . , tk)zⁿ, fn(t1, . . . , tk)∈C[[t1, . . . , tk]], then

χ_S(f;t₁, . . . , t_k) =X

p∈S

α_pt^p₁¹· · ·t^p_k^k =f₀(t₁, . . . , t_k).

The next lemma is a slight generalization of the original approach of Elliott.

Lemma. Let P ⊂N^k₀, let f(t1, . . . , t_k) =X

p∈P

αpt^p₁¹· · ·t^p_k^k ∈C[[t1, . . . , t_k]], p= (p1, . . . , p_k), αp∈C, be a formal power series, and let S be the solutions in P of the

Diophantine inequality

a₁x₁+· · ·+a_kx_k+a≥0, a_i, a∈Z.

If

ξ_S(t₁, . . . , t_k, z) =z^af(t₁z^a1, . . . , t_kz^ak)

=

∞

X

n=−∞

X

p∈P

αpt^p₁¹· · ·t^p_k^kzⁿ=

∞

X

n=−∞

fn(t1, . . . , tk)zⁿ, f_n(t₁, . . . , t_k)∈C[[t₁, . . . , t_k]],n=a₁p₁+· · ·+a_kp_k+a, then

χ_S(f;t₁, . . . , t_k) = X

(p1,...,pk)∈S

α_pt^p₁¹· · ·t^p_k^k =

∞

X

n=0

f_n(t₁, . . . , t_k).

Now the problem is how to findχ_S(f;t₁, . . . , t_k) iff(t₁, . . . , t_k) is an explicitly given power series which converges to a rational function and we

Combining the results from

V. Drensky, C.K. Gupta, Constants of Weitzenb¨ock derivations and invariants of unipotent transformations acting on relatively free algebras, J. Algebra, 292 (2005), 393-428

with ideas from

F. Benanti, S.Boumova, V. Drensky, G. Genov, P. Koev, Computing with rational symmetric functions and applications to invariant theory and PI-algebras, Serdica Math. J., 38 (2012), 137-188

for the formal power series f(t₁, t₂) = t₁−t₂

1−(t₁+t₂) = (t₁−t₂)

∞

X

n=0

(t₁+t₂)ⁿ,

S ={(p₁, p2)∈N²0|p1≥p2}, we obtain χS(f;t1, t2) = 1−√

1−4t1t2

2t2−(1−√

1−4t1t2).

The third idea of Elliott The following definition is given by Berele.

A. Berele, Applications of Belov’s theorem to the cocharacter sequence of p.i. algebras, J. Algebra, 298 (2006), 208-214.

A nice rational function is a rational function with denominator which is a product of monomials of the form(1−t^α₁¹· · ·t^α_k^k).

Nice rational functions appear in many places of mathematics. For example, the Hilbert series of any finitely generated multigraded commutative algebra is of this form.

The main theorem of Elliott

The following theorem was proved by Elliott. The proof also gives an algorithm how to find the characteristic seriesχS(t1, . . . , tk) of the set S of solutions.

Theorem. Let

a11x1+· · ·+a_1kx_k= 0

· · ·

a_m1x₁+· · ·+a_mkx_k= 0,

where a_ij ∈Z,i= 1, . . . , m,j = 1, . . . , k, be a system of homogenous linear Diophantine equations. Then the characteristic seriesχ_S(t₁, . . . , t_k) of the setS of the solutions of the system inN^k0 is a nice rational function.

Idea of the proof. We start with the characteristic series of the setN^k₀ of all points with nonnegative integers coordinates

χ_Nk

0(t1, . . . , t_k) =

k

Y

i=1

1

1−t_i = X

pi≥0

t^p₁¹· · ·t^p_k^k. Let the first equation of the system be of the form

a1x1+· · ·+adxd−ce+1xe+1− · · · −ckxk= 0, ai>0, cj >0, d≤e, and let it have a set of solutions S. Then the characteristic series

χS(t1, . . . , tk) is equal to the coefficientf0(t1, . . . , tk) of the Laurent seriesξ_S=ξ_S(t₁, . . . , t_k, z)

=χ_Nk

0(t1z^a1, . . . , t_dz^ad, t_d+1, . . . , te, te+1z^−ce+1, . . . , t_kz^−ck)

Clearly, ξS has the form ξ_S =

d

Y

i=1 e

Y

j=d+1 k

Y

m=e+1

1

(1−tiz^ai)(1−tj)(1−tmz^−cm). If the Diophantine equation contains both both positiveai and negative cm coefficients, then we shall use the Elliott tricky equality

1

(1−Az^a)(1−Cz^−c) = 1 1−ACz^a−c

1

1−Az^a + 1

1−Cz^−c−1

, where a, c∈NandA, C are again monomials int1, . . . , tk. Applying the equality of Elliott to a pair of ai andcm, we shall replace the productξ_S by a sum of three similar products with numerators±1. Continuing with the application of the equality to each of the three expressions in several steps we shall obtain a sum of products with denominators containing factors which do not depend on zand factors with only positive or only negative degrees of z.

If a summand of ξS has the form

d

Y

i=1 k

Y

j=d+1

1

(1−Aiz^ai)(1−Bj) =

k

Y

j=d+1

1 1−Bj



1 +X

n≥1

D_nzⁿ



,

D_n∈Q[[t₁, . . . , t_k]], then it gives a contribution

k

Y

j=d+1

1 1−B_j to χ_S(t1, . . . , t_k). Similar arguments work in the case with negative degrees of z in the denominator of the summand. Finally, ξ_S(t₁, . . . , t_k) is a sum of the fractions which do not depend on z.

The following theorem is a modification of the theorem of Elliott for systems of Diophantine inequalities.

Theorem. Let aij, ai ∈Z,i= 1, . . . , m,j= 1, . . . , k, be arbitrary integers. Then the characteristic seriesχ_S(t₁, . . . , t_k) of the setS of the solutions in nonnegative integers of the system of Diophantine inequalities

a11x1+· · ·+a1kxk+a1 ≥0

· · ·

a_m1x₁+· · ·+a_mkx_k+a_m≥0 is a nice rational function.

The proof is based again on the method of Elliott. We start with the characteristic series

χ_Nk

0 =

k

Y

i=1

1 1−t_i, then form the function

ξ_S =χ

N^k0(t₁z^a1, . . . , t_dz^ad, t_d+1, . . . , t_e, t_e+1z^−ce+1, . . . , t_kz^−ck).

Using the equality of Elliott we present ξ_S as a sum of fractions

±Y 1

(1−Az^a)(1−B), a >0, and ±Y 1

(1−Cz^−c)(1−D), c >0, where A, B, C, Dare monomials in t₁, . . . , t_k. Then χ_S(t₁, . . . , t_k) is the sum of all fractions

The algorithm of Xin

The algorithm proposed by Xin uses the partial fraction decomposition of rational functions.

G. Xin, A fast algorithm for MacMahon’s partition analysis, Electron. J.

Comb., 11 (2004), No. 1, Research paper R58.

We shall illustrate the algorithm with a small example: The equation x₁+ 2x₂−3x₃= 0.

We start with χ_N³

0(t₁, t₂, t₃) = 1

(1−t₁)(1−t₂)(1−t₃), ξ_S=ξ_S(t1, t2, t3, z) =χ_N³

0(t1z, t2z², t3z⁻³) and decompose ξ_S as a sum of partial fractions with respect toz:

ξS = A

1−t₁z +B0+B1z

1−t₂z² +C0+C1z+C2z² z³−t₃ ,

A= t²₁

(t²₁−t2)(1−t³₁t3),

(1 +t t t + (t +t²t )z)t

We have to presentξ_S in the form ξs=

+∞

X

n=−∞

fn(t1, t2, t3)zⁿ

and to compute the coefficient f0. From the expansion of the third summand as a Laurent series

C0+C1z+C2z² z³−t3

= C0+C1z+C2z² z³(1−_z^t3₃) =

−1

X

n=−∞

fn(t1, t2, t3)zⁿ we conclude that this summand does not contribute to f0 and hence

χS(t1, t2, t3) =f0=A+B0 = 1 +t1t2t3+ (t1t2t3)² (1−t³₁t₃)(1−t³₂t²₃)

= 1−(t₁t₂t₃)³

(1−t₁t₂t₃)(1−t³₁t₃)(1−t³₂t²₃).

This easily implies that the minimal solutions of the equation x₁+ 2x₂−3x₃ = 0

are (1,1,1),(3,0,1),(0,3,2). If we want to make a system adding to the equationx1+ 2x2−3x3 = 0the equation

2x₁−3x₂+x₃= 0, applying a similar procedure to χS(t1z²,t2

z³, t3z) we obtain the characteristic series of the system

χ(t₁, t₂, t₃) = 1 1−t1t2t3

,

Our solution of the problem of Robles-P´erez and Rosales We shall illustrate the method of Elliott and the algorithm of Xin on the example of the Diophantine transport problem which was one of the two main motivations of the present project

n ≥ 4x+ 5y+ 3 n+ 1 ≤ 3x+ 6y

We want to make the first inequality homogeneous, rewrite it in the form

−4x−5y+n−3u≥0 and shall solve it for u= 1.

We start with ξ_N⁴

0(x₁, y₁, n₁, u₁) = 1

(1−x₁)(1−y₁)(1−n₁)(1−u₁), ξ_N⁴

0(x1

z⁴,y1

z⁵, n1z, u z³) =

+∞

X

k=−∞

hk(x1, y1, n1, u1)z^k and using the algorithm of Xin we obtain

h(x1, y1, n1, u1) =X

k≥0

hk(x1, y1, n1, u1)

= 1

.

Rewriting h(x1, y1, n1, u1) in the form h(x1, y1, n1, u1) =X

l≥1

p_l(x1, y1, n1)u^p₁

the solution of −4x−5y+n−3u≥0 foru= 1 corresponds to the coefficient

p1(x1, y1, n1) = n³₁

(1−x₁n⁴₁)(1−y₁n⁵₁)(1−n₁). This implies that the solutions of the first inequality are

(x, y, n) = (0,0,3) +a₁(1,0,4) +a₂(0,1,5) +a₃(0,0,1), a₁, a₂, a₃∈N0.

Now we replace the second inequality 3x+ 6y−n−1≥0 with 3x+ 6y−n−v≥0and working with

ξ = 1

1−v₁p1(x1, y1, n1)

we obtain the characteristic series of the solutions S forv = 1in the form χ_S(x₁, y₁, n₁) = y₁⁴n²³₁

(1−y₁n⁵₁)(1−y₁n⁶₁)(1−x₁y₁n⁹₁) and the solutions of our system are

(x, y, n) = (0,4,23) +b1(0,1,5) +b2(0,1,6) +b3(1,1,9), b1, b2, b3∈N0.

Applications to algebra

The details for the results (including their authorship) can be found in:

F. Benanti, S. Boumova, V. Drensky, G.K. Genov, P. Koev, Computing with rational symmetric functions and applications to invariant theory and PI-algebras, Serdica Math. J. 38 (2012), Nos 1-3, 137-188.

V. Drensky, E. Hristova, Invariants of symplectic and orthogonal groups acting onGL(n,C)-modules, arXiv:1707.05893v2 [math.AC].

V. Drensky, E. Hristova, Noncommutative invariant theory of symplectic and orthogonal groups, Linear Algebra and its Applications 581 (2019), 198-213.

E. Hristova, Hilbert series and invariants in exterior algebras, C.R. Acad.

Sci. Bulg., 73 (2020), No. 2, 153-162.

M. Domokos, V. Drensky, Constructive noncommutative invariant theory, Transformation Groups (to appear), arXiv:1811.06342 [math.RT].

Applications to PI-algebras

Till the end of the talk we assume that K is a field of characteristic 0. Let KhXi=Khx₁, x2, . . .i be the free associative algebra (with or without 1) and let

Pn=span{x_σ(1)· · ·x_σ(n) |σ∈Sn}

be the multilinear component of degree nin KhXi. (Here Sn is the symmetric group of degree n.) IfR is an (associative) PI-algebra with T-ideal T(R)⊂KhXi, then T(R)is generated by its multilinear elements

[

n≥1

(P_n∩T(R)). The vector space P_n∩T(R)is quite big and it is more convenient to study the multilinear polynomials modulo the polynomial identities of R:

Pn(R) =Pn/(Pn∩T(R)).

The symmetric group Sn acts from the left on Pn by

σ:x_i1· · ·x_in →x_σ(i1)· · ·x_σ(in), x_i1· · ·x_in ∈P_n, σ∈S_n, and as an S_n-module P_n∼=KS_n (the group algebra ofS_n as a left S_n-module). ThenP_n(R) is anS_n-factor module ofP_n.

Problem. For a given PI-algebraR describe theSn-module structure of Pn(R),n= 1,2, . . ..

Weaker version of the problem. Compute the codimensions ofR cn(R) = dim(Pn(R)), n= 1,2, . . . ,

The weakest and most promising version of the problem. Find the asymptotics of the codimension sequence of R.

We shall consider the first problem only.

The irreducible representations of S_nare described with partitions of n λ= (λ₁, . . . , λ_k)`n, λ₁ ≥ · · · ≥λ_k≥0, λ₁+· · ·+λ_k=n, with Sn-characterχ_λ. Then

χ_n(R) =χ_Sn(P_n(R)) =X

λ`n

m_λ(R)χ_λ

is the n-th cocharacter of Rwherem_λ(R)∈N0 is the multiplicity of χ_λ in χ_n(R).

Problem. Compute the cocharacter sequence χ_n(R),n= 1,2, . . ..

Cocharacters of PI-algebras

and Hilbert series of relatively free algebras

The general linear group acts canonically on the d-dimensional vector space with basis X_d={x₁, . . . , x_d} and this action is extended diagonally on the free d-generated associative algebraKhX_di=Khx₁, . . . , x_di:

g:f(x₁, . . . , x_d)→f(g(x₁), . . . , g(x_d)), g∈GL_d(K), f(X_d)∈KhX_di.

As in the case of the action ofSn onPn the relatively free algebra Fd(R) =KhX_di/(KhX_di ∩T(R))

is a GL_d(K)-factor module ofKhX_di. Then F_d(R) is a direct sum of irreducible polynomail GL (K)-modules.

The irreducible polynomial GL_d(K)-modules W_d(λ) are described by partitions λin ≤dparts. Then

Fd(R) =X

n≥0

X

λ`n

mλ(R)Wd(λ), λ= (λ1, . . . , λd).

By a theorem of Berele and Drensky in the early 1980s the multiplicities m_λ(R),λ= (λ₁, . . . , λ_d) are the same as the multiplicities in the cocharacter sequence of R.

Approach to the computing of the cocharacters of R. Compute the multiplicity series

M(Fd(R), t1, . . . , td) =X

λ

mλ(R)t^λ₁¹· · ·t^λ_d^d.

Since the base field is infinite, the relatively free algebraF_d(R) is N^d₀-graded with respect to the grading counting the degree of each variable in the monomials. Its Hilbert series is

H(F_d(R), t₁, . . . , t_d) = X

ni≥0

dim(F_d^(n1,...,nd)(R))tⁿ₁¹· · ·tⁿ_d^d

where F^(n1,...,nd)(R) is the homogeneous component of degree

The Hilbert series can also be expressed in terms of Schur functions:

H(F_d(R), t₁, . . . , t_d) =X

λ

m_λ(R)S_λ(t₁, . . . , t_d) where S_λ(t₁, . . . , t_d) is the Schur function corresponding to λ= (λ₁, . . . , λ_d).

Schur functions can be expressed as fractions of Vandermonde type determinants

S_λ(T_d) = V(λ+δ, T) V(δ, T) ,

where λ= (λ₁, . . . , λ_d),δ= (d−1, d−2, . . . ,2,1,0), and for µ= (µ₁, . . . , µ_d)

V(µ, T) =

t^µ₁¹ t^µ₂¹ · · · t^µ_d¹ t^µ₁² t^µ₂² · · · t^µ_d² ... ... . .. ...

.