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Algebras of Linear Growth

the Dynamical Mordell�Lang Conjecture

Dmitri [email protected]

Higher School of Economics,

Moscow, Russia

Spa, 2017

Dmitri Piontkovski (HSE) Algebras of Linear Growth and the Dynamical Mordell�Lang ConjectureSpa, 2017 1 / 17

Graded algebras

We call an associative algebra A graded if

A = A0 ⊕A1 ⊕A2 ⊕ . . . ,

where A0 = k is a basic �eld, dim kAi <∞. All ouralgebras are graded.Hilbert function: hA(n) = dimAnHilbert series:HA(z) =

∑n≥0 z

ndimAn =∑

n≥0 znhA(n).

Graded algebras

A = A0 ⊕A1 ⊕A2 ⊕ . . . ,

∑n≥0 z

ndimAn =∑

n≥0 znhA(n).

Graded algebras

A = A0 ⊕A1 ⊕A2 ⊕ . . . ,

∑n≥0 z

ndimAn =∑

n≥0 znhA(n).

Graded algebras

A = A0 ⊕A1 ⊕A2 ⊕ . . . ,

∑n≥0 z

ndimAn =∑

n≥0 znhA(n).

Rationality

An algebra A is called �nitely presented if it is de�nedby a �nite number of generators and relations.

Theorem (Govorov, 1972)

If the relations of a �nitely presented algebra A are

monomials in generators then HA(z) is a rational

function.

Conjecture (Govorov)

For each �nitely presented algebra A the Hilbert series

HA(z) is a rational function.

First counterexamples: Ufnarovski, 1978 and Shearer,1980.Open questions: Govorov conjecture for Noetherianalgebras and for Koszul algebras.

Rationality

function.

Rationality

function.

Rationality

function.

Rationality

function.

Rationality in the linear growth caseAn algebra has linear growth, if GK�dimA ≤ 1, that is,for some c > 0 we have hA(n) = dimAn < c.

Example

Let A = 〈x, y|x2, yxy, xy2tx for all t ≥ 0〉. ThenAn = k{yn, xyn−1, yn−1x, xyn−2x} for n 6= 2t + 2 orAn = k{yn, xyn−1, yn−1x} otherwise.We have HA(z) = 1 + 2z + 4z2/(1− z)− z2

∑t≥0 z

Problem (Govorov conjecture for algebras of lineargrowth, GALG)

Suppose that an algebra A of linear growth is �nitely

presented. Is HA(z) a rational function?

For such algebras, HA(z) is rational i� hA(n) iseventually periodic, that is, ∃n0, T > 0 such thathA(n) = hA(n+ T ) for all n > n0.

Example

∑t≥0 z

Example

∑t≥0 z

Automaton algebras

Let X be a �nite generating set of an algebra A.Consider a multiplicative ordering `<' of the set of allwords in X. A word on X is called normal in A if it isnot a linear combination of less words. The set N of allnormal words is a linear basis of A.

De�nition (Ufnarovski)

An algebra A is called automaton if N is a regularlanguage.

Recall that a language is regular i� it is recognized by a�nite automaton.

Automaton algebras

Property (generalized Govorov theorem): If A is graded

automaton (that is, it is graded and automaton with

homogeneous X and a degree-compatible ordering `<'),

then HA(z) is a rational function.

Conjecture (Ufnarovski, 1990)

A �nitely presented algebra of linear growth is

automaton.

Conjecture (Ufnarovski conjecture for gradedalgebras, UGA)

A graded �nitely presented algebra of linear growth is

graded automaton.

UGA implies GALG.

Automaton algebras

automaton.

graded automaton.

UGA implies GALG.

Automaton algebras

automaton.

graded automaton.

UGA implies GALG.

Automaton algebras

automaton.

graded automaton.

UGA implies GALG.

The �nite characteristic case

Theorem

Suppose that the �eld k has a �nite characteristic. Then

both Govorov conjecture for algebra of linear growth and

Ufnarovski conjecture for graded algebras hold if and

only if k is an algebraic extension of its prime sub�eld.

`If' part: essentially, the case of �nite �eld.'Only if' part (counterexamples to GALG): later.

Theorem

The case of in�nite �eldWhat about the case char k = 0?

Example (Fermat algebras)

For α, β ∈ k×, let A = Aα,β be generated bya, b, c, x, y, z subject to 26 relations xc− αcx, yb− βcyand others. Then hA(n+ 3) is 10 or 11 according towhether the Fermat equality αn + βn = 1 holds. So, ithas no nonzero solution in k× for each n ≥ 3 if and onlyif hA(i) = 10 for all i ≥ 6 and each A = Aα,β .

Theorem

Let g ≥ 5 be an integer. If the �eld k is in�nite, then

there are in�nitely many (periodic) sequences hA for

g-generated quadratic k-algebras of linear growth. If If,in addition, k contains all primitive roots of unity, then

both the length d of the initial non-periodic segment

and the period T of hA can be arbitrary large.Dmitri Piontkovski (HSE) Algebras of Linear Growth and the Dynamical Mordell�Lang ConjectureSpa, 2017 8 / 17

Theorem

Skolem�Mahler�Lech theorem

Theorem (Skolem�Mahler�Lech)

If char k = 0 and an = c1an−1 + · · ·+ cdan−d is a linear

recurrence over k, then the zero set {n ≥ 0|an = 0} isthe �nite union of several arithmetic progressions and a

�nite set.

We refer to a set of nonnegative integers which is theunion of a �nite set and a �nite collection of arithmeticprogressions as SML.All known proofs essentially use p-adic analysis.

Example (Lech, 1953)

If the �eld k has characteristic p > 0 and t ∈ k istranscendental over the prime sub�eld Fp then thesequence an = (t+ 1)n − tn − 1 satis�es an = 0 i�n = pm with m ≥ 0.

�nite set.

The dynamical Mordell�Lang conjecture:

formulation

A challenging generalization:

Conjecture (The dynamical Mordell�Langconjecture)

Let V be a quasiprojective variety over k (of

characteristic zero), let Φ : V → V be any morphism,

and let α ∈ V. Then for each subvariety Y ⊂ V, the set

{n ≥ 0|Φn(α) ∈ Y } is SML.

See the book (which is unfortunately absent in ourlibrary...) :Jason P. Bell, Dragos Ghioca, Thomas J.Tucker, The Dynamical Mordell-lang Conjecture, AMS,2016 (Mathematical Surveys and Monographs, vol. 210)

some results

Some important cases are known.

Theorem (Bell, 2006)

Let char k = 0. The conjecture is true provided that Vis an a�ne variety and Φ is a polynomial automorphism.

Note that this corollary implies theSkolem�Mahler�Lech theorem.

Theorem

For any �eld k, GALG implies the linear dynamical

Mordell�Lang conjecture.

So, we have implicationsUGA=⇒GALG=⇒Linear ML=⇒ Skolem�Mahler�Lech.

some results

Theorem

some results

Theorem

some results

Theorem

Generalized dynamical Mordell�Lang

conjecture

Does dML implies UGA or (at least) GALG?Let A is generated and related in degrees at most D.Let V = mod-A≤D be the category of graded moduleswith generators and relations in degrees 0..D. ThenF : M 7→M≥1[1] is an endofunctor of V . For m > 0,let Vm = {M ∈ Ob V |dimM0 = m}.Then GALG is equivalent to the following categoricalversion of dML: Given a �nitely presented graded

algebra A of linear growth as above and a positive

integer m, is the set {n ≥ 0|Fn(A) ∈ Vm} always SML?

conjecture

Thank you!

Proof of the main theorem`Only if' k an algebraic extension of Fp =⇒ UGA =⇒GALGFor example, let us show: k is �nite =⇒ GALGProof. There are �nite number (say, N) of isomorphismclasses of M ∈ mod-A≤D such that dimM0 ≤ C(where dimAn ≤ C for all n ≥ 0). Then the sequencehA(n) = dimFn(A)0 is periodic with both period andnon-periodic segment of length at most N .`If'k contains Fp[t] =⇒ GALG fails.Using Lech example (an = (t+ 1)n − tn − 1): there is a6-generated algebra with generators a, b, c, x, y, z withrelations xc− (t+ 1)cx, yc− tcy, zc− cz and 23 others(the same as for Fermat algebras). Then

hA(n+ 3) =

{11, n = pm for some m ≥ 0,10, otherwise.

hA(n+ 3) =

Normal words in algebras of linear growthThe next Proposition follows from [Belov and others,1997].

Proposition

If a (non-graded) algebra A of linear growth is

generated by a �nite set S, then there are U, V,W ⊂ S∗

such that each normal word in A has the form

w = acnb, where a ∈ U, b ∈ V, c ∈W,n ≥ 0.

Proposition

Suppose a language L consists of the words of the

above form. Then L is regular i� for each

a ∈ U, b ∈ V, c ∈W the set

Na,b,c = {n ∈ Z+|acnb ∈ L}

is SML.Dmitri Piontkovski (HSE) Algebras of Linear Growth and the Dynamical Mordell�Lang ConjectureSpa, 2017 15 / 17

Proposition

Normal words in algebras of linear growth

Corollary

Suppose that the algebra A is automaton. Then there

are a generating set 1 ∈ S ⊂ A and an ordering such

that for some Q ⊂ S3 the set of normal words in A is

{acnb|n ≥ 0, (a, b, c) ∈ Q}.

Thank you very much!

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