On cogrowth function of algebras and its logarithmical gap

Let $A \cong k\langle X \rangle / I$ be an associative algebra. A finite word over alphabet $X$ is $I${\it-reducible} if its image in $A$ is a $k$-linear combination of length-lexicographically lesser words. An {\it obstruction} in a subword-minimal $I$-reducible word. A {\em cogrowth} function is number of obstructions of length $\le n$. We show that the cogrowth function of a finitely presented algebra is either bounded or at least logarithmical. We also show that an uniformly recurrent word has at least logarithmical cogrowth.


Cogrowth of associative algebras
Let A be a finitely generated associative algebra over a field k. Then A ∼ = k〈X 〉/I , where k〈X 〉 is a free algebra with generating set X = {x 1 , . . . , x s } and I is a two-sided ideal of relations. Further we assume the generating set is fixed. Let "≺" be a well-ordering of X , x 1 ≺ · · · ≺ x s . This order can be extended to a linear order on the set 〈X 〉 of monomials of k〈X 〉, i.e. finite words in alphabet X : u 1 ≺ u 2 if |u 1 | < |u 2 | or |u 1 | = |u 2 | and u 1 < l ex u 2 . Here | · | denotes the length of a word, i.e. the degree of a monomial, and < l ex is the lexicographical order. For f ∈ k〈X 〉 we denote by f its leading (with respect to ≺) monomial. An algebra k〈X 〉/I is said to be finitely presented if I is a finitely generated ideal.
We call a monomial w ∈ 〈X 〉 I -reducible if w = f for some relation f ∈ I . In the opposite case, we call w I -irreducible. Denote the set of all monomials of degree at most n by 〈X 〉 ≤n . Let A n ⊆ A be the image of 〈X 〉 ≤n under the canonical map. The growth V A (n) is the dimension of the linear span of A n . It is easily shown that V A (n) is equal to the number of I -irreducible monomials in 〈X 〉 ≤n .
We call a monomial w ∈ 〈X 〉 an obstruction for I if w is I -reducible, but any proper subword of w is I -irreducible. The cogrowth of algebra A is defined as the function O A (n), the number of obstructions of length n.
The celebrated Bergman gap theorem says that the growth function V A (n) is either constant, linear of no less than (n + 1)(n + 2)/2 [2]. In this section we give a non-trivial bound on the cogrowth function for finitely presented algebras.

Theorem 1. Let A be a finitely presented algebra. Then the cogrowth function O A (n) is either constant or no less than logarithmic: O
The constant C depends only on the maximal length of a relation.
Recall that a Gröbner basis of an ideal I is a subset G ⊆ I such that for any f ∈ I there exists g ∈ G such that the leading monomial of f contains the leading monomial of g as a subword. One of Gröbner bases can be obtained by taking for each obstruction u a relation f u ∈ I such that f u = u.
If f and g are two elements of k〈X 〉, g ∈ I and the word g is a subword of f , then f can be replaced by f such that f − f ∈ I and f ≺ f . This operation is called a reduction.
Let f and g be two elements of k〈X 〉. If u 1 u 2 = f and u 2 u 3 = g for some u 1 , u 2 , u 3 ∈ 〈X 〉, then the word u 1 u 2 u 3 is called a composition of f and g , and the normed element f u 3 − u 1 g is the result of this composition.

Lemma 2 (Diamond Lemma [3]).
Let two-sided ideal I be generated by a subset U of a free associative algebra k〈X 〉. Suppose that (i) there are no f , g ∈ U such that g is a proper subword of f , and (ii) for any two elements f , g ∈ U the result of any their composition can be reduced to 0 after finitely many reductions with elements from U . Let u, v ∈ S and let h be the result of some composition of f u and f v . It is clear that the leading monomial of h has length less then 2N . We start reducing h with elements from { f w | w ∈ S}. After finally many steps we obtain either 0 or an element h such that h does not contain subwords from S. But since there are no obstructions from [N , 2N ], the second case is impossible.
The word problem for a finitely presented algebra, i.e. the question whether a given element f ∈ k〈X 〉 lies in I , is undecidable in the general case. But if I has a finite Gröebner basis G, then A has a decidable word problem. Note also that the problem whether a given element in a finitely presented associative algebra is a zero divisor (or is it nilpotent) is undecidable, even if we are given a finite Gröebner basis [6]. But if the ideal of relations is generated by monomials and has a finite Gröebner basis, the nilpotency problem is algorithmically decidable [2].

Colength of a period
A monomial algebra is a finitely generated associative algebra whose defining relations are monomials. Let u be a finite word in alphabet X and let A u be the algebra k〈X 〉/I , where I is generated by the set of monomials that are not subwords of the periodic sequence u ∞ . Such algebras A u play important role in the study of monomial algebras [2].
Let W be a sequence on alphabet X , i.e. a map X N . A finite word v is an obstruction for W if v is not a subword of W but any proper subword v of v is a subword of W . If u is a finite word, the number of obstructions for u ∞ is always finite. We call this number the colength of the period u. We say that the period is defined by the set of obstructions.
In [5], G. R. Chelnokov proved that a sequence of minimal period n cannot be defined by fewer than log 2 n + 1 obstructions. G. R. Chelnokov also gave for infinitely many n i an example of a binary sequence with minimal period n i and colength of the period log ϕ n i , where ϕ = 5+1 2 . P. A. Lavrov found the precise lower estimation for colength of period.

Theorem 4 (cf. [7]). Let A = {a, b} be a binary alphabet. Let u be a word of length n and colength c,
The case of an arbitrary alphabet was considered in [8] by P. A. Lavrov and independently in [4] by I. I. Bogdanov and G. R. Chelnokov.

Cogrowth function for an uniformly recurrent sequence
A sequence of letters W on a finite alphabet is called uniformly recurrent (u.r. for brevity) if for any finite subword u of W there exists a number C (u,W ) such that any subword of W having length C (u,W ) contains u as a subword. This property can be considered as a generalization of periodicity [9].
For a sequence of letters W denote by A W the algebra k〈X 〉/I W , where I W is generated by the set of monomials that are not subwords of W . A monomial algebra A is called almost simple if each of its proper factor algebras B = A/I is nilpotent. In [2] it was shown that almost simple monomial algebras are algebras of the form A W , where W is an u.r. sequence.
Again, a finite word u is an obstruction for W if it is not a subword of W but any its proper subword is a subword of W . The cogrowth function O W (n) is the number of obstructions with length n.

Theorem 5. Let W be an u.r. non-periodic sequence on a binary alphabet. Then
A factorial language is a set U of finite words such that for any u ∈ U all subwords of u also belong to U . Denote by U k the words of U having length k. A finite word u is called an obstruction for U if u ∈ U , but any proper subword belongs to U . Denote the factorial language consisting of all subwords of a given sequence W by L (W ). To prove Theorem 5 we will assume the contrary and construct an infinite factorial language that is a proper subset of L (W ).
Let U be a factorial language and k be an integer. The Rauzy graph R k (U ) of order k is the directed graph with vertex set U k and edge set U k+1 . Two vertices u 1 and u 2 of R k (U ) are connected by an edge u 3 if and only if u 3 ∈ U , u 1 is a prefix of u 3 , and u 2 is a suffix of u 3 .
For a sequence W we denote the graph R k (L (W )) by R k (W ). Further the word graph will always mean a directed graph, the word path will always mean a directed path in a directed graph. The length |p| of a path p is the number of its vertices, i.e. the number of edges plus one. If a path p 2 starts at the end of a path p 1 , we denote their concatenation by p 1 p 2 . Recall that a directed graph is strongly connected if for every pair of vertices {v 1 , v 2 } it contains a directed path from v 1 to v 2 and a directed path from v 2 to v 1 . It is clear that any Rauzy graph of an u.r. non-periodic sequence is a strongly connected digraph and is not a cycle. Let U be a factorial language and let m n. A word a 1 . . . a m ∈ U m corresponds to a path of length m − n + 1 in R n (U ), this path visits vertices a 1 . . . a n , a 2 . . . a n+1 , . . . , a m−n+1 . . . a m . The graph R m (U ) can be considered as a subgraph of L m−n (R n (U )). Moreover, the graph R n+1 (U ) is obtained from L(R n (U )) by removing edges that correspond to obstructions of length n + 1.
We call a vertex v of a directed graph H a fork if v has out-degree more than one. Furthermore we assume that all forks have out-degrees exactly 2 (this is the case of a binary alphabet). For a directed graph H we define its entropy regulator: er (H ) is the minimal integer such that any directed path of length er(H ) in H contains at least one vertex that is a fork in H .

Proposition 6. Let H be a strongly connected digraph that is not a cycle. Then er(H ) < ∞.
Proof. Assume the contrary. Let n be the total number of vertices in H . Consider a path of length n + 1 in H that does not contain forks. Note that this path visits some vertex v at least twice. This means that starting from v it is possible to obtain only vertices of this cycle. Since the graph H is strongly connected, H coincides with this cycle.

Lemma 7. Let H be a strongly connected digraph, er(H ) = K . Then er(L(H )) = K .
Proof. The forks of the digraph L(H ) are edges in H that end at forks. Consider K vertices forming a path in L(H ). This path corresponds to a path of length K + 1 in H . Since er(H ) ≤ K , there exists an edge of this path that ends at a fork.

Lemma 8. Let H be a strongly connected digraph, er(H ) = K , let v be a fork in H , the edge e starts at v. Let the digraph H * be obtained from H by removing the edge e. Let G be a subgraph of H * that consists of all vertices and edges reachable from v. Then G is a strongly connected digraph. Also G is either a cycle of length at most K , or er(G) ≤ 2K .
Proof. First we prove that the digraph G is strongly connected. Let v be an arbitrary vertex of G, then there is a path in G from v to v . Consider a path p of minimum length from v to v in H . Such a path exists, for otherwise H is not strongly connected. The path p does not contain the edge e, for otherwise it could be shortened. This means that p connects v to v in the digraph G. From any vertex of G we can reach the vertex v, hence G is strongly connected.
Consider an arbitrary path p of length 2K in the digraph G, suppose that p does not have forks. Since er(H ) = K , then in p there are two vertices v 1 and v 2 which are forks in H and there are no forks in p between v 1 and v 2 . The out-degrees of all vertices except v coincide in H and G. If v 1 = v or v 2 = v, then we find a vertex of p that is a fork in G. If v 1 = v 2 = v, then there is a cycle C in G such that |C | ≤ K and C does not contain forks of G. Since G is a strongly connected graph, it coincides with this cycle C . Corollary 9. Let W be a binary u.r. non-periodic sequence, then for any n Proof. We prove this by induction on n. The base case n = 0 is obvious. Let er(R n−1 (W )) = K and suppose W has exactly a obstructions of length n + 1. These obstructions correspond to paths of length 2 in the graph R n−1 (W ), i.e. edges of the graph H := L(R n−1 (W )). From Lemma 7 we have that er(H ) = K . The graph R n (W ) is obtained from the graph H by removing some edges e 1 , e 2 , . . . , e a . Since W is a u.r. sequence, the digraphs H and H − {e 1 , e 2 , . . . , e a } are strongly connected. This means that the edges e 1 , . . . , e a start at different forks of H . We also know that R n (W ) is not a cycle. The graph R n (W ) can be obtained by removing edges e i from H one by one. Applying Lemma 8 a times, we show that er(R n (W )) ≤ 2 a K , which completes the proof. Proof. Consider in H the path p u of length k + 2, corresponding to u. Divide first k vertices of p u into three subpaths of length at least K . Since er(H ) = K , each of these subpaths contains a fork (some of these forks can coincide). Next, we consider three cases.

Case 1.
Assume that the path p u visits at least two different forks of H . Then p u contains a subpath of the form pe, where p is a path connecting two different forks v 1 and v 2 (and not containing other forks) and e is an edge starting at v 2 . It is clear that the length of p 1 does not exceed K + 1. Lemma 8 implies that there is a strongly connected subgraph G of H such that G contains the vertex v 2 but does not contain the edge e 2 .
If G is not a cycle, then er(G) ≤ 2K . Hence, the graph B := L k (G) is a subgraph of L k (H ), and from Lemma 7 we have er(B ) ≤ 2K . It is also clear that the digraph B does not contain the edge u.
If G is a cycle, we denote it by p 1 and denote its first edge by e 1 (we assume that v 2 is the first and last vertex of p 1 ). The length of p 1 does not exceed K . Among the vertices of p 1 there are no forks of H besides v 2 . Therefore, v 1 ∈ p 1 . Call a path t in H good, if t does not contain the subpath pe. Let us show that for any good path s in H there are two different paths s 1 and s 2 starting at the end of s such that |s 1 | = |s 2 | = 3K and the paths ss 1 , ss 2 are also good.
It is clear that for any good path we can add an edge such that the new path is also good. There is a path t 1 , |t 1 | < K such that st 1 is a good path and ends at some fork v. If v = v 2 , then two edges e i , e j start at v, the paths st 1 e i and st 2 e j are good, and each of them can be prolonged further to a good path of arbitrary length. If v = v 2 , then the paths st 1 p 1 e and st 1 p 1 e 1 are good and can be extended.
Consider in L k (H ) a subgraph that consists of all vertices and edges that are good paths in H , let B be a strongly connected component of this subgraph. It is clear that er(B ) ≤ 3K and the digraph B does not contain the edge u.

Case 2.
Assume that the path p u visits exactly one fork v 1 (at least 3 times), but there are forks besides v 1 in H . There are two edges e 1 and e 2 that start at v 1 . Starting with these edges and moving until forks, we obtain two paths p 1 and p 2 . The edge e 1 is the first edge of p 1 , the edge e 2 is the first of p 2 , and |p 1 |, |p 2 | ≤ K . We can assume that p 1 is a subpath of p u . Then p 1 ends at v 1 (and is a cycle) and p 2 ends at some fork v 2 = v 1 (if v 1 = v 2 , then v 1 is the only fork reachable from v 1 ). We complete the proof as in the previous case: p 1 e 1 is a subpath of p u . We call a path good if it does not contain p 1 e 1 . As above, we can show that if s is a good path in H , then there are two different paths s 1 and s 2 such that |s 1 | = |s 2 | = 3L and the paths ss 1 , ss 2 are also good.
As above, B will be a strongly connected component in the subgraph of L k (H ) that consists of vertices and edges corresponding to good paths in H .

Case 3.
Assume that there is only one fork v in H . Then there are two cycles p 1 and p 2 of length ≤ K that start and end at v. Let e 1 be the first edge of p 1 and let e 2 be the first edge of p 2 . The path p u contains one of the following subpaths: p 1 e 1 , p 2 e 2 , p 1 p 1 e 2 or p 2 p 2 e 1 . Denote this path by t . Call a path good if it does not contain t . A simple check shows that we can complete the proof as in the previous cases.
Proof of Theorem 5. Arrange all the obstructions u i of the u.r. binary sequence W by their length in non-descending order. If lim k→∞ log 3 |u k | k ≤ 1, then the statement of the Theorem holds. If lim k→∞ log 3 |u k | k > 1 then the sequence |u k |/3 k tends to infinity. Hence, there exists n 0 such that |u n 0 |/3 n 0 > 10 and |u n |/3 n > |u n 0 |/3 n 0 for all n > n 0 . In this situation, |u n 0 +k | > |u n 0 | + 4 · 2 n 0 · 3 k for any k > 0.
Let v i = u i if 1 ≤ i ≤ n 0 and let v i be a subword of u i of length |u n 0 | + 4 · 2 n 0 · 3 i −n 0 if i > n 0 . Denote by U the set of all finite binary words that do not contain subwords from {v i }. It is clear that U is a proper subset of L (W ). We get a contradiction with the uniform recurrence of W if we show that the language U is infinite. The Rauzy graph R u n 0 −1 (U ) is equal to R u n 0 −1 (W ), and from Corollary 9 we have er(R u n 0 −1 (L )) ≤ 2 n 0 .
By induction on n we show that for all n ≥ n 0 the graph R |v n |−1 (U ) contains a strongly connected subgraph H n such that er(H n ) ≤ 3 n−n 0 · 2 n 0 . We already have the base case n = n 0 . The graph R |v n+1 |−1 (U ) is obtained from L |v n+1 |−|v n | (R |v n |−1 ) by removing at most one edge. Note that |v n+1 | − |v n | > 3 · er(H n ), so we can use Lemma 10 for the digraph H n and k = |v n+1 | − |v n |. This completes the inductive step.
All the graphs R |v n |−1 (U ) are nonempty and, therefore, the language U is infinite.
For a sequence W over an alphabet A = {a 1 , . . . , a k } of size k, we replace in W each letter a i by 0 i 1 and obtain a binary sequence W . If W is u.r. and non-periodic, then W is also u.r. and non-periodic. It is clear that all long enough obstructions of W correspond to some of the obstructions of W , so we obtain Corollary 11. Let W be an u.r. non-periodic sequence on a finite alphabet. Then lim n→∞ O W (n)/log 3 n ≥ 1.
Example. Consider a finite alphabet {0, 1} and the sequence of words u i , defined recursively as u 0 = 0, u 1 = 01, u k = u k−1 u k−2 for k ≥ 2. Since u i is a prefix of u i +1 , the sequence (u i ) has a limit, called a Fibonacci word F = 0100101001001 . . . . In Example 25 of [1] the set {11, 000, 10101, 00100100, . . .} of obstructions of F is described. These words have lengths equal to Fibonacci numbers. Since the Fibonacci word is u.r., in Theorem 5 we cannot replace the constant 3 by a number smaller than 5+1 2 .