Values of the length function for nonassociative algebras

We study realizable values of the length function for unital possibly nonassociative algebras of a given dimension. To do this we apply the method of characteristic sequences and establish sufficient conditions of realisability for a given value of length. The proposed conditions are based on binary decompositions of the value and algebraic constructions that allow to modify length function of an algebra. Additionally we provide a classification of unital algebras of maximal possible length in terms of their basis.


Introduction
Let F be an arbitrary field.In this paper A denotes a finite dimensional unital not necessarily associative F-algebra with the multiplication (•) usually denoted by the concatenation.Let S = {a 1 , . . ., a k } be a finite generating set of A. Any product of a finite number of elements from S is a word in S. The length of the word w, denoted l(w), equals to the number of letters in the corresponding product.We consider 1 as a word in S with the length 0.
The set of all words in S of lengths less than or equal to i is denoted by S i , here i ≥ 0. Note that similarly to the associative case, m < n implies that S m ⊆ S n .The set L i (S) = S i is the linear span of the set S i (the set of all finite linear combinations with coefficients belonging to F).We write L i instead of L i (S) if S is clear from the context.It should be noted that for unital algebras L 0 (S) = 1 = F for any S, and for non-unital algebras L 0 = ∅.We denote L(S) = ∞ i=0 L i (S).
Since the set S is generating for A, the equality A = L(S) holds.
Definition 1.1.The length of a generating set S of a finite-dimensional algebra A is defined as follows: l(S) = min{k ∈ Z + : L k (S) = A}.Definition 1.2.The length of an algebra A is l(A) = max{l(S) : L(S) = A}.
The problem of the associative algebra length computation was first discussed in [12,13] for the algebra of 3 × 3 matrices in the context of the mechanics of isotropic continua.
It is straightforward to see that the length of a unital associative algebra is strictly less than its dimension, and this bound is sharp.Namely, one-generated unital associative algebra of the dimension d has length d − 1.The first non-trivial result in this direction is going back to Paz [11].More results on the length of abstract associative algebras can be found, for example, in [6,9,10] and their bibliography.
In general most of the known results on the length function are just bounds that are not sharp.Even the sharp upper bound for the length of the matrix algebra is not known through these bounds are important in the number of applications, see [9] and references therein.In particular, the knowledge of the universal upper bound for the length provides the maximal size of products in generators that we need to consider to verify a certain property, for example, unitary similarity, [1].Moreover, a great deal of work has been done investigating the related notion of length for given generating sets of matrices as in Definition 1.1, see [5,7,8] and references therein.
The results on the lengths of nonassociative algebras were obtained in the works [2,3,4].The common tool to work without associativity was the method of characteristic sequences which are defined as follows.
3. Let for some r > 0, k > 1 the elements m 1 , . . ., mr be already defined and the sets L 1 (S), . . ., L k−1 (S) are considered.Then we inductively continue the process in the following way.Denote Such sequences are directly connected to the dimension of algebra and the length of the respective generating set.Combinatorial properties of these sequences, in turn, allow us to establish various facts about lengths.In particular, using characteristic sequences it is possible to provide a strict upper bound on length of a general nonassociative unital algebra.
The above bound is strict, see [4,Example 2.8].To compare associative and nonassociative cases let us observe that the maximal possible length of an associative algebra of dimension n is considerably smaller, namely it is equal to (n − 1).Moreover, in the present paper we show that for all n and any positive integer l ≤ n − 1 there are associative algebras of dimension n and length l, see Example 3.1.
Therefore the following question appears naturally.What values of length between 1 and 2 n−2 can be obtained as lengths of general nonassociative algebras of dimension n?
As it was shown in [2], the nonassociative situation is quite different and a gap in the set of values of length was discovered.In particular, already for the value l = 2 n−2 − 1 there are no algebras of length l.
In the present paper we investigate the set of realizable values for the length function on nonassociative algebras.Our main purpose is to provide an answer to the above question.As another application of the proposed method we characterize all algebras of the maximal length in terms of their basis.
Instead of the previously known combinatorial criteria based on characteristic sequences, which was nontrivial to check, see [2,Theorem 3.19], we provide an efficient numerical sufficient condition for feasibility of length values.Our condition is stated in terms of the binary decomposition of an integer under consideration and is easy to check.
In particular, we show that for algebras of dimension 2, 3, 4 all values of length are realizable, and for any dimension n ≥ 5 there are gaps of non-realizable values.If the value is bigger than a half of the maximal length value, then all the gaps are completely determined.Namely, we prove that only realizable values for k > 2 n−3 are of the form 2 n−3 + 2 q , and moreover, all integers of the type 2 h and 2 As a corollary a lower bound for the first non-realizable value is obtained.Namely, it is proved that all values between 1 and 2 ⌈ n 2 ⌉ are realizable for n ≥ 6.We also compute the total number of realizable values in each interval of the form [ Sufficient condition is provided for a given integer to be realizable as a value of length.Namely, if l < 2 n−k for some k such that the binary decomposition of l contains less than or equal to k entries 1, then l is realizable.An example is given that this condition is not necessary.A lower bound for the number of realizable values is obtained and several new gaps in the set of values of the length function are found.
Additionally, we expand upon connection of characteristic sequence to respective generating set.This is achieved by constructing a special basis W = {e 0 , . . ., e n−1 } of words corresponding to the characteristic sequence M = {m 0 , . . ., m n−1 } such that the multiplication low of the elements of W is defined by the addition in M .As an application of this technique we characterize nonassociative algebras of maximal length in terms of their basis.
Our paper is organized as follows.In Section 2 we provide previously established results and notions, relevant for further proofs.In Section 3 we introduce basic numerical results for possible length values examining proto-characteristic sequences.Section 4 covers the application of binary decomposition of a given value in checking its feasibility as a length of an algebra.Throughout Sections 5 and 6 we classify algebras of maximal length in terms of their basis.

Basic results and notions
To formulate our results let us start by introducing basic notions and tools in the nonassociative situation.Definition 2.1.[2, Definition 3.1] We call the sequence M = (m 0 , . . ., m n−1 ) proto-characteristic if it satisfies the following properties: 3. The sequence (m 0 , . . ., m n−1 ) is non-decreasing.

l(
Proposition 2.10.[4, Proposition 3.7] Let A be an F-algebra, dim A = n > 2. Assume that S is a generating set for A and (m 0 , m 1 , . . ., m n−1 ) is the characteristic sequence of S.
Then for each h satisfying m h ≥ 2 it holds that there are indices 0 < t 1 ≤ t 2 < h such that Proposition 2.11.[4, Theorem 3.8] Let A be an F-algebra, dim A = n, n > 2. Let also S be a generating set for A, (m 0 , m 1 , . . ., m n−1 ) be the characteristic sequence of S. Then for each positive integer h ≤ n − 1 it holds that m h ≤ 2 h−1 .

General observations
As it was already mentioned in the introduction, the length of a unital associative algebra is strictly less than its dimension and the corresponding bound is sharp.Moreover, we note that there are unital associative algebras of a given dimension n and any length between 1 and (n − 1) as the following example shows.
Example 3.1.Let us consider an arbitrary field F and an associative F-algebra A l with the basis {e 0 = 1, e 1 , . . ., e n−1 } and the multiplication determined by the following rule: It is straightforward to check that A l is associative and dim (A l ) = n.
Consider the generating set S 0 = {e 1 , e l+1 , e l+2 , . . ., e n−1 } of A l .Since e l = e l 1 is an irreducible word of the length l, we have l(A l ) ≥ l(S 0 ) = l.
Let S be an arbitrary generating set of A l satisfying l(S) = l(A l ).We consider the set S ′ , obtained from S by nullifying coefficients at e 0 in the expansions of elements of S via the basis {e 1 , . . ., e n−1 }.The equality L 1 (S) = L 1 (S ′ ) holds.Hence by Lemma 2.8 it follows that L k (S) = L k (S ′ ) for every positive integer k.In particular this allows to conclude that S ′ is a generating set and l(S) = l(S ′ ).However, we have l(S ′ ) ≤ l as a product of any l + 1 elements of S ′ is zero due to the fact that a product of any l + 1 non-unit basis elements e i 1 , . . ., e i l+1 is zero.Thus l(A l ) ≤ l, which implies l(A l ) = l.
Below A is again nonassociative algebra and we discuss some general properties of the values of its length.Proposition 3.2.Let A be a unital algebra over a field F such that dim A = n and l(A) = l.Then there exists a unital algebra A ′ such that dim A ′ = n + 1 and l(A ′ ) = l.
Proof.Let S be a generating set of A with maximal length, M = (m 0 , . . ., m n−1 ) be its generating sequence.By Lemma 2.7 we have m n−1 = l.
Note that by Proposition 2.2 the sequence M is proto-characteristic with certain functions t 1 and t 2 .
Consider the sequence . It is also protocharacteristic: Properties 1-3 of the Definition 2.1 can be straightforwardly checked, and for Property 4 we can define functions t ′ 1 (k) and t ′ 2 (k) as t 1 (k) + 1 and t 2 (k) + 1.To do this we observe that for all j satisfying m ′ j > 1 it holds that m ′ j = m j−1 .By Theorem 2.3 there exists a unital algebra of dimension n+1 and length m ′ n = m n−1 = l.Proposition 3.3.Let A be a unital algebra over a field F such that dim A = n and l(A) = l.Then there exists a unital algebra A ′ such that dim A ′ = n + 1 and l(A ′ ) = 2l.
Proof.Let S be a generating set of A with maximal length, M = (m 0 , . . ., m n−1 ) be its generating sequence.By Lemma 2.7 we have m n−1 = l.
Note that by Proposition 2.2 the sequence M is proto-characteristic with certain functions t 1 and t 2 .
Consider the sequence . It is also proto-characteristic: Properties 1-3 of the Definition 2.1 can be straightforwardly seen, and for Property 4 we can define functions t ′ 1 (k) and t ′ 2 (k) to be equal to t 1 (k) and t 2 (k), respectively, for k < n and t ′ 1 (n) = t ′ 2 (n) = n − 1.By Theorem 2.3 there exists a unital algebra of dimension n+1 and length m ′ n = 2m n−1 = 2l.Proposition 3.4.Let A be a unital algebra over a field F such that dim A = n and l(A) = l.Then there exists a unital algebra A ′ such that dim A ′ = n + 1 and l(A ′ ) = l + 1.
Proof.Let S be generating set of A with maximal length, M = (m 0 , . . ., m n−1 ) be its generating sequence.By Lemma 2.7 holds m n−1 = l.
Note that by Proposition 2.2 M satisfies Properties 1-4 of Definition 2.1 with certain functions t 1 and t 2 .
Consider the sequence . It is also proto-characteristic: Properties 1-3 of the Definition 2.1 can be seen easily, and for Property 4 we can define the functions t ′ 1 (k) and t ′ 2 (k) to be equal t 1 (k) and t 2 (k), respectively, for k < n and t Proof.Consider the following sequence (m 0 , . . ., m n−1 ) = (0, 1, . . ., 1).It falls under criteria of Theorem 2.3, thus the statement follows.
To construct a concrete example we consider the following algebra.
Example 3.6.For every natural n ≥ 2 let us consider an arbitrary field F and F-algebra A with the basis {e 0 = 1, e 1 , . . ., e n−1 } and the multiplication determined by the rule: for all p, q: 1 ≤ p, q ≤ n − 1 epeq = 0.
The length of this algebra is 1, as for all u, v ∈ A it holds that uv ∈ u, v , i.e. there can be no irreducible words of length 2 or greater.
Observe that there are no gaps in feasible length values for small dimensions, namely dim A ∈ {1, 2, 3, 4}.It is shown in [4] that for each n ≥ 5 the gaps in the set of values of the length function do exist.To investigate their structure we need the following technical statements.Proposition 3.8.Let n ≥ 2, h ≤ n − 2 be integers.Then there exists a unital F-algebra A with dim A = n and l(A) = 2 h .Proof.We will prove this statement using an induction on n.
The base.For n = 2 the only possible value of h is 0 and the statement holds by Proposition 3.5.
The step.Assume the statement holds for n = N − 1, N ≥ 3, and consider n = N .The case h = 0 holds by Proposition 3.5.For the case 1 ≤ h ≤ N − 2 by the induction hypothesis there exists an algebra of dimension N − 1 and length 2 h−1 as h − 1 is non-negative and h − 1 ≤ N − 3. Thus by Proposition 3.3 there exists an algebra of dimension N and length 2 • 2 h−1 = 2 h .Lemma 3.9.Let A be an F-algebra, dim A = n > 4. Assume S is a generating set for A and (m 0 , m 1 , ..., m n−1 ) is the characteristic sequence of S. Then for all h such that 2 If m h ≥ 2, we have m h = m i + m j for some i, j ≤ h − 1 by Proposition 2.10.Since the characteristic sequence is monotonically non-decreasing, we have Proposition 3.10.Let A be a unital F-algebra, dim A = n, n > 4, S be a generating set of A, (m 0 , m 1 , . . ., m n−1 ) be the characteristic sequence of S. Then

Moreover, in the case
for all 2 ≤ k ≤ h, then m h = 2 h−1 and Item 2 is straightforward.
Otherwise consider the first k, 2 ≤ k ≤ h such that m k = 2m k−1 , i.e. by Lemma 3.9 2. If m k ≥ 2 we have m h = m i + m j for some 1 ≤ i, j ≤ k − 1 by Proposition 2.10.Note that i, j cannot be equal to k − 1 simultaneously due to the choice of k.Thus 4 Binary decomposition

Gap bounds
In this subsection we further investigate the set of realizable values of the length function.In particular, we use here a new method to determine whether certain value is a feasible length of an algebra of dimension n.This method is based on binary decomposition of n.
The theorem below generalizes the results of Proposition 3.8.The converse statement is also true and provides a generalization for Proposition 3.10.
The base.For h = 4 according to Proposition 3.10, the value m 4 = 7 is impossible.Thus by Proposition 2.9, if m The step.If the statement holds for h = 4, . . ., H − 1, let us prove it for h = H ≥ 5. Assume the contrary, i.e. that there exists Q ∈ {0, . . ., H − 3} such that 2 There are the following two possibilities. 1.
x, thus the equality is impossible.
These contradictions conclude the proof.
Observe that Theorem 4.1 can be generalized even further.
Proposition 4.3.Let n and k be integers such that n > k > 1. Assume that l is a positive integer such that the following two conditions are satisfied 2. there are no more than k elements 1 in the binary decomposition of l.
Then there exists an algebra A of dimension n and length l.
Proof.We will prove this statement using a double induction on n and k.The step for k.Assume that we have proven the statement for k = 2, . . ., K − 1 where 2 < K < N .Consider l such that l < 2 N−K and there are no more than K 1s in its binary decomposition.If there are k 1 < K 1s, the statement follows from the induction hypothesis for k, since l < 2 N−K < 2 N−k 1 .If there are exactly K 1s (it should be noted that it means that K < N − 1 in this case, since if K = N − 1, then l would be equal to 1), then we have two possibilities again.
1. l is odd.Note that l − 1 has K − 1 elements 1 and 1) .By the induction hypothesis for n there exists an algebra of dimension N − 1 and length l − 1.Thus, by Proposition 3.4 there exists an algebra A of dimension N and length l.
2. l is even.Note that l 2 has K elements 1 and l 2 < 2 N −K 2 = 2 (N−1)−K .By the induction hypothesis for n there exists an algebra of dimension N − 1 and length l 2 .Thus, by Proposition 3.3 there exists an algebra A of dimension N and length l.
The following set of realizable values is also a consequence from Theorem 4.1.
Proposition 4.4.Let n > 4 be integer.Then for each p ∈ {2, . . ., n − 3} and q ≤ p there exists a unital F-algebra A with dim A = n and l(A) = 2 p + 2 q .Proof.By Theorem 4.1 there exists algebra A of dimension p + 3 and length 2 p + 2 q .If p + 3 < n, we can apply Proposition 3.2 n − p − 3 times and get an algebra of dimension n and the same length.Proposition 4.5.Let n, k be integers satisfying 1 < k < n.
1.There are at least 1 as we have n − k digits with the leading digit equal to 1 and we can place remaining h − 1 elements 1 into other n − k − 1 digits arbitrarily.Thus, by summing all binomial coefficients for all possible h we can infer that there are at least 1 Observe that the converse statement to Proposition 4.3, in the same sense as Theorem 4.2 is the 'converse' to Theorem 4.1, is not true.In particular, there exist algebras of length x which has k 1s in its binary decomposition, but x ≥ 2 n−k , and Proposition 4.5 provides only a lower bound on the number of feasible values in specified intervals.
We conclude this subsection by the list of all realizable values in the top half of the possible length values.They appear to be rather rare and more or less isolated.Namely, the corollary below demonstrates that all realizable values of length that are bigger than the half of the maximal value have the form 2 n−3 + 2 p , where 1 ≤ p ≤ n − 3.This implies that the all the intervals [  Proof.Consider a generating set S of A such that l(A) = l(S) and its characteristic sequence (m 0 , m 1 , . . ., m n−1 ).Since m n−1 = l(A) > 2 n−3 , by Theorem 4.2 we have m n−1 = 2 n−3 +2 p , p ∈ {0, . . ., n − 3}.

Bounds for the subsequent values of the length function
Now we may investigate the upper bound for the sequence of consequent values of lengths for algebras of a given dimension, which is the same as the lower bound for the first non-feasible value.
Definition 4.8.Let n ≥ 2 be a natural number.By B(n) we understand the maximal integer l > 0 such that there exist algebras of dimension n and lengths 1, . . ., l.
Proposition 4.9.We have Proof.The first three values follow from Proposition 3.7, since these are the maximal possible values of length for n ∈ {2, 3, 4}.For n = 5 note that 7 is not a feasible length value by Corollary 4.7 as 7 = 2 5−3 + 3 = 2 are also realizable since 1) l 1 < 2 h = 2 2h−1−(h−1) and 2) l 2 = 2 h − 1 is the only integer with more than h − 1 elements 1 in its binary decomposition satisfying l 2 < 2 h .To show that l 2 is a feasible length value we use the induction on h.
(as F ∩ S ′ = ∅, words of length 1 cannot be reduced), and, since in W 1 there are no elements which are words of length at least 2 in S ′ , Item 4 holds as well.The step.Assume W 1 , . . ., W K−1 are already constructed.Consider an irreducible word w of length K ≤ l(S).Note that w = w ′ • w ′′ , where w ′ has length j, 1 ≤ j < K, and w ′′ has length K − j , which is also less than K.As w ′ ∈ L j (S ′ ) = W j and w ′′ ∈ L K−j (S ′ ) = W K−j , we have w ∈ W j • W K−j .From this it follows that all of the irreducible words of length K belong to i=1,...,K−1 W i W K−i .Thus we can expand W K−1 with irreducible words belonging to W i W K−i for some indices i to obtain the basis of L K (S ′ ).We name the resulting set W K , as it satisfies Items 2-4: i) It is indeed a basis by construction.
ii) Only irreducible words of length K in S ′ were added, as any word of length less than K belongs to W K−1 already.
iii) Each added element is a product of two elements from W K−1 , which, by construction, is a subset of W K .Lemma 5.5.Let A be an F-algebra, dim (A) = n ≥ 2, S be a generating set of A. Let M = (m 0 , . . ., m n−1 ) be the characteristic sequence of S and W be an S-graded basis of A with associated set S ′ ⊂ S and sequence of sets W 0 ⊂ W 1 ⊂ . . .⊂ W l(S) = W . Then there exists a numbering e i , i = 0, . . ., n − 1, of the elements from W such that 1. e 0 = 1 2. For all r such that 1 ≤ r ≤ l(S) it holds that Wr = {e 0 , . . ., e s 1 +...+sr }, where sr = dim Lr(S) − dim L r−1 (S).Additionally, the length of e j as a word in S ′ is equal to m j for j = 0, . . ., s 1 + . . .+ sr.
3. Under this numbering for all r such that 1 ≤ r ≤ l(S) there exist functions t b.For all h 1 , h 2 such that s 1 < h 1 < h 2 ≤ s 1 +. ..+sr at least one of the inequalities t Proof.We will construct this numbering and respective functions using induction on r.The base.For r = 1 set e 0 = 1 and e 1 , . . ., es 1 to be elements of S ′ in an arbitrary order.This guarantees Item 1 and Items 2 and 3 for r = 1 with t with Item 3 holding by the induction hypothesis.
2. If s R > 0, then consider all the elements of W which are the words of length R in S ′ (note that all of them belong to W R by Item 3 of Lemma 5.4).Since by Item 4 of Lemma 5.4 they are equal to products of two shorter, already numbered elements of W , we can establish a correspondence between them and pairs of indices: a word w of length R would correspond to (j 1 , j 2 ), where e j 1 e j 2 = w.Since all the words in W are distinct, all the pairs in the resulting correspondence are also distinct.
By sorting these pairs in lexicographic order we can continue the numbering of elements of W .We name the one with the lexicographic minimal corresponding pair e s 1 +...+s R−1 +1 , the next e s 1 +...+s R−1 +2 and continue in this way.Item 2 will hold for r = R as there are exactly s R words of length R in W . Since W R−1 is a basis of L R−1 (S), W R is a basis of L R (S) and W R \ W R−1 consists of irreducible words of length R. For Item 3 we define t Note that there is only one way for (m 0 , . . ., m n−1 ) = (0, . . ., 2 n−2 ) to define functions t 1 , t 2 from the set {2, . . ., n − 1} to {1, . . ., n − 1} so they would satisfy a.For k such that 1 < k < n the equality m t 1 (k) + m t 2 (k) = m k and inequalities t 1 (k), t 2 (k) < k hold.
This means that by Proposition 5.6 applied to the generating set S there is a basis {e 0 , . . ., e n−1 } such that e 0 = 1, S = {e 1 } (as e 1 is a word of length 1 in singleton S), e i is a word of length m i = 2 i−1 for i = 1, . . ., n − 1 in S and e 2 j = e j+1 for j = 1, . . ., n − 2 as t 1 (j + 1) = t 2 (j + 1) = j.In particular, this basis is long.
Additionally, e i is an irreducible word.For i = 0, 1 this is evident.To prove this for i ≥ 2, note that the dimension of L 2 i−2 (S) is equal to the number of elements of the characteristic sequence less than or equal to 2 i−2 , i.e. i, which means that i linearly independent words e 0 , . . ., e i−1 which belong to L 2 i−2 (S) form its basis.Also by Lemma 2.6 there are no irreducible words of lengths 2 i−2 +1, . . ., 2 i−1 −1 as there are no such elements in the characteristic sequence of S. This allows us to conclude that L 2 i−2 (S) = L 2 i−1 −1 (S) = e 0 , . . ., e i−1 , from which follows e i ∈ L 2 i−1 −1 (S).
The following is immediate from the two propositions above.Theorem 6.5.A unital F-algebra A of dimension n > 2 has maximal length 2 n−2 if and only if it has a long basis E = {e 0 , e 1 , . . ., e n−1 } such that epeq ∈ e 0 , . . ., e max(p,q) for p = q.Corollary 6.6.For every unital F-algebra A with dim (A) = n > 2 and l(A) = 2 n−2 there exist elements f (j) p,q and f (0) , . . ., f (n−1) in F, p, q ∈ {1, . . ., n − 1}, p = q, j = 0, . . ., max(p, q) such that A is isomorphic to an algebra defined by generators x 0 , . . ., x n−1 and relations Proof.By the theorem above, A has a long basis E = {e 0 , e 1 , . . ., e n−1 } such that epeq ∈ e 0 , . . ., e max(p,q) for p = q.Using this basis as well as the fact that e 2 n−1 ∈ A we can find the elements f (j) p,q and f (0) , . . ., f (n−1) ∈ F such that epeq = max(p,q) j=0 f (j) p,q e j , e 2 n−1 = n−1 i=0 This allows to construct the desired isomorphism simply by mapping e i onto x i .
3 there exists a unital algebra of dimension n+1 and length m ′ n = 2m n−1 = l + 1.Let us demonstrate that for a given dimension n there indeed exist algebras with the minimal possible length values.An example of the algebra of the maximal possible value is given in [4, Example 2.8].Proposition 3.5.There exists a unital F-algebra of dimension n ≥ 2 and length 1.
The base for n.If n = 3 then the only possible values of k and l are 2 and 1 respectively and the statement is straightforward.The step for n.Assume the statement holds for n = 3, . . ., N − 1 with N > 3 and consider n = N .The base for k.If k = 2 then there are two possibilities: 1. l = 2 h , where 0 ≤ h ≤ N − 3. It is the statement of Proposition 3.8.2. There exist positive integers h 1 , h 2 such that h 1 < h 2 < N − 2 and l = 2 h 2 + 2 h 1 .If h 2 = N − 3, then the statement follows directly from Theorem 4.1.If h 2 < N − 3, then by Theorem 4.1 there exist an algebra of length l and dimension h 2 + 3. It follows by Proposition 3.2 that there is an algebra of length l and dimension N .
there are exactly 2 n−k−1 values in the interval.By Proposition 4.3 for a given h such that 0 ≤ h ≤ min(k − 1, n − k − 1) all of the values below 2 n−k with h + 1 elements 1 in binary decomposition are feasible.The number of such values in the interval [2 then the binomial sum is equal to 2 n−k−1 which means that all of the values in the interval are feasible.3.If k = 2, then by Theorems 4.1 and 4.2 only 1 + n − 3 = n − 2 values in the interval are feasible.
are gaps in the the set of values of the length function.
0 and S 1 are its finite subsets such that L 1 Proposition 3.7.For algebras of dimension n = 2, 3, 4 each value of length between 1 and 2 n−2 is attainable.Proof. 1.Let n = 2. Then the statement is straightforward as 2 n−2 = 1. 2. Let n = 3.Then the statement follows from Example 3.6 and [4, Example 2.8]. 3.In the case n = 4 the values that require justification are only 2 and 3, since 1 can be attained by Example 3.6 and 4 can be attained by [4, Example 2.8].