Identities in upper triangular tropical matrix semigroups and the bicyclic monoid

Transcription

1 Identities in upper triangular tropical matrix semigroups and the bicyclic monoid Marianne Johnson (University of Manchester) Joint work with Laure Daviaud and Mark Kambites AAA94 + NSAC 2017, 17th June 2017

2 Semigroup identities for matrices? Let Σ be an alphabet, and consider two words w, v Σ +. Say that w = v is a semigroup identity for S if ϕ(w) = ϕ(v) for all semigroup morphisms ϕ : Σ + S. Does the semigroup M n (X) satisfy an identity?

3 Semigroup identities for matrices? Let Σ be an alphabet, and consider two words w, v Σ +. Say that w = v is a semigroup identity for S if ϕ(w) = ϕ(v) for all semigroup morphisms ϕ : Σ + S. Does the semigroup M n (X) satisfy an identity? Clearly depends upon X!

4 Semigroup identities for matrices? Let Σ be an alphabet, and consider two words w, v Σ +. Say that w = v is a semigroup identity for S if ϕ(w) = ϕ(v) for all semigroup morphisms ϕ : Σ + S. Does the semigroup M n (X) satisfy an identity? Clearly depends upon X! There can be LOTS of identities (i.e. no finite basis)... M. Sapir/Volkov, 1980s: M n (K) for n > 1, K a finite field is not finitely based.

5 Semigroup identities for matrices? Let Σ be an alphabet, and consider two words w, v Σ +. Say that w = v is a semigroup identity for S if ϕ(w) = ϕ(v) for all semigroup morphisms ϕ : Σ + S. Does the semigroup M n (X) satisfy an identity? Clearly depends upon X! There can be LOTS of identities (i.e. no finite basis)... M. Sapir/Volkov, 1980s: M n (K) for n > 1, K a finite field is not finitely based. Or none at all... Golubchik and Mikhalev, 1978: M n (K) for n > 1, K an infinite field does not satisfy a non-trivial semigroup identity.

6 The tropical semifield Let T denote the tropical semifield (R { },, ), where and denote two binary operations defined by: x y := max(x, y), x y := x + y. Addition is idempotent. x x = x. 0 is one : x 0 = 0 x = x. is zero : x = x = x, x = x =

7 The tropical semifield Let T denote the tropical semifield (R { },, ), where and denote two binary operations defined by: x y := max(x, y), x y := x + y. Addition is idempotent. x x = x. 0 is one : x 0 = 0 x = x. is zero : x = x = x, x = x = The semigroup M n (T) is the set of all n n tropical matrices, with multiplication defined in the obvious way: ( ) ( ) = ( )

8 The tropical semifield Let T denote the tropical semifield (R { },, ), where and denote two binary operations defined by: x y := max(x, y), x y := x + y. Addition is idempotent. x x = x. 0 is one : x 0 = 0 x = x. is zero : x = x = x, x = x = The semigroup M n (T) is the set of all n n tropical matrices, with multiplication defined in the obvious way: ( ) ( ) = ( We write UT n (T) to denote the subsemigroup of all upper triangular tropical matrices. )

9 Tropical matrix identities Does M n (T) satisfy a non-trivial semigroup identity? Yes, for n = 2, (Izhakian and Margolis, 2010). Yes, for n = 3 (Shitov, 2014). Open for n > 3.

10 Tropical matrix identities Does M n (T) satisfy a non-trivial semigroup identity? Yes, for n = 2, (Izhakian and Margolis, 2010). Yes, for n = 3 (Shitov, 2014). Open for n > 3. Finitely generated subsemigroups of M n (T) have polynomial growth (d Alesandro and Pasku, 2003).

11 Tropical matrix identities Does M n (T) satisfy a non-trivial semigroup identity? Yes, for n = 2, (Izhakian and Margolis, 2010). Yes, for n = 3 (Shitov, 2014). Open for n > 3. Finitely generated subsemigroups of M n (T) have polynomial growth (d Alesandro and Pasku, 2003). Why do we care? Tropical matrices can provide a tool for studying other interesting semigroups. Bicyclic monoid embeds into UT 2 (T) (Izhakian and Margolis, 2010).

12 Tropical matrix identities Does M n (T) satisfy a non-trivial semigroup identity? Yes, for n = 2, (Izhakian and Margolis, 2010). Yes, for n = 3 (Shitov, 2014). Open for n > 3. Finitely generated subsemigroups of M n (T) have polynomial growth (d Alesandro and Pasku, 2003). Why do we care? Tropical matrices can provide a tool for studying other interesting semigroups. Bicyclic monoid embeds into UT 2 (T) (Izhakian and Margolis, 2010). Plactic monoid of rank 3 embeds into a direct product of copies of UT 3 (T) (Izhakian, 2017; Cain, Klein, Kubat, Malheiro, Okniński, 2017).

13 Tropical matrix identities Does M n (T) satisfy a non-trivial semigroup identity? Yes, for n = 2, (Izhakian and Margolis, 2010). Yes, for n = 3 (Shitov, 2014). Open for n > 3. Finitely generated subsemigroups of M n (T) have polynomial growth (d Alesandro and Pasku, 2003). Why do we care? Tropical matrices can provide a tool for studying other interesting semigroups. Bicyclic monoid embeds into UT 2 (T) (Izhakian and Margolis, 2010). Plactic monoid of rank 3 embeds into a direct product of copies of UT 3 (T) (Izhakian, 2017; Cain, Klein, Kubat, Malheiro, Okniński, 2017). Free monogenic inverse monoid embeds into UT 3 (T).

14 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity?

15 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016)

16 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016) How can we tell if w = v is an identity for UT n (T)?

17 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016) How can we tell if w = v is an identity for UT n (T)? We provide a necessary and sufficient condition (in terms of certain tropical polynomials) for an identity w = v to hold in UT n (T).

18 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016) How can we tell if w = v is an identity for UT n (T)? We provide a necessary and sufficient condition (in terms of certain tropical polynomials) for an identity w = v to hold in UT n (T). This yields a polynomial time algorithm to verify whether a given identity is satisfied by UT n (T).

19 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016) How can we tell if w = v is an identity for UT n (T)? We provide a necessary and sufficient condition (in terms of certain tropical polynomials) for an identity w = v to hold in UT n (T). This yields a polynomial time algorithm to verify whether a given identity is satisfied by UT n (T). What is the variety generated by UT n (T)?

20 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016) How can we tell if w = v is an identity for UT n (T)? We provide a necessary and sufficient condition (in terms of certain tropical polynomials) for an identity w = v to hold in UT n (T). This yields a polynomial time algorithm to verify whether a given identity is satisfied by UT n (T). What is the variety generated by UT n (T)? For n = 2 we show that UT 2 (T) generates the same semigroup variety as the bicyclic monoid.

21 Upper triangular tropical matrix identities Does UT n (T) satisfy a non-trivial semigroup identity? Yes! A couple of different constructions known. (Izhakian 2014, Okniński 2015, Taylor 2016) How can we tell if w = v is an identity for UT n (T)? We provide a necessary and sufficient condition (in terms of certain tropical polynomials) for an identity w = v to hold in UT n (T). This yields a polynomial time algorithm to verify whether a given identity is satisfied by UT n (T). What is the variety generated by UT n (T)? For n = 2 we show that UT 2 (T) generates the same semigroup variety as the bicyclic monoid. Several variants of these two also generate the same variety.

22 The case n = 2: Tropical polynomials For w Σ, s Σ define a formal tropical polynomial in variables x t for t Σ by: f w s = w i =s t Σ x λw t (i 1) t, where λ w t (k) = #t s in the first k letters of w.

23 The case n = 2: Tropical polynomials For w Σ, s Σ define a formal tropical polynomial in variables x t for t Σ by: f w s = w i =s t Σ x λw t (i 1) t, where λ w t (k) = #t s in the first k letters of w. Theorem: Let w, v Σ +. The identity w = v holds on UT 2 (T) if and only if f w s (x) = f v s (x) for all s Σ and all x R Σ.

24 The case n = 2: Two letter identities In the case where w and v are words on a two letter alphabet, it suffices to compare four pairs of tropical polynomials in just one variable: Corollary: Let w, v Σ +, where Σ = {a, b}. The identity w = v holds on UT 2 (T) if and only if for all x R we have: f w a (x, 1) = f v a (x, 1), f w a (x, 1) = f v a (x, 1), fb w(x, 1) = f b v (x, 1), and fb w(x, 1) = f b v (x, 1).

25 The case n = 2: Example fs u = u i =s t Σ Let w := abba ab abba and v = abba ba abba. Then for example, x λu t (i 1) t fa w (x a, x b ) = x 0 a x 0 b x1 a x 2 b x2 a x 2 b x3 a x 3 b x4 a x 5 b max(0, x a + 2x b, 2x a + 2x b, 3x a + 3x b, 4x a + 5x b ) f w a (x, 1) = max(0, x + 2, 2x + 2, 3x + 3, 4x + 5) max(0, x + 2, 3x + 3, 4x + 5) fa v (x a, x b ) = x 0 a x 0 b x1 a x 2 b x2 a x 3 b x3 a x 3 b x4 a x 5 b max(0, x a + 2x b, 2x a + 3x b, 3x a + 3x b, 4x a + 5x b ) f v a (x, 1) = max(0, x + 2, 2x + 3, 3x + 3, 4x + 5) max(0, x + 2, 3x + 3, 4x + 5)

26 The case n = 2: Example fs u = u i =s t Σ Let w := abba ab abba and v = abba ba abba. Then for example, x λu t (i 1) t fa w (x a, x b ) = x 0 a x 0 b x1 a x 2 b x2 a x 2 b x3 a x 3 b x4 a x 5 b max(0, x a + 2x b, 2x a + 2x b, 3x a + 3x b, 4x a + 5x b ) f w a (x, 1) = max(0, x + 2, 2x + 2, 3x + 3, 4x + 5) max(0, x + 2, 3x + 3, 4x + 5) fa v (x a, x b ) = x 0 a x 0 b x1 a x 2 b x2 a x 3 b x3 a x 3 b x4 a x 5 b max(0, x a + 2x b, 2x a + 3x b, 3x a + 3x b, 4x a + 5x b ) f v a (x, 1) = max(0, x + 2, 2x + 3, 3x + 3, 4x + 5) max(0, x + 23x + 3, 4x + 5)

27 The case n = 2: Example fs u = u i =s t Σ Let w := abba ab abba and v = abba ba abba. Then for example: x λu t (i 1) t fa w (x a, x b ) = x 0 a x 0 b x1 a x 2 b x2 a x 2 b x3 a x 3 b x4 a x 5 b max(0, x a + 2x b, 2x a + 2x b, 3x a + 3x b, 4x a + 5x b ) f w a (x, 1) = max(0, x + 2, 2x + 2, 3x + 3, 4x + 5) max(0, x + 2, 3x + 3, 4x + 5) fa v (x a, x b ) = x 0 a x 0 b x1 a x 2 b x2 a x 3 b x3 a x 3 b x4 a x 5 b max(0, x a + 2x b, 2x a + 3x b, 3x a + 3x b, 4x a + 5x b ) f v a (x, 1) = max(0, x + 2, 2x + 3, 3x + 3, 4x + 5) max(0, x + 2, 3x + 3, 4x + 5)

28 Connection to the bicyclic monoid Izhakian and Margolis... Gave an embedding of the bicyclic monoid B into UT 2 (T). So, any identity satisfied by UT 2 (T) is also satisfied by B. Proved that UT 2 (T) satisfies Adjan s identity: abba ab abba = abba ba abba, thus exhibiting a minimal length identity for UT 2 (T). Asked: Do B and UT 2 (T) generate the same variety?

29 Connection to the bicyclic monoid Izhakian and Margolis... Gave an embedding of the bicyclic monoid B into UT 2 (T). So, any identity satisfied by UT 2 (T) is also satisfied by B. Proved that UT 2 (T) satisfies Adjan s identity: abba ab abba = abba ba abba, thus exhibiting a minimal length identity for UT 2 (T). Asked: Do B and UT 2 (T) generate the same variety? We use our tropical polynomials to answer this question...

30 Connection to the bicyclic monoid Do B and UT 2 (T) generate the same variety?

31 Connection to the bicyclic monoid Do B and UT 2 (T) generate the same variety? Some observations about these two(?) varieties... Checking whether an identity holds amounts to checking certain linear inequalities. For B: Pastijn, For UT 2 (T): DJK, 2017.

32 Connection to the bicyclic monoid Do B and UT 2 (T) generate the same variety? Some observations about these two(?) varieties... Checking whether an identity holds amounts to checking certain linear inequalities. For B: Pastijn, For UT 2 (T): DJK, The variety generated is not finitely based. For B: Shneerson, (B has infinite axiomatic rank.) For UT 2 (T): Chen, Hu, Luo, O. Sapir, 2016.

33 Connection to the bicyclic monoid Do B and UT 2 (T) generate the same variety? Some observations about these two(?) varieties... Checking whether an identity holds amounts to checking certain linear inequalities. For B: Pastijn, For UT 2 (T): DJK, The variety generated is not finitely based. For B: Shneerson, (B has infinite axiomatic rank.) For UT 2 (T): Chen, Hu, Luo, O. Sapir, Theorem: B and UT 2 (T) generate the same variety.

34 Connection to the bicyclic monoid Do B and UT 2 (T) generate the same variety? Some observations about these two(?) varieties... Checking whether an identity holds amounts to checking certain linear inequalities. For B: Pastijn, For UT 2 (T): DJK, The variety generated is not finitely based. For B: Shneerson, (B has infinite axiomatic rank.) For UT 2 (T): Chen, Hu, Luo, O. Sapir, Theorem: B and UT 2 (T) generate the same variety. We show that if w v in UT 2 (T), then there is a morphism from Σ + to the image of B in UT 2 (T) that falsifies the identity.

35 Variants Let T = R, Q, Z. Can also consider: T max := (T { },, ). Define the semigroup B T := T T via the product (a, b) (c, d) = (a b + max(b, c), d c + max(b, c)). For each T as above we have B = B N0 B T UT 2 (T max ) UT 2 (T). Since B and UT 2 (T) generate the same semigroup variety, it follows that each of the intermediate variants above must satisfy exactly the same semigroup identities as these two.

36 Generalisation Given a set Γ equipped with a reflexive and transitive relation R, define a semigroup: Γ(T) = {A T Γ Γ : A is bounded above and A i,j irj}, with product (A B)(i, j) = k Γ A(i, k) B(k, j). Examples to keep in mind... Γ = {1,..., n}, R the complete relation: M n (T). Γ = {1,..., n}, R the natural order: UT n (T). If Γ is a poset in with a global bound on the length of chains, then checking whether an identity holds in Γ(T) is easy...

37 Bounded chains and tropical polynomials Theorem: Γ a poset with maximum chain length n. w = v holds in Γ(T) if and only if f w u,ρ(x) = f v u,ρ(x) for all: words u Σ of length at most n 1; chains ρ = (ρ 0 ρ 1 ρ u ) in Γ; and values x R Σ Γ. Formal tropical polynomials in variables x(s, i), (s Σ, i Γ): f w u,ρ := s Σ u x(s, ρ k ) βw s (α k,α k+1 ), k=0 where the sum ranges over all sequences: 0 = α 0 < α 1 < < α u < α u +1 = w + 1 in which w αk = u k, and where β w s (α k, α k+1 ) = s strictly between w αk and w αk+1.

38 Bounded chains and tropical polynomials Corollary: Γ a poset with maximum chain length n. Γ(T) and UT n (T) generate the same variety. Γ(T) satisfies a non-trivial semigroup identity. The polynomials that must agree for w = v to hold in UT n (T) are, up to a change of variables, the same as those that must agree for w = v to hold in Γ(T). Since a non-trivial identity holds in each UT n (T), the result follows.

39 Example: The free monogenic inverse monoid with product I = {(i, j, k) Z 3 : i, j 0, j k i}. (i, j, k) (i, j, k ) = (max(i, i + k), max(j, j k), k + k ) Taking Γ = {1, 2, 3} with partial order 1, 2 3 can see that I Γ(T) UT 3 (T) k i (i, j, k) k j. 0 Follows that I satisfies all identities satisfied by UT 2 (T) (and hence B). Can show that I does not embed in UT 2 (T).

40 Further connections to the bicyclic monoid? Izhakian and Margolis, 2010: Proved that... UT 2 (T) satisfies Adjan s identity: M 2 (T) satisfies the identity: abba ab abba = abba ba abba. a 2 b 2 b 2 a 2 a 2 b 2 a 2 b 2 b 2 a 2 = a 2 b 2 b 2 a 2 b 2 a 2 a 2 b 2 b 2 a 2. Does squaring an identity for UT 2 (T) always yield an identity for M 2 (T)? No! E.g. ab 2 ab ab a 2 b = ab 2 a 2 b 2 a 2 b holds in UT 2 (T). But a 2 b 4 a 2 b 2 a 2 b 2 a 4 b 2 = a 2 b 4 a 4 b 4 a 4 b 2 doesn t hold in M 2 (T).

41 Tropical matrix identities are long Izhakian and Margolis, 2010: M 2 (T) satisfies the identity: a 2 b 2 b 2 a 2 a 2 b 2 a 2 b 2 b 2 a 2 = a 2 b 2 b 2 a 2 b 2 a 2 a 2 b 2 b 2 a 2. Does a shorter identity hold for M 2 (T)? DJ, 2017: Yes! Although not much shorter... List of necessary conditions for identities over a two letter alphabet; rules out all words of length < 17. Explicit calculation yields a 2 b 3 a 3 ba ba b 3 a 2 = a 2 b 3 ab ab a 3 b 3 a 2.

42 The case n = 2: How the polynomials arise Consider morphisms ϕ : Σ + UT 2 (T) of the form: ( xs x ( ) ϕ(s) = ) s, for all s Σ. 0

43 The case n = 2: How the polynomials arise Consider morphisms ϕ : Σ + UT 2 (T) of the form: ( xs x ( ) ϕ(s) = ) s, for all s Σ. 0 For all w Σ +, can see that: ϕ(w) 1,1 = t Σ w t x t, ϕ(w) 2,1 =, ϕ(w) 2,2 = 0

44 The case n = 2: How the polynomials arise Consider morphisms ϕ : Σ + UT 2 (T) of the form: ( xs x ( ) ϕ(s) = ) s, for all s Σ. 0 For all w Σ +, can see that: ϕ(w) 1,1 = w t x t, ϕ(w) 2,1 =, ϕ(w) 2,2 = 0 t Σ ( ( )) ϕ(w) 1,2 = max s Σ max w=usv u t x t + x s + 0 t Σ

45 The case n = 2: How the polynomials arise Consider morphisms ϕ : Σ + UT 2 (T) of the form: ( xs x ( ) ϕ(s) = ) s, for all s Σ. 0 For all w Σ +, can see that: ϕ(w) 1,1 = w t x t, ϕ(w) 2,1 =, ϕ(w) 2,2 = 0 t Σ ( ( )) ϕ(w) 1,2 = max s Σ max w=usv u t x t + x s + 0 t Σ ( )) = max s Σ (x s + max wi =s λ w t (i 1)x t t Σ where w t = the number of occurrences of the letter t in w, and λ w t (k) = #t s in the first k letters of w.