On semisimple twisted finite category algebras

Michał Stolorz, University of Aberdeen

June 2011

Abstract We will show that a twisted category algebra k_{α}C, where C is a finite category in which all idempotent morphisms in C commute is semisimple if and only if C is an inverse category. In addition, we present a method for explicitly obtaining nilpotent elements of twisted periodic semigroup algebras of non-inverse periodic semigroups.

A semigroup S is called an inverse semigroup if for every element s in S exists the unique element ŝ ∈ S such that s ∘ ŝ ∘ s = s and ŝ ∘ s ∘ ŝ = ŝ. This element will be called the quasi-inverse of s. Similarly we say that a category C is inverse if for every morphism φ in C there exists a unique morphism φ̂ such that φ̂ ∘ φ ∘ φ̂ = φ̂ and φ ∘ φ̂ ∘ φ = φ.

Theorem 1. Let k be a field of characteristic zero and k_{α}C be a twisted category algebra where C is a finite category such that for all objects X ∈ C any two idempotents in End_{C}(X) commute. Then the following are equivalent:

(i) The algebra k_{α}C is semisimple;

(ii) The category C is inverse.

Before proving the theorem, we collect some basic facts. Let C be a finite category, k a commutative ring and α a 2-cocycle map, that is α is a k^{×} valued map defined for any two morphisms such that s∘t is defined, moreover if t∘u is also defined, then α satisfies α(s, t∘u)α(t, u) = α(s∘t, u)α(s, t). Then the twisted category algebra k_{α}C is an algebra with basis Mor(C) and multiplication defined on the basis elements by s • t = α(s, t)s ∘ t if the composition s ∘ t is defined in C and s • t = 0 otherwise. Note that for two elements x, y ∈ C it is clearly indicated whether we mean the composition in C or multiplication in k_{α}C, the first one is denoted by x ∘ y and the latter by x • y. To avoid possible confusion when taking power, we will write x^{n} in the case of composition in C and x^{•n} when we mean the multiplication in k_{α}C.

Lemma 2. Let C be a category, A an abelian group, e an idempotent in C, α: S ×S → A a 2-cocycle map and x, y any morphisms in C such that e∘x = x and y ∘e = y. Then α(e, x) = α(e, e) = α(y, e).

Proof. By applying the 2-cocycle property we get: α(e, x) = α(e, e ∘ x)α(e, x)α(e, x)^{−1} = α(e ∘ e, x)α(e, e)α(e, x)^{−1} = α(e, e) α(y, e) = α(y ∘ e, e)α(y, e)α(y, e)^{−1} = α(y, e ∘ e)α(e, e)α(y, e)^{−1} = α(e, e)

A morphism φ: X → Y in a category C is called split if there exists a morphism ψ: Y → X such that φ = φ ∘ ψ ∘ φ. We can also assume that ψ and φ satisfy the following property ψ = ψ ∘ φ ∘ ψ (we can always take ψ ∘ φ ∘ ψ instead of ψ).

Lemma 3. Let A be a semisimple finite dimensional algebra over a field k. Then every morphism in the category mod(A) of finite dimensional A-modules is split.

Lemma 4. Let k be a field, C a finite category and α: C × C → k^{×} be a 2-cocycle map. If k_{α}C is semisimple then every morphism in C is split.

Proof. Let φ: X → Y be a morphism in C. Let φ̄: k_{α}C • id_{X} → k_{α}C • id_{Y} be a k_{α}C-module homomorphism given by the right composition with φ. Since mod(k_{α}C) is semisimple, then there is a k_{α}C-homomorphism ψ̄: k_{α}C • id_{Y} → k_{α}C • id_{X} such that φ̄ = φ̄ψ̄φ̄. Take ψ = ∑_{σ∈Hom(Y,X)} a_{σ}σ in k_{α}Hom_{C}(Y, X) such that ψ̄ is induced by right multiplication with ψ. Then x•φ = ∑_{σ} a_{σ}x•φ•σ•φ for all x ∈ k_{α}C, in particular for x = id_{X}:

α(id_{X}, φ)φ = ∑_{σ} a_{σ}id_{X} • φ • σ • φ

Hence: φ = ∑_{σ} b_{σ}φ ∘ σ ∘ φ where b_{σ} = a_{σ}α(φ, σ)α(φ ∘ σ, φ). This means there is some σ ∈ Hom_{C}(Y, X) such that b_{σ} ≠ 0 and φ = φ ∘ σ ∘ φ as required.

Proof of Theorem 1. (ii)⇒(i) Because C is a finite inverse category, kC is Morita equivalent to ∏_{e} kC_{e} where C_{e} are finite groups. For every C_{e}, the group H^{2}(C_{e}, k^{×}) is finite and every element has an order which divides |C_{e}|. Take any element ᾱ ∈ H^{2}(C_{e}, k^{×}) and let α be a representative of ᾱ in Z^{2}(C_{e}, k^{×}) such that α(x, y)^{|C_{e}|} = 1 for all x, y ∈ C_{e}. Let A_{e} be the group of all |C_{e}|-th roots of unity in k. Then α ∈ Z^{2}(C_{e}, k^{×}). Let N be the least common multiple of all |C_{e}|, the group A of all N-th roots of unity in k is finite and every A_{e} is a subset of A. Hence by [3, 4.3], for ᾱ ∈ H^{2}(C, k^{×}) there is a representative α such that α ∈ Z^{2}(C, A). Consider now the extension category Ĉ of the category C by the group A:

Ob(Ĉ) = Ob(C)  and  Hom_{C}(X, Y) = Hom_{C}(X, Y) × A

For any two morphisms (φ, a) ∈ HomĈ(X, Y) and (ψ, b) ∈ HomĈ(Y, Z) the composition is defined in the following way:

(ψ, a) ∘ (φ, b) = (ψ ∘ φ, baα(ψ, φ)) ∈ HomĈ(X, Z).

Then Ĉ is a finite inverse category therefore kĈ is semisimple [3] and hence k_{α}C is also semisimple. (i)⇒(ii) Suppose k_{α}C is semisimple. Let φ: X → Y be a morphism in C. Then by lemma 4 φ is split and there is some morphism ψ: Y → X such that φ ∘ ψ ∘ φ = φ and ψ ∘ φ ∘ ψ = ψ. Suppose that τ: Y → X is some other element which also satisfies φ ∘ τ ∘ φ = φ and τ ∘ φ ∘ τ = τ. Then φ ∘ ψ, φ ∘ τ ∈ End_{C}(X) and ψ ∘ φ, τ ∘ φ ∈ End_{C}(Y) so they must commute. Then:

φ ∘ τ = φ ∘ ψ ∘ φ ∘ τ = φ ∘ τ ∘ φ ∘ ψ = φ ∘ ψ
τ ∘ φ = τ ∘ φ ∘ ψ ∘ φ = ψ ∘ φ ∘ τ ∘ φ = ψ ∘ φ

But this implies:

ψ = ψ ∘ φ ∘ ψ = ψ ∘ φ ∘ τ = τ ∘ φ ∘ τ = τ

We state now some immediate consequences (some of them are already well known.)

Corollary 5. Let k be a field of characteristic zero and C be a finite category such that End_{C}(X) is commutative for all objects X in C. Then the following are equivalent:

(i) The algebra kC is semisimple.

(ii) The category C is inverse.

(iii) There is a family of finite abelian groups {G_{i}}_{1≤i≤n} such that mod(kC) ≅ mod(∏_{i=1}^{n} kG_{i}).

Proof. (i) ⇒ (ii) This is a consequence of theorem 1. (ii) ⇒ (iii) Follows from corollary 4.2 in [3]. (iii) ⇒ (i) Maschke’s theorem.

Corollary 6. Let k be a field of characteristic 0 and S be a finite semigroup in which all idempotents are central. Then the following are equivalent:

(i) S is an inverse semigroup.

(ii) S is a separative semigroup.

(iii) kS is semisimple

Proof. (ii)⇒ (i) By assumption all idempotents in S commute, hence to show that S is inverse it is sufficient to show that every element is split (von Neumann regular). Let s ∈ S and t be a positive integer such that s^{t} is an idempotent. If t = 1 we have s ∘ s ∘ s = s and we are done. Suppose now t > 1. We will show that s ∘ s^{t−1} ∘ s = s. Let t′ ≥ 0 be an integer such that t + t′ = 2^{m} < 2t for some m. We will show by induction that for all 0 ≤ n ≤ m we have s^{2^{m}} = s^{t+2^{m}}. This is clearly true for n = m:

s^{2^{m}} = s^{t+t′} = s^{t+t+t′} = s^{t+2^{m}}

suppose now that for some n > 0 we have: s^{2^{n}} = s^{t+2^{n}}, then:

(s^{2^{n−1}})^{2} = s^{2^{n}} = s^{t+2^{n}} = s^{t+2^{n−1}} ∘ s^{2^{n−1}} and also: s^{2^{n}} = s^{t+2^{n}} = s^{2t+2^{n}} = (s^{t+2^{n−1}})^{2}

hence:

(s^{2^{n−1}})^{2} = s^{t+2^{n−1}} ∘ s^{2^{n−1}} = (s^{t+2^{n−1}})^{2} .

Since S is separative, we get s^{2^{n−1}} = s^{t+2^{n−1}}. In particular, we can take n = 0 to get: s = s^{t+1} = s ∘ s^{t−1} ∘ s.

(i) ⇒ (ii) Take any s, t ∈ S such that s^{2} = s ∘ t = t^{2} then, because all idempotents are in the centre of S, s = s ∘ ŝ ∘ s = ŝ ∘ s ∘ s = (ŝ ∘ s) ∘ t and similarly t = (t̂∘ t) ∘ s. Therefore s ≤ t and t ≤ s hence s = t. The relation ≤ is the natural partial order on inverse semigroup, for details see [2]. The implication (i) ⇒ (iii) has been shown by W. D. Munn, see [5] and also [7]. The implication (iii)⇒(i) is an easy consequence of theorem 1.

The equivalence of (ii) and (iii) is slightly more general than results by W. D. Munn (see[4], [6]) where commutativity of S is assumed. It is well known, that some of the above implications hold in a greater generality. For example, (i) ⇒ (iii) as well as (i) ⇒ (ii) do not need S to be finite (but it is required for (ii) ⇒ (i), as a counter example consider the semigroup (ℕ, +) which is not inverse but is separative). It is also tempting to relax the assumption about central idempotents and following Theorem 1 assume only that any two idempotents commute. But then (ii) ⇒ (i) is no longer true. The smallest counter example is the semigroup S_{1} given by the multiplication table:

S_{1} = \begin{matrix}∘ & a & b & c & d & e \\ a & a & a & a & a & a \\ b & a & a & d & a & b \\ c & a & e & a & c & a \\ d & a & b & a & d & a \\ e & a & a & c & a & e \end{matrix}

The semigroup S_{1} is the unique (up to isomorphism or anti-isomorphism) example of a semigroup of order 5 which is inverse (hence all idempotents commute) but is not separative. The dihedral group D_{3} is the smallest example of a non commutative semigroup which is inverse and separative. It is also the only (up to isomorphism or anti-isomorphism) semigroup of order 6 with these properties. Computations have been performed using the GAP4 with the Smallsemi package [1].

We conclude this note with an explicit construction of nilpotent elements in twisted semigroup algebras of non-inverse periodic (i.e. all of its elements are of finite order) semigroup:

Proposition 7. Let k be an algebraically closed field and S be a periodic semigroup which is not inverse. If α ∈ Z^{2}(S, k^{×}), then the twisted algebra k_{α}S contains some non-zero nilpotent elements of the form σ − ξτ for some σ, τ ∈ S and ξ ∈ k^{×}.

Lemma 8. Let S = ⟨s⟩ be a finite cyclic semigroup, k be algebraically closed field and α: S×S → k^{×} be a 2-cocycle map. Then there exists an isomorphism between kS and k_{α}S which sends s to μs for some μ ∈ k^{×}.

Proof. Denote by m be the smallest positive integer such that there is some positive integer r with the property s^{m} = s^{m+r}. Let μ ∈ k be such that μ^{r} = λ, where λ satisfies s^{•m+r} = λs^{•m}. Then the map k_{α}S → kS which sends s to μs extends to an algebra isomorphism φ: k_{α}S → kS. It is easy to see that φ is multiplicative (in particular φ(s^{•m+r}) = φ(s^{•m})) and bijective (λ is invertible).

Lemma 9. Let S be a finite semigroup. Then there exists some positive integer t such that for any element s in S, s^{t} is idempotent.

Lemma 10. Let S be a finite semigroup. Then there exists some T ∈ ℕ_{+} such that s^{T} ∘ s ∘ s^{T} = s^{T} for all s ∈ S.

Proof. Take t as described in Lemma 9, i.e. such that s^{2t} = s^{t} for all s ∈ S. Consider T = 2t − 1. Then: s^{T} ∘ s ∘ s^{T} = s^{2T +1} = s^{4t−1} = s^{2t−1} = s^{T} .

The failure of S to be inverse can be caused by two reasons: for some s there are two elements s_{1}, s_{2} ∈ S which satisfy s ∘ s_{i} ∘ s = s and s_{i} ∘ s ∘ s_{i} = s_{i}, or for some element s ∈ S there is no s_{1} ∈ S such that both properties hold: s ∘ s_{1} ∘ s = s and s_{1} ∘ s ∘ s_{1} = s_{1}. According to these two cases, the proof of proposition 7 will be split into two parts:

Proof of Proposition 7. Case (a). Let s, s_{1}, s_{2} ∈ S where s_{1} ≠ s_{2} satisfy the following properties:

s ∘ s_{1} ∘ s = s,  s_{1} ∘ s ∘ s_{1} = s_{1},  s ∘ s_{2} ∘ s = s,  s_{2} ∘ s ∘ s_{2} = s_{2}

Note that we must have s ∘ s_{1} ≠ s ∘ s_{2} or s_{1} ∘ s ≠ s_{2} ∘ s. Otherwise we would get:

s_{1} = (s_{1} ∘ s) ∘ s_{1} = s_{2} ∘ (s ∘ s_{1}) = s_{2} ∘ s ∘ s_{2} = s_{2}.

Suppose s ∘ s_{1} ≠ s ∘ s_{2} then s ∘ s_{1} − s ∘ s_{2} ∈ kS is a non zero element. Then: (s ∘ s_{1} − s ∘ s_{2})^{•2} = α(s ∘ s_{1}, s ∘ s_{1})s ∘ s_{1} ∘ s ∘ s_{1} − α(s ∘ s_{1}, s ∘ s_{2})s ∘ s_{1} ∘ s ∘ s_{2}

− α(s ∘ s_{2}, s ∘ s_{1})s ∘ s_{2} ∘ s ∘ s_{1} + α(s ∘ s_{2}, s ∘ s_{2})s ∘ s_{2} ∘ s ∘ s_{2}

Using lemma 2 we get that the latter is equal to

α(s ∘ s_{1}, s ∘ s_{1})(s ∘ s_{1} − s ∘ s_{2} − s ∘ s_{1} + s ∘ s_{2}) = 0.

Similarly, if s_{1} ∘ s ≠ s_{2} ∘ s then s_{1} ∘ s − s_{2} ∘ s ∈ kS is non zero and (s_{1} ∘ s − s_{2} ∘ s)^{•2} = 0.

Case (b). Suppose there exists some element s ∈ S for which there is no s_{1} satisfying s_{1} ∘s∘s_{1} = s_{1} and s ∘ s_{1} ∘ s = s. By the lemma 10, we know there is some T ∈ ℕ_{+} such that s^{T} ∘ s ∘ s^{T} = s^{T}. Since s^{T} cannot be the quasi inverse for s, we get: T > 1 and s ∘ s^{T} ∘ s ≠ s. Let U be the cyclic semigroup generated by s. Because S is periodic, U is finite and by lemma 8 there is an algebra isomorphism φ: kU → k_{α}U such that φ(s) = as for some a ∈ k^{×}. Therefore it is sufficient to show that s − s^{T +2} is nilpotent in kU, because then φ(s − s^{T +2}) = as − bs^{T +2} hence also s − \frac{b}{a}s^{T +2} are nilpotent in k_{α}U ⊂ k_{α}S.

We will show by induction that for all n ∈ ℕ (s − s^{T +2})^{n} = s^{n} − s^{T +n+1}. Clearly this is true for n = 1. Suppose now, for some n > 0 we have:

(s − s^{T +2})^{n} = s^{n} − s^{T +n+1}

Then for n + 1: (s − s^{T +2})^{n+1} = (s − s^{T +2})(s − s^{T +2})^{n} = (s − s^{T +2})(s^{n} − s^{T +n+1}) = s^{2T +n+3} − 2s^{T +n+2} + s^{n+1}. Using the property S^{2T +1} = s^{T} we get that the last expression is equal to:

s^{T +n+2} − 2s^{T +n+2} + s^{n+1} = s^{n+1} − s^{T +n+2}.

The element (s − s^{T +2}) is nilpotent in kU. Indeed, by taking its (T + 1)st power we get:

(s − s^{T +2})^{T +1} = s^{T +1} − s^{2T +2} = s^{T +1} − s^{T +1} = 0.

Corollary 11. Let k be a commutative ring and S be a finite semigroup which is not inverse. Then the k-algebra kS contains some non-zero nilpotent elements of the form σ − τ for some σ, τ ∈ S.

Corollary 12. Let S be a finite commutative inverse semigroup. Then for any element s ∈ S the quasi-inverse for s is of the form ŝ = s^{2t−1} where t is an integer such that s^{t} is an idempotent.

Proof. Suppose s^{2t−1} is not the inverse for s. Then following the proof of the case (b) we know that s − s^{T +2} is a non-zero nilpotent element in kS for all commutative rings k. In particular, also for k = ℂ. But this is impossible since ℂS is isomorphic to a direct product of some number of copies of ℂ, hence contains no non-zero nilpotent elements.

References

[1] A. Distler and J. D. Mitchell. Smallsemi — a GAP package, version 0.6.2, 2010.

[2] Mark V. Lawson. Inverse Semigroups: The Theory of Partial Symmetries. World Scientific Publishing Company, January 1999.

[3] Markus Linckelmann. On inverse categories and transfer in cohomology. preprint (2010).

[4] W.D. Munn. Semiprimitivity of inverse semigroup algebras. Proc. Roy. Soc. Edinburgh A, 93:83–98, 1982.

[5] W.D. Munn. On commutative semigroup algebras. Proc. Cambridge Phil. Soc., 93:237–246, 1983.

[6] W.D. Munn. Inverse semigroup algebras. In Gregory Karpilovsky, editor, Group and Semigroup Rings, Centro de Brasileiro de Pesquisas Fisicas Rio de Janeiro and University of Rochester, volume 126 of North-Holland Mathematics Studies, pages 197 – 223. North-Holland, 1986.

[7] Benjamin Steinberg. Möbius functions and semigroup representation theory. J. Comb. Theory Ser. A, 113:866–881, July 2006.