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I came across the following semi-group and the associated finite dimensional semi-group algebras over a field $K$ (which are Nakayama algebras) as they have very nice homological properties. My question is whether those semi-groups appeared before in the literature.

Fix a binary necklace with colours black and white with $n \geq 2$ points. Number the points from $0$ to $n-1$ in the necklace and think of them as elements in $\mathbb{Z}/\mathbb{Z}n$. The basis elements of the semi-group algebra together with the zero associated to the necklace are the zero element 0 with $0x=0=x0$ for all other elements $x$, idempotents $e_i$ for any point $i$ with $e_i e_j = \delta_{i,j} e_i$ and intervals starting at a point $i$ in the necklace and ending at a point $j$. When the starting point $i$ is white, the intervals have the form $[i,j]$ for $j=i+1,i+2,\dotsc,i+n-1$ and when the starting point $i$ is black, the intervals have the form $[i,j]$ for $j=i+1,i+2,\dotsc,i+n=i$. So there is one more interval for a black point than for a white point, namely the interval $[i,i]$. Let the length $l([i,k])$ of an interval be equal to the number of points it contains, where we set the length of $e_i$ equal to 1 and the length of $[i,i]$ equal to $n+1$.

Now the multiplication in the semigroup is given by $e_i [j,k]= \delta_{i,j} [j,k]$, $[j,k] e_i = [j,k] \delta_{k,i}$ and $[i,k][s,t]= \delta_{k,s} [i,t]$ (this means we can glue two intervals together if $[i,k]$ has the end point equal to the start point of $[s,t]$), where we set this to be $0$ if $l([i,k])+l([s,t])>n$ (meaning that the result would be no element anymore) in case $i$ is a white point and $0$ if $l([i,k])+l([s,t])>n+1$ if $i$ is a black point.

For example, when all points are white and the necklace has $n$ points, the semi-group algebra is isomorphic as an algebra to the Taft algebra $K\langle C,X\rangle/\langle C^n-1,X^n,XC- u CX\rangle$ when $u$ is a primitive $n$-th root of unity in $K$. The Taft algebra is also a Hopf algebra, so one might hope that there is at least a bialgebra structure for the more general class of those necklace algebras (maybe assuming the field has a certain characteristic or other properties).

Question 1: Does this semi-group appear somewhere in the literature already or is it a special case of a class of semigroups?

I found a lot of nice homological properties of the semigroup algebras but maybe there are also nice structural properties.

Question 2: Does the semigroup algebra of a binary necklace have nice structural properties like a bialgebra structure ?

It has no (ungraded) Hopf algebra structure in most cases as the algebras are rarely Frobenius (precisely when all points are white or all points are black).

Some nice representation-theoretic/homological properties of those algebras are that they have the double centraliser propertiy with a smaller such necklace algebra and they are Iwanaga–Gorenstein.

Recall that a cyclic Nakayama algebra is just a quiver algebra whose quiver is an oriented cycle (which we can identify with a necklace). The Kupisch series $[c_0,c_1,...,c_{n-1}]$ of a Nakayama algebra is just the vector space dimension of the indeomcposable projective module $e_i A$ at point $i$ in the quiver and describes the Nakayama algebra uniquely. The necklace algebra is isomorphic to the cyclic Nakayama algebra with $n$ simples with Kupisch series $[c_0,...,c_{n-1}]$ where $c_i=n$ in case $i$ is a white point and $c_{i+1}=n+1$ in case $i$ is a block point.

Mare
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    When you write $e_i e_j = \delta_{i,j} e_i$, what exactly does this mean? Interpreting $\delta_{i,j}$ as the Kronecker delta it seems you want $e_i e_j = e_i$ whenever $j \neq i$, but what would $e_i e_i$ be equal to in this case? – Carl-Fredrik Nyberg Brodda Aug 22 '20 at 09:15
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    @Carl-FredrikNybergBrodda It means e_i^2=e_i and e_i e_j =0 if $i \neq j$. So we need a zero element as well it seems. I added that. Thank you. – Mare Aug 22 '20 at 09:16
  • I actually never worked with semi-groups before but it seems we need a zero here but in the semigroup algebra the zero should be equal to the zero of the semigroup. That is a bit confusing. – Mare Aug 22 '20 at 09:19
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    English note: 'interval' (one 'l'); TeX note: please use $K\langle C,X\rangle/\langle C^n-1,X^n,XC- u CX\rangle$ $K\langle C,X\rangle/\langle C^n-1,X^n,XC- u CX\rangle$ rather than $K<C,X>/(C^n-1,X^n,XC- u CX>$ $K<C,X>/(C^n-1,X^n,XC- u CX>$ (https://mathoverflow.net/a/366101). I have edited accordingly. – LSpice Aug 22 '20 at 13:17
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    @LSpice Thanks. – Mare Aug 22 '20 at 13:28
  • Do you want $e_i$ to be different from $[i,i]$ to be different than $e_i$ if $i$ is black? Because they are both idempotents that seem to imitate each other. Also are your endpoints for intervals taken mod n – Benjamin Steinberg Aug 22 '20 at 21:47
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    If you want to identify the zero of the semigroup and the algebra it is called the contracted semigroup algebra. – Benjamin Steinberg Aug 22 '20 at 21:50
  • @BenjaminSteinberg $[i,i]$ is not an idempotent, we have $[i,i]^2=0$ because of the length condition. – Mare Aug 22 '20 at 21:51
  • @BenjaminSteinberg I added in my question the description as a quiver algebra directy. It is a Nakayama algebra with easy Kupisch series. But I was hoping that a pictorial or semigroup description might reveal some more structure, like a nice coalgebra or even bialgebra structure. Those algebras have many nice homological properties, so that it seems likely that they come from something nice. Maybe there is a more general class of semigroups behind them. – Mare Aug 22 '20 at 21:54
  • @BenjaminSteinberg Those are in fact exactly the Nakayama algebra with symmetric Cartan matrix and Loewy length at most $n+1$ (the Loewy length condition can be easily removed by allowing "larger" intervalls, which I did omit here for simplicity). Maybe there is a nice class of finite dimensional semigroup algebras having symmetric Cartan matrix that generalise those algebras. Symmetric Cartan matrix seems to be rare and attracts nice homological properties. – Mare Aug 22 '20 at 22:00
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    Ok I thought [i,i] would have length 0. – Benjamin Steinberg Aug 22 '20 at 22:05
  • @BenjaminSteinberg I added now that $[i,i]$ has length n. That was indeed not so clear. – Mare Aug 22 '20 at 22:08

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