Let $S$ be a finite semigroup and $K$ a field of characteristic 0 (we can assume the complex numbers for simplicity).



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> Question: Is there a characterization when the center of the semigroup algebra $KS$ is one-dimensional and a basis is given by the idendity of $KS$ (meaning $KS$ has an identity)?
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If it helps, we can assume that $S$ is a band, meaning that $x^2=x$ for all $x \in S$.


Having trivial center seems to be rare but there are still many examples.
Here the number of semigroups with trivial center with $n \geq 2$ elements for small $n$:
0,5,9,251,1478.


Here the number of bands with trivial center with $n \geq 2$ elements for small $n$:
0,3,4,49,142.



Semigroup algebras with one dimensional center
Let me answer what goes on with bands. I'll assume the band is a monoid for simplicity. I think everything works so long as the band is unital but this can get messier to describe. Details of what I'm saying can be found in [Margolis, S., & Steinberg, B. (2012). Quivers of monoids with basic algebras. Compositio Mathematica, 148(5), 1516-1560.](https://www.cambridge.org/core/journals/compositio-mathematica/article/quivers-of-monoids-with-basic-algebras/39F94B0FF1357609C65687DBA4D307D4)  and some of our later papers.


The algebra of a band is split basic over every field so it is isomorphic to the algebra of a bound quiver.


A left regular band is a semigroup satisfying the identities $x^2=x, xyx=xy$. These identities say that repetition is useless. For monoids, the first identity is a consequence of the second. Left regular bands are precisely the bands $S$ for which $aS=bS$ implies $a=b$ (so left divisibility is an order). Therefore, a left regular band is ordered by $a\leq b\iff aS\subseteq bS\iff ba=a$.


If $M$ is a left regular band, then the principal left ideals of $M$ are closed under intersection and hence form a lattice $\Lambda(M)$. The vertices of the quiver of $M$ (if you work with left modules) is $\Lambda(M)$. If $Ma,Mb$ are two principal left ideals, then there are no arrows from $Ma$ to $Mb$ unless $Ma\subsetneq Mb$. If $Ma\subsetneq Mb$, then the number of edges from $Ma$ to $Mb$ is one less than the number of connected components of the restriction of the Hasse diagram of $M$ to those elements $m\in M$ with $Ma\subseteq Mm$ and $m<b$. So as you can see this is already a bit messy to say when you get a connected quiver.


A right regular band is defined dually: $x^2=x$ and $xyx=yx$ for all $x,y$. The quiver of a right regular band is computed dually. The principal right ideals form the vertices, there are no arrows $aM$ to $bM$ unless $bM\supsetneq aM$, in which case you count the number of connected components minus one of the Hasse diagram for the order $a\leq b$ if $ab=a$ of the set of $m$ with $m<b$ and $aM\subseteq mM$.


A regular band is a band satisfying $xyxzx=xyzx$. In other words $a\mapsto xax$ is a retraction from $M$ to $xMx$. It turns out that if $M$ is a band and $K$ is a field, then the map $M\to KM\to KM/J(KM)^2$ (where $J(KM)$ is radical) factors through a regular band $M'$ (the universal regular band image of $M$). Hence the quiver of $KM$ and $KM'$ are the same, and so one is reduced to the case of regular bands.


Every regular band $M$ embeds in a direct product $M\_l\times M\_r$ where $M\_l$ is a left regular band that has lattice of principal left ideals isomorphic to the lattice of principal two-sided ideals of $M$ and $M\_r$ is a right regular band that has lattice of principal right ideals isomorphic to the lattice of principal two-sided ideals of $M$. The quiver of $M$ is obtained by superimposing the quivers of $M\_l$ and $M\_r$ on each other (identifying their vertices via the bijections above). So the vertices of the quiver can be identified with the two-sided principal ideals of $M$. Arrows going up in the order are computed via the left regular band image $M\_l$ and arrows going down in the order are computed via the right regular band image $M\_r$. In particular, the quiver need not be acyclic.


How you would every turn this into a simple characterization of connectedness of the quiver is beyond me.


Here is a good example. Let $S$ be the semigroup $\{1,2\}\times \{1,2\}$ with multiplication $(i,j)(k,l)=(i,l)$. Let $M=S\cup I$ where $I$ is an identity. Then the quiver of $M$ has two vertices $v,w$ and edges $x\colon v\to w$ and $y\colon w\to v$. There is a single relation $yx=0$ (where I compose paths from right to left because I do left modules).


**Added.** In a paper with Margolis and Saliola, we showed the path algebra of an acyclic quiver is always the monoid algebra of a left regular band.


