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sniffnoySo, I have recently been reading/thinking a bit about central groupoids. This is a quite misnamed class of object; they're not groupoids, the term comes from an old use of "groupoid" to mean "magma". (Blech.) They're magmas (sets with a binary operation) satisfying the central identity, (ab)(bc)=b.
You may recall my project regarding equational probabilities in algebraic structures. I deliberately limited this to algebraic structures I considered sufficiently interesting. Central groupoids did not initially make the list. But after seeing them mentioned on Terry Tao's blog recently for the second time, I decided to check them out. Turns out, they're a worthy inclusion, and I've added them now! Because they have rather more structure than I expected.
Like, central groupoids can be alternately characterized in terms of "central digraphs" -- directed graphs where, for any x and y, there's a unique vertex z such that x→z→y. Finding this z is the multiplication in your central groupoid. Going the other way, x→y means that y is a right-multiple of x, or equivalently, that x is a left-multiple of y. And then considering the adjacency matrix of such a digraph gives us a third way of looking at things; it's a 0-1 matrix, and the centrality condition is equivalent to saying that its square is the all-ones matrix! Neat.
It turns out that if you have a finite central groupoid, it must have a square number of elements, call it k². This can be seen by considering the degree of the vertices. It turns out that each vertex has the same indegree and the same outdegree, and that these are moreover equal to each other; this constant is k. (If you look at, say, the right-multiples of x, and want to biject them to the left-multiples of y, you simply right-multiply by y! And then this proves the rest.) If you fix some x, look at its right-multiples, and then look at *their* right-multiples (obviously this all works on the left as well), these secondary sets of right-multiples partition the set; this shows that the overall number of elements is k².
There's another significance to k, too. Consider what happens when you square an element x of your central groupoid. Either x²=x, i.e. x is an idempotent, meaning it has a self-loop; or, since (x²)²=x, they form a 2-cycle (this is the only way to have 2-cycles). How many of each are there? Obviously if k is odd there must be at least one idempotent, but that's pretty weak. Well, it turns out there's always exactly k idempotents! The thing is, I haven't seen an algebraic/combinatorial proof of this -- the only proof I've seen goes through the matrix interpretation! See, the number of idempotents is the trace of the adjacency matrix M, and M²=J, the all-ones matrix, whose eigenvalues are k² and a bunch of 0s. So the eigenvalues of M are k and a bunch of 0s, and therefore its trace is k.
Still, I had to wonder -- if there's k idempotents, and k² elements... could it be that each element can be written uniquely as a product of two idempotents? That would certainly explain things and be cool! Unfortunately, it isn't true. It's true of the "standard" central groupoids; these are constructed by taking a set S, and then defining an operation on S×S via (x1,x2)(y1,y2=(x2,y1). Unfortunately, it's not true in general; there are counterexamples with k=3. (With k=2, there's only the standard one.)
(EDIT 12/14: Worth noting also, if you were hoping for the opposite, that only the standard ones have this property, well, that's also false.)
So, in that way they fail. And they also fail in that -- like inverse-property loops -- despite having enough structure to be "sink-resistant", they still don't have an equational probabilities gap. There's probably lots of equations you could use to demonstrate this, but the one I used was (x(xx))²=x(xx). In a standard central groupoid, x(xx) and (xx)x are always idempotents, but in general this needn't be so; and if you check out the example I wrote up on my page that I linked, you can see that I was able to construct a central groupoid which yielded a probability of 1-n/(n+1)², approaching 1 from below. No gap!
So, those are central groupoids. Kind of neat, but probably not something I'm going to think about any further.
You may recall my project regarding equational probabilities in algebraic structures. I deliberately limited this to algebraic structures I considered sufficiently interesting. Central groupoids did not initially make the list. But after seeing them mentioned on Terry Tao's blog recently for the second time, I decided to check them out. Turns out, they're a worthy inclusion, and I've added them now! Because they have rather more structure than I expected.
Like, central groupoids can be alternately characterized in terms of "central digraphs" -- directed graphs where, for any x and y, there's a unique vertex z such that x→z→y. Finding this z is the multiplication in your central groupoid. Going the other way, x→y means that y is a right-multiple of x, or equivalently, that x is a left-multiple of y. And then considering the adjacency matrix of such a digraph gives us a third way of looking at things; it's a 0-1 matrix, and the centrality condition is equivalent to saying that its square is the all-ones matrix! Neat.
It turns out that if you have a finite central groupoid, it must have a square number of elements, call it k². This can be seen by considering the degree of the vertices. It turns out that each vertex has the same indegree and the same outdegree, and that these are moreover equal to each other; this constant is k. (If you look at, say, the right-multiples of x, and want to biject them to the left-multiples of y, you simply right-multiply by y! And then this proves the rest.) If you fix some x, look at its right-multiples, and then look at *their* right-multiples (obviously this all works on the left as well), these secondary sets of right-multiples partition the set; this shows that the overall number of elements is k².
There's another significance to k, too. Consider what happens when you square an element x of your central groupoid. Either x²=x, i.e. x is an idempotent, meaning it has a self-loop; or, since (x²)²=x, they form a 2-cycle (this is the only way to have 2-cycles). How many of each are there? Obviously if k is odd there must be at least one idempotent, but that's pretty weak. Well, it turns out there's always exactly k idempotents! The thing is, I haven't seen an algebraic/combinatorial proof of this -- the only proof I've seen goes through the matrix interpretation! See, the number of idempotents is the trace of the adjacency matrix M, and M²=J, the all-ones matrix, whose eigenvalues are k² and a bunch of 0s. So the eigenvalues of M are k and a bunch of 0s, and therefore its trace is k.
Still, I had to wonder -- if there's k idempotents, and k² elements... could it be that each element can be written uniquely as a product of two idempotents? That would certainly explain things and be cool! Unfortunately, it isn't true. It's true of the "standard" central groupoids; these are constructed by taking a set S, and then defining an operation on S×S via (x1,x2)(y1,y2=(x2,y1). Unfortunately, it's not true in general; there are counterexamples with k=3. (With k=2, there's only the standard one.)
(EDIT 12/14: Worth noting also, if you were hoping for the opposite, that only the standard ones have this property, well, that's also false.)
So, in that way they fail. And they also fail in that -- like inverse-property loops -- despite having enough structure to be "sink-resistant", they still don't have an equational probabilities gap. There's probably lots of equations you could use to demonstrate this, but the one I used was (x(xx))²=x(xx). In a standard central groupoid, x(xx) and (xx)x are always idempotents, but in general this needn't be so; and if you check out the example I wrote up on my page that I linked, you can see that I was able to construct a central groupoid which yielded a probability of 1-n/(n+1)², approaching 1 from below. No gap!
So, those are central groupoids. Kind of neat, but probably not something I'm going to think about any further.
