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Is there a name for the algebra (and its tensor products) given by generators $U_{j}$, $j \in \mathbb{Z}_{n}$

under the conditions $U_{j} = (1 - U_{j-1})(1-U_{j+1})$? There is no restriction on the commutativity of $U_{j}$. I am interested in structures for all possible cases for $U_{j}$.

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How do I interpret your relation when $j=1$ or $n$? – David Hill Apr 9 '13 at 14:28
For $j=1$, $j-1$ will be $n$ and for $j=n$, $j+1$ will be $1$. – Turbo Apr 9 '13 at 14:37
In that case you can just renumber indices $i \to i-1$ and write $U_i$, $i\in \mathbb{Z}_n$. Adding some context would be helpful. Are the variables $U_i$ (anti)commutative? – Vít Tuček Apr 10 '13 at 8:53
I don't know if this is helpful, but when $n=2$, $U_1$ can be written in terms of $U_0$, so we can regard the algebra as a quotient of a polynomial algebra: $k[x]/\langle x(x^3-4x^2+2x-1)\rangle$. Perhaps it would be easier for someone to recognize this algebra. – David Hill Apr 11 '13 at 19:39
The behaviour of this algebra is very strange. It is finite-dimensional for $n=2,3,4$ (dimensions are $4$, $9$, $25$ respectively), and infinite-dimensional for $n=5$ (since already its quotient by the commutator ideal is infinite-dimensional). I don't know what happens for $n=6$. In any case, it doesn't look like anything I saw before. In what context did you encounter it? – Vladimir Dotsenko Dec 22 '15 at 16:01

As requested, I elaborate on my comment.

First of all, let me make a change of variables $a_i=U_i-1$. The relations then become $a_i+1=a_{i-1}a_{i+1}$.

For $n=2,3,4$ I used the Magma online calculator. The commands

F<x,y> := FreeAlgebra(RationalField(),2); B := [x^2-y-1,y^2-x-1]; GroebnerBasis(B);

give the result

[ x^2 - y - 1, x*y - y*x, y^2 - x - 1 ]

so the algebra has a basis $1,x,y,yx$, so is four-dimensional.

The commands

F<x,y,z> := FreeAlgebra(RationalField(),3); B := [x*z-y-1,y*x-z-1,z*y-x-1]; GroebnerBasis(B);

give the result

[ x*y*z - z*x*y + y*z - z*x - x + y, y*z*x - z*x*y - x*y + y*z - x + z, y*z^2 + y*z - z*x - x - z - 1, z^2*x - y*z + z*x - y - z - 1, z^3 - x*y + z^2 - x - y - 1, x^2 - z^2 + x - z, x*z - y - 1, y*x - z - 1, y^2 - z^2 + y - z, z*y - x - 1 ]

so the algebra has a basis $1,x,y,z,xy,yz,zx,z^2,zxy$, so is nine-dimensional.

A similar computation for $n=4$ gives a Gröbner basis which is a bit too long to format properly, and a basis for the algebra $1,x,y,z,t,xy,yz,zt,tx,tz,x^2,y^2,z^2,t^2,xyz,y^2z,yzt,z^2t,ztx,tx^2,txy,t^2x,t^3,txyz,t^2x^2$, so the algebra is 25-dimensional.

For $n=5$, the calculator spits more and more elements as the degree grows, so it might even be that the Gröbner basis is infinite. However, if we consider the abelianisation of this algebra, the command

F := PolynomialRing(RationalField(),5); B := [x*z - y-1, y*t-z-1, z*u-t-1, t*x-u-1, u*y-x-1]; GroebnerBasis(B);

produces the result

[ x - y*u + 1, y*t - z - 1, z*u - t - 1 ]

which defines an infinite-dimensional algebra (since the associated monomial algebra is defined by the relations $x=0$, $yt=0$, $zu=0$), so the original algebra must for sure be infinite-dimensional.

[Alternatively, we can note that for commuting elements $a_i$, the equation $a_{i-1}a_{i+1}=a_i+1$ is the celebrated "pentagon recurrence" related to cluster algebras of type $A_2$, and we have $a_{n+5}=a_n$ for all choices of $a_0,a_1$ for which the sequence is uniquely defined, so the corresponding abelianisation corresponds to something 2-dimensional geometrically, and the algebra is infinite-dimensional.]

Overall, it is not quite clear if we should hope that these algebras obey a nice pattern or are easily recognisable, but they do look interesting.

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At $n=5$ you say 'so the corresponding abelianisation corresponds to something 2-dimensional geometrically' could you say whether at $n=2^k+1$ where $k\in\Bbb N_{>0}$ this is something $k$-dimensional geometrically and for other $n=2t+1$ where $t\in\Bbb N_{>0}$ and not of form $n=2^k+1$ where $k\in\Bbb N_{>0}$ can we tell anything at all? – Turbo Dec 22 '15 at 22:14
One more query what is the dimensions if $n=2t$ form where $t\in\Bbb N$ ? – Turbo Dec 22 '15 at 22:39
I gave all numerical information that I have at the moment. I think that your questions about $2^t+1$ etc. are a bit far fetched. Looking at the commutative case, I would expect qualitative difference depending on whether or not $n$ is divisible by 5. Powers of 2 are a red herring. – Vladimir Dotsenko Dec 22 '15 at 22:42
It might turn out interesting if $2^k+1$ are indeed special that way (I am not sure if this has anything to do with divisibility by $5$). I am also thinking may be for even $n$ something constant should suffice. – Turbo Dec 22 '15 at 22:45
I do not understand what you mean by "something constant" in your last sentence. Divisibility by 5 matters simply because for all $n$ divisible by 5 the algebra is automatically infinite-dimensional. (Abelianization for such $n$ is still geometrically 2-dimensional). – Vladimir Dotsenko Dec 22 '15 at 23:49

Reminds of a Hecke algebra.

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could you elaborate? – Turbo May 2 '13 at 18:47

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