14
$\begingroup$

  1. Is there a name for the algebra (and its tensor products) given by generators $M_{j}$, $j \in \mathbb{Z}_{n}$ under the conditions $$M_{j} = (1 - M_{j-1})(1-M_{j+1})$$ where $j=1\implies j-1=n$ and $j=n\implies j+1=1$?

  2. There is no restriction on the commutativity of $M_{j}$ and so is there a matrix representation of this algebra for all possible cases for $M_{j}$?

By tensor product I mean generalization of following extension to multiple indices:

For single index $$M_{j} = \prod_{\substack{i\in\{-1,0,1\}\\i\neq j}}(1 - M_{j+i})$$ which is same as above and for two indices $$M_{j,j'} = \prod_{\substack{i,i'\in\{-1,0,1\}\\i\neq j\wedge i'\neq j}}(1 - M_{j+i,j'+i'})$$ relation holds.

Update Like Lena I too thought it is similar to a Hecke algebra however since I am not familiar I could not pin down details.

Improve this question
$\endgroup$
9
  • 3
    $\begingroup$ How do I interpret your relation when $j=1$ or $n$? $\endgroup$
    – David Hill
    Apr 9, 2013 at 14:28
  • $\begingroup$ For $j=1$, $j-1$ will be $n$ and for $j=n$, $j+1$ will be $1$. $\endgroup$
    – Turbo
    Apr 9, 2013 at 14:37
  • 3
    $\begingroup$ 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? $\endgroup$
    – Vít Tuček
    Apr 10, 2013 at 8:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – David Hill
    Apr 11, 2013 at 19:39
  • 3
    $\begingroup$ 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? $\endgroup$
    – Vladimir Dotsenko
    Dec 22, 2015 at 16:01

2 Answers 2

Reset to default
7
$\begingroup$

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<x,y,z,t,u> := 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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Turbo
    Dec 22, 2015 at 22:14
  • $\begingroup$ One more query what is the dimensions if $n=2t$ form where $t\in\Bbb N$ ? $\endgroup$
    – Turbo
    Dec 22, 2015 at 22:39
  • $\begingroup$ 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. $\endgroup$
    – Vladimir Dotsenko
    Dec 22, 2015 at 22:42
  • $\begingroup$ 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. $\endgroup$
    – Turbo
    Dec 22, 2015 at 22:45
  • $\begingroup$ 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). $\endgroup$
    – Vladimir Dotsenko
    Dec 22, 2015 at 23:49
-1
$\begingroup$

Reminds of a Hecke algebra.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ could you elaborate? $\endgroup$
    – Turbo
    May 2, 2013 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.