What is this finite dimensional algebra?

Question

Fix a field $k$. Consider the (non-commutative, associative) $k$-algebra $A$ with generators $x$, $y$ subject to the relations

\begin{align*} x^2&=x\\ y^2&=y\\ x-xy-yx+y&=1 \end{align*} This algebra is four dimensional: using the third relation, one can always reorder so all occurrences of $y$ are on the right, then using the first two relations, the powers on $x$ and $y$ can be reduced to $0,1$. So everything is a linear combination of $1,x,y,xy$. It isn't hard to see that these are linearly independent too, so $\dim_k(A)=4$.

The multiplication table is \begin{array}{c|cccc} & 1 & x & y & xy \\ \hline 1 & 1 & x & y & xy \\ x & x & x & xy & xy\\ y & y & x+y-xy-1 & y & 0\\ xy & xy & 0 & xy & 0 \end{array}

The algebra is not $M_2(k)$, because it has a two-sided ideal, $I=\langle xy,yx\rangle$. $A/I\cong k\times k$.

It is not a domain, as $xyx=yxy=0$, but $x,xy,yx,y\ne 0$.

I am wondering if there is anything more we can say? Is there a neater description?

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. — Shaun, Nov 14 '24 at 19:13

Erik D · Answer 1 · 2024-11-14T19:00:08.837

For $\mathrm{char}k \ne 2$, this is the radical, self-injective Nakayama algebra with two simples and Loewy length 2.

Set $e_1 = \frac{1}{2}(1+x-y)$, $e_2 = \frac{1}{2}(1+y-x)$, $a = xy$ and $b = yx$. Then $e_1e_2=0=e_2e_1$, $e_1a=a=ae_2$, $e_2a= 0=ae_1$, $e_2b=b=be_1$, $e_1b= 0=be_2$, and $ab=0=ba$. The identity element is $1 = e_1+e_2$, the radical $\mathrm{rad} A = \langle a,b\rangle$, and $\mathrm{rad}^2 A = 0$.

As a quiver with relations, $e_1$ and $e_2$ represent the vertices, $a$ an arrow from $e_2$ to $e_1$, and $b$ arrow from $e_1$ to $e_2$ (or the other way around, depending on your convention for compositing paths), with relations $ab = ba =0$.

Anixx · Answer 2 · 2024-11-15T03:17:59.500

Here is a matrix representation:

$1=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$

$x=\left( \begin{array}{cccc} \frac{1}{2} & \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 1 \\ 0 & 0 & \frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & -\frac{1}{2} & \frac{1}{2} \\ \end{array} \right)$

$y=\left( \begin{array}{cccc} \frac{1}{2} & -\frac{1}{2} & 1 & 0 \\ -\frac{1}{2} & \frac{1}{2} & 0 & 1 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ \end{array} \right)$

One can see that all your conditions are satisfied.

$x^2=x$
$y^2=y$
$x+y-xy-yx=1$
$xy=\left( \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$
$yx=\left( \begin{array}{cccc} 0 & 0 & 1 & -1 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)=x+y-xy-1$
$xyx=yxy=0$

I suggest the following basis so to make it more familiar:

$j=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ \end{array} \right)$

$\varepsilon =\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$

This way, $x=1/2+j/2+\varepsilon$ and $y=1/2-j/2+\varepsilon$, $j^2=1$, $\varepsilon^2=0$.

Thus, the algebra is $\mathbb{R}\langle j,\varepsilon\rangle/(j^2-1,\varepsilon^2,j\varepsilon+\varepsilon j)$

OP did not specify what field $k$ they work over. Your presentation is great, and it works not just over $k=\mathbb R$ (which you assume in your last line), but over any field, as long as that field's characteristic is not $2$ (in which case, to begin with, one cannot divide by 2). That's all I wanted to point out. — Torsten Schoeneberg, Nov 16 '24 at 23:09

Dietrich Burde · Answer 3 · 2024-11-14T19:51:54.733

5

The comments have been moved, so let me mention that there is a complete classification of complex and real $4$-dimensional associative algebras, commutative and non-commutative ones. A list (of unital associative algebras) is also given at this site, see here:

Classification of 4 dimensional real associative unital algebra

In our paper we classify $4$-dimensional Novikov algebras, and as a special case the commutative, associative algebras in dimension $4$, not necessarily unital. For a full classification in dimension $4$ see this paper, Theorem $16$ on page $6-8$, and the references given. Actually, one of the references is this paper by Rakhimov et al.

edited Nov 14 '24 at 19:51

answered Nov 14 '24 at 19:21

Dietrich Burde

140,055

2

Can we identify, which one is this? – Anixx Nov 14 '24 at 19:39
@Anixx Yes, we can take Rakhimov's list, and since it is not nilpotent, and it is unital, there are not many possibilities. Furthermore, some structural properties are given in this classification. – Dietrich Burde Nov 14 '24 at 19:42
From the table in the linked post it seems it should be No 11? – Anixx Nov 14 '24 at 19:42
$\mathbb{R}\langle x,y\rangle/(x^2-1,y^2,xy+yx)$? – Anixx Nov 14 '24 at 19:46
The variables should be different. – Anixx Nov 14 '24 at 19:52
Okay, so I found the matrix representation, see my answer. – Anixx Nov 15 '24 at 02:55
Indeed, turned out to be No 11. – Anixx Nov 15 '24 at 03:15
@Anixx Very good! – Dietrich Burde Nov 15 '24 at 10:29

score 5 · Accepted Answer · answered Nov 16 '24 at 10:23

Another characterization of $A$ is that it is the Clifford algebra of the binary quadratic form $q(x,y)=x^2$. Indeed, writing $(e_1,e_2)$ for the canonical basis of $k^2$, which is orthogonal for (the bilinear form associated to) $q$, this Clifford algebra is a $4$-dimensional algebra with basis $1$, $e_1$, $e_2$, $e_1e_2$, with $$\begin{align*} e_1^2 &= q(e_1)=1 \\ e_2^2 &= q(e_2)=0 \\ e_1e_2 &= -e_2e_1 \end{align*}$$ which is the presentation given in Annix's answer.

To be complete, I repeat the result of their computation, showing that we can take $x=1/2+u$ and $y=1/2+v$ with $u=e_1/2+e_2$ and $v=-e_1/2+e_2$to retrieve your generators. Indeed, $x^2=x$ and $y^2=y$ then follow from $u^2=v^2=q(u)=q(v)=1/4$, and $xy+yx = x+y-1=u+v$ follows from $$xy+yx = 1/4 + \frac{u+v}{2}+uv + 1/4 + \frac{u+v}{2} + vu = 1/2+u+v+2b(u,v)$$ where $b$ is the bilinear form associated to $q$, so $b(u,v)=b(e_1/2+e_2,-e_1/2+e_2)=-1/4$.

Just to be precise, we assume $char(k)\neq 2$ here. – Torsten Schoeneberg Nov 16 '24 at 21:53 — Torsten Schoeneberg, Nov 16 '24 at 21:53

Oeyvind Solberg · Answer 5 · 2024-12-05T08:14:35.443

Just to complete the picture with a characteristic 2 example. Here we will use GAP/QPA to compute a presentation of this algebra.

Q := Quiver( 1, [[1,1,"x"],[1,1,"y"]]);
k := GF(2^8);
kQ := PathAlgebra(k,Q);
AssignGeneratorVariables(kQ);
relations := [ x-x^2, y-y^2, x-x*y-y*x + y - v1 ];
I := Ideal( kQ, relations );
gb := [];
while Length(els) > 0 do
  for el in els do
    el_tip := Tip(el);
    Add(gb, el/TipCoefficient(el_tip));
  od;
n := Length(gb);
  els := [];
for i in [1..n] do
    x := TipWalk(gb[i]);
    k := Length(x);
for j in [1..n] do
  y := TipWalk(gb[j]);
  l := Length(y);

  for r in [Maximum(0, k-l)..k-1] do
    if x{[r+1..k]} = y{[1..k-r]} then
      b := x{[1..r]};
      c := y{[k-r+1..l]};

      overlap := gb[i]*Product(c, One(kQ)) - Product(b, One(kQ))*gb[j];
      remainder := RemainderOfDivision(overlap, gb, kQ);

      if not IsZero(remainder) then
        AddSet(els, remainder);
      fi;
    fi;
  od;
od;

od;
od;
gb := TipReducedList( gb, kQ );
gb := ReducedListQPA( gb, kQ );
GroebnerBasis( I, gb );
A := kQ / I;
B := AlgebraAsQuiverAlgebra(A);
gap> AdjacencyMatrixOfQuiver(QuiverOfPathAlgebra(B[1]));
[ [ 0, 1 ], [ 1, 0 ] ]
gap> B[2];
[ [(Z(2)^0)*x], [(Z(2)^0)v1+(Z(2)^0)x], [(Z(2)^0)xy],  [(Z(2)^0)v1+(Z(2)^0)x+(Z(2)^0)y+(Z(2)^0)x*y] ]
gap> RelationsOfAlgebra(B[1]);
[ (Z(2)^0)a1a2, (Z(2)^0)a2a1 ]

From this output, you can read that the underlying quiver is that of a selfinjective Nakayama algebra with 2 vertices. The vertices correspond to $\{ x,1+x\}$ and the arrows correspond to $\{ xy, 1 + x + y + xy\}$. This is probably an isomorphism for all fields of characteristic $2$. Finally, having a quiver $\Gamma$ with two vertices and two arrows $a\colon 1\to 2$ and $b\colon 2\to 1$ and relations $\{ ab, ba\}$, we would have $$A \simeq k\Gamma/\langle ab,ba\rangle$$

What is this finite dimensional algebra?

5 Answers5

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