2
$\begingroup$

Let $A$ be an algebra, say over $\mathbb{C}$ and finite-dimensional, but not necessary semisimple. I have the strong feeling, which I would like to prove and use, about the following rather natural approach.

Consider the set $S$ of central idempotents in $A$, then I want to turn the set of central idempotents $\Delta$ into an abstract simplicial complex. Two idempotents $e_1,e_2$ are connected iff $e_1\in e_2A$, simplices consist of completely connected subset and and the highest-dimensional simplices hence are flags associated to a decomposition of $A$ into central primitive idempotents. This is much how one defines buildings as simplicial complexes e.g. of subspaces of a finite vector space....

Is this OK? I'm not familiar with calculating with central prinimtive idempotents - especially I'm concerned, if any decomposistion into central primitive idempotents has the same cardinality??

..if yes,it seems so natural, something similar ought to exist (surely more advanced ;-) - any reference? I would probably also suffice with a related construction and/or I want to find out more established facts about such an "algebraic simplicial complex / building"...Thanks for your help in advance.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

The central idempototents of a finite dimensional algebra form a finite Boolean algebra with 1 as the max and the central primitive idempotents as the atoms. The decomposition of 1 into central idempotents is thus unique. So the central idempotents are the face lattice of a simplex. The order is $e\leq f$ if $e\in fA$. Your simplicial complex would be the order complex of this Boolean algebra and so would be the barycentric subdivision of a simplex.

Added details. The boolean algebra operations are given by $e\wedge f=ef$, $e\vee f=e+f-ef$ and $\neg e=1-e$. The finiteness follows, for example, because one can look at the regular representation of the algebra $A$ by matrices and observe that the central idempotents form a commutative semigroup of idempotent matrices. Such a semigroup is simultaneously diagonalizable and there are only $2^n$ diagonal idempotent $n\times n$ matrices.

Improve this answer
$\endgroup$
2
  • $\begingroup$ GREAT, THANX, THAT HELPS! I've looked up "order complexes" and these indeed nice :-)...well, I didn't even fully "believe" I could just on/off central primitive idempotents, but embedding them into matrix rings certainly boosted my intuition (shame!) What is the argument/theory that descibes the complex then as a subdevided simplex? $\endgroup$
    – Simon Lentner
    Commented Apr 15, 2012 at 16:51
  • $\begingroup$ If F is the face poset of a simplicial complex, then it is straightforward to verify the order complex of F is the barycentric subdivision. A vertex of the order complex is an element f of F which corresponds to the barycenter of f. It is easy to check under this correspondence chains are simplices of the barycentric subdivision. Try it out on a 2-simplex. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 15, 2012 at 23:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .