I'm interested in finding an algebra with elements x,y which are identified by every finite-dimensional module. I'm primarily interested in the case that everything is over the complex field, but answers over other fields would also be interesting.
$\begingroup$
$\endgroup$
asked Sep 10, 2011 at 3:38
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
The Weyl algebra, generated by $x$ and $y$ subject to the single relation $xy-yx=1$, is an example. A silly one, one may add: it does not have any finite dimensional representation!
There are more interesting examples: let $\mathfrak g$ be the Lie algebra with basis $x_i$, $y_i$, with $i\in\mathbb Z$, and $z$, such that the only non-zero brackets are $[x_i,y_i]=z$, for all $i\in\mathbb Z$. If $A=\mathcal U(\mathfrak g)$ is the enveloping algebra of $\mathfrak g$, then $z$ acts by zero in every finite dimensional module.
There are finitely-generated examples, too.
answered Sep 10, 2011 at 3:39
$\begingroup$
$\endgroup$
4
A similar example to Mariano's is the algebra generated by x,y subject to the relation $xy=1$. In a finite dimensional representation, yx must of course map to the same element as 1. However, there is a faithful infinite dimensional rep where x,y go to a unilateral shift and its adjoint so yx is not 1.
answered Sep 10, 2011 at 19:26
-
$\begingroup$ There is always a faithful representation: the regular one :) $\endgroup$ Sep 10, 2011 at 22:14
-
1$\begingroup$ Yes, but it is not immediate from a presentation that the relation xy=1 doesn't imply yx=1. One needs to give an explicit representation showing that xy=1 is possible with yx not equal to 1. $\endgroup$ Sep 10, 2011 at 23:53
-
$\begingroup$ Ah. Sure. What I had in mind is this: Bergman's diamond lemma immediately (there are no ambiguitities at all!) implies that your algebra has a basis of monomials and, in particular, that x and y are different and that yx is not one. $\endgroup$ Sep 11, 2011 at 0:14
-
2$\begingroup$ By the way my answer is a special case of the following fact: if M is a finitely generated monoid which is not residually finite and x,y in M are elements of M not separable in a finite image, then the corresponding elements of the monoid algebra are not separable by finite dimensional reps. This follows from Malcev's theorem that finitely generated linear monoids are residually finite. $\endgroup$ Sep 11, 2011 at 3:39