most recent 30 from http://mathoverflow.net 2013-05-25T10:25:47Z http://mathoverflow.net/feeds/question/120112 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120112/deformations-and-dimensions-q-deform-finite-to-infinite Dyke Acland 2013-01-28T13:58:14Z 2013-01-28T22:18:36Z

<p>Let $A$ be a (complex) (finitely generated) algebra with some type of $q$-deformation $A_q$, where $A_1 = A$. Moreover, let $V_q$ be a vector subspace of $A_q$, such that $V_1$ is a finite-dimensional subspace of $A$. What conditions do we need to assume on the $q$-deformation for this to imply that $V_q$ is also finite-dimensional? In other words, for which type of deformations will an infinite dimensional subspace always be infinite dimensional in the $q \to 1$ limit? Or do such a set of conditions exist?</p> <p>The specific example I'm interested in here is quantum-$SU_N$, the so-called coordinate algebra quantum group (as opposed to the quantized enveloping algebra.) As far as I understand this is a Poisson algebra type deformation, with the associated properties.</p>

http://mathoverflow.net/questions/120112/deformations-and-dimensions-q-deform-finite-to-infinite/120140#120140 Peter Samuelson 2013-01-28T19:06:05Z 2013-01-28T19:06:05Z

<p>(This question is fairly open-ended and probably doesn't have a good answer, so this is more of a suggestion.) One thing that might be useful is a PBW-type basis for $A_q$. In other words, many ``quantum" algebras occurring in practice have subalgebras $B_q, C_q, D_q \subset A_q$ such that the multiplication map $B_q \otimes_{\mathbb C} C_q \otimes_{\mathbb C} D_q \to A_q$ is a vector space isomorphism. Often the algebras $B, C, D$ are much simpler than $A$, so this gives a natural basis for $A_q$, which makes studying subspaces of $A_q$ easier.</p>