# Deformations and Dimensions: $q$-Deform Finite to Infinite?

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?

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.

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I don't quite see how this is answerable in the general context without giving a very specific sense to what «some type of $q$-deformation» means, and in what sense the $V_q$ deform $V_1$. Even in the specific example you mention, you'd probably need to give more details. (Notice that the algebra structure on $A$ and $A_q$ seems to play no role in your question) – Mariano Suárez-Alvarez Jan 28 '13 at 19:16

(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.