most recent 30 from http://mathoverflow.net 2013-05-20T07:16:03Z http://mathoverflow.net/feeds/question/25771 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25771/is-there-a-notion-of-tensor-product-of-perfect-bases-of-representations-of-lie-al Ben Webster 2010-05-24T13:43:54Z 2011-01-30T00:20:15Z

<p><a href="http://arxiv.org/abs/math.QA/0601391" rel="nofollow">Berenstein and Kazhdan</a> define <strong>perfect bases</strong> as an "unquantized" version of crystal bases. A perfect basis is roughly a basis with a crystal structure such that $E_i\cdot v=\mathbb{C}\cdot \tilde{e}_iv+\cdots$ where the $\cdots$ indicates terms in basis vectors killed by $\tilde{e}_i^{\epsilon_i(v)-2}$ (here $E_i$ is an element of the Lie algebra, and $\tilde{e}_i$ is a Kashiwara operator).</p> <p>The cool theorem is that any given finite-dimensional representation only has one possible crystal attached to it.</p> <p>Note that many of the "nicest" crystal bases (in particular, the global crystal basis) are perfect bases when specialized at q=1, this is far from universally true. In particular, taking the tensor product of perfect bases in the naive sense doesn't result in a new perfect basis. </p> <blockquote> <p>Does anyone know of a way of fixing this, and getting in a canonical perfect basis on the tensor product from perfect bases on the factors? </p> </blockquote> <p>What I particularly want is a natural bijection from the basis in the tensor product to the product of the original bases, sending the induced crystal structure to the crystal tensor product.</p>