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It's a theorem of Lusztig that if one takes two (or more) representations of a quantum group at a non-root-of-unity choice of $q$, and look at the canonical basis of their tensor product, then each isotypic component is spanned by a subset of the canonical basis (i.e. each basis vector lies in an isotypic component) [EDIT: I mean the canonical basis of the tensor product in the sense of Lusztig's paper "Canonical bases on tensor products" not the tensor product of the canonical bases].

When one reduces this tensor product at a root of unity instead it's no longer necessarily semi-simple. But if we assume that each of the tensor factors does remain simple, it will be a tilting module, and have a canonical direct sum decomposition in "isotypic" components corresponding to tilting modules. These are quite different from the isotypic components at a generic value of $q$. Are these still spanned by canonical basis vectors?

EDIT: I would be basically equally happy if there were a split (but not canonically) filtration whose individual spaces were spanned by canonical basis vectors such that the successive quotients where tilting modules.

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    $\begingroup$ I had thought that the canonical basis of a tensor product does not refine the isotypic direct sum decomposition. Rather, my understanding was that it refines the induced descending filtration, using the partial ordering on isotypics induced by the root-step partial ordering on dominant weights. Some form of your question could still survive even if this is true. $\endgroup$ Commented Jan 27, 2010 at 22:37
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    $\begingroup$ Ben: there is no canonical decomposition of tilting module into direct sum of isotypic component. For example trivial module (which is tilting) might be a submodule of another tilting module (e.g. of its own projective cover); the direct sum of these two does not split canonically.. $\endgroup$ Commented Jan 27, 2010 at 23:07
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    $\begingroup$ Maybe you could still ask whether the basis filtration-refines some decomposition into tilting modules, even if it isn't a canonical decomposition. $\endgroup$ Commented Jan 27, 2010 at 23:46
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    $\begingroup$ I think part of the confusion here is that there is more then one possible notion of canonical basis for the tensor product. One can consider the tensor product of the canonical basis of each factor. This does not restrict to a basis of each isotypic component. On the other hand there is a notion of a canonical basis of the tensor product itself, which I think does have this property. The distinction is explained for the simplest case in arxiv.org/pdf/math/0511467 section 5.2, but I think it has been defined in greater generality (probably by Lusztig and/or Kashiwara) $\endgroup$ Commented Feb 7, 2010 at 16:31
  • $\begingroup$ That's what I thought. It's in Lusztig's book. $\endgroup$
    – Ben Webster
    Commented Feb 7, 2010 at 18:24

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Have you tried looking at this question using web bases instead of Lusztig's (dual) canonical basis? These have only been worked out for rank one, rank two (Greg Kuperberg) and $B_3=Spin(7)$, so far ... .

I also agree that these web bases were introduced to study the invariant tensors. However they do also give bases for representations.

I also agree that this is a different question. However these two bases have similar properties. This alternative question may be easier and may (or may not) suit your purpose.

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