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Consider the Hecke algebra $H_n$ of type $A_{n-1}$ with standard basis $T_w$, $w \in S_n$ with the quadratic relations $(T_s - u) (T_s + u^{-1}) = 0$ and braid relations. The unsigned canonical basis $C'_w$, $w \in S_n$ gives rise to a basis for the irreducible $H_n$-module $M_\lambda$ of shape $\lambda$: fix some SYT $T$ of shape $\lambda$; then this basis is {$C'_w : P(w) = T$} after quotienting by cells lower down in the Kazhdan-Lusztig preorder and $P(w)$ is the insertion tableau of $w$.

Similarly, $M_\lambda$ has a basis coming from the signed canonical basis $C_w$, $w \in S_n$.

What is known about the transition matrix between these two bases? Does it become the identity matrix at $u=0$?

This seems to be trickier to understand than the transition matrix between all the $C$s and all the $C'$s. I expect I will be able to prove the second question on my own, but I'd rather cite it if it's in the literature somewhere.

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$C'_w \equiv T_w \equiv C_w$ mod $u$ by definition, so "Yes" to your second question. –  Sammy Black Mar 26 '10 at 0:20
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up vote 2 down vote accepted

Not much seems to be known about this matrix in general, and it does become the identity at $u=0$. I show this in the paper Quantum Schur-Weyl duality and projected canonical bases http://arxiv.org/abs/1102.1453 using quantum Schur-Weyl duality and its compatibility with canonical bases. This only proves it in type A. I do not know what happens in other types.

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You probably know all of this already, but here goes...

Write $C'_w = T_w + \sum_{x < w} p_{x,w} T_x$ where $p_{x,w} \in u\mathbb{Z}[u]$. Now, the other basis can be defined by applying the involutive automorphism $b: \mathcal{H}_n \to \mathcal{H}_n$, given by $b(T_w)=T_w$ and $b(u)=-u^{-1}$.

Define $C_w := b(C'_w)$.

Since, $b$ commutes with the bar involution, this basis is bar invariant as well.

Explicitly, $C_w = T_w + \sum_{x < w} (-1)^{\ell(w)+\ell(x)} \bar p_{x,w} T_x$.

So $C_w = \bar{P}^{-1} P C'_w$ which seems hard to compute in general.

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Don't you mean $C_w=\overline{P}P^{-1}C_w'$ ? –  Johannes Hahn Sep 11 '11 at 12:49
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