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I'm trying to decompose the Kazhdan-Lusztig C' basis element associated to the longest word in $S_n$, $C'_{w_0}$ into products and sums of elements $C'_w$ where $w < w_0$ in the Bruhat order. For example I know that $C'_{s_1s_2s_1}=C'_{s_1}C'_{s_2}C'_{s_1}-C'_{s_1}$.

The question I have is the following: Are such decompositions known as a general fact, or would I have to go through the process of changing the basis back to the standard basis and mapping back to the KL basis after the fact? Of course the latter can be done by hand (or computer for higher n) but I was just curious if there was a better way to approach this that I've overlooked as a non-expert.

Thanks!

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I think that this would be very difficult in general, as it depends on choices of reduced expression. Ben Elias has found a nice answer for dihedral groups in terms of representations of sl_2. However you are asking about $S_n$, so these results probably aren't so useful. –  Geordie Williamson Apr 19 '13 at 5:43
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Is the expansion $$C'_w = C'_{ws}C'_s - \sum_{z\leq ws;\, zs < z} \mu(z,ws) C'_z$$ for $ws < w$ what you're looking for? ($s$ is a simple reflection.) Here, $\mu(z,ws)$ is the coefficient of $q^{(\ell(ws)-\ell(z)-1)/2}$ in the Kazhdan-Lusztig polynomial $P_{z,ws}(q)$. Perhaps there is a simplification for the specific case of $w=w_0$, but I don't know it. For general $w$, a simpler formula would be tantamount to having a handle on the combinatorially elusive $\mu$-coefficients.

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It seems like that will be helpful, thanks! This is much simpler than the method I mentioned earlier... –  Michael Abel Apr 19 '13 at 2:23
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