Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let G be a connected complex algebraic group G and X=G/B, where B is the Borel subgroup. $ M=G\times C^{*}$, where $C^*$ acts on X trivially. Let $K_M(X)$ be the equivariant K-theory. Let $s\in S$ be a simple reflection in the Weyl group of G, $P_s=B\cup BsB$ be the parabolic subgroup, and $\pi_{s}:X\rightarrow G/P_{s}$ be the natual map. Then there is a unique endomorphism $T_s$ on $K_M(X)$ such that $E+T_s E=\pi_s^{*}(\pi_s)_{*}(E)-\pi_s^{*}(\pi_s)_{*}(E\otimes \Omega_s^1) $, where $\Omega_s^1$ is the line bundle on X of holomorphic differential 1--forms along the fibre of $\pi_s$ and $C^{*}$ acts on it by multiplication on each fibre.

Let $Z[q,q^{-1}]$ be the group algebra of $C^{*}$, P be the lattice of weights on T, the maximal torus in B, and let $\mathfrak{R}_0$ be the group ring of P over $Z[q,q^{-1}]$. Note that $K_M(X)\cong K_G(X)\otimes Z[q,q^{-1}]$, and $\{L_p\}_{p\in P}$ form a Z-basis of $K_G(X)$, where $L_p$ is the line bundle on X associated to $p\in P$.

Now my question is, for $\lambda\in P$, how do I deduce the formula (8.1): $T_s(\lambda)=\frac{\lambda-s(\lambda)}{\alpha_s-1}-q\frac {\lambda-s(\lambda)\alpha_s}{\alpha_s-1}$ ? Thank you.

share|improve this question
    
I'd really like motivation for this question - is there something you're planning to use this for? I'd also like to get some sense of your background - from which viewpoint are you approaching this subject? (This is a case where having a real name for you would really help - even if I haven't heard of you at all I could at least figure out whom your advisor is/was, or at least whom you might have learned about this subject from!) –  Alexander Woo Sep 26 '12 at 20:24
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.