Quantum Serre relations and braided commutator.

Question

I am reading the lecture notes. On page 21, it is said that when $a_{ij}=-1$, we have \begin{align} ad_c(x_i)^{1-a_{ij}}(x_j)=x_i^2x_j - (q+q^{-1})x_ix_jx_i+x_jx_i^2. \quad (1) \end{align} Here $ad_c(x_i)(x_j)=x_ix_j - q^{a_{ij}}x_jx_x$. I am trying to verify $(1)$. We have \begin{align} & ad_c(x_i)^{1-a_{ij}}(x_j)\\ & =ad_c(x_i)^{2}(x_j) \\ & =ad_c(x_i)(x_ix_j - q^{-1}x_jx_i) \\ & = x_i^2 x_j -q^{-1}x_ix_jx_i -q^{-1}( x_ix_j - q^{-1}x_jx_i )x_i \\ & = x_i^2 x_j -2q^{-1}x_ix_jx_i + q^{-2}x_jx_i^2. \end{align} But I didn't get $x_i^2x_j - (q+q^{-1})x_ix_jx_i+x_jx_i^2$. I don't know where I made a mistake. Thank you very much.

Answer 1

When $a_{ij}=-1$, \begin{align} & ad_\Psi(x_i)^{1-a_{ij}}(x_j) \\ & = ad_c(x_i)^2(x_j) \\ & = ad_c(x_i)(x_i \otimes x_j - K_i.(x_j) \otimes x_i) \\ & = ad_c(x_i)(x_i \otimes x_j - q^{-1} x_j \otimes x_i) \\ & = x_i \otimes x_i \otimes x_j - K_i.(x_i \otimes x_j) \otimes x_i - q^{-1} ( x_i \otimes x_j \otimes x_i - K_i.(x_j \otimes x_i) \otimes x_i) \\ & = x_i \otimes x_i \otimes x_j - K_i.x_i \otimes K_i.x_j \otimes x_i - q^{-1} ( x_i \otimes x_j \otimes x_i - K_i.x_j \otimes K_i.x_i \otimes x_i) \\ & = x_i \otimes x_i \otimes x_j - q^2 x_i \otimes q^{-1} x_j \otimes x_i - q^{-1} ( x_i \otimes x_j \otimes x_i - q^{-1} x_j \otimes q^2 x_i \otimes x_i) \\ & = x_i \otimes x_i \otimes x_j - q x_i \otimes x_j \otimes x_i - q^{-1} x_i \otimes x_j \otimes x_i + x_j \otimes x_i \otimes x_i \\ & = x_i \otimes x_i \otimes x_j - (q+q^{-1}) x_i \otimes x_j \otimes x_i + x_j \otimes x_i \otimes x_i. \end{align}

Answer 2

I think you are making a mistake in the fourth line of your computation: you compute $-q^{-1}ad_c(x_i)\big(x_j\big)x_i \ $, instead of the correct: $\ -q^{-1}ad_c(x_i)\big(x_jx_i\big)$. However, it seems that it does not affect the final result. Here's what I get: \begin{align} & ad_c(x_i)^{1-a_{ij}}(x_j)=\\ & =ad_c(x_i)^{2}(x_j)= \\ & =ad_c(x_i)(x_ix_j - q^{-1}x_jx_i)= \\ & = x_i^2 x_j -q^{-1}x_ix_jx_i -q^{-1}ad_c(x_i)\big(x_jx_i\big)= \\ & = x_i^2 x_j -q^{-1}x_ix_jx_i-q^{-1}\big(x_ix_jx_i-q^{-1}x_jx_i^2\big)= \\ & = x_i^2 x_j -q^{-1}x_ix_jx_i-q^{-1}x_ix_jx_i+q^{-2}x_jx_i^2 = \\ & = x_i^2 x_j -2q^{-1}x_ix_jx_i+q^{-2}x_jx_i^2 \end{align} but it's still different from the result of the paper. Maybe there's some typo in the paper.

It is interesting to note that the two expressions $$ x_i^2 x_j -2q^{-1}x_ix_jx_i+q^{-2}x_jx_i^2 $$ and $$ x_i^2x_j - (q+q^{-1})x_ix_jx_i+x_jx_i^2 $$ coincide if $q=q^{-1}$. (However, in such cases, $q$ is customarily considered not to be a root of unity).

Hope that helps a bit.