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Let $G$ be a simply connected Lie group with Lie algebra $\mathfrak{g}$ and $r \in \bigwedge^2 \mathfrak{g}$. For $x \in G$ let $\lambda_x$ denote the left multiplication by $x$. Let $[\cdot, \cdot]$ denote the Schouten bracket for multivector fields. Then show that $$\left [\left ( d_e \lambda_x \right )^{\otimes 2} (r), \left (d_e \lambda_x \right )^{\otimes 2} (r) \right ] = \left (d_e \lambda_x \right )^{\otimes 3} \left ([[r,r]] \right )$$ where $[[r,r]] = \left [r_{12}, r_{13} \right ] + \left [r_{12}, r_{23} \right ] + \left [r_{13}, r_{23} \right ]$, i.e., if $r = \sum_i x_i \otimes y_i$, then $$[[r,r]] = \sum_{i,j} [x_i, x_j] \otimes y_i \otimes y_j + \sum_{i,j} x_i \otimes [y_i, x_j] \otimes y_j + \sum_{i,j} x_i \otimes x_j \otimes [y_i, y_j] \,.$$

This amounts to say that the left action respects the Schouten bracket. I have asked the same question in MSE and here's the link but unfortunately the question is closed. This is the place I got stuck and hence I am seeking help. Could anyone give some suggestion or insight as to how to come up with this equality?

Thanks for your time.

Source $:$ Etingof lecture notes on compact quantum groups (Alternative intuitive approach to Theorem $3.1$ in page no. $28$).

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  • $\begingroup$ If your question was closed on MSE for lack of context, you should really rewrite it to include more context rather than posting it verbatim to MO. $\endgroup$
    – Callum
    Commented Jan 8, 2023 at 15:51
  • $\begingroup$ @Callum$:$ I don't know how to show the equality. The computation seems complicated to me. That's why I am seeking for help. Don't know how to show the equality in a convenient way. $\endgroup$ Commented Jan 8, 2023 at 16:15
  • $\begingroup$ @Callum$:$ What I noticed is that the bivector field given by $x \mapsto (d_e \lambda_x)^{\otimes 2} (r)$ is a left invariant bivector field; hence locally it is the sum of wedge product of two left invariant vector fields. Don't know whether it will be of any importance here. $\endgroup$ Commented Jan 8, 2023 at 16:18
  • $\begingroup$ @Callum$:$ I think I have found a way to compute it. $\endgroup$ Commented Jan 9, 2023 at 14:19

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