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Definition $:$ A Lie group $G$ is said to be a Poisson Lie group if it is equipped with a Poisson structure and the multiplication map $m : G \times G \longrightarrow G$ is a Poisson map i.e. for any $f, g \in C^{\infty} (G)$ and $x, y \in G$ the following equality holds $:$

$$\{f, g\} (xy) = \{f \circ \rho_{y}, g \circ \rho_{y} \} (x) + \{f \circ \lambda_{x}, g \circ \lambda_{x} \} (y)$$ where $\{\cdot, \cdot \}$ is the Poisson bracket on $G$ and $\rho_{y} : G \longrightarrow G$ and $\lambda_{x} : G \longrightarrow G$ are respectively the right translation by $y$ and the left translation by $x.$

Pavel Etingof, in his lecture notes on quantum groups, pointed out that in terms of the Poisson bivector $\Pi$ associated to $G$ the above relation takes the following form $:$ $$\Pi (xy) = (d_{x} (\rho_{y}) \otimes d_{x} (\rho_{y})) \Pi (x) + (d_{y} (\lambda_{x}) \otimes d_{y} (\lambda_{x})) \Pi (y).$$

But I can't get his point. Could anyone please shed some light on it?

Thanks for your time.

EDIT $:$ I think I have to first find out the equivalence of two definitions of bivector fields. The first one being the section of two fold wedge product of the tangent bundle and another one is given in the form of $C^{\infty} (G)$-bilinear, skew symmetric map from $\Omega^{1} (G) \times \Omega^{1} (G) \longrightarrow C^{\infty} (G).$

Anil Bagchi.
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  • This answer may be helpful to you (what you're asking is Proposition 1 in Part 2): https://math.stackexchange.com/questions/4443798/how-to-show-that-the-inversion-map-is-a-anti-poisson-map/4586194#4586194 – T.P. Dec 03 '22 at 16:32
  • @T.P.$:$ Sorry it's my fault. Both the RHS and the LHS are in $T_{xy} G \otimes T_{xy} G.$ – Anil Bagchi. Dec 03 '22 at 16:47
  • @T.P.$:$ I don't understand your notation. What is $\wedge^{2} T_{g} R_{h}\ $? Is it $d_{g} (R_{h}) \otimes d_{g} (R_{h})\ $? – Anil Bagchi. Dec 03 '22 at 16:51
  • I use $\pi_{xy}$ for your $\Pi(xy)$; $T_gR_h$ are the differentials, so your $d_x\rho_y$; the maps $\wedge^2T_gR_h:=(T_gR_h)\wedge (T_gR_h)$ are your $d_g\rho_y \otimes d_g \rho_y$. – T.P. Dec 03 '22 at 16:56
  • @T.P.$:$ All these are pretty unconventional notations except $\pi.$ Which book have you got these notations in? Usually by $T_{g}$ most authors mean the tangent space at $g$ when the underlying manifold is understood. – Anil Bagchi. Dec 04 '22 at 05:08
  • @T.P.$:$ The definition of Poisson map you are using seems different than that of mine. – Anil Bagchi. Dec 04 '22 at 05:26
  • @T.P.$:$ What is your $\Pi.$ You mean $\mu_{\ast} \Pi = \pi.$ What is it? Is it the pushforward? But I don't know how to pushforward multivector fields. – Anil Bagchi. Dec 04 '22 at 05:49
  • $Tf$ instead of $df$ in this setting is a very common notation and I personally prefer it for these two reasons: 1) because it distinguishes between the differential of forms and the differential of smooth maps viewed as a map between tangent manifolds, 2) $T$ is a functor that acts on objects (smooth manifolds) as $TM$ and of morphisms as $Tf$.

    The pushforward in this case is $\phi_*\Pi = \pi \Leftrightarrow (d_x\phi \otimes d_x\phi)\Pi_x = \pi_{\phi(x)}$

     – T.P. Dec 04 '22 at 09:14
  • @T.P.$:$ I have few questions. $(1)$ How to show that Poisson bivectors are related by Poisson maps? – Anil Bagchi. Dec 04 '22 at 09:23
  • That follows easily by unraveling the definitions. Oh and I forgot, a lot of these things can be found e.g. in the https://link.springer.com/book/10.1007/978-3-642-31090-4 – T.P. Dec 04 '22 at 09:25

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