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The context is that of coboundary Lie bialgebras discussed in "Lie bialgebras, Poisson Lie groups and dressing transformations" by Y. Kosmann-Schwarzbach.

In section 4.2, she defines objects like $r^\rho$ and $r^\lambda$ where $r$ is an element of $\Lambda^2\frak g$ and the superscripts $\rho$ and $\lambda$ should stand for the right - and left-translates of the element $r$ that we could write as $$r= \sum_{i,j} r_{ij}T_i\wedge T_j \,,$$ for $T_i$ a basis of the Lie algebra $\frak g$.

However I can't find any explantation to what this notation, e $r^{\lambda/\rho}$, actually means...

On a group element $h$ the left and right translation are evident: $$\lambda_g(h)=gh\,, \quad \rho_g(h)=hg,$$

where $g,h$ are in the connected Lie group whose algebra is $\frak g$. But I don't understand how a left- or right-translation is actually computed for a two-form like above.


EDIT:

More handson I would like to understand how for example the Poisson structure constructed out of $r$ by $$\pi = r^\lambda - r^\rho \,,$$ looks in terms of the decomposition of $r$ in terms of the components $r_{ij}$ and the generators $T_i$. That is can we express the components $\pi_{ij}$ i.t.o. of the group element $g$ and $r_{ij}$?

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What is meant is that you defined the action of $r$ on general tangent vectors by tranlating the vectors to the neutral element via a left or right translation. So for $g\in G$ and tangent vectors $\xi,\eta\in T_gG$, one puts \begin{gather} r^{\lambda}(g)(\xi,\eta):=r(T_g\lambda_{g^{-1}}\cdot\xi,T_g\lambda_{g^{-1}}\cdot\eta))\\ r^{\rho}(g)(\xi,\eta):=r(T_g\rho_{g^{-1}}\cdot\xi,T_g\rho_{g^{-1}}\cdot\eta)). \end{gather}

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  • $\begingroup$ @Andres Cap Thanks that already helps! But I would like to understand how the action of $g$ can be expressed at the level of the components rather than in abstract notation. I added a more detailed question above. Thanks in advance! $\endgroup$ – Anne O'Nyme Mar 25 at 7:58

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