# Left-/right-translate of a two-form

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 $$r_{ij}$$?

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 $$\begin{gather} r^{\lambda}(g)(\xi,\eta):=r(T_g\lambda_{g^{-1}}\cdot\xi,T_g\lambda_{g^{-1}}\cdot\eta))\\ r^{\rho}(g)(\xi,\eta):=r(T_g\rho_{g^{-1}}\cdot\xi,T_g\rho_{g^{-1}}\cdot\eta)). \end{gather}$$
• @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! – Anne O'Nyme Mar 25 at 7:58