TITLE: Is there a definition of a dual Lie algebra?
QUESTION [18 upvotes]: Let $L$ be a Lie algebra. For vector spaces, modules, Banach spaces, etc. we have the notion of a dual. 
Question: Is it possible to define naturally a Lie algebra $L^*$ that is in some sense dual to $L$? 
If $L$ is semisimple over an algebraically closed field of characteristic zero, then $L$ is uniquely determined by its root system, and there is a good definition of a dual root system. So if $L$ has root system $\Phi$, we could define $L^*$ to be the semisimple Lie algebra with dual root system $\Phi^\vee$. Does this make sense? Is this a good definition? Is there a definition of a dual that would be consistent with this?
Looking up "dual of a Lie algebra" on the internet brings up pages about Lie coalgebras, which seems to be the dual of the category of Lie algebras. I also found that we can define a Poisson manifold structure of the dual vector space $L^*$. The definition of a Poisson  manifold looks similar to that of a Lie algebra, but I do not really know anything about this, and the material on that wikipedia page goes way over my head.

REPLY [6 votes]: I would like to add a word on the Lie bialgebra thing.
In your wording this can be seen as follows: say that you have a Lie algebra structure on $L$ and a Lie algebra structure on the dual $L^*$ (coming out of the blue) how can you require them to be compatible in any possible sense? You have a couple of equivalent ways: consider the relation between $L$ and its dual $L^*$ as a pairing 
$\langle X,\xi\rangle=\xi(X)$ for every $X\in L$, $\xi\in L^*$; 
this extends to an inner product on $L\oplus L^*$. Then you can require your bracket to extend to a Lie bracket on $L\oplus L^*$ invariant under this pairing, i.e.
$\langle [A,B],C\rangle=-\langle B,[A,C]\rangle\, \forall A,B,C\in L\oplus L^*$
This is the requirement that your Lie algebra is a Lie bialgebra and admits a Lie cohomology interpretation.
Is this canonical? No. On any Lie algebra the null Lie bracket on its dual defines a Lie bialgebra. Furthermore on a given Lie algebra you have in principle many inequivalent Lie bialgebra sturctures on it (even difficultto classify).
What if $L$ is semisimple and you have at your disposal all the bla-blas of root systems... well, things get interesting, then. Let  $K$ be a compact Lie group, such that $L=Lie(K)$ and let us say we are in the case in which $L_\mathbb C$ the complexification is simple. Fix a Cartan decomposition of  $L_\mathbb C$ with $H_1,\ldots, H_n$ basis of Cartan and  $X^\pm_\alpha$ positive and negative root vectors. then $L$ is $\mathbb R$--linearly spanned by
$\imath H_i$, $E_\alpha=X^+_\alpha-X^-_\alpha$, $F_\alpha=\imath(X^+_\alpha+X^-_\alpha)$ 
there is a unique map $\delta:L\to L\otimes L$ extending
$\delta(H_i)=0, \delta(E_{\alpha_i})=d_i E_{\alpha_i}\otimes \imath H_i,
\delta(F_{\alpha_i })=d_i F_{\alpha_i}\otimes \imath H_i$
such that $\delta^*:L^*\otimes L^*\to L^*$ defines a Lie bracket and this forms a Lie bialgebra.. So, in a way, this specific Lie bialgebra structure is quite canonical. Is $L^*$ semisimple as well? Not at all. It turns out that the Lie bracket on $L\oplus L^*$ is isomorphic to the one on $L_\mathbb C$ (which is indeed complex semisimple) but $L^*$ is a solvable Lie algebra. At the level of Lie groups this integrates to the Iwasawa decomposition
$G=KAN$
where $G$ is the complex semisimple integrating $L_\mathbb C$ and $AN$ is the real solvable integrating $L^*$.