Is there a definition of a dual Lie algebra?

Question

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.

Answer 1

In some sense. The opposite Lie algebra $L^{op}$ has bracket the negative of the bracket of $L$. (When you exponentiate this up you'll see that this is compatible with the notion of opposite group.)

Taking the dual root system is more sophisticated and is related to taking the Langlands dual, which is really an operation on certain Lie groups and is not defined for all Lie algebras.

The vector space dual $L^{\ast}$ does indeed have a Poisson manifold structure, but the thing that resembles a Lie algebra is not the Poisson manifold itself but the algebra of functions on it. Here the algebra of polynomial functions on the dual is just $\text{Sym}(L)$, which does indeed have a structure extending the Lie bracket called a Poisson bracket making it a Poisson algebra (the sort of thing that's the algebra of functions on a Poisson manifold).

You can also define the dual of a Lie bialgebra by taking vector space duals, which is again a Lie bialgebra in the finite-dimensional case.

But if your question was actually "is there a natural Lie algebra structure on the vector space dual?" then I am fairly confident that the answer is no. There's no reason to expect this sort of thing a priori, e.g. if you have an associative algebra there isn't a natural associative algebra structure on its vector space dual either.

Answer 2

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 $L^*$.