How much can you say of a Lie algebra knowing the Weyl Group?

Question

The question is exactly the one stated in the title: "How much can you say of a semisimple Lie algebra knowing just the Weyl Group?".

Then if you prefer I have few more restrictions:

1) Knowing that the algebra is simple;

2) Knowing that the ground field is of complex numbers.

If you have some references for the reconstructing procedures from Weyl Group to the Lie algebra they're welcomed. Thank you

Answer 1

My favorite source for this staff is Bourbaki "Lie Groups and Lie Algebras", Vol. 4-6. In short, there is a bijection between finite-dimensional complex semisimple Lie algebras and finite reduced root systems. An irreducible root system is almost uniquely determined by the corresponding finite reflection group (the Weyl group). The sole exception is the fact that the Weyl groups $W$ of the root systems $B_n$ and $C_n$ happen to be isomorphic (as reflection groups, meaning that an abstract isomorphism sends reflections to reflections), while $B_n$ is not isomorphic to $C_n$ if $n\ge 3$. Other than that, non-isomorphic irreducible root systems have non-isomorphic Weyl groups. The proof of the latter is simply by inspection (looking through the list of Dynkin diagrams).

One more thing: If you do not insist on isomorphisms sending reflections to reflections then you get some weird examples such as $G_2\cong A_2\times A_1$ (the dihedral group of the order 12 is isomorphic to the product of the dihedral group of order 6 and with ${\mathbb Z}_2$). On the other hand, of course, the Lie algebra ${\mathfrak g}_2$ is not isomorphic to $sl(3)\oplus sl(2)$.

Edit. As requested, one more "exotic" isomorphism between finite Weyl groups is: $$ B_{2k+1}\cong A_1 \times D_{2k+1}, $$ everything else can be traced to these two examples. (I am sure this was known long time ago.) See section 2.3 of

B. Baumeister, T. Gobet, K. Roberts, P. Wegener, On the Hurwitz action in finite Coxeter groups, Journal of Group Theory, Vol. 20 (2017).

As for infinite finitely generated Coxeter groups, much is known about "diagram rigidity", but no complete answer (yet).

Answer 2

As the other answer deals with the case of a split Lie algebra, which is the only case if your ground field $k$ is algebraically closed, I just want to add some words about what happens if you drop that assumption. But let's stay in characteristic $0$.

You would have to say what you mean by "the Weyl group" because there are (at least) two different groups now: The "absolute" Weyl group, which is the one of a scalar extension of your Lie algebra to an algebraically closed field (basically coming from the roots of a maximal torus); and the "relative" Weyl group, which is the one corresponding to the so-called $k$-rational root system of your Lie algebra over $k$ (basically the roots of a maximal split torus). It turns out the relative Weyl group is a subquotient of the absolute one.

However, none of them individually gives you full information, and even knowing both of them and how they are related will not, in general, determine the Lie algebra up to isomorphism.

Examples:

• The absolute Weyl group does not suffice: Over most fields that are not algebraically closed, there are various non-isomorphic simple Lie algebras that become isomorphic after scalar extension to an algebraic closure. (So-called forms of a certain type). All forms of one type have the same absolute Weyl group, but usually there are several non-isomorphic ones. This happens over $\mathbb{R}$ as well as over $p$-adic, or number fields. (Actually, one easy sufficient criterion to get some non-isomorphic forms at least for some types is for the ground field to have a quadratic or cubic extension.)
• The relative Weyl group does not suffice: Over $\mathbb{R}$, every compact (semisimple) Lie algebra has trivial relative Weyl group; but there is one of these forms for each type, and of course Lie algebras of different types are very non-isomorphic.
• Even knowing both does not suffice, if $k$ is a number field: Let's take $\mathbb{Q}$ for example. Let $p \neq q$ be two distinct primes, both congruent 3 modulo 4. The generalised quaternion algebras $\displaystyle \left(\frac{-1, p}{\mathbb{Q}}\right)$ and $\displaystyle \left(\frac{-1, q}{\mathbb{Q}}\right)$ are both division algebras with centre $\mathbb{Q}$. Consider them as Lie algebras with the commutator as Lie bracket. The "pure quaternions" of each of them (which form the derived Lie algebra; or isomorphically, the quotient modulo the centre) are three-dimensional Lie algebras which are forms of $\mathfrak{sl}_2$ (i.e. of type $A_1$). Hence, they have the same absolute Weyl group $S_2$. Further, they are both "compact" / "anisotropic", hence have trivial relative Weyl groups. However, they are not isomorphic. Similar examples exist for every number field.
• Knowing both also does not suffice for $p$-adic fields: Over such a field $k$, for each natural number $d$, there are $\phi(d)$ (Euler's $\phi$) different $k$-central skew fields of degree $d$ up to isomorphism, and $\lfloor (\phi(d)+1)/2\rfloor$ ones up to isomorphism or anti-isomorphism. The derived Lie algebras of these last classes are mutually non-isomorphic. But scalar extension of any of them with any field extension $K|k$ of degree $d$ gives $\mathfrak{sl}_d(K)$. Consequently, they all have absolute Weyl group $S_d$, and it is true that they also all have trivial relative Weyl group (again they are "compact" or "anisotropic" forms of type $A_{d-1}$). But e.g. for $d=5$, there are two non-isomorphic ones.

As for good news, I think that over $\mathbb{R}$, knowledge of both the absolute and the relative Weyl group might be enough to determine the isomorphism class of the Lie algebras (modulo the issue at the end of Moishe Cohen's answer). That is essentially due to a case-by-case check and the fact that over the reals, there is a unique compact/anisotropic forms of each type. As said above, that is not true e.g. over number fields or $p$-adic fields, and makes the above counterexample possible.