As described here, we have a category of root systems, where a morphism from a root system $\Phi$ in a Euclidean space $E$ to a root system $\Phi'$ in $E'$ is given by a linear map $f: E \to E'$ such that $f(\Phi) \subseteq \Phi'$ and $f(\alpha) \dashv f(\beta) = \alpha \dashv \beta$ for all $\alpha, \beta \in \Phi$, where $\alpha \dashv \beta \mathrel{:=} 2 \frac{\langle \alpha | \beta \rangle}{\langle \beta | \beta \rangle} \in \mathbb{Z}$ and $\langle ~ |~ \rangle$ denotes the inner product on $E$. For isomorphisms of root systems, this last condition is in fact redundant, and follows automatically from the first two. This seems to be the "correct" notion of morphism for root systems: we have finite coproducts of root systems (which are oddly denoted by a product in most texts) and an initial object (the empty root system). Moreover, the Weyl group construction is a covariant functor on root systems.

Now, given a complex semisimple Lie algebra $\mathfrak{g}$, we can assign to it a root system $\Phi$, given by the set of nonzero weights of the representation of a Cartan subalgebra $\mathfrak{h} \leq \mathfrak{g}$ on $\mathfrak{g}$ via the adjoint action.

This assignment of a root system to a complex semisimple Lie algebra is an invariant, in the sense that isomorphic Lie algebras have isomorphic root systems (and conversely). But is this construction functorial? That is given a morphism of semisimple Lie algebras $f: \mathfrak{g} \to \mathfrak{g}'$, do we obtain a morphism of their root systems? If not a covariant functor, might this be a contravariant functor?

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One point that was raised by Mariano Suárez-Alvarez in the comments is that the choice of Cartan subalgebra is not functorial. Nevertheless, in the end the construction of a root system does not depend on the choice of Cartan subalgebra, which leaves me hope that this construction might still be functorial.

If it is indeed functorial, then the classification of semisimple complex Lie algebras would have an elegant description in terms of this "root system" functor: it is strong monoidal, conservative, and reflects simple objects (irreducible root systems being the simple objects in the category of root systems). Furthermore, since the Weyl group of a root system is functorial, then the assignment of the Weyl group to a Lie algebra would simply be the composite of the two functors.

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    $\begingroup$ Two comments. (a) It is better not to make the inner product part of the definition of a root system (see Serre, Complex Semisimple Lie Algebras), especially when you want to think functorially. (b) For a reductive group G over an algebraically closed field, Deligne and Lusztig 1976 (Annals), 1.1, define "the" maximal torus, "the" Borel subgroup, etc. of G, and hence "the" root system of G. $\endgroup$ – zeno Jul 23 '16 at 20:03

For everything you're requesting, it seems reasonable only to consider this question using maps that are isomorphisms, since: (i) the zero map between semisimple Lie algebras has no reasonable "associated" map between root systems, (ii) inclusions such as ${\rm{Sp}}_{2n} \rightarrow {\rm{SL}}_{2n}$ have no reasonable associated map either way between the root systems (the latter is simply laced of rank $2n-1$ and the former has two root lengths and rank $n$).

After one restricts the morphisms under considerations to be isomorphisms (which is where all of the substance lies anyway), the answer is "yes". It comes down to a trick, but there is some real content underlying the trick. I have no idea what "strongly monoidal" means, but if one thinks about how mathematicians use semisimple Lie algebras, semisimple Lie groups, and semisimple algebraic groups (after they're done being classified) then such a classification that is functorial with respect to isomorphisms is not only elegant but also useful to do important things.

More to the point, the need for such a version of the classification theorem (really in a form with split connected semisimple groups over general fields in place of semisimple Lie algebras over $\mathbf{C}$) already arose back in the 1960's and especially 1970's, as part of the structure theory of connected semisimple groups over general fields (to classify Galois-twisted forms, especially over fields of arithmetic interest and over $\mathbf{R}$) and in the definition of the Langlands dual group beyond the split case. It is sometimes attributed to Kottwitz, but the main idea can already be found in SGA3 in the discussion of the "scheme of Dynkin diagrams" (see Exp. XXIV, 3.1--3.4) and it has probably been rediscovered multiple times (e.g., by Tits for his notion of $\ast$-action of Galois groups on diagrams when formulating classification theorems over general fields without split hypotheses).

So finally the question comes down to this: is there a way to associate a root system $\Phi(\mathfrak{g})$ to a complex semisimple Lie algebra $\mathfrak{g}$ in a manner that is functorial with respect to isomorphisms? Sure! Let $G = G(\mathfrak{g}) = {\rm{Aut}}_{\mathfrak{g}/\mathbf{C}}^0$ be the identity component of the automorphism scheme of $\mathfrak{g}$. It is a basic fact from the structure theory of connected semisimple groups and semisimple Lie algebras over $\mathbf{C}$ that the groups $G(\mathfrak{g})$ are exactly the connected semisimple algebraic groups with trivial center, and the natural map $${\rm{Lie}}({\rm{Ad}}_G): {\rm{Lie}}(G) \rightarrow {\rm{Lie}}({\rm{GL}}(\mathfrak{g})) = {\rm{End}}(\mathfrak{g})$$ is an isomorphism onto $\mathfrak{g}$ (embedded via ${\rm{ad}}_{\mathfrak{g}}$). It is also a general fact that the natural action of $G(\mathbf{C})$ on $\mathfrak{g}$ is transitive on the set of pairs $(\mathfrak{h}, \mathfrak{b})$ consisting of Cartan subalgebras $\mathfrak{h} \subset \mathfrak{g}$ and Borel subalgebras $\mathfrak{b} \subset \mathfrak{g}$ containing $\mathfrak{h}$, and that the stabilizer in $G(\mathbf{C})$ of such a pair is the subgroup $T(\mathbf{C})$ where $T \subset G$ is the unique maximal torus whose Lie algebra is $\mathfrak{h}$ (inside $\mathfrak{g}$). For such a pair, let $(\Phi, \Delta)$ be the associated "based root datum" consisting of the root system for $(\mathfrak{g}, \mathfrak{h})$ and the root basis attached to $\mathfrak{b}$ (i.e., the simple roots in the positive system of roots consisting of those whose root lines are contained in $\mathfrak{b}$).

Note that the adjoint action on $T(\mathbf{C})$ on $\mathfrak{h}$ is trivial. Hence, if $(\mathfrak{h}', \mathfrak{b}')$ is another such pair in $\mathfrak{g}$, with $(\Phi', \Delta')$ the associated based root datum, then there is a canonical isomorphism $(\Phi, \Delta) \simeq (\Phi', \Delta')$: the adjoint action on $\mathfrak{g}$ arising from any element $g \in G(\mathbf{C})$ that carries $\mathfrak{h}$ onto $\mathfrak{h}'$ and carries $\mathfrak{b}$ onto $\mathfrak{b}'$ (such $g$ is unique modulo $T(\mathbf{C})$, so this isomorphism between based root data is independent of the choice of such $g$).

We define the canonical based root datum $(\Phi(\mathfrak{g}), \Delta(\mathfrak{g}))$ to be the inverse limit of all such $(\Phi, \Delta)$'s along the canonical isomorphisms just specified. More concretely, elements of $\Phi(\mathfrak{g})$ are exactly the compatible systems of roots with respect to those isomorphisms, and likewise for $\Delta(\mathfrak{g})$. It is clear from the construction that this pair is functorial with respect to isomorphisms among such $\mathfrak{g}$'s. Now compose this functorial construction (functorial with respect to isomorphisms) with the "forgetful" functor that drops the data of the root basis!

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    $\begingroup$ Very nice explanation! $\endgroup$ – Venkataramana Jul 23 '16 at 3:18
  • $\begingroup$ A trivial observation on a lovely answer: of course the rank of the root system of $\mathrm{SL}_{2n}$ is $2n - 1$. $\endgroup$ – LSpice Jul 26 '16 at 19:51
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    $\begingroup$ "inclusions such as ${\rm{Sp}}_{2n} \rightarrow {\rm{SL}}_{2n}$ have no reasonable associated map either way between the root systems" -- you might call this "not reasonable", but with $\alpha_i$ the usual basis, projecting along $\alpha_i - \alpha_{2n-i} = 0$ (so to say, the coinvariants of the non-trivial Dynkin diagram aut.), $A_{2n-1}$ projects onto $C_n$. I for my part have wondered about the inclusions $SO(2n) \subset SO(2n+1) \subset SO(2n+2)$ for some time: The former corresponds ("covariantly") to $D_n \subset B_n $, the latter ("contravariantly") to $D_{n+1}\twoheadrightarrow B_n$. $\endgroup$ – Torsten Schoeneberg Feb 1 '17 at 5:49
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    $\begingroup$ @TorstenSchoeneberg, it is actually precisely the projection you mention that allowed me to prove a result, which allowed me to reduce a problem for $Sp_{2n}$ to a problem for $SU(2n)$. Where can I find please a classification of all such maps between root systems? Should I open a new post? It is somewhat related to the OP's original question though. $\endgroup$ – Malkoun Mar 2 at 18:28
  • $\begingroup$ @Malkoun: I am delighted to hear that. I am not aware of a comprehensive treatment of this "folding" (projecting) of root systems; in the theory of $k$-rational forms it comes up a lot, but corresponding to Galois-semilinear maps; I would think that it also gives information in the linear case, but I've never seen that treated. I've brought it up in a comment in math.stackexchange.com/questions/1144878/…, and the special folding $E_6 \twoheadrightarrow F_4$ is used in mathoverflow.net/q/203295/27465 and mathoverflow.net/q/83747/27465. $\endgroup$ – Torsten Schoeneberg Mar 3 at 19:52

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