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  • For regular subalgebras $h$ of some group's Lie algebra $g$, $$ h \subset h $$ the root system of the subalgebra is a subset of the root system of the original's group algebra.
  • Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras

Nevertheless, there must be some map from the root system of the subalgebra to the root system of the original algebra because that's how embedding a subalgebra is defined. My problem is finding this map.

In other words: Given a set of roots for the original algebra $g$ and a subset of this root system related in some way to the root system of the special subalgebra. This subset is, by definition of a special subalgebra, not directly the root system of the subalgebra, but there must be some map to the correct root system of the special subalgebra. How can I find the corresponding map?

Torsten Schoeneberg
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jak
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    Are you taking some special kind of subalgebra? Or what do you mean by root system? – Tobias Kildetoft Jul 09 '15 at 08:15
  • @TobiasKildetoft I'm currently looking at special subalgebras of $E_6$. My subalgebra appears in the (physical) context of symmetry breaking. After symmetry breaking only some of the generators (roots) of $E_6$ remain unbroken and generate a new algebra, i.e. the subalgebra. By definition, for special subalgebras the root system of the subalgebra is not a subset of the original root system of $E_6$. – jak Jul 09 '15 at 10:20
  • But you need to assume something about the subalgebra to even be able to speak about the root system of it. You have not actually specified what a "special" subalgebra is (it might be a common term, I just have not seen it before). – Tobias Kildetoft Jul 09 '15 at 10:24
  • @TobiasKildetoft a special subalgebra is defined to be a subalgebra that is not regular. For every Lie algebra $g$ we can write down the corresponding roots = root system of the algbera. If some subset of the roots of $g$ generate a new algebra, let's call it $h$, we have found a regular subalgebra $ h\subset g$. In physics we have a method to compute "symmetry breaking", which means that some generators (roots) get "broken". The remaining, unbroken, generators are a subset of the original ones. Equivalently the unbroken roots are a subset of the original set of roots. – jak Jul 09 '15 at 10:31
  • @TobiasKildetoft The task is then to figure out which algebra these generators generate. If we're dealing with a regular subalgebra we can see it immediately, because the subset of roots coincides with the root system of another algebra. Then we have found the regular subalgebra $h$ that remains after symmetry breaking. Now, if the remaining root systems does NOT coincide with some the root system of another algebra we're dealing with a special subalgebra after symmetry breaking. I'm trying to figure out how to determine which special subalgebra we're dealing with. – jak Jul 09 '15 at 10:33
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    I am really not sure what you mean by writing down the roots of an arbitrary Lie algebra. You mean take a maximal toral subalgebra and consider non-zero weight spaces for the action of this? But this need not give a root system. – Tobias Kildetoft Jul 09 '15 at 10:36
  • @TobiasKildetoft roots are the weights of the adjoint representation. We can define every representation by writing down the corresponding weights. The weights are labelled by the eigenvalues of the Cartan (=diagonal) generators. For every algebra, we can write down the weights of the adjoint representation and these are the root system. – jak Jul 09 '15 at 10:44
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    Those will not in general form a root system (in the abstract sense). And the roots in this sense will not be elements of the Lie algebra itself, but in the dual of the maximal toral subalgebra. – Tobias Kildetoft Jul 09 '15 at 10:49

1 Answers1

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  1. As Tobias Kildetoft points out, the question does not make sense in this generality. Root systems are reasonably defined only for semisimple, or at best reductive, Lie algebras. But even a simple Lie algebra generally has tons of subalgebras which are far from being reductive. Or what would you say is "the root system of $\mathfrak{h}:=\pmatrix{ * & \dots & * & * \\ 0 & \ddots & & * \\ 0 & 0 & \ddots & * \\ 0 & 0 & 0 & *} \subset \mathfrak{sl}_n(k)$"?

  2. Even if we restrict to semisimple subalgebras $\mathfrak{h} \subset \mathfrak{g}$, in general there is no reasonable map $R_\mathfrak{h} \rightarrow R_{\mathfrak g}$ between their respective root systems. For example look at the inclusion $\mathfrak{so}_5(\mathbb C) \subset \mathfrak{sl}_5(\mathbb C)$. Good luck finding a map from the root system $B_2$ to the root system $A_4$. So your statement

Nevertheless, there must be some map from the root system of the subalgebra to the root system of the original algebra because that's how embedding a subalgebra is defined.

must be based on a misunderstanding. Maybe you mean something the other way around, like "if there is an embedding of root systems $R_1 \hookrightarrow R_2$, and $\mathfrak{g}_1, \mathfrak{g}_2$ are split (say, complex) semisimple Lie algebras having the respective root systems, then there's an embedding $\mathfrak{g}_1 \hookrightarrow \mathfrak{g}_2$. (I'm not even sure if that is literally true, in any case it's non-trivial.)

But think of this: If you have an automorphism of $\mathfrak{g}$, then its fixed points form a subalgebra $\mathfrak{h}$ of $\mathfrak{g}$. In some nice examples this subalgebra is reductive or even semisimple. If the automorphism was chosen so as to stabilise a certain Cartan subalgebra, then one can get the root system of $\mathfrak{h}$ as (subsystem of) a "folding" (quotient) of the root system of $\mathfrak g$. Compare e.g. Semisimple complex Lie algebra of type $A_3$ contains a Lie subalgebra of type $B_2$ as fixed points of an automorphism., Folding and realization of type B Lie group as a subgroup of $GL_n$., Understanding $G_2$ inside Spin(7)? (EDIT: problem solved), and the comments to https://mathoverflow.net/a/244895/27465.

The main reason I bring up this class of examples is that here one does have a map between the root systems, but here it is the other way around: In the first question linked above, the inclusion $$\mathfrak{sp}_4(\mathbb C) \hookrightarrow \mathfrak{sl}_4(\mathbb C)$$ corresponds "contravariantly" to a quotient map $$A_3 \twoheadrightarrow C_2.$$

And sometimes it is even more intricate: "Folding" of $A_4$ gives the non-reduced root system $BC_2$, which has $B_2$ as a reduced subsystem. That's the connection of the root systems $B_2$ and $A_4$ that governs my above example $\mathfrak{so}_5(\mathbb C) \subset \mathfrak{sl}_5(\mathbb C)$:

$$\matrix{A_4 &\twoheadrightarrow & BC_2 \\ && \cup \\ && B_2}$$

You cannot express this with one map between $A_4$ and $B_2$ in either direction. So if there is any way in which one turns root systems into a category such that inclusions of semisimple Lie algebras would correspond to morphisms between root systems, these morphisms would not be maps in the standard sense: It's at least something like above, taking subsystems of quotients. (I am not even sure if that would "catch" all subalgebras. I do believe that in principle such an assignment would need to be contravariant though, i.e. to a morphism $\mathfrak{h} \rightarrow \mathfrak{g}$ would correspond a morphism $R_{\mathfrak{g}} \rightarrow R_{\mathfrak{h}}$.)

  1. On the positive side, one finds certain very special kinds (in particular, as you say "regular" -- a nomenclature which unfortunately is totally overused for c. 200 different things) of subalgebras by looking at (special kinds of) sub-root systems. That's the content of Borel-de Siebenthal and related theories ("Branching Rules"). You say you are interested in $E_6$, maybe this helps: https://mathoverflow.net/q/314025/27465. (On the other hand, folding gives something like a ghost of a group of type $F_4$ inside $E_6$ groups, cf. https://mathoverflow.net/q/203295/27465.)
Torsten Schoeneberg
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