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Where does the name "toral" come from in "toral subalgebra"?

I know a little (very little) Lie groups theory, so I guess it could be related to a Lie group whose Lie algebra is the toral one. Is this Lie group a torus or toral in any sense?

Edit: I am not the OP, but let me clarify what I think the original question was getting at (and the one I'm interested in). A toral subalgebra of a Lie algebra is one in which the adjoint map of every element is diagonalizable. This isn't obviously related to tori in any way, hence the question. A comment claims that these are the Lie algebras of tori, "mostly, ... but not strictly...". So I'm looking for a clarification of what's going on here.

WillG
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Ugo Iaba
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    This link may be helpful. Apparently a torus in the sense of Lie groups is isomorphic to an $n$-torus. – Viktor Vaughn May 21 '14 at 17:49
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    A toral subalgebra is a subalgebra tangent to a torus, mostly. That's not strictly true, but it is true often. – Mariano Suárez-Álvarez Dec 07 '22 at 20:50
  • @WillG: The definition in that link is not the standard one in sources I know, for non-algebraically closed fields. One has to additionally assume the subalgebra is abelian. Otherwise the real Lie algebra $\mathfrak{su}_2$ would count as toral (all of its elements are semisimple). That page seems to not be a good source. – Torsten Schoeneberg Dec 07 '22 at 23:42
  • @TorstenSchoeneberg I see. One book I've been using, by Claudio Procesi, seems to use this definition too. He may be considering only complex Lie algebras, though. – WillG Dec 08 '22 at 04:06
  • This definition also appears in Humphreys, §8.1, where again the field is assumed algebraically closed. – WillG Dec 08 '22 at 04:36
  • @WillG: As I said in my comment, for algebraically closed fields, this definition is OK. As soon as the field is not algebraically closed, it is not. (Cf. also my own blunder from nine years ago, https://math.stackexchange.com/q/624932/96384). – Torsten Schoeneberg Dec 08 '22 at 17:17

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I think there is no need for deep research here. The adjective "toral" refers to the noun "torus".

$$\text{ Toral subalgebras : Lie algebras = Tori : Lie (or Algebraic) Groups}$$

Of course if one becomes technical enough, this might not be true verbatim in every situation (for starters one could restrict to the ground field $\mathbb C$ and to the semisimple case, up to isogeny or something), but the general idea can only be that toral subalgebras play the same role in Lie algebras as tori play in Algebraic Groups / Lie Groups. I can vouch for that at least in the semisimple case, where on the Lie algebra level one fiddles with split, and maximal, and split maximal, and maximal split toral subalgebras -- just as in the group setting one fiddles with split, and maximal, and split maximal, and maximal split tori. Cf. my answer to Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?.

Added: The analogy I mean above goes, very roughly and ignoring technicalities, as follows:

The two basic algebraic groups over a field $K$ which one "knows" are the multiplicative group $G_m=(K^\times, \cdot)$ and the additive group $G_a=(K, +)$. Now, to understand a general reductive group $G$, it is very worthwhile to find a maximal torus inside it, that is a subgroup that is just a product of copies $T = G_m \times ... \times G_m$. Why? It turns out such a thing is kind of the "backbone / spine" of the entire group. Namely, there are a bunch of other subgroups in $G$ each of which is a copy of $G_a$ ("one-parameter unipotent subgroups"); together with $T$ they make up (almost) the entire group; and there is a combinatorial/geometric object called the "root system" which tells us (not everything but) a lot about 1) how our $T$ operates on each of those $G_a$ and 2) how those $G_a$ behave towards each other (they do not in general commute with each other, but the root system tells us a lot about how they fail to commute).

The inspirational example is the matrix group $G=GL_n(K)$ in which one particularly nice maximal torus are the diagonal matrices $T = \pmatrix{\ast & \dots &0 \\ \vdots & \ddots & \vdots\\ 0& \dots &\ast}$, and the one-parameter subgroups are the $Id_n + K\cdot E_{ij}$ (that is all, the matrices that look like $\pmatrix{1 & * &0 \\ 0 & 1 & 0\\ 0& 0 &1}$ or $\pmatrix{1 & 0 &0 \\ 0 & 1 & 0\\ *& 0 &1}$ or ... .)

Compare to:

The most basic Lie algebra over a field $K$ which one "knows" is just $K$ itself with trivial Lie bracket. Now, to understand a general reductive Lie algebra $\mathfrak g$, it is very worthwhile to find a maximal toral subalgebra inside it, that is a subalgebra that is just an abelian $\mathfrak h = K \times ... \times K$ which operates in particularly nice way on the rest of $\mathfrak g$. Why? It turns out such a thing is kind of the "backbone / spine" of the entire group. Namely, there are a bunch of other subalgebras in $\mathfrak g$, each of which in itself again is just $K$ ("root spaces"); together with $\mathfrak h$, they make up (almost) the entire Lie algebra; and there is a combinatorial/geometric object called the "root system" which tells us (not everything but) almost everything about 1) how our $\mathfrak h$ operates on each of those root spaces and 2) how those root spaces behave towards each other (they do not in general commute with each other, but the root system tells us almost everything about how they fail to commute).

The inspirational example is the matrix Lie algebra $\mathfrak g =\mathfrak{gl}_n(K)$ in which one particularly nice maximal toral subalgebra are the diagonal matrices $\mathfrak h = \pmatrix{\ast & \dots &0 \\ \vdots & \ddots & \vdots\\ 0& \dots &\ast}$, and the root spaces are the $K\cdot E_{ij}$ (that is all, the matrices that look like $\pmatrix{0 & * &0 \\ 0 & 0 & 0\\ 0& 0 &0}$ or $\pmatrix{0 & 0 &0 \\ 0 & 0 & 0\\ *& 0 &0}$ or ... .)

It should be noted that in a reductive Lie algebra, the maximal toral subalgebras are exactly the Cartan subalgebras; and many sources use that name instead for the above $\mathfrak h$ when presenting the above structure theory.


So the follow-up question is why tori of Algebraic Groups or Lie Groups are called tori. I am sure the Lie Group terminology predates and inspired the one in Algebraic Groups; from there, I guess compact Lie Groups were a primary subject of interest, and compact tori are products of copies of $S^1$. Which, in the only interesting imaginable dimension, namely $S^1 \times S^1$, "is" the geometric shape called by the Latin word "torus" (maybe since antiquity: https://mathshistory.st-andrews.ac.uk/Miller/mathword/t/). Confer also https://hsm.stackexchange.com/q/6687/6514.

  • Can you clarify the reasoning behind your analogy "Toral subalgebras : Lie algebras = Tori : Lie groups"? How are these similar? – WillG Dec 07 '22 at 22:46
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    Toral subalgebras (of a Lie algebra) are those in which every element is diagonalizable. Is this condition equivalent to being the Lie algebra of a torus? I'm not interested in the etymology of the word "torus," but rather in the connection between toral subalgebras (as defined here) and tori. – WillG Dec 07 '22 at 22:48
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    What do you know about the structure of semisimple Lie algebras and semisimple Lie groups, in particular relating to root systems? Once you learn that, you will see the connection. The roots themselves are characters of a (split maximal) torus (in the group setting) or of a (split maximal) toral subalgebra (in the algebra setting). – Torsten Schoeneberg Dec 07 '22 at 23:06
  • I've been going through the classification of Lie algebras recently, but am only partway through. So I don't see the connection yet, but I'll keep an eye out for the what you mentioned. – WillG Dec 08 '22 at 04:00