Hope the title and premise of my question is correct...

Suppose I have two linearly independent elements $g_1, g_2$ of a real finite-dimensional Lie algebra $\mathfrak{g}$. I wish to find the maximal minimal Lie subalgebra they generate (is this the right term)? What I mean is that I would want to keep forming Lie brackets $[g_1,g_2], [g_1,[g_1,g_2]]$ etc. and find the largest set of linearly-independent elements (over reals) generated from this procedure which are eventually closed under the bracket. This set is then a basis for my desired Lie subalgebra; I can get its dimension etc.

Would it make sense if I 'complexify' the elements, such as defining $h_1 = g_1 + ig_2$ and $h_2 = g_1 - ig_2$ and playing the same game of applying the Lie bracket over and over again to find the largest set that is linearly independent and closed? I guess I am a little unsure because I am not even sure I am able to say $h_1, h_2$ is linearly-independent (now over the complex numbers), let alone their brackets? or will complexifying give the same result for the dimension?

Note added: I am interested really in the case of $\mathfrak{g}=\mathfrak{su}(d)$, so maybe my question only makes sense there.

An example is $\mathfrak{su}{(3)}$: there are 8 linearly-independent (over reals) basis vectors, for example specified by the Gell-mann matrices $\lambda_i$. If $g_1 = \lambda_1$ and $g_2=\lambda_2$ then the Lie subalgebra they generate is $\{ \lambda_1, \lambda_2, \lambda_3\}$ since $[\lambda_1,\lambda_2] \propto \lambda_3$ and $[\lambda_3, \lambda_1] \propto \lambda_2$ and $[\lambda_2, \lambda_3] \propto \lambda_1$. So the dimension is 3.

However I can define $\lambda_{\pm} = \lambda_1 \pm i \lambda_2$ which a physicist would recognize as the raising and lowering operators and one can work out $[\lambda_+, \lambda_-] = \lambda_3$ and $[\lambda_{\pm}, \lambda_3] = \pm \lambda_{\pm}$ so that seems to give 3 linearly independent elements $\{ \lambda_+, \lambda_-, \lambda_3\}$ again (but now over $\mathbb{C}$). However, $\lambda_{\pm}$ is no longer an element of $\mathfrak{su}(3)$ (it is not hermitian).

Nevertheless I would still be able to say dim = 3 in both cases (though one is real and one is complex).

But I guess my question is, is such an equivalence a lucky coincidence or is there something more general at play?

[Title edited from Maximal to Minimal]


1 Answer 1


It might be helpful to split up what you ask into two questions:

Question A. Given any Lie algebra $\mathfrak g$ (over any field $K$), and a set of elements $S = \{x_1, ..., x_n\} \subset \mathfrak g$, how to determine the Lie subalgebra $A(S, \mathfrak g)$ of $\mathfrak g$ generated by $S$, i.e. the smallest Lie subalgebra of $\mathfrak g$ which contains $S$?

Answer: Not easy except in trivial cases (e.g. if all elements of $S$ commute with each other, in particular if $\lvert S \rvert =1$). In principle, the formal description of all elements of $A(S, \mathfrak g)$ as linear combinations of finite nested commutators gives an algorithm, but this might quickly become computationally bothersome. Cf. Computational procedure to find the basis of a Lie algebra generated by a finite collection of operators, open question Fastest way to find the dimension of a Lie algebra generated by a finite set of operators?.

I do not see how specifying $K= \mathbb R$ would significantly alter or simplify this problem compared to any other infinite field. With the big exception of there probably being more known classification in the literature.

Question B. Does complexification, or more generally changing the base field by scalar extension or scalar restriction, behave well w.r.t. forming such subalgebras?

Answer: Depends on what one means, and unfortunately people are sloppy in this complexification business. Example:

  1. Viewing $\mathfrak g := \mathfrak{gl}_n(\mathbb C)$ as a real Lie algebra, for any non-zero element $x \in \mathfrak g$, the set $S = \{x, ix\}$ generates a two-dimensional real (abelian) Lie subalgebra. However, in $\mathfrak{gl}_n(\mathbb C)$ viewed as a complex Lie algebra, that same set is linearly dependent and generates a one-dimensional complex Lie subalgebra. Note that this process is not what should be called complexification. Rather, we are dealing with scalar restriction here. (The true complexification of this $\mathfrak g$ is not the complex $\mathfrak{gl}_n(\mathbb C)$ it happens to sit in, rather it would be isomorphic to $\mathfrak{gl}_n(\mathbb C) \oplus \mathfrak{gl}_n(\mathbb C)$.)

  2. By "true complexification" as above I mean the scalar extension by tensors, i.e. $\mathfrak{g}_{(E)} := \mathfrak{g} \otimes_K E$ for any field extension $E |K$ (for convenience, take $\mathbb C | \mathbb R$). In this case, it does follow from the description mentioned in the answer to Question A that forming subalgebras is well-behaved in the sense that for any set $S \subset \mathfrak g$ as above, and denoting $S_{(E)}$ the image of $S$ in the scalar extension i.e. $\{x_1 \otimes 1, ..., x_n \otimes 1\}$, then (up to canonical iso) the scalar extension of the subalgebra generated by the set is the subalgebra of the scalar extension generated by the (scalar extension of the) set, or more succinctly:

$$ A(S, \mathfrak g)_{(E)} = A(S_{(E)}, \mathfrak g_{(E)})$$

Of course this implies in particular that the dimensions over the respective ground fields are equal.

Now I suspect that when you wrote things like $g_1 + ig_2$ you had not made up your mind as to whether you were thinking of the real Lie algebra as a scalar restriction, or the complex one as a scalar extension, which as just explained can make a big difference. But you are lucky: The reason that in your case things work, is that in your example, $\mathfrak g = \mathfrak{su}(d)$, the complexification happens to be the set $\mathfrak{sl}_d(\mathbb C)$ you most probably imagine your matrices in to begin with, so in hindsight your notation is adapted to the good second case although it looks like the bad first one. I would like to note though that when replacing $g_1, g_2$ by $h_1 =g_1+ig_2$ and $h_2=g_1-ig_2$ (which I understand in that "knowing those elements from elsewhere" makes computations easier) you are also using a little linear algebra in saying that with complex scalars, $g_1, g_2$ span the same space as $h_1, h_2$.

  • $\begingroup$ Thank you! This is indeed the answer I'm looking for. Just a small minor clarifying question though -- if $\{g \}$ is a basis for $\mathfrak{g}$ why is $\{g \otimes 1\}$ a basis for $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$? I take the extension to mean tensoring a complex space and allowing for arbitrary real linear superpositions; I would thus have imagined $\{ g \otimes 1, g \otimes i \}$ be to a basis? Maybe I have it mixed up. $\endgroup$
    – nervxxx
    Commented Mar 31, 2023 at 3:51
  • $\begingroup$ Edit: I think what I wrote in the comment above is correct if the field is real; otherwise, $\{ g\otimes 1\}$ is a basis over complex numbers. Would this be right? $\endgroup$
    – nervxxx
    Commented Mar 31, 2023 at 4:15
  • $\begingroup$ Yes, that seems right. $\endgroup$ Commented Mar 31, 2023 at 5:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .