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I read (and hopefully understood) the classification of semi-simple complex Lie algebras, construction of their root spaces and their representation theory (represented as complex matrices). As an example, I am interested in the difference between $\mathfrak{sl}(2)$ and $\mathfrak{su}(2)$.

The standard approach for a complex semi-simple Lie algebra $\mathfrak{g}$ is to construct a Cartan subalgebra $\mathfrak{h}\subset\mathfrak{g}$ and then compute root spaces spanned by eigenvectors of the Cartan subalgebra in the adjoint representation. So far, so good.

Now I want to consider a real semi-simple Lie algebra $\mathfrak{g}$. My understanding was that I can still construct a Cartan subalgebra $\mathfrak{h}\subset\mathfrak{g}$ spanned by a maximal commuting set. By construction (?), the adjoint representation $\mathrm{ad}_H(A)=[H,A]$ for $H\in\mathfrak{h}$ will be anti-symmetric with respect to the Killing form (maybe I'm wrong, because in the real basis the Killing form may not be definite?) which leads to purely imaginary (or vanishing) eigenvalues and generally complex eigenvectors. This means the root spaces cannot be spanned by elements of the real Lie algebra, we need to complexify to get $E_{\pm\alpha}$.

When I now consider a complex representation $\rho$ of the real Lie algebra $\mathfrak{g}$, I would just proceed as for the complex case. I have my complex $E_{\pm\alpha}$ from which I can construct real objects according to $Q_{\alpha}=E_{\alpha}+E_{-\alpha}$ and $P_{\alpha}=i(E_{\alpha}-E_{-\alpha})$. At least for finite dimensional (or discrete) representations, I can use the weight eigenspaces as orthonormal basis of the complex representation vector space $\mathcal{H}$. When I then represent my real Lie algebra element, i.e., Cartan elements $\rho(H)$ or general $\rho(Q_{\alpha})$ and $\rho(P_{\alpha})$, I can decompose them into the $\rho(E_{\pm\alpha})$ by analytic continuation, where I know exactly how $\rho(E_{\pm\alpha})$ raises/lowers the weight...

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    $\begingroup$ There are indeed some changes. Lie's Theorem need not hold for real Lie algebras, for example. The classification is more complicated and involves real forms. $\endgroup$ Mar 4, 2020 at 16:10
  • $\begingroup$ Thanks a lot! If I understand correctly, Lie's theorem applies to solvable Lie algebras, while I'm mostly interested in real semi-simple Lie algebras. I understand that real forms are required for the classification. But do I need them to choose a real Cartan subalgebra (within the real Lie algebra) and then to compute roots. Are there problems with the Killing form, when it's non-definite (non-compact real forms). $\endgroup$
    – LFH
    Mar 4, 2020 at 16:17
  • $\begingroup$ I thought real forms correspond to different ways to embed a real subalgebra into a complex algebra. For example su(2) and sl(2) have isomorphic complexifications, but correspond to different embeddings as real subalgebras of their common complexification. However, if I start with a real algebra, I have already chosen it's real form or am I missing something? I really want to focus on the construction of possible weights etc. for a representation of a fixed real Lie algebra (probably non-compact). $\endgroup$
    – LFH
    Mar 4, 2020 at 16:20
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    $\begingroup$ In the classification of complex semsimple Lie algebras several corollaries of Lie's theorem are used, e.g., see how Corollary $5.12$ is applied in the classification here, e.g., in Proposition $8.4$. $\endgroup$ Mar 4, 2020 at 16:21
  • $\begingroup$ Ah, ok, thanks a lot! - this makes sense. In the real case, you can't choose a basis, such that the Killing form is definite and so proposition 8.4 (4) may be false. Can you recommend a reference that discusses the differences between real and complex case in a comprehensive way? My focus is really not the classification, but merely the construction of roots and weight vectors when starting from a real Lie algebra. In particular: Can I choose the Cartan subalgebra to be real and construct weight vectors from it? $\endgroup$
    – LFH
    Mar 4, 2020 at 16:37

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If I understand correctly, you are interested in representations of a semisimple real Lie algebra $\mathfrak{g}$ on complex vector spaces. These are in equivalence with the complex representations of its complexification $\mathfrak{g}_{\mathbb C}$, cf. Obtaining representation of a real Lie algebra from the complexification "by restriction" , In what sense are complex representations of a real Lie algebra and complex representations of the complexified Lie algebra equivalent?, How does one define weights for a semisimple Lie group?.

In particular, e.g. the representations of both $\mathfrak{su}_2$ and $\mathfrak{sl}_2(\mathbb R)$ are direct sums of irreducible ones, each of which is described up to isomorphism by its highest weight, which are parametrized by $\mathbb Z_{\ge 0}$. They are indeed the irreps of $\mathfrak{sl}_2(\mathbb C)$, which I assume you know well, restricted to the respective real subalgebras.

Note however that this equivalence of categories misses some subtler points, cf. https://math.stackexchange.com/a/3258221/96384. Also, the subsection "a slightly different example" there goes through explicit representations of all three real forms of the complex Lie algebra $\mathfrak{sl}_3(\mathbb C)$ and might be helpful to get a feeling for what's going on.

For the theory so far, it does not matter whether you choose a Cartan subalgebra in the real Lie algebra or in its complexified version, because for the construction of the roots and weights you (should) only use the latter anyway; note that in a real Lie algebra, a CSA plays the role of a maximal torus, but in a complex Lie algebra, that of a maximal split torus, cf. Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?; also note that e.g. in $\mathfrak{su}_2$, every $1$-dimensional subspace is a CSA, but there are no roots -- the root spaces only appear in the complexified version.

Finally, in your last paragraph you seem to be mixing up things or start doing something else, namely construct representations on real vector spaces. That is something which needs much more subtle considerations. For the compact real forms, there is a combinatorial criterion about which of the complex representations have "a real structure" (i.e. come from a representation on a real vector space) and which don't, cf. https://math.stackexchange.com/a/2774741/96384 -- e.g. for $\mathfrak{su}_2$, the irreps with even dimension are "truly complex", whereas the ones with odd dimension restrict to acting on real vector spaces. For the non-split and non-compact forms, the story is more intricate. I tried to outline the way I understand it here: https://math.stackexchange.com/a/3298058/96384 (where I think that the accepted answer to the question contains a mistake), and applied that to an example here: https://math.stackexchange.com/a/3298176/96384.


Added: As to what you try in your last paragraph, that would seem to work for the Lie algebras of compact forms, but not in general. A crucial part in the classification is how complex conjugation acts on the roots (and then more refined, on the root spaces). In the compact case, it sends each $\alpha$ to $-\alpha$, and maybe you can choose the $E_\alpha$ so that it sends $E_\alpha$ to $E_{-\alpha}$, meaning your proposed $P_\alpha$ and $Q_\alpha$ are indeed elements of your real Lie algebra. However, the action of complex conjugation on non-compact forms can be very different. (Well for the split forms it's trivial, but ...) To see what can happen, look at the quasi-split form of $\mathfrak{sl}_5(\mathbb C)$. This consists of those matrices in $\mathfrak{sl}_5(\mathbb C)$ where $a_{ij}=-\overline{a_{6-j,6-i}}$ ("antihermitian to the secondary diagonal"). If you call $E_{\alpha_i} := E_{i,i+1}$ for $i=1,…,4$, then observe that complex conjugation transposes $E_{\alpha_1} \leftrightarrow E_{\alpha_4}$ and $E_{\alpha_2} \leftrightarrow E_{\alpha_3}$. (The conjugation operates as the outer automorphism on the Dynkin diagram of $A_4$). So now of course you can still look at "refined" operators playing the role of your $P$'s and $Q$'s (denoting complex conjugation by $\sigma$)

$$E_{\alpha} + \sigma(E_\alpha)$$

$$ iE_{\alpha} + \sigma(i E_{\alpha})= iE_{\alpha} – i \sigma(E_{\alpha})$$

and maybe you can get something like $\sigma(E_\alpha) = \pm E_{\sigma(\alpha)}$ but what exactly happens further depends on what exact $\alpha$ you’re looking at; here, $\alpha_3$ needs different treatment than the other $\alpha$'s. Also, all those refined $P$'s and $Q$'s together might not yet give a basis for the real Lie algebra! Further, what do we do with the $H_\alpha$? Not saying this is not doable, just that it might be quite intricate.

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  • $\begingroup$ Thank you so much for the detailed answers and all the cross references. I read your "slightly different example" which was quite helpful. Regarding real vs. complex representations: Yes, I"m interested in complex representations. If I understand correctly, both the CSA and the root vector spaces are truly complex. What I was hoping (but may be wrong) is that the CSA can always be chosen to be real, such that the root basis vector (as elements of the complexified Lie algebra) are complex conjugate. I then tried to construct real elements $Q_{\alpha}$ and $P_{\alpha}$ forming a real basis. $\endgroup$
    – LFH
    Mar 5, 2020 at 11:53
  • $\begingroup$ Hi Torsten, After going carefully through math.stackexchange.com/a/3258221/96384, I now understand the big picture much better. I'm also reading Knapp's book at the moment. I now also see that what I really need is an understanding how I can express elements of the Cartan subalgebra of the root spaces (both subspaces of the complexification) in terms of real objects or vice versa, i.e., having a weight vector of the representation, I have a complex basis whose action is fully controlled. However, how does it relate to the real basis, I started with? $\endgroup$
    – LFH
    Mar 5, 2020 at 13:41
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    $\begingroup$ Ah, I think now I understand what you want in the last paragraph. I will think about it. $\endgroup$ Mar 5, 2020 at 16:27
  • $\begingroup$ Thanks a lot. My understanding is that the classification of real algebras with equivalent complexification somehow does exactly this: It encodes how a given real algebra is embedded in its complexification, for which the representation theory is known. I just don't understand yet, if one can choose a completely real CSA or what other conditions there may be. I only started reading Knapp's book this week. $\endgroup$
    – LFH
    Mar 5, 2020 at 18:13

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