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For example, $SO(4, \mathbb{C})$ is a 6 complex dimensional complex lie algebra, and its Dynkin diagram looks like $A_2$. It has real forms (6 real dimensions) at least of:

$\mathfrak{so}(4) \cong \mathfrak{su}(2)\oplus\mathfrak{su}(2) $,

$\mathfrak{so}(1,3) \cong \mathfrak{sl}(2,\mathbb{C})_{|\mathbb{R}}$,

$\mathfrak{so}(2,2)\cong \mathfrak{sl}(2,\mathbb{R})\oplus\mathfrak{sl}(2,\mathbb{R}) $.

Given the Coxeter-Dynkin diagram of the complexified lie-agebra, how can I build the Satake-Tits diagram for these real forms, and understand how conjugation and duality transforms affect the diagrams? References are fine, but most texts I have found on this are not beginner friendly.

Edit: I was bothered by a lack of resources on the quaternionic real forms of $D_n$, and so have compiled the scattered results and definitions into one place for simple reference: https://arxiv.org/pdf/2504.04396

Craig
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    Those are not the real forms of $\mathfrak{sl}_2$, they are the real forms of $\mathfrak{so}_4 \cong\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$. The real forms of $\mathfrak{sl}_2$ are simply $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{su}(2)$ – Callum Mar 23 '23 at 08:02
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    The basic idea of a Satake diagram is take a maximal split toral subalgebra $\mathfrak{a}$ and extend it to a full maximal toral subalgebra $\mathfrak{h}$ (i.e a Cartan subalgebra) Then make the Dynkin diagram of $\mathfrak{h}^\mathbb{C}\leq\mathfrak{g}^\mathbb{C}$ and colour the nodes black when their roots are 0 on $\mathfrak{a}$ and draw arrows between complex conjugate pairs – Callum Mar 23 '23 at 08:09
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    Classifying all the diagrams that can occur is a bit trickier but you can find a table in the appendices of Helgason's Differential Geometry and here is a good discussion about the possibilities – Callum Mar 23 '23 at 08:21
  • Oh and I think there is one more real form missing from your list for $\mathfrak{so}_4$ which is $\mathfrak{sl}(2,\mathbb{R}) \oplus \mathfrak{su}(2)$ which I believe is isomorphic to $\mathfrak{so}^*(4)$ the subalgebra of $\mathfrak{so}_4$ invariant for a quaternionic structure on $\mathbb{C}^4$. (Someone correct me if that is wrong) – Callum Mar 23 '23 at 13:10
  • Thanks for these comments. They're helpful. Also: Interesting, is that real form distinct from $\mathfrak{so}(1,3)$? – Craig Mar 23 '23 at 14:34
  • Yes it is. In general $\mathfrak{so}_n$ has real forms $\mathfrak{so}(p,q)$ for $p+q=n$ and $\mathfrak{so}^*(n)$ (as long as $n$ is even). The real form $\mathfrak{so}(3,1)$ is instead isomorphic to $\mathfrak{sl}_2$ viewed as a 6-dimensional real Lie algebra which you can think of as two white vertices connected by an arrow but no edge – Callum Mar 23 '23 at 16:42

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I've written enough comments that I think I should just turn them into an answer:

To make a Satake diagram we first fix a Cartan involution $\tau$ of our real Lie algebra $\mathfrak{g}$ and write $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ for the $\pm1$ eigenspaces of $\tau$. Note $\mathfrak{k}$ here is just a maximal compact subalgebra.

Take a maximal toral subalgebra $\mathfrak{a} \leq \mathfrak{p}$ (We call such subalgebras split -- alternatively split means diagonalisable over $\mathbb{R}$). Extend $\mathfrak{a}$ to a full Cartan subalgebra $\mathfrak{h}\leq \mathfrak{g}$ such that $\tau \mathfrak{h} = \mathfrak{h}$. Now we can complexify and make the root system for $\mathfrak{h}^\mathbb{C}\leq \mathfrak{g}^\mathbb{C}$ and draw its Dynkin diagram.

Note that both the real structure (a.k.a complex conjugation) and our Cartan involution give natural actions on this root system since they both preserve $\mathfrak{h}$ so we can divide our roots up into imaginary, real and complex ones. Alternatively we can call roots which are $0$ on $\mathfrak{a}$ "compact" and these turn out to be exactly the imaginary ones. The other roots ("non-compact") are either fixed by the real structure or swapped with another root. So on our Dynkin diagram we colour the compact roots in and draw arrows between the roots swapped by the real structure.

In the example given the complex Lie algebra $\mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C}) \cong \mathfrak{so}_4$ has Dynkin diagram given by 2 unconnected nodes. So we have real forms given by:

  • $\mathfrak{sl}(2,\mathbb{R}) \oplus \mathfrak{sl}(2,\mathbb{R}) \cong \mathfrak{so}(2,2)$, the split form - diagram looks like the Dynkin diagram (always true for the split form)
  • $\mathfrak{su}(2) \oplus \mathfrak{su}(2) \cong \mathfrak{so}(4)$, the compact form - both nodes are black (always true for the compact form)
  • $\mathfrak{sl}(2,\mathbb{C}) \cong \mathfrak{so}(3,1)$, viewed as a real Lie algebra - two white nodes joined by an arrow. Think of this as the two complex $\mathfrak{sl}(2,\mathbb{C})$ being complex conjugate to each other.
  • $\mathfrak{sl}(2,\mathbb{R}) \oplus \mathfrak{su}(2) \cong \mathfrak{so}^*(4)$, the quaternionic one - diagram is $1$ black node and $1$ white.

In fact this is all the possible Satake diagrams we could have naively tried so we are certainly done.

In general not all possible diagrams give actual real forms (and over different fields the valid Satake-Tits diagrams will be different even though the Dynkin diagrams are the same) but there are some rules such as the diagram being invariant under the "opposition involution" which cuts down the possibilities. You can read more about that here. The opposition involution is exactly the thing that encodes duality.

Tables of Satake diagrams can be found in Helgason's Differential Geometry, Lie Groups and Symmetric Spaces as Well as Onishchik and Vinberg's Lie Groups (volume III, I think).

Callum
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    This matches https://math.stackexchange.com/a/3749142/96384 (only the first three cases appear here), great. As for that "quaternionic one", do I assume correctly that such only appear in $D_{2n}$ i.e. $\mathfrak{so}{4n}$, and look like this: one of the "horns" is black, the other white, their immediate link is white, and before that they're alternatingly black and white? (For $D_4$ that is accidentally isomorphic to $\mathfrak{so}(6,2)$ but otherwise it is distinct from any $\mathfrak{so}(p,q)$.) That form is made possible by the opposition involution not switching the "horns" in $D{2n}$! – Torsten Schoeneberg Mar 24 '23 at 03:31
  • Yes that's the family I had in mind. But you are right the diagram for even and odd n is different so it perhaps does not fit as I had thought it might. Does the opposition involution definitely swap the two nodes in $D_2$? Can we define it for non simple Lie algebras? – Callum Mar 24 '23 at 07:54
  • No that's the joke, it is defined but is the identity, i.e. does not swap the two points, $2$ is even. – Torsten Schoeneberg Mar 24 '23 at 15:25
  • @TorstenSchoeneberg Ah I just fully misread your comment. I never see $\mathfrak{so}^*(4)$ mentioned anywhere so I was concerned it might not actually be valid. – Callum Mar 24 '23 at 16:33
  • Might be different nomenclature: I think e.g. Onishchik Vinberg call $\mathfrak{u}_n^(\mathbb H)$ what you seem to call $\mathfrak{so}^(2n)$. Also, contrary to what I claimed above, these also exist for odd $n$, then the two "horns" are white and connected by an arrow, their link is black, and then it goes on white-black-white-black etc. – Torsten Schoeneberg Mar 24 '23 at 18:12
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    @TorstenSchoeneberg I call it that to align with the notation on Wikipedia's List of simple Lie groups but also because I find it helpful to refer to a real Lie algebra in terms of its complexification. However on that list it only considers $D_n$ for $n\geq 4$, presumably to avoid double counting but left me with a small worry that it didn't actually work in the slightly peculiar case that is $D_2$. – Callum Mar 26 '23 at 11:46
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    I was bothered by a lack of resources on the quaternionic real forms of $D_n$, and so have compiled the scattered results and definitions into one place for simple reference: https://arxiv.org/pdf/2504.04396 – Craig Apr 08 '25 at 01:45