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I've been trying to teach myself the theory of Lie groups. The sources I've been reading reference Lie algebras in the context of Dynkin and Satake diagrams, but not Lie groups (which I am more interested in). I've been trying to translate these results to Lie groups, but I'm almost sure I'm getting my facts wrong. (I am in particular confused about what is true over $\mathbb{C}$, and what is true over $\mathbb{R}$ because some sources don't mention this explicitly.)

It would be very helpful for me to get some guidance so that I know how to think about the big story before continuing with my study.

Question

Is the following story correct? (And if not, what is the correct story?)

  1. Semi-simple real Lie algebras correspond to disjoint unions of Satake diagrams.

  2. Semi-simple real Lie algebras are always the Lie algebra of some real compact Lie group. This Lie group is not unique, but it is unique if we assume that it is connected and simply connected.

  3. Any connected compact real Lie group with a semi-simple (real) Lie algebra is semi-simple. Namely, it is reductive with finite center.

  4. Any compact real Lie group is a maximal compact closed subgroup of its complexification. (Although there may be other maximal compact subgroups.)

  5. A complex Lie group is reductive iff it is the complexification of a compact real Lie group.

  6. The maximal closed compact subgroups of a complex reductive Lie group are all real Lie groups.

  7. The maximal closed compact subgroups of a complex reductive Lie group correspond to disjoint unions of Dynkin diagrams.

Semi-simple complex Lie algebras correspond to disjoint unions of Dynkin diagrams.

$\,\,\,$8. A semi-simple complex Lie algebra is always the Lie algebra of a reductive complex Lie group. (This group may not be unique. I am not sure whether it is unique if we assume that it's simply connected as a manifold.)

EDIT: the point below did not appear in the original question

$\,\,\,$9. Non-isomorphic real (resp. complex) connected and simply connected Lie groups have non-isomorphic real (resp. complex) Lie algebras.

Additional Question

In addition to my confusion about the story above, I am also confused about the way in which this helps classify real Lie groups in general. Namely, if I understand correctly, Satake diagrams only classify connected compact simply connected real Lie groups; and the connected compact real Lie groups are known because they are quotients of connected compact simply connected real Lie groups by central subgroups. But what about non-connected compact real Lie groups? Are they completely unrelated to the Satake diagrams story? Is there any way to classify them?

EDIT: The following is another point of confusion that did not appear in the original question

Let's say we are given a semi-simple complex Lie algebra, and let's say that it's the Lie algebra of some reductive connected and simply connected complex Lie group. Since Dynkin diagrams are a coarser tool than Satake diagrams, I take from that the this complex Lie algebra is potentially the complexification of several non-isomorphic real Lie algebras. However, if I understand correctly, a complex Lie group is the complexification of at most one (up to iso.) real Lie group. So how could it be that the complex Lie algebra has several real forms, but the complex Lie group has only one? What are these real Lie groups that correspond to the various real forms of the complex Lie algebra?

Never mind, I got it. Satake diagrams correspond to not-nec.-compact, connected, simply connected, real Lie groups.

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    $\begingroup$ In any compact group G, the connected component of the identity is normal. The quotient is profinite. (Please let's assume finite, and that the id. component N is finite-dimensional). To get a start on classifying disconnected groups, try to understand the possible ways G/N can act on N, i.e. the finite subgroups of Aut(N). There's a little bit of information contained in the Dynkin diagram: Aut(N) is itself a compact group, the id. component is basically N again (anyway same Lie algebra) and the finite component group is naturally identified with the automorphisms of the Dynkin diagram. $\endgroup$
    – ya-tayr
    Commented Jan 3, 2014 at 6:13
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    $\begingroup$ #1 is correct, #2 is correct if you remove "compact", #3 is correct, #4 is correct with the parenthetical removed (it is the unique one up to conjugation), #5 is correct with "connected" impose on both sides, #6 is probably not what you meant to ask (as every closed subgroup of every real Lie group is a real Lie subgroup, #7 is ambiguous (there is one conjugacy class when the ambient group has finite component group), #8 is correct if the group is required to be connected and simply connected. $\endgroup$
    – user76758
    Commented Jan 3, 2014 at 7:14
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    $\begingroup$ A good example to think about is that the complex Lie group $O(n)$ is the complexification of the real Lie group $O(p,q)$ for any $p+q=n$, because any real symmetric form with signature $(p,q)$ is equivalent, over the complex numbers, to any other nondegenerate symmetric form of the same dimension. $\endgroup$
    – Will Sawin
    Commented Jan 3, 2014 at 23:52

1 Answer 1

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2 is false. The smallest counterexample is $\mathfrak{sl}_2(\mathbb{R})$. A necessary and sufficient condition for a semisimple real Lie algebra to be the Lie algebra of a compact Lie group is that the Killing form is negative definite (compact Lie algebra). I think it is known that every semisimple complex Lie algebra has a unique compact real form.

But what about non-connected compact real Lie groups? Are they completely unrelated to the Satake diagrams story? Is there any way to classify them?

This is at least as hard as classifying finite groups.

Edit: Regarding 9, a much stronger statement is true. Taking tangent spaces at the identity induces an equivalence of categories between the category of connected, simply connected real Lie groups and the category of finite-dimensional real Lie algebras. I think this statement is still true with "real" replaced by "complex."

Never mind, I got it. Satake diagrams correspond to not-nec.-compact, connected, simply connected, real Lie groups.

By the above, classifying these is equivalent to classifying finite-dimensional real Lie algebras, and this classification is hopeless already for nilpotent Lie algebras of some specific small dimension that I can't remember right now. Statement 1 is still correct.

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  • $\begingroup$ Qiaochu, so do Satake diagrams correspond to connected, reductive, simply connected real Lie groups? $\endgroup$
    – Sam D.
    Commented Jan 4, 2014 at 0:21
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    $\begingroup$ @Sam: I don't think so, e.g. $\mathbb{R}$ is reductive but not semisimple. If "semisimple Lie group" is a term people use for Lie groups whose Lie algebras are semisimple then Satake diagrams classify connected, simply connected, semisimple (real) Lie groups. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 4, 2014 at 0:48
  • $\begingroup$ I think "semisimple Lie group" usually means reductive with finite center. I guess the point is that $\mathbb{R}$-anisotropic semisimple groups must be compact, but Satake diagrams classify semisimple (in the sense of being reductive with finite center) real Lie groups that have a copy of $\mathbb{G}_m$? $\endgroup$
    – Sam D.
    Commented Jan 4, 2014 at 1:01
  • $\begingroup$ Sorry, I mean that may have a copy of $\mathbb{G}_m$. $\endgroup$
    – Sam D.
    Commented Jan 4, 2014 at 1:15
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    $\begingroup$ @QiaochuYuan: It is generally not a good idea (or at least will tend to lead to confusion) to speak of "reductive" beyond the algebraic group setting; e.g., calling $\mathbf{R}$ a "reductive" Lie group isn't a useful thing to do. $\endgroup$
    – user76758
    Commented Jan 4, 2014 at 10:22

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