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By Lie's third fundamental theorem we know that for each finite dimensional Lie algebra $\mathfrak{g}$ there is a unique simply connected and connected Lie group $G$ such that $T_eG \simeq \mathfrak{g}$.

Without knowing anything else about $G$, what can we use the Lie algebra for in order to get as much information about $G$

In particular, I want to classify the adjoint representation and consequently the adjoint orbits, of $G$, given a Lie algebra.

For example take $\mathfrak{g}=\mathfrak{so}(3)$. Now I want to pretend I know nothing about the Lie group and classify the adjoint orbits. How can I proceed? What exponential can I use if I know nothing about $G$?

I am trying to do this in as general a way as possible.

Math1000
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Matthew
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2 Answers2

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One thing to realize is that the group 'actually acting' in the adjoint representation is often not the group itself, but one of its, not-simply connected quotients. Your example is excellent: the simply connected group is $SU(2)$ but the group acting on $\mathbb{R}^3$ in the adjoint representation is $SO(3)$: the group elements $I$ and $-I$ in $SU(2)$ act the same.

Related example: the Universal Cover of $SL_{2}(\mathbb{R})$ has no faithful finite-dimensional representation so the group acting on $\mathbb{R}^3$ in the adjoint representation is $PSL(2, \mathbb{R})$.

However this can be a blessing in disguise. The group that acts in the adjoint representation can be found by using the ordinary matrix exponential on the adjoint representation of the Lie algebra.

Vincent
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  • Your example is very insightful, thank you! Would you know of a reference or a proof of your final statement as I would like to use it as an argument in the construction of the adjoint orbits! – Matthew Jan 13 '20 at 17:32
  • Wikipedia, https://en.wikipedia.org/wiki/Adjoint_representation, in discussing the relatation between the adjoint rep ad of the Lie algebra and the adjoint rep Ad of the group makes some remarks on the role of the exponential map in this process - I think one could extract a proof from there. But frankly I expect that this works more generally for every representation (not just the adjoint one). I will think a bit more about it. – Vincent Jan 13 '20 at 20:21
  • Would that generalise to say that if two lie groups have the same lie algerba then their adjoint representation is the same? – Matthew Jan 14 '20 at 11:38
  • Yes! That is to say: the rep is a homorphism from the group into the group of linear transformations of $\mathbb{g}$ so strictly speaking these are not the same since they have different domains, but their images, the subgroups of $End(\mathbb{g})$ that you get are the same. They are in some sense the smallest group with that Lie algebra, where (in the same sense) the simply connected one is the biggest group with that Lie algebra. The other answer gives a different way of realizing this 'smallest' group – Vincent Jan 14 '20 at 11:45
  • Ah yes. And as their image is the same we would get the same orbit! Thanks that's very clear now. – Matthew Jan 14 '20 at 11:56
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This might not be entirely satisfactory, but at least for a semisimple Lie algebra $\mathfrak{g}$ over $\mathbb R$, take its automorphism group $G_0 := Aut(\mathfrak{g})$. That is the adjoint Lie group sitting over $\mathfrak{g}$. If you want the simply connected one, take its universal cover $\widetilde{G_0}$.

I do not know, though, how efficiently one could actually compute or determine the group structure of the automorphisms of, say, $\mathfrak{so}_3$, when pretending to know nothing about the matrix group $SO_3$; and then the same problem when going to the simply connected cover.

Torsten Schoeneberg
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