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How do we specify the embedding of a Lie group $G_1$ as a subgroup into a larger Lie group $G_2$, with $G_1 \subset G_2$ that agree with a constraint on the mapping between their representations?

By specifying the embedding, I mean that we can determine precisely the way how the Lie group embedding $G_1 \subset G_2$ is fixed as the embedding between two differentiable manifolds (since Lie groups are differentiable manifolds).

For example, is it enough to give some irreducible representation (irrep) of $G_1$ called $\mathbf{R}_{1,j}$ and some irrep of $G_2$ called $\mathbf{R}_2$, then we dictate the map $$ \mathbf{R}_{1} = \bigoplus_j \mathbf{R}_{1,j} \text{ in } G_1 \mapsto \mathbf{R}_2 \text{ in } G_2, \text{ and } G_1 \subset G_2 \tag{1} $$ would the above be precisely enough to specify the embedding? Is this a necessary and sufficient condition? If not, what else data is needed?

Of course, as Lie group embedding, $G_1 \subset G_2$, the two groups must share the common identity element $\mathbf{1}$. So their group identity element $$ \mathbf{1}_{G_1} =\mathbf{1}_{G_2} \tag{2} $$ are the same point on two manifolds.

In particular, we can foresee the case that the reducible rep $\mathbf{R}_{1}$ is decomposed as a irrep via $\bigoplus_j \mathbf{R}_{1,j}$. Is there a way that we can enumerate distinct kinds of embedding, given that we restrict the embedding data of eq.(1) and eq.(2)? What will be sufficient to give such an embedding classification?

  • I propose that a continuous deformation of embedding may be regarded as the same class of embedding (if there is a continuous way to deform the $G_1$ within $G_2$, then it is the same embedding). A discrete deformation of embedding may be regarded as a different class of embedding.

  • I suppose it should be some homotopy class data like $[G_1,G_2]$, or the way that the $G_1$ can wrap around $G_2$. But I am not completely sure whether such classification gives an integer value class of distinct maps.

p.s. To illustrate my question further concretely, I will give an example in the next post but a new different post: here.

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    $\begingroup$ Maybe you're looking for the statement that a homomorphism of Lie groups is completely determined by its derivative at the identity, as a map of Lie algebras? $\endgroup$
    – LSpice
    Commented Aug 17, 2021 at 17:26
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    $\begingroup$ (But notice that even completely understanding the resulting map $\operatorname{Rep}(G_2) \to \operatorname{Rep}(G_1)$ cannot determine the map $G_1 \to G_2$, since it is insensitive to conjugation on the target by $\operatorname{Aut}(G_2)$ (and hence to conjugation on the source by $G_1$).) $\endgroup$
    – LSpice
    Commented Aug 17, 2021 at 17:27
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    $\begingroup$ Are you maybe asking about the possibility of constructing a Lie group embedding satisfying certain conditions? $\endgroup$
    – Fernando Muro
    Commented Aug 18, 2021 at 11:24
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    $\begingroup$ Two homomorphism which are same representations are related to each other by a conjugation by an element of a target group $G_2$. Let's call this element $g$. There is a homotopy between $g$ and $e$ (the unit element of $G_2$, which should give a "continuous deformation of embedding", but I am not sure exactly what you mean by "continuous deformation of embedding". $\endgroup$
    – user43326
    Commented Aug 18, 2021 at 16:56
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    $\begingroup$ @user43326, is it obvious that two homomorphisms that do the same thing on representations differ by conjugation? I'm not even sure I see it for $G$ compact. $\endgroup$
    – LSpice
    Commented Aug 18, 2021 at 19:02

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