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I'd like to know whether there exists a standard way of classifying homomorphisms between two given Lie Groups, at least for some class of Lie groups, e.g. matrix groups. For instance, suppose that I want to classify all Lie group homomorphisms $\varphi\colon S^1\to GL(n,\mathbb{R})$. What can I say about it? Thank you.

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There are a few viewpoints towards solving a problem like this.

For example, suppose one wants to classiy all $f:H\rightarrow G$. We'll start by making a simplifying assumption that $H$ is simply connected. Now note that any such $f$ induces a map $f_\ast:\mathfrak{h}\rightarrow\mathfrak{g}$.

Now, $f_\ast$ is a linear map between $\mathfrak{h}$ and $\mathfrak{g}$ which preserves the bracket (in the sense that $[f_\ast v,f_\ast w] = f_\ast[v,w]$). Since we're talking about linear maps, one has a hope of classifying these. Since $H$ is simply connected, every Lie algebra map $\mathfrak{h}\rightarrow \mathfrak{g}$ is of the form $f_\ast$ for some $f:H\rightarrow G$ and further, at least if $H$ is connected, $f_\ast = g_\ast$ implies $f=g$.

For nonsimply connected $H$, first figure out the answer for the universal cover $\tilde{H}$, then try to figure out which maps descend back to $H$.

Another approach is via representation theory. This works especially well when the target space $G$ is something "nice" like $GL_n$, or $U(n)$, $SO(n)$, etc. If the target is something like, say, $U(3)/Z(U(3))$ or $G_2$, then it's slightly harder to apply representation theory.

Here, the goal is to classify linear $H$ actions on a vector space. Two such actions are equivalent if there is an isomorphism of the vector space intertwining the two actions. In many cases (but not all!) such an isomorphism is necessarily induced by conjugation by an element of $G$. What representation theory often gives you is a classification of maps $f:H\rightarrow G$ up to conjugation.

In the example you mentioned, the final answer (which I currently don't have time to type up all the details to), is that, up to conjugation, every map $S^1\rightarrow GL(n,\mathbb{R})$ sends $\theta \in S^1$ to the block diagonal matrix $$\operatorname{diag}(R(k_1\theta), R(k_2\theta),..., R(k_{\lfloor n/2 \rfloor}\theta),1)$$ if $n$ is odd or $$\operatorname{diag}(R(k_1\theta), r(k_2\theta),..., R(k_{\lfloor n/2 \rfloor}\theta))$$ if $n$ is even. Here, $R(m\theta) = \begin{bmatrix} \cos(m\theta) & \sin(m\theta)\\ -\sin(m\theta) & \cos(m\theta)\end{bmatrix}$ with $m$ an integer.

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It is better to use \ast instead of * inside LaTeX, because the interpretation of *...* as delimiters for italics tends to confuse the engine. I've fixed it because the post didn't render correctly for me. –  t.b. Apr 3 '12 at 12:52
@t.b.: I wasn't aware t.b., thanks! –  Jason DeVito Apr 3 '12 at 12:55
Out of curiosity, Jason, what do you mean by "nice" - I think I recall reading that $G_2$ has a 14-dimensional matrix representation, why is it different? And how would you use representation theory in the case of a group like $G_2$? I'm a relative newcomer to the field and would appreciate the clarification. Thanks! –  Erik Pan Feb 6 at 2:16
@Erik: "Nice" means "standard matrix group". The nice thing about the standard matrix groups is we have a really good understanding of when an element is conjugate to an element in one of them. For example. Suppose you have an n-dimensional real representation of $H$. This can, of course, be thought of as a homomorphism $H\rightarrow Gl_n(\mathbb{R})$. Now, a natural question is whether or not the image of $H$ is actually conjugate to a subgroup of $O(n)$ or not. It turns out, there is a well understood necessary and sufficient condition: (continued) –  Jason DeVito Feb 6 at 2:48
$H$ is conjugate to a subgroup of $O(n)$ iff $H$ preserves an inner product on $\mathbb{R}^n$. Now, suppose $n=14$. One can ask whether or not the image of $H$ is conjugate to a subgroup of $G_2$, but how do you even begin to answer a question like that? (I wouldn't be surprised if some experts know how to answer this question, but I don't). Let me give an actual example (which I still don't understand). You're right that $G_2$ has a 14 dim real representation, but it actually has a smaller one. (cont) –  Jason DeVito Feb 6 at 2:50

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