Linear representations of compact groups I was reading about linear representations of compact groups in the book of Jean-Pierre Serre: "Linear Representations of Finite Groups", and he states that most of the properties for representations of finite groups carry over to representations of compact groups, but i didn't understand why those properties are valid specific in compact groups, but not in any kind of infinite groups.
My understanding was because in compact groups we can replace the expressions "$(1/g)(\sum_{g \in G} f(t)$" by "$\int_G f(t) dt$" and by doing so, we can carry over the results of finite group representations, is that right?
To give some context in my knowledge: i don't have a strong background in measure theory and topology yet.
 A: As other commenters have pointed out, this is a very broad question. However here is one aspect of where the finite-dimensional representation theory of compact groups mirrors finite groups: semisimplicity of representations.
Suppose that $V$ is a finite-dimensional complex representation of a finite group $G$, so we have a homomorphism $\rho_V \colon G \to \operatorname{GL}(V)$. One way to prove that $V$ is semisimple is to put a $G$-invariant inner product on $V$. We can do that by taking any inner product $H \colon V \times V \to \mathbb{C}$ and averaging it over the group: define a new inner product $\langle -, - \rangle \colon V \times V \to \mathbb{C}$ by the formula
$$ \langle u, v \rangle = \sum_{g \in G} H(\rho_V(g) u, \rho_V(g) v). $$
Then you can check that this is indeed an inner product, and that each operator $\rho_V(g)$ acts by unitary transformations under the new inner product, and therefore whenever we have a subrepresentation $U \subseteq V$ the orthogonal complement $U^\perp$ is also a subrepresentation, with $U \oplus U^\perp = V$. (Everything above also works for real representations).
If $V$ is a finite-dimensional complex representation of a compact group $G$, then you can do exactly the same thing, replacing the sum over group elements with an integral over the group using the Haar measure. The fact that the Haar measure is translation-invariant means that the new inner product $\langle -, - \rangle$ makes the operators $\rho_V(g)$ unitary, and so again representations are semisimple.
This does not work for groups that are not compact. For example, take the Lie group $G = (\mathbb{R}, +)$, and consider the representation into $2 \times 2$ matrices
$$ \rho \colon \mathbb{R} \to \operatorname{GL}_2(\mathbb{R}), \quad t \mapsto \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}. $$
We can see that $\rho$ is not semisimple, since it contains a subrepresentation spanned by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ but no complementary subrepresentation.
