Are compact & connected Lie Groups in correspondence with semi-simple Lie groups? I think there is a condition on the center (discrete?) but I'm not sure.
[Math] Compact connected iff semi-simple for Lie Groups
lie-groups
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About the question of uniqueness of maximal compact subgroups, up to conjugation, I propose a Riemannian geometric approach:
Theorem (É. Cartan). A compact group of isometries of a nonpositively curved complete simply-connected Riemannian manifold has a fixed point.
Proof. (Rough sketch) Consider any orbit. It is compact. By convexity of distance functions (the curvature of the ambient is nonpositive), its center of mass can be defined and it is plainly a fixed point. QED
Now one uses the fact that the symmetric space of non-compact type $G/K$ is nonpositively curved. If $H$ is a compact subgroup of $G$ then it has a fixed point $gK$ by the theorem, so $g^{-1}Hg$ fixes the basepoint $1K$ and hence is contained in $K$.
Edit: on suggestion of Ben, I complete my answer as follows. Let $\mathfrak g = \mathfrak k + \mathfrak p$ be the decomposition of the Lie algebra of $G$ into the eigenspaces of the involution. Since $M$ is complete and nonpositively curved, the Hadamard-Cartan theorem says that the map $\varphi:\mathfrak p \to G/K$ given by $\varphi(X)=(\exp X)K$ is a smooth covering. Next one sees that $\varphi$ is injective (assume $\varphi(X)=\varphi(Y)$, apply $\mathrm{Ad}$ to both sides and use the uniqueness of polar decomposition of a matrix, and the semisimplicity of $\mathfrak g$, to deduce that $X=Y$). Hence there are diffeomorphisms $\mathbf R^n\cong\mathfrak p\cong G/K$. Now it follows rather easily that $\mathfrak p\times K\to G$, $(X,k)\mapsto(\exp X)k$ is a diffeomorhism ($G=\exp[\mathfrak p]\cdot K$ in general for a symmetric space).
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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The answer to your title is "no"; lots of semi-simple Lie groups are not compact (for example, $SL_2(\mathbb{R})$). You're getting this mixed up with the fact that a complex semi-simple Lie group has a unique compact real form, and that this is a bijection to semi-simple compact Lie groups. (Complex reductive groups are in bijection with general compact Lie groups; this allows torus factors on both sides).