Lie group compactness from generators Let us consider the Poincaré algebra, characterized by the following commutators:
\begin{align}
[H,P_i]&=0\\
[H,K_i]&=P_i\\
[P_i,P_j]&=0\\
[K_i,K_j]&=-\epsilon_{ijk}J_k\\
[P_i,K_j]&=\delta_{ij}H\\
[J_i,J_j]&=\epsilon_{ijk}J_k\\
[J_i,K_j]&=\epsilon_{ijk}K_k\\
[J_i,P_j]&=\epsilon_{ijk}P_k\\
[J_i,H]&=0
\end{align}
How could I know - using the algebra only - if the subgroup, generated by the $K_i$ generators, is compact or not? Is there a criteria for establishing compactness?
My understanding of a compact group is related to the notion of bounded and connected sets. For example the Lorentz group has four disconnected pieces, so it's a non-compact group.
 A: As Cosmas Zachos hints at in the comments, a non-Abelian Lie algebra belongs to a compact Lie group if its Killing form $K(X;Y) = \mathrm{tr}(\mathrm{ad}_X\circ \mathrm{ad}_Y)$ is negative-definite, cf. also compact Lie algebra where you can find a full list of all compact Lie algebras. The reason for this is that a non-degenerate Killing form induces a Levi-Civita connection $\nabla_X Y = \frac{1}{2}[X,Y]$ on the Lie group with Ricci curvature $-\frac{1}{4}K(X,Y)$, which is bounded below if the Killing form is negative-definite and therefore the Lie group is compact by Bonnet-Myers. Note that a negative-semidefinite Killing form, i.e. one which is degenerate, may or may not belong to a compact Lie group.
(Dis)connectedness has nothing to do with his - the Lorentz group is non-compact and has four connected components, but already the identity component, the proper orthochronous Lorentz group, is non-compact. Compactness and connectedness are different and unrelated topological properties.
A: There is already a good answer by ACuriousMind. Here we want to stress some important facts.


*

*Let there be given an $n$-dimensional real Lie algebra $$\mathfrak{g}~=~{\rm span}_{\mathbb{R}}\{t_a\mid a=1,\ldots, n\}, \tag{M1}$$
where$^1$
$$[t_a,t_b] ~=~\underbrace{f_{ab}{}^{c}}_{\in\mathbb{R}}~ t_c.\tag{M2}$$

*Let us assume that the $t_a$'s are the generators of a faithful finite-dimensional linear representation of the Lie algebra, cf. Ado's theorem.

*Lie's third theorem (more precisely Lie-Cartan's theorem) guarantees the existence of a corresponding connected & simply-connected Lie group $G$, such that its Lie algebra is $\mathfrak{g}$. In a neighborhood of the identity, the Lie group is reconstructed by the exponential map
$$ \exp(\mathfrak{g})~\subseteq~ G \tag{M3}.$$

*If the real Lie algebra $\mathfrak{g}$ is semisimple, it has a Cartan decomposition $$\mathfrak{g}~=~\mathfrak{k}+\mathfrak{p}.\tag{M4}$$ Then $K/Z_G$ is compact and $\exp(\mathfrak{p})$ non-compact, cf. Wikipedia.

*Pure Lie group/algebra theory does not introduce the notion of hermitian conjugation. However, this additional structure is often present in physics. If $t_a=-t^{\dagger}_a$ is anti-Hermitian, it corresponds to a compact direction; while if $t_a=t^{\dagger}_a$ is Hermitian, it corresponds to a non-compact direction.
--
$^1$ Be aware that in much of the physics literature, there is an extra factor of the imaginary unit $i$ in various places, e.g.
$$[t_a,t_b] ~=~i~\underbrace{f_{ab}{}^{c}}_{\in\mathbb{R}}~ t_c,\tag{P2}$$
and
$$ \exp(i\mathfrak{g})~\subseteq~ G \tag{P3}.$$ 
In particular, if $t_a=t^{\dagger}_a$ is Hermitian, it corresponds to a compact direction; while if $t_a=-t^{\dagger}_a$ is anti-Hermitian, it corresponds to a non-compact direction.
