Lie group compactness from generators

Question

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.

Answer 1

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.

Answer 2

There is already a good answer by ACuriousMind. Here we want to stress some important facts.

1. 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$^1$ $$[t_a,t_b] ~=~\underbrace{f_{ab}{}^{c}}_{\in\mathbb{R}}~ t_c.\tag{M2}$$

2. 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.

3. 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 $$ \exp(\mathfrak{g})~\subseteq~ G \tag{M3}.$$

4. If the real Lie algebra $\mathfrak{g}$ is semisimple, the Killing form is non-degenerate (Cartan's criterion), and it has a Cartan decomposition $$\mathfrak{g}~=~\mathfrak{k}\oplus\mathfrak{p}.\tag{M4}$$ Then $K/Z_G$ is compact and $\exp(\mathfrak{p})$ non-compact, cf. Wikipedia. We can construct a real semi-simple Lie algebra $$\mathfrak{g}_{\rm compact}~:=~\mathfrak{k}\oplus i\mathfrak{p}\tag{M5}$$ with a negative definite Killing form. The corresponding connected & simply-connected Lie group $G_{\rm compact}$ is compact, cf. Ref. 1.

5. Pure Lie group/algebra theory does not introduce a notion of hermitian conjugation. However, if $G_{\rm compact}\subseteq GL(V,\mathbb{C})$, where $V$ is a finite-dimensional complex vector space, then we can use Weyl's Unitarian_trick to construct a sesquilinear form such that $G_{\rm compact}\subseteq U(V)$.

6. A sesquilinear form 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.

References:

1. M.M. Alexandrino & R.G. Bettiol, Introduction to Lie groups, isometric and adjoint actions and some generalizations, arXiv:0901.2374; Theorem 2.28.

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$^1$ 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.