# 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.

• Computed the Killing form? Oct 21, 2017 at 15:18
• There is no subgroup generated by the K's, because their algebra doesn't "close". Oct 21, 2017 at 15:35
• Yes, its true. But the minimal subgroup, which contains this K as one of its generators? Oct 21, 2017 at 15:50
• I would recommend the following: Contrast the Killing form of SO(3), a sphere, to that of SO(2,1), a hyperboloid. Then, trash H and P to stick to the Lorentz algebra above, (after correcting your error in the [K,K]). Associate it to SO(3,1), and observe the 5d hypersurface linked to the Killing form. Oct 25, 2017 at 1:47

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.

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.

• Notes for later: Math convention: $\quad GL(n,\mathbb{R})$ $[SL(n,\mathbb{R})]$ has $\frac{n(n-1)}{2}$ compact and $\frac{n(n+1)}{2}[-1]$ non-compact generators, respectively. $\quad SO^+(p,q;\mathbb{R})$ has $\frac{p(p-1)}{2}+\frac{q(q-1)}{2}$ compact and $pq$ non-compact generators. $\quad Sp(2n,\mathbb{R})=\left\{\begin{pmatrix} a & b_S \cr c_S & -a^T\end{pmatrix} \right\}$ has $n^2$ compact and $n(n+1)$ non-compact generators. RaS=aH=C. RS=H=nC. A Lie group whose Lie algebra is a complex vectors space has an equal number of compact and non-compact generators. May 1, 2023 at 11:53
• Notes for later: $\quad U(p,q)$ [$SU(p,q)$] has $p^2+q^2[-1]$ compact and $2pq$ non-compact generators, respectively. $\quad Sp(p,q)$ has $p(2p+1)+q(2q+1)$ compact and $4pq$ non-compact generators. aH=C. H=nC. Cartan involution: $\quad\theta(x)=-x^t$. $\quad\theta([x,y])=[\theta(x),\theta(y)]$. The invariant bilinear form $\quad \beta(x,y)={\rm tr}(xy)$ has positive (negative) signature for Hermitian (anti-Hermitian) elements, respectively. May 2, 2023 at 8:26