# How does the parameter space or the group manifold of $SO^+(3,1)$ differ from that of $SO(3)$?

For $SO(3)$, the parameter space is $\mathbb{S}^3$ with diametrically opposite points identified. It is a simple exercise to show that $R(\hat{\textbf{n}}, \psi)\in SO(3)$ can be parameterized as $$R_{jk}(\hat{\textbf{n}},\psi)=\cos\psi\delta_{jk}+(1-\cos\psi)n_jn_k-\sin\psi\epsilon_{jkl}n_l$$ with $i,j,k=1,2,3$. Now $\hat{\textbf{n}}$ can be directed towards any point from the origin. One can check from the formula above, that the rotation through $\psi$ about an axis is geometrically equivalent to the rotation through $(2\pi-\psi)$ about the opposite axis and therefore, the parameter space is doubly connected.

$\bullet$ How does the parameter space or the group manifold of $SO^+(3,1)$ differ from that of $SO(3)$?

$\bullet$ Is it possible to find out or guess the topological connectivity property of 3-D Euclidean space and Minkowski space itself from the respective metrics?

• Concerning the first subquestion (v3), see e.g. Wikipedia: $SO^+(3,1)/SO(3)\cong \mathbb{R}^3$ corresponding to 3 non-compact boost parameters. – Qmechanic Nov 26 '16 at 8:14

A general element of $\mathrm{O}(1,3)$ may be written as $$\begin{pmatrix} \epsilon & 0 \\ 0 & Q\end{pmatrix}\begin{pmatrix} \sqrt{1 + \lambda^T \lambda} & \lambda^T \\ \lambda & \sqrt{\mathbf{1}_3 + \lambda\lambda^T}\end{pmatrix}$$ where $\epsilon\in\{\pm 1\}, Q\in\mathrm{O}(3),\lambda\in\mathbb{R}^3$. So as a manifold, we have $\mathrm{O}(1,3)\cong \{\pm 1\} \times\mathrm{O}(3)\times \mathbb{R}^3$, and the connected component of the identity is isomorphic to $\mathrm{SO}(3)\times\mathbb{R}^3 = \mathbb{R} P ^3\times\mathbb{R}^3$.

3D space (any space, really) has the same connectivity (homotopy groups) for all metrics because connectivity does not depend on the metric - the homotopy groups are a topological invariant.

As ACuriousMind says, one cannot infer global connectedness properties from the metric, if by "metric" for a Lie group I assume you mean symmetric billinear form on the Lie algebra. A very simple way to see this is to note that $SO(3)$ and $SU(2)$ have the same Lie algebra, whereas, as you know, the former is doubly connected, the latter simply so. So nothing that the metric can know will detect this fact.

However, one global topological property that is encoded in the metric, and which is a key difference between $SO^+(3, 1)$ and $SO(3)$ is a group's compactness or otherwise. One looks at the Killing form on the Lie algebra for this test. For we have the following:

Theorem A Lie group which is centerless or at most has a discrete center is compact if and only if its Killing form on its Lie algebra is negative definite.

See, for a proof, and further information:

S. Helgason "Differential geometry Lie groups and symmetric spaces" Chap. II, section 6, proposition 6.6.

So you compute the function $K:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R};\;K(X,\,Y) = \mathrm{tr}(\mathrm{ad}(X)\,\mathrm{ad}(Y))$ and test it. We get the following results from this relevant to your question, and some other simple and interesting ones besides.

1. If the Killing form is nondegenerate, then the Lie group is semisimple;
2. If the Killing form is degenerate, but its kernel is the same as the center of the algebra, then the Lie group is reductive, i.e. the direct product of an Abelian group and a semisimple one. Abelian groups cannot have their compactness inferred from the Lie algebra (witness $\mathbb{R}$ as compared with $U(1)$), but we may be able to infer compactness of the semisimple part because:
3. Iff the Killing form is negative definite for the semisimple part, then the semisimple part is compact.

Beyond Metrics and the Tangent Space

Some further thoughts about your question are as follows. From your Rodrigue's formula for $SO(3)$ you seem to be saying that one can construct the whole Lie group from any Lie algebra member, and that therefore you should, in principle, be able to deduce the global topological properties. This statement last statement is true for $SO(3)$, but it is not true for $SO^+(1,\,3)$ (as far as parameterizations grounded on the matrix representation of the Lie algebra are concerned). Let's work through this assertion, and how it fits with my Lie theoretic comments in the section above now i.e. that you can't use properties of the metric defined on the tangent spaces to do this as discussed above (since the tangent spaces are all simply left-translated copies of the Lie algebra kitted with an inner product, wontedly the Killing form).

The Rodrigues formula arises as follows. The characteristic equation for a $3\times 3$, real-element, skew-symmetric matrix $H$ (i.e. a member of the $3\times3$ matrix representation of the Lie algebra $\mathfrak{so}(3)$) is:

$$H^3 + H =0\tag{1}$$

This equation is then used to simplify the exponential's Taylor series: owing to the linear dependence expressed in (1), there can only be terms in $\mathrm{id},\,H$ and $H^2$. On gathering together all the terms involving these three, your formula results.

A different Rodrigues formula arises for the exponentiation of the $2\times2$ matrix representation of the same Lie algebra. For a member $J\in\mathfrak{su}(2)$ of the set of $2\times 2$ skew-Hermitian, complex element matrices, we have the characteristic equation:

$$J^2 + \mathrm{id}=0\tag{2}$$

and so, when we gather terms in the now two linearly independent entities $\mathrm{id}$ and $J$ using this characteristic equation we get the different Rodrigues formula:

$$\exp J = \cos\|J\|\,\mathrm{id} + \frac{\sin\|J\|}{\|J\|}\,J\tag{3}$$

which is de Moivre's formula for quaternions.

Thus we see that the exponential, mapping the Lie algebra into the Lie group, is different for $SO(3)$ and $SU(2)$. Because the exponential map is onto in these cases, we can parameterize the whole Lie group by elements of the relevant Lie algebra matrices. Because we can construct the whole group thus, we can infer topological properties as you have already observed.

For $SO^+(1,\,3)$, however, we meet a snag and this procedure fails:

$\exp:\mathfrak{so}(1,\,3)\to SO^+(1,\,3)$ is not surjective. There are elements of $SO^+(1,\,3)$ which are not the exponential of any Lie algebra member.

In principle we can find a Rodrigues formula for $SO^+(1,\,3)$ by the same procedure as above: use the characteristic equation to simplify the exponential series. This would answer your first subquestion. I have even done it in the past but it is horrendously complicated and I can't find my Mathematica file wherein i made the calculations. But now the Rodrigues formula will not help us, because our parameterization will not reach all members of the Lie group. So we cannot determine global topological properties from the parameterization and Rodrigues formula.

The crucial group property here is compactness. It is known that if a Lie group's identity connected component is compact, then $\exp$ is always surjective between a Lie algebra and the identity component. But, whilst a sufficient condition for surjectivity, compactness is not necessary, as the following two examples show. It fails to be surjective for $SO^+(1,\,3)$. But it is surjective for the identity component of the noncompact Euclidean groups (rotations and translations). As far as I know, the surjectivity of $\exp$ and when it holds is still an open research question.

Other Global Parameterizations

So much for parameterizations grounded on matrix representations of the Lie algebra. There are other parameterizations that do allow one to represent a general group member, even for noncompact Lie groups. However, they are not always practicable to compute.

Any connected matrix group will admit a Polar Decomposition, which is closely related to the Cartan Decompositon (which works for semisimple groups, whereas the Polar Decomposition is in theory general). ACuriousMind has shown you the Polar Decomposition for $SO^+(1,\,3)$ (the formula for $\epsilon=1$ is the polar decomposition of the identity connected component). In general, a square complex matrix $A$ can be decomposed as

$$A = U\,P$$

where $U$ is a unitary matrix, $P = \sqrt{A^\dagger\,A}$ is a positive semidefinite Hermitian matrix and so $U = A\,P^{-1}$. The square root is not always tractably computable as in general a singular value decomposition needs to be done (this of course works numerically, but it won't in general get you an analytic parameterization); ACuriousMind's answer shows how it can be done analytically in $SO^+(1,\,3)$.