redtree said:
Per post #16,... such as ##O(1,3)## with subgroups classified by ##det \pm 1## and ##\Lambda_0^0 > 1 \lor <1##, cannot have a single set of generators spanning the entire group.
1) There are no subgroup
s. I have already said in #16, O(1,3) has
one, and only one, Lie subgroup, and that is SO^{\uparrow}(1,3). This is because SO^{\uparrow}(1,3) is the
only component of O(1,3) that
contains the group
identity element e (i.e., the identity matrix e = I_{4}). The identity element I_{4} is
not contained in anyone of the other components \Lambda_{P}SO^{\uparrow}(1,3), \Lambda_{T}SO^{\uparrow}(1,3) and \Lambda_{PT}SO^{\uparrow}(1,3). So these components are
not groups, let alone Lie subgroups. In general, if G is a
real matrix Lie group, then only a
closed subgroup of G is a
smooth manifold. This is the
reason why O(1,3) and SO^{\uparrow}(1,3) share the
same Lie algebra \mathfrak{o}(1,3) = \mathfrak{so}(1,3): any smooth (time-like) curve in O(1,3) passing through the identity element
must lie entirely in SO^{\uparrow}(1,3) because the determinant (which is continuous in \mathbb{R}^{4 \times 4}) cannot
jump from +1 to -1. This means that SO^{\uparrow}(1,3) and O(1,3) share the same tangent space at the identity, T_{e}\left(SO^{\uparrow}(1,3)\right) = T_{e}\left(O(1,3) \right), i.e., the same Lie algebra \mathfrak{o}(1,3) = \mathfrak{so}(1,3).
2) Of course, the exponential map \mbox{exp}: \mathfrak{o}(1,3) \to O(1,3) is well-defined but it is not surjective because there are matrices in O(1,3) with determinant -1, whereas (symbolically) \mbox{det}\left( e^{\mathfrak{o}(1,3)}\right) \equiv \mbox{det}\left( e^{\mathfrak{so}(1,3)}\right) = e^{\mbox{Tr}\left( \mathfrak{so}(1,3)\right)} = 1. So,
exponentiation (of the algebra \mathfrak{o}(1,3) = \mathfrak{so}(1,3)) leads to the group SO^{\uparrow}(1,3) instead of O(1,3). Of course, that is not a problem because the other 3 components of O(1,3) can be obtained by multiplying SO^{\uparrow}(1,3) with \Lambda_{P}, \ \Lambda_{T} and \Lambda_{PT}.
3) For every Lie group G, the definition \mathfrak{g} = T_{e}(G) gives
uniquely a Lie algebra. Given a Lie algebra, however, there may be
more than one Lie group for which it is
the Lie algebra. And, in infinite dimensions, the group may not exist. For example, it is known that no Lie group corresponds to the complexified Virasoro algebra (without central extension). Groups with the same Lie algebra are said to be
locally isomorphic. Examples of locally isomorphic groups : SO(3) and SU(2); SO(4) and SU(2) \times SU(2); SO^{\uparrow}(1,3) and SL(2 , \mathbb{C}). All locally isomorphic groups can be obtained by factoring the
simply connected covering group by various discrete normal subgroups of its fundamental group. For example SO^{\uparrow}(1,3) \cong SL(2, \mathbb{C}) / \mathbb{Z}_{2}.
4) Finally, I leave you to think about the following
Exercise (i). In the textbooks, you often read the following sentence: “The full Lorentz group O(1,3) is a real Lie group consisting of four
disconnected components (or pieces)”. What does the word “
disconnected” mean in that sentence, when we know for sure that the component SO^{\uparrow}(1,3) is a real and
connected Lie group?
Exercise (ii). Show that the exponential map \mbox{exp} : \mathfrak{so}(1,3) \to SO^{\uparrow}(1,3) is well-defined and “surjective”. Or, if you don’t understand the mathematical jargons, show that
any Lorentz matrix \Lambda \in SO^{\uparrow}(1,3) can be written as \Lambda = \mbox{exp} \left( \frac{1}{2}\omega^{\rho \sigma}M_{\rho \sigma} \right), where \omega^{\mu\nu} = - \omega^{\nu\mu} are the real parameters, and the matrices M_{\rho \sigma}, whose elements are given by (M_{\rho \sigma})^{\mu}{}_{\nu} = \delta^{\mu}_{\rho} \ \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \ \eta_{\rho \nu} , form a basis of the Lie algebra \mathfrak{so}(1,3) of the Lorentz group SO^{\uparrow}(1,3), i.e., \big[M_{\mu\nu} , M_{\rho \sigma} \big] = \eta_{\mu \sigma} M_{\nu\rho} - \eta_{\mu\rho} M_{\nu\sigma} + \eta_{\nu\rho} M_{\mu\sigma} - \eta_{\nu\sigma} M_{\mu\rho} .
