# The representation of Lorentz group

by karlzr
Tags: lorentz, representation
 P: 86 The lorentz group SO(3,1) is isomorphic to SU(2)*SU(2). Then we can use two numbers (m,n) to indicate the representation corresponding to the two SU(2) groups. I understand (0,0) is lorentz scalar, (1/2,0) or (0,1/2) is weyl spinor. What about (1/2, 1/2)? I don't get why it corresponds to lorentz vector. Thanks a lot.
HW Helper
PF Gold
P: 2,602
 Quote by karlzr The lorentz group SO(3,1) is isomorphic to SU(2)*SU(2).
This is a common mistake made in many places but it is not true. What is true is that the Lorentz algebra ##so(3,1)## is isomorphic to ##su(2)\oplus su(2)##. The Lorentz group ##SO(3,1)## is noncompact, which can be seen by noting that the elements obtained from the boosts, ##\exp(i a L_i)##, have a parameter ##a## which is valued in the reals, rather than a finite or periodic interval. Since ##SU(2)\times SU(2)## is compact, the groups cannot be isomorphic. What is true is that ##SO(4) = [SU(2)\times SU(2)]/\mathbb{Z}_2##.

Since these are matrix groups, we can derive the representations of the group from those of the algebra. So the representation theory of ##SO(3,1)## corresponds in a particular way to that of ##SU(2)\times SU(2)##, even though the groups aren't isomorphic.

 Then we can use two numbers (m,n) to indicate the representation corresponding to the two SU(2) groups. I understand (0,0) is lorentz scalar, (1/2,0) or (0,1/2) is weyl spinor. What about (1/2, 1/2)? I don't get why it corresponds to lorentz vector. Thanks a lot.
To see the mapping, note that we can make a 4-vector from the identity matrix and the Pauli matrices as ##\sigma^\mu = (1, \sigma^i)##. In terms of the spinors, ##(\sigma^\mu)_{\alpha\dot{\alpha}}## has an index ##\alpha## corresponding to ##(1/2,0)## and an index ##\dot{\alpha}## corresponding to ##(0,1/2)##. These matrices are also invertible. We can therefore map any 4-vector ##V^\mu## to a matrix ## V_{\alpha\dot{\alpha}} = (\sigma_\mu)_{\alpha\dot{\alpha}}V^\mu##. The object ##V_{\alpha\dot{\alpha}}## transforms as the ##(1/2, 1/2)## representation. Since the mapping was invertible, this shows that the ##(1/2, 1/2)## representation is isomorphic to the 4-vector representation.
P: 2,470
 Quote by fzero Lorentz group ##SO(3,1)## is noncompact, which can be seen by noting that the elements obtained from the boosts, ##\exp(i a L_i)##, have a parameter ##a## which is valued in the reals, rather than a finite or periodic interval.
I'm just trying to fill in the gaps and my Group Theory is rusty. Does non-compactness follow from the fact that there can be chosen a sequence ##\{a_n\}##, such as one diverging to ##\pm \infty##, so that ##\displaystyle \lim_n \exp(i a_n L_i)## does not belong to ##SO(3, 1)##?

 Mentor P: 6,248 The representation of Lorentz group Matrices are subsets of ##\mathbb{R}^N## for some appropriate ##N##. Consequently, by the Heine-Borle theorem, compactness is equivalent to closed and bounded. Consider the the subgroup of the Lorentz group ##SO(3,1)## that consists of boosts along a fixed axis (say, the x-axis) in a particular inertial reference frame. This group is homeomorphic to ##\mathbb{R}##, which is neither closed nor bounded.
 Sci Advisor Thanks P: 4,160 For a continuous group G, one defines a volume element in the group space, dv = dξ1 dξ2... dξn, which is invariant under group operations. Then V = ∫G dv is the total volume, and G is compact or noncompact depending on whether V is finite or infinite.
 Sci Advisor P: 2,470 But can I simply say that SO(3, 1) does not contain its limit points, and therefore, by definition, not compact? Or is there a nuance I'm missing?
Mentor
P: 6,248
 Quote by K^2 But can I simply say that SO(3, 1) does not contain its limit points, and therefore, by definition, not compact? Or is there a nuance I'm missing?
Yes, but see a more precise and explicit formulation below.

Limit point compactness and (actual) compactness are equivalent for second countable Hausdorff spaces, which is the situation here.

##A \subset X## is limit point compact if every infinite subset of ##A## has a limit point. Negating this gives ##A \subset X## is not limit point compact if there exists an infinite subset of ##A## that does not have a limit point.

Take ##A=SO(3,1)## and ##X=\mathbb{R}^{16}## (4x4 matrices). It's not that ##A## doesn't contain all its limit points, it's that there are infinite subsets of ##A## that don't have any limits points.

 Quote by George Jones Matrices are subsets of ##\mathbb{R}^N## for some appropriate ##N##. Consequently, by the Heine-Borle theorem, compactness is equivalent to closed and bounded. Consider the the subgroup of the Lorentz group ##SO(3,1)## that consists of boosts along a fixed axis (say, the x-axis) in a particular inertial reference frame. This group is homeomorphic to ##\mathbb{R}##, which is neither closed nor bounded.
I thought you (K^2) might want to see this directly for ##A##. My example illustrates this. More explicitly, consider the infinite subset of ##A## that consists of matrices of the form

$$\begin{pmatrix} \cosh n & \sinh n & 0 & 0\\ \sinh n & \cosh n & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$

for all integers ##n##. There exist neighbourhoods of each such matrix that don't contain any other matrices of the same form. Thus ##A## does not *any* limits points (even ones that don't "live in" in ##A##).

Consequently, ##A = SO(3,1)## is not compact.