So I understand that the Lie algebra of $SO(1,3)$ is isomorphic to the Lie algebra of $SU(2)\oplus SU(2)$, and the Lie algebra of $SO(3)$ is isomorphic to one copy of $SU(2)$ (at the group level we have $SO(3)\cong SU(2)/Z_2$).

I guess my question is in two parts:

  1. In ”Physics from Symmetry” by Schwichtenberg, he says that the covering group is more fundamental - i.e. by considering $su(2)$ instead of $SO(3)$ we get spin.

  2. As far as I am aware, if we take the exponential map of a Lie algebra homomorphism/representation then we get a representation of the group. This would imply $SO(3)$ is a representation of $SU(2)$, but how can there exist a homomorphism from $SU(2)$ to $SO(3)$, if each element in $SU(2)$ gets identified with two elements of $SO(3)$?

  • 1
    $\begingroup$ The homomorphism (isomorphism) only exists around the unit element of the group. $\endgroup$
    – p6majo
    Dec 26 '19 at 9:48
  • 2
    $\begingroup$ This Q&A of mine discusses in technical detail how quantum mechanics leads to the usage of covering groups. $\endgroup$
    – ACuriousMind
    Dec 26 '19 at 12:18
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    $\begingroup$ You want to be careful here. The complexification of $SO(3,1)$ is isomorphic to $SU(2)\oplus SU(2)$. $\endgroup$ Dec 26 '19 at 14:38
  • $\begingroup$ This article in WP is more than helpful: en.wikipedia.org/wiki/Wigner%27s_theorem $\endgroup$
    – DanielC
    Dec 29 '19 at 17:49

$SO(3)$ irreps (irreducible representations) naturally only contain spin $0,1,2,\dots$

$SU(2)$ introduces half integer irreps.

The fundamental representation of $SO(3)$ is a $\bf 3$, consisting of the rotations around the three axes:

$\left(\begin{array} {c c c} \cos(\phi)& -\sin(\phi) &0\\ \sin(\phi) & \cos(\phi) & 0 \\ 0 & 0 & 1\end{array}\right)$

$\left(\begin{array} {c c c} \cos(\psi)& 0 & \sin(\psi) \\ 0 & 1 & 0 \\-\sin(\psi) & 0 & \cos(\psi) \end{array}\right)$

$\left(\begin{array} {c c c} 1 & 0 & 0 \\ 0 & \cos(\chi)& -\sin(\chi) \\ 0 & \sin(\chi) & \cos(\chi) \end{array}\right)$

It corresponds to $j=1$, the spin-1-representation. Tensor products generate all other representations with $j\in \mathbb{N}$.

$SU(2)$ is more fundamental, since it contains besides representations with $j\in \mathbb{N}$ also half-integer spins $j\in \frac{\mathbb{N}}{2}$.

The fundamental representation $\bf 2$ of $SU(2)$ is represented by the $2\times 2$-complex-valued matrices, obtained by exponentiating the Pauli-spin matrices. It corresponds to $j=\frac{1}{2}$. From it, all integer and half-integer representations can be generated.

  • $\begingroup$ With help of those three matrices we can safely recalculate a given vector of any length from one reference frame to another. In order to speak of spin, we must deal with wave functions, not just with vectors in 3D space. $\endgroup$ Dec 29 '19 at 17:45

For myself, I answered part 1 of your question empirically. There are objects (eg: electrons) in the world which only return to their original state after rotation by $4\pi$ (not $2\pi$). These "half integer spin" objects also have an even number of states (ie: z projections of angular momentum). No representations of $SO(3)$ provide these properties, but representations of $SU(2)$ do and also provide representations to rotate all the integer spin particles too. Thus $SU(2)$ more completely represents nature (ie: is more fundamental).


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