I'd be grateful if someone could check that my exposition here is correct, and then venture an answer to the question at the end!

$SO(3)$ has a fundamental representation (spin-1), and tensor product representations (spin-$n$ for $n\in\mathbb{Z})$.

$SO(3)$ has universal covering group $SU(2)$. The fundamental representation of $SU(2)$ and its tensor product representations descend to projective representations of $SO(3)$. We call these representations spin representations of $SO(3)$ (spin-$n/2$ for $n\in \mathbb{Z}$).

The complex vector space $\mathbb{C}^2$ has elements called spinors, which transform under a rotation $R$ according to the relevant representative $D(R)$. The natural generalisation of a spinor is called a pseudotensor, and lives in the tensor product space.

We can repeat the analysis for the proper orthochronous Lorentz group $L_+^\uparrow$. We find that the universal covering group is $SL(2,\mathbb{C})$ and we get two inequivalent spin-$1/2$ projective representations of $L_+^\uparrow$, namely the fundamental and conjugate representations of $SL(2,\mathbb{C})$.

Now when we pass to the full Lorentz group, somehow the projective representations disappear and become genuine representations. Why, morally and mathematically, is this? If it's possible to give an answer without resorting to the Lie algebra, and just working with representations of the group I'd be delighted!

Many thanks in advance.


1 Answer 1


Now when we pass to the full Lorentz group, somehow the projective representations disappear and become genuine representations.

I don't think this is true. Some but not all of the spinor representations of the proper orthochronous Lorentz group extend to representations of the full Lorentz group; you just add parity reversal and time reversal. But the new representations are still projective.

  • $\begingroup$ You are right: the Lorentz group does have projective representations. The full argument is given in Weinberg I ch. 2.7: if a group has no projective reps, then (1) its Lie algebra must have no central charge(s) and (2) the group must be simply connected. (1) is okay for the Lorentz group, but (2) not: the group is isomorphic to $SL(2,\mathbb{C})/\mathbb{Z}_2$, which is not simply connected. However, the 'phase' appearing under group multiplication can only be $\pm 1$ (if the state in question is a boson/fermion), and there can be no mixing between the two. $\endgroup$
    – Vibert
    Commented Dec 21, 2012 at 22:47
  • $\begingroup$ @user1504 - Many thanks! So when people talk about half spin representations of the full Lorentz group they really mean projective representations then? Perhaps it's because I come from a mathematical background, but I don't like it when people talk about a "representation" where $D(I)\neq I$! $\endgroup$ Commented Dec 22, 2012 at 0:20
  • 1
    $\begingroup$ @EdwardHughes: Yes, that's correct. Projective reps of $G$ are reps of the universal cover of $G$. They aren't actually representations of $G$. But in quantum mechanics, since we don't care about any constants multiplying a state, symmetries can be realized as projective representations. $\endgroup$
    – user1504
    Commented Dec 22, 2012 at 1:25
  • $\begingroup$ @user1504: it's careless to say that we don't care about constants multiplying a state, since normally in QM phases matter. In this case we don't care, because by unitarity the phase cannot depend on the state in question, so you cannot measure it, even in superpositions. $\endgroup$
    – Vibert
    Commented Dec 22, 2012 at 16:09
  • $\begingroup$ @Vibert So the Wikipedia article here is wrong, and should say projective representation everywhere it says representation then? $\endgroup$ Commented Dec 23, 2012 at 0:39

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