Why do we always classify states under covering group representations instead of the group itself? For example see the following picture I lifted from 'Symmetry in physics' by Gross

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So in the first example, why not classify states under SO(3) instead of SU(2)?

EDIT: From a physicists perspective, I know enough that we consider projective representations of groups in quantum mechanics because those are the most general thing that keeps $ |\langle \phi|\psi\rangle|^2 $ invariant. But Gross above seems to indicate that representations of the covering group are a better way to go. Furthermore, covering groups are always simply connected, to where-as groups with projective representations are in general not simply connected - is this property of (universal) covering groups related to why we use them in these sorts of cases?

  • $\begingroup$ A group is an abstract mathematical object. Measurements/Experiments are in the real field. So, if you want to draw meaningful conclusions that you can verify experimentally, you have to work with operator representations (in $\mathbb{R}$) that are appropriate. Ultimately, if you cannot prove something (or its consequences) that can be experimentally verified, it is worthless. While this does not directly answer your question, I think it is a useful reminder on why we work with representations instead of abstract objects. Hence the comment. $\endgroup$ Commented Dec 28, 2012 at 1:22
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    $\begingroup$ The reason we use covering groups is because finding projective representations of a Lie group $G$, is equivalent to finding linear representations of its universal cover $G'$, and we usually prefer to work with single valued objects. For example half-integer spin representations are linear wrt. $SU(2)$, but projective wrt. $SO(3)$. However $G$ and $G'$ have isomorphic Lie algebras, so one can alternatively consider representations of the Lie algebra, which is what physicist usually prefer to do. $\endgroup$
    – Heidar
    Commented Dec 28, 2012 at 11:25
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    $\begingroup$ @Heidar that could (and, I think, should) be an answer, or at least the start of one. $\endgroup$
    – David Z
    Commented Dec 29, 2012 at 5:06

1 Answer 1


Your edit shows just the right idea: we want to classify the projective representations.

There are two ways a projective representation can fail to lift to an ordinary linear representation: algebraically, or topologically.

The first refers to the Lie algebra: there are nontrivial projective representations if the algebra admits a nontrivial central extension. This is not the case for the Poincaré or so(3) algebras, so we don't have to worry about it, but it does happen, for example, for the Galilean group. In this case, we can add a 'mass' operator that commutes with everything, but can be appended to commutators of other operators.

The second possibility is that the group may fail to be simply connected, and this is what happens with Poincaré and rotation groups. The projective representations are then classified by representations on the universal covering group. The symmetry group you start with is then indistinguishable from the covering group, along with superselection rules (no superposition of a single boson and a single fermion can be prepared, for example).


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