Why is it interesting?

If we only do "naive" representation theory of the Lorentz group $SO(3,1)$ we miss some very important representations. These representations that we miss are not representations in the usual sense, but so-called projective representations. These can be used in a quantum framework where the vector space our operator act on is the Hilbert space, because it makes no observable difference if we multiply all states with a global phase. For the Lorentz group, the projective representations that we would miss if we only consider the usual representations are the ones that describe spin $1/2$ particles, i.e. fermions. We can find these, if we derive the "naive" representations of the universal cover of the Lorentz group $SL(2,\mathbb{C})$. The projective representations of $SO(3,1)$ correspond to regular representations of $SL(2,\mathbb{C})$.

Layman

Student

"a projective unitary representation is just an ordinary unitary representation with an extra phase factor that prevents it from being an honest homomorphism. Working with projective representations isn't as easy as working with ordinary representations since they have the pesky phase factor cc, so we try to look for ways of avoiding them. In some cases, this can be achieved by noting that the projective representations of a group $G$ are equivalent to the ordinary representations of $G′$ its universal covering group, and in this case, we therefore elect to examine the representations of the universal cover instead." http://physics.stackexchange.com/questions/96045/idea-of-covering-group

>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." http://physics.stackexchange.com/questions/47740/why-do-we-classify-states-under-covering-groups-instead-of-the-group-itself

Usually, a representation $R$ is defined as a map that maps the group elements onto linear operators over some vector space and satisfies $$ R(gh) = R(g) R(h). \tag{1} $$ Here $g$ and $h$ are two group elements and a representation $R(g)$ usually simply means that we get have a matrix that represents the element $g$.

In quantum mechanics, there is an additional freedom that we can make use of. The vector space in quantum mechanics is the Hilbert space of states $|\Psi \rangle$ and we have the freedom to rephase states as we wish:

$$ |\Psi \rangle \rightarrow \mathrm{e}^{i\theta} |\Psi \rangle . $$ The reason for this is, that "a transformation of [the Hilbert space] by an overall phase is unobservable" (source). We only care about probability amplitudes $ \langle \Phi | \Psi \rangle $ and note about the states themselves. Therefore, a global rephasing $$|\Psi \rangle \rightarrow \mathrm{e}^{i\theta} |\Psi \rangle , \qquad |\Phi \rangle \rightarrow \mathrm{e}^{i\theta} |\Phi \rangle , \qquad \ldots . $$ of all states has no influence.

Therefore, we can introduce a slightly less strict condition to define the type of linear operators that we allow. Instead of the condition in Eq. 1, we now relax the condition and define the new type of map as $$ R(gh) = R(g) R(h) \mathrm{e}^{i\theta(g,h) }. \tag{2}$$ This is allowed because the operators $R(g)$, $R(h)$ and $R(gh)$ act on elements of the Hilbert space of states, i.e. on states $$ R(gh) |\Psi \rangle = R(g) R(h) \mathrm{e}^{i\theta(g,h) } |\Psi \rangle .$$ A map that satisfies the condition in Eq. 2 is called a projective representation.

Researcher

Examples

Projective Representations of the Lorentz Group

The Lorentz group is not simply connected and therefore, we can find projective representations. (It is doubly connected).

However, instead of working with projective representations, we can work with regular representations of the corresponding universal covering group. For the Lorentz group $SO(3,1)$ the universal covering group is $SL(2,\mathbb{C})$, which is simply connected. It's fundamental representation acts on Weyl spinors. The fundamental representation of the Lorentz group, which acts on four-vectors can be seen as the "next-higher" representation (and therefore non-fundamental) representation of $SL(2,\mathbb{C})$. A four-vector is a rank two spinor, which means that it carries two spinor indices.

"The point here is that we’re looking at projective representations of the Lorentz group (quantum mechanics says that we’re allowed to transform up to a phase). The existence of a projective representation of a group is closely tied to its topology (whether or not it is simply connected); the Lorentz group is not simply connected, it is doubly connected. The objects with projective phase -1 (i.e. that pick up a minus sign after a 360 degree rotation) are precisely the half-integer spinor representations, i.e. the fermions." http://www.quantumdiaries.org/2011/08/23/the-spin-of-gauge-bosons-vector-particles/

"The conclusion of all of this is that, when trying to find all irreducible projective representations of the Poincare group, we can instead look for all irreducible representations of $SL(2, C) \ltimes \mathbb{R}(3,1)$ on ordinary Hilbert spaces instead" http://web.science.uu.nl/itf/Teaching/2011/2012/Daniel%20Prins.pdf

FAQ

How can we derive Projective Representations?

This is very difficult and one usually uses that one can derive the regular representations of the universal covering group instead. However this procedure does not always work:

"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)." http://physics.stackexchange.com/questions/47740/why-do-we-classify-states-under-covering-groups-instead-of-the-group-itself

For an explicit derivation of projective representations, see section 4.3.3 the Bargmann Group in Symmetries in Fundamental Physics by K. Sundermeyer. To quote from it:

"finding the representations of the Galilei group is rather laborious. This is the reason why there are only a few textbooks which deal with this topic. Here, I only sketch the line of argumentation, leaving out the technical details. This is done in order to indicate how delicate the determination of projective representations of non-simple groups can be; more details in [340]. "

Ref. 340 is Lévy-Leblond, J.-M.: Galilei group and nonrelativistic quantum mechanics. J. Math. Phys. 4, 776–788 (1963)

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