Projective Representations and Multipliers

A projective representation implements a group law up to scalar phases while implementing the induced action on rays exactly. Its mathematical data include a multiplier, whose identities encode associativity, and an equivalence under rephasing the implementers. This page develops those identities, including the conjugation required by antiunitary elements; the physical introduction is Projective Representations.

Required background. Wigner’s Theorem supplies the lifts and their phase uniqueness. Helpful background. The symmetry introduction explains their physical meaning.

Associativity constrains a unitary multiplier

Let $G$ be a group and choose unitaries $U_g$ with $U_e=I$ satisfying

$$U_gU_h=\omega(g,h)U_{gh}, \qquad \omega(g,h)\in U(1).$$

The identity normalization gives $\omega(e,g)=\omega(g,e)=1$. Associativity of the operator product requires

$$\omega(g,h)\omega(gh,k) =\omega(h,k)\omega(g,hk).$$

Indeed, expanding $(U_gU_h)U_k$ gives the left side multiplying $U_{ghk}$; expanding $U_g(U_hU_k)$ gives the right side. A function satisfying this normalized identity is a multiplier, or a normalized two-cocycle.

The relation is exact. “Projective” describes the scalar freedom, not an approximation to the group law. Nor does an arbitrary function of two group elements define a valid representation: it must satisfy this compatibility condition, and finding operators that realize it is a further problem.

Rephasing changes the representative multiplier

Choose $\beta:G\to U(1)$ with $\beta(e)=1$, and set $U'_g=\beta(g)U_g$. Then

$$\omega'(g,h) =\frac{\beta(g)\beta(h)}{\beta(gh)}\omega(g,h).$$

The prefactor is a coboundary. Thus the individual values of $\omega$ depend on phase choices, whereas its equivalence class under these transformations does not.

A multiplier can be removed exactly when such a rephasing produces $\omega'=1$. It is insufficient to remove phases for a few generators while ignoring their relations. The obstruction belongs to the full composition law.

For a topological group, continuity or measurability requirements must also be imposed in the appropriate category. A continuous ray action need not possess one globally continuous phase choice for its lifts. The formulas here are algebraic; global topological lifting questions are treated through Central Extensions.

A commuting group with anticommuting lifts

Take $G=\mathbb Z_2\times\mathbb Z_2$, with labels $(a,b)$ and addition modulo two. On $\mathbb C^2$ use Pauli matrices $X,Z$ and set

$$U_{(a,b)}=X^aZ^b.$$

Since $ZX=-XZ$,

$$\omega((a,b),(c,d))=(-1)^{bc}.$$

The two group generators commute, but their lifts satisfy $XZ=-ZX$. Rephasing the two lifts cannot change the scalar commutator

$$XZX^{-1}Z^{-1}=-I.$$

If these lifts could be rephased into an ordinary representation of the abelian group, they would commute. Hence the multiplier is nontrivial. This is a finite example of detecting an obstruction without computing an entire cohomology group.

The sign does not prevent the actions on rays from commuting: multiplying a vector by $-1$ leaves its ray unchanged. It does constrain how the representation combines with other sectors or subsystems.

Antiunitary elements twist the cocycle identity

Now allow $U_g$ to be unitary or antiunitary. Let $s(g)=+1$ in the unitary case and $s(g)=-1$ in the antiunitary case. Composition requires $s(gh)=s(g)s(h)$, and for a phase $z$ write

$$z^{s(g)}= \begin{cases}z,&s(g)=+1,\\\overline z,&s(g)=-1.\end{cases}$$

Moving a scalar past $U_g$ gives $U_gz=z^{s(g)}U_g$. Associativity therefore yields the twisted identity

$$\omega(g,h)\omega(gh,k) =\omega(h,k)^{s(g)}\omega(g,hk).$$

Rephasing similarly gives

$$\omega'(g,h) =\frac{\beta(g)\beta(h)^{s(g)}}{\beta(gh)} \omega(g,h).$$

Using the untwisted formulas for a group containing antiunitary operations can erase a genuine invariant. For example, let an order-two group element $r$ be represented by an antiunitary $\Theta$. Then $\Theta^2=\omega(r,r)I$, and the twisted identity forces $\omega(r,r)$ to be real, hence $\pm1$. Rephasing $\Theta$ does not change that sign, because $\beta(r)\overline{\beta(r)}=1$.

Exercise: why a unitary square is different

An order-two ray symmetry has a unitary lift $V$ with $V^2=e^{i\alpha}I$. Can its square be changed to $I$ by rephasing? Compare with an antiunitary lift whose square is $-I$.

Solution

For the unitary lift, $V'=e^{-i\alpha/2}V$ satisfies $(V')^2=I$. For an antiunitary lift, $(e^{i\beta}\Theta)^2=e^{i\beta}e^{-i\beta}\Theta^2=\Theta^2$, so the sign cannot be changed. This comparison concerns the order-two relation; other group relations can still obstruct a simultaneous choice of phases.

References

• V. Bargmann, “On Unitary Ray Representations of Continuous Groups”, Annals of Mathematics 59, 1–46, 1954 — multipliers and ray representations.
• K. Landsman, Foundations of Quantum Theory: From Classical Concepts to Operator Algebras, Fundamental Theories of Physics 188, Springer, 2017, Chapter 5 — symmetry groups and projective implementations.
• K.-H. Neeb and G. Ólafsson, “Antiunitary representations and modular theory”, arXiv:1704.01336 [math.RT], 2017 — group representations including antiunitary operations.