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I am reading the QFT book by Weinberg. The first volume. On page 82, he discussed projective representation.

$$ U(T_2) U(T_1) = \exp(i \phi(T_2, T_1)) U(T_2 T_1 ) .\tag{2.7.1} $$

Here $\phi(T_2, T_1 )$ is the phase factor.

It is strange (at least to me) that he invoked the associativity condition

$$U(T_3) (U(T_2) U(T_1))= (U(T_3)U(T_2))U(T_1)$$

to impose the condition on $\phi$:

$$ \phi(T_2, T_1) + \phi(T_3, T_2 T_1) = \phi(T_3, T_2) + \phi(T_3 T_2, T_1). \tag{2.7.2}$$

Why is the associative law $$U(T_3) (U(T_2) U(T_1))= (U(T_3)U(T_2))U(T_1)~?$$ As you are now discussing projective representation, and you do not care about a phase factor, the associative law should be

$$ U(T_3) (U(T_2) U(T_1))\sim (U(T_3)U(T_2))U(T_1), $$

i.e., the two sides can differ up to a phase factor. There should be no requirement on $\phi$ at all.

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    $\begingroup$ I wonder if the group-representation context is diverting attention from something more basic. Are you familiar with the fact that the identity $U_3(U_2U_1)=(U_3U_2)U_1$ holds for all unitary operators $U_1,U_2,U_3$? $\endgroup$ Nov 28, 2021 at 19:06
  • $\begingroup$ @ChiralAnomaly Indeed! what a stupid question. $U_3(U_2 U_1) = (U_3 U_2) U_1$ necessarily. There is no phase factor here. $\endgroup$
    – S. Kohn
    Nov 28, 2021 at 23:27

3 Answers 3

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  1. Weinberg is discussing a projective unitary representation $P: G \to PU({\cal H})$ of some symmetry group $G$, so that $$\forall T_1,T_2\in G:~~ P(T_1)\ast P(T_2) ~=~ P(T_1T_2).\tag{A}$$

  2. For each element $T\in G$, the image $P(T)=U(1) U(T)$ is a coset, where $U(1)$ is the group of phase factors, and $U(T)\in U({\cal H})$ is a unitary linear operator: ${\cal H}\to {\cal H}$.

  3. Alternatively, we can describe the projective unitary representation as a map: $U: G\to U({\cal H})$ such that $$\begin{align}&\forall T_1,T_2\in G~\exists e^{i\phi( T_1,T_2)}\in U(1):~~\cr & U(T_1)\circ U(T_2) ~=~ e^{i\phi(T_1,T_2)} U(T_1T_2)\end{align}\tag{2.7.1}$$ by introducing a Schur multiplier.

  4. Now let us address OP's question. Composition $\circ$ of maps is always manifestly associative: $$\begin{align}&\forall T_1,T_2,T_3\in G:~~\cr &(U(T_3)\circ U(T_2))\circ U(T_1) ~=~ U(T_3)\circ (U(T_2)\circ U(T_1)),\end{align}\tag{B}$$ not just modulo a phase factor as OP seems to suggest.

  5. Sketched proof of eq. (B): Apply both sides of eq. (B) to a vector $v\in{\cal H}$. $\Box$

  6. Eq. (B) implies the 2-cocycle condition (2.7.2) modulo $2\pi$.

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  • $\begingroup$ Thanks a lot! I think your point 4 is the same as what "Chiral Anomaly" suggested. $\endgroup$
    – S. Kohn
    Nov 28, 2021 at 23:33
  • $\begingroup$ Yes, it appears so. $\endgroup$
    – Qmechanic
    Nov 29, 2021 at 21:50
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Recall that an operation $\circ : A \times A \to A$ on a set $A$ is called associative iff $$ (a \circ b ) \circ c = a \circ (b \circ c)$$ for $a,b,c \in A$. In your example, we demand that multiplication of elements of the projective representation is still associative (so that, in the end, we know that the $U(T_i)$ are elements of some group, e.g. unitary matrices). This a demand which should be satisfied independent of the fact that the representation is projective.

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Recall what a group representation is: a homomorphism between the group being represented and the group of automorphisms of the representation space. Therefore, the representation operators still form a group, among whose definition's requirements is the associativity, alongside the existence of an inverse and of a neutral element.

All that's changed to a (normal) group representation is the "representation property", but this novel property is still imposed onto some group elements (i.e. operators which form a group).

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