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A symmetry group $\mathcal{G}$ may be represented on the physical Hilbert space by unitary operators $U(g)$ such that it satisfies the composition rule $$U(g_1)U(g_2)=e^{i\phi(g_1,g_2)}U(g_1g_2).\tag{1}$$ If $\phi(g_1,g_2)$ is of the form $$\phi(g_1,g_2)=\alpha(g_1g_2)-\alpha(g_1)-\alpha(g_2)\tag{2}$$ the projective representations in (1) with such a phase (2) can be replaced by an ordinary representation by replacing $U(g)$ with $$\tilde{U}(g)=e^{i\alpha(g)}U(g).\tag{3}$$ If I understand it correct, Weinberg says in his QFT book that $\phi=0$ satisfies relation (2), and is therefore, a trivial two-cocycle. I don't understand why $\phi=\pi$ cannot satisfy the relation (2). It is satisfied by choosing $$\pi+\alpha(g_1g_2)=\alpha(g_1)+\alpha(g_2)\tag{4}.$$

Therefore, projective representations corresponding to $\phi=\pi$ (or any constant value of $\phi$), can be eliminated by defining a new representation of the form (3). Do I misunderstand something?

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  • $\begingroup$ @NorbertSchuch (4) is not from Weinberg. My question is why $\phi=0$ is said to satisfy (2) but not some other value of $\phi$ (say, $\phi=\pi$)? $\endgroup$ – SRS Feb 2 '17 at 7:53
  • $\begingroup$ @NorbertSchuch Can you explain in what sense $\phi=0$ satisfies (2)? I think the way I'm understanding it may not be correct. And I edited the question and hope it is less confusing now? $\endgroup$ – SRS Feb 2 '17 at 8:19
  • $\begingroup$ You say "I don't understand why ϕ=π cannot satisfy the relation (2)." But $\phi=\pi$ can satisfy (2) -- I gave you an example. Is this really your question? IMO, the question I would find interesting would be "is any proj. representation with $\phi=\pi$ of the form (2)". Can you clarify this? --- Regarding my other comment: For $\phi=0$, just choose $\alpha(g)\equiv 0$. Then, (2) holds. $\endgroup$ – Norbert Schuch Feb 2 '17 at 9:31
  • $\begingroup$ Ok, it is indeed (quite trivially) generally true. I have added an answer. $\endgroup$ – Norbert Schuch Feb 2 '17 at 9:36
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You are right: Any projective representation $U(g)$ with $\phi(g_1,g_2)=\pi$ can be made a linear representation by replacing $$ \tilde U(g) = -U(g)\ . $$ This corresponds to choosing $\alpha(g)=\pi$ in Eq. (2). (Note that while this satisfies (4), it is not clear whether Eq. (4) itself is a good definition -- it is not clear that $\alpha(g)$ for each $g$ is uniquely defined this way.)

Indeed, this is generally true when $\phi(g_1,g_2)=\vartheta$ is constant: In each of these cases, you can define $\tilde U(g)=e^{-i\vartheta}U(g)$, or $\alpha(g)=-\vartheta$.

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    $\begingroup$ Dear @Norbert Schuch This is what I understand from your answer. Correct me if I'm wrong. A phase $\phi(g_1,g_2)$ which does not satisfy (2) cannot be a mere constant because all constant $\phi$ can be made to satisfy (2). Is that correct? Therefore, does it not mean that all projective representations with constant $\phi$, can be replaced by ordinary representation? And for non-trivial projective representation, one must have a phase $\phi(g_1,g_2)$ which is different from a constant? $\endgroup$ – SRS Feb 3 '17 at 5:23
  • $\begingroup$ Precisely. $\quad$ $\endgroup$ – Norbert Schuch Feb 3 '17 at 6:35
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No one makes any claim that it doesn't work for $\phi = \pi$ as well! The whole point of the relation between projective and linear representation is that a single projective representation can have more than one pair of $(U,\phi)$ related to it.

I discuss the relation between projective representations and unitary representations at length in this answer of mine. The relevant point to your question is the coboundary condition: Two functions $\phi(g_1,g_2)$ and $\phi'(g_1,g_2)$ define the same projective representation if there exists another function $\alpha(g)$ such that $$ \mathrm{e}^{\phi'(g_1,g_2)}\mathrm{e}^{\alpha(g_1g_2)} = \mathrm{e}^{\alpha(g_1)}\mathrm{e}^{\alpha(g_2)}\mathrm{e}^{\phi(g_1,g_2)},$$ which is eq. (2) from my other answer written in your exponential notation. Your eq. (2) is exactly this for $\phi' = 0$, meaning the statement looks kind of vacuous from this viewpoint: Of course $\phi=0$ and $\phi'=0$ define the same representation!

You are correct that this works for any constant functions $\phi,\phi'$, they all belong to the same projective representation. This is kind of the point of the whole construction: When you choose a pair $(U,\phi)$ to turn a projective representation into a "twisted" linear representation on the original Hilbert space, this choice is not unique and two choices ("cocycles") fulfilling certain relations with each other (the "coboundary" conditions) always come from the same projective representation. For more detail, see my other answer.

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Projective representations aren’t faithful by their very construction. If this was a ‘normal’ faithful representation (to the same target space), you’d have a group homomorphism, i.e. no phase factor.

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