I'm reading the following article "Symmetry protected topological orders and the group cohomology of their symmetry group" https://arxiv.org/abs/1106.4772

There it is defined a projective representation of a group $G$. A projective representation $u$ has the property for $g_1$ and $g_2$ in $G$

$$u(g_1)u(g_2) = \omega(g_1,g_2) u(g_1 g_2)\tag1$$

Where $\omega$ is called factor system. Then it is asserted that a if the representation is multiplied by a factor such as $\beta(g) u(g)$ then the new factor system is

$$\omega'(g_1,g_2) = \frac{\beta(g_1 g_2)}{\beta(g_1)\beta(g_2)} \omega(g_1, g_2)\tag2$$

I am not able to deduce this last property so any help would be appreciated. What I do is the following:

$$ u'(g_1)u'(g_2) = \beta(g_1)\beta(g_2)u(g_1)u(g_2) = \beta(g_1)\beta(g_2) \omega(g_1, g_2) u(g_1 g_2)\tag3$$


$$\omega'(g_1, g_2)u'(g_1g_2) = \omega'(g_1,g_2)\beta(g_1 g_2)u(g_1 g_2)\tag4$$

Thus we have

$$\omega'(g_1,g_2)\beta(g_1 g_2)u(g_1 g_2) = \beta(g_1)\beta(g_2) \omega(g_1, g_2) u(g_1 g_2)\tag5$$


$$\omega'(g_1,g_2) = \frac{\beta(g_1)\beta(g_2)}{\beta(g_1 g_2)}\omega(g_1, g_2)\tag6$$

As I said, any help would be appreciated, including the reason why my derivation is wrong.

EDIT: After the discussion below I realized how to get the correct result in the article yet I can't see the mistake in my method.

Since $$u'(g_1)u'(g_2) = \omega'(g_1,g_2) u'(g_1 g_2) $$

what I did was to replace the left hand side with the right hand of $(3)$ and replace the left hand side with the right hand side of $(4)$ thus I obtained $(5)$

  • $\begingroup$ Thanks for the corrections in my question. Equation (5) comes equationg right side of equation (4) with the right side of equation (3). I equate them because $u'(g_1)u'(g_2) = \omega'(g_1,g_2) u'(g_1 g_2)$ $\endgroup$ – Pam Oct 24 '17 at 21:35
  • $\begingroup$ As I mentioned, its because I think $u'(g_1)u'(g_2) = \omega'(g_1,g_2) u'(g_1 g_2)$ For now I don't see the mistake in this but I'll think about it $\endgroup$ – Pam Oct 24 '17 at 21:37
  • $\begingroup$ So now I see how can I get the result in the article, yet I can't see why the method I used is wrong. I'm feeling very stupid now lol. What do you exactly mean with "equation $4$ has an $\omega$ in the left hand side"? (maybe my english is failing here) $\endgroup$ – Pam Oct 24 '17 at 21:44
  • $\begingroup$ @AccidentalFourierTransform The two sides of the equation $u^\prime(g_1) u^\prime(g_2) = \omega^\prime(g_1, g_2) u^\prime(g_1 g_2)$ are respectively equal to the LHS of equation (3) and the LHS of equation (4). So the RHS of equation (3) and the RHS of equation (4) should be the same, right? $\endgroup$ – higgsss Oct 25 '17 at 3:02
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    $\begingroup$ I think your derivation is correct and that equation (2) from the paper should be changed to equation (6). But I suppose that correcting this kind of error won't change the conclusion of the paper? I guess you are just overthinking about this. $\endgroup$ – higgsss Oct 25 '17 at 3:15

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