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I have read that the unitary group is somehow given by the direct product $U(N)=U(1)*SU(N)$ and it follows that for $N$ going to zero we get just $U(1)$. How it can be possible? What does it mean $SU(0)$ then?

(This topic comes from the calculation of the renormalization constant for the coupling at 1 loop level in QCD. My book says that to get the "QED limit" we should compute the limit of that constant for $N$ going to $0$, that is $\zeta_g^{(1)}=-\frac{11N-2n_f}{24\epsilon}+O(\epsilon^0)$ in QCD becames $\zeta_g^{(1)}=\frac{1}{6\epsilon}+O(\epsilon^0)$ in QED when $N=0$. Here $n_f$ is the number of quark flavours, that in QED is set equal to $1$.)

Thank you!

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    $\begingroup$ $\mathrm{SU}(1)\equiv 1$, and therefore $\mathrm U(N)|_{N=1}=\mathrm U(1)\times\mathrm{SU}(N)|_{N=1}=\mathrm U(1)\times 1=\mathrm U(1)$. In other words, you get $\mathrm U(1)$ when $N$ goes to one, not to zero. But QCD deals with $\mathrm{SU}(N)$, not with $\mathrm U(N)$, and so it's a different story. (Note also that the correct statement is $\mathrm U(N)=(\mathrm{SU}(N)\times \mathrm U(1))/\mathbb Z_N$, although this is kind of irrelevant here). $\endgroup$ – AccidentalFourierTransform Jun 9 '18 at 16:37
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    $\begingroup$ $\uparrow$ Which book? $\endgroup$ – Qmechanic Jun 9 '18 at 16:42
  • $\begingroup$ Suggestion: Replace 0 with 1 in various places, cf. @AccidentalFourierTransform's above comment. $\endgroup$ – Qmechanic Jun 9 '18 at 16:44
  • $\begingroup$ Related: physics.stackexchange.com/q/169087/2451 $\endgroup$ – Qmechanic Jun 9 '18 at 16:47
  • $\begingroup$ That's the point: it's not for $N$ going to $1$ that we get the expected QED group $U(1)$, but it says that it's for $N$ going to $0$ that we find $U(1)=U(1)*SU(0)(?)$. What does it means? $\endgroup$ – MariNala Jun 10 '18 at 21:39

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