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Dec 31, 2023 at 11:19 comment added SethK You're right, the fact that we only have one photon is also needed there.
Dec 31, 2023 at 9:41 comment added Jules Lamers Of course any number of copies of $U(1)$, i.e. $U(1)\times \dots \times U(1)$, is abelian and compact
Dec 31, 2023 at 8:56 comment added CBBAM Thank you, this was exactly the kind of answer I was looking for!
Dec 31, 2023 at 8:56 vote accept CBBAM
Dec 31, 2023 at 6:25 history answered SethK CC BY-SA 4.0