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Why is the principal bundle with $U(1)$ as its base and ${\mathbb Z}_2$ as a fiber still isomorphic to $U(1)\,$?

Stated alternatively, why is it that $U(1) = U(1)/{\mathbb Z}_2\,$?

I would greatly appreciate a simple explanation understandable to a physicist.

Michael_1812
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    It's the same reason that $S^1$ mod antipodes is still $S^1$. (Or, why $\mathbb{R}P^1$ is homeo to $S^1$.) – Randall May 19 '24 at 16:00
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    Try this related question: https://math.stackexchange.com/questions/2688271/is-bbb-r-bbb-z-isomorphic-to-bbb-r-2-bbb-z/2688288#2688288 – Randall May 19 '24 at 16:03

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Not any such principal bundle has that property, for example the trivial bundle would be isomorphic to the group product $\mathbb Z_2\times U(1).$ A principal bundle is not completely defined by specifying the underlying manifold and the fibre: you need an action of the group on the bundle.

The situation that you describe can be arrived at by the projection

$$U_1\to U_1:\exp(i\theta)\mapsto\exp(2i\theta)$$

and the action

$$\mathbb Z_2\times U_1\to U_1:(z,\exp(i\theta))\mapsto \exp(i(\theta+z\pi))$$

where an element of $\mathbb Z_2$ is interpreted as a residue class of 0 or 1 modulo 2, i.e., $z$ is either an even or an odd integer; the result of the exponential only depends on the parity of $z.$

In your alternative formulation, $\mathbb Z_2$ can be modeled by the subgroup $\{\exp(i0\pi),\exp(i1\pi)\}=\{1,-1\}$ of $U(1).$

Lieven
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  • Dear Lieven, many thanks for your answer. Is my understanding correct that in your 2nd formula $z$ is either $1$ or $-1$ ? Could you please also clarify in what sense my formulation is `alternative'? I indeed implied ${\mathbb Z}_2 = \left{ 1,,-,1 \right},$. – Michael_1812 May 19 '24 at 18:55
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    Sorry, I meant 0 and 1. Will make this explicit. – Lieven May 19 '24 at 19:07
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    With "alternative formulation" I refer to your second sentence that starts with "Stated alternatively...". In that case $\mathbb Z_2$ consists of the complex numbers ${\exp(i0\pi),\exp(i1\pi)}={1,-1}.$ – Lieven May 19 '24 at 19:13
  • And if we use ${\mathbb Z}_2 = \left{ 1,,-,1 \right},$, how will your formulae read? – Michael_1812 May 19 '24 at 19:13
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    Then you should just put the number $z$ (1 or -1) in front of the exponential: $(z,u)\mapsto zu.$ – Lieven May 19 '24 at 19:15
  • Dear @Lieven, I have understood everything in your story, except one last detail: why do we at all need the formula $U(1)\to U(1):~\exp(i\theta)\mapsto\exp(2i\theta),$? And why do you call it projection? I used to think that the canonical projection $\pi$ maps the entire fiber to a point on the base. Smth like $\pi:~({\mathbb Z}_2,,e^{i\theta})\mapsto e^{i\theta}$. I guess my perception of principal bundle is superficial, and I am missing something big here. Sorry for the silly question. – Michael_1812 May 21 '24 at 18:39
  • Further to my previous question. My understanding is that an arbitrary point of the bundle is $\exp(i(\theta+z\pi))$, where $z = 0,,1$, and $\exp{i\theta}$ is the base point to which the fiber is attached. (Please correct me if I am wrong.) To fully define the bundle, we need to nominate a group that keeps the fibers intact. What will be that group, and how exactly will it be acting on an arbitrary point $\exp(i(\theta+z\pi)),$? Will it be simply $\exp iz\pi$ with $z = 0,,1,$? I do appreciate your generous patience. – Michael_1812 May 21 '24 at 19:28
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    I need to correct you on the base point: the bundle point $\exp(i(\theta+z\pi))$ projects to the base point $\exp(2i(\theta+z\pi))=\exp(2i\theta).$ – Lieven May 21 '24 at 21:29
  • You probably imply that while the canonical projection is $\mathbb Z_2\times U(1)\to U(1):(z,\exp(i\theta))\mapsto \exp(i(\theta+z\pi))$, it also matters how exactly we set an isomorphism between $U(1)$ and $U(1)/{\mathbb Z}_1,$. Is this what you have in mind? I am still struggling to understand the motivation for your first formula. – Michael_1812 May 22 '24 at 02:50
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    The mapping from $\mathbb Z_2\times U(1)$ to $U(1)$ is not the projection but the group action. The projection is the mapping from $U(1)$ to $U(1)$ that doubles the angle. I cannot do drawings here, but imagine a more or less horizontal double loop (think of the edge of a Möbius strip) hovering over a single loop. In the double loop angles are measured so that you need $2\pi$ to go twice around, i.e., cover the full double loop. In the single loop angles are measured "normally". – Lieven May 22 '24 at 06:42
  • Once again, many thanks. Have to cogitate on this. – Michael_1812 May 22 '24 at 16:10