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What is the total space of the universal bundle over $B\mathbb{Q}$, i.e. what is $E\mathbb{Q}$ for $B\mathbb{Q}=E\mathbb{Q}/\mathbb{Q}$ where $B\mathbb{Q}$ is the classifying space?

Thoughts/Attempt: $B\mathbb{Q}=E\mathbb{Q}/\mathbb{Q}$ and $B\mathbb{Q}=K( \mathbb{Q},1)$ so $E\mathbb{Q}=K( \mathbb{Q},1)\rtimes\mathbb{Q}$. Now we just need $K(\mathbb{Q},1).$ The classifying space of $\mathbb{Z}$ is a canonical example, but I haven't seen it for $\mathbb{Q}$.

Thanks in advance! I would be grateful for any help.

Sergio Charles
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    $\mathbb{Q}$ is a certain filtered colimit of copies of $\mathbb{Q}$, so you can describe $B\mathbb{Q}$ as a certain filtered colimit of circles. It's known as the "rational circle." – Qiaochu Yuan Jul 27 '18 at 21:46
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    @QiaochuYuan: a filtered colimit copies of $\mathbb{Z}$, you mean? Or have I misunderstood? (To the OP: the construction Qiaochu describes is also explained in this answer: https://math.stackexchange.com/a/84467/73817) – Alex Wertheim Jul 27 '18 at 21:49
  • Whoops, yeah, that. – Qiaochu Yuan Jul 27 '18 at 21:51
  • Could you write this as an answer? @QiaochuYuan – Sergio Charles Jul 27 '18 at 23:34
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    @Jake, true, but since $\Omega \mathbb{Q}\simeq\ast$ regardless of its topology, the resulting classifying space $B\mathbb{Q}$ will be a $K(\mathbb{Q},1)$ anyway. – Tyrone Jul 28 '18 at 09:10

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