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The classifying space $BS^1$ for $S^1$-bundles can be taken to be the colimit of $\mathbb{CP}^n$ which are smooth manifolds and the inclusions $\mathbb{CP}^n \hookrightarrow \mathbb{CP}^{n+1}$ are smooth embeddings.

Now consider another compact Lie group $G$. There is a classifying space $B_G S^1$ for $G$-equivariant $S^1$-bundles which is an infinite CW complex so that for any $G$-space $X$ and $G$-equivairant line bundle $L \to X$ there is a $G$-equivariant classifying map $c: X \to B_G S^1$.

Question: can the classifying space $B_G S^1$ be approximated by manifolds like $\mathbb{CP}^n$ with smooth $G$-actions?

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    $\begingroup$ For every integer $m\geq 1$, for the unitary group $U(m)\subset \textbf{GL}_m(\mathbb{C})$, the classifying space is a colimit of the natural inclusions of the complex Grassmannians $\text{Grass}(m,\mathbb{C}^{m+n})\hookrightarrow \text{Grass}(m,\mathbb{C}^{m+n+1})$. So the result holds for $U(m)$. Every compact Lie group $G$ has a faithful unitary representation of finite dimension $m$, i.e., there is an embedding of compact Lie groups $G\hookrightarrow U(m)$. So the classifying space $BG$ is realized as the total space of a smooth fiber bundle over $BU(m)$ with fibers $U(m)/G$. $\endgroup$
    – Jason Starr
    Commented Mar 28, 2023 at 23:33
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    $\begingroup$ @JasonStarr But this is about classifying G-equivariant line bundles. The classifying space at least should have some G-action. $\endgroup$
    – UVIR
    Commented Mar 29, 2023 at 0:34
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    $\begingroup$ @JasonStarr Here are two groups involved, a group G and $S^1$. When $G$ is trivial the answer should be $ES^1$ or $BS^1$. $\endgroup$
    – UVIR
    Commented Mar 29, 2023 at 1:06
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    $\begingroup$ What is any other description of $B_G(S^1)$? Is it the $Y=\mathbb P(U)$, where $U$ is a complete universe (a sum of infinitely many copies of all irreps of $G$). That has a manifold approximation $\mathbb P(V)$, where $V$ is parameterized by the directed set of finite dimensional reps. You can choose a cofinal sequence if you like. This space has the property that $Y^H$ is a disjoint union of copies of $BS^1$, parameterized by homomorphisms from $H\to S^1$, which sounds right according to this. $\endgroup$
    – Ben Wieland
    Commented Mar 29, 2023 at 2:25
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    $\begingroup$ @BenWieland Yes, your suggestion is definitely correct, and well-known in the equivariant stable homotopy literature, although I am not sure what would be the most convenient reference. $\endgroup$
    – Neil Strickland
    Commented Mar 29, 2023 at 11:50

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