I'm merely trying to understand the following statement I came across while reading.

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The text doesn't provide a reference or a recap of the construction that explains this fact. I'm adequately versed in the definitions of representation, Lie group, and vector bundle, so probably I don't understand the classifying space $BG$ sufficiently well to know how this would work. If anyone could elucidate I would be grateful.

up vote 2 down vote accepted

For a Lie group $G$, one has the principal $G$-bundle $G\hookrightarrow EG\to BG$ over the classifying space $BG$. In this case, $EG$ is a right $G$-space, and a representation $\rho:G\to\operatorname{GL}_n(\mathbb{C})$ gives $\mathbb{C}^n$ a left action of $G$ by: $$g\cdot v := (\rho(g))(v).$$ Now, form the associated bundle $EG\times_G\mathbb{C}^n\to BG$ as follows: $EG\times_G\mathbb{C}^n$ is defined as the quotient space: $$EG\times_G\mathbb{C}^n := \frac{EG\times\mathbb{C}^n}{(x,v)\sim (x\cdot g,g^{-1}\cdot v)}$$ The map $EG\times\mathbb{C}^n\xrightarrow{\ \pi_1 \ }EG\xrightarrow{\ p \ }BG$ factors through $EG\times_G\mathbb{C}^n$ giving a map: $$\bar{p}:EG\times_G\mathbb{C}^n\to BG,\quad \bar{p}[x,v] = p(x).$$ In this answer I explain how to obtain the trivializations.

As a quick example, if you take $U(1)=S^1$ to be your Lie group, then the classifying space $BU(1)$ is the Grassmanian $\operatorname{Gr}_1(\mathbb{C}^{\infty})=\mathbb{CP}^{\infty}$, and $EU(1)$ is the Stiefel manifold $V_1(\mathbb{C}^{\infty})$. If we take our representation to be the inclusion $\rho:U(1)\hookrightarrow\operatorname{GL}_1(\mathbb{C})$, then the associated bundle $EU(1)\times_{U(1)}\mathbb{C}$ is precisely the canonical line bundle over $\mathbb{CP}^{\infty}$. In fact, the pullback of $EU(1)\to \mathbb{CP}^{\infty}$ over the inclusion $i:\mathbb{CP}^n\hookrightarrow\mathbb{CP}^{\infty}$ is the canonical line bundle over $\mathbb{CP}^n$.

  • 1
    Okay, I think I was forgetting that there's an $EG$ over $BG$ which is a principal bundle (the associated bundle construction I'm perfectly fine with). I think maybe I should read more about what these universal objects actually are, though I seem to recall Milnor's construction is some kind of infinite join and that's about the point I stopped reading (this was many years ago now, when I was less wise...). – Brian Klatt Apr 19 at 19:16
  • 1
    @BrianKlatt If you'd like a reference these notes are good. – user171308 Apr 20 at 2:17

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