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Quick question:-Sometimes when you have a principal $G$-bundle $p:P\rightarrow B$ and another topological space $F$ upon which $G$ acts, it's said that one can define the associated bundle $P\times_{B}F$, but what does that mean when you have the base space like that?

I know how $P\times_{G}F$ is defined, but when it's written $P\times_{B}F$ how should I interpret it?

EDIT: Maybe comes from assuming that $B$ must be a Lie group? In that case the definition is always the same... are there any particular other definitions?

I'm saying that cause the definition of $Spinc^{\mathbb{C}}$-structure is the following:

"Let $p:P\rightarrow B$ be a principal $SO(p,q)$-bundle. A $Spin^{\mathbb{C}}$-structure is a principal $Spin^{\mathbb{C}}(p,q)$-bundle $\widetilde{p}:\widetilde{P}\rightarrow B$ and a principal $U(1)$-bundle $p':P'\rightarrow B$ such that

(i)There is a bundle map $\widetilde{r}:\widetilde{P}\rightarrow P\times_{B}P'$ that is a double cover;

(ii)The diagram

$$\begin{array}{ccc} \widetilde{P}&\times& Spin^{\mathbb{C}}(p,q) & \longrightarrow &\widetilde{P}\\ \downarrow{\widetilde{r}} & & \downarrow{r^{\mathbb{C}}} & &\downarrow{\widetilde{r}}\\ (P\times_{B}P')&\times&SO(p,q)\times U(1)&\longrightarrow&P\times_{B}P'\\ \end{array}$$ commute. As you can see we're assuming nothing about B, unless the definition I have is not complete.

Filippo
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    note I know nothing about spin structures, but in general if you have two fiber bundles over the same base $(X,\pi,B)$ and $(X’,\pi’,B)$, then you can form the fiber product $(X\times_BX’,\tilde{\pi},B)$, where $X\times_BX’={(x,x’)\in X\times X’,:, \pi(x)=\pi’(x’)}$. This is once again a fiber bundle over $B$, where the fiber over each $b\in B$ is the cartesian product of the fibers, $X_b\times X_b’$. If you want, you can also think of this as the pullback along the diagonal $\delta:B\to B\times B$ of the product fiber bundle $(X\times X’,\pi\times\pi’,B\times B)$. – peek-a-boo Jul 08 '23 at 15:08
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    Yup, I think this makes sense because $P$ has an action of $SO(p,q)$ and $P’$ has an action of $U(1)$, so the product group $SO(p,q)\times U(1)$ can naturally act on a space where the fibers are the products of the two fibers above. This is the only obvious way the commutative diagram makes sense formally. – peek-a-boo Jul 08 '23 at 15:12
  • Ahh yes you're right, I forgot about that option. Thank you – Filippo Jul 08 '23 at 15:20

1 Answers1

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Say $B$ is a group (say a Lie group, so its both a nice topological space (a manifold) and a group ) and $P \to B$ is a principal $B$-bundle. That means the base space is $B$ but also the group acting on $P$ is $B$.

Then there is an action of $B$ on both $P$ and $F$ and then the associated bundle $P \times_B F \to B$ is defined as the set of equivalence classes $$ (p,f) \sim (bp,b \cdot F) $$ for $b \in B$, which is also a group. By $bp$ I refer to the action of $B$ on $P$ and by $b \cdot F$ I mean the action of $b$ on $F$.

That being said, I wouldnt be surprised if all such bundles end up being trivial, but thats another questions.

F. Conrad
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  • Ok, so in the definition of $Spin^{\mathbb{C}}$-structure all base spaces are Lie groups? – Filippo Jul 08 '23 at 14:23
  • @Filippo No. This answer applies only when the base space is a Lie group, but the correct notion, described in peek-a-boo's comments under the question, applies much more generally. – Andreas Blass Jul 08 '23 at 15:44
  • I didnt see your edit when I answered. But the comment from peek-a-boo is 100% correct and is probably what you were looking for in the $spin^C$ context. – F. Conrad Jul 08 '23 at 17:05