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We consider a smooth fiber bundle $p:E \rightarrow B$ with fiber $F$. I want to understand how such a fiber bundle is classified by a map $B\rightarrow $BDiff$(F)$. Namely, we should have a bijection $$ [B,\text{BDiff}(F)] \rightarrow \{\text{smooth $F$-bundles }E\to B\}/\text{iso} $$ given by the pullback of some universal bundle EDiff$(F)\to$BDiff$(F)$. However, i can't seem to figure out the details or locate them in the literature.

Furthermore, how does the topology of Diff$(F)$ come into play? I know that we topologize Diff$(F)$ with the Whitney $C^\infty$-topology, but why is it exactly this topology that makes BDiff$(F)$ classifying such $F$-bundles? Also, in the case where $F$ is not compact, we have two different non-equivalent versions of the Whitney $C^\infty$-topology on Diff$(F)$ namely the strong and weak one. Which one of these are used or does it matter?

Any help or helpful references would be appreciated.

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    $\begingroup$ You can obtain a smooth fiber bundle $F \to E \to B$ from a principal $\operatorname{Diff}(F)$-bundle over $B$ via the associated bundle construction. You need $\operatorname{Diff}(F)$ to be a topological group in order to form $B\operatorname{Diff}(F)$. $\endgroup$ Jan 19 '21 at 18:15
  • $\begingroup$ Is this correspondance between smooth fiber bundles $F \to E \to B$ and principal Diff$(F)$-bundles over $B$ bijective (modulo isomorphisms)? In this case, this answers my first question. Regarding the second comment; i know that we use the topology on Diff$(F)$ to get the classifying space, but i still don't know why it is the Whitney topology, that is the correct choice. For example, once could give Diff$(F)$ the discrete topology, in which case BDiff$(F)$ becomes the Eilenberg-Maclane space $K($Diff$(F),1)$ but i know this is not the correct choice of topology. $\endgroup$
    – Frederik
    Jan 19 '21 at 18:33
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    $\begingroup$ Yes, it is. As for the topology, you need the transition functions of a smooth $F$-bundle to be smooth functions (if the topology was discrete, that would force the transition functions to be locally constant). I'm not sure how this works in the infinite-dimensional case though ($\operatorname{Diff}(F)$ is infinite dimensional). $\endgroup$ Jan 19 '21 at 18:42
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Since asking the question i have learned that the canonical reference for this is the very detailed book "The convenient setting of global analysis" by Kriegl and Michor. In this book, they introduce the theory of infinite dimensional smooth manifolds. Let me sketch the argument.

They work with convinient vector spaces which are locally convex vector space satisfying a condition called $c^\infty$-completeness and a notion of smooth mappings between them. It turns out that one has to introduce a new topology called the $c^\infty$-topology on the locally convex vector spaces, because in the locally convex topology, smooth mappings are not necessarily continuous. Then infinite dimensional manifolds are defined analogously to ordinary smooth manifolds using convenient vector spaces equipped with the $c^\infty$-topology instead of Euclidian space. The set $C^\infty(M,N)$ equipped with the Whitney $C^\infty$-topology becomes an infinite dimensional manifold, and $\text{Emb}(M,N)\subset C^\infty(M,N)$ and $\text{Diff}(F)\subset C^\infty(F,F)$ are submanifolds, and $\text{Diff}(F)$ is a Lie group.

Let $F$ be a closed smooth finite dimensional manifold, and let $\ell^2(\mathbb{N})$ denote the Hilbert space of sequences bounded in the norm $\lvert \lvert \cdot \rvert \rvert_2$. The Lie group $\text{Diff}(F)$ acts on the embedding space $\text{Emb}(F,\ell^2(\mathbb{N}))$ by composition and yields a principal $\text{Diff}(F)$-bundle $$ \text{Emb}(F,\ell^2(\mathbb{N})) \rightarrow \text{Emb}(F,\ell^2(\mathbb{N}))/\text{Diff}(F). $$ The total space is contractible which implies that the base space is homotopy equivalent to $\text{BDiff}(F)$. Denote the base space with $B_\infty(F)$. As they mention in the book, the base space can be thought of as the nonlinear version of the infinite Grassmannian. Consider the associated bundle $$ \text{Emb}(F,\ell^2(\mathbb{N})) \times_{\text{Diff}(F)}F \rightarrow B_\infty(F), $$ and denote its total space by $E_\infty(F)$. This associated bunde has fiber $F$ and is the universal $F$-bundle (Theorem 44.24 in the book). Since $B_\infty(F)\simeq B\text{Diff}(F)$, we get the wanted bijection in the question.

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