Automatically trivial fibrations Dalhousie University Halifax, 19.4.2010

Every numerable fibre space over a contractible base is trivial. Relevant definitions: Numerability is a technical condition that allows glueing of maps and spaces. Numerably trivial fibre space is a fibration, all fibres are homotopy equivalent to some F.

f*E B E A The pull-back construction is homotopy invariant: f inclusion  f*E=p-1(A), f constant  f*E=BEb A f The pull-back construction is homotopy invariant: What about the converse: if every numerable fibre space over B is trivial, is then B contractible? However, the path fibration is usually not numerable... The converse is still true, but I don’t know any proof of this fact that avoids the classification theorem.

G ≤ Homeo(F) for bundles with fibre F Let us turn to fibres: is every numerable fibre space with contractible fibres trivial? Non-trivial vector bundles have contractible fibres... ... equivalence of vector bundles involves the action of general linear group, which doesn’t allow the contraction of fibres All vector bundles are trivial as (numerable) fibre spaces – even those that are non-trivial as bundles. G ≤ Homeo(F) for bundles with fibre F Bundles are determined by their structure groups. aut F for numerable fibre spaces with fibre F For every G there is a classifying bundle EBG, which contains all possible G-bundles with fibre F, i.e. every G-bundle with fibre F is a pull-back of the classifying bundle. (Generalizes the familiar classification of n-dimensional vector bundles by maps to the Grasmannian - the space of all n-dimensional subspaces in . ) { homotopy classes of maps BBG } pull-back of the classifying bundle { equivalence classes of G-bundles with base B } is a bijection

{ equivalence classes of numerable fiber spaces over B } Extension to fiber spaces: instead of G take aut F, the space of self-homotopy equivalences of F. Construction of B(autF) is considerably complicated by the fact that it is not a group (there is no continuous choice of homotopy inverses). Still, it can be done, and there is a classifying numerable fibre space FE B(autF) { homotopy classes of maps BB(autF) } pull-back of the classifying fibre space { equivalence classes of numerable fiber spaces over B } is a bijection Corollaries: Every numerable fibre space with contractible fibre is trivial. Every numerable fibre space with base B is trivial  B is contractible The construction of BH works for topological monoids, as well. For H=B. It yields a numerable fibre space B  E  B (B) B, with E*. Triviality implies B B  E, therefore B*. Every numerable fibre space with fibre F is trivial  B(aut F) is contractible

Computation of B(aut F): Assume F has only one non-trivial homotopy group, F =K(G,n).

Such groups are called complete. Examples: symmetric groups Sn, except S2 (commutative) and S6 (has an outer automorphism) many simple groups all Aut(G) when G is simple non-abelian Other F: finitely many homotopy groups  B(aut F)* (Didierjean spectral sequence argument) H- or coH-space  B(aut F)* (aut F  F has a section) Open problem: is there a non-contractible finite complex F such that B(aut F)*?