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It is well known that the diffeomorphism group of there sphere $\operatorname{Diff}(S^n)$ has the homotopy type of a product $X:=O(n+1)\times \operatorname{Diff}_{\partial D^n}(D^n)$ of the orthogonal group and the diffeomorphism of the disk $D^n$ with are the identity in a neighborhood of the identity.

Does the classifying space $B\operatorname{Diff}(S^n)$ have the homotopy type of $BX$? If not, do we at least have an isomorphism on the level of cohomology: $H^\ast(B\operatorname{Diff}(S^n);\mathbb{Z})\cong H^\ast(BX;\mathbb{Z})?$

Of course, if there was a homotopy equivalence $\phi:X\rightarrow \operatorname{Diff}(S^n)$ which is simultaneously a group homomorphism, the answer would be "yes". But the only homotopy equivalence I'm aware of (which comes from the link at the start of this post) is not obviously homotopic to a homomorphism.

The linked homotopy equivalence has the property that $\phi$ restricted to either factor is a homomorphism. Using this, it's easy to show that $H^\ast(B\operatorname{Diff}(S^n);\mathbb{Z})$ surjects onto the cohomlogy of each factor of $BX$, but I've been unable to piece this together to get what I want.

Motivation: I'm trying to understand characteristic classes for smooth sphere bundles whose structure group does not reduce to $O(n+1)$. It's clear that, except for a few small $n$, one should get "extra" characteristic classes, but I'm trying to understand if one also gets an analogue of Stiefel-Whitney, Euler, and Pontryagin classes satisfying all the "usual" properties, e.g., the mod $2$ reduction of the Euler class should be the top Stiefel-Whitney class; Stiefel-Whitney classes satisfy the Wu-formula, etc. I'd also be happy with pointers to the literature.

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    $\begingroup$ Rationally, stable characteristic classes exist and satisfy the same formulas. This is because we have maps $B\operatorname{Diff(S^n)} \rightarrow B\operatorname{Top}(n+1)\rightarrow B\operatorname{Top} $ and the latter is rationally equivalent to $BO$ because $\operatorname{Top}/O$ has only rational homotopy groups. In the torsion case I have no clue. I would ask Watanabe if he has ever calculated the characteristic classes of the bundles he constructed in dimension 4. $\endgroup$ Jun 29 at 4:16
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    $\begingroup$ Some references: M. Weiss'es arxiv.org/pdf/1507.00153.pdf and Igusa's "Pontrjagin classes and higher torsion of sphere bundles", researchgate.net/publication/…. $\endgroup$ Jun 29 at 13:18
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    $\begingroup$ @ConnorMalin: what can be said rationally about the map $Diff(S^n)\to Top(n+1)$? Could you point me to a reference dealing with the issue? I assume the map takes a diffeomorphism of $S^n$ to an origin-preserving homeomorphism of $\mathbb R^{n+1}$ via an open cone; if not, how is it defined? $\endgroup$ Jun 29 at 13:42
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    $\begingroup$ @IgorBelegradek Yes, just extend via cone. The right hand side is studied by Weiss here. $\endgroup$ Jun 29 at 13:53
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    $\begingroup$ @IgorBelegradek: Thanks for the references! $\endgroup$ Jun 29 at 14:14

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For many values of $n$, the answer to both questions is no. Since the fundamental groups of $BX$ and $B\mathrm{Diff}(S^n)$ are finite for $n\ge5$ (This uses that $\pi_0\mathrm{Diff}_\partial(D^n)$ is isomorphic to the group of homotopy $(n+1)$-spheres which is finite by a famous result of Kervaire--Milnor.), their first cohomology groups vanishes, so the first degree in which one could hope to detect a cohomological difference is $2$. And there is indeed a difference in degree 2 in many cases.

To see this, assume $n\ge5$ and note first that since $BX$ and $B\mathrm{Diff}(S^n)$ have finite fundamental groups, if the second cohomology groups were isomorphic, then the same would hold for the first homology group (this uses that $H_1$ and $H_2$ are finitely generated and that $H_1$ is torsion), so it suffices to show that $$H_1(B\mathrm{Diff}(S^n))\cong \pi_0\mathrm{Diff}(S^n)^{\mathrm{ab}}\quad\text{and}\quad H_1(BX)\cong \pi_0X^{\mathrm{ab}}$$ are not isomorphic. Since $\pi_0O(n+1)=\pm1$ and $\pi_0\mathrm{Diff}_\partial(D^n)$ are abelian, we have $$\pi_0X^{\mathrm{ab}}\cong \pi_0\mathrm{Diff}_\partial(D^n)\oplus \pm1.$$ Testing whether a diffeomorphism is orientation preserving gives an extension $$0\rightarrow \pi_0\mathrm{Diff}_\partial(D^n)\rightarrow \pi_0\mathrm{Diff}(S^n) \rightarrow\pm1\rightarrow 0$$ which admits a splitting induced by the standard $O(n+1)$-action on $S^n$, so we get $$\pi_1B\mathrm{Diff}(S^n)^{\mathrm{ab}}\cong\Big(\pi_0\mathrm{Diff}_\partial(D^n)\rtimes \pm1\Big)^{\mathrm{ab}}\cong \pi_0\mathrm{Diff}_\partial(D^n)_{\pm 1}\oplus \pm 1$$ where $(-)_{\pm 1}$ denotes taking coinvariants with respect to the action induced by the extension.

This shows that $H^2(BX)$ and $H^2(B\mathrm{Diff}(S^n))$ cannot be isomorphic as long as the $\pm 1$-action on the finite abelian group $\pi_0\mathrm{Diff}_\partial(D^n)$ is nontrivial. This action is given by conjugation with a reflection in one of the coordinates, which can be seen to induce multiplication with $-1$ on $\pi_0\mathrm{Diff}_\partial(D^n)$, so the action is nontrivial iff $\pi_0\mathrm{Diff}_\partial(D^n)$ is not annihilated by multiplication with $2$. As mentioned above, this group agrees with the group of homotopy $(n+1)$-spheres, which is known to not have this property in many cases (see e.g. the tables on wikipedia).

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  • $\begingroup$ This makes perfect sense, thank you! Do you happen to know how to modify this if we restrict to orientation preserving diffeomorphisms and use $SO(n+1)$? Then your argument, at least, implies isomorphisms on $H^1$ and the torsion part of $H^2$. $\endgroup$ Jun 29 at 14:14

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