1
$\begingroup$

I am confused with the following argument. I know I am doing something wrong but I can't find my mistake. On one hand, one knows that if $M$ is a Lie group, then $$\mathrm{Diff}(M)\simeq M\times\mathrm{Diff}(M, \mathrm{rel}\,x_0),$$ where $x_0$ is the identity element in $M$. One way to check this is considering the fibration $\mathrm{Diff}(M)\rightarrow M$ that maps $\phi$ to $\phi(x_0)$. Then the fiber is $\mathrm{Diff}(M, \mathrm{rel}\,x_0)$. In order to check the fibration is trivial, one can find a global section in a straightforward way by using the lie group structure of $M$.

Thus, if we particularize $M=\mathbb{S}^3$ we get that $$\mathrm{Diff}(\mathbb{S}^3)\simeq\mathbb{S}^3\times \mathrm{Diff}(\mathbb{S}^3, \mathrm{rel}\,x_0)$$ whereas we know by Hatcher's theorem that $$\mathrm{Diff}(\mathbb{S}^3)\simeq O(4).$$ What am I missing? I will appreciate any comment or suggestion. Thanks in advance.

Improve this question
$\endgroup$
5
  • 10
    $\begingroup$ Why would there be a contradiction? You also have a fiber bundle $O(3) \to O(4) \to S^3$ and it's trivial for similar reasons. So already you know $O(4) \simeq S^3 \times O(3)$. I'm not an expert, but why wouldn't it be true that $\operatorname{Diff}(S^3,x_0) \simeq O(3)$? $\endgroup$
    – Najib Idrissi
    Commented Mar 5, 2019 at 11:45
  • 2
    $\begingroup$ What is the problem? The conclusion from your argument should be that Diff(S^3relx_0) should have the homotopy type of O(3). I’m afraid that product structures are not compatible with homotopy equivalences, so you can not conclude that O(4) is a product S^3xO(3). $\endgroup$
    – ThiKu
    Commented Mar 5, 2019 at 11:46
  • 1
    $\begingroup$ I was thinking that $Diff(\mathbb{S}^3 rel x_0)$ was homotopy equivalent to $Diff(\mathbb{D}^3, rel \partial)$, which has the homotopy type of a point. Therefore I concluded that $O(4)$ should be h.e. to $\mathbb{S}^3$, which is clearly not true. Thank you for your answers. $\endgroup$
    – X1921
    Commented Mar 5, 2019 at 14:37
  • 1
    $\begingroup$ There is a further fibration still, whose projection is $\text{Diff}(S^3, x_0) \to GL(T_{x_0} S^3)$. The fiber is homotopy equivalent to the diffeomorphisms of the disc rel boundary. $\endgroup$
    – mme
    Commented Mar 5, 2019 at 14:41
  • $\begingroup$ @Mike Miller yes, thank you very much for the nice observation. $\endgroup$
    – X1921
    Commented Mar 5, 2019 at 15:04

0

Reset to default

You must log in to answer this question.