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Let us consider a standard metric and differentiable structure on $S^d$.

We first consider the diffeomorphism group and isometry group of $S^2$ or $S^d$ on itself.

1. What is the diffeomorphism group of $S^2$ or $S^d$?

According to Wiki, the diffeomorphism group of $S^2$ has the homotopy-type of the subgroup $O(3)$. This was proven by Steve Smale.

  • What does exactly "the homotopy-type of the subgroup $(3)$" mean?

So is the answer $$Diff(S^2)=O(3)?$$

Is the answer true in general for $$Diff(S^d)=O(d+1)?$$ at least for the standard sphere not the exotic spheres?

2. What is the isometry group of $S^2$ or $S^d$? Every isometry group of a metric space is a subgroup of isometries. According to Isometries of the sphere $\mathbb{S}^{n}$, it seems that $$Isometry(S^d)=O(d+1)?$$ in general.

Therefore, is that $$Diff(S^d)=Isometry(S^d)?$$

annie marie cœur
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    The diffeomorphism and isometry groups are never the same in dimension $>0$. The diffeomorphism group is much larger (infinite dimensional, in fact). That they have the same homotopy type just means that there is a homotopy equivalence between them. – Kajelad May 22 '21 at 18:05
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    "The homotopy-type of $X$" means "the equivalence class of $X$ under the equivalence relation of homotopy equivalence". – Lee Mosher May 23 '21 at 15:26
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    See my answer to your earlier question for a proof that for all Riemannian manifolds $X$ of dimension $\ge 1$, $\text{Diff}(X) \ne \text{Isometry}(X)$. – Lee Mosher May 23 '21 at 15:28

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