I want to understand why the projective space $\mathbb RP^n$ is diffeomorophic to $SO(n+1)/O(n)$? and why we can write the latter as $O(n+1)/O(n)\times O(1)$?

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    $\begingroup$ What precisely is $O(1)$? What is $O(n+1)/O(1)$ isomorphic to? $\endgroup$ – Ted Shifrin Jun 23 '16 at 2:25

Consider the canonical action of $SO(n+1)$ on $R^{n+1}-\{0\}$. It defines an action of $SO(n+1)$ on $RP^n$ as follows: let $x$ be an element of $R^{n+1}-\{0\}$, denote by $[x]$ the class of $x$ in $RP^{n+1}$, for $g\in SO(n+1)$, write $g([x])=[g(x)]$. This action is transitive, since for $[x], [y]\in RP^n$, you can assume $\|x\|=\|y\|=1$, there exists $g\in SO(n+1)$ such that $g(x)=y$, it results that $g([x])=[y]$.

Let $x\in RP^n$, the stabilizer $G_x\subset SO(n+1)$ of $[x]$, is the set of elements $g\in S0(n+1)$ such that $g$ fixes the line generates by $x$, this implies that $g$ preserves the orthogonal $H_x$ of $x$, the restriction $r:G_x\rightarrow O(H_x)=O(n)$ is bijective. It is injective, if the restriction of $g$ and $h$ on $H_x$ coincide, if $g(x)=x$ then $det(g)=det(r(g))=1$, this implies that $h(x)=x$, otherwise $h(x)=-x$ and $det(h)=-1$. Contradiction since $h\in SO(n+1)$. If $g\in O(H_x)$, we construct the element $h\in SO(n+1)$ defined by $h_{\mid H_x}=g$ and $h(x)=x$ if $det(g)=1$, otherwise $h_{\mid H_x}=g$ and $h(x)=-x$ if $det(g)=-1$.

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    $\begingroup$ Could you please elaborate on what you mean by the "orthogonal $H_x$" and how the restriction $r: G_x \rightarrow O(H_x)$ is precisely defined? Thanks in advance! $\endgroup$ – ComplexFlo Jun 7 '17 at 14:41

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