Isometry group of $\Bbb RP^3$ Is there a reference for a proof of the statement that the isometry group of the real projective space $\Bbb RP^3$ is isomorphic to $SO(3)\times SO(3)$? How does this group acts on $\Bbb RP^3$?
 A: John's argument establishes that the collection of orientation preserving isometries of $\mathbb{R}P^3$ has the structure of an $SO(3)$ bundle over $SO(3)$.  But, as mentioned in the comments, it does not immediately follow that such a bundle is Lie isomorphic to $SO(3)\times SO(3)$.
However, I claim that this is, nonetheless true.  That is, if $G$ is a Lie group diffeomorphic to the total space of some $SO(3)$ bundle over $SO(3)$, then $G$ must, in fact, be isomorphic as a Lie group to $SO(3)\times SO(3)$.
To see this, begin with the bundle $SO(3)\rightarrow G\rightarrow SO(3)$, the long exact sequence of homotopy groups together with the fact that $\pi_2 = 0$ for any Lie group gives an exact sequence $$0\rightarrow \pi_1(SO(3))\rightarrow \pi_1(G)\rightarrow \pi_1(SO(3))\rightarrow 0.$$
From this we conclude that either $\pi_1(G) \cong C_4$ or $C_2\times C_2$ (with $C_k$ being the cyclic group of order $k$).  In either case, the fundamental group is order $4$.
Now, the universal cover of $G$, $\tilde{G}$, is a compact simply connected Lie group of dimension $6$.  Simply connected compact Lie groups are completely classified:  They are all products of simple Lie groups, which are all classified.  The only simple Lie group in dimension at most $6$ is $SU(2)$.
It follows that $\tilde{G}$ is isomorphic as a Lie group to $SU(2)\times SU(2)$.  Now, every Lie covering, including $\tilde{G}\rightarrow G$, is given by modding out $\tilde{G}$ by a subgroup of its center.  Since $Z(SU(2)) = C_2$ (generated by $\pm I$), $Z(\tilde{G}) = C_2\times C_2$.
Since $\pi_1(G)$ has order $4$, we must therefore conclude that $G\cong \tilde{G}/Z(\tilde{G}) = SU(2)/\{\pm I\} \times SU(2)/\{\pm I\}\cong SO(3)\times SO(3).$
A: First, you can think of $RP^3$ as the set of all unit tangent vectors to the unit sphere; I'll denote that $T_0 S^2$. That corresponds to a point $v \in S^2$ (which I'll regard as a unit vector) and a unit tangent vector at the "tip" of $v$, which is a unit vector $w$ that's orthogonal to $v$. So it really amounts to a pair of perpendicular unit vectors $(v, w)$ in 3-space.
Why is this the same as $\Bbb RP^3$? I'll say something at the end of my answer.
Suppose that $f$ is an isometry of $P = \Bbb RP^3$. It takes the pair
$(n, e) = (\pmatrix{0\\0\\1}, \pmatrix{1\\0\\0})$, i.e., an eastward-pointing vector at the north pole) to some other pair $(a, b)$. Extending this pair to a triple $(a, b, a \times b)$ using the cross product, we can read that triple as a matrix $M$ which is (you should check this) orthogonal, and has det = +1, so it's in $SO(3)$. If we look at $g(x) = M^{-1}f(x)$, we get a new isometry of $P$ that fixes the point $(n, e)$.
So: every isometry is a combination of a rotation (my $M$) and an isometry fixing the point $(n, e) \in P$. So now the question is "Why are isometries fixing a single point of $P$ isomorphic to $SO(3)$?"
Let's lift such an isometry $g: P \to P$ to $S^3$; I'm going to assume that $(n, e)$ lifts to $\pmatrix{0\\0\\0\\1}$.
That gets us an isometry $G: S^3 \to S^3$ that preserves antipodal pairs, and by composing with the antipodal map if necessary, we can assume that $G$ actually fixes the point $\pmatrix{0\\ 0 \\ 0\\ 1}$ rather than sending it to its antipode.
Well, if you fix that point (and of course its antipode) then your isometry must fix (setwise) the orthogonal great-sphere, which is just the set of points $(x,y,z,0)$, where $x^2 + y^2 + z^2 = 1$, i.e., it's the standard 2-sphere. And each rotation of that 2-sphere (i.e. each element of $SO(3)$) gives you an isometry of that 2-sphere. (And those are the only ones!)
So...I guess that's the correspondence: one element $SO(3)$ to take $f(n, e)$ back to $(n, e)$, and (after lifting) another element of $SO(3)$ to rotate the orthogonal 2-sphere in 3-space.
I don't have a reference for this, but I suppose that it's probably in Steenrod's Fiber Bundles book somewhere in disguise.
Wait! I promised an explanation of why $P = \Bbb RP^3$ is the same as $T_0 S^2$, didn't I? Well, the double cover of $P$ is certainly $S^3$, by the definition of the projective plane. And I've shown you the correspondence $(v, w) \mapsto (v, w, v \times w)$ between $T_0 S^2$ and $SO(3)$. And $SO(3)$ is also double-covered by $S^3$ (that actually is described in Steenrod), and some theorem involving universal covers or something shows that any two things double-covered by $S^3$ must be homeomorphic. But it's been almost 40 years since I had to know this stuff, so I cannot cite chapter and verse, alas.
