Elementary reference for the isometry group of $\mathbb{RP}^2$ Endow the real projective plane with the distance defined by $d(L,L')$ := "the angle between the lines $L$ and $L'$ ".
It is the case that every isometry from $RP^2$ onto $RP^2$ is induced by an element of the orthogonal group $O(3)$.
I want to use this fact in a research paper. But it turns out that all my considerations are so elementary that, apart from this fact, this paper is readable by a 3rd year student at the University. I would like to have a reference for this result that would be readable by such a student (without resorting to Riemannian structures).
But I have no reference at all for this result actually, so a reference with more fancy technology would already be valuable.
Thank you very much.
 A: You are not assuming, a priori that the isometry is differentiable, so appeals to Riemannian geometry wouldn't do you any good anyway.  Likewise, appeals to double covers and the idea that one could 'lift' the distance function on $\mathbb{RP}^2$ to $S^2$ somehow are likely to get you into problems if you want to do it by purely elementary means.
I think that a reasonable approach is the following one:  If you think of $\mathbb{RP}^2$ as the space of lines through the origin in $\mathbb{R}^3$ and denote by $[u]\in\mathbb{RP}^2$ the line spanned by a nonzero vector $u\in\mathbb{R}^3$, then the function $f:\mathbb{RP}^2\times \mathbb{RP}^2\to \mathbb{R}$ defined by
$$
f\bigl([u],[v]\bigr) = \frac{(u\cdot v)^2}{(u\cdot u)(v\cdot v)}\tag1
$$
is well-defined and is equivalent to specifying your distance function (since $\sin^2\bigl(\cos^{-1}r\bigr)= 1-r^2$).  Thus, you are asking to show that any map $\phi:\mathbb{RP}^2\to \mathbb{RP}^2$ that satisfies $$f\bigl(\phi([u]),\phi([v])\bigr) = f\bigl([u],[v]\bigr)\tag2$$ for all $u,v\in\mathbb{RP}^2$ is of the form $\phi([u]) = [Au]$ for some $A\in\mathrm{SO}(3)$.  (Since $[-u]=[u]$, you don't need $\mathrm{O}(3)$, as it does not act effectively on $ \mathbb{RP}^2$.)
Now, since $\mathrm{SO}(3)$ already acts transitively on triples of orthogonal lines and preserves $f$, it follows that it's enough to understand the isometries that fix three given points $[u_1],[u_2],[u_3]$ that satisfy $f\bigl([u_i],[u_j]\bigr)= \delta_{ij}$, and you might as well assume that these points are $[u_1] = [1,0,0]$, $[u_2]=[0,1,0]$, and $[u_3]=[0,0,1]$.  Since there are 4 points that are at equal distance from all three of these points, namely $[x_1,x_2,x_3]$,
where $x_i^2 = 1/3$, we can, by applying a diagonal element of $\mathrm{SO}(3)$, further assume that $\phi$ fixes the point $[u_4] = [x,x,x]$ where $x^2 = 1/3$.  
Now, the task is to show that this forces such a $\phi$ to fix all points.  
Take any point $[v] = [x,y,z]\in\mathbb{RP}^2$ and assume, as we can, that $x^2+y^2+z^2 = 1$.  Let $[w] = \phi([v]) = [p,q,r]$, where $p^2+q^2+r^2=1$.  Since $\phi$ fixes $[u_i]$ and satisfies (2), we know that 
$$
p^2 = x^2,\quad q^2 = y^2,\quad r^2 = z^2,\quad (p+q+r)^2=(x+y+z)^2.\tag3
$$
If, say $x=0$, then $p=0$, and the remaining equations $q^2=y^2$, $r^2=z^2$
and $(q+r)^2=(y+z)^2$ easily imply that $(q+ir)^2 = (y+iz)^2$, so  $[0,q,r]=[0,y,z]$, so $\phi([v])=[v]$.
A similar argument shows that, if either $y=0$ or $z=0$, then $\phi([v]) = [v]$.
Since points of the form $[\cos\theta,\sin\theta,0]$ are now known to be fixed by $\phi$, it follows that $(\cos\theta\,x+\sin\theta\,y)^2= (\cos\theta\,p+\sin\theta\,q)^2$ for all $\theta$, and hence, $xy=pq$.  Similarly, $xz=pr$ and $yz=qr$.  This, coupled with the above equations, implies that $(p,q,r) = \pm(x,y,z)$, i.e., that $\phi([v])=[v]$, as was to be shown.
N.B.:  Note that one doesn't need to assume that $\phi$ is invertible (or even surjective) in order for this proof to work.  (While these are not hard to show directly, they would be extra steps that would require some care and a higher level of sophistication, since they seem to require notions of continuity and compactness that are, themselves, above the merely algebraic.) 
A: Take the double cover by $2$-dimensional sphere $S\to RP^2$. The metric pulls back
to the sphere as the usual spherical metric and $O(3)$ is the isometry group of this metric, (almost) by definition of $O(3)$. 
A: The following argument is short enough so you can actually include it in your paper (I would do that in the following sketchy form, but that is a matter of style).
Say that a triple of points is orthogonal if the distance between each pair is $\pi/2$. Consider configurations of four points in $\mathbb{P}^2(\mathbb{R})$ such that the first three form an orthogonal triple and the fourth is equidistant form each point in that triple. 
Observe that $\text{O}(3)$ acts transitively on the set of such configurations. Observe further that any isometry that fixes the three axis lines and the line $[1,1,1]$ is trivial. Deduce that the isometry group is $\text{O}(3)$.
A: Robert Bryant's post provides a neat proof: it is a good reference for the result I am interested in. I found another (more conventional: published book vs MO answer) elementary reference for it:  9.7.1 and 19.1.2.2 in Berger's Geometry. This proof is less trickily efficient than that of Robert Bryant, but it is also interesting (geometric insight). I have decided to cite both.
