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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.

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  • $\begingroup$ Hi. What you need is that every isometry extends to a linear transformation of $\mathbb{R}^{n+1}$. This could be seen quite easily directly, but for a quick reference (if you don′t find another) you could use "the fundamental theorem of projective geometry" that says that if a projective line goes to a proj line then the transformation must be linear (up to a field automorphism, but there aren′t any over $\mathbb{R}$). $\endgroup$
    – Uri Bader
    May 8, 2016 at 18:26
  • $\begingroup$ @UriBader: To use this argument, you first have to prove that isometries carry lines to lines, which is not trivial. $\endgroup$ May 8, 2016 at 19:04
  • $\begingroup$ @RobertBryant, three near by points are colinear iff the triange inequality is an equality. Further apart colinear points could be seen to be so using the metric by adding some more points on the same line. This is easy enough. $\endgroup$
    – Uri Bader
    May 8, 2016 at 19:16
  • $\begingroup$ @RobertBryant, and I gave this argument and not a direct one because this question is a reference request, and the FTOPG is the best I could think of. But I didn't post it as an answer, as indeed it is not direct enough and I suppose someone will give an exact reference soon. $\endgroup$
    – Uri Bader
    May 8, 2016 at 19:23
  • $\begingroup$ @UriBader: I think that, when you write out the details, you will find that it is not very clean. Also, 'the fundamental theorem of projective geometry' is not so trivial either, as evidenced by the fact that you have to appeal to the triviality of field automorphisms of $\mathbb{R}$. $\endgroup$ May 8, 2016 at 19:23

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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.)

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    $\begingroup$ This is clean and elementary, +1. But surely it does not qualify as an answer to the OP's request for a cite-able reference? $\endgroup$ May 8, 2016 at 19:53
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    $\begingroup$ @ViditNanda: What's wrong with citing MathOverflow answers? :) Seriously, though, it probably would be better to cite a book or an article, and I don't doubt that the above argument can be found somewhere in the literature, but I confess that I don't know where to look. That said, it probably would have been better to have asked this question on MathStackExchange, because it's not really a research question, and the above argument is so elementary that it must be known somewhere or at least be an exercise somewhere. Probably, it's in some undergraduate book on non-Euclidean geometry. $\endgroup$ May 8, 2016 at 20:03
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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 $O(3)$.

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  • $\begingroup$ You don't know that the isometry is differentiable, and it is not clear how you are going to 'pull back' the given distance function on $\mathbb{RP}^2$ to define a distance function on the $2$-sphere. $\endgroup$ May 8, 2016 at 19:06
  • $\begingroup$ @Robert Bryant: You don't know whether the angle between two lines is differentiable?? $\endgroup$ May 9, 2016 at 2:54
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    $\begingroup$ In fact, the distance function on $\mathbb{RP}^2\times\mathbb{RP}^2$ that gives the angle between two lines is not differentiable along the diagonal, but that's not the issue. The issue is whether an isometry $\phi:\mathbb{RP}^2\to\mathbb{RP}^2$ for this distance function is necessarily differentiable. (Of course it is, but that statement would require proof, if you really wanted to give a complete argument along the lines you propose, and this would involve bringing in considerably more machinery than the simple algebra that is necessary.) $\endgroup$ May 9, 2016 at 8:55
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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 $\text{O}(3)$.

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  • $\begingroup$ You wrote, "Observe further that an isometry that fixes the three axis lines is trivial." But it's not true that an isometry that fixes the three axis lines is trivial: Consider the mapping $\phi\bigl([x,y,z]\bigr) = [-x,-y,z]$, which fixes the three axis lines but does not fix $[1,1,1]$. $\endgroup$ May 8, 2016 at 21:00
  • $\begingroup$ Admittedly, "sketchy form" calls for mistakes... $\endgroup$
    – Uri Bader
    May 8, 2016 at 21:36
  • $\begingroup$ Now your argument is a sketch of the proof that I gave. $\endgroup$ May 8, 2016 at 21:56
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    $\begingroup$ @RobertBryant, I believe this question is NOT about a proof. It is NOT about original math. It is primary a reference request, and secondary (my own interpretation) a request for an advice regarding the presentation of a mathematical fact the OP seems to know well. I am NOT competing your math. I made an advice regarding the presentation. It is slightly different than yours. Possibly worse, that a matter of taste. My own opinion is that it answers the question better - otherwise I wouldn't write it down. No offence. $\endgroup$
    – Uri Bader
    May 8, 2016 at 22:04
  • $\begingroup$ (and frankly, I didn't read your answer. Too many symbols.) $\endgroup$
    – Uri Bader
    May 8, 2016 at 22:08
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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.

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