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Consider four points, $\sigma_i$ i=1,2,3,4 on the line $\mathrm{Im}(z) = 0$ in the complex plane $\mathbb{C}$. Does it exist a rational function of these four points which is $\mathrm{SL}(2,\mathbb{R})$ invariant, but not $\mathrm{SL}(2,\mathbb{C})$ invariant?

For sure, you can build an invariant (not rational) with this property: for instance suppose that the geodesic $\gamma$ (with respect to the natural hyperbolic metric of the halfplane) joining $1$ and $3$ with the geodesic $\delta$ joining $2$ and $4$ intersect at a point $P$. Choose an $\mathrm{SL}(2,\mathbb{R})$ element so that $P$ is sent to $i$ and $\sigma_1$ to $0$. It follows that $\sigma_4 = - 1/\sigma_2$. Since the group $\mathrm{SL}(2,\mathbb{R})$ has now been completely fixed, the location of $\sigma_2$ in this frame is an invariant.

Note however, that the cross ratio of these four points, computed in this frame, is a function quadratic in $\sigma_2$ therefore can not distinguish wether $\sigma_2$ is greater or smaller than 0! If we go to the Poincarè disk model, the two cases are mapped to each other by an inversion which is not an element of $\mathrm{Aut}(D)$, but it is an element of $\mathrm{SL}(2,\mathbb{C})$ therefore the cross-ratio cannot distinguish the two cases, which are however not equivalent under any $\mathrm{SL}(2,\mathbb{R})$ transformation!

The question is: is there a rational function of $\sigma_i$ that does the job?

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    $\begingroup$ The group $H=SL(2,R)$ is ZAriski dense in $G=SL(2,C)$ so any rational function on the projective line which is $H$ invariant is also $G$ invariant. $\endgroup$ – Venkataramana Dec 13 '18 at 4:31
  • $\begingroup$ Similarly related question, for which I guess your answer still apply: Suppose you have two points in the real line as above, and one in the half-space Im(z)>0. Again, you can build sl(2,R) invariants for these three points, but not SL(2,C) and the question is: Can you build SL(2,R) invariants which are rational functions of these points? I guess no, for the same reason $\endgroup$ – giulio bullsaver Dec 13 '18 at 6:16
  • $\begingroup$ Yes, for the same reasons. $\endgroup$ – Venkataramana Dec 13 '18 at 7:21
  • $\begingroup$ Thanks! For the sake of completeness I will add that I discovered that if you allow also complex conjugates you can find one, but then, more precisely you cannot find meromorphic SL(2,R) invariant functions of three points as above. $\endgroup$ – giulio bullsaver Dec 13 '18 at 8:19

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