# SL(2,R) invariant which are not SL(2,C) invariants

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?

• 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. – Venkataramana Dec 13 '18 at 4:31
• 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 – giulio bullsaver Dec 13 '18 at 6:16
• Yes, for the same reasons. – Venkataramana Dec 13 '18 at 7:21
• 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. – giulio bullsaver Dec 13 '18 at 8:19