# Lines in upper half-space

I'm teaching a tour-of-classical-geometry class this semester, and we are soon to introduce hyperbolic geometry. I am very inexpert in this subject, and I have a question about a compatibility of a pair of basic viewpoints on the hyperbolic plane.


In light of the "complex upper half-plane" model, there is a second natural way one might arrive at this object: $\RP^1$ has a natural embedding into $\CP^1$ as the fixed points of the action of $\Gal(\C/R)$. The other points in $\CP^1$ appear in Galois-conjugate pairs which, without any loss of generality, can be identified with the half of the pair that happens to lie in the upper half-plane. This makes it clear that automorphisms of $\RP^1$ extend to automorphisms of $\CP^1$: these are exactly the automorphisms of $\CP^1$ (i.e., the Möbius transformations, or linear fractional transformations) which fix $\RP^1$ (i.e., which have real coefficients).

What's not obvious from the second presentation is how to start doing plane geometry with $(\CP^1 \setminus \RP^1) \; //\; \Gal(\C/\R)$. The original presentation of the hyperbolic plane comes equipped with a definition of a line: these are the geodesics for the metric, and one computes that these are given by circles in the complex half-plane with center lying on the real axis. Explicitly, these are solution sets of equations like $$z \bar z - b (z + \bar z) + c = 0$$ where $b$ and $c$ are real and the solution set is nonempty. This expression is evidently Galois-invariant, but I can't guess where it would come from in the Galois version of the story.

Question: Is there a purely Galois perspective on why this equation deserves to be called anything like a "line" in $\CP^1\;//\;\Gal(\C/\R)$?

It seems likely to me that this question, if answerable, is a little tricky. For instance, I imagine that it rests on $[\C:\R] = 2$ — one of life's happy accidents, but something obscuring in a question like this.

P.S.: I'm happy to be told I'm looking for something that isn't there. Higher dimensional hyperbolic spaces probably don't arise like this, and maybe the idea of doing plane geometry in this object is totally a consequence of the accidental isomorphism with the hyperbolic plane. That's OK too!

• A minor point, of terminology: it is not accurate to say an isometry of the hyperbolic plane extends to give an isometry of $\mathbb{R}P^1$. Instead, an (orientation preserving) isometry of the hyperbolic plane extends to a projective transformation of $\mathbb{R}P^1$ (also known, as you indicated, as a Mobius transformation of $\mathbb{R}P^1$, also known as a fractional linear transformation). – Lee Mosher Mar 31 '16 at 20:26
• Here's a thought: take $p$, $q$ distinct on $\R$, set $c = (p+q)/2$ and $r = (p-q)/2$, so that the correct hyperbolic line meeting $p$ and $q$ is given by $$z \bar z - c(z + \bar z) + (c^2 - r^2) = 0.$$ Lifting the naive real polynomial $(x - p)(x - q)$ vanishing only at $p$ and $q$ to the Galois-invariant expression $((z + \bar z)/2 - p)((z + \bar z)/2 - q)$, this differs from the "correct" line equation by a scalar multiple of $(z - \bar z)^2$. But this difference vanishes identically on $\RP^1$... Maybe this is a clue? – Eric Apr 4 '16 at 20:31
• Ah, you can get it come out exactly: $$\frac{(z - p)(\bar z - q) + (\bar z - p)(z - q)}{2} = z \bar z - c(z + \bar z) + (c^2 - r^2).$$ Still not clear why shuffling the Galois action through the $\RP^1$–based expression like this is a reasonable thing to do. – Eric Apr 4 '16 at 22:40

You can rediscover the geometry if you assume that the isometries of the geometry are given by some homographies.

The group of orientation-preserving homographies who leave $\Bbb P^1(\Bbb R)$ invariant is $G = PGL_2^+(\Bbb R) = GL_2^+(\Bbb R) / \Bbb R^*I_2$ (the $+$ indicates we only keep matrices with a positive determinant).
Its Lie algebra is $A = (M_2(\Bbb R), +) / \langle I_2 \rangle$ coming with the natural map $\exp : A \to G$

Given a point $z \in \Bbb C$ and an element $a \in A$, we get an orbit $\{\exp(ta).z \mid t \in \Bbb R\}$. Those orbits can be a line (or a segment), a circle (or an arc), or a kind of spiral.

Lines and circles can be symmetric with respect to complex conjugation, but spirals can't, so we can only look for lines and circles.

First, stand at a point $z$ in the upper half plane, and rotate around yourself. What you should be seeing should be given by a homography with a simple fixpoint at $z$ (and at its conjugate). The map $f \mapsto f'(z)$ is an isomorphism between the group of complex homographies fixing $z,\bar z$ and $\Bbb C^*$ ; the subgroup of real homographies then corresponds to the unit circle, and they are going to be our group of rotations around $z,\bar z$. So we decide they will be isometries of our geometry.

Looking at the orbits of the points around you through that circle group, you should see the upper half plane partitioned into lots of circles.
The circles can be characterized by the property that the image of $z$ by the inversion through them gives $\bar z$. (looking at the equations of those circles, they form a projective line, and all those circles are concurrent in a pair of conjugate points of the complexified real plane, which can be interpreted to be the conjugate pair $\{z,\bar z\}$)

Also, any circle in the upper half plane is such a circle (just ... look at the complex intersection points of the circle with the real line to get its "hyperbolic center")

What should an infinite line look like ?

Among complex homographies with a fixed point at a conjugate pair $\{z, \bar z\}$ there is another natural subgroup to consider, those with a derivative in $\Bbb R_{>0}$ (caution : those homographies are NOT real ! however, they satisy $f \circ \bar f = id$ ; so their orbits are still invariant by conjugation, so it is not that bad)

Those orbits will naturally make right angles with all our circles, so this gives another definition of a line as an object that makes right angles with all the circles centered around a point.

In particular they will make right angles with $\Bbb P^1(\Bbb R)$, so they are the vertical lines and circles with a center on the real axis.

Another way to define a line is as the bissecting line between two points $z_1,z_2$. Since we can judge when two points are at equal distance with another point, we can talk about such a bissector : $z$ is on the line bissecting $z_1$ and $z_2$ if there is a circle of "center" $z$ going through both $z_1$ and $z_2$. With this you obtain the circle (or vertical line) normal to the real line which inverts $z_1$ to $z_2$.

Finally, to get the distance between $2$ points, we look at the line joining them. It has two intersection points $r_1,r_2$ with the horizon (the real axis), so let us look at real homographies fixing $r_1,r_2 \in \Bbb P^1(\Bbb R)$. Again, looking at their (real) derivative at $r_1$ (the one at $r_2$ will be its multiplicative inverse) gives an isomorphism between this group and $(\Bbb R_{>0}, \times)$. Among this group, there is a particular homography sending $z_1$ to $z_2$. Taking the absolute value of the natural logarithm of that derivative is the only possible choice to define distance travelled by that homography, and so the distance between $z_1$ and $z_2$.

The simplest numerical example is when the line joins $0$ and $i\infty$. Then the group is simply $\Bbb R_{>0}$ acting by homothety. Given $iy_1$ and $iy_2$ on the line, their distance is therefore $|\log(y_2/y_1)| = |\log y_2 - \log y-1| = \int_\gamma \frac 1 y |dl|$.

Since rotations are supposed to be isometries, the length of a segment near $z$ doesn't depend on its direction : the length of an infinitesimal segment of euclidiean length $dl$ at $x+iy$ has to be $\frac 1y dl$, from which we recover the $ds^2 = (dx^2 + dy^2)/y^2$.