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The cross ratio of 4 points $A,B,C,D$ in the plane is defined by

$$(A,B,C,D) = \frac{AC}{AD} \frac{BD}{BC}$$

And it's a ratio which is preserved under projections, inversions and in general, by Möbius-Transformations.

Although I can see it's utility and power, I cannot see a geometric definition or intuition for the cross ratio. Can someone give me any insights about this?

EDIT: To be more specific, I'd like to express the cross-ratio as the length of some segment constructible with straightedge-and-compass.

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Draw the circle passing through $B,C,D$ and the circle passing through $A,C,D$ (or the line for three collinear points). The argument of the cross-ratio $\arg(A,B,C,D)$ is the angle between the two circles* where they meet at $C$.

To work this out, notice that the construction and the answer you get are invariant under Möbius maps, so you can make everything simple before you calculate by putting $D$ at $\infty$, $C$ at $0$, and $B$ at $1$.

To work out what the modulus of the cross-ratio should be, just swap some of the points over and repeat. For example, $(A,C,B,D)=1-(A,B,C,D)$ so $\arg[1-(A,B,C,D)]$ gives the angle at $B$ between the circles passing through $A,B,D$ and $B,C,D$.

There are several other angles that you could measure, but they are all related by various bits of spherical geometry (I find it easiest to think about this stuff on a sphere, by stereographic projection) in such a way that knowing two of them tells you all the rest.


$\ast$ There's some ambiguity in which angle to take, since there are two choices, and what counts as positive or negative. This can be resolved by checking for each circle the order the three marked points come in, and whether the fourth point is inside or outside the circle; details left as an exercise for the reader (translation: I can't be bothered to write them down).

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  • $\begingroup$ Thank you very much! That's exactly the kind of solution I was looking for. $\endgroup$ – Jonas Gomes Jan 31 '15 at 20:01
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What you have written is the "classical" cross-ratio of projective geometry, which is defined not for any old 4-tuple of points in the plane but only for a 4-tuple of points that lie on some line. The geometric meaning is that the cross-ratio is a complete "projective" invariant. For example, if you are looking in perspective at 4 collinear points $A,B,C,D$, and if you wish to know whether there is another perspective from which those 4 points are equally spaced in order along a line, then that other perspective exists if and only if the cross ratio of $A,B,C,D$ equals the cross-ratio of four equally spaced points on the line such as $1,2,3,4$, namely $\frac{1-3}{1-4} \frac{2-4}{2-3} = \frac{4}{3}$.

Our mammalian brains know this: if our eye looks at a perspective picture of a line with marks labelled by the integers in order, it won't look like a number line to us unless the cross-ratios of all successive 4-tuples are equal to $\frac{4}{3}$. There's a nice explanation of this in John Stillwell's book "The Four Pillars of Geometry".

But for a general 4-tuple of points in the plane, your formula is not correct. Instead you should think of $A,B,C,D$ as being four points in the complex plane $\mathbb{C}$, and an expression like $AC$ should be replaced by the complex valued difference, and then one uses complex valued multiplication and division to obtain the complex valued cross-ratio formula $$\frac{A-C}{A-D} \cdot \frac{B-D}{B-C} $$

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  • $\begingroup$ Thanks for you answer, I should've written more carefully the expression for the cross ratio, indeed. I understand the cross ratio is a complete projective invariant, but I can't find a geometrical interpretation for the number $(A,B,C,D)$. For example, given three points $P,Q,R$ with $P-Q-R$ we can define the geometric mean of $P,Q,R$ by $\sqrt{PQ * QR}$ and a geometric meaning for this would be the height of the right triangle of hypotenuse $PR$ and with projections $PQ$ and $QR$. That's the sort of interpretation I'm looking for. $\endgroup$ – Jonas Gomes Nov 15 '14 at 21:46
  • $\begingroup$ @JonasGomes: One geometric meaning for the cross-ratio, in somewhat the sense that you are asking, is that there is a unique Mobius transformation that takes $A \to 0$, $B \to \infty$, and $D \to 1$, and that unique Mobius transformation takes $C$ to the cross-ratio. $\endgroup$ – Lee Mosher Nov 16 '14 at 3:47
  • $\begingroup$ And can you give a more geometrical meaning? Maybe the measure of some segment constructible with edge-and-compass? $\endgroup$ – Jonas Gomes Nov 16 '14 at 4:48
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    $\begingroup$ @JonasGomes: I don't know how to do anything like that. Cross-ratio is a conformal invariant. The group of conformal transformations (Mobius transformations) is quite a bit larger than, say, the group of Euclidean isometries or even the group of Euclidean similarities. So the invariants of conformal geometry---such as cross-ratio---are expected to be quite a bit less rigid than isometry invariants or similarity invariants. $\endgroup$ – Lee Mosher Nov 16 '14 at 14:42
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Let's consider the cross-ratio in the complex plane so that it looks like $$(A,B,C,D)=\frac{A-C}{A-D}\cdot\frac{B-D}{B-C}$$

Given trhee point $A,B,C$ in $\widehat{\mathbb C}=\mathbb C\cup\{\infty\}$ there is a unique Moebius transformation $z\mapsto \frac{az+b}{cz+d}$ that sends $A\to 1, D\to 0, C\to\infty$ and this is exactly $$f(z)= \frac{A-C}{A-D}\cdot\frac{z-D}{z-C}$$

Thus $(A,B,C,D)$ as defined above is just $f(B)$. In other words, the cross-ratio of four points is the image of one point once three of them are mapped to $0,1,\infty$ via a Moebius map.

Thus, any geometric quantity that you can "compute" from the four points $0,1,\infty,z$ by means that are invariant under Moebius maps, is the same quantity that you compute for any four points with $(A,B,C,D)=z$. For instance, the three angles of the triangle with vertices $0,1,z$.

Moreover, if you consider $\widehat{\mathbb C}$ as the boundary of the hyperbolic space $\mathbb H^3$, then the points $A,B,C,D$ are the vertices at infinity of a so-called ideal tetrahedron. The cross-ratio is a complete isometry invariant of geodesic ideal tetrahedra. Thus any metric (hyperbolic metric) quantity you compute (e.g. distances between opposite edges) depends only on the cross-ratio. You can find more about this viewpoint in any book of hyperbolic geometry.

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It was my Conjecture.

If a right circular cone in 3-space of semi-vertical angle $ \alpha $ has a pencil of four

rays/generators (OA,OB,OC, OD)through its vertex O, then,

the cross-ratio of their projection on an arbitrary plane is a projective invariant, is some trigonometric or other function of $ \alpha $.

Earlier thoughts ( Drexel Univ. Math Forum geometry.puzzles) :

http://mathforum.org/kb/thread.jspa?forumID=129&threadID=2578843&messageID=9148204#9148204

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  • $\begingroup$ Since the question was about the plane, how is this relevant? $\endgroup$ – Rory Daulton Jan 31 '15 at 13:38
  • $\begingroup$ @RoryDaulton Going out of the plane in projective geometry could be very relevant. $\endgroup$ – Pp.. Jan 31 '15 at 14:16
  • $\begingroup$ @Pp.. Thanks, could not be taken to finish line, hope someone finds out a 3D object whose projected property is the cross-ratio. $\endgroup$ – Narasimham Jan 31 '15 at 14:21

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