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The one quantity every Möbius map leaves unchanged, read two ways: as a test for four points lying on a circle, and as the map that sends any three points to zero, one and infinity — which is how the map carrying one region to another is written down rather than solved for.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute a cross ratio, use it to write the Möbius map sending three points to zero, one and infinity, build the map sending three given points to three given targets, test four points for lying on one circle, handle the point at infinity by deleting brackets, and say why exactly one such map exists.
Möbius transformations, and the fact that one other than the identity has at most two fixed points. That fact is about to be turned round: a map with three prescribed images exists, and there is only one of it.
The cross ratio of four distinct points is $(z, z_1; z_2, z_3) = \dfrac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$, with the convention that a bracket containing the point at infinity is dropped. A quantity is invariant under a family of maps when every map of the family leaves it unchanged. Four points are concyclic when one circle or line contains all of them.
The invariance. For any Möbius map $T$ and any four distinct points,
$$(T z, T z_1; T z_2, T z_3) = (z, z_1; z_2, z_3).$$
It is checked by verifying it for the three building blocks — translation, scaling, inversion — which is a short computation each time, and then composing.
Why that is useful. Read the cross ratio as a function of its first argument: $z \mapsto (z, z_1; z_2, z_3)$ is a Möbius map, and by inspection it sends $z_1 \mapsto 0$, $z_2 \mapsto 1$ and $z_3 \mapsto \infty$. So the map sending any three points to the standard triple is not solved for; it is written down.
And that builds any map you want. To send $z_1, z_2, z_3$ to $w_1, w_2, w_3$, set the two cross ratios equal,
$$(w, w_1; w_2, w_3) = (z, z_1; z_2, z_3),$$
and solve for $w$. The result is the unique Möbius map doing it — unique because two of them would compose into a map fixing three points, which must be the identity.
Concyclicity. Four distinct points lie on one circle or line exactly when their cross ratio is real. The reason is the invariance again: send three of them to $0$, $1$ and $\infty$, which lie on the real axis; the circle or line through the three goes to the real axis; and the fourth point is on that circle exactly when its image is real, which is exactly when the cross ratio is real.
A convention for infinity. If one of the four points is $\infty$, delete the two brackets containing it. So $(z, z_1; z_2, \infty) = \dfrac{z - z_1}{z_2 - z_1}$, which is the formula that comes out of taking a limit and is much easier to remember as a deletion.
Another way: picture
Three points are exactly enough to pin down a generalised circle, and exactly enough to pin down a Möbius map. Those two facts are the same fact: send any three points to $0$, $1$ and $\infty$ and the circle through them becomes the real axis. Every question about circles in this unit can be moved to the real axis first, where it is usually obvious.
Another way: steps
A map is usually wanted to carry a region somewhere, not three points — and three points is how it is done, in three moves.
Pick three points on the boundary. A Möbius map sends generalised circles to generalised circles, so the boundary circle or line through the three chosen points is carried to the boundary circle or line through their three images. Choosing where the boundary goes is choosing where three of its points go.
Respect the order. A Möbius map preserves orientation, so if you walk along the boundary through the three points in order, the region on your left goes to the region on the left of the image walk. Getting the order backwards produces the map onto the complementary region — the outside of the disc instead of the inside — and the algebra gives no warning.
Check with one interior point. Take any convenient point of the region, push it through the map, and see whether it lands in the intended target. One substitution settles it, because a Möbius map cannot send part of a region one way and part of it the other.
The standard example is the Cayley transform. Send $0, 1, \infty$ on the real axis to $-1, -i, 1$ on the unit circle: the real axis goes to the unit circle, and the map comes out as $w = \dfrac{z - i}{z + i}$. Checking the interior point $i$ gives $0$, which is inside the disc — so the upper half plane is the side that goes to the inside.
It is a function of an ordered quadruple. Permuting the four points changes the value — there are six possible values in general, related by $\lambda$, $1/\lambda$, $1 - \lambda$ and so on — so the cross ratio of these four points is an incomplete phrase until the order is fixed.
What is invariant is the cross ratio of a quadruple under a Möbius map applied to all four, in the same order. That is a different statement from being independent of the order, and conflating them produces answers that are wrong by a reciprocal or by a subtraction from one.
Two other habits worth watching. Conventions differ: some books write the cross ratio with the roles of the second and fourth points exchanged, so a value quoted from elsewhere may be the reciprocal of the one this lesson computes. Fix one convention and check it against a case you know.
And the point at infinity is handled by deletion, not by substitution. Putting a symbol for infinity into the formula and cancelling is an abuse that happens to work; the honest statement is that the two brackets containing it are dropped, which is what the limit produces.
Send $0, 1, \infty$ to $-1, -i, 1$. The left side is $(w, -1; -i, 1)$ and the right side, with $\infty$ deleted, is $\dfrac{z - 0}{1 - 0} = z$.
Delete the brackets containing the point at infinity.
So $\dfrac{(w + 1)(-i - 1)}{(w - 1)(-i + 1)} = z$, and solving for $w$ gives $w = \dfrac{z - i}{z + i}$.
One equation, one unknown.
That is the Cayley transform, and it maps the upper half plane onto the unit disc — as the interior point $i$ going to $0$ confirms.
Check with one interior point.
Are $1$, $i$, $-1$, $-i$ concyclic? Compute $(1, i; -1, -i) = \dfrac{(1 - i)(-1 + i)}{(1 + i)(-1 - i)}$.
Write out the four differences.
The top is $-(1 - i)^{2} = 2i$ and the bottom is $-(1 + i)^{2} = -2i$, so the ratio is $-1$.
Simplify.
Real, so yes — they lie on one circle, which is of course the unit circle. No geometry was used.
A geometric fact from an arithmetic test.
It is the cross ratio with those three: $T(z) = \dfrac{(z - 1)(2 - 3)}{(z - 3)(2 - 1)}$.
Write it down rather than solving for it.
So $T(z) = \dfrac{-(z - 1)}{z - 3} = \dfrac{1 - z}{z - 3}$, and substituting $1$, $2$, $3$ checks it.
Compute the cross ratio of the four points $1$, $6$, $7$, $12$, in that order, as $\dfrac{(1 - 7)(6 - 12)}{(1 - 12)(6 - 7)}$. Fill in the table.
| Value | |
|---|---|
| The first difference on top | |
| The second difference on top | |
| The first difference underneath | |
| The second difference underneath | |
| The cross ratio |
$T$ is the Möbius map sending $3$ to $0$, $4$ to $1$ and $8$ to infinity. What is $T(10)$? Give a fraction.
Answer:
Is this true: the whole plane maps conformally onto the unit disc?
Put the steps of building the Möbius map that sends three given points to three given targets in order.
Number the steps in order (write the number in the box):
Match each generalised circle to its image under $w = 1/z$.
| The unit circle, with inside and outside exchanged | The real axis again | The circle of radius one over $3$ | A circle through the origin | |
|---|---|---|---|---|
| The unit circle | ||||
| The real axis | ||||
| The circle $|z| = 3$ | ||||
| The vertical line through $1$ |
Three distinct points and three distinct targets are given. How many Möbius maps send the first three to the second three, in that order?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Compute the cross ratio of the four points $3$, $5$, $8$, $10$, in that order, as $\dfrac{(3 - 8)(5 - 10)}{(3 - 10)(5 - 8)}$. Fill in the table.
| Value | |
|---|---|
| The first difference on top | |
| The second difference on top | |
| The first difference underneath | |
| The second difference underneath | |
| The cross ratio |
You can build the Möbius map with three prescribed images and can test four points for concyclicity. Say in your own words why three points is the right amount of information. Next: putting the standard maps together to carry one region onto another.
9. Your turn: the map sending $1, 2, 3$ to $0, 1, \infty$, step 2