Geometry · Groups
Translation, rotation, reflection, and glide reflection — every motion of the plane falls into one of three classes. The three-nails theorem explains why, and Chasles' theorem gives the full classification. Below: interactive models instead of long proofs.
A motion of the plane is a transformation that preserves distances: if the distance between points $A$ and $B$ was $x$, then after the motion the distance between their images $A'$ and $B'$ is still $x$ — for every pair of points.
The simplest example is a translation: every point shifts by the same vector, so the distances between them don't change. It will turn out that there really aren't many "simple" kinds of plane motion — and together they exhaust the whole list.
Switch tabs and drag the handles — the flag shows immediately what each motion does. The pale outline is the original position, the blue flag is the image.
If a motion leaves three points $A$, $B$, $C$ fixed and they don't lie on one line, it leaves absolutely everything fixed — that is, it's the identity ($\mathrm{id}$). Drag $A$, $B$, $C$, and a test point $P$: the three circles of radii $|AP|$, $|BP|$, $|CP|$ around the nails meet at exactly one point — $P$ itself. So no other point can be the image of $P$ under a motion that fixes $A,B,C$.
Since three fixed points off one line always mean $\mathrm{id}$, let's work through the more modest cases in decreasing order.
Two fixed points. This isn't a coincidence from the degenerate case above but a general fact — the two-point lemma: if a motion leaves distinct points $A$ and $B$ fixed, it leaves the whole line $AB$ fixed — for a point $C$ on that line the equality of lengths ($|AC|+|CB|=|AB|$) must be preserved, while for a point off the line it would turn into a strict inequality. And there can't be a third fixed point off that line either: otherwise, by the theorem above, everything would be fixed. So if a motion isn't the identity but at least a pair of points is fixed, exactly the whole line through them is fixed and nothing more — that's a reflection about it.
One fixed point. The motion can't fix a second one — by the same lemma, that would fix an entire line. This is a rotation about it — you already saw this in §2.
No fixed points at all. We already have one example — a nonzero translation. But, as it turns out next, that's not the only answer: there's one more, fourth, kind of motion.
The composition of a reflection and a shift along its own axis is also a motion with no fixed point whatsoever, yet it can't be reduced to a translation, a rotation, or a plain reflection. This is the promised fourth, independent kind — glide reflection.
Every motion of the plane belongs to exactly one of three classes:
$\rightrightarrows$ translation (including the zero one — the identity motion);
$\circlearrowleft$ rotation about an arbitrary center;
$\leftharpoonup\leftharpoondown$ glide reflection — including a plain reflection as the case of zero shift.
The identity motion is a translation by the zero vector; a plain reflection is a glide reflection with zero shift along the axis. No other motions of the plane exist.
The picture shows an original flag (pale outline) and its image (blue) under one specific motion. Before opening the answer — which class is it: translation, rotation, or glide reflection?
Answer: it's a glide reflection — the flag both got reflected (flipped its "side") and shifted along the axis of reflection. Neither a translation nor a rotation alone could do that: a translation doesn't change the flag's orientation, and a rotation leaves its own center fixed.
What you get if you first perform the motion from the row, then the one from the column (the "row∘column" order, same as in the composition table for the line). Two cells on the diagonal are ambiguous: their exact outcome depends on the parameters, not just the class.
| $\rightrightarrows$ translation | $\circlearrowleft$ rotation | $\leftharpoonup\leftharpoondown$ glide reflection | |
|---|---|---|---|
| $\rightrightarrows$ translation | $\rightrightarrows$ translation | $\circlearrowleft$ rotation | $\leftharpoonup\leftharpoondown$ glide reflection |
| $\circlearrowleft$ rotation | $\circlearrowleft$ rotation | $\rightrightarrows$ translation or $\circlearrowleft$ rotation | $\leftharpoonup\leftharpoondown$ glide reflection |
| $\leftharpoonup\leftharpoondown$ glide reflection | $\leftharpoonup\leftharpoondown$ glide reflection | $\leftharpoonup\leftharpoondown$ glide reflection | $\rightrightarrows$ translation or $\circlearrowleft$ rotation |
The composition of two glide reflections is a translation if their axes are parallel, and a rotation (by twice the angle between the axes) if the axes cross — exactly as for two plain reflections: the glide (shift) component of both cancels out along the way. The same effect shows up for rotation∘rotation: a translation if the sum of the angles is a multiple of $2\pi$, otherwise a rotation by that sum.
Treating the plane as the field of complex numbers, every motion is written as a single formula. Adding a number is a translation; multiplying by a number with $|a|=1$ is a rotation about zero; conjugation $\bar z$ is a reflection about the real axis. Putting these operations together gives the whole family:
$f(z) = az + b$, $|a|=1$ — orientation-preserving motions (translation, rotation): $a=1$ — a translation by vector $b$; $a\ne1$ — a rotation by angle $\arg a$ about the point $z_0=b/(1-a)$.
$f(z) = a\bar z + b$, $|a|=1$ — orientation-reversing motions (reflection, glide reflection).
Set up a motion and watch its formula — first how it's computed from the geometric parameters, then how it looks expanded out.
Set up $g$ (applied first) and $f$ (applied second) — $f\circ g$ appears right below: the class of the result, its formula, and parameters computed from the algebra, not eyeballed.
If a motion is given not by the geometric parameters of its axes and vectors but by the images of specific points, its complex formula can be recovered algebraically. Two or three points are enough.
When it's already known whether the motion preserves orientation or reverses it, two distinct points $z_1, z_2$ and their images $w_1, w_2$ are enough to pin it down uniquely.
For an orientation-preserving motion ($w = az + b$) the coefficients come from a linear system:
$$a = \frac{w_2 - w_1}{z_2 - z_1}, \quad b = \frac{w_1 z_2 - w_2 z_1}{z_2 - z_1}$$For an orientation-reversing motion ($w = a\bar{z} + b$) conjugation shows up in the denominators:
$$a = \frac{w_2 - w_1}{\bar{z}_2 - \bar{z}_1}, \quad b = \frac{w_1 \bar{z}_2 - w_2 \bar{z}_1}{\bar{z}_2 - \bar{z}_1}$$Two points aren't enough in this case: they leave an ambiguity. There is exactly one orientation-preserving and exactly one orientation-reversing motion sending the directed segment $z_1z_2$ to $w_1w_2$. A third point $z_3$, not on the line through the first two, is needed, along with its image $w_3$.
The check is simple: from $z_1, z_2$ compute the coefficients $a, b$ assuming the motion is orientation-preserving. If the equality $w_3 = a z_3 + b$ holds for the third point, the guess is confirmed. If the identity doesn't hold, the motion is guaranteed to be orientation-reversing, and the coefficients need to be recomputed with the formulas for $\bar{z}$.