What, exactly, is a glide reflection? Well, here's how the U of Chicago defines it:
Let r be the reflection in line m and T be any translation with nonzero magnitude and direction parallel to m. Then G, the composite of T and r, is a glide reflection.
Just as reflections, rotations, and translations have nicknames -- "flips," "slides," and "turns," respectively -- glide reflections have the nickname "walks." The U of Chicago gives the example of the isometry mapping the right footprint to the left footprint while walking as a glide reflection. Another name for glide reflection is "transflection," since it is the composite of a reflection and a translation.
Why do we need to worry about glide reflections? Well, we can look at the types of isometries that exist in different geometries. First of all, in 1D geometry, there are only two isometries -- reflections and translations. You can't perform a rotation on a one-dimensional line.
For two dimensions, the number of isometries depends on whether we are using Euclidean, hyperbolic, or spherical (elliptic) geometry. In particular, we have:
- spherical: rotation, reflection, glide reflection
- Euclidean: translation, rotation, reflection, glide reflection
- hyperbolic: horolation, translation, rotation, reflection, glide reflection
Recall that a translation is defined to be the composite of reflections in parallel lines -- so we need to know how parallel lines work in order to know what isometries there are. In Euclidean geometry there is one type of parallel, but in hyperbolic geometry there are two types of parallel -- horoparallel and ultraparallel -- and so there are two different isometries, horolation and translation, depending on whether the mirrors are horoparallel or ultraparallel, respectively. But in spherical geometry, there are no types of parallel lines, so there are no translations at all (much less horolations).
Horolations, translations, and rotations all preserve orientation, since they are the composite of two reflections (the first mirror switching orientation and the second switching it back). But what seems counter-intuitive is that there are only two isometries that reverse orientation in all three types of geometry -- reflections and glide reflections. In hyperbolic geometry, the composite of a reflection and a translation is a glide reflection, but what about the composite of a horolation and a reflection? I suspect what happens is that if an isometry is the composite of a horolation and a reflection, one can find a translation and a (different) reflection whose composite is also that isometry, so it is really a glide reflection after all. (One cannot use the Glide Reflection Theorem mentioned in this chapter to find the translation or reflection, as its proof requires Playfair and so fails in hyperbolic geometry.)
In spherical geometry, the situation is every worse. There are no translations in spherical geometry, so how can there be glide reflections? Well, this link explains it well:
Glide-Reflections: Like Euclidean geometry, the combination of a reflection and a translation is a new kind of symmetry. We saw above that translations on the sphere are really rotations, and hence a glide-reflection could also be called a rotation-reflection.
Recall how I wrote that what appears to be a translation on the sphere is really a rotation.
Returning to Euclidean geometry, we see the above discusses the composite of a rotation and a reflection, so what happens when we compose these in Euclidean geometry? For that matter, what is the composite of a rotation and a translation? As it turns out, neither of these produces any new isometries in two dimensions. (Three dimensions is another matter.)
The reason for this is that in plane geometry, three mirrors suffice. That is, not only is any isometry the composite of just reflections, but that they are the composite of at most three reflections. The U of Chicago alludes to this fact, and shows a hierarchy where all isometries are the composite of one, two, or three reflections. The text doesn't prove this, but we can prove it here easily.
First, suppose we have three noncollinear points, A, B, and C, and another point P. Now suppose we let P' be any point other than P such that AP = AP', BP = BP', and CP = CP'. So, by the Converse of the Perpendicular Bisector Theorem, A, B, and C are all on the perpendicular bisector of
PP' -- that is, they are collinear, a contradiction since these points are assumed to be noncollinear. Therefore, the conclusion is that there is no point other than P such that AP = AP', BP = BP', and CP = CP' -- that is, if P' is a point with the same distances to A, B, and C as P, then P' is the same point as P.
Now we take any isometry that maps three noncollinear points A, B, C to D, E, F. Now we begin by performing a sequence of reflections. The first mirror is the perpendicular bisector of AD. This reflection maps A to D, and B, C to new points B', C'. The second mirror is the perpendicular bisector of
B'E. Notice that D is on this line, since DE = AB = A'B' = DB' (isometries preserve distance and the Converse of the PBT). So D reflects to itself, B' reflects to E, and C' reflects to some C". Now the third mirror is the perpendicular bisector of C"F. By similar reasoning, both D and E are already on this line, and so D and E reflect to themselves and C" reflects to F.
So the composite of three reflections maps A, B, C to D, E, F. Now let O be any point in the plane, P its image under the original isometry, and P' its image under the three reflections. Since isometries preserve distance, AP = AP', BP = BP', and CP = CP'. Then P' is the same point as P, and since this is true for any point in the plane, the composite of three reflections is the original isometry. Thus any isometry is the composite of at most three reflections. QED
The "at most" part is there because not all three reflections are necessary. If A and D are the same point then the first reflection isn't necessary, and if B' and E are the same point then the second reflection isn't necessary. There's actually a 50-50 chance that the third mirror isn't necessary -- since D and E are on the reflecting line and reflections preserve angle measure, by the Two Sides of Line Assumption part of the Angle Measure Postulate, C" must be one of only two points, one of which is F and the other is its reflection across line DE.
And so the hierarchy listed in the U of Chicago is complete. Three mirrors suffice, and all of the possibilities are listed in the chart. In particular, if one were to take two congruent triangles and toss them into the air, when they land, the first can be mapped to the second by one of the four possible isometries -- reflection, rotation, translation, or glide reflection. Chances are that it will be either a rotation or a glide reflection. Think about it -- there's a 50-50 chance that the figures will have the same orientation, which would mean that it's either a translation or a rotation. But we have to be extremely lucky for it to be a translation -- that would mean that the triangles are "facing the same way," with corresponding sides exactly parallel. If the sides deviated even by, say 1 degree, from being parallel, a translation is impossible and it must be a rotation -- the magnitude of that rotation would in fact be that same 1 degree. A rotation is said to have more degrees of freedom than a translation, which is why a rotation is much more likely.
Similarly, a glide reflection has more degrees of freedom than a mere reflection, in case the two triangles have opposite orientation. Therefore, a random isometry will almost surely be either a rotation or a glide reflection. Notice that degrees of freedom and dimension are very closely related -- that's why the only two isometries possible in 1D, reflections and translations, have fewer degrees of freedom than the two isometries that require 2D.
I once tutored a geometry student who had a worksheet on glide reflections. The student had to use a coordinate plane to perform the glide reflections, which were given as the composite of a reflection and a translation. But the problem was that on the worksheet, the direction of the translation wasn't always parallel to the reflecting line! In fact, in one of the problems the translation was perpendicular to the reflecting line. That would mean that the resulting composite wasn't truly a glide reflection at all, but just a mere reflection!