summary of notes on glides
you seem not to get the basic idea of whether a map is or is not orientation preserving. the composition of two reflections is orientation preserving, hence is never a glide.
one challenge in reading these notes will be figuring out what the goofy symbols were really supposed to be.
BASIC RESULT:
Theorem: Let f = RSoRToRU, be a composition of three reflections. If all three lines S,T,U have a common point, or if all three are parallel, then f is a reflection. In every other case f is a glide.
522/722 notes 8: Glides: the only isometries, other than translations, with no fixed points.
Definition: A ``glide`` is a composition †«RL or RL«† of a non trivial translation † and a reflection RL in a line L which is invariant for †. Equivalently, a glide is a composition of three reflections © = RL«RT«RS, or RT«RS«RL,where S ± T are parallel, and L is perpendicular to S,T. [Note: if the translation † is trivial, we do not get a glide, but a reflection instead.]
We will prove that glides have no fixed points, exactly one invariant line, and are the only isometries with just one invariant line. We will also prove that a composition of three reflections whose lines have no common point, but are not all parallel, is always a glide. It will follow that a glide is a new isometry, and that every isometry is either a glide, a rotation, a translation, or a reflection. Thus there are no more new isometries to be found, (in the plane).
Lemma: A glide has no fixed points.
proof: If © = RL«† is a glide, where L is an invariant line for †, then every point x of L is mapped by † to another point y = †(x) ± x of L, and then y is left fixed by R, so no point x of L is fixed by R«†. If M ± L is parallel to L, then every point of M is mapped to a different point of M by †, and then M itself is mapped by RL to a completely different line RL(M) parallel to M, but not equal to it, as we know from notes ò6. So no point of M is left fixed by R«† either. Since every point of the plane lies on some line parallel to L, we have proved there are no fixed points for R«†.
Exercise: Prove a glide of form †«R has no fixed points. QED.
Lemma: If f is any isometry with two invariant lines L and M which are not parallel, then L and M meet at a fixed point p of f. In particular, an isometry with no fixed points cannot have two invariant lines which are not parallel.
proof: If L, M are invariant for f, and L€M = {p}, then f(p) lies on f(L) = L, and on f(M) = M, hence {f(p)} = L€M = {p}, so f(p) = p. QED.
Lemma: If † is a translation along L, the glides RL«† and †«RL both have L as their only invariant line.
proof: Since L is invariant for both † and R, L is invariant for the compositions R«† and †«R. If M ± L is a line parallel to L, then †(M) = M, and R(M) ± M. Hence (R«†)(M) = R(M) ± M, so M is not invariant for R«†. Since R(M) is parallel to M, R(M) is invariant for †, so (†«R)(M) = †(R(M)) = R(M) ± M, and M is not invariant for †«R either. Since a glide has no fixed points, there are no invariant lines not parallel to L by the previous lemma. Thus L is the only invariant line for R«†, and for †«R. QED.
Since a glide has no fixed points, and only one invariant line, a glide is a new isometry. [A translation has no fixed points, but many invariant lines.] To characterize glides just by their invariant lines, we need to understand the invariant lines of rotations too.
Definition: A rotation ® about p is called a "half turn" if it can be written as a composition of two perpendicular reflections, i.e. if and only if ® = RM«RL, where L and M are perpendicular and L€M = {p}.
Remark: It does not follow from the definition that every way of writing a halfturn as a composition of two reflections involves perpendicular reflections, but this can be proved [see exercise below].
Theorem: A "half turn" about p has for invariant lines exactly the set of all lines containing p. It restricts on each of these lines to a reflection in the point p. A non trivial rotation about p which is not a halfturn has no invariant lines.
proof: If ® ± id is a rotation about p, and L a line not containing p, we claim L is not an invariant line. To see this note there is a 1-1 correspondence between lines not containing p, and points of the plane different from p as follows. If x ± p is a point, consider the line M perpendicular to the segment Äxp at x. Then x is the unique closest point of M to p. Conversely, if M is any line not containing p, there is a unique closest point x of M to p, found by dropping a perpendicular from p to M, and taking the point x where the perpendicular hits M. Thus M determines x and conversely x determines M. If M is a line not containing p, and x the point of M nearest p, then since an isometry preserves lengths and angles, ®(x) is the point of ®(M) nearest p. Since ®(x) ± x, then ®(M) ± M. Thus a line M not containing p is never invariant for ®.
Now let ® = RM«RL, be a half turn about p, where L and M are perpendicular and L€M = {p}. I claim every line through p is invariant for ®, and that ® restricts to a reflection in p on every such line. This is clear for the lines L and M themselves. I.e. since RL is the identity on L, ® acts on L like RM, which restricts to reflection of L in the point p. The same is true (in the other order) for M. Now let K be any other line through p. We want to show that if x is any point on K, that ®(x) also lies on K. Since we know the lengths d(p,x) and d(p,®(x)) are equal, and that ®(x) ± x, this will prove our claim for the line K. Look at this picture:
Let x ± p be a point on K, and draw the lines A,B through x parallel to L and M. Let c be the point where A meets M, and b the point where B meets L. Then ® maps A to the line parallel to L and passing through ®(c), as in the picture. Also ® maps B to the line throguh ®(b) and parallel to M as shown. Since x is the unique point on A€B, ®(x) is the unique point on ®(A)€®(B) as shown. We claim ®(x) is on K. Since ® is an isometry, triangles xpc and ®(x)p®(c) are congruent, so angles xpc and ®(x)p®(c) are equal. Thus angle xp®(x) is a straight angle. I.e. ®(x) lies on the line K, which is thus an invariant line for ®, and ® restricts on K to the reflection in the point p.
Now let ® = RM«RL, where L€M = {p}, be a rotation which is not a halfturn. Thus by definition there is no way to write ® using perpendicular reflections. Now let K be any line through p, and write ® = RU«RK where U contains p. Then U is not perpendicular to K, so K is invariant for RK but not for RU, so ®(K) = (RU«RK)(K) = RU(K) ± K, and K is not invariant for ®. QED.
Exercise: Every way of writing a halfturn as a composition of two reflections uses perpendicular reflections.
Lemma: If an isometry g has no fixed points and leaves all lines parallel to L invariant, then g is a translation along L.
proof: Let's force a fixed point on L, by composing with a translation along L. I.e. then there is a translation † along L such that †«g fixes a point x on L. Then †«g leaves all lines parallel to L invariant, since g and † do and †«g has a fixed point. Then †«g cannot be a rotation by what we know about the invariant lines of rotations. If †«g were a reflection, it would equal RV, where V is the line perpendicular to L at x. But then g = †-1«RV is a composition of three parallel reflections, hence a reflection, contradicting g having no fixed points. Thus †«g = id, which implies g = †-1 is a translation. QED.
Theorem: An isometry f with only one invariant line L is a glide.
proof: If M ± L is parallel to L, then f(M) is parallel to f(L) = L, and the same distance from L as M, [why?]. Since f(M) ± M, we must have f(M) = RL(M). Thus RL«f leaves invariant all lines parallel to L. If RL«f has a fixed point, then from what we know of the invariant lines of reflections and rotations, RL«f must be a reflection in a line U perpendicular to L. But if RL«f = RU, then f = RL«RU fixes the point where L meets U, a contradiction. So RL«f has no fixed points and then RL«f = † is a translation along L by the previous lemma. Thus f = RL«† is a glide. QED.
Exercise: If L is an invariant line for a non trivial translation †, the glides †«RL and RL«† are equal.
Since we know every isometry is a compositions of at most three reflections, and since we understand compositions of two or fewer reflections, the next result which will allow us to classify all isometries of the plane.
Theorem: Let f = RS«RT«RU, be a composition of three reflections. If all three lines S,T,U have a common point, or if all three are parallel, then f is a reflection. In every other case f is a glide.
proof: First, to realize some of the possibilities for arrangements of the three lines, consider the following pictures.
The following two arrangements of the lines both give reflections:
On the other hand all the following arrangements give glides:
Note that if exactly two lines are parallel, it does not matter whether they are the first two, the last two, or the first and third, and it does not matter whether the non parallel line is perpendicular to them or not, you still get a glide. Also if none of the three lines are parallel, you again get a glide.
Now let's prove it. We already know that if the three lines are "concurrent", i.e. have a common fixed point, or if all three lines are parallel, then their composition of the three reflections is a reflection. Next we prove that in the picture above of seven other cases, we only need to consider the bottom one.
So assume the three lines are neither concurrent nor all parallel.
If f = RL«RM«RU, where L,M are not parallel but meet at p, and if U does not contain p, we can rewrite the rotation ® = RL«RM as ® = RS«RT where T is the unique line through p parallel to U and S contains p. Then f = RS«RT«RU, where T,U are parallel, but S is not parallel to T,U since S and T meet at p. Moreover, since U does not contain p, T ± U.
If f = RS«RT«RU, where S ± T are parallel and U is not parallel to S,T, suppose U meets T at p and S at q (as in the picture above). Then rewrite the rotation ® = RT«RU as ® = RURV where V contains p. Since T ± U, ® ± id, so then U ± V, hence V does not contain q. Then f = RS«RU«RV where S and U are not parallel but meet at q which is not on V. Then by argument in the previous paragraph we may rewrite f with the last two lines parallel and distinct, and the first line not parallel to them. Thus we need only consider that case.
Thus we may assume f = RS«RT«RU where T ± U are parallel and S is not parallel to T,U. If S is perpendicular to T,U, f is a glide by definition. So assume S is not perpendicular to T,U. To prove f is a glide, we will show that f has exactly one invariant line.
First if K is a line which is not parallel to S, we claim K is not invariant for f. If † = RT«RU, and K is invariant for † then since S is neither parallel nor perpendicular to T,U, then K is not invariant for RS, so f(K) = (RS«†)(K) = RS(K) ± K. If K is not invariant for †, then †(K) is parallel to K but not equal to it. If K meets S at x, then †(K) meets S at some other point y ± x. In particular, y is not on K. Since y is on †(K), and RS(y) = y, y is on f(K) = RS(†(K)). Since y is on f(K) but not on K, f(K) ± K.
Now if K is parallel to S, it is possible for K to be invariant for f, but there is only one such invariant K. I.e. then †(K) is parallel to K hence also to S, and K will be invariant for f if and only if RS takes †(K) back to K. This happens if and only if K and †(K) are equidistant from S, i.e. if and only if S is exactly halfway between K and †(K). That there is exactly one such line K will follow from the next lemma.
Lemma: Every point of the plane is moved exactly the same distance by †. [See picture below.]
proof: Take a point x, let y = †(x), and let ∂ = d(x,y). Then take any other point a not on the line Äxy. If å = †(a), the lines Äxy and Äaå are parallel (since both are invariant for †), and also lines Äxa and Äyå are parallel, since we know † takes the line Äxa to a line parallel to Äxa, while obviously † takes Äxa to Äyå. Thus xaåy forms a parallelogram, so ∂ = d(x,y) = d(a,å). Thus every point a not on line Äxy is moved the same distance by †, namely ∂. Now by the same argument, but using a and å instead of x and y, we can prove every point not on line Äaå is also moved the same distance ∂. This completes the proof by showing all points on the line Äxy are also moved a distance ∂. QED.
In the following picture, meant to illustrate the previous lemma, the vertical lines are invariant lines for †.
We have noted that a line K is f-invariant if and only if the line S is halfway between K and †(K). By the lemma then, the only possible f -invariant lines, are the lines K and ëK in the following picture.
I.e. the only posssible invariant lines parallel to S are found as follows: let x be any point and y = †(x), and let ∂ = d(x,y). Then choose any point o on S and construct the two points a, å on the same †-invariant line W as o, with both distances d(a,o) = ∂/2 = d(å,o). Let K be the unique line through a, parallel to S, and ëK the unique line through å, parallel to S. Then K and ëK are the only possible f-invariant lines. K is f-invariant if and only if ëK = †(K), and ëK is f-invariant if and only if K = †(ëK).
What we have to prove is that either ëK = †(K) or K = †(ëK), but not both. It suffices to prove that either å = †(a) or a = †(å), but not both. It is impossible for both of these statements to be true since if †(a) = å, then †-1(å) = a. Since a translation is determined by its action on one point, the only way we can have †(å) = a also, is if † = †-1. But if † = RT«RU, then †-1 = RU«RT, and when T ± U, it is easy to check these translations act differently on T for example. It remains to show one of the equations å = †(a) or a = †(å) holds.
To prove it, consider the picture above. In the picture above, since segments Äxy and Äaå are equal and parallel, they form the vertices of a parallelogram taken in some order. For example suppose that segments Äxa and Äyå are parallel, as shown in the picture. Since the line L = Äxa is taken by † to the unique line †(L) parallel to L and containing y, we must have †(L) = Äyå. So †(a) lies on Äyå and on the †-invariant line W containing a. The only point on that line is å, so †(a) = å. Consequently if K is the line parallel to S and containing a, then †(K) = the line parallel to S and containing å. Since S is halfway between K and †(K), hence f(K) = RS(†(K)) = K, and we have an invariant line for f.
On the other hand if segments Äxå and Äya are parallel, then †(å) = a, and the f - invariant line is the line parallel to S and containing å. We cannot have both pairs of segments, Äxa and Äyå as well as Äxå and Äya, parallel to each other, since if one pair is parallel, the other pair form the diagonals of the parallelogram, and the diagonals of a parallelogram meet. On the other hand at least one of these pairs must be parallel. I.e. suppose segments Äxå and Äya are not parallel but meet, as in the picture above. Then x and å are on opposite sides of the line Äya. Since all points of the segment Äyå except y are on the same side of that line as å, and all points of the segment Äxa, except a, are on the same side of that line as x, the segments Äxa and Äyå do not meet. Hence the vertices xaåy, in that order do form a parallelogram. Thus there is precisely one choice of parallel segments, so we do have either †(a) = å or †(å) = a, but not both. So either the line K parallel to S and containing a is f-invariant, or the line ëK parallel to S and containing å is f-invariant, but not both. Convince yourself that no other lines can be f-invariant. QED.
Corollary: Every isometry of the plane is either the identity, a non trivial rotation, a non trivial translation, a reflection, or a glide.
proof: We know every isometry of the plane can be written as a composition of at most three reflections. We already knew that a composition of two reflections was either a rotation or a translation, and now we know a composition of three reflections is either a reflection or a glide. QED.
Exercise: A composition of an even number of reflections cannot equal a composition of an odd number of reflections. A composition of an even number of reflections is a rotation or a translation, and a composition of an odd number of reflections is a reflection or a glide.
Remarks: One can make an analogous study of isometries of Euclidean three space, and there turn out to be six kinds, rotation about an axis (a fixed line), reflection in a fixed plane, translation, reflection in a plane ° followed by rotation about a line perpendicular to °, reflection in a plane ° followed by translation along a line in °, and "screw" motions: translation along a line L followed by rotation about L.
One can also study more general transformations of the plane, such as "collineations", transformations of the plane which always take lines to lines. For any two triangles in the plane there is a unique collineation taking the vertices of one to those of the other, in a given order. Thus the group of all collineations is in bijection with the set of all plane triangles. It is convenient to represent plane collineations using 2≠2 matrices and plane vectors. Since collineations can be used to map every triangle to any other, any property that is invariant under all collineations is true for all triangles. This gives one pretty way to prove the medians of every triangle are concurrent.
