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Isometry question.

  1. May 16, 2008 #1
    I need to show that every isometry in the plane is a composition of at most three reflections.

    Now every isometry in the plane is one of the four:
    1.translation.
    2.rotation.
    3.reflection.
    4.glide.

    now obviously refelction is itself.
    now for translation i showed with a picture that we need only two reflections.
    for rotation we need something like 3 reflections, and also for glide, which answers another question is it possible for glide to be a composition of just two reflection, i don't think so, but don't see how to show this.
    Is any of what i wrote make sense?

    thanks in advance.
     
  2. jcsd
  3. May 16, 2008 #2
    #4 should be "glide reflection," i.e. the composition of a glide (i.e., translation) and a reflection.

    Actually, for the orientation-preserving isometries -- translations and rotations -- you need only two reflections.

    For the orientation-reversing isometries -- reflections and glide reflections -- you need 1 and 3 reflections, respectively.

    What's unclear to me is what you've proven before. Have you already shown that an isometry of the Euclidean plane has to be one of those four? Or are you just assuming that, i.e., using that as your definition of isometry?
     
  4. May 17, 2008 #3
    we proved this.

    Yes you are correct for tranlation we need only 2 reflections.

    but what with the glide, how do you show 2 isn't enough?
     
  5. May 17, 2008 #4
    Well, I know that I'm correct, because I have taught this very same subject many times.

    You need to re-read my message: A glide is the same thing as a translation; as any adept middle school student here in Ohio will tell you, they are synonymous words.

    Therefore, your list is incorrect because
    (1) you repeat yourself (as you wrote it: #1=#4); and
    (2) you left out an isometry, notably, the glide reflection, which is the isometry defined as the composition of a glide and a reflection.

    You now have all the information you need to solve this. I am done with it.
     
  6. May 17, 2008 #5
    the definition our lecturer gave us is that:
    glide is the composition of reflection with translation, sorry that the ohio's terminology is different. ToS.

    So to rephrase my question so it fits your terminology how to show that a glide reflection cannot be composed by less than 3 reflections?
     
  7. May 17, 2008 #6
    thanks in advance.
     
  8. May 18, 2008 #7
    first, you need to talk to your lecturer and point out that in every other geometry course, a glide is a translation is a slide is as a glide. Look it up in H.S.M. Coxeter's Introduction to Geometry and Geometry Revisited. Or look up "glide reflection" in Wikipedia.

    second, and most importantly, if you've shown that a translation needs at least two reflections, then you've already shown that you need at least three reflections. To show that you can't use less than that, you need to show that the composition of any two reflections gives either a translation or a rotation; and finally that a glide reflection is never a translation, rotation, or reflection (hint: look at how many fixed points each type of isometry will have).
     
  9. May 18, 2008 #8

    mathwonk

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    the key in doodle bobs reply is the concept of orientation. ask yourself whether two reflections reverse or preserve orientation?

    as he said, to understand isometries, focus on fixed points.

    1) an isometry with three non collinear fixed points is the identity.

    hence any two isometries sending three non collinear points to the same destination are equal.

    thus to show some unknown isometry equals a specifically constructed one, just construct one that agrees with it on three general points.

    e.g. given an isometry with no fixed points but taking P to X, just translate X back to P, and now the compositions fixes P. Now try to understand an isometry fixing P. if it has another fix point Q, then what? if not, then suppose it takes Q to Y, and rotate Y back to Q, with P as center. Now you have an isometry composed with a translation and a rotation, that fixes both P and Q. so???
     
  10. May 18, 2008 #9
    I don't quite follow your hint, i.e what do you mean by fixed point, a point which the isometry maps it to itself?
    if so a rotation of 2pi can map each point on the circle to itself.
    a non trivial translation doesn't have a fixed point.
    and a reflection has also as many fixed point on the line wrt which we are reflecting the points, cause those points on the line stay put.
    now a glide reflection i don't think that it has a fixed point, i.e if you are at the "mirror" line (my phrase, sorry fo inaccuracy) you only get translated, and you can't get back to the same point.

    am i missing something here?
     
  11. May 18, 2008 #10

    mathwonk

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    have you read my more detailed hints?
     
  12. May 19, 2008 #11
    well for 1), I proved something similar that the composition of three rotations of a triangle is the identity.

    so i need to show that glide reflcetion can't be composed with the other three in order to give the identity.
    well with translation, obviously when with composition of them you can't get back to the same point, also with reflection.
    as for rotation, i think you can go back, it actually depends on the angle and the radius of the circle of rotation, but rotation can be composed of two reflections, am i missing something here?
     
  13. May 19, 2008 #12

    mathwonk

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    translation have no fixed points and preserve orientation, while glide reflections have no fixed points but reverse orientation.

    of course you have to know orientation makes sense, and for that you probably have to prove no odd number of reflections can equal any even number of reflections.
     
  14. May 19, 2008 #13

    mathwonk

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    some class notes on isometries

    522/722 notes 3. Reflections, rotations, and translations

    We have defined isometries and proved their most basic properties. Now we want to introduce the most familiar examples of isometries of the plane: reflections, rotations, and translations, and prove some of their properties. Some of them are a little subtle, and deserve careful discussion. The primary subtlety involves the concepts of "direction" or "orientation". This is the ability to know a "clockwise" from a "counterclockwise" rotation, or to speak about rotations through negative angles, and the concept of the direction of a translation.

    For a rotation, we are tempted to think intuitively of having something like a protractor, that one can move around the plane and use to measure off angles. In particular one may suppose a rotation is determined by knowing a point P in the plane (the center of rotation), and a real number ø, the measure of the angle of rotation. I.e. one might suppose that given a center P and an angle measure ø, and given any other point X, that one can measure off an angle XPY, proceeding either in the "clockwise" or in the "counter clockwise" direction from the segment PX, and having angle measure equal to ø. In particular one might think it makes sense also to speak of an angle of negative measure -ø, which simply meant that the angle XPY was measured off in the clockwise direction from PX.

    But the official definition of a "protractor" geometry only says that if an angle measure ø is given, i.e. a positive number ø, say between 0 and 180, and a segment PX, and if you choose a side of the line PX, then there is a unique ray PY with Y lying in the chosen side, so that angle XPY has measure ø. It does not say that choosing ø positively tells you how to pick the "counterclockwise" side, nor that choosing ø negatively tells you how to pick a clockwise side.

    By contrast, in the Cartesian plane R2 we use in calculus, you have more flexibility: you just go to P, and draw a unit circle centered at P. Then you mark the point A where this circle meets a ray in the direction of the positive x axis, and the point B where it meets a ray in the direction of the positive y axis. Then you indicate a direction arrow on the circle pointing from PA to PB. Then given a point X, you join it to P and mark where it meets the unit circle, say by x. Then you know which side of the line PX is the counterclockwise one, namely the one the direction arrow is pointing to. Then given a real number ø say between 0 and π, (thinking of radian measure), you lay off an arc xy of length exactly ø on the unit circle, proceeding in the direction of your arrow, i.e. in the direction from PA towards PB.
    Then measure off on the ray Py a length equal to that of the segment PX, and the end point is Y, the result of rotating X about P through an angle ø.
    Now we could always make an assumption that in our geometry the same procedure is possible, namely that given a segment PX and a positive number ø between 0 and 180, that there is determined also a particular side of the line PX, a "counterclockwise side", and that this choice of a side is determined "in the same way" throughout the plane. But it is not so easy even to tell what we mean by "in the same way" throughout the plane. Moreover it is very instructive to examine how to introduce a notion of orientation into a geometry rather than just assuming one is already there.

    Hence rather than simply impose a solution to the problem by saying we have all the structure usually assumed in R2, it will be well worth our time to study how the concept of orientation is developed, and how it can be introduced into a geometry only satisfying the axioms we already have. Thus for the time being at least we must give up the assumptions we usually make in R2, namely that a rotation is determined by a point P and a number ø, and that the rotation at P associated to ø+ß is the composition of the rotations at P associated to ø and ß. These require a notion of orientation which we do not wish to assume we have. So we need to examine carefully just what information is needed to determine a rotation.

    Suppose we try defining rotations as follows: We can determine an "oriented" angle either by three ordered non collinear points XPY, or by (i) a segment PX and (ii) a positive number between 0 and 180 (thinking of degrees) and (iii) a choice of a side of the line PX, on which to construct the third point Y. Then suppose we are given an oriented angle say XPY, where the segments PX and PY have the same length. Then we probably believe there is unique rotation ® about P taking X to Y. If so, and if Z is another point, how do we describe the point W = ®(Z)? I.e. PZ and PX are different lines. So how does knowing the side of the line PX that Y is on, tell us how to choose which side of PZ to put W on?

    Looking at the picture, it appears to me that we should say that as long as Z and X are on opposite sides of the line PY then the new point W should be on the opposite side of PZ from Y. Does that seem correct to you? I.e. does it work in all possible configurations of the points X,Y,Z,P, such that P,Z are on opposite sides of line PY? If Z is in the interior of the angle XPY then apparently W should be on the same side of PZ as Y. Where should W be if if X is in the interior of angle ZPY?

    Now what about another question. Suppose we are given an oriented angle XPY somewhere in the plane, and another point Q. Does the oriented angle XPY determine a rotation centered at Q?

    I.e. we have to construct an oriented angle at Q which is not only the same size as XPY but also similarly oriented. Looking at the picture I suppose we start with the lines L,M through Q which are parallel to PX, and to PY. But how do we decide which of the four angles they form is the oriented one we want? I.e. if we want the ray QZ to have the same direction as PX, how do we say where Z is? I.e. which side of M is it on? Please think about this question, and let me know your ideas. Of course what we really would like is to have a concept of translation, so we can just translate the whole angle from P to Q. So the idea of a rotation determined by a remotely given angle seems to involve the concept of translation. But recall we haven't defined translation yet either.

    Here is a naive attempt at defining a translation: Let P≠Q be two distinct points of the plane and define the isometry T = "translation along the segment PQ" as follows: If X is any point not on the line PQ, construct the line L = XP. Then Q is not on L so is on one side of L or the other. Consider the unique line M parallel to PQ and containing X. There are exactly two points on this line whose distance from X equals d(P,Q), one on each side of XP. Define f(X) to be the one on the same side of XP as Q. If X is on the line PQ again there are exactly two points of this line whose distance from X is d(P,Q). We must say which one is f(X). We use the fact that both points P and X divides the line PQ into two non empty disjoint "sides". Thus we can speak of whether two points are or are not on "the same side" of P or of X. If X is on the same side of P as Q is, then take f(X) on the opposite side of X from P. If X is on the opposite side of P from Q, then take f(X) on the same side of X as P.

    The previous clumsy attempts at defining rotations and translations seem much too complicated to be useful, so we will take another approach, the one found in the standard books. I.e. we know from experience that there are at least three important and useful isometries, namely reflections, rotations, and translations. Only one of them, the reflection, is relatively easy to define precisely. Hence one way to proceed is to try to define rotations and translations in terms of reflections. Although this is somewhat unintuitive, we have seen the difficulties involved in trying to follow intuition too strictly. So first recall:

    Definition: A line L in the plane determines a unique reflection RL in the following way: given any point P on L, RL(P) = P. Given any point P not on L, Q = RL(P) if and only if L is the perpendicular bisector of the segment PQ.

    Remark: If P is not on L and if Q = RL(P), then Q lies on the unique line through P perpendicular to L, and on the opposite side of L from P, and at the same distance from L as P. In particular the only points left "fixed" by RL are the points of L. To construct RL(P) = Q, choose two points X,Y on L and construct the two circles centered at X and Y, and passing through P. Since these two circles meet once at P which is not on their line of centers, they meet exactly once more on the opposite side of L, at Q. That point Q is R(P).
    Note that RoR = id, for any reflection R.

    Theorem: Every reflection is an isometry of the plane.
    proof: Let A,B be any two points, and let X = R(A), Y = R(B). We must prove that d(A,B) = d(X,Y). If both A,B lie on L this is immediate, since then X = A, B = Y. If only one of the points say A lies on L, it is also easy, since then X = A, and the triangles AOB and AOY in the picture below are congruent by SAS, hence segments AB and AY are equal.

    If both A,B are off L, but on the same side, we generally have this picture:

    Here ACDX is a rectangle (AC and XD are both parallel to L by construction, and AX and BY are both perpendicular to L hence parallel to each other and perpendicular to AC and XD), hence triangles ACB and XDY are congruent by SAS. Thus d(A,B) = d(X,Y).
    Finally, if A,B are on opposite sides of L, we generally have the picture below:

    Here we have to prove that the picture really looks like this. I.e. if we draw the two segments AB and XY, how do we know they meet on L? To see this let's draw just the segment AB, and let O be the point where it meets L. Then connect X to O and Y to O, and try to prove that XOY is a straight angle. We know that angles AOL and BOQ are both equal say to ø, since they are opposite angles formed by two straight lines AB and L. We also know that triangle AOL is congruent to triangle XOP by SAS. Hence angle XOP also equals ø. Similarly, triangle YOQ is congruent to BOQ and angle YOQ also equals ø. Then since we know AOB is a straight angle, then angle XOB is supplementary to angle AOX, (i.e. they add up to 180). Since angles AOX and YOB both equal 2ø, they are equal. Hence XOB is also supplementary to YOB, i.e. XOY is a straight angle. Now by congruent triangles, we have segments AO and XO are equal, and also segments BO and YO are equal. Hence segments AB and XY are equal. QED.

    Exercise: Prove the cases of the theorem above when the segment AB is perpendicular to the line L.

    Now we want to characterize reflections by their fixed points. It follows from the two and three fixed point principles that an isometry ≠ identity has either one fixed point, or a fixed line.

    Theorem: If f is an isometry of the Euclidean plane with a fixed line L, and no other fixed points, then f is reflection in L.
    proof: Assume P is a point off L. We must show L is the perpendicular bisector of the segment Pf(P). Let Y,Z be any two points of L and draw the two circles F,G centered at Y,Z, and passing through P. These circles meet at P which is not on their line of their centers L. Hence by theorems from Euclidean geometry stated above they meet again at another point Q on the opposite side of L. By another Euclidean theorem stated above, the perpendicular bisector of the chord PQ contains the centers Y,Z of both circles F,G hence equals L. We claim Q = f(P). Since d(Y,P) = d(f(Y),f(P)) = d(Y,f(P)), f(P) must be on the circle F. Similarly f(P) is on the circle G. Thus f(P) is either P or Q, but P is not fixed by f so f(P) = Q. QED.


    Now we show how to define rotations in terms of reflections.
    Definition: A rotation is a composition of two reflections ® = RMoRL, where L and M are intersecting lines. If L = M, then ® = id. If L,M intersect in just one point P, we say the rotation is non trivial, and has center P.
    Note this deals with the problem of oriented rotations, without even mentioning it. I.e. suppose we have three non collinear points XPY:

    There should intuitively be two rotations about P determined by this picture, both with angle measure equal to the angle XPY. Namely one rotation "from PX to PY" or counterclockwise as it appears in the picture, and one "from PY to PX", which in this picture appears to be clockwise. How do these both arise from the new definition?

    We consider the two lines L = PX and M = PY. These determine two reflections RL and RM, and therefore they also determine two rotations depending on which order we compose them in. I.e. the two rotations ® = RLoRM, and ®’ = RMoRL, are (usually) different rotations, both centered at P, and both through angles of the same size, but the angles are oppositely oriented.

    If you try out composing the two reflections above, applying them say to a point on PX = L, or on PY = M, you will see that RMoRL is the rotation in the direction "from L to M", i.e. in the apparent counterclockwise direction, and that RLoRM is rotation in the opposite direction, i.e. "clockwise" in this picture. You will also verify that the rotation is not one through the angle XPY, but through twice that angle. Thus if we wanted to get a rotation centered at P, through the angle XPY, and oriented from PX to PY, we would first have to construct the angle bisector for that angle as below.

    Then if K is the line PZ (and L is the line PX), the desired rotation would be the composition RKoRL.

    Question: For what lines L ≠ M do we have RLoRM = RMoRL?

    Theorem: Any rotation ® = RLoRM is an isometry.
    proof: We know that the composition of two isometries is an isometry, and that a reflection is an isometry. QED.

    Now let's try to prove our "rotations" do have at least one property we expect from a rotation.
    Theorem: A non trivial rotation has exactly one fixed point.
    proof: If ® = RLoRM where L meets M only at P, then P is fixed by both RL and RM hence also by their composition. If Q ≠ P is any other point such that RLoRM(Q) = Q, then composing both sides with RL we get RM(Q) = RL(Q) = X. Since Q does not lie on both lines L,M it cannot be fixed by both RM and RL, so Q ≠ X. By definition of a reflection, since RL(Q) = X then L is the perpendicular bisector of the segment QX. But since RM(Q) = X, then M is also the perpendicular bisector of the segment QX. Hence L = M, and ® is actually the trivial rotation id = RLoRL. This contradiction proves there are no fixed points of ® other than P. QED.

    Theorem: An isometry with exactly one fixed point is a rotation.
    proof: If f fixes P but not A, and B = f(A), let L be the perpendicular bisector of segment AB. Then RL fixes P and takes B to A. Hence RLof fixes both P and A, hence is either the identity or reflection in the line M = PA. But if RLof = id, then RLoRLof = RL, and since RLoRL = id, we get f = RL, contradicting the assumption f has exactly one fixed point. Hence RLof = RM, and thus f = RLoRM. Since L,M meet at P, f is by definition a rotation. QED.

    Theorem: The inverse of a rotation is a rotation.
    proof: Since the inverse of an isometry ® fixes a point if and only if ® fixes it, the inverse of a non trivial rotation has exactly one fixed point, hence is also a rotation. A second proof: if ® = RLoRM, then it is easy to check by multiplying that ®-1 = RMoRL, also a rotation. QED.

    Theorem: If A,B are two points having the same distance from P, then there is a rotation ® about P taking A to B.
    proof: If APB is a straight angle, then let L be the line they span and M the line perpendicular to L and passing through P. Then the rotation RMoRL takes A to B. Otherwise draw the circle centered at P and passing through A and B, and construct the point X halfway between A and B on the smaller arc they span by bisecting the angle APB. Then X is the point where the bisector meets the circle.
    Now let L be the perpendicular bisector of chord AX and M the perpendicular bisector of chord XB. Then ® = RMoRL takes A to B, and both L, M pass through P. QED.

    Now we must finally pay the bill for taking this "easy way" of defining rotations. The rotation in the previous theorem should certainly be unique, but why? If two rotations ®, ®’ about P which agree at A ≠ P, surely ® and ®’ are equal, since ®-1o ®’ should be a rotation fixing both P and A, and since a non trivial rotation fixes only one point this should be the identity. But how do we know the composition of two rotations is a rotation?? Two rotations compose to give an isometry composed of four reflections! We must prove that this isometry can also be expressed as a composition of just two reflections. We do know at least that the composition of two rotations about P fixes P, so the composition must be either a rotation or a reflection. To finish off this proof we must show there is only one fix point. So we do not know yet that the rotations about a point P form a group, much less an abelian group. Do you have any ideas on how to do this?

    Theorem: Any isometry is a composition of ≤ three reflections.
    proof: If f is the identity then f = RoR where R is any refelction. If f has a fixed line then f is itself a reflection. If f has exactly one fixed point, then f is a rotation, hence a composition of two reflections. If f has no fixed points and f(A) = B ≠ A, let L be the perpendicular bisector of segment AB. Then RLof fixes A, hence equals a composition of one or two reflections. If RLof = RM, then f = RLoRM, while if RLof = RMoRK, then f = RLoRMoRK. QED.

    That tells us the composition of two rotations can be written as the composition of one, two, or three reflections. Hence a complete understanding of isometries will rest on a study of compositions of three reflections. We postpone that and instead define translations.

    Definition: A "translation" is a composition of two reflections, T = RLoRM where the lines L,M are parallel.

    Again note that RMoRL translates in the direction "from L to M", while RLoRM translates the same distance but in the opposite direction. Intuitively, the "length" of the translation, i.e. the distance a point moves, is twice the distance between the lines L,M.

    Theorem: A translation T ≠ id has no fixed points.
    proof: If T = RMoRL fixes A, then RMoRL(A) = A, so composing with RM, implies RL(A) = RM(A) = B. If A = B, then A is fixed by both RM and RL so A lies on both lines L, M. This is impossible since those lines are parallel, and different so A ≠ B. Then the fact that RL(A) = RM(A) = B, implies that the perpendicular bisector of the segment AB must be both L and M. Then L = M, and T = id, a contradiction. QED.

    Theorem: The inverse of a translation is a translation.
    proof: The inverse of T = RMoRL is T-1 = RLoRM. QED.

    Theorem: If A,B are any two points, there is a translation T taking A to B.
    proof: If A = B, let L = M, and then T = id works. If A ≠ B, let M be the perpendicular bisector of the segment AB, and let L be the line through A parallel to M. Then RMoRL works. QED.

    Although intuition tells us that the T in the previous theorem is unique, we do not know how to prove this yet. The key point is to show that a composition of two translations is a translation. Like rotations about one point, translations form a group, as does the set of all translations and all rotations taken together, but how to prove this? We can also ask whether translations are characterized by their fixed points, as reflections and rotations are; i.e. are there any other isometries with no fixed points? What do you think?
     
  15. May 20, 2008 #14
    i don't have a problem with translation, but with rotation, i mean i think if you compose rotation with glide reflection then you can have a fixed point, it depends on the rotation's radius and anlge of turn.

    with translation as i said you can't go back to the same point when composed with glide reflection.
     
  16. May 20, 2008 #15
    The first sentence is correct, but the second one is not: given a particular glide reflection, you can compose it with a specific translation so that the resulting transformation has an entire line of fixed points.

    You really have way more than enough information at this point to solve this problem. I would suggest that now is the time to sit down by yourself with a blank piece of paper and to work the problem out by yourself.
     
  17. May 20, 2008 #16
    but if the first sentence is correct, then by the theorem quoted by mathwonk, glide reflection and rotation can be composed by the same number of reflections, which we said a rotation can be composed by 2 reflections, but we want to show that glide reflection cannot be a composition of below 3 reflections.
     
  18. May 20, 2008 #17
    ok i see the answer, sorry for my thickness.
    glide reflection is a composition of translation and reflection and translation can be composed of two reflections, now if tranlation cannot be composed by two reflections im done.
    my problem is that i think i can represent translation by one reflection, i mean if we reflect a point by one vertical line then it's translated to the other point, is it not.

    for example let us first reflect a point through a vertical line and afterwards through a horizontal line, isn't this a glide reflection?
     
  19. May 20, 2008 #18

    mathwonk

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    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.
     
  20. May 21, 2008 #19
    in our first week we didn't discuss orientation, just today we spoke of this, ofcourse now it make sense.
     
  21. Sep 9, 2008 #20
    I am trying to prove that every rotation is an isometry using congruent triangles. How is the best way to go about that? Do I need to take 2 congruent triangles and rotate them around several times to 360 degrees and prove they are still congruent?

    Thanks
     
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