What possible value is there in teaching ring theory to high school students, particularly, if by doing so, topics vital to the preparation for calculus are abandoned?
Well, I was discussing the wording of the Common Core Standard, where the strange wording "the polynomials form a system analogous to the integers" appears. I found a commenter on another blog who had something to say about this particular standard, from four years ago:
http://tcmfa.blogspot.com/2011/03/april-6.html
With an eye towards my own high school experience, it's interesting to see a standard like A-APR.1: "Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials." Are students really being asked to see the underlying ring structure, or is it enough for them simply to demonstrate that addition, etc. of two polynomials will produce another? I'm sure I could have done the latter, but explicit, verbal knowledge of the "system" way of seeing things definitely came later.
I decided to check the practice PARCC exams for Algebra I and II to see whether any questions about ring theory actually appear on the PARCC. And, as I expected, there are no questions that ask about this "system analogous to the integers" on the PARCC (and I doubt that there are any such questions on the SBAC either) -- which is all that really matters. So standard A-APR.1 might as well have omitted all but the last five words of the standard:
CCSS.MATH.CONTENT.HSA.APR.A.1
[A]dd, subtract, and multiply polynomials.
The polynomial questions that I saw on that practice PARCC were on the difficult side, and so I definitely feel that some of these questions are flawed. But fortunately, our fears that the Common Core would expect students to master ring theory are unfounded, and we don't have to worry about ring theory taking time away from topics that prepare students
I only brought up ring theory because I wanted to tie the Common Core Standards to the Google Doodle celebrating the birthday of mathematician Emmy Noether, and since it was during a spring break post, I wanted to discuss topics that aren't directly related to what should be taught in an actual high school classroom.
And that takes us into our next spring break topic. I mentioned in the last post that Emmy Noether studied symmetry -- including translational and rotational symmetry -- as applied to physics. We already discussed translations and rotations for Common Core Geometry, but those transformations applied to the two-dimensional plane. The physical world has three dimensions, and so it's the 3D transformations that actually apply to physics.
Section 9-5 of the U of Chicago text is on reflections in space. I find this to be an interesting topic, but since it doesn't appear on the PARCC, it would be a waste of time to cover it in class. There will be no worksheet for this lesson, since it's not intended to be taught in class. This is why I waited until spring break to blog about this topic.
The text begins by defining what it means for a plane in 3D, rather than a line in 2D, to be a perpendicular bisector:
"In general, a plane M is the perpendicular bisector of a segment
Now we can definition 3D reflections almost exactly the same way we define 2D reflections:
"For a point A which is not on a plane M, the reflection image of A over M is the point B if and only if M is the perpendicular bisector of
If you think about it, when you (a 3D figure!) look at yourself in a mirror, the mirror itself isn't a line, but rather a plane. Mirrors in 2D are lines, while mirrors in 3D are planes. So now we can define what it means for two 3D figures to be congruent:
"Two figures F and G in space are congruent figures if and only if G is the image of F under a reflection or composite of reflections."
Most texts don't actually define what it means for two 3D figures to be congruent. We know that the traditional textbook definition, that congruent figures have corresponding sides and angles congruent, only applies to polygons. It doesn't even apply to circles, much less 3D figures. But we can simply use the Common Core definition -- two figures are congruent if and only if there exists an isometry (i.e., a composite of reflections) mapping one to the other -- and it instantly applies to all figures, polygons, circles, and 3D figures.
In 2D there are only four isometries -- reflections, translations, rotations, and glide reflections. An interesting question is, how many isometries are there in 3D?
Well, for starters, translations and rotations exist in 3D. We can define both of these exactly the same way that we do in 2D -- a translation is the composite of two reflections in parallel planes, while a rotation is the composite of two reflections in intersecting planes.
Notice that every 2D rotation has a center -- the point of intersection of the reflecting lines. The same thing happens in 3D, except that the intersection of two planes isn't a point, but a line. Therefore, a 3D rotation has an entire line as its center -- every point on this line is a fixed point of the rotation. But usually, instead of calling the line the center of the rotation, we call it the axis of the rotation. One 3D object that famously rotates is the earth, and this rotation has an axis -- the line that passes through the North and South Poles. Confusingly, the mirror of a 2D reflection is often called an axis -- but in some ways, these two definitions are related. One can perform a 2D reflection by taking a 2D figure and rotating it 180 degrees about the axis in 3D.
Glide reflections also exist in 3D -- although these are often called glide planes in 3D. A glide reflection is the composite of a reflection and a nontrivial translation parallel to the mirror. Notice that there are infinitely many directions to choose from for our translation in 3D, whereas if this were a 2D glide reflection there are only two possible directions for a translation that's parallel to the mirror.
But are there any other isometries in 3D? Well, we notice that a glide reflection is the composite of two other known isometries, a reflection and a translation. So the next natural possibility to consider is, what if we find the composite of the other two combinations? What is the composite of a reflection and a rotation, or a translation and a rotation?
In 2D, the composite of a translation and a nontrivial rotation is another rotation. This is possible to prove, as follows: let T be a translation and R be a nontrivial rotation, and G be the composite of T following R. Both translations and rotations preserve orientation, and so their composite G must preserve orientation as well. In 2D, three mirrors suffice -- that is, every isometry is the composite of at most three reflections. Since G preserves orientation, it must be the composite of an even number of rotations. Therefore G is the composite of two reflections (or the identity transformation -- the transformation that maps every point to itself), and so G is either a translation or a rotation.
Let's try an indirect proof -- assume that G is a translation. That is:
T o R = G
Since T is a translation, it has a translation vector t, and as we're assuming that G is a translation, it must also have a translation vector g. Now let U be the inverse translation of T -- that is, the translation whose vector is -t, the additive inverse of t. We now compose U on both sides:
U o T o R = U o G
Now since U and T are inverses, U o T must be I, the identity transformation:
I o R = U o G
Since I is the identity transformation, I o R must be R:
R = U o G
Notice that G and U are both translations, whose vectors are g and -t, respectively. Then their composite must be another translation V whose vector is g - t. So now we have:
R = V
that is, a rotation equals a translation. Now no rotation can equal a translation (as the former has a fixed point, while the latter has no fixed point) unless both are the identity -- which contradicts the assumption that R is a nontrivial rotation (i.e., not the identity). Therefore, the composite of a translation and a nontrivial rotation isn't a translation, so it must be a rotation. QED
But this proof is invalid in 3D. This is because the proof uses a step that only works in 2D -- namely that three mirrors suffice. We must show how many mirrors suffice in 3D.
Let's recall why mirrors suffice in 2D. Let G be any 2D isometry, and let A, B, and C be three noncollinear points whose images under G are A', B', and C'. The first mirror maps A to A', the second mirror fixes A' and maps B to B'. It could be that the image of C under both mirrors is already C', otherwise a third mirror maps it to C'. Notice that the proof of the existence of these mirrors is nontrivial and depends on theorems such as the Perpendicular Bisector Theorem, since reflections are defined using perpendicular bisectors.
As it turns out, four mirrors suffice in 3D. To prove this, we let G be any 3D isometry, and let A, B, C, and D be four noncoplanar points. The first mirror maps A to A', the second mirror fixes A' and maps B to B', the third mirror fixes A' and B' and maps C to C', and the fourth, if necessary, fixes A', B', and C' and maps D to D'.
And so this opens the door for there to be a new transformation in 3D, one that is the composite of a translation and a rotation as well as the composite of four reflections. We can imagine twisting an object like a screw. A screwdriver rotates the screw about its axis, but then it's being translated into the wall in the same direction as that axis. And because of this, this new transformation is often called a screw motion.
We still have one last combination, the composite of a reflection and a rotation. It is subtle why the composite of a reflection and a rotation in 2D is usually a glide reflection -- why should the composite of a reflection and a rotation equal the composite of a (different) reflection and a translation? And it's even subtler why the composite of a reflection and a rotation may be a new transformation in 3D.
But the simplest example of this roto-reflection is the inversion map. In 3D coordinates, we map the point (x, y, z) to its opposite point (-x, -y, -z). This map is the composite of three reflections -- the mirrors are the three coordinate planes (xy, xz, and yz). As the composite of an odd number of reflections, it must reverse orientation. Yet it can't be a reflection, since it has only a single fixed point (0, 0, 0) and not an entire plane. Similarly, it can't be a glide plane because glide planes don't have any fixed points at all.
Roto-reflections are formed when the axis of the rotation intersects the reflecting plane in a single point -- and of course, this single point is the only fixed point of the roto-reflection.
So now we have six isometries in 3D -- reflections, translations, rotations, glide planes, screw motions, and roto-reflections. Are there any others? As it turns out, these six are all of them -- and the proof depends on the fact that four mirrors suffice in 3D.
Returning to the U of Chicago text, we have the definition of a reflection-symmetric figure:
"A space figure is F is a reflection-symmetric figure if and only if there is a plane M such that the reflection of F in M is F."
Similarly, a figure can be rotation-symmetric, as well as roto-reflection-symmetric. In the text, figures such as the right cylinder have reflection, rotation, and roto-reflection symmetry.
But a figure can't have translation symmetry unless it's infinite, as translations lack fixed points. So likewise, figures that have glide reflection or screw symmetry must also be infinite, as these transformations are based on translations. The translational symmetry mentioned by Noether refers to infinite space.
Of course, there exist dilations in 3D space as well. There is very little discussion of similarity in 3D, except to compare the surface areas and volumes of 3D figures.
Thus concludes my spring break post. I return to posting for our Common Core Geometry course on Monday, March 30th, which will be Day 137 on our calendar. The topic will be the surface areas of pyramids and cones, from Section 10-2 of the U of Chicago text.
No comments:
Post a Comment