The taxicab norm on is defined by setting where Show that this is indeed a norm. As usual, we can define a distance function by setting Find 2 non-congruent, non-degenerate triangles with sides of length 1, 1, and 2.

This exercise is related to an 80-year old open problem of Paul Erdös. Consider a unit square . Inscribe in exactly squares with no common interior point. (The squares do not need to cover all of .) Denote by the side lengths of these squares, and define

Show that , and that equality holds iff is a perfect square.

(Erdös problem is to find all for which . Currently, it is only known that for all , , , and that if , then is a perfect square.)

(This exercise comes from Jerry Shurman, Geometry of the quintic, Wiley-Interscience, 1997.)A rigid motion of is a function such that

for all vectors (Here, is the usual inner product on .) The goal of this exercise is to show that these maps are precisely the functions of the form where is an orthogonal matrix and .

By definition, a matrix with real entries is orthogonal iff where is the transpose of . Show that this is equivalent to stating that preserves inner products, in the sense that

for all .

Show that if where is orthogonal, then is indeed a rigid motion.

To prove the converse, assume now that is rigid.

Show first that is a bijection.

Let and set Eventually, we want to show that is a linear transformation. Begin by checking that and that for all .

Show that for all , and conclude that preserves addition.

Argue similarly that preserves scalar multiplication, and is therefore linear.

It follows that for some matrix . Conclude the proof by checking that is indeed orthogonal.

Solve exercises 2-7, 10-14 from Chapter 6 of the textbook.

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Dr. Caicedo,
On question #1, v_1 and v_1 are vectors right? The sum of their magnitudes gives another vector’s length… bold v. I’m just verifying that they’re not scalars…

Also, #2: the max of a sum… does that mean that you take the largest value for the possible sums?

1. What I meant was is a vector in It therefore has two coordinates (with respect to the standard basis), and I am calling the first and the second (So, yes, are scalars.)

2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.

We have not defined “non-congruent” yet. Are we talking about using the standard distance measure and using trig to show they have different angle measures?

Let’s be as generous as it is reasonable, and take congruent to mean same lengths, angles, areas. What I mean is, we want two “obviously different” triangles whose sides have the same lengths nonetheless. (And non-degenerate is intended to eliminate the example of a segment with a point in the middle.)

Dr. Caicedo,
In the assigned problem 3 #5, To show A is orthogonal, do we prove A^T(A)=I? If so, these two matrices can get very messy with 3X3 real entries… is there another way?

On Assigned problem 3 #1, what is required to show 1 to 1 and onto here?
Thanks,
Amy

Hi Amy,
For 3.5, do you have another characterization of orthogonality? (Perhaps one proved earlier somewhere in problem 3?)
I am not sure I understand your question about 3.1. Can you tell me what you mean?

Hi, I am not sure how to show R is a bijection. I suppose I am still unclear what a rigid motion is. I keep going through the motions of showing one thing after another, but am not sure what the big picture is for R.
Thanks,
Amy

The definition of being rigid is the equation at the beginning, for all

What do you get if ? This says that preserves distances. It implies (very easily) that is 1-1 and that it is continuous.

You may remember from vector calculus that where is the angle between the vectors and . Since preserves distances, the equation defining rigid motions then says that the angle between and is the same as the angle between and

In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of with those two properties.

The most direct argument for surjectivity that I have thought of is as follows; there are several claims I make along the way that need to be justified, of course, but this is the skeleton:

Let we want to find some such that Pick points in the range of say

Consider the tetrahedron There are only two tetrahedrons in that are congruent to and have corresponding to Say, and Then must be the image of either or

A better layout of the exercise would have been to show first that preserves inner products and then show that bijects. To show that surjects one could also work in coordinates:

Let be the standard orthonormal basis in Since for all with and then is an orthonormal basis.

Next, for any vector and

So to make hit any point, express the point as and then let Then

Maybe this immediately shows that is linear, further shortening the exercise?

Very neat solution. The solution I suggested above and the one below depend a bit too much on geometric ideas that don’t seem relevant to the problem, so I like your approach much better.

Here is another way of proving surjectivity. It was suggested by my friend and colleague Ramiro de la Vega. I think that Dr. Shurman‘s solution above is the best one, but all the solutions we have exhibit different ideas which may prove useful in different contexts.

For this approach, first prove that the image under of a straight line is a straight line. This is the key observation.

Surjectivity is now easy: Show now that the image under of a plane is a plane. For this, consider three noncollinear points in a plane and note that is the union of the lines that go through one of these three points and the opposite side in the triangle they determine.

Finally, consider a tetrahedron, and note that any point in space is in one of the lines going through a vertex and the opposite face.

The only reference I know for precisely these matters is the handbook chapter MR2768702. Koellner, Peter; Woodin, W. Hugh. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, 1951–2119, Springer, Dordrecht, 2010. (Particularly, section 7.) For closely related topics, see also the work of Yong Cheng (and of Cheng and Schindler) on Harr […]

As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are related in deep ways, actually, through large cardinals. (Under enough large cardinals, $L({\mathbb R})$ of $V$ is a model of determinacy and (something stronge […]

Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory. How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$? Really, even pointers on how to say anything […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

Very briefly: Yes, there are several programs being developed that can be understood as pursuing new axioms for set theory. For the question itself of whether pursuing new axioms is a reasonably line of inquiry, see the following (in particular, the paper by John Steel): MR1814122 (2002a:03007). Feferman, Solomon; Friedman, Harvey M.; Maddy, Penelope; Steel, […]

This is a very interesting question and the subject of current research in set theory. There are, however, some caveats. Say that a set of reals is $\aleph_1$-dense if and only if it meets each interval in exactly $\aleph_1$-many points. It is easy to see that such sets exist, have size $\aleph_1$, and in fact, if $A$ is $\aleph_1$-dense, then between any tw […]

Say that the triangle is $ABC$. The vector giving the median from $A$ to $BC$ is $(AC+AB)/2$. Similarly, the one from $B$ to $AC$ is $(BA+BC)/2$, and the one from $C$ to $BA$ is $(CB+CA)/2$. Adding these, we get zero since $CB=-BC$, etc.

The usual definition of a series of nonnegative terms is as the supremum of the sums over finite subsets of the index set, $$\sum_{i\in I} x_i=\sup\biggl\{\sum_{j\in J}x_j:J\subseteq I\mbox{ is finite}\biggr\}.$$ (Note this definition does not quite work in general for series of positive and negative terms.) The point then is that is $a< x

Dr. Caicedo,

On question #1, v_1 and v_1 are vectors right? The sum of their magnitudes gives another vector’s length… bold v. I’m just verifying that they’re not scalars…

Also, #2: the max of a sum… does that mean that you take the largest value for the possible sums?

I apologize, I meant v_1 and v_2 on my first question.

Thanks,

Amy

Hi Amy,

1. What I meant was is a vector in It therefore has two coordinates (with respect to the standard basis), and I am calling the first and the second (So, yes, are scalars.)

2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.

Hey Dr. Caicedo,

I think you meant to say in your last comment that .

Nick

[*Sigh* Yes, thanks. Corrected. -A]We have not defined “non-congruent” yet. Are we talking about using the standard distance measure and using trig to show they have different angle measures?

Let’s be as generous as it is reasonable, and take congruent to mean same lengths, angles, areas. What I mean is, we want two “obviously different” triangles whose sides have the same lengths nonetheless. (And

non-degenerateis intended to eliminate the example of a segment with a point in the middle.)Dr. Caicedo,

In the assigned problem 3 #5, To show A is orthogonal, do we prove A^T(A)=I? If so, these two matrices can get very messy with 3X3 real entries… is there another way?

On Assigned problem 3 #1, what is required to show 1 to 1 and onto here?

Thanks,

Amy

Hi Amy,

For 3.5, do you have another characterization of orthogonality? (Perhaps one proved earlier somewhere in problem 3?)

I am not sure I understand your question about 3.1. Can you tell me what you mean?

Hi, I am not sure how to show R is a bijection. I suppose I am still unclear what a rigid motion is. I keep going through the motions of showing one thing after another, but am not sure what the big picture is for R.

Thanks,

Amy

The definition of being rigid is the equation at the beginning, for all

What do you get if ? This says that preserves distances. It implies (very easily) that is 1-1 and that it is continuous.

You may remember from vector calculus that where is the angle between the vectors and . Since preserves distances, the equation defining rigid motions then says that the angle between and is the same as the angle between and

In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of with those two properties.

I’ll write about surjectivity a bit later.

The most direct argument for surjectivity that I have thought of is as follows; there are several claims I make along the way that need to be justified, of course, but this is the skeleton:

Let we want to find some such that Pick points in the range of say

Consider the tetrahedron There are only two tetrahedrons in that are congruent to and have corresponding to Say, and Then must be the image of either or

A better layout of the exercise would have been to show first that preserves inner products and then show that bijects. To show that surjects one could also work in coordinates:

Let be the standard orthonormal basis in Since for all with and then is an orthonormal basis.

Next, for any vector and

So to make hit any point, express the point as and then let Then

Maybe this immediately shows that is linear, further shortening the exercise?

HTML ate my inner product symbols.

[Edited: I fixed it. Thanks! -A.]Hi Jerry,

Thanks a lot!

Very neat solution. The solution I suggested above and the one below depend a bit too much on geometric ideas that don’t seem relevant to the problem, so I like your approach much better.

There are a few statements that probably need more details (if this were to be turned in as homework), but this is really an effective proof.

Thanks for sharing.

Here is another way of proving surjectivity. It was suggested by my friend and colleague Ramiro de la Vega. I think that Dr. Shurman‘s solution above is the best one, but all the solutions we have exhibit different ideas which may prove useful in different contexts.

For this approach, first prove that the image under of a straight line is a straight line. This is the key observation.

Surjectivity is now easy: Show now that the image under of a plane is a plane. For this, consider three noncollinear points in a plane and note that is the union of the lines that go through one of these three points and the opposite side in the triangle they determine.

Finally, consider a tetrahedron, and note that any point in space is in one of the lines going through a vertex and the opposite face.