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.

Typeset using LaTeX2WP. Here is a printable version of this post.

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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
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 answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

A simple example is the permutation $\pi$ given by $\pi(n)=n+2$ if $n$ is even, $\pi(1)=0$, and otherwise $\pi(n)=n−2$. It should be clear that $\pi$ is computable and has the desired property. By the way, regarding the footnote: if a bijection is computable, so is its inverse, so $\pi^{-1}$ is computable as well. In general, given a computable bijection $\s […]

The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may a […]

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All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

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

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.