Rolling Circles and Balls (Part 3)

In Part 1 and Part 2 we looked at the delightful curves you get by rolling one circle on another. Now let’s see what happens when you roll one circle inside another!

Four times as big

If you roll a circle inside a circle that’s 4 times as big, we get an astroid:

Puzzle 1. How many times does the rolling circle turn as it rolls all the way around?

By the way: don’t confuse an astroid with an asteroid. They both got their names because someone thought they looked like stars, but that’s where resemblance ends!

You can get an astroid using this funny parody of the equation for a circle:

x^{2/3} + y^{2/3} = 1

Or, if you don’t like equations, you can get a quarter of an astroid by letting a ladder slide down a wall and taking a time-lapse photo!

In other words, you get a whole astroid by taking the envelope of all line segments of length 1 going from some point on the x axis to some point on the y axis!

Three times as big

If the fixed circle is just 3 times as big as the one rolling inside it, we get an deltoid:

Puzzle 2. Now how many times does the rolling circle turn as it rolls all the way around?

By the way: it looks like we’re back to naming curves after body parts… but we’re not: both this curve and the muscle called a deltoid got their names because they look like the Greek letter delta:

Puzzle 3. Did the Greek letter delta get that name because it was shaped like a river delta, or was it the other way around?

As you might almost expect by now, if you’ve been reading this whole series, there are weird relations between the deltoid and the astroid.

For example: take a deltoid and shine parallel rays of light at it from any direction. Then the envelope of these rays is an astroid!


We summarize this by saying that the astroid is a catacaustic of the deltoid. This picture is by Xah Lee, who has also made a nice movie of what happens as you rotate the light source:

• Xah Lee, Deltoid catacaustic movie.

I don’t completelly understand the rays going through the deltoid, in either the picture or the movie. It looks like those rays are getting refracted, but that would be a diacaustic, not a catacaustic. ,I think they’re formed by continuing reflected rays to straight lines that go through the deltoid. If you didn’t do that you wouldn’t get a whole astroid, just part of one.

Anyway, at the very least you get part of an astroid, which you can complete to a whole one. And then, as you rotate the light source, the astroid you get rolls around the deltoid in a pleasant manner! This is nicely illustrated here:

• Deltoid catacaustic, Wolfram Mathworld.

You can also get a deltoid from a deltoid! Draw all the osculating circles of the deltoid—that is, circles that match the deltoid’s curvature as well as its slope at the points they touch. The centers of these circles lie on another, larger deltoid:

We summarize this by saying that the evolute of a deltoid is another deltoid.

There are also fancier ways to get deltoids if you know more math. For example, the set of traces of matrices lying in the group SU(3) forms a filled-in deltoid in the complex plane!

This raises a question which unfortunately I’m too lazy to answer myself. So, I’ll just pose it as a puzzle:

Puzzle 4. Is the set of traces of matrices lying in SU(4) a filled-in astroid? In simpler terms, consider the values of a + b + c + d that we can get from complex numbers a,b,c,d with |a| = |b| = |c| = |d| = 1 and abcd = 1. Do these values form a filled-in astroid?

Twice as big

But now comes the climax of today’s story: what happens when we let a circle roll inside a circle that’s exactly twice as big?

Now you can see the rolling circle turn around once as it rolls all the way around the big one.

More excitingly, if we track a point on the rolling circle, it traces out a straight line! Can you see why? Later I’ll give a few proofs.

This gadget is called a Tusi couple. You could use it to convert a rolling motion into a vibrating one using some gears. Greg Egan made a nice animation showing how:

The Tusi couple is named after the Persian astronomer and mathematician Nasir al-Din al-Tusi, who discovered it around 1247, when he wrote a commentary on Ptomely’s Almagest, an important astronomical text:

He wrote:

If two coplanar circles, the diameter of one of which is equal to half the diameter of the other, are taken to be internally tangent at a point, and if a point is taken on the smaller circle—and let it be at the point of tangency—and if the two circles move with simple motions in opposite direction in such a way that the motion of the smaller is twice that of the larger so the smaller completes two rotations for each rotation of the larger, then that point will be seen to move on the diameter of the larger circle that initially passes through the point of tangency, oscillating between the endpoints.

I don’t quite understand why he was interested in this, but it has something to do with using epicycles to build linear motion out of circular motion. It also has something to do with the apparent motion of planets between the Earth and the Sun.

Later Copernicus also studied the Tusi couple. He proved that the moving point really did trace out a straight line:

However, many suspect that this was not a true rediscovery: al-Tusi had also proved this, and some aspects of Copernicus’ proof seem too similar to al-Tusi’s to be coincidence:

• George Saaliba, Whose science is Arabic science in Renaissance Europe?, Section 2: Arabic/Islamic science and the Renaissance science in Italy.

• I. N. Veselovsky, Copernicus and Nasir al-Din al-Tusi, Journal for the History of Astronomy 4 (1973), 128–130.

In fact, the Tusi couple goes back way before al-Tusi. It was known to Proclus back around 450 AD! Apparently he wrote about it in his Commentary on the First Book of Euclid. Proclus is mainly famous as a philosopher: people think of him as bringing neo-Platonism to its most refined heights. Given that, it’s no surprise that he also liked math. Indeed, he said:

Wherever there is number, there is beauty.

And of course this is what I’ve been trying to show you throughout this series.

Why the Tusi couple works

So, here are three proofs that a Tusi couple really does trace out a straight line. I posed this as a puzzle on Google+, and here are my favorite three answers. I like them because they’re very different. Since people have thought about Tusi couples since 450 AD, I doubt any of these proofs are original to the people I’m mentioning here! Still, they deserve credit.

The first, due to Omar Antolín Camarena, is in the style of traditional Euclidean geometry.

Let O be the center of the big circle. Let A be the position of the traced point at the instant t when it’s the point of tangency between the circles, let B be its position at some future time t′ and let C be the point of tangency at that same time t′. Let X be the center of the small circle at that future time t′.

We want to prove A, B and O lie on a line. To do this, it suffices to show that the angle AOC equals the angle BOC:

Want: ∠BOC = ∠AOC

The arc AC of the big circle has the same length as the arc BC of the small circle, since they are both the distance rolled between times t and t′. But the big circle is twice as, so the angle BXC on the little circle must be twice the angle AOC on the big circle:

Know: ∠BXC = 2 ∠AOC

But it’s a famous fact in Euclidean geometry that the angle BXC is twice the angle BOC:

Know: ∠BXC = 2 ∠BOC

From the two equations we know, the one we want follows!

Here’s the proof of that ‘famous fact’, in case you forgot it:

The second proof is due to Greg Egan. He distilled it down to a moving picture:

It’s a ‘proof without words’, so you may need to think a while to see how it shows that the Tusi couple traces out a straight line.

The third proof is due to Boris Borcic. It uses complex numbers. Let the big circle be the unit circle in the complex plane. Then the point of contact between the rolling circle and the big one is:

e^{i t}

so the center of the rolling circle is:

\displaystyle{ \frac{e^{it}}{2} }

Since the rolling circle turns around once clockwise as it rolls around the big one, the point whose motion we’re tracking here:

is equal to:

\displaystyle{ \frac{e^{it}}{2} + \frac{e^{i(\pi - t)}}{2} = \frac{e^{i t} - e^{-i t}}{2} = i \sin t = i \; \mathrm{Im}(e^{it}) }

So, this point moves up and down along a vertical line, and its height equals the height of the point of contact, e^{it}.

But the same sort of argument shows that if we track the motion of the opposite point on the rolling circle, it
equals:

\displaystyle{ \frac{e^{it}}{2} - \frac{e^{i(\pi - t)}}{2} = \frac{e^{i t} + e^{-i t}}{2} = \cos t = \mathrm{Re}(e^{it}) }

So, this opposite point moves back and forth along a horizontal straight line… and its horizontal coordinate equals that of the point of contact!

You can see all this clearly in the animation Borcic made:

The point of contact, the tip of red arrowhead, is e^{it}. The two opposite points on the rolling circle are \cos t and i \sin t. So, the rectangle here illustrates the fact that

e^{it} = \cos t + i \sin t

It’s interesting that this famous formula is hiding in the math of the Tusi couple! But it shouldn’t be surprising, because the Tusi couple is all about building linear motion out of circular motion… or conversely, decomposing a circular motion into linear motions.

Credits

Few of these pictures were made by me, and none of the animations. For most, you can see who created them by clicking on them. The animation of the astroid and deltoid as envelopes come from here:

• Envelope, Math Images Project.

where they were made available under a GNU Free Documentation License. Here’s another animation from there:

This is a way of creating a deltoid as the envelope of some lines called ‘Wallace-Simson lines’. The definition of these lines is so baroque I didn’t dare tell it to you it earlier. But if you’re so bored you’re actually reading these credits, you might enjoy this.

Any triangle can be circumscribed by a circle. If we take any point on this circle, say M, we can drop perpendicular
lines from it to the triangle’s three sides, and get three points, say P, Q and R:

Amazingly, these points lie on a line:

This is the Wallace–Simson line of M. If we move the point M around the circle, we get lots of Wallace–Simson lines… and the envelope of these lines is a deltoid!

By the way: the Wallace–Simson line is named after William Wallace, who wrote about it, and Robert Simson, who didn’t. Don’t confuse it with the Wallace line! That was discovered by Alfred Russel Wallace.

This entry was posted on Tuesday, September 11th, 2012 at 9:18 am and is filed under astronomy, mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

70 Responses to Rolling Circles and Balls (Part 3)

  1. anitachowdry says:
    11 September, 2012 at 11:15 am

    Fascinating and informative, and I love all the different references. I really enjoy opening up my e-mails and finding your articles.

    Reply
  2. Peter Grunberg says:
    11 September, 2012 at 11:59 am

    Thank you for another dazzling and clearly presented post. When I was a teenager in Australia, it was the senior maths teacher at our high school, a maniac for arcane but spectacular Euclidean geometry, who first brought the class’s attention to the existence of the Wallace-Simson line. In the absence of Wikipedia I spent several confused days wondering whether Edward VIII’s beloved Wallis Simpson (later the Duchess of Windsor) was an amateur geometer.

    Reply
  3. Greg Egan says:
    11 September, 2012 at 12:53 pm

    I haven’t proved Puzzle 4, but here’s an animation that convinces me it’s true:

    The circling red dot is one of the three independent phases. Each of the coloured lines is due to varying the second phase, while the progression from line to line such that they sweep out a filled-in deltoid is due to varying the third phase.

    So the filled-in deltoid is what you get as the trace when you fix one of the phases, and as you vary that phase, the deltoid sweeps out an astroid!

    Reply
  4. Greg Egan says:
    11 September, 2012 at 2:11 pm

    As one tiny step towards proving Puzzle 4, I can show that if you fix two of the independent phases, the third one will make the trace oscillate back and forth along a straight line.

    As Tobias said, if the independent phases are e^{i \alpha}, e^{i \beta} and e^{i \gamma}, the trace is:

    T = e^{-i (\alpha +\beta +\gamma )}+e^{i \alpha }+e^{i \beta }+e^{i \gamma }

    If we fix \alpha and \beta, we have:

    T (1+e^{i (\alpha +\beta )}) \\ = e^{-i (\alpha +\beta +\gamma )}+e^{i (\alpha +\beta +\gamma )}+e^{i (2 \alpha +\beta )}+e^{i (\alpha +2 \beta )}+e^{i \alpha }+e^{i \beta }+e^{-i \gamma }+e^{i \gamma } \\ = 2\cos(\alpha +\beta +\gamma )+e^{i (2 \alpha +\beta )}+e^{i (\alpha +2 \beta )}+e^{i \alpha }+e^{i \beta }+2\cos(\gamma)

    Since the imaginary part of this is independent of \gamma, as \gamma varies the overall value just moves back and forth along a horizontal line, and so the trace itself (with the other phases fixed) must simply move back and forth along some line.

    Reply
  5. Greg Egan says:
    11 September, 2012 at 3:26 pm

    It’s not hard to show that as you vary \gamma with the other two phases fixed, the trace moves along a line segment that, if extended, would intercept the real axis at 2(\cos \alpha + \cos \beta) and the imaginary axis at 2(\sin \alpha + \sin \beta) i.

    The distance between the real and imaginary intercepts is:

    \sqrt{8(1+\cos(\alpha-\beta))}

    which is at most 4. If we focus on the case where \alpha=\beta and so the distance is exactly 4, we get something like the “sliding ladder” construction of the astroid.

    Reply
  6. John Baez says:
    11 September, 2012 at 4:17 pm

    If you have a matrix in \mathrm{SU}(n+1), its eigenvalues are n+1 unit complex numbers that multiply to 1, and its trace is the sum of these numbers. We can take n of them to e^{i \theta}, and then remaining one has to be e^{-i n \theta}. Then their sum is

    n e^{i \theta} + e^{-i n \theta}

    and I bet this is precisely the curve traced out by a circle of radius 1 rolling inside a circle of radius n+1. This sort of curve is called a hypocycloid, and the formula for it looks right.

    Then we just need to show that any point on a line segment between 0 and

    n e^{i \theta} + e^{-i n \theta}

    is also a possible trace of a matrix in \mathrm{SU}(n+1), and we’ll know that the set of possible traces is a filled-in hypocycloid with n+1 cusps.

    Reply
    • John Baez says:
      11 September, 2012 at 5:08 pm

      Yes, if you look at this you can see it’s

      3 e^{i \theta} + e^{-3 i \theta}

      All the 3’s are precisely what’s needed to make this shape with 4-fold symmetry! You can see the center of the little circle is moving around a circle that’s 3 times as big, and the little circle is rotating with 3 times the angular speed as its center is revolving—in the opposite direction (that gives the minus sign).

      Reply
  7. Greg Egan says:
    11 September, 2012 at 4:21 pm

    Here’s the SU(5) version:

    The five-pointed outline is the curve you get by setting all four independent phases to the same value. To keep the image from getting too cluttered, I’ve only shown lines within the rolling astroid where two of the phases are identical, giving rise to “sliding ladders”.

    Reply
    • John Baez says:
      11 September, 2012 at 5:20 pm

      So, combining everything said so far, we’re seeing very clearly how the traces of matrices in \mathrm{SU}(n) are confined to lie inside the hypocycloid with n cusps… and we can use this idea to get an hypocycloid with n-1 cusps to roll inside an epicycloid with n cusps.

      So we could have a Tusi couple rolling inside a deltoid rolling inside an astroid rolling inside the 5-cusped hypocycloid you show here, ad infinitum… and all this would be related to the groups \mathrm{SU}(2), \mathrm{SU}(3), \mathrm{SU}(4), and so on. Epicycles within epicycles!

      Reply
      • Shanth says:
        12 September, 2012 at 4:25 pm

        Following along what’s been said, it’s easy to show that :

        \mathrm{Tr[SU}(n+1)] = n e^{i\theta} \mathrm{Tr[SU}(n)] + e^{-in\theta}

        where \mathrm{Tr[SU}(n)] is all the possible values the trace of an \mathrm{SU}(n) matrix could have. The really interesting observation is that the expression n e^{i\theta} z + e^{-in\theta} is the interior of the appropriate hypocycloid if z is in the interior of the polygon formed by the n-th roots of unity. All values of z on the boundary of this polygon will generate the hypocycloid, and for z in the interior of this polygon we will cover the interior of the polygon.

        Now, of course taking z \in [0,1] is sufficient to generate the interior of the relevant hypocycloid, after all two parameters are all we need. But since instead of the line segment we take a 2-d region, it would seem that the polygon would be the natural choice, but the region appearing SU(n+1) trace seems to be the hypocycloid inscribed in that polygon. I wonder what’s going on in that excluded area and if anyone here has something interesting that to say about this.

        Oh, and those animations by Greg Egan are just lovely!

        Reply
      • Greg Egan says:
        12 September, 2012 at 11:14 pm

        Shanth wrote:

        \mathrm{Tr[SU}(n+1)] = n e^{i\theta} \mathrm{Tr[SU}(n)] + e^{-in\theta}

        That’s a great inductive formula! But I think there’s a tiny mistake in the scaling, and it ought to be:

        \mathrm{Tr[SU}(n+1)] = e^{i\theta} \mathrm{Tr[SU}(n)] + e^{-in\theta}

        If the n eigenvalues of a diagonalised element of \mathrm{SU}(n) are:

        e^{i\phi_1}, ... , e^{i\phi_{n-1}}, e^{-i(\phi_1+\phi_2+...+\phi_{n-1})}

        then we can get a set of n+1 eigenvalues of a diagonalised element of \mathrm{SU}(n+1) as:

        e^{i\theta}e^{i\phi_1}, ... , e^{i\theta}e^{i\phi_{n-1}}, e^{i\theta}e^{-i(\phi_1+\phi_2+...+\phi_{n-1})}, e^{-in\theta}

        Of course you do need the factor of n if you want to map things on the scale of the unit circle to the set \mathrm{Tr[SU}(n+1)], with:

        z \to n e^{i\theta} z + e^{-in\theta}

        Reply
      • Greg Egan says:
        13 September, 2012 at 12:44 am

        Shanth wrote:

        All values of z on the boundary of this polygon [formed by the n-th roots of unity] will generate the hypocycloid …

        I don’t think that’s correct. Under the map:

        \theta \to n e^{i\theta} z + e^{-in\theta}

        I think z would need to be on the boundary of the n-cusped hypocycloid inscribed in the unit circle, in order for this map to generate the (n+1)-cusped hypocycloid inscribed in the circle of radius n+1.

        For example, set n=3, z=\frac{1}{2}(1+e^{2\pi i/3}), which lies halfway between two cube roots of unity, and \theta=\frac{5\pi}{12}. This maps to:

        \frac{5}{2 \sqrt{2}} (-1+i)

        The Cartesian formula for the solid hypocycloid with 4 cusps, inscribed in the circle of radius 4, is:

        (\frac{x}{4})^{2/3} + (\frac{y}{4})^{2/3} \le 1

        But 2(\frac{5}{8 \sqrt{2}})^{2/3} \approx 1.1604.

        Reply
      • Greg Egan says:
        13 September, 2012 at 1:05 am

        My guess that putting z anywhere on the boundary of the n-cusped hypocycloid will map \theta \in [0,2\pi] to the boundary of the (n+1)-cusped hypocycloid was wrong. This is only true if z lies on one of the cusps.

        Reply
      • John Baez says:
        13 September, 2012 at 1:50 am

        I don’t think Shanth should have brought the polygon whose vertices are nth roots of unity into the game; this contains the hypocycloid, but properly, except in the case of the Tusi couple.

        But the formula

        \displaystyle{ \mathrm{Tr[SU}(n+1)] = \bigcup_{\theta \in [0,2\pi]} e^{i\theta} \mathrm{Tr[SU}(n)] + e^{-in\theta} }

        looks like a good way to start organizing an inductive proof that

        \mathrm{Tr[SU}(n)] = \{ \mathrm{tr}(g) : g \in \mathrm{SU}(n) \}

        is an n-cusped hypocycloid, and also that we can roll an n-cusped hypocycloid inside an (n+1)-cusped hypocycloid in such a way that every point gets hit.

        What remains obscure to me is the fact shown in this picture by Xah Lee:

        Namely, that the catacaustic of the n-cusped hypocycloid is the (n+1)-cusped hypocycloid… well, least for n = 3… and that as we rotate the parallel beams of light forming the catacaustic, the (n+1)-cusped hypocycloid rolls around the n-cusped one, as shown in this movie!!!

        For all I know, this amazing extra stuff is only true for n = 3. It sure doesn’t seem true for n = 2.

        Reply
        • Aaron Wolbach says:
          26 September, 2012 at 6:09 pm

          (I apologize in advance for the long post.)

          Here is another idea that might possibly lead to a proof of the claim that

          \mathrm{TR}[\mathrm{SU}(n)] = \{ \mathrm{TR}(A) \; | \; A \in \mathrm{SU}(n) \; \}

          is the region bounded by an n-cusped hypocycloid. But it also uses some deeper ideas. Maybe it could be used to attack some of the other problems as well.

          We can easily construct a Lie group action of the circle group U(1) on SU(n) with the property that the U(1) orbit of the identity matrix I is the n-cusped hypocycloid in question. Define

          \begin{array}{ccl} \Psi : U(1) &\longrightarrow \mathrm{Diff}(\mathrm{SU}(n) ) \\ \theta &\mapsto \Psi_{\theta} \end{array}

          where the diffeomorphism \Psi_\theta is given by left multiplication by the matrix

          \begin{pmatrix} e^{i \theta} & & & \\ & \ddots & & \\ & & e^{i \theta} & \\ & & & e^{-i (n-1) \theta} \end{pmatrix}

          It’s very easy to check that the orbit of the identity matrix is indeed the hypocycloid – which is to say that the image of this orbit under the trace map

          \mathrm{TR} : \mathrm{SU}(n) \longrightarrow \mathbb{C}

          is the hypocycloid with n-cusps.

          There are lots of interesting things to say about this. First, observe that the matrices in SU(n) which map to a point on the hypocycloid (via the trace map) are the conjugates of [\Psi_{\theta}]. That is,

          \mathrm{TR}^{-1}(z) = \{ A[\Psi_{\theta}]A^{-1} | A \in \mathrm{SU}(n) \}.

          So there is a one-to-one correspondence between the points of the hypocycloid and these conjugacy classes. In particular, something interesting is happening to the conjugacy classes that correspond to the cusps.

          Second, the action I’ve described above is really just the action of SU(n) on itself but restricted to a specific embedding of U(1) into SU(n). At each point of U(1), we can decompose the tangent space of SU(n) into the direct sum of the tangent space of U(1) and the normal space to U(1). In particular, since the Lie algebra of SU(n) is isomorphic to the tangent space at the identity, we have a decomposition

          \mathfrak{su}(n) = \mathbb{R} \oplus N

          We can actually compute a generator for the first factor from the definition of the U(1) embedding above. Flowing from a point in SU(n) along that left invariant vector field is equivalent to acting on SU(n) by U(1) as defined above. What happens when we flow along a left invariant vector field which is normal to U(1)?

          In order to show that the traces of special unitary matrices are bounded by the hypocycloid, it would be enough to show that for any matrix A in SU(n), and any left invariant vector field X which is normal to U(1) in the sense above, we have

          |\mathrm{TR}(A)| \geq |\mathrm{TR}(\exp(\epsilon X)A)|

          for small epsilon.

          I haven’t actually done the computation, but this at least seems doable.

        • Aaron Wolbach says:
          26 September, 2012 at 6:21 pm

          My Latex was apparently broken […]

        • John Baez says:
          26 September, 2012 at 8:19 pm

          Great comment! It will take me a while to absorb it.

          I’ve tried to fix the LaTeX. Your main mistake was using

          $\latex      $

          or just

          $      $

          when you should enclose math formulas in

          $latex      $

          This is a common error. Also, you tried to use \begin{align}, which alas doesn’t work here. You have to use \begin{array}.

      • Greg Egan says:
        13 September, 2012 at 5:09 am

        The catacaustics under parallel illumination of the hypocycloids with other than 3 cusps aren’t hypocycloids.

        In Cartesian coordinates, with L=k-1 where k is the number of cusps, the hypocycloid has the parametric form:

        (x,y) = (L \cos (t)+\cos (L t), L \sin (t)-\sin (L t))

        while its catacaustic when illuminated with parallel rays at an angle of \theta has the parametric form:

        (x,y) = \frac{1}{2(L-1)}\\ (L (-\cos (2 \theta +M t)+\cos (2 \theta +Q t)+Q \cos (t))+M \cos (L t),\\ L (\sin (2 \theta +M t)-\sin (2 \theta +Q t)+Q \sin (t))-M \sin (L t))

        where M=L-2, Q=2L-1.

        Something special happens when k=3, L=2, M=0, Q=3; the catacaustic becomes:

        (x,y)=(-\cos (2 \theta)+\cos (2 \theta +3 t)+3 \cos (t), \sin (2 \theta )-\sin (2 \theta +3 t)+3 \sin (t))

        If we subtract off the rotating central point (-\cos (2 \theta ), \sin (2 \theta )), reparameterise the curve with u=t+\theta/2, and rotate by \theta/2, we get:

        (3 \cos (u)+\cos (3 u),3 \sin (u)-\sin (3 u))

        or the astroid in standard position.

        Reply
    • Blake Stacey says:
      11 September, 2012 at 8:15 pm

      Wicked cool!

      Random thought: what about the other famous series of groups? Does anything interesting happen if we look at the traces of all matrices in,say, \hbox{Sp}(2n, \mathbb{C})?

      Reply
      • John Baez says:
        13 September, 2012 at 1:32 am

        Good question! I feel the question will be the most fun if we take a compact Lie group G, an irreducible representation \rho and form the set

        S = \{ g \in G : \mathrm{tr}(\rho(g)) \}

        If G is compact this set will be compact, which somehow feels more pleasant to me.

        Someone must have studied these sets, but I’ve never seen any work on them—except the Wikipedia article that said for the fundamental representation of \mathrm{SU}(3) we get the deltoid, with–tut tut!—no source for this claim.

        Here’s one easy thing. The sets for the fundamental representations of \mathrm{SU}(n) have an appealing n-fold cyclic symmetry, and it’s easy to see where that comes from, and generalize that idea. It comes from the fact that the center of \mathrm{SU}(n) is the cyclic group \mathbb{Z}/n.

        Let Z be the center of \mathrm{SU}(n). By Schur’s Lemma \rho(z) is a complex multiple of the identity operator for any z \in Z. Say it’s \alpha(z) times the identity operator:

        \rho(z) = \alpha(z) I

        So,

        \mathrm{tr}(\rho(zg)) = \alpha(z) \mathrm{tr}{\rho(g)}

        Thus, if we have some point p \in S, the point \alpha(z) p will also be in S.

        So, there’s some abelian group

        A = \{ \alpha(z) : z \in Z \}

        that acts on our set S. We can assume without loss of generality that \rho is a unitary representation, and then these numbers \alpha(z) will be unit complex numbers. So, A acts as rotations on S. This makes it easy to see that A is a cyclic group, too.

        Now, the center Z is known for any compact Lie group you might care to name, since any such group is either a product of a torus and a bunch of compact simple Lie groups, or a quotient of such a group by a discrete subgroup.

        But of all the compact simple Lie groups, it’s the groups \mathrm{SU}(n) that get to have the biggest centers. So, I guess this little theory I just developed is the most potent in the case we already understand, where we don’t really need it.

        Reply
  8. arch1 says:
    11 September, 2012 at 4:53 pm

    Are people taking requests from the cheap seats?

    I’d love to see an animation showing the continuum along which line segment, deltoid, astroid etc. lie.

    One approach to this would be to depict, at each time t>0, the track produced by a circle rolling once around the inside of another circle which is (t+1) times as big.

    I imagine that this would be very pretty, run backwards to the right music:-)

    Reply
    • chuckbukowski says:
      11 September, 2012 at 8:28 pm

      For most times (the irrational ones) you would get a filled annulus with outer radius t+1 and inner radius t, so the thing gets a bit boring, except for a countable subset of t’s.

      Reply
      • chuckbukowski says:
        11 September, 2012 at 8:30 pm

        sorry, the inner radius is t-1

        Reply
      • arch1 says:
        12 September, 2012 at 4:02 pm

        Chuck, I think your comment applies if you delete the word ‘once’ from my suggestion.

        But if you display only the track resulting from the inner circle rolling *just once* around the inside of the other one, I think the resulting animation would be considerably more interesting.

        Reply
        • chuckbukowski says:
          12 September, 2012 at 4:11 pm

          True. Sorry for the misreading!

        • arch1 says:
          12 September, 2012 at 6:01 pm

          Having thought this through a little more carefully, I now think that tracks generated by the inner circle rolling n times around the inside of the larger circle would be a) uninteresting for n= infinity (as Chuck points out), b) only very mildly interesting for the n=1 value represented by my initial request, c) possibly quite interesting for moderate values of n (10-50, say). So if I were implementing this I would now include a slider bar to control n.

          Given these changing requirements, and the fact that I am no longer certain that even I’d be entranced by the results, I hereby downgrade my heartfelt request to an idle thought which just may capture someone’s fancy.

    • John Baez says:
      13 September, 2012 at 1:28 pm

      arch1 wrote:

      I’d love to see an animation showing the continuum along which line segment, deltoid, astroid etc. lie.

      While the programs here don’t quite do what you want, they’re still lots of fun.

      Reply
  9. Qiaochu Yuan says:
    11 September, 2012 at 5:35 pm

    Aha! We were wondering what shape the traces of an element of \text{SU}(3) describe over at MO… that’s an interesting answer!

    Reply
  10. Michelle Beissel says:
    12 September, 2012 at 10:44 am

    Goodness, I understood about 75% of this post! I am still struck by the sensuality of the forms and their interrelationships and think if more laypeople were given a chance they would see beauty and become interested in the world that geometry opens up. Perhaps describing geometry as promiscuous and incestuous, interest can be drummed up? :)

    Puzzle 1: 3 turns for the inner circle
    Puzzle 2: 2 turns for the inner circle
    Puzzle 3: River deltas were named after the Greek letter delta which was derived from the Phoenician alphabet which was restricted to straight, angular forms because of the medium used.
    Puzzle 4: John, you are not a miracle worker, I haven’t the foggiest idea.

    Reply
    • John Baez says:
      12 September, 2012 at 4:47 pm

      Congratulations, Michelle: that’s three out of four correct! Well, technically all 4 are correct, including this:

      John, you are not a miracle worker…

      I didn’t know the answer to the fourth one, but now I think I do, thanks to the discussion here… and it’s quite exciting.

      I am still struck by the sensuality of the forms and their interrelationships and think if more laypeople were given a chance they would see beauty and become interested in the world that geometry opens up.

      A lot of math is just as delicious as this, and plenty of it is even more mind-blowing… but much of it is harder to convey directly to the senses. So, after spending years going wherever my mind would take me, I’ve decided it’s good to spend some time taking people on tours that are easy to enjoy.

      Perhaps describing geometry as promiscuous and incestuous, interest can be drummed up? :)

      In France I’m sure it’d work—but I’m not sure it’s wise in parts of the US. I hear osculating circles have been banned in three states.

      (There’s an amusing misspelling on that page.)

      Reply
  11. John Baez says:
    13 September, 2012 at 2:16 am

    People who enjoy hypocycloids will enjoy the webpage, pointed out to me by William Rutiser, on Mathew Murray’s steam-powered Hypocycloidal Pumping Engine. This picture by Rich Carlstedt reminds me of Greg Egan’s animated gears:

    Reply
  12. John Baez says:
    13 September, 2012 at 1:25 pm

    Check out this rolling circle program made by Rahul Sidhu! You’ll start out seeing

    radiusRatio = 0.3

    which means the radius of the little circle is 0.3 times that of the big one. But you can adjust this. If you set

    radiusRatio = 1/3

    (just by typing it in), you’ll get a deltoid.

    Puzzle. What happens when you set

    radiusRatio = 2/3 ?

    Try to guess before you run the program to check your answer! This question was raised by Allen Knutson over on G+.

    Once you’ve figured the answer to this puzzle, or run the program to see the answer, try this program, also made by Rahul Sidhu.

    Reply
    • mircea says:
      13 September, 2012 at 1:47 pm

      Conjecture (beware of implicit spoiler): If the ratio is 2/3 is it true that the line giving the moving radius (line joining the moving point which describes the deltoid and the center of the moving circle) stays always tangent to the deltoid?

      Reply
  13. Dennis Des Chene says:
    18 September, 2012 at 4:54 am

    A very thorough article by A. I. Sabra in Perspectives on Science relates some of the history of the Tusi couple. Some historians have thought that the generation of rectilinear motion from circular motions meant trouble for Aristotelian cosmology (in which only circular motions occur in the heavens) by undermining, in this limiting case, the distinction between circular and rectilinear motion. Sabra argues against that claim, and to some extent against Saliba. The Tusi couple was problematic but occurred in the course of “normal science” intended to extend the system of Ptolemy, not to overturn it.

    Reply
    • John Baez says:
      19 September, 2012 at 4:17 am

      That’s interesting—thanks! It’s interesting to me how this old way of reconciling linear and circular motion turns out to become

      e^{i t} = \cos t + i \sin t

      in more modern mathematics.

      Reply
  14. Greg Egan says:
    29 September, 2012 at 2:48 am

    Here’s an inductive proof that the set of traces of \mathrm{SU}(n) is the filled hypocycloid with n cusps in the complex plane, created by rolling a circle of radius 1 inside a circle of radius n.

    The parameterised border of the hypocycloid with n cusps is:

    H_n(t)=(n-1)e^{i t}+e^{-i (n-1)t}, 0\le t\le2\pi

    This is the trace of a diagonalised element of \mathrm{SU}(n) with entries:

    e^{i t}, e^{i t}, ..., e^{i t}, e^{-i (n-1)t}

    Suppose we change all but one of the e^{i t} entries to e^{i s}, and change the last entry accordingly to keep the determinant equal to 1:

    e^{i t}, e^{i s}, ..., e^{i s}, e^{-i ((n-2)s + t)}

    The trace is:

    K_n(t,s)=e^{i t}+(n-2)e^{i s}+e^{-i ((n-2)s + t)} \\ = e^{i t}+ e^{-\frac{i t}{n-1}} ((n-2)e^{i (s+\frac{t}{n-1})}+e^{-i ((n-2)(s+\frac{t}{n-1}))}) \\ = e^{i t}+ e^{-\frac{i t}{n-1}} H_{n-1}(s+\frac{t}{n-1})

    If we let 0\le s\le2\pi while keeping t fixed, K_n(t,s) is the border of a hypocycloid with n-1 cusps, translated and rotated from the standard position.

    We will write the filled n-cusped hypocycloid in standard position as simply H_{n}, and the filled (n-1)-cusped hypocycloid as K_n(t). Formally:

    H_n = \{r H_n(t): 0\le r\le 1, 0\le t\le 2\pi\}

    K_n(t) = e^{i t}+ e^{-\frac{i t}{n-1}} H_{n-1}

    The two hypocycloids intersect at s=t, and in fact are tangent there; this is easy to check by computing the derivatives with respect to t and s respectively then setting
    s=t:

    \partial_t H_n(t) = i (n-1) \left(e^{i t}-e^{-i (n-1) t}\right)

    \partial_s K_n(t,s)|_{\;s=t} = i (n-2) \left(e^{i t}-e^{-i (n-1) t}\right)

    The two hypocycloids also intersect at the cusps of the smaller hypocycloid. If we set s=\frac{2\pi k - t}{n-1} for k=0,...,n-2, we have:

    K_n(t,\frac{2\pi k - t}{n-1})=e^{i t}+ e^{-\frac{i t}{n-1}} H_{n-1}(\frac{2\pi k}{n-1})\\ = e^{i t}+ e^{-\frac{i t}{n-1}} (n-1) e^{\frac{2\pi k i}{n-1}}\\ = (n-1) e^{\frac{2\pi k - t}{n-1} i} + e^{i t}\\ = H_n(\frac{2\pi k - t}{n-1})

    We can see that the filled smaller hypocycloid K_n(t) will sweep over every point of the filled larger hypocycloid H_n as follows. For some point z in the interior H_n, assume wlog that \mathrm{arg} z \in [0,\frac{2\pi}{n}], so z is in the wedge between the first and second cusps. If z \in K_n(0) then we’re done, so suppose z \notin K_n(0). The border of K_n(0) intersects the border of H_n at the first cusp of H_n, and as we increase t the border of K_n(t) intersects the border of H_n at H_n(t). The region in the first wedge of H_n that lies outside K_n(t) is divided into two pieces by the point of intersection when t\in(0,\frac{2\pi}{n}), so to escape K_n(t) our point z must always be in one piece or the other. But the first of these pieces is empty at t=0 and the second piece is empty at t=\frac{2\pi}{n}, so there must be some value of t when z intersects the border of K_n(t).

    So:

    \displaystyle{ H_n = \bigcup_{\;t \in [0,2\pi]} K_n(t)\\ = \bigcup_{\;t \in [0,2\pi]} e^{-\frac{i t}{n-1}} H_{n-1} + e^{i t} }

    Now, given any diagonalised matrix in \mathrm{SU}(n-1), if we multiply the whole matrix by some phase, e^{i\theta}, and then append a phase of e^{-i(n-1)\theta} to the list of diagonal entries to keep the determinant equal to 1, we get a diagonalised element of \mathrm{SU}(n). This gives us the inductive formula for the trace that Shanth derived:

    \displaystyle{ \mathrm{Tr}[\mathrm{SU}(n)] = \bigcup_{\;-(n-1)\theta \in [0,2\pi]} e^{i\theta} \mathrm{Tr}[\mathrm{SU}(n-1)] + e^{-i(n-1)\theta} }

    We only need to let the added phase, e^{-i(n-1)\theta}, vary over all of the unit circle, rather than the multiplying phase e^{i\theta}, because, by our inductive hypothesis for n-1, \mathrm{Tr}[\mathrm{SU}(n-1)] has (n-1)-fold rotational symmetry around the origin, so extending the range of \theta would be redundant.

    Setting \theta = -\frac{t}{n-1}, we have:

    \displaystyle{ \mathrm{Tr}[\mathrm{SU}(n)] = \bigcup_{\;t \in [0,2\pi]} e^{-\frac{i t}{n-1}} \mathrm{Tr}[\mathrm{SU}(n-1)] + e^{i t} }

    So if our hypothesis is true for n-1, i.e. the filled hypocycloid H_{n-1} is
    \mathrm{Tr}[\mathrm{SU}(n-1)], then it will be true for n.

    Reply
  15. Aaron Wolbach says:
    29 September, 2012 at 2:02 pm

    Greg that’s fantastic! I have actually constructed a different proof of this fact which is more Lie-theoretic. I will post it later this weekend.

    Reply
  16. Greg Egan says:
    29 September, 2012 at 4:01 pm

    Thanks Aaron. I’ll look forward to seeing your proof!

    Actually, fun as all the rotating lower-order hypocycloid business is, I’m wondering if I haven’t missed a much simpler argument.

    SU(n) is simply connected, so the loop of SU(n) elements we’ve found whose traces are the border of the hypocycloid can be continuously contracted to any point in SU(n). Suppose Z is any matrix in SU(n) with a trace of 0. Then the homotopy that takes the loop in SU(n) to Z, composed with the trace map, ought to take the border of the hypocycloid to the origin of the complex plane. So the entire filled hypocycloid must belong to Tr[SU(n)].

    Reply
    • Aaron Wolbach says:
      29 September, 2012 at 7:01 pm

      That’s actually pretty clean Greg. My proof was more algebraic. Basically, the rotations and translations we are seeing in the plane are coming from the Lie group actions of several independent circle subgroups of SU(n). My idea of proof was to show first that these circle subgroups generate the diagonal subgroup of SU(n). And then second – trivially – that the image of the diagonal subgroup under the trace map is the entire image of SU(n) under the trace map.

      It turns out that the trace map is equivariant with respect to two U(1) actions. One U(1) acts on a copy of SU(n) embedded in SU(n+1) upstairs – giving a 1-parameter family of embeddings – and the other U(1) acts on plane (\mathbb{C}) downstairs. When we watch the hypocycloid roll around in the plane, we’re seeing a shadow of this Lie group action and in fact, Shanth’s recursion formula tells us that the circle action on \mathbb{C} is

      \theta \cdot z = e^{i \theta} z + e^{-i n \theta}

      Maybe I won’t post my whole proof. But the geometry behind it is very pretty, and once my kids go to bed I’ll probably put up an exposition about that.

      Reply
    • John Baez says:
      30 September, 2012 at 1:13 am

      That’s slick. Technically you’re using this famous result: if you have a map from the disk to itself that maps the boundary to the boundary in a way that has winding number 1, it has to map the disk onto the disk. You’re using it in the case where ‘the disk’ is actually two different disks: 1) the region of the plane whose boundary is the hypocycloid, and 2) the disk in SU(n) formed by taking a loop of SU(n) elements whose traces are the border of the hypocycloid, and continuously contracting it to a point.

      This famous result is usually stated in this form: “there’s no retraction of the disk onto its boundary”. And the usual proof uses a bit of algebraic topology, though Milnor found one using calculus.

      This famous result is often used as a lemma to prove Brower’s fixed point theorem.

      Reply
  17. Greg Egan says:
    29 September, 2012 at 4:20 pm

    Ah, one thing I forgot to show is that Tr[SU(n)] isn’t larger than the filled hypocycloid. But if you compute the derivative of the trace with respect to any one of the individual n-1 independent phases, at the point where they’re all identical (i.e. on the border of the hypocycloid), the derivative is always tangent to the border: it’s just \frac{1}{n-1} of the derivative with respect to the curve parameter t.

    Everywhere but the cusps, some open neighbourhood of the tangent line lies in the interior of the filled hypocycloid, so no change in the phases can take you out of it. And at the cusps, moving along the tangent out of the hypocycloid would take you out of the disk of radius n, which the traces must lie within by the triangle inequality.

    Reply
  18. John Baez says:
    30 September, 2012 at 12:35 am

    These proofs concerning hypocycloids and traces of elements of SU(n) are great! I’ve been too busy with the beginning of classes to join in, but I’ll try to find out if these results are known. If not, they should be published. I’m sorry the connection to caustics fizzles out after SU(3) and SU(4)… I’m itching to write a paper entitled ‘Caustic remarks’, and I think just that contribution alone—the title—would justify me being a coauthor.

    Reply
  19. Greg Egan says:
    30 September, 2012 at 10:22 am

    I just found a paper that deals with the SU(n) traces, and some generalisations! I’ll give the link and abstract rather than trying to summarise:

    • N. Kaiser, Mean eigenvalues for simple, simply connected, compact Lie groups, 28 September 2006.

    Abstract: We determine for each of the simple, simply connected, compact and complex Lie groups SU(n), Spin(4n+2) and E6 that particular region inside the unit disk in the complex plane which is filled by their mean eigenvalues. We give analytical parameterizations for the boundary curves of these so-called trace figures. The area enclosed by a trace figure turns out to be a rational multiple of $\pi$ in each case. We calculate also the length of the boundary curve and determine the radius of the largest circle that is contained in a trace figure. The discrete center of the corresponding compact complex Lie group shows up prominently in the form of cusp points of the trace figure placed symmetrically on the unit circle. For the exceptional Lie groups G2, F4 and E8 with trivial center we determine the (negative) lower bound on their mean eigenvalues lying within the real interval [-1,1]. We find the rational boundary values -2/7, -3/13 and -1/31 for G2, F4 and E8 respectively.

    Reply
    • John Baez says:
      30 September, 2012 at 10:42 pm

      Great! Unfortunately (from the viewpoint of a mathematician trying to publish another paper), Kaiser has already worked out the connection between hypocycloids and the traces of matrices in SU(n):

      Fortunately (from the viewpoint of someone just wanting the truth to be known), he seems to have gone further and done traces of all the compact simply-connected Lie groups in certain famous irreps (though not all irreps). It’s neat how the traces of Spin(4n+2) elements in the spinor rep lie in a curve a bit like an astroid, but which gets skinnier and skinnier as n increases:

      It’s also neat how E6 in one of its two complex 27-dimensional reps gives traces lying in a curve a bit like a deltoid, but skinnier:

      Reply
    • John Baez says:
      30 September, 2012 at 10:59 pm

      I know what you’re asking:

      “What about E7 and E8?”

      and

      “Why did Kaiser draw pictures for Spin(4n+2) but not other spin groups?”

      The answer can be found here:

      • John Baez, Division algebras and quantum theory.

      The author here recalls that every irreducible representation of a group is either ‘real’, ‘complex’ or ‘quaternionic’ in a certain sense: this is Dyson’s ‘three-fold way’. And he writes:

      SU(2) is not the only compact Lie group with the property that all its irreducible continuous unitary representations on complex Hilbert spaces are real or quaternionic. For a group to have this property, it is necessary and sufficient that every element be conjugate to its inverse. All compact simple Lie groups have this property except those of type An for n > 1, Dn with n odd, and E6 (see Bourbaki [20]).

      If a representation of a Lie group is ‘real’, the trace of any group element in that representation is a real number, so we don’t get a pretty picture of the set of allowed traces: it’s just some interval in the real line.

      If it’s ‘quaternionic’ I bet the trace of any group element is again real. I don’t see why right off the bat, but I know an example. The spin-1/2 representation of SU(2) is ‘quaternionic’, and the trace of an SU(2) element in this representation is a real number lying on the interval [-2,2]. Indeed, this interval is the line segment traced out (pardon the pun) by the Tusi couple!

      So, it’s only the truly ‘complex’ representations where the traces give a pretty picture, and—translating notations—the only compact simple Lie groups with representations of this sort are SU(n) for n > 2, Spin(4n+2) for n > 1, E6.

      That’s why we see SU(6), Spin(14) and E6 in the pictures above, but not Spin(12) or G2 or F4 or E7 or E8.

      Reply
    • Greg Egan says:
      1 October, 2012 at 5:30 am

      One thing that impressed me was how Kaiser derives the equation for the hypocycloids as the boundary of the trace figure so easily! Essentially, he writes the trace for SU(n) as:

      \mathrm{Tr} = z_1+z_2+...+z_{n-1}+\frac{1}{z_1 z_2 ... z_{n-1}}

      and then says that we can find the boundary by eliminating all but one variable with the “zero-derivate conditions”: that the partial derivatives of the trace with respect to z_1, z_2, ... z_{n-2} are all zero.

      \partial_{z_j} \mathrm{Tr} = 1 - \frac{1}{z_j (z_1 z_2 ... z_{n-1})} = 0, j=1,2,...,n-2\\ z_1 z_2 ... z_{n-1} = z_j^{-1}, j=1,2,...,n-2

      It follows that the first n-2 of the z_j are equal (say to z_1), and for the remaining two terms we have:

      z_{n-1} = z_1^{-(n-1)}\\ \frac{1}{z_1 z_2 ... z_{n-1}} = z_1

      which lets us write the trace as:

      \mathrm{Tr} = (n-1) z_1 + z_1^{-(n-1)}

      For z_1 a phase this is just the familiar formula for the hypocycloid with n cusps.

      It’s worth stressing that in all the discussions we’ve had about the boundary, we’ve set all the independent phases to be equal there, while the final phase — the product of their reciprocals — is different, whereas Kaiser ends up with the same overall collection of phases but the last two of them are swapped compared to our scheme.

      But why is Kaiser allowed to choose the partial derivates of the trace with respect to z_1, z_2, ... z_{n-2} to be zero on the boundary? He uses this condition without explaining it, but I guess the idea is that we have a compact manifold without boundary of real dimension n-1 being projected onto the complex plane, and where the manifold projects to the boundary of its shadow the linearised map has to change from having an (n-3)-dimensional kernel to an (n-2)-dimensional kernel, so you can choose coordinates there such that n-2 of the coordinate vectors lie in the kernel.

      Reply
    • Greg Egan says:
      1 October, 2012 at 8:29 am

      Nice as it is, I think Kaiser’s method for deriving the boundary curve for the SU(n) trace could do with a bit more exposition and justification. It’s certainly not the case that given any function from the diagonal subgroup of SU(n) to the complex plane you can blithely declare that you’ve set the partial derivatives with respect to n-2 of the phases equal to zero, and then expect that to give you the boundary of the image. Generically there will be some choice of coordinates where the derivatives on the boundary vanish for all but one coordinate, but for a more general function than the trace that coordinate system need not line up with the phases.

      So he’s exploiting a lot of nice symmetries of the problem, but I wish he’d given a more careful account of the things he’s relying on to obtain the result.

      Reply
      • John Baez says:
        1 October, 2012 at 6:52 pm

        I guess he’s a physicist! They have an irksome tendency to get the right answer before mathematicians do, while leaving us to justify all their reasoning. Paraphrasing Kevin McCrimmon, “the lions get the fresh kill, and they leave the spoils for the hyenas.”

        There are some more interesting things to do, however. For example, there’s a unique probability measure on SU(n) that’s translation-invariant. This pushes forward via the trace map to a probability measure on the interior of the n-cusped hypocycloid. This gives the ‘probability distribution on traces of matrices in SU(n)’.

        The real part or imaginary part of the trace then becomes a real-valued random variable, and we can compute its moments: its mean, its variance, and so on. There will be a nice combinatorial way to do this.

        The case of SU(3) was discussed on Mathoverflow by Qiaochu Yuan. In this case the moments of twice the real part of the trace form this sequence:

        1, 0, 2, 2, 12, 30, 130, 462, 1946, 7980, 34776, 153120, 694056, 3194334, 14971242, …

        which is sequence A151366 in the Online Encyclopedia of Integer Sequences. The nth term here is the number of walks with n steps start and end at the origin and stay within \mathbb{N}^2 where at each step you either go one step north, south, east, west, northwest or southeast. In other words, you can move any of these ways:

        (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0)

        Qiaochu also works out the moments of the imaginary part of the trace.

        So, this just goes to show that there’s a lot of math here…

        Reply
      • Jonathan Rayner says:
        1 May, 2017 at 11:52 pm

        @Greg Egan Did you ever fully patch up Kaiser’s method to be more clear/rigorous? I’m working on a generalization of this problem and the method of arguing that all but one of the partial derivatives must vanish on points on the boundary is very useful, but I haven’t been able to complete the argument myself.

        I’m referring to your comments:

        “I guess the idea is that we have a compact manifold without boundary of real dimension n-1 being projected onto the complex plane, and where the manifold projects to the boundary of its shadow the linearised map has to change from having an (n-3)-dimensional kernel to an (n-2)-dimensional kernel, so you can choose coordinates there such that n-2 of the coordinate vectors lie in the kernel.”

        “Nice as it is, I think Kaiser’s method for deriving the boundary curve for the SU(n) trace could do with a bit more exposition and justification. It’s certainly not the case that given any function from the diagonal subgroup of SU(n) to the complex plane you can blithely declare that you’ve set the partial derivatives with respect to n-2 of the phases equal to zero, and then expect that to give you the boundary of the image. Generically there will be some choice of coordinates where the derivatives on the boundary vanish for all but one coordinate, but for a more general function than the trace that coordinate system need not line up with the phases.”

        Reply
      • Greg Egan says:
        2 May, 2017 at 5:11 am

        @Jonathan Rayner: I haven’t really thought about this much since 2012, but re-reading these comments and doing a few new calculations, it strikes me that the following might be a simple argument for the shape of the boundary.

        Suppose we parameterise the diagonal subgroup of \mathrm{SU}(n) with n-1 phase angles, \phi_i, which we will use as real coordinates of a compact real manifold without boundary. If we take the trace map and then identify the complex plane with \mathbb{R}^2, at any point in the manifold we can use the trace to map the coordinate tangent vectors to vectors in \mathbb{R}^2:

        \displaystyle{\mathrm{Tr}^*: \partial_{j} \to (-\sin(\phi_j) - \sin(\phi_1+...\phi_{n-1}), \cos(\phi_j) - \cos(\phi_1+...\phi_{n-1}))}

        Generically, these n-1 vectors in \mathbb{R}^2 will have rank 2, but along the boundary of the trace figure they will have rank 1. Now, there will actually be lots of different loops in the diagonal subgroup that map to the boundary, but one obvious case is the loop where all the \phi_j are equal:

        \displaystyle{\phi_j(s) = s}

        and hence all the \mathrm{Tr}^*(\partial_j) are equal:

        \displaystyle{ \mathrm{Tr}^*(\partial_j) = (-\sin(s) - \sin((n-1)s), \cos(s) - \cos((n-1)s)) }

        Every single point of that loop must map to the boundary, so (along with a topological argument for the boundary having only a single connected component), we’re perfectly entitled to use the image of that loop to determine the shape of the boundary.

        Kaiser just finds another loop in the manifold that also maps to the boundary; he puts:

        \displaystyle{\phi_j(s) = s, j=1,...,n-2 \\ \phi_{n-1}(s) = -(n-1)s}

        For this loop, we have:

        \displaystyle{ \phi_1+...\phi_{n-1} = -s }

        so the first n-2 of the vectors in \mathbb{R}^2 are zero, while the last one is:

        \displaystyle{ \mathrm{Tr}^*(\partial_{n-1}) = (\sin((n-1)s) + \sin(s), \cos((n-1)s) - \cos(s)) }

        Clearly the set of vectors in the plane has rank 1 everywhere on this loop, so it too can be used to determine the shape of the boundary.

        Reply
        • Jonathan Rayner says:
          2 May, 2017 at 6:24 am

          Thanks, this is quite a clear argument! One thing to make it watertight in my mind:

          Is it obvious that points where the differential/pushforward is decreased rank (rank 1) must lie on the boundary? I understand there’s a proof that regular points lie only in the interior in this situation and so the boundary will only have critical points, but can critical points $only$ be on the boundary?

          Sorry if the answer is obvious (I’m a grad student self-teaching)

        • Greg Egan says:
          2 May, 2017 at 8:37 am

          In general, there would be no reason why a critical point couldn’t be mapped to the interior of the projection; you can imagine, say, taking a spherical balloon, poking your finger into one side, then projecting the deformed surface onto the ground. Near the tip of your finger the surface would be vertical, but it would map to the interior of the projection.

          So it does take an extra argument to see why the curve we get in the plane from either of the loops I mentioned must lie entirely on the boundary of the projection. If we could show that every critical point must project to the same hypocycloid, I think that would do it, but I can’t see an easy proof of that off the top of my head.

        • Greg Egan says:
          3 May, 2017 at 5:13 am

          It turns out that things are much more complicated than I thought! It’s not true that all critical points project to the boundary.

          Let’s try to characterise the general case. At a given point in the diagonal subgroup manifold, the projections of the coordinate tangent vectors onto the plane are:

          \mathrm{Tr}^*: \partial_{j} \mapsto

          (-\sin(\phi_j) - \sin(\phi_1+ \cdots + \phi_{n-1}), \cos(\phi_j) - \cos(\phi_1+ \cdots + \phi_{n-1}))

          These vectors will all lie on the same unit circle in the plane, whose centre is displaced from the origin by a unit vector. So that circle passes through the origin.

          If this set of projections is rank-1, then all the vectors must lie on a line through the origin, but since they also lie on a circle that passes through the origin, there are only two possibilities: zero, and some non-zero vector, say v, that gives the other point of intersection of the line with the circle.

          For any vectors that are equal to v, their \phi coordinates must all be the same; say their common value is s. And whenever the vector for coordinate j is zero, we must have:

          \displaystyle{\phi_j = -(\phi_1+ c\dots +\phi_{n-1})}

          We will write t = \phi_1+ \cdots +\phi_{n-1}, so for these zero vectors, \phi_j = -t.

          Suppose k of the n-1 vectors are equal to v and n-1-k are equal to zero. Then we have:

          \displaystyle{t = k s - (n-1-k) t}

          \displaystyle{t = \frac{k}{n-k} s}

          The projection of the curve containing the points is then:

          \displaystyle{C(s) = k \exp(i s) + (n-k) \exp(-i \frac{k}{n-k} s)}

          This gives the boundary hypocycloid when k=n-1 or k=1, but for other values of k it gives curves that lie in the interior, except at the cusps.

          For example, for n=5, the blue curve in the image below comes from k=1,4, while the orange curve comes from k=2,3.

        • Greg Egan says:
          3 May, 2017 at 8:22 am

          I should add that in the simplest case for which the trace is a 2-dimensional set, SU(3), there are only two kinds of curves of critical points, and they both map to the boundary deltoid.

          I wish I could visualise how the projection of the (topological) 2-torus that is the SU(3) diagonal subgroup ends up being a deltoid, but it’s a struggle to picture how it sits in \mathbb{R}^6, which I think is the smallest Euclidean space in which it can be embedded in such a way that the trace is an orthogonal projection.

        • Greg Egan says:
          3 May, 2017 at 12:54 pm

          I’ve put an animation of SU(3)’s diagonal subgroup as a torus here:

          https://plus.google.com/113086553300459368002/posts/BuWJ9eR9Qnw

        • John Baez says:
          4 May, 2017 at 12:55 am

          Very nice! – and it’s also nice that the animation shows up here. I don’t think many mathematicians or physicists have this mental picture of the maximal torus of SU(3), though it’s something a lot of us have thought about.

          (The word “the” here is a bit irksome, but in a compact Lie group all maximal tori are conjugate to each other, so it’s tolerable. You’re using the maximal torus that consists of all diagonal matrices in SU(3).)

        • Jonathan Rayner says:
          8 July, 2018 at 11:18 am

          Hi there, sorry for the radio silence. I actually wrote up something in detail a few months ago using a different approach to you Greg, although based on the same idea of studying the critical points. Thanks for sharing yours!

          I’m now actually thinking that what I wrote should be publicly available, (and perhaps would be good for my applications to math PhD programs). Of course I’d cite the convo here and a conversation I had about this on stackexchange, as well as provide a link in the comments here for anyone following in future.

          Do you or John have any advice on what the appropriate medium would be? I asked someone in the mathematics department at my university and they said that a colleague told them that it’s a well-known result, but I still haven’t seen a proof in any pure math journals (or physics or applied math for that matter).

          Perhaps it’s considered too simple? Maybe the arXiv or a blog post is the correct setting. What do you think?

        • John Baez says:
          9 July, 2018 at 11:49 pm

          Jonathan wrote:

          Do you or John have any advice on what the appropriate medium would be?

          These days the easiest way to publicize results is on a blog. It takes a lot of work to create a blog that many people actually read! But if you’re just trying to get a result out there, you don’t need a big readership.

          Another easy way is to ask a question on MathOverlow and then answer it. This may seem a bit odd, but they actually suggest that as a reasonable thing to do.

          A bit harder than these, but better for more substantial results, is to write a paper and put it on the arXiv.

          Harder still, but even better if you’re trying for an academic career, is to publish a paper in a journal. Putting it on the arXiv is a natural step to take before this.

          I asked someone in the mathematics department at my university and they said that a colleague told them that it’s a well-known result…

          Don’t let it rest there! Find out who this colleague is, and talk to them, and find out why he thinks it’s “well-known”. This is part of doing research—talking to people, tracking down rumors, etc.

        • Todd Trimble says:
          10 July, 2018 at 11:09 am

          John wrote

          “Another easy way is to ask a question on MathOverlow and then answer it. This may seem a bit odd, but they actually suggest that as a reasonable thing to do.”

          While it’s true that you can find some such pronouncement written by “them”, this is more of a global StackExchange.com sentiment, and the reality is that the reaction to such a course of action on MathOverflow is likely to be less than welcoming unless handled delicately, to avoid seeming like you’re self-promoting. A few general pointers: carefully explain the mathematical context and why you think this would be is of general interest to the mathematical research community. (If you can link to a related MO discussion, all the better.) Get in there somewhere the fact that this arose during a discussion at John Baez’s blog Azimuth and that he mentioned the possibility of bringing this to broader attention via MathOverflow, and say something like, “I apologize if this is not an appropriate use of MathOverflow.” Finally — and this is important — flag your question for the moderators, asking them to make it Community Wiki. Having it CW prevents you from gaining (or losing) reputation points, thus presenting your post as a selfless act that is not in view of boosting your MO reputation. This CW would then also apply to any answer that appears and they too would not result in change in reputation, although both question and answer can still be voted on. I can’t promise that the thread would survive even with these precautions, but I’m reasonably certain your chances would be much improved by following these pointers.

        • nad says:
          10 July, 2018 at 2:25 pm

          @Todd Trimble

          As Peter Krautzberger says

          Please always remember XKCD 1357.

        • Todd Trimble says:
          10 July, 2018 at 8:54 pm

          Yes, nad, I know about and subscribe to that interpretation of free speech. In fact I consider it obvious. But I don’t understand how it’s a propos of what I wrote.

  20. Rolling Circles and Balls (Part 4) « Azimuth says:
    2 January, 2013 at 12:38 am

    In Part 3 we rolled a circle inside a circle that was 2, 3, or 4 times as big […]

    Reply
  21. Russell says:
    24 February, 2013 at 7:11 am

    On seeing the diagram of the Tusi couple, I started to wonder if this is the same principle – KUNDEL Magnetic Coupling Device (http://www.kundelmagnetics.com/) . It has a 2:1 pole ratio, and translates rotation motion into reciprocal motion and vice versa. I am trying to imaging how to represent the magnetic fields as a tusi couple (without success so far).

    Reply
  22. airving2013 says:
    17 March, 2015 at 12:43 am

    This is fantastic – great work!

    Reply

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