mathbabe

I’m back home from HCSSiM (see my lecture notes here). Yesterday I took the bus from Amherst to New York and slept all the way, then got home and took a nap, and then after putting my 3-year-old to bed crashed on the couch until just now when I woke up. That’s about 13 hours of sleep in the past 20, and I’m planning to take a nap after writing this. That’s just an wee indication of how sleep deprived I became at math camp.

Add to that the fact that my bed there was hard plastic, that I completely lost touch with the memory of enjoying good food (taste buds? what are those?), and that I was pitifully underpaid, and you might think I’m glad to be home.

And I am, because I missed my family, but I’m already working feverishly to convince them to let me go again next year, and come with me next time if that would be better. Because I’m so in love with those kids and with those junior staff and with Kelly and with Hampshire College and with the whole entire program.

Just so you get an idea of what there is to love, check out one of the students talking about his plan for his yellow pig day shirt which I luckily captured on my phone:

And here’s a job description which always makes me laugh (and cry) (and I only worked there the first half):

When people haven’t experienced HCSSiM, we worry about being able to explain adequately the unusual commitment required by the exploitative position. The workday is long and challenging; it is also exciting and rewarding. A senior faculty and 2 junior faculty members actively participate in the morning classes (8:30 – 2:30, M-S) and evening problem sessions (7:30 – 10:30, M-F) of each of the c. 14-student Workshops (7/2 – 7/20 = days); they prepare (and duplicate) daily problem sets; they proofread notes and program journal articles, and they write evaluations; they offer constructive criticism; they attend the afternoon Prime Time Theorems (a 51-minute math-club type talk, over half given by visitors) and give 1 or 2. The staffing and most of those teaching opportunities (chores) apply to the 2nd half of the program when students take a Maxi-course (8:30-11, M-S, and 7:30 – 10:30, M-F, 7/23 – 8/10). During the 2nd half of the program, students also take, consecutively, 2 Mini-courses, which meet from 11:17 until 12:30 for 7 days and which have no attached problem sessions; many minis are created or co-created by junior staff. Except for Kelly and Susan (who are on call) the staff live in in the dorm (Enfield this year), join students for meals and recreational activities, provide transportation and counseling and supervision for students, and help to get the program to sleep around 11:17. There is virtually no hope of getting any research done or of maintaining an outside social life. In spite of (with some because of) the preceding, the job is exhilarating as well as exhausting; we have repeaters, and there are a lot of good math teachers out there who credit HCSSiM with teaching them to teach.

This is a continuation of this, where I take notes on my workshop at HCSSiM.

Magic Squares

First Elizabeth Campolongo talked about magic squares. First she exhibited a bunch, including these classic ones which I found here:

Then we noted that flipping or rotating a magic square gives you another one, and also that the trivial magic square with all “1”‘s should also count as a crappy kind of magic square.

Then, for 3-by-3 magic squares, Elizabeth showed that knowing 3 entries (say the lowest row) would give you everything. This is in part because if you add up all four “lines” going through the center, you get the center guy four times and everything else once, or in other words everything once and the center guy three times. But you can also add up all the horizontal lines to get everything once. The first sum is 4C, if C is the sum of any line, and the second sum is 3C, so subtracting we get that C is the the center three times, or the center is just C/3.

This means if you have the bottom row, you can also infer C and thus the center. Then once you have the bottom and the center you can infer the top, and then you can infer the two sides.

After convincing them of this, Elizabeth explained that the set of magic cubes was actually a vector space over the rational numbers, and since we’d exhibited 3 non-collinear ones and since we’d proved we can have at most 3 degrees of freedom for any magic square, we actually had a basis.

Finally, she showed them one of the “all prime magic squares”:

The one with the “17” in the corner of course. She exhibited this as a sum of the basis we had written down. It was very cool.

The game of Set

My man Max Levit then whipped out some Set cards and showed people how to play (most of them already knew). After assigning numbers mod 3 to each of the 4 dimensions, he noted that taking any two cards, you’d be able to define the third card uniquely so that all three form a set. Moreover, that third card is just the negative of the sum of the first two, or in other words the sum of all three cards in a set is the vector

Next Max started talking about how many cards you can have where they don’t form a set. He started in the case where you have only two dimensions (but you’re still working mod 3). There are clearly at most 4, with a short pigeon hole argument, and he exhibited 4 that work.

He moved on to 3 and 4 dimensions and showed you could lay down 9 in 3 and 20 in 4 dimensions without forming a set (picture from here), which one of our students eagerly demonstrated with actual cards:

Finally, Max talked about creating magic squares with sets, tying together his awesome lecture with Elizabeth’s. A magic square of sets is also generated by 3 non-collinear cards, and you get the rest from those three placed anywhere away from a line:

Probability functions on lattices

Josh Vekhter then talked about probability distributions as functions from posets of “event spaces” to the real line.  So if you role a 6-sided die, for example, you can think of the event “it’s a 3, 4, or 6” as being above the event “it’s a 3”. He talked about a lattice as a poset where there’s always an “or” and an “and”, so there’s always a common ancestor and child for any two elements.

Then he talked about the question of whether that probability function distributes in the way it “should” with respect to “and” and “or”, and explained how it doesn’t in the case of the two slit experiment.

He related this lack of distribution law of the probability function to the concept of the convexity of the space of probability distributions (keeping in mind that we actually have a vector space of possibly probability functions on a given lattice, can we find “pure” probability distributions that always take the values of 0 or 1 and which form a kind of basis for the entire set?).

This is not my expertise and hopefully Josh will fill in some details in the coming days.

King Chicken

I took over here at the end and discussed some beautiful problems related to flocks of chickens and pecking orders, which can be found in this paper. It was particularly poignant for me to talk about this because my first exposure to these problems was my entrance exam to get into this math program in 1987, when I was 14.

Notetaking algorithm

Finally, I proved that the notetaking algorithm we started the workshop with 3 weeks before actually always works. I did this by first remarking that, as long as it really was a function from the people to the numbers, i.e. it never gets stuck in an infinite loop, then it’s a bijection for sure because it has an inverse.

To show you don’t get stuck in an infinite loop, consider it as a directed graph instead of an undirected graph, in which case you put down arrows on all the columns (and all the segments of columns) from the names to the numbers, since you always go down when you’re on a column.

For the pipes between columns, you can actually squint really hard and realize there are two directed arrows there, one from the top left to the lower right and the other from the top right to the lower left. You’ve now replaces the “T” connections with directed arrows, and if you do this to every pipe you’ve removed all of the “T” connections altogether. It now looks like a bowl of yummy spaghetti.

But that means there is no chance to fall into an infinite loop, since that would require a vertex. Note that after doing this you do get lots of spaghetti circles falling out of your diagram.

This is a continuation of this, where I take notes on my workshop at HCSSiM.

Two days ago Benji Fisher came to my workshop to talk about group laws on rational points of weird things in the plane. Here are his notes.

Degenerate Elliptic Curves in the plane

Conics in the plane

Pick . Consider the line given by . Where does intersect the y-axis? Where does it intersect the unit circle,

Substitute into the equation for the circle to get

After you do it the hard way, notice that you already know one root: .

The sum of the roots is and their product is . Either way, you get

.

From  you get .

Do not forget that if you are given and , then .

This gives you a 1-1 correspondence between the points of the circle (conic) and the points of the -axis (including ). The formula for Pythagorean triples also falls out. So do the formulae for the tangent-of-the-half-angle substitution, which is useful when integrating rational functions of and : set and

There are several ways you can generalize this. You could project a sphere onto a plane. I want to consider replacing the circle with a cubic curve. The problem is that the expected number of intersections between a line and a cubic is 3, so you get a 2-to-1 correspondence in general. That is interesting, too, but for now I want to consider the cases where the curve has a double point and I choose lines through that point. Such lines should intersect the cube in one other point, giving a 1-1 correspondence between the curve (minus the singular point) and a line (minus a small number of points).

cubic with a cusp

Let be the curve defined by

,

which has a sharp corner at the origin. This type of singularity is called a cusp. Let denote the line through the origin and the point , which has slope .

1. Sketch the curve . Does Mathematica do a good job of this?
2. The line meets the curve at the origin and in one other point, . Find formulae for and in terms of and for in terms of and .
3. You almost get a digestion (bijection) between and the line . What points do you have to omit from each in order to get a digestion?
4. Three points , and  on are collinear. What condition does this impose on the corresponding values of ?

The calculations are easier than for the circle: , and

You have to remove the point from the curve and the point  from the line. Well, you do not have to remove them, but the formula for in terms of and is fishy if you do not. The point at infinity (the third point of intersection between the curve and the -axis) corresponds to itself.

The condition for colliniarity is

Plug in the expressions in terms of the coordinates, chug away, and you should get

.

If we let , then is the natural coordinate on the line . (Maybe I should use that line to start with instead of .)

cubic with a node

This problem deals with the curve defined by

,

which intersects itself at the origin. This type of singularity is called a node. Let denote the line through the origin and the point , which has slope .

1. Sketch the curve . Does Mathematica do a good job of this?
2. The line meets the curve at the origin and in one other point, ). Find formulae for and in terms of and for in terms of and .
3. You almost get a digestion (bijection) between and the line . What points do you have to omit from each in order to get a digestion?
4. Three points , and  on are collinear. What condition does this impose on the corresponding values of ?

Once again, . The usual method gives  and . In order to get a 1-1 correspondence, you need to delete the singular point  from the curve and the points and from the line.

The lines through the origin with slope are tangent to the curve. If you plug away, you should find that the condition for colliniarity is:

 .

Remember our curve (not to be Maximum Confused with )? It’s the one whose equation is The condition for 3 points to be collinear on is:

Claim: in terms of , the condition is

(Hint: In terms of , .)

If you start with the known equation and replace the ‘s with ‘s, it takes some work to get down to the condition .

If you start with the LHS of the desired equation, there is a shortcut:

.

But then we have

Note that the change-of-variable formulae are fractional linear transformations. Geometrically, is the natural coordinate on the line  and is the natural coordinate on the line .

To get from one line to the other, just draw a line through the origin.

One interpretation of our results for the curves , , is that it gives us a way to add points on the line (with coordinate ) and multiply points on the line (with coordinate ) using just a straightedge, provided that we are allowed to draw lines through the point at infinity.

In other words, we are allowed to draw vertical lines. I will continue to refer to this as “straightedge only.” I forgot to mention: you need to have the curve provided as well. Explicitly, this is the rule. Given two points on the line, find the corresponding point on the curve. Draw a line through them. The third point of intersection with the curve will be the (additive or multiplicative) inverse of the desired sum. Draw a vertical line through this point: the second point of intersection (or the third, if you count the point at infinity as being the second) will be the desired sum/product.

More precisely, it is the point on the curve corresponding to the desired sum/product, so you have to draw one more line.

Another interpretation is that we get a troupe (or karafiol) structure on the points of the curve, excluding the singular point but including the point at infinity. The point at infinity is the identity element of the troupe (group). The construction is exactly the same as in the previous paragraph, except you leave off the parts about starting and ending on the line.

smooth cubic

Similarly, we get a troupe (group) structure on any smooth cubic curve. For example, consider the curve defined by

Start with two points on the curve, say  and . The equation for the line through these two points is  $

Solving for gives the equation where 

Plug this into the equation defining and you get, after a little algebra,

Luckily, we know how to solve the general cubic. (Just kidding! Coming back to what we did with the circle, we observe that and are two of the roots, so we can use either of the relations  or .)

The result is:

where the final form comes from squaring and using the relations and .

At this point, a little patience (or a little computer-aided algebra) gives

Do not forget the final step:  is the third point of intersection, but to get the sum we need to draw a vertical line, or reflect in the -axis:

Now, I give up. With a lot of machinery, I could exlpain why the group law is associative. (The identity, as I think I said above, is the point at infinity. Commutativity and inverses are clear.) What I can do with a different machine (Mathematica or some other computer-algebra system) is verify the associative law. I could also do it by hand, given a week, but I do not think I would learn anything from that exercise.

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This is a continuation of this, where I take notes on my workshop at HCSSiM.

Aaron was visiting my class yesterday and talked about Sandpiles. Here are his notes:

Sandpiles: what they are

For fixed , an beach is a grid with some amount of sand in each spot. If there is too much sand in one place, it topples, sending one grain of sand to each of its neighbors (thereby losing 4 grains of sand). If this happens on an edge of the beach, one of the grains of sand falls off the edge and is gone forever. If it happens at a corner, 2 grains are lost. If there’s no toppling to do, the beach is stable. Here’s a 3-by-3 example I stole from here:

Do stable beaches form a group?

Answer: well, you can add them together (pointwise) and then let that stabilize until you’ve got back to a stable beach (not trivial to prove this always settles! But it does). But is the sum well-defined?

In other words, if there is a cascade of toppling, does it matter what order things topple? Will you always reach the same stable beach regardless of how you topple?

Turns out the answer is yes, if you think about these grids as huge vectors and toppling as adding other 2-dimensional vectors with a ‘-4’ in one spot, a ‘1’ in each of the four spots neighboring that, and ‘0’ elsewhere. It inherits commutativity from addition in the integers.

Wait! Is there an identity? Yep, the beach with no sand; it doesn’t change anything when you add it.

Wait!! Are there inverses? Hmmmmm….

Lemma: There is no way to get back to all 0’s from any beach that has sand.
Proof. Imagine you could. Then the last topple would have to end up with no sand. But every topple adds sand to at least 2 sites (4 if the toppling happens in the center, 3 if on an edge, 2 if on a corner). Equivalently, nothing will topple unless there are at least 4 grains of sand total, and toppling never loses more than 2 grains, so you can never get down to 0.

Conclusion: there are no inverses; you cannot get back to the ‘0’ grid from anywhere. So it’s not a group.

Try again

Question: Are there beaches that you can get back to by adding sand?

There are: on a 2-by-2 beach, the ‘2’ grid (which means a 2 in every spot) plus itself is the ‘4’ grid, and that topples back to the ‘2’ grid if you topple every spot once. Also, the ‘2’ grid adds to the grid and gets it back.

Wow, it seems like the ‘2’ grid is some kind of additive identity, at least for these two elements. But note that the ‘1’ grid plus the ‘2’ grid is the ‘3’ grid, which doesn’t topple back to the ‘1’ grid. So the ‘2’ grid doesn’t work as an identity for everything.

We need another definition.

Recurrent sandpiles

A stable beach C is recurrent if (i) it is stable, and (ii) given any beach A, there is a beach B such that C is the stabilization of A+B. We just write this C = A+B but we know that’s kind of cheating.

Alternative definition: a stable beach C is recurrent if (i) it is stable, and (ii) you can get to C by starting at the maximum ‘3’ grid, adding sand (call that part D), and toppling until you get something stable. C = ‘3’ + D.

It’s not hard to show these definitions are equivalent: if you have the first, let A = ‘3’. If you have the second, and if A is stable, write A + A’ = ‘3’, and we have B = A’ + D. Then convince yourself A doesn’t need to be stable.

Letting A=C we get a beach E so C = C+E, and E looks like an identity.

It turns out that if you have two recurrent beaches, then if you can get back to one using a beach E then you can get back to to the other using that same beach E (if you look for the identity for C + D, note that (C+D)+E = (C+E) + D = C+D; all recurrent beaches are of the form C+D so we’re done). Then that E is an identity element under beach addition for recurrent beaches.

Is the identity recurrent? Yes it is (why? this is hard and we won’t prove it). So you can also get from A to the identity, meaning there are inverses.

The recurrent beaches form a group!

What is the identity element? On a 2-by-2 beach it is the ‘2’ grid. The fact that it didn’t act as an identity on the ‘1’ grid was caused by the fact that the ‘1’ grid isn’t itself recurrent so isn’t considered to be inside this group.

Try to guess what it is on a 2-by-3 beach. Were you right? What is the order of the ‘2’ grid as a 2-by-3 beach?

Try to guess what the identity looks like on a 198-by-198 beach. Were you right? Here’s a picture of that:

We looked at some identities on other grids, and we watched an app generate one. You can play with this yourself. (Insert link).

The group of recurrent beaches is called the m-by-n sandpile group. I wanted to show it to the kids because I think it is a super cool example of a finite commutative group where it is hard to know what the identity element looks like.

You can do all sorts of weird things with sandpiles, like adding grains of sand randomly and seeing what happens. You can even model avalanches with this. There’s a sandpile applet you can go to and play with.

This is a continuation of this, where I take notes on my workshop at HCSSiM.

We switched it up a bit and I went to talk about Rubik’s Cubes in Benji Fisher’s classroom whilst Aaron Abrams taught in mine and Benji taught in Aaron’s. We did this so we could meet each other’s classes and because we each had a presentation that we thought all the classes might want to see.

In my class, besides what Aaron talked about (which I hope to blog about tomorrow), we talked about representing groups with generators and relations and we looked at how to create fractals on the complex plane using Mathematica.

Solving the Rubik’s Cube with group theory

First I talked about the position game, so ignoring orientation of the pieces. In other words, if all the pieces are in the correct position but are twisted or flipped, that’s ok.

Consider the group acting on the set which is the Rubik’s cube. We can fix the centers, and then this group is clearly generated by 90 degree clockwise turns on each of the 6 faces (clockwise when you look straight at the center of the face). It has a bunch of relations too, of course, including (for example) that the “front” move commutes with the “back” move and that any of these generators is trivial if you do it 4 times in a row.

Next we noted that the 8 corners always move to other corners on any of these moves, and that the 12 edges always move to edges. So we could further divide the “positions game” into an “edges game” and a “corners game.” Furthermore, because of this separation, any action can be realized as an element of the group

But there’s more: any one of these generators is a 4-cycle on both edges and corners, which is odd in both cases, so so is a product of the generators. This means the group action actually lands inside

where I denote by the odd permutations of

Next I stated and proved that the 3-cycles generate , by writing any product of two transpositions as a product of two or three 3-cycles.

The reason that’s interesting is that, once I show that I can act by arbitrary 3-cycles on the edges or corners, this means I get that entire group up there of the form (even, even) or (odd, odd).

Next I demonstrated a 3-cycle on three “bad pieces” where one needs to go into the position of one of the other two. Namely:

1. Put two “bad” pieces on the top and one below, including the piece which is the destination of the third.
2. Do some move U which moves the third bad piece below to its position on the top without changing anything else.
3. Move the top so you decoy the new piece with the other bad piece on top. Call this decoy move V.
4. Undo the U move.
5. Undo the V move.

Seen from the perspective of the top, all that happens is that suddenly it has one piece in the right place (after U), then it gets rid of a piece it didn’t want anyway. From the perspective of the bottom, it sent up a piece it didn’t want to the top, then unmessed itself by taking back another piece from the top.

That’s a 3-cycle on the pieces, on the one hand (and a commutator on the basic moves), and we can perform it on any three corners or edges by first performing some move to get 2 out of the 3 pieces on the same face and the third not on that face. This is essentially acting by conjugation (and it’s still a commutator).

Moreover, the overall effect on the puzzle is that it took 3 pieces that were messed up and got at least 1 in the right spot.

The overall plan now is that you solve the corners puzzle, then the edges puzzle, and you’re done. Specifically, I get to the (even, even) situation by turning one face once if necessary, and then I can perform arbitrary 3-cycles to solve my problem. If I have fewer than three messed up corners (resp. edges), then I have either 0, 1, or 2 messed up corners (resp. edges). But I can’t have exactly 1, and exactly 2 would be odd, so I have 0.

That’s the position game, and I’ve given an algorithm to solve it (albeit not efficient!).

What about the orientations?

Shade the left and right side of a solved cube. Both the position (cubicle) and the piece of plastic (cubie) can be considered as shaded. Also shade a strip  on the front and on the back, so we get the top and bottom edge of the front and back shaded.

Convince yourself that every edge and every corner position has exactly one shaded face.

When the cube is messed up, the shaded face of the position may not coincide with the shaded face of the cubie in that position. For edges it’s either right or wrong – assign each a value of 0 (if it’s right) or 1 (if it’s wrong). These can be considered orientation values modulo 2.

For corners, it can be wrong in two ways. The cubie’s shaded face could coincide with the position’s shaded face (give it a value 0), or it could be clockwise 120 degrees (value 1), or counterclockwise 120 degrees (value -1). These numbers are modulo 3.

Next, convince yourself that all of the generators of the group of actions on the rubik’s cube preserve the fact that the sum of the orientation numbers is 0, considered either mod 3 for the corners or mod 2 for the edges.

In particular, this means we can’t have exactly one edge flipped but everything in the right position. If you see this on a Rubik’s cube then someone took it apart and put it back together wrong.

However, you can twist one corner clockwise and another one counterclockwise, which means you can get any configuration of orientation numbers on the corners that satisfy that their sum is 0. Same with edges – it’s possible to flip two.

Finally, I exhibit how to do each of these basic orientation moves by 2 3-cycles. First I choose a pivot corner and perform a 3-cycle on those three corners, and then I do another 3-cycle, which uses the same 3 corners but is slightly different and results in everything being back in the right position but two corners twisted. I can do the same thing for edges, and I have thus totally solved the cube.

This method of 3-cycles is slow but it generalizes to all Rubik’s puzzles, so for example the 7 by 7 cube:

Or, with some work, something else like this:

p.s. I learned all of this when I was 15 from Mike Reid at HCSSiM 1987.

p.p.s. This was what we ate after singing our Yellow Pig Carols last night:

This is a continuation of this, where I take notes on my workshop at HCSSiM.

Permutation notation

Using a tetrahedron as inspiration, we found the group of rotational symmetries as a subgroup of the permutation group We then spent a lot of time discussing how to write good notation for permutations, coming up finally with the standard form but complaining about how multiplication is in a weird direction, stemming from the composition rule

Quadratic Reciprocity

Today we proved quadratic reciprocity, which totally rocks. In order to get there we needed three lemmas, which were essentially a bunch of different formulas for computing the Jacobi symbol.

The first was that, modulo we have:

The second one was that we could write as -1 to the , where is the number of negative numbers you get when you write the numbers modulo with a number of absolute value less than

In the third one we assume is odd (we deal with separately). We prove we can write as -1 to the , where is (also) the sum of the greatest integers for

Each of these lemmas, in other words, gave us different ways to compute the symbol . They are each pretty easy to prove:

The first one uses the fact that if is not a square, then the product of all nonzero numbers mod can be paired up into pairs of different numbers whose product is , so we get altogether but on the other hand we already know by Wilson’s Theorem that that product is -1.

The second one just uses the fact that, ignoring negative signs, is the same list as but paying attention to negative signs we can take out an and also a Since we already know the first lemma we’re done.

The third lemma follows from repeatedly applying the division algorithm to and , so writing

for all . We replace the remainders by the previous form of “small representatives” which we call   modulo , making the remained positive or negative but smaller than this replacement requires that we add . We add all the division formulas up and realize that we only need to care about the parity of so in other words we work modulo 2. It looks like this:

Since is odd, we can ignore it modulo 2. Since is odd same there. We get:

Up to signs, so we can conclude our lemma.

Incidentally, using the formula above (before we ignore ) it’s also easy to see that

Finally, we show that the sum is just the number of lattice points in the region above the x-axis, to the left of the line and below the line

But, letting some other odd prime, we can do the same exact thing and we find that is just the number of lattice points in the region above the x-axis, to the left of the line and below the line Flipping the second triangle over the line demonstrates that we have two halves of the rectangle with lattice points. Here’s the picture for and (ignore the dotted lines):

Therefore we’ve proved:

Alternating Group

We introduced cyclic notation, showing everything is a product of transpositions, then we introduced the sign function and showed that can only be written as a product of an even number of transpositions. This means the sign function is well defined, and it’s easy to see then that it’s a homomorphism, so we can define its kernel to be the alternating group.

Transformation of the Plane

We discussed rotations, shears, shrinks, reflections, and translations of the plane and demonstrated them in Mathematica using this.

Tomorrow is Yellow Pig Day, which is a yearly tradition here at HCSSiM during which we celebrate our mascot the yellow pig and our favorite number, 17.

In particular, there will be an hour and a half lecture during which we will hear many surprising and elegant 17 facts (examples: the longest time anybody has ever sat in a tub of ketchup? 17 hours. The average adolescent male has a sexually related thought how often? Every 17 seconds).

Some time after the 17 lecture we get together to sing Yellow Pig Carols. These are usually set to the tune of some song, with the lyrics changed to refer to math, in particular the number 17, and yellow pigs, Weird Al Yankovic- style. Check out this list for a taste.

But there’s a problem, which is that the songs are really old and many of the tunes are unknown to this crop of kids. In desperate need of a revamp, I got my workshop to write a new song, which I’m super proud of. We started out by voting on a song (winner: Somewhere Over the Rainbow) for which we wrote new lyrics. Here they are:

Some July Seventeenth

Some july seventeenth
when pigs fly
we’ll see patches of yellow
scattered across the sky

Some “f” over the reals
satisfy
f of x plus f of y
is f-of-x plus y

someday I’ll find another g
besides f of t is k t or zer-o
and then I will compose the two
and get solutions that are new
they’ll appear-o

Some groups they are abelian
they commute
but some have commutators
whose actions are not moot

The action on the complex plane
by matrices is so in-sanely dum-ber
than even that of conjugation
whose equivalence relation pairs the num-bers

Some july seventeenth
when pigs fly
we’ll see patches of yellow
scattered across the sky

Honestly I thought it would stop there, but people around here have been on a tear. My hilarious junior staff Maxwell Levit has written a brilliant song based on Gotye’s “Somebody I Used to Know”. If you haven’t been living under a rock, you will have heard that song (and if you haven’t heard that song, please go ahead and do so now), about a guy wondering why this woman has left him and won’t talk to him, and then she comes in and tells her story which is how much of a manipulative creep he really was. There’s a dramatic video featuring nakedness and body paint which adds to the drama and to the song.

Well Max just turned that shit around and now it’s Fermat singing to Fermat’s Last Theorem, wondering where he went wrong, and then the theorem talks back and tells us the real deal. Plus he uses the word “marginalia,” which is in itself awesome. Here it is:

A Theorem that I used to know

(Fermat)
Now and then I think of Diophantine equations
Like how Pythagoras showed the case for n=2
Told myself that I understood,
And didn’t write down what I thought I would
Remember when I looked back at my marginalia

You can get addicted to a certain type of hubris
Assuming you don’t need to use elliptic curves
So when I found my proof did not make sense,
I knew it wasn’t my incompetence
But I’ll admit I was confused to say the least.

But you didn’t have to be so hard,
Make out like my intuitive method was for nothing
I don’t even need to know
But they treat you like Wiles solved you and that feels so rough

No you didn’t have to stoop so low
Elliptic curves and modular forms lack in imagination
I guess I don’t need you though,
Now you’re just some theorem that I used to know.

Now you’re just some theorem that I used to know.
Now you’re just some theorem that I used to know.

(theorem)
Now and then I think of when you said you’d solved me.
Part of me believing I had some marvelous proof.
But I don’t really work that way.
Adhering to everything you say.

You said that you could let it go,
And you shouldn’t get too hung up on a theorem that you used to know!

(Fermat)
But you didn’t have to be so hard,
Make out like my intuitive method was for nothing
I don’t even need to know
But they treat you like Wiles solved you and that feels so rough

No you didn’t have to stoop so low
Elliptic curves and modular forms lack in imagination
I don’t even need you though,
Now you’re just some theorem that I used to know.

[x2]
Some Theorem!
(I used to know)
Some Theorem!
(Now you’re just some theorem that I used to know)

(I used to know)
(That I used to know)
(I used to know)
Some Theorem!

We’re gonna make Devin Ivy, a fantastically funny junior staff here as well as a photographer, play Gotye/ Fermat in the video, with yellow pigs getting continually plastered all over his body. He’s Gotye’s spitting image:

This is a continuation of this, where I take notes on my workshop at HCSSiM.

We originally defined as and now we re-introduce it as the set of matrices of the form with the obvious map where is sent to:

After we reminded people about matrix addition and multiplication, we showed this was an injective homomorphism under addition and also jived with the multiplication that we knew coming from Overall our lesson was not so different from this one.

Then we talked about actions on the plane including translations and scaling and showed that under the above map, “multiplication by ” or by its corresponding matrix gives us the same thing.

Platonic Solids

We went over the 5 platonic solids from yesterday- we’d proved it was impossible to have more than 5, and now it was time to show all 5 are actually possible! That’s when we whipped out Wayne Daniel’s “all five” puzzle:

We then introduced the concept of dual graph, and showed which platonic solids go to which under this map. We saw an example of a toy which flips from one platonic solid (cube) to its dual (octahedron) when you toss it in the air, the Hoberman Flip Out.

Here is it in between states:

Finally, we talked about symmetries of regular polyhedra and saw how we could embed an action into the group of symmetries on its vertices. So symmetries on tetrahedra is a subgroup of . It’s a lot easier to understand how to play with 4 numbers than to think about moving around a toy, so this is a good thing. Although it’s more fun to play with a toy.

Then one of our students, Milo, showed us how he projected a tiling of the plane onto the surface of a sphere using the language “processing“. Unbelievable.

After that we went to an origami workshop to construct yellow pigs as well as platonic solid type structures.

This is a continuation of this, where I take notes on my workshop at HCSSiM.

Lagrange

We reminded people that a finite group is a finite set with an associative multiplication law “*”, and an identity and inverses with respect that law. A subgroup is a subset which is a group in its own right with the same multiplication law. Given a subgroup of a group we defined the cosets of to be, for some of the form:

We proved that “being in the same coset as” is an equivalence relation and that each coset has the same size. Altogether the cosets form a partition of the group, so the size of the group is the product of the size of any coset and the number of cosets.

One of the students decided that you could form a group law on the set of cosets, inheriting the multiplication operation from When it didn’t quite work out, he postulated that he could do it if the group is commutative. Pretty smart kid.

Primitive Roots modulo p

Next we spent more time than you’d think proving there’s a nonzero number modulo whose powers generate all of the nonzero elements modulo In other words, that there is an isomorphism

The left group is a group under multiplication, and the right one is a group under addition, which means this can be seen as a kind of logarithm in finite groups. Indeed an element on the left is mapped to that power of such that So it’s very much like a logarithm.

To prove such an exists, we actually show that such roots exist, and that in fact there are th roots of unity for any which divides Another case where it helps to prove something harder than what you actually want.

To do that, we have a few lemmas. First, that there are at most roots to a degree polynomial modulo unless it’s the “zero” polynomial. Second, that if there’s a polynomial which achieves this maximum, it must divide the “Fermat’s Little Theorem” polynomial These two lemmas aren’t hard.

Next we proved that there are at most primitive th roots of unity for any by showing that, if we had one, then we’d take certain powers to get all of them, and then if we hadn’t counted one then we’d be able to produce too many roots of the polynomial

Finally, we wrote out boxes labeled with the divisors of and we labeled balls the non-zero numbers mod We put a ball with label into the box with its order (the smallest positive number so that ). Since there are at most in each box, and since we need to put every ball in some box, there must be exactly balls in each box since we proved a few days ago that:

Platonic Solids

Riffing on our proof of Euler’s formula for planar graphs, we convinced the kids that we could just as well consider a graph to be on a sphere, using a stereographic projection.

Then, using graphs on spheres as guides, we looked at how many examples we could come up with for a regular polytope, where each face has the same number of edges and each vertex has the same number of edges leaving it. We came up with the five platonic solids.

This is a continuation of this, where I take notes on my workshop at HCSSiM.

Quadratic equations modulo p

We looked at the general quadratic equation modulo an odd prime (assume ) and asked when it has roots. We eventually found the solution to be similar to the one we recognize over the reals, except you need to multiply by instead of dividing by . In particular, we achieved the goal, which was to reduce this general form to a question of when a certain thing (in this case ) is a square modulo , in preparation for quadratic reciprocity.

We then defined the Jacobi symbol and proved that or .

Cantor

We then reviewed some things we’d been talking about in terms of counting and cardinality, we defined the notation for sets: we define to mean “exists an injection from to “. We showed it is a partial ordering using Cantor-Schroeder-Bernstein.

Then we used Cantor’s argument to show the power set of a set always has strictly bigger cardinality than the set.

Euler and planar graphs

At this point the CSO (Chief Silliness Officer) Josh Vekhter took over class and described a way of assessing risk for bank robbers. You draw out a map of a bank as a graph with vertices in the corners and edges as walls. It can look like a tree, he said, but it would be a pretty crappy bank. But no judgment. He decided that, from the perspective of a bank robber, corners are good (more places to hide), rooms are good (more places for money to be stashed) but walls are bad (more things you need to break through to get money.

With this system, and assigning a -1 to every wall and a 1 to every corner or room, the overall risk of a bank’s map consistently looks like 2. He proved this is always true using induction on the number of rooms.

A continuation of this, where I take notes on my workshop at HCSSiM.

First we had a few proofs from the previous night’s problem set, including a proof of Hall’s Marriage Problem using Dilworth’s Theorem.

Counting stuff

We then proved an uncountable set union a countable set is uncountable, with the help of Lior’s comment from yesterday.

Then we proved the Cantor-Schroeder-Bernstein Theorem, which states that if you have two injective maps and in the opposite directions: and then you can construct a bijection between and , and in particular the two sets will have the same cardinality. It’s not that hard – consider the orbits of points in X and Y under repeated applications of the two injective maps , and if possible, by pulling back by and

It doesn’t take much thinking to convince yourself that these orbits come in three forms: an infinite list in both directions, a finite loop, or an infinite forward path but a finite backwards path, where at some point you can’t pull back any more. In the last case you could get “stuck” in either or Since the orbits form a partition of all of the points, you can independently decide how to define the bijection depending on what that orbit looks like. Namely, it takes an element in to unless it’s an orbit that gets going backwards in in which case you take to

Now that I think of it, I’m pretty sure this proof uses the axiom of choice, and according to the wikipage on this theorem it doesn’t need to, but I don’t know a proof which avoids it. The truth is I can never tell. Please explain to me if you can, how you can verify if you’re using the axiom of choice in an argument where you make infinitely many decisions.

Homomorphisms

We defined a homomorphism of groups and wrote it in both additive and multiplicative notation, because it turns out some of the students were getting confused. We proved very basic properties that follow such as the fact that the identity element goes to the identity element under a homomorphism, and the image of the inverse of an element is the inverse of the image. We ended by asking them to consider the set of homomorphisms between a cyclic group of 6 elements and itself, or between a cyclic group of 6 elements to a cyclic group of 7 elements.

Fermat and Wilson

Next we whipped out some theorems modulo a prime We looked at the action defined on numbers mod 17 when you multiply them by 3, and noticed it just scrambles the non-zero ones up (and of course sends 0 to 0). We proved that this is true in general. But this means that the product of all the non-zero numbers mod p is the same if we pre-multiply by any , which means that product is equal to itself times and since it’s invertible that means That’s just a hop skip and jump away from Fermat’s Little Theorem, which states that  for every number .

Next, we wondered, what was that product of all those non-zero numbers mod ? It turns out that each of those nonzero guys is invertible, so if you pair each up with its inverse their contribution to the product is just 1, and the leftovers are just the guys who are their own inverse, which is only 1 or -1 (which we proved, and which is most definitely not true modulo 8). So the whole product is -1. That’s Wilson’s Theorem, but we called it Wilevson’s Theorem since Lev came up with the argument.

A continuation of this, where I take notes on my workshop at HCSSiM.

We first saw presentations from the students from last night’s problem set. One was a four line proof that the last two digits of are 61. The other was a beautiful proof that the only real numbers that have more than one decimal representation are rationals of the form for some integer

Dillworth’s Theorem Revisited

After going over examples of chains and antichains, and making sure we knew that there must be at least as many chains as there are elements in an anti-chains if we want the chains to cover our set, we set up a proof by induction on the number of elements in our set. The base case is easy (one element, one chain, one element in a maximal anti-chain) and then to reduce to a smaller case we remove any maximal element . Note this just means there’s nothing above it. But now the inductive hypothesis holds, and we cover the remaining set with chains and moreover we define to be the largest element in such that it is contained in some maximal anti-chain.

We then stated two lemmas. The first is that any maximal anti-chain must have a unique element in each chain $C_i$, and the second is that, after defining the elements as above, they form a maximal anti-chain. We treat these lemmas as black boxes for the proof (the first we did yesterday, the second is on problem set tonight).

Now we put back the removed point There are two cases, either is incomparable to any of the ‘s or it is comparable to at least one of them.

In the first case, we have an anti-chain of size namely the ‘s plus , and we can form a st chain consisting of just itself.

In the second case, is comparable to some Since is maximal, and since is maximal in its chain with respect to being in a maximal antichain, we can form a new chain which is just the same thing as for and below, and goes straight up to from Note we may be missing some stuff that used to be above in But that doesn’t matter, because if we remove this new chain, we see that the maximal anti-chains leftover is only of size so by Strong Induction we can cover it with chains, and then bring back this chain to get an overall covering with chains.

Dihedral Groups

After reviewing the formal definition of a Karafiol (group), we used rotations and flips of a folder and then of a regular 5-gon to come up with a non-commutative group. We defined a subgroup and explored the different subgroups of the dihedral group as well as other examples we’d seen coming from modular arithmetic.

Euler’s Totient function

We introduced the number of positive integers less than relatively prime to as a function of called and wrote out a table up to 17. We observed that and that, from the Chinese Remainder Theorem, we also could infer that for we have Writing out a prime factorization for we realized we’d have a formula for if we just knew for any prime and any positive We wrote out a picture and decided

We ended by proving Euler’s formula by induction on the number of distinct prime factors:

A continuation of this, where I take notes on my workshop at HCSSiM.

The real numbers are uncountable

Today we used Cantor’s diagonal argument to prove that the real numbers aren’t countable. Namely, we assumed they were, and that we had a bijection and then proved it didn’t contain the real number whose th digit we defined to be 1 if the th digit of is 7, and 7 otherwise. One of our students pointed out that there are some numbers that have more than one decimal expansion, so the general argument that a certain decimal expansion isn’t in the list isn’t a total proof. But we decided that we’d be able to prove on homework that the only numbers that have more than one representative in decimal expansions are rational with their denominators divisible only by 2 and 5, which is not the case for the number YP we’d been talking about.

Then we wanted to show that has the same cardinality as We tried to use a “splicing argument” where we create a new decimal expansion out of two decimal expansions (where the odd digits correspond to the first and the even to the second), but then we decided this map isn’t well defined, since we still have this “overlap” problem where 0.999… = 1 but the pair doesn’t get mapped by the splicing map to the same decimal expansion as the pair

So then we decided to instead consider the set of decimal expansions, which is a bit bigger than the reals, and there the splicing map works. We reduced it to this case and are leaving it to homework to prove that the set of decimal expansions has the same cardinality as the reals. Although I don’t actually know how to do this without Cantor-Schroder-Bernstein, which we haven’t proven yet.

The Chinese Remainder Theorem 

We stated and proved the Chinese (Llama) Remainder Theorem. It was a nice example of a constructive proof that assumes it’s possible and then, when you follow your nose, it’s possible to completely characterize what the solution must look like, and then it falls out. We showed it described a bijection of sets We haven’t talked about homomorphisms of groups yet though so we can’t prove it’s an isomorphism of groups.

Dilworth’s Theorem on chains and anti-chains in posets

We started the proof of Dilworth’s Theorem but didn’t finish yet (to be continued tomorrow). Turns out that this problem is really hard and that, moreover, there are lots of ways to convince yourself it’s true but be wrong about. The inductive proof in wikipedia, for example, is incomplete, but can be extended to a complete proof.

As an example, try to cover the following poset on the power set of five letters with 10 chains:

In particular you can see we solve a matching problem on the way, between subsets of size two and subsets of size 3 in the set of five letters, which isn’t the normal “take the complement” match, but rather is an inclusion of one in the other. I’m still wondering if there’s a more direct way to do this.