A trailer for p-adic analysis, first half: USA TST 2003

I think this post is more than two years late in coming, but anywhow…

This post introduces the -adic integers , and the -adic numbers . The one-sentence description is that these are “integers/rationals carrying full mod information” (and only that information).

The first four sections will cover the founding definitions culminating in a short solution to a USA TST problem.

In this whole post, is always a prime. Much of this is based off of Chapter 6B from Straight from the Book.

1. Motivation

Before really telling you what and are, let me tell you what you might expect them to do.

In elementary/olympiad number theory, we’re already well-familiar with the following two ideas:

• Taking modulo a prime or prime , and
• Looking at the exponent .

Let me expand on the first point. Suppose we have some Diophantine equation. In olympiad contexts, one can take an equation modulo to gain something else to work with. Unfortunately, taking modulo loses some information: (the reduction is far from injective).

If we want finer control, we could consider instead taking modulo , rather than taking modulo . This can also give some new information (cubes modulo , anyone?), but it has the disadvantage that isn’t a field, so we lose a lot of the nice algebraic properties that we got if we take modulo .

One of the goals of -adic numbers is that we can get around these two issues I described. The -adic numbers we introduce is going to have the following properties:

1. You can “take modulo for all at once”.

   In olympiad contexts, we are used to picking a particular modulus and then seeing what happens if we take that modulus. But with -adic numbers, we won’t have to make that choice. An equation of -adic numbers carries enough information to take modulo .

2. The numbers form a field, the nicest possible algebraic structure: makes sense. Contrast this with , which is not even an integral domain.
3. It doesn’t lose as much information as taking modulo does: rather than the surjective we have an injective map .
4. Despite this, you “ignore” some “irrelevant” data. Just like taking modulo , you want to zoom-in on a particular type of algebraic information, and this means necessarily losing sight of other things.

   Let me draw an analogy. We know that the equation

   

   has no integer solutions, because, well, squares are nonnegative. But you will find that this equation has solutions modulo any prime , because once you take modulo you stop being able to talk about numbers being nonnegative. The same thing will happen if we work in -adics: the above equation has a solution in for every prime .

So, you can think of -adic numbers as the right tool to use if you only really care about modulo information, but normal isn’t quite powerful enough.

To be more concrete, I’ll give a poster example now:

Example 1 (USA TST 2002/2)

For a prime , show the value of

does not depend on .

Here is a problem where we clearly only care about -type information. Yet it’s a nontrivial challenge to do the necessary manipulations mod (try it!). The basic issue is that there is no good way to deal with the denominators modulo (in part is not even an integral domain).

However, with -adic analysis we’re going to be able to overcome these limitations and give a “straightforward” proof by using the identity

Such an identity makes no sense over or for converge reasons, but it will work fine over the , which is all we need.

2. Algebraic perspective

We now construct and . I promised earlier that a -adic integer will let you look at “all residues modulo ” at once. This definition will formalize this.

2.1. Definition of

Definition 2 (Introducing )

A -adic integer is a sequence

of residues modulo for each integer , satisfying the compatibility relations for .

The set of -adic integers forms a ring under component-wise addition and multiplication.

Example 3 (Some -adic integers)

Let . Every usual integer generates a (compatible) sequence of residues modulo for each , so we can view each ordinary integer as -adic one:

On the other hand, there are sequences of residues which do not correspond to any usual integer despite satisfying compatibility relations, such as

which can be thought of as .

In this way we get an injective map

which is not surjective. So there are more -adic integers than usual integers.

(Remark for experts: those of you familiar with category theory might recognize that this definition can be written concisely as

where the inverse limit is taken across .)

Exercise 4

Check that is an integral domain.

2.2. Base expansion

Here is another way to think about -adic integers using “base ”. As in the example earlier, every usual integer can be written in base , for example

More generally, given any , we can write down a “base ” expansion in the sense that there are exactly choices of given . Continuing the example earlier, we would write

and in general we can write

where , such that the equation holds modulo for each . Note the expansion is infinite to the left, which is different from what you’re used to.

(Amusingly, negative integers also have infinite base expansions: , corresponding to .)

Thus you may often hear the advertisement that a -adic integer is an “possibly infinite base expansion”. This is correct, but later on we’ll be thinking of in a more and more “analytic” way, and so I prefer to think of this as a “Taylor series with base ”. Indeed, much of your intuition from generating functions (where is a field) will carry over to .

2.3. Constructing

Here is one way in which your intuition from generating functions carries over:

Proposition 5 (Non-multiples of are all invertible)

The number is invertible if and only if . In symbols,

Contrast this with the corresponding statement for : a generating function is invertible iff .

Proof: If then , so clearly not invertible. Otherwise, for all , so we can take an inverse modulo , with . As the are themselves compatible, the element is an inverse.

Example 6 (We have )

We claim the earlier example is actually

Indeed, multiplying it by gives

(Compare this with the “geometric series” . We’ll actually be able to formalize this later, but not yet.)

Remark 7 ( is an integer for )

The earlier proposition implies that (among other things); your intuition about what is an “integer” is different here! In olympiad terms, we already knew made sense, which is why calling an “integer” in the -adics is correct, even though it doesn’t correspond to any element of .

Fun (but trickier) exercise: rational numbers correspond exactly to eventually periodic base expansions.

With this observation, here is now the definition of .

Definition 8 (Introducing )

Since is an integral domain, we let denote its field of fractions. These are the -adic numbers.

Continuing our generating functions analogy:

This means is “Laurent series with base ”, and in particular according to the earlier proposition we deduce:

Proposition 9 ( looks like formal Laurent series)

Every nonzero element of is uniquely of the form

Thus, continuing our base analogy, elements of are in bijection with “Laurent series”

for . So the base representations of elements of can be thought of as the same as usual, but extending infinitely far to the left (rather than to the right).

(Fair warning: the field has characteristic zero, not .)

Remark 10 (Warning on fraction field)

This result implies that you shouldn’t think about elements of as (for ) in practice, even though this is the official definition (and what you’d expect from the name ). The only denominators you need are powers of .

To keep pushing the formal Laurent series analogy, is usually not thought of as quotient of generating functions but rather as “formal series with some negative exponents”. You should apply the same intuition on .

(At this point I want to make a remark about the fact , connecting it to the wish-list of properties I had before. In elementary number theory you can take equations modulo , but if you do the quantity doesn’t make sense unless you know . You can’t fix this by just taking modulo since then you need to get , ad infinitum. You can work around issues like this, but the nice feature of and is that you have modulo information for “all at once”: the information of packages all the modulo information simultaneously. So you can divide by with no repercussions.)

3. Analytic perspective

3.1. Definition

Up until now we’ve been thinking about things mostly algebraically, but moving forward it will be helpful to start using the language of analysis. Usually, two real numbers are considered “close” if they are close on the number of line, but for -adic purposes we only care about modulo information. So, we’ll instead think of two elements of or as “close” if they differ by a large multiple of .

For this we’ll borrow the familiar from elementary number theory.

Definition 11 (-adic valuation and absolute value)

We define the -adic valuation in the following two equivalent ways:

• For we let be the largest such that (or if ). Then extend to all of by .
• Each can be written uniquely as for , . We let .

By convention we set . Finally, define the -adic absolute value by

In particular .

This fulfills the promise that and are close if they look the same modulo for large ; in that case is large and accordingly is small.

3.2. Ultrametric space

In this way, and becomes a metric space with metric given by .

Exercise 12

Suppose is continuous and for every . Prove that .

In fact, these spaces satisfy a stronger form of the triangle inequality than you are used to from .

Proposition 13 ( is an ultrametric)

For any , we have the strong triangle inequality

Equality holds if (but not only if) .

However, is more than just a metric space: it is a field, with its own addition and multiplication. This means we can do analysis just like in or : basically, any notion such as “continuous function”, “convergent series”, et cetera has a -adic analog. In particular, we can define what it means for an infinite sum to converge:

Definition 14 (Convergence notions)

Here are some examples of -adic analogs of “real-world” notions.

• A sequence , \dots converges to a limit if .
• The infinite series converges if the sequence of partial sums , , \dots, converges to some limit.
• \dots et cetera \dots

With this definition in place, the “base ” discussion we had earlier is now true in the analytic sense: if then

Indeed, the th partial sum is divisible by , hence the partial sums approach as .

While the definitions are all the same, there are some changes in properties that should be true. For example, in convergence of partial sums is simpler:

Proposition 15 ( iff convergence of series)

A series in converges to some limit if and only if .

Contrast this with in . You can think of this as a consequence of strong triangle inequality. Proof: By multiplying by a large enough power of , we may assume . (This isn’t actually necessary, but makes the notation nicer.)

Observe that must eventually stabilize, since for large enough we have . So let be the eventual residue modulo of for large . In the same way let be the eventual residue modulo , and so on. Then one can check we approach the limit .

Here’s a couple exercises to get you used to thinking of and as metric spaces.

Exercise 16 ( is compact)

Show that is not compact, but is. (For the latter, I recommend using sequential continuity.)

Exercise 17 (Totally disconnected)

Show that both and are totally disconnected: there are no connected sets other than the empty set and singleton sets.

3.3. More fun with geometric series

While we’re at it, let’s finally state the -adic analog of the geometric series formula.

Proposition 18 (Geometric series)

Let with . Then

Proof: Note that the partial sums satisfy , and as since .

So, is really a correct convergence in . And so on.

If you buy the analogy that is generating functions with base , then all the olympiad generating functions you might be used to have -adic analogs. For example, you can prove more generally that:

Theorem 19 (Generalized binomial theorem)

If and , then for any we have the series convergence

(I haven’t defined , but it has the properties you expect.) The proof is as in the real case; even the theorem statement is the same except for the change for the extra subscript of . I won’t elaborate too much on this now, since -adic exponentiation will be described in much more detail in the next post.

3.4. Completeness

Note that the definition of could have been given for as well; we didn’t need to introduce it (after all, we have in olympiads already). The big important theorem I must state now is:

Theorem 20 ( is complete)

The space is the completion of with respect to .

This is the definition of you’ll see more frequently; one then defines in terms of (rather than vice-versa) according to

(Remark for experts: is a field with a non-Arcihmedian valuation; then is its valuation ring.)

Let me justify why this definition is philosophically nice.

Suppose you are a numerical analyst and you want to estimate the value of the sum

to within . The sum consists entirely of rational numbers, so the problem statement would be fair game for ancient Greece. But it turns out that in order to get a good estimate, it really helps if you know about the real numbers: because then you can construct the infinite series , and deduce that , up to some small error term from the terms past , which can be bounded.

Of course, in order to have access to enough theory to prove that , you need to have the real numbers; it’s impossible to do serious analysis in the non-complete space , where e.g. the sequence , , , , \dots is considered “not convergent” because . Instead, all analysis is done in the completion of , namely .

Now suppose you are an olympiad contestant and want to estimate the sum

to within mod (i.e. to within in ). Even though is a rational number, it still helps to be able to do analysis with infinite sums, and then bound the error term (i.e. take mod ). But the space is not complete with respect to either, and thus it makes sense to work in the completion of with respect to . This is exactly .

4. Solving USA TST 2002/2

Let’s finally solve Example~1, which asks to compute

Armed with the generalized binomial theorem, this becomes straightforward.

Using the elementary facts that and , this solves the problem.

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New oly handout: Constructing Diagrams

I’ve added a new Euclidean geometry handout, Constructing Diagrams, to my webpage.

Some of the stuff covered in this handout:

• Advice for constructing the triangle centers (hint: circumcenter goes first)
• An example of how to rearrange the conditions of a problem and draw a diagram out-of-order
• Some mechanical suggestions such as dealing with phantom points
• Some examples of computer-generated figures

Enjoy.

Revisiting Arc Midpoints in Complex Numbers

1. Synopsis

One of the major headaches of using complex numbers in olympiad geometry problems is dealing with square roots. In particular, it is nontrivial to express the incenter of a triangle inscribed in the unit circle in terms of its vertices.

The following lemma is the standard way to set up the arc midpoints of a triangle. It appears for example as part (a) of Lemma 6.23.

Theorem 1 (Arc midpoint setup for a triangle)

Let be a triangle with circumcircle and let , , denote the arc midpoints of opposite , opposite , opposite .

Suppose we view as the unit circle in the complex plane. Then there exist complex numbers , , such that , , , and

Theorem 1 is often used in combination with the following lemma, which lets one assign the incenter the coordinates in the above notation.

Lemma 2 (The incenter is the orthocenter of opposite arc midpoints)

Let be a triangle with circumcircle and let , , denote the arc midpoints of opposite , opposite , opposite . Then the incenter of coincides with the orthocenter of .

Unfortunately, the proof of Theorem 1 in my textbook is wrong, and I cannot find a proof online (though I hear that Lemmas in Olympiad Geometry has a proof). So in this post I will give a correct proof of Theorem 1, which will hopefully also explain the mysterious introduction of the minus signs in the theorem statement. In addition I will give a version of the theorem valid for quadrilaterals.

2. A Word of Warning

I should at once warn the reader that Theorem 1 is an existence result, and thus must be applied carefully.

To see why this matters, consider the following problem, which appeared as problem 1 of the 2016 JMO.

Example 3 (JMO 2016, by Zuming feng)

The isosceles triangle , with , is inscribed in the circle . Let be a variable point on the arc that does not contain , and let and denote the incenters of triangles and , respectively. Prove that as varies, the circumcircle of triangle passes through a fixed point.

By experimenting with the diagram, it is not hard to guess that the correct fixed point is the midpoint of arc , as seen in the figure below. One might be tempted to write , , , and assert the two incenters are and , and that the fixed point is .

This is a mistake! If one applies Theorem 1 twice, then the choices of “square roots” of the common vertices and may not be compatible. In fact, they cannot be compatible, because the arc midpoint of opposite is different from the arc midpoint of opposite .

In fact, I claim this is not a minor issue that one can work around. This is because the claim that the circumcircle of passes through the midpoint of arc is false if lies on the arc on the same side as ! In that case it actually passes through instead. Thus the truth of the problem really depends on the fact that the quadrilateral is convex, and any attempt with complex numbers must take this into account to have a chance of working.

3. Proof of the theorem for triangles

Fix now, so we require , , . There are choices of square roots , , we can take (differing by a sign); we wish to show one of them works.

We pick an arbitrary choice for first. Then, of the two choices of , we pick the one such that . Similarly, for the two choices of , we pick the one such that . Our goal is to show that under these conditions, we have again.

The main trick is to now consider the arc midpoint , which we denote by . It is easy to see that:

Lemma 4 (The isosceles trapezoid trick)

We have (both are perpendicular to the bisector). Thus is an isosceles trapezoid, and so .

Thus, we have

Thus

as desired.

From this we can see why the minus signs are necessary.

Exercise 5

Show that Theorem 1 becomes false if we try to use , , instead of , , .

4. A version for quadrilaterals

We now return to the setting of a convex quadrilateral that we encountered in Example 3. Suppose we preserve the variables , , that we were given from Theorem 1, but now add a fourth complex number with . How are the new arc midpoints determined? The following theorem answers this question.

Theorem 6 ( setup)

Let be a convex quadrilateral inscribed in the unit circle of the complex plane. Then we can choose complex numbers , , , such that , , , and:

• The opposite arc midpoints , , of triangle are given by , , , as before.
• The midpoint of arc not including or is given by .
• The midpoint of arc not including or is given by .
• The midpoint of arc is and the midpoint of arc is .

This setup is summarized in the following figure.

Note that unlike Theorem 1, the four arcs cut out by the sides of do not all have the same sign (I chose to have coordinates ). This asymmetry is inevitable (see if you can understand why from the proof below).

Proof: We select , , with Theorem 1. Now, pick a choice of such that is the arc midpoint of not containing and . Then the arc midpoint of not containing or is given by

On the other hand, the calculation of for the midpoint of follows by applying Lemma 4 again. (applied to triangle ). The midpoint of is computed similarly.

In other problems, the four vertices of the quadrilateral may play more symmetric roles and in that case it may be desirable to pick a setup in which the four vertices are labeled in order. By relabeling the letters in Theorem 6 one can prove the following alternate formulation.

Corollary 7

Let be a convex quadrilateral inscribed in the unit circle of the complex plane. Then we can choose complex numbers , , , such that , , , and:

• The midpoints of , , , cut out by the sides of are , , , .
• The midpoints of and are and .
• The midpoints of and are and .

To test the newfound theorem, here is a cute easy application.

Example 8 (Japanese theorem for cyclic quadrilaterals)

In a cyclic quadrilateral , the incenters of , , , are the vertices of a rectangle.

An Apology for HMMT 2016

Median Putnam contestants, willing to devote one of the last Saturdays before final exams to a math test, are likely to receive an advanced degree in the sciences. It is counterproductive on many levels to leave them feeling like total idiots.

— Bruce Reznick, “Some Thoughts on Writing for the Putnam”

Last February I made a big public apology for having caused one of the biggest scoring errors in HMMT history, causing a lot of changes to the list of top individual students. Pleasantly, I got some nice emails from coaches who reminded me that most students and teams do not place highly in the tournament, and at the end of the day the most important thing is that the contestants enjoyed the tournament.

So now I decided I have to apologize for 2016, too.

The story this time is that I inadvertently sent over 100 students home having solved two or fewer problems total, out of 30 individual problems. That year, I was the problem czar for HMMT February 2016, and like many HMMT problem czars before me, had vastly underestimated the difficulty of my own problems.

I think stories like this are a lot worse than people realize; contests are supposed to be a learning experience for the students, and if a teenager shows up to Massachusetts and spends an entire Saturday feeling hopeless for the entire contest, then the flight back to California is going to feel very long. Now imagine having 100 students go through this every single February.

So today I’d like to say a bit about things I’ve picked up since then that have helped me avoid making similar mistakes. I actually think people generally realize that HMMT is too hard, but are wrong about how this should be fixed. In particular, I think the common approach (and the one I took) of “make problem 1 so easy that almost nobody gets a zero” is wrong, and I’ll explain here what I think should be done instead.

1. Gettable, not gimme

I think just “easy” is the wrong way to think about the beginning problems. At ARML, the problem authors use a finer distinction which I really like:

• A problem is gettable if nearly every contestant feels like they could have gotten the problem on a good day. (In particular, problems that require knowledege that not all contestants have are not gettable, even if they are easy with it.)
• A problem is a gimme if nearly every contestant actually solves the problem on the contest.

The consensus is always that the early problems should be gettable but not gimme’s. You could start every contest by asking the contestant to compute the expected value of 7, but the contestants are going to notice, and it isn’t going to help anyone.

(I guess I should make the point that in order for a problem to be a “gimme”, it would have to be so easy to be almost insulting, because high accuracy on a given problem is really only possible if the level of the problem is significantly below the level of the student. So a gimme would have to be a problem that is way easier than the level of the weakest contestant — you can see why these would be bad.)

In contrast, with a gettable problem, even though some of the contestants will miss it, they’ll often miss it for a reason like 2+3=6. This is a bit unfortunate, but it is still a lot better if the contestant goes home thinking “I made a small arithmetic error, so I have to be more careful” than “there’s no way I could have gotten this, it was hopeless”.

But that brings to me to the next point:

2. At the IMO 33% of the problems are gettable

At the IMO, there are two easy problems (one each day), but there are only six problems. So a full one-third of the problems are gettable: we hope that most students attending the IMO can solve either IMO1 or IMO4, even though many will not solve both.

If you are writing HMMT or some similar contest, I think this means you should think about the opening in terms of the fraction 1/3, rather than problem 1. For example, at HMMT, I think the czars should strive instead to make the first three or four out of ten problems on each individual test gettable: they should be problems every contestant could solve, even though some of them will still miss it anyways. Under the pressure of contest, students are going to make all sorts of mistakes, and so it’s important that there are multiple gettable problems. This way, every student has two or three or four real chances to solve a problem: they’ll still miss a few, but at least they feel like they could do something.

(Every year at HMMT, when we look back at the tests in hindsight, the first reflex many czars have is to look at how many people got 0’s on each test, and hope that it’s not too many. The fact that this figure is even worth looking at is in my opinion a sign that we are doing things wrong: is 1/10 any better than 0/10, if the kid solved question 1 quickly and then spent the rest of the hour staring at the other nine?)

3. Watch the clock

The other thing I want to say is to spend some time thinking about the entire test as a whole, rather than about each problem individually.

To drive the point: I’m willing to bet that an HMMT individual test with 4 easy, 6 medium, and 0 hard problems could actually work, even at the top end of the scores. Each medium problem in isolation won’t distinguish the strongest students. But put six of them all together, and you get two effects:

• Students will make mistakes on some of the problems, and by central limit theorem you’ll get a curve anyways.
• Time pressure becomes significantly more important, and the strongest students will come out ahead by simply being faster.

Of course, I’ll never be able to persuade the problem czars (myself included) to not include at least one or two of those super-nice hard problems. But the point is that they’re not actually needed in situations like HMMT, when there are so many problems that it’s hard to not get a curve of scores.

One suggestion many people won’t take: if you really want to include some difficulty problems that will take a while, decrease the length of the test. If you had 3 easy, 3 medium, and 1 hard problem, I bet that could work too. One hour is really not very much time.

Actually, this has been experimentally verified. On my HMMT 2016 Geometry test, nobody solved any of problems 8-10, so the test was essentially seven problems long. The gradient of scores at the top and center still ended up being okay. The only issue was that a third of the students solved zero problems, because the easy problems were either error-prone, or else were hit-or-miss (either solved quickly or not at all). Thus that’s another thing to watch out for.

A story of block-ascending permutations

I recently had a combinatorics paper appear in the EJC. In this post I want to brag a bit by telling the “story” of this paper: what motivated it, how I found the conjecture that I originally did, and the process that eventually led me to the proof, and so on.

This work was part of the Duluth REU 2017, and I thank Joe Gallian for suggesting the problem.

1. Background

Let me begin by formulating the problem as it was given to me. First, here is the definition and notation for a “block-ascending” permutation.

Definition 1

For nonnegative integers , …, an -ascending permutation is a permutation on whose descent set is contained in . In other words the permutation ascends in blocks of length , , …, , and thus has the form

for which for all .

It turns out that block-ascending permutations which also avoid an increasing subsequence of certain length have nice enumerative properties. To this end, we define the following notation.

Definition 2

Let denote the set of -ascending permutations which avoid the pattern .

(The reason for using will be explained later.) In particular, if .

Example 3

Here is a picture of a permutation in (but not in , since one can see an increasing length subsequence shaded). We would denote it .

Now on to the results. A 2011 paper by Joel Brewster Lewis (JBL) proved (among other things) the following result:

Theorem 4 (Lewis 2011)

The sets and are in bijection with Young tableau of shape .

Remark 5

When , this implies , which is the set of -avoiding permutations of length , is in bijection with the Catalan numbers; so is which is the set of -avoiding zig-zag permutations.

Just before the Duluth REU in 2017, Mei and Wang proved that in fact, in Lewis’ result one may freely mix and ‘s. To simplify notation,

Definition 6

Let . Then denotes where

Theorem 7 (Mei, Wang 2017)

The sets are also in bijection with Young tableau of shape .

The proof uses the RSK correspondence, but the authors posed at the end of the paper the following open problem:

Problem

Find a direct bijection between the sets above, not involving the RSK correspondence.

This was the first problem that I was asked to work on. (I remember I received the problem on Sunday morning; this actually matters a bit for the narrative later.)

At this point I should pause to mention that this notation is my own invention, and did not exist when I originally started working on the problem. Indeed, all the results are restricted to the case where for each , and so it was unnecessary to think about other possibilities for : Mei and Wang’s paper use the notation . So while I’ll continue to use the notation in the blog post for readability, it will make some of the steps more obvious than they actually were.

2. Setting out

Mei and Wang’s paper originally suggested that rather than finding a bijection for any and , it would suffice to biject

and then compose two such bijections. I didn’t see why this should be much easier, but it didn’t seem to hurt either.

As an example, they show how to do this bijection with and . Indeed, suppose . Then is an increasing sequence of length right at the start of . So had better be the largest element in the permutation: otherwise later in the biggest element would complete an ascending permutation of length already! So removing gives a bijection between .

But if you look carefully, this proof does essentially nothing with the later blocks. The exact same proof gives:

Proposition 8

Suppose . Then there is a bijection

by deleting the st element of the permutation (which must be largest one).

Once I found this proposition I rejected the initial suggestion of specializing . The “easy case” I had found told me that I could take a set and delete the single element from it. So empirically, my intuition from this toy example told me that it would be easier to find bijections whee and were only “a little different”, and hope that the resulting bijection only changed things a little bit (in the same way that in the toy example, all the bijection did was delete one element). So I shifted to trying to find small changes of this form.

3. The fork in the road

3.1. Wishful thinking

I had a lucky break of wishful thinking here. In the notation with , I had found that one could replace with either or freely. (But this proof relied heavily on the fact the block really being on the far left.) So what other changes might I be able to make?

There were two immediate possibilities that came to my mind.

• Deletion: We already showed could be changed from to for any . If we can do a similar deletion with for any , not just , then we would be done.
• Swapping: If we can show that two adjacent ‘s could be swapped, that would be sufficient as well. (It’s also possible to swap non-adjacent ‘s, but that would cause more disruption for no extra benefit.)

Now, I had two paths that both seemed plausible to chase after. How was I supposed to know which one to pick? (Of course, it’s possible neither work, but you have to start somewhere.)

Well, maybe the correct thing to do would have to just try both. But it was Sunday afternoon by the time I got to this point. Granted, it was summer already, but I knew that come Monday I would have doctor appointments and other trivial errands to distract me, so I decided I should pick one of them and throw the rest of the day into it. But that meant I had to pick one.

(I confess that I actually already had a prior guess: the deletion approach seemed less likely to work than the swapping approach. In the deletion approach, if is somewhere in the middle of the permutation, it seemed like deleting an element could cause a lot of disruption. But the swapping approach preserved the total number of elements involved, and so seemed more likely that I could preserve structure. But really I was just grasping at straws.)

3.2. Enter C++

Yeah, I cheated. Sorry.

Those of you that know anything about my style of math know that I am an algebraist by nature — sort of. It’s more accurate to say that I depend on having concrete examples to function. True, I can’t do complexity theory for my life, but I also haven’t been able to get the hang of algebraic geometry, despite having tried to learn it three or four times by now. But enumerative combinatorics? OH LOOK EXAMPLES.

Here’s the plan: let . Then using a C++ computer program:

• Enumerate all the permutations in .
• Enumerate all the permutations in .
• Enumerate all the permutations in .

If the deletion approach is right, then I would hope and look pretty similar. On the flip side, if the swapping approach is right, then and should look close to each other instead.

It’s moments like this where my style of math really shines. I don’t have to make decisions like the above off gut-feeling: do the “data science” instead.

3.3. A twist of fate

Except this isn’t actually what I did, since there was one problem. Computing the longest increasing subsequence of a length permutation takes time, and there are or so permutations. But when , we have , which is a pretty big number. Unfortunately, my computer is not really that fast, and I didn’t really have the patience to implement the “correct” algorithms to bring the runtime down.

The solution? Use instead.

In a deep irony that I didn’t realize at the time, it was this moment when I introduced the notation, and for the first time allowed the to not be in . My reasoning was that since I was only doing this for heuristic reasons, I could instead work with and probably not change much about the structure of the problem, while replacing , which would run times faster. This was okay since all I wanted to do was see how much changing the “middle” would disrupt the structure.

And so the new plan was:

• Enumerate all the permutations in .
• Enumerate all the permutations in .
• Enumerate all the permutations in .

I admit I never actually ran the enumeration with , because the route with and turned out to be even more promising than I expected. When I compared the empirical data for the sets and , I found that the number of permutations with any particular triple were equal. In other words, the outer blocks were preserved: the bijection

does not tamper with the outside blocks of length and .

This meant I was ready to make the following conjecture. Suppose , . There is a bijection

which only involves rearranging the elements of the th and st blocks.

4. Rooting out the bijection

At this point I was in a quite good position. I had pinned down the problem to a finding a particular bijection that I was confident had to exist, since it was showing up to the empirical detail.

Let’s call this mythical bijection . How could I figure out what it was?

4.1. Hunch: preserves order-isomorphism

Let me quickly introduce a definition.

Definition 9

We say two words and are order-isomorphic if if and only . Then order-isomorphism gives equivalence classes, and there is a canonical representative where the letters are ; this is called a reduced word.

Example 10

The words , and are order-isomorphic; the last is reduced.

Now I guessed one more property of : this should order-isomorphism.

What do I mean by this? Suppose in one context changed to ; then we would expect that in another situation we should have changing to . Indeed, we expect (empirically) to not touch surrounding outside blocks, and so it would be very strange if behaved differently due to far-away numbers it wasn’t even touching.

So actually I’ll just write

for this example, reducing the words in question.

4.2. Keep cheating

With this hunch it’s possible to cheat with C++ again. Here’s how.

Let’s for concreteness suppose and the particular sets

Well, it turns out if you look at the data:

• The only element of which starts with and ends with is .
• The only element of which starts with and ends with is .

So that means that is changed to . Thus the empirical data shows that

In general, it might not be that clear cut. For example, if we look at the permutations starting with and , there is more than one.

• and are both in .
• and are both in in .

Thus

but we can’t tell which one goes to which (although you might be able to guess).

Fortunately, there is lots of data. This example narrowed down to two values, but if you look at other places you might have different data on . Since we think is behaving the same “globally”, we can piece together different pieces of data to get narrower sets. Even better, is a bijection, so once we match either of or , we’ve matched the other.

You know what this sounds like? Perfect matchings.

So here’s the experimental procedure.

• Enumerate all permutations in and .
• Take each possible tuple , and look at the permutations that start and end with those particular four elements. Record the reductions of and for all these permutations. We call these input words and output words, respectively. Each output word is a “candidate” of for a input word.
• For each input word that appeared, take the intersection of all output words that appeared. This gives a bipartite graph , with input words being matched to their candidates.
• Find perfect matchings of the graph.

And with any luck that would tell us what is.

4.3. Results

Luckily, the bipartite graph is quite sparse, and there was only one perfect matching.

246|1357 => 2467|135
247|1356 => 2457|136
256|1347 => 2567|134
257|1346 => 2357|146
267|1345 => 2367|145
346|1257 => 3467|125
347|1256 => 3457|126
356|1247 => 3567|124
357|1246 => 1357|246
367|1245 => 1367|245
456|1237 => 4567|123
457|1236 => 1457|236
467|1235 => 1467|235
567|1234 => 1567|234

If you look at the data, well, there are some clear patterns. Exactly one number is “moving” over from the right half, each time. Also, if is on the right half, then it always moves over.

Anyways, if you stare at this for an hour, you can actually figure out the exact rule:

Claim 11

Given an input , move if is the largest index for which , or if no such index exists.

And indeed, once I have this bijection, it takes maybe only another hour of thinking to verify that this bijection works as advertised, thus solving the original problem.

Rather than writing up what I had found, I celebrated that Sunday evening by playing Wesnoth for 2.5 hours.

5. Generalization

5.1. Surprise

On Monday morning I was mindlessly feeding inputs to the program I had worked on earlier and finally noticed that in fact and also had the same cardinality. Huh.

It seemed too good to be true, but I played around some more, and sure enough, the cardinality of seemed to only depend on the order of the ‘s. And so at last I stumbled upon the final form the conjecture, realizing that all along the assumption that I had been working with was a red herring, and that the bijection was really true in much vaster generality. There is a bijection

which only involves rearranging the elements of the th and st blocks.

It also meant I had more work to do, and so I was now glad that I hadn’t written up my work from yesterday night.

5.2. More data science

I re-ran the experiment I had done before, now with . (This was interesting, because the elements in question could now have either longest increasing subsequence of length , or instead of length .)

The data I obtained was:

246|13578 => 24678|135
247|13568 => 24578|136
248|13567 => 24568|137
256|13478 => 25678|134
257|13468 => 23578|146
258|13467 => 23568|147
267|13458 => 23678|145
268|13457 => 23468|157
278|13456 => 23478|156
346|12578 => 34678|125
347|12568 => 34578|126
348|12567 => 34568|127
356|12478 => 35678|124
357|12468 => 13578|246
358|12467 => 13568|247
367|12458 => 13678|245
368|12457 => 13468|257
378|12456 => 13478|256
456|12378 => 45678|123
457|12368 => 14578|236
458|12367 => 14568|237
467|12358 => 14678|235
468|12357 => 12468|357
478|12356 => 12478|356
567|12348 => 15678|234
568|12347 => 12568|347
578|12346 => 12578|346
678|12345 => 12678|345

Okay, so it looks like:

• exactly two numbers are moving each time, and
• the length of the longest run is preserved.

Eventually, I was able to work out the details, but they’re more involved than I want to reproduce here. But the idea is that you can move elements “one at a time”: something like

while preserving the length of increasing subsequences at each step.

So, together with the easy observation from the beginning, this not only resolves the original problem, but also gives an elegant generalization. I had now proved:

Theorem 12

For any , …, , the cardinality

does not depend on the order of the ‘s.

6. Discovered vs invented

Whenever I look back on this, I can’t help thinking just how incredibly lucky I got on this project.

There’s this perpetual debate about whether mathematics is discovered or invented. I think it’s results like this which make the case for “discovered”. I did not really construct the bijection myself: it was “already there” and I found it by examining the data. In another world where did not exist, all the creativity in the world wouldn’t have changed anything.

So anyways, that’s the behind-the-scenes tour of my favorite combinatorics paper.

Joyal’s Proof of Cayley’s Tree Formula

I wanted to quickly write this proof up, complete with pictures, so that I won’t forget it again. In this post I’ll give a combinatorial proof (due to Joyal) of the following:

Theorem 1 (Cayley’s Formula)

The number of trees on labelled vertices is .

Proof: We are going to construct a bijection between

• Functions (of which there are ) and
• Trees on with two distinguished nodes and (possibly ).

This will imply the answer.

Let’s look at the first piece of data. We can visualize it as points floating around, each with an arrow going out of it pointing to another point, but possibly with many other arrows coming into it. Such a structure is apparently called a directed pseudoforest. Here is an example when .

You’ll notice that in each component, some of the points lie in a cycle and others do not. I’ve colored the former type of points blue, and the corresponding arrows magenta.

Thus a directed pseudoforest can also be specified by

• a choice of some vertices to be in cycles (blue vertices),
• a permutation on the blue vertices (magenta arrows), and
• attachments of trees to the blue vertices (grey vertices and arrows).

Now suppose we take the same information, but replace the permutation on the blue vertices with a total ordering instead (of course there are an equal number of these). Then we can string the blue vertices together as shown below, where the green arrows denote the selected total ordering (in this case ):

This is exactly the data of a tree on the vertices with two distinguished vertices, the first and last in the chain of green (which could possibly coincide).

Positive Definite Quadratic Forms

I’m reading through Primes of the Form , by David Cox (link; it’s good!). Here are the high-level notes I took on the first chapter, which is about the theory of quadratic forms.

(Meta point re blog: I’m probably going to start posting more and more of these more high-level notes/sketches on this blog on topics that I’ve been just learning. Up til now I’ve been mostly only posting things that I understand well and for which I have a very polished exposition. But the perfect is the enemy of the good here; given that I’m taking these notes for my own sake, I may as well share them to help others.)

1. Overview

Definition 1

For us a quadratic form is a polynomial , where , , are some integers. We say that it is primitive if .

For example, we have the famous quadratic form

As readers are probably aware, we can say a lot about exactly which integers can be represented by : by Fermat’s Christmas theorem, the primes (and ) can all be written as the sum of two squares, while the primes cannot. For convenience, let us say that:

Definition 2

Let be a quadratic form. We say it represents the integer if there exists with . Moreover, properly represents if one can find such and which are also relatively prime.

The basic question is: what can we say about which primes/integers are properly represented by a quadratic form? In fact, we will later restrict our attention to “positive definite” forms (described later).

For example, Fermat’s Christmas theorem now rewrites as:

Theorem 3 (Fermat’s Christmas theorem for primes)

An odd prime is (properly) represented by if and only if .

The proof of this is classical, see for example my olympiad handout. We also have the formulation for odd integers:

Theorem 4 (Fermat’s Christmas theorem for odd integers)

An odd integer is properly represented by if and only if all prime factors of are .

Proof: For the “if” direction, we use the fact that is multiplicative in the sense that

For the “only if” part we use the fact that if a multiple of a prime is properly represented by , then so is . This follows by noticing that if (and ) then .
Tangential remark: the two ideas in the proof will grow up in the following way.

• The fact that “multiplies nicely” will grow up to become the so-called composition of quadratic forms.
• The second fact will not generalize for an arbitrary form . Instead, we will see that if a multiple of is represented by a form then some form of the same “discriminant” will represent the prime , but this form need not be the same as itself.

2. Equivalence of forms, and the discriminant

The first thing we should do is figure out when two forms are essentially the same: for example, and should clearly be considered the same. More generally, if we think of as acting on and is any automorphism of , then should be considered the same as . Specifically,

Definition 5

Two forms and said to be equivalent if there exists

such that . We have and so we say the equivalence is

• a proper equivalence if , and
• an improper equivalence if .

So we generally will only care about forms up to proper equivalence. (It will be useful to distinguish between proper/improper equivalence later.)

Naturally we seek some invariants under this operation. By far the most important is:

Definition 6

The discriminant of a quadratic form is defined as

The discriminant is invariant under equivalence (check this). Note also that we also have .

Observe that we have

So if and (thus too) then for all . Such quadratic forms are called positive definite, and we will restrict our attention to these forms.

Now that we have this invariant, we may as well classify equivalence classes of quadratic forms for a fixed discriminant. It turns out this can be done explicitly.

Definition 7

A quadratic form is reduced if

• it is primitive and positive definite,
• , and
• if either or .

Exercise 8

Check there only finitely many reduced forms of a fixed discriminant.

Then the big huge theorem is:

Theorem 9 (Reduced forms give a set of representatives)

Every primitive positive definite form of discriminant is properly equivalent to a unique reduced form. We call this the reduction of .

Proof: Omitted due to length, but completely elementary. It is a reduction argument with some number of cases.

Thus, for any discriminant we can consider the set

which will be the equivalence classes of positive definite of discriminant . By abuse of notation we will also consider it as the set of equivalence classes of primitive positive definite forms of discriminant .

We also define ; by the exercise, . This is called the class number.

Moreover, we have , because we can take for and for . We call this form the principal form.

3. Tables of quadratic forms

Example 10 (Examples of quadratic forms with , )

The following discriminants have class number , hence having only the principal form:

• , with form .
• , with form .
• , with form .
• , with form .
• , with form .

This is in fact the complete list when .

Example 11 (Examples of quadratic forms with , )

The following discriminants have class number , hence having only the principal form:

• , with form .
• , with form .
• , with form .
• , with form .
• , with form .
• , with form .
• , with form .
• , with form .

This is in fact the complete list when .

Example 12 (More examples of quadratic forms)

Here are tables for small discriminants with . When we have

• , with forms and .
• , with forms and .
• , with forms and .
• , with forms and .
• , with forms and .
• , with forms and .

As for we have

• , with forms and .
• , with forms and .
• , with forms and .
• , with forms , and .

Example 13 (Even More Examples of quadratic forms)

Here are some more selected examples:

• has forms , and .
• has forms and .
• has forms , and .

4. The Character

We can now connect this to primes as follows. Earlier we played with , and observed that for odd primes , if and only if some multiple of is properly represented by .

Our generalization is as follows:

Theorem 14 (Primes represented by some quadratic form)

Let be a discriminant, and let be an odd prime. Then the following are equivalent:

• , i.e. is a quadratic residue modulo .
• The prime is (properly) represented by some reduced quadratic form in .

This generalizes our result for , but note that it uses in an essential way! That is: if , we know is represented by some quadratic form of discriminant \dots but only since do we know that this form reduces to .

Proof: First assume WLOG that and . Thus , since otherwise this would imply . Then

hence .

The converse direction is amusing: let for integers , . Consider the quadratic form

It is primitive of discriminant and . Now may not be reduced, but that’s fine: just take the reduction of , which must also properly represent .

Thus to every discriminant we can attach the Legendre character (is that the name?), which is a homomorphism

with the property that if is a rational prime not dividing , then . This is abuse of notation since I should technically write , but there is no harm done: one can check by quadratic reciprocity that if then . Thus our previous result becomes:

Theorem 15 ( consists of representable primes)

Let be prime. Then if and only if some quadratic form in represents .

As a corollary of this, using the fact that one can prove that

Corollary 16 (Fermat-type results for )

Let be a prime. Then is

• of the form if and only if .
• of the form if and only if .
• of the form if and only if .

Proof: The congruence conditions are equivalent to , and as before the only point is that the only reduced quadratic form for these is the principal one.

5. Genus theory

What if ? Sometimes, we can still figure out which primes go where just by taking mods.

Let . Then it represents some residue classes of . In that case we call the set of residue classes represented the genus of the quadratic form .

Example 17 (Genus theory of )

Consider , with

We consider the two elements of :

• represents .
• represents .

Now suppose for example that . It must be represented by one of these two quadratic forms, but the latter form is never and so it must be the first one. Thus we conclude that

• if and only if .
• if and only if .

The thing that makes this work is that each genus appears exactly once. We are not always so lucky: for example when we have that

Example 18 (Genus theory of )

The two elements of are:

• , which represents exactly the elements of .
• , which also represents exactly the elements of .

So the best we can conclude is that OR if and only if This is because the two distinct quadratic forms of discriminant happen to have the same genus.

We now prove that:

Theorem 19 (Genii are cosets of )

Let be a discriminant and consider the Legendre character .

• The genus of the principal form of discriminant constitutes a subgroup of , which we call the principal genus.
• Any genus of a quadratic form in is a coset of the principal genus in .

Proof: For the first part, we aim to show is multiplicatively closed. For , we use the fact that

For , we instead appeal to another “magic” identity

and it follows from here that is actually the set of squares in , which is obviously a subgroup.

Now we show that other quadratic forms have genus equal to a coset of the principal genus. For , with we can write

and thus the desired coset is shown to be . As for , we have

so the desired coset is also , since was the set of squares.

Thus every genus is a coset of in . Thus:

Definition 20

We define the quotient group

which is the set of all genuses in discriminant . One can view this as an abelian group by coset multiplication.

Thus there is a natural map

(The map is surjective by Theorem~14.) We also remark than is quite well-behaved:

Proposition 21 (Structure of )

The group is isomorphic to for some integer .

Proof: Observe that contains all the squares of : if is the principal form then . Thus claim each element of has order at most , which implies the result since is a finite abelian group.

In fact, one can compute the order of exactly, but for this post I Will just state the result.

Theorem 22 (Order of )

Let be a discriminant, and let be the number of distinct odd primes which divide . Define by:

• if .
• if and .
• if and .
• if and .
• if and .

Then .

6. Composition

We have already used once the nice identity

We are going to try and generalize this for any two quadratic forms in . Specifically,

Proposition 23 (Composition defines a group operation)

Let . Then there is a unique and bilinear forms for such that

• .
• .
• .

In fact, without the latter two constraints we would instead have and , and each choice of signs would yield one of four (possibly different) forms. So requiring both signs to be positive makes this operation well-defined. (This is why we like proper equivalence; it gives us a well-defined group structure, whereas with improper equivalence it would be impossible to put a group structure on the forms above.)

Taking this for granted, we then have that

Theorem 24 (Form class group)

Let , be a discriminant. Then becomes an abelian group under composition, where

• The identity of is the principal form, and
• The inverse of the form is .

This group is called the form class group.

We then have a group homomorphism

Observe that and are inverses and that their images coincide (being improperly equivalent); this is expressed in the fact that has elements of order . As another corollary, the number of elements of with a given genus is always a power of two.

We now define:

Definition 25

An integer is convenient if the following equivalent conditions hold:

• The principal form is the only reduced form with the principal genus.
• is injective (hence an isomorphism).
• .

Thus we arrive at the following corollary:

Corollary 26 (Convenient numbers have nice representations)

Let be convenient. Then is of the form if and only if lies in the principal genus.

Hence the represent-ability depends only on .

OEIS A000926 lists 65 convenient numbers. This sequence is known to be complete except for at most one more number; moreover the list is complete assuming the Grand Riemann Hypothesis.

7. Cubic and quartic reciprocity

To treat the cases where is not convenient, the correct thing to do is develop class field theory. However, we can still make a little bit more progress if we bring higher reciprocity theorems to bear: we’ll handle the cases and , two examples of numbers which are not convenient.

7.1. Cubic reciprocity

First, we prove that

Theorem 27 (On )

A prime is of the form if and only if and is a cubic residue modulo .

To do this we use cubic reciprocity, which requires working in the Eisenstein integers where is a cube root of unity. There are six units in (the sixth roots of unity), hence each nonzero number has six associates (differing by a unit), and the ring is in fact a PID.

Now if we let be a prime not dividing , and is coprime to , then we can define the cubic Legendre symbol by setting

Moreover, we can define a primary prime to be one such that ; given any prime exactly one of the six associates is primary. We then have the following reciprocity theorem:

Theorem 28 (Cubic reciprocity)

If and are disjoint primary primes in then

We also have the following supplementary laws: if , then

The first supplementary law is for the unit (analogous to ) while the second reciprocity law handles the prime divisors of (analogous to .)

We can tie this back into as follows. If is a rational prime then it is represented by , and thus we can put for some prime , . Consequently, we have a natural isomorphism

Therefore, we see that a given is a cubic residue if and only if .

In particular, we have the following corollary, which is all we will need:

Corollary 29 (When is a cubic residue)

Let be a rational prime, . Write with primary. Then is a cubic residue modulo if and only if .

Proof: By cubic reciprocity:

Now we give the proof of Theorem~27. Proof: First assume

Let be primary, noting that . Now clearly , so done by corollary.

For the converse, assume , with primary and . If we set for integers and , then the fact that and is enough to imply that (check it!). Moreover,

as desired.

7.2. Quartic reciprocity

This time we work in , for which there are four units , . A prime is primary if ; every prime not dividing has a unique associate which is primary. Then we can as before define

where is primary, and is nonzero mod . As before , we have that is a quartic residue modulo if and only if thanks to the isomorphism

Now we have

Theorem 30 (Quartic reciprocity)

If and are distinct primary primes in then

We also have supplementary laws that state that if is primary, then

Again, the first law handles units, and the second law handles the prime divisors of . The corollary we care about this time in fact uses only the supplemental laws:

Corollary 31 (When is a quartic residue)

Let be a prime, and put with primary. Then

and in particular is a quartic residue modulo if and only if .

Proof: Note that and applying the above. Therefore

Now we assumed is primary. We claim that

Note that since was is divisible by , hence divides . Thus

since is odd and is even. Finally,

From here we quickly deduce

Theorem 32 (On )

If is prime, then if and only if and is a quartic residue modulo .

Some Thoughts on Olympiad Material Design

(This is a bit of a follow-up to the solution reading post last month. Spoiler warnings: USAMO 2014/6, USAMO 2012/2, TSTST 2016/4, and hints for ELMO 2013/1, IMO 2016/2.)

I want to say a little about the process which I use to design my olympiad handouts and classes these days (and thus by extension the way I personally think about problems). The short summary is that my teaching style is centered around showing connections and recurring themes between problems.

Now let me explain this in more detail.

1. Main ideas

Solutions to olympiad problems can look quite different from one another at a surface level, but typically they center around one or two main ideas, as I describe in my post on reading solutions. Because details are easy to work out once you have the main idea, as far as learning is concerned you can more or less throw away the details and pay most of your attention to main ideas.

Thus whenever I solve an olympiad problem, I make a deliberate effort to summarize the solution in a few sentences, such that I basically know how to do it from there. I also make a deliberate effort, whenever I write up a solution in my notes, to structure it so that my future self can see all the key ideas at a glance and thus be able to understand the general path of the solution immediately.

The example I’ve previously mentioned is USAMO 2014/6.

Example 1 (USAMO 2014, Gabriel Dospinescu)

Prove that there is a constant with the following property: If are positive integers such that for all , then

If you look at any complete solution to the problem, you will see a lot of technical estimates involving and the like. But the main idea is very simple: “consider an table of primes and note the small primes cannot adequately cover the board, since ”. Once you have this main idea the technical estimates are just the grunt work that you force yourself to do if you’re a contestant (and don’t do if you’re retired like me).

Thus the study of olympiad problems is reduced to the study of main ideas behind these problems.

2. Taxonomy

So how do we come up with the main ideas? Of course I won’t be able to answer this question completely, because therein lies most of the difficulty of olympiads.

But I do have some progress in this way. It comes down to seeing how main ideas are similar to each other. I spend a lot of time trying to classify the main ideas into categories or themes, based on how similar they feel to one another. If I see one theme pop up over and over, then I can make it into a class.

I think olympiad taxonomy is severely underrated, and generally not done correctly. The status quo is that people do bucket sorts based on the particular technical details which are present in the problem. This is correlated with the main ideas, but the two do not always coincide.

An example where technical sort works okay is Euclidean geometry. Here is a simple example: harmonic bundles in projective geometry. As I explain in my book, there are a few “basic” configurations involved:

• Midpoints and parallel lines
• The Ceva / Menelaus configuration
• Harmonic quadrilateral / symmedian configuration
• Apollonian circle (right angle and bisectors)

(For a reference, see Lemmas 2, 4, 5 and Exercise 0 here.) Thus from experience, any time I see one of these pictures inside the current diagram, I think to myself that “this problem feels projective”; and if there is a way to do so I try to use harmonic bundles on it.

An example where technical sort fails is the “pigeonhole principle”. A typical problem in such a class looks something like USAMO 2012/2.

Example 2 (USAMO 2012, Gregory Galperin)

A circle is divided into congruent arcs by points. The points are colored in four colors such that some points are colored Red, some points are colored Green, some points are colored Blue, and the remaining points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.

It’s true that the official solution uses the words “pigeonhole principle” but that is not really the heart of the matter; the key idea is that you consider all possible rotations and count the number of incidences. (In any case, such calculations are better done using expected value anyways.)

Now why is taxonomy a good thing for learning and teaching? The reason is that building connections and seeing similarities is most easily done by simultaneously presenting several related problems. I’ve actually mentioned this already in a different blog post, but let me give the demonstration again.

Suppose I wrote down the following:

You can tell what each of the ‘s, ‘s, ‘s have in common by looking for a few moments. But what happens if I intertwine them?

This is the same information, but now you have to work much harder to notice the association between the letters and the numbers they’re next to.

This is why, if you are an olympiad student, I strongly encourage you to keep a journal or blog of the problems you’ve done. Solving olympiad problems takes lots of time and so it’s worth it to spend at least a few minutes jotting down the main ideas. And once you have enough of these, you can start to see new connections between problems you haven’t seen before, rather than being confined to thinking about individual problems in isolation. (Additionally, it means you will never have redo problems to which you forgot the solution — learn from my mistake here.)

3. Ten buckets of geometry

I want to elaborate more on geometry in general. These days, if I see a solution to a Euclidean geometry problem, then I mentally store the problem and solution into one (or more) buckets. I can even tell you what my buckets are:

1. Direct angle chasing
2. Power of a point / radical axis
3. Homothety, similar triangles, ratios
4. Recognizing some standard configuration (see Yufei for a list)
5. Doing some length calculations
6. Complex numbers
7. Barycentric coordinates
8. Inversion
9. Harmonic bundles or pole/polar and homography
10. Spiral similarity, Miquel points

which my dedicated fans probably recognize as the ten chapters of my textbook. (Problems may also fall in more than one bucket if for example they are difficult and require multiple key ideas, or if there are multiple solutions.)

Now whenever I see a new geometry problem, the diagram will often “feel” similar to problems in a certain bucket. Exactly what I mean by “feel” is hard to formalize — it’s a certain gut feeling that you pick up by doing enough examples. There are some things you can say, such as “problems which feature a central circle and feet of altitudes tend to fall in bucket 6”, or “problems which only involve incidence always fall in bucket 9”. But it seems hard to come up with an exhaustive list of hard rules that will do better than human intuition.

4. How do problems feel?

But as I said in my post on reading solutions, there are deeper lessons to teach than just technical details.

For examples of themes on opposite ends of the spectrum, let’s move on to combinatorics. Geometry is quite structured and so the themes in the main ideas tend to translate to specific theorems used in the solution. Combinatorics is much less structured and many of the themes I use in combinatorics cannot really be formalized. (Consequently, since everyone else seems to mostly teach technical themes, several of the combinatorics themes I teach are idiosyncratic, and to my knowledge are not taught by anyone else.)

For example, one of the unusual themes I teach is called Global. It’s about the idea that to solve a problem, you can just kind of “add up everything at once”, for example using linearity of expectation, or by double-counting, or whatever. In particular these kinds of approach ignore the “local” details of the problem. It’s hard to make this precise, so I’ll just give two recent examples.

Example 3 (ELMO 2013, Ray Li)

Let be nine real numbers, not necessarily distinct, with average . Let denote the number of triples for which . What is the minimum possible value of ?

Example 4 (IMO 2016)

Find all integers for which each cell of table can be filled with one of the letters , and in such a way that:

• In each row and column, one third of the entries are , one third are and one third are ; and
• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are , one third are and one third are .

If you look at the solutions to these problems, they have the same “feeling” of adding everything up, even though the specific techniques are somewhat different (double-counting for the former, diagonals modulo for the latter). Nonetheless, my experience with problems similar to the former was immensely helpful for the latter, and it’s why I was able to solve the IMO problem.

5. Gaps

This perspective also explains why I’m relatively bad at functional equations. There are some things I can say that may be useful (see my handouts), but much of the time these are just technical tricks. (When sorting functional equations in my head, I have a bucket called “standard fare” meaning that you “just do work”; as far I can tell this bucket is pretty useless.) I always feel stupid teaching functional equations, because I never have many good insights to say.

Part of the reason is that functional equations often don’t have a main idea at all. Consequently it’s hard for me to do useful taxonomy on them.

Then sometimes you run into something like the windmill problem, the solution of which is fairly “novel”, not being similar to problems that come up in training. I have yet to figure out a good way to train students to be able to solve windmill-like problems.

6. Surprise

I’ll close by mentioning one common way I come up with a theme.

Sometimes I will run across an olympiad problem which I solve quickly, and think should be very easy, and yet once I start grading I find that the scores are much lower than I expected. Since the way I solve problems is by drawing experience from similar previous problems, this must mean that I’ve subconsciously found a general framework to solve problems like , which is not obvious to my students yet. So if I can put my finger on what that framework is, then I have something new to say.

The most recent example I can think of when this happened was TSTST 2016/4 which was given last June (and was also a very elegant problem, at least in my opinion).

Example 5 (TSTST 2016, Linus Hamilton)

Let be a positive integers. Prove that we must apply the Euler function at least times before reaching .

I solved this problem very quickly when we were drafting the TSTST exam, figuring out the solution while walking to dinner. So I was quite surprised when I looked at the scores for the problem and found out that empirically it was not that easy.

After I thought about this, I have a new tentative idea. You see, when doing this problem I really was thinking about “what does this operation do?”. You can think of as an infinite tuple

of prime exponents. Then the can be thought of as an operation which takes each nonzero component, decreases it by one, and then adds some particular vector back. For example, if then is decreased by one and each of and are increased by one. In any case, if you look at this behavior for long enough you will see that the coordinate is a natural way to “track time” in successive operations; once you figure this out, getting the bound of is quite natural. (Details left as exercise to reader.)

Now when I read through the solutions, I found that many of them had not really tried to think of the problem in such a “structured” way, and had tried to directly solve it by for example trying to prove (which is false) or something similar to this. I realized that had the students just ignored the task “prove ” and spent some time getting a better understanding of the structure, they would have had a much better chance at solving the problem. Why had I known that structural thinking would be helpful? I couldn’t quite explain it, but it had something to do with the fact that the “main object” of the question was “set in stone”; there was no “degrees of freedom” in it, and it was concrete enough that I felt like I could understand it. Once I understood how multiple operations behaved, the bit about almost served as an “answer extraction” mechanism.

These thoughts led to the recent development of a class which I named Rigid, which is all about problems where the point is not to immediately try to prove what the question asks for, but to first step back and understand completely how a particular rigid structure (like the in this problem) behaves, and to then solve the problem using this understanding.

Holomorphic Logarithms and Roots

In this post we’ll make sense of a holomorphic square root and logarithm. Wrote this up because I was surprised how hard it was to find a decent complete explanation.

Let be a holomorphic function. A holomorphic th root of is a function such that for all . A logarithm of is a function such that for all . The main question we’ll try to figure out is: when do these exist? In particular, what if ?

1. Motivation: Square Root of a Complex Number

To start us off, can we define for any complex number ?

The first obvious problem that comes up is that there for any , there are two numbers such that . How can we pick one to use? For our ordinary square root function, we had a notion of “positive”, and so we simply took the positive root.

Let’s expand on this: given (here ) we should take the root to be

such that ; there are two choices for , differing by .

For complex numbers, we don’t have an obvious way to pick . Nonetheless, perhaps we can also get away with an arbitrary distinction: let’s see what happens if we just choose the with .

Pictured below are some points (in red) and their images (in blue) under this “upper-half” square root. The condition on means we are forcing the blue points to lie on the right-half plane.

Here, for each , and we are constraining the to lie in the right half of the complex plane. We see there is an obvious issue: there is a big discontinuity near the point and ! The nearby point has been mapped very far away. This discontinuity occurs since the points on the negative real axis are at the “boundary”. For example, given , we send it to , but we have hit the boundary: in our interval , we are at the very left edge.

The negative real axis that we must not touch is is what we will later call a branch cut, but for now I call it a ray of death. It is a warning to the red points: if you cross this line, you will die! However, if we move the red circle just a little upwards (so that it misses the negative real axis) this issue is avoided entirely, and we get what seems to be a “nice” square root.

In fact, the ray of death is fairly arbitrary: it is the set of “boundary issues” that arose when we picked . Suppose we instead insisted on the interval ; then the ray of death would be the positive real axis instead. The earlier circle we had now works just fine.

What we see is that picking a particular -interval leads to a different set of edge cases, and hence a different ray of death. The only thing these rays have in common is their starting point of zero. In other words, given a red circle and a restriction of , I can make a nice “square rooted” blue circle as long as the ray of death misses it.

So, what exactly is going on?

2. Square Roots of Holomorphic Functions

To get a picture of what’s happening, we would like to consider a more general problem: let be holomorphic. Then we want to decide whether there is a such that

Our previous discussion when tells us we cannot hope to achieve this for ; there is a “half-ray” which causes problems. However, there are certainly functions such that a exists. As a simplest example, should definitely have a square root!

Now let’s see if we can fudge together a square root. Earlier, what we did was try to specify a rule to force one of the two choices at each point. This is unnecessarily strict. Perhaps we can do something like the following: start at a point in , pick a square root of , and then try to “fudge” from there the square roots of the other points. What do I mean by fudge? Well, suppose is a point very close to , and we want to pick a square root of . While there are two choices, we also would expect to be close to . Unless we are highly unlucky, this should tells us which choice of to pick. (Stupid concrete example: if I have taken the square root of and then ask you to continue this square root to , which sign should you pick for ?)

There are two possible ways we could get unlucky in the scheme above: first, if , then we’re sunk. But even if we avoid that, we have to worry that we are in a situation, where we run around a full loop in the complex plane, and then find that our continuous perturbation has left us in a different place than we started. For concreteness, consider the following situation, again with :

We started at the point , with one of its square roots as . We then wound a full red circle around the origin, only to find that at the end of it, the blue arc is at a different place where it started!

The interval construction from earlier doesn’t work either: no matter how we pick the interval for , any ray of death must hit our red circle. The problem somehow lies with the fact that we have enclosed the very special point .

Nevertheless, we know that if we take , then we don’t run into any problems with our “make it up as you go” procedure. So, what exactly is going on?

3. Covering Projections

By now, if you have read the part of algebraic topology. this should all seem very strangely familiar. The “fudging” procedure exactly describes the idea of a lifting.

More precisely, recall that there is a covering projection

Let . For , we already have the square root . So the burden is completing .

Then essentially, what we are trying to do is construct a lifting for the following diagram: Our map can be described as “winding around twice”. From algebraic topology, we now know that this lifting exists if and only if

is a subset of the image of by . Since and are both punctured planes, we can identify them with .

Ques 1

Show that the image under is exactly once we identify .

That means that for any loop in , we need to have an even winding number around . This amounts to

since has no poles.

Replacing with and carrying over the discussion gives the first main result.

Theorem 2 (Existence of Holomorphic th Roots)

Let be holomorphic. Then has a holomorphic th root if and only if

for every contour in .

4. Complex Logarithms

The multivalued nature of the complex logarithm comes from the fact that

So if , then any complex number is also a solution.

We can handle this in the same way as before: it amounts to a lifting of the following diagram. There is no longer a need to work with a separate since:

Ques 3

Show that if has any zeros then possibly can’t exist.

In fact, the map is a universal cover, since is simply connected. Thus, is trivial. So in addition to being zero-free, cannot have any winding number around at all. In other words:

Theorem 4 (Existence of Logarithms)

Let be holomorphic. Then has a logarithm if and only if

for every contour in .

5. Some Special Cases

The most common special case is

Corollary 5 (Nonvanishing Functions from Simply Connected Domains)

Let be continuous, where is simply connected. If for every , then has both a logarithm and holomorphic th root.

Finally, let’s return to the question of from the very beginning. What’s the best domain such that we can define ? Clearly cannot be made to work, but we can do almost as well. For note that the only zero of is at the origin. Thus if we want to make a logarithm exist, all we have to do is make an incision in the complex plane that renders it impossible to make a loop around the origin. The usual choice is to delete negative half of the real axis, our very first ray of death; we call this a branch cut, with branch point at (the point which we cannot circle around). This gives

Theorem 6 (Branch Cut Functions)

There exist holomorphic functions

satisfying the obvious properties.

There are many possible choices of such functions ( choices for the th root and infinitely many for ); a choice of such a function is called a branch. So this is what is meant by a “branch” of a logarithm.

The principal branch is the “canonical” branch, analogous to the way we arbitrarily pick the positive branch to define . For , we take the such that and the imaginary part of lies in (since we can shift by integer multiples of ). Often, authors will write to emphasize this choice.

Example 7

Let be the complex plane minus the real interval . Then the function by has a holomorphic square root.

Corollary 8

A holomorphic function has a holomorphic th root for all if and only if it has a holomorphic logarithm.

Facts about Lie Groups and Algebras

In Spring 2016 I was taking 18.757 Representations of Lie Algebras. Since I knew next to nothing about either Lie groups or algebras, I was forced to quickly learn about their basic facts and properties. These are the notes that I wrote up accordingly. Proofs of most of these facts can be found in standard textbooks, for example Kirillov.

1. Lie groups

Let or , depending on taste.

Definition 1

A Lie group is a group which is also a -manifold; the multiplication maps (by ) and the inversion map (by ) are required to be smooth.

A morphism of Lie groups is a map which is both a map of manifolds and a group homomorphism.

Throughout, we will let denote the identity, or if we need further emphasis.

Note that in particular, every group can be made into a Lie group by endowing it with the discrete topology. This is silly, so we usually require only focus on connected groups:

Proposition 2 (Reduction to connected Lie groups)

Let be a Lie group and the connected component of which contains . Then is a normal subgroup, itself a Lie group, and the quotient has the discrete topology.

In fact, we can also reduce this to the study of simply connected Lie groups as follows.

Proposition 3 (Reduction to simply connected Lie groups)

If is connected, let be its universal cover. Then is a Lie group, is a morphism of Lie groups, and .

Here are some examples of Lie groups.

Example 4 (Examples of Lie groups)

• under addition is a real one-dimensional Lie group.
• under addition is a complex one-dimensional Lie group (and a two-dimensional real Lie group)!
• The unit circle is a real Lie group under multiplication.
• is a Lie group of dimension . This example becomes important for representation theory: a representation of a Lie group is a morphism of Lie groups .
• is a Lie group of dimension .

As geometric objects, Lie groups enjoy a huge amount of symmetry. For example, any neighborhood of can be “copied over” to any other point by the natural map . There is another theorem worth noting, which is that:

Proposition 5

If is a connected Lie group and is a neighborhood of the identity , then generates as a group.

2. Haar measure

Recall the following result and its proof from representation theory:

Claim 6

For any finite group , is semisimple; all finite-dimensional representations decompose into irreducibles.

Proof: Take a representation and equip it with an arbitrary inner form . Then we can average it to obtain a new inner form

which is -invariant. Thus given a subrepresentation we can just take its orthogonal complement to decompose .
We would like to repeat this type of proof with Lie groups. In this case the notion doesn’t make sense, so we want to replace it with an integral instead. In order to do this we use the following:

Theorem 7 (Haar measure)

Let be a Lie group. Then there exists a unique Radon measure (up to scaling) on which is left-invariant, meaning

for any Borel subset and “translate” . This measure is called the (left) Haar measure.

Example 8 (Examples of Haar measures)

• The Haar measure on is the standard Lebesgue measure which assigns to the closed interval . Of course for any , for .
• The Haar measure on is given by

  

  In particular, . One sees the invariance under multiplication of these intervals.

• Let . Then a Haar measure is given by

  

• For the circle group , consider . We can define

  

  across complex arguments . The normalization factor of ensures .

Note that we have:

Corollary 9

If the Lie group is compact, there is a unique Haar measure with .

This follows by just noting that if is Radon measure on , then . This now lets us deduce that

Corollary 10 (Compact Lie groups are semisimple)

is semisimple for any compact Lie group .

Indeed, we can now consider

as we described at the beginning.

3. The tangent space at the identity

In light of the previous comment about neighborhoods of generating , we see that to get some information about the entire Lie group it actually suffices to just get “local” information of at the point (this is one formalization of the fact that Lie groups are super symmetric).

To do this one idea is to look at the tangent space. Let be an -dimensional Lie group (over ) and consider the tangent space to at the identity . Naturally, this is a -vector space of dimension . We call it the Lie algebra associated to .

Example 11 (Lie algebras corresponding to Lie groups)

• has a real Lie algebra isomorphic to .
• has a complex Lie algebra isomorphic to .
• The unit circle has a real Lie algebra isomorphic to , which we think of as the “tangent line” at the point .

Example 12 ()

Let’s consider , an open subset of . Its tangent space should just be an -dimensional -vector space. By identifying the components in the obvious way, we can think of this Lie algebra as just the set of all matrices.

This Lie algebra goes by the notation .

Example 13 ()

Recall is a Lie group of dimension , hence its Lie algebra should have dimension . To see what it is, let’s look at the special case first: then

Viewing this as a polynomial surface in , we compute

and in particular the tangent space to the identity matrix is given by the orthogonal complement of the gradient

Hence the tangent plane can be identified with matrices satisfying . In other words, we see

By repeating this example in greater generality, we discover

4. The exponential map

Right now, is just a vector space. However, by using the group structure we can get a map from back into . The trick is “differential equations”:

Proposition 14 (Differential equations for Lie theorists)

Let be a Lie group over and its Lie algebra. Then for every there is a unique homomorphism

which is a morphism of Lie groups, such that

We will write to emphasize the argument being thought of as “time”. Thus this proposition should be intuitively clear: the theory of differential equations guarantees that is defined and unique in a small neighborhood of . Then, the group structure allows us to extend uniquely to the rest of , giving a trajectory across all of . This is sometimes called a one-parameter subgroup of , but we won’t use this terminology anywhere in what follows.

This lets us define:

Definition 15

Retain the setting of the previous proposition. Then the exponential map is defined by

The exponential map gets its name from the fact that for all the examples I discussed before, it is actually just the map . Note that below, for a matrix ; this is called the matrix exponential.

Example 16 (Exponential Maps of Lie algebras)

• If , then too. We observe (where ) is a morphism of Lie groups . Hence

  

• Ditto for .
• For and , the map given by works. Hence

  

• For , the map given by works nicely (now is a matrix). (Note that we have to check is actually invertible for this map to be well-defined.) Hence the exponential map is given by

  

• Similarly,

  

  Here we had to check that if , meaning , then . This can be seen by writing in an upper triangular basis.

Actually, taking the tangent space at the identity is a functor. Consider a map of Lie groups, with lie algebras and . Because is a group homomorphism, . Now, by manifold theory we know that maps between manifolds gives a linear map between the corresponding tangent spaces, say . For us we obtain a linear map

In fact, this fits into a diagram

Here are a few more properties of :

• , which is immediate by looking at the constant trajectory .
• , i.e. the total derivative is the identity. This is again by construction.
• In particular, by the inverse function theorem this implies that is a diffeomorphism in a neighborhood of , onto a neighborhood of .
• commutes with the commutator. (By the above diagram.)

5. The commutator

Right now is still just a vector space, the tangent space. But now that there is map , we can use it to put a new operation on , the so-called commutator.

The idea is follows: we want to “multiply” two elements of . But is just a vector space, so we can’t do that. However, itself has a group multiplication, so we should pass to using , use the multiplication in and then come back.

Here are the details. As we just mentioned, is a diffeomorphism near . So for , close to the origin of , we can look at and , which are two elements of close to . Multiplying them gives an element still close to , so its equal to for some unique , call it .

One can show in fact that can be written as a Taylor series in two variables as

where is a skew-symmetric bilinear map, meaning . It will be more convenient to work with than itself, so we give it a name:

Definition 17

This is called the commutator of .

Now we know multiplication in is associative, so this should give us some nontrivial relation on the bracket . Specifically, since

we should have that , and this should tell us something. In fact, the claim is:

Theorem 18

The bracket satisfies the Jacobi identity

Proof: Although I won’t prove it, the third-order terms (and all the rest) in our definition of can be written out explicitly as well: for example, for example, we actually have

The general formula is called the Baker-Campbell-Hausdorff formula.

Then we can force ourselves to expand this using the first three terms of the BCS formula and then equate the degree three terms. The left-hand side expands initially as , and the next step would be something ugly.

This computation is horrifying and painful, so I’ll pretend I did it and tell you the end result is as claimed.
There is a more natural way to see why this identity is the “right one”; see Qiaochu. However, with this proof I want to make the point that this Jacobi identity is not our decision: instead, the Jacobi identity is forced upon us by associativity in .

Example 19 (Examples of commutators attached to Lie groups)

• If is an abelian group, we have by symmetry and from . Thus in for any abelian Lie group .
• In particular, the brackets for are trivial.
• Let . Then one can show that

  

• Ditto for .

In any case, with the Jacobi identity we can define an general Lie algebra as an intrinsic object with a Jacobi-satisfying bracket:

Definition 20

A Lie algebra over is a -vector space equipped with a skew-symmetric bilinear bracket satisfying the Jacobi identity.

A morphism of Lie algebras and preserves the bracket.

Note that a Lie algebra may even be infinite-dimensional (even though we are assuming is finite-dimensional, so that they will never come up as a tangent space).

Example 21 (Associative algebra Lie algebra)

Any associative algebra over can be made into a Lie algebra by taking the same underlying vector space, and using the bracket .

6. The fundamental theorems

We finish this list of facts by stating the three “fundamental theorems” of Lie theory. They are based upon the functor

we have described earlier, which is a functor

• from the category of Lie groups
• into the category of finite-dimensional Lie algebras.

The first theorem requires the following definition:

Definition 22

A Lie subgroup of a Lie group is a subgroup such that the inclusion map is also an injective immersion.

A Lie subalgebra of a Lie algebra is a vector subspace preserved under the bracket (meaning that ).

Theorem 23 (Lie I)

Let be a real or complex Lie group with Lie algebra . Then given a Lie subgroup , the map

is a bijection between Lie subgroups of and Lie subalgebras of .

Theorem 24 (The Lie functor is an equivalence of categories)

Restrict to a functor

• from the category of simply connected Lie groups over
• to the category of finite-dimensional Lie algebras over .

Then

1. (Lie II) is fully faithful, and
2. (Lie III) is essentially surjective on objects.

If we drop the “simply connected” condition, we obtain a functor which is faithful and exact, but not full: non-isomorphic Lie groups can have isomorphic Lie algebras (one example is and ).