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

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.

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.

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.)

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.

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.

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?

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.

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.

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.

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.

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.

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.

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Suppose you are a math PhD student at MIT. Officially, this “costs” $50K a year in tuition. Fortunately this number is meaningless, because math PhD students serve time as teaching assistants in exchange for having the nominal sticker price waived. MIT then provides a stipend of about $25K a year for these PhD student’s living expenses. This stipend is taxable, but it’s small and you’d pay only $1K-$2K in federal taxes (about 6%).

The new GOP tax proposal strikes 26 U.S. Code 117(d) which would cause the $50K tuition waiver to *also* become taxable income: the PhD student would pay taxes on an “income” of $75K, at tax brackets of 12% and 25%. If I haven’t messed up the calculation, for our single PhD student this means **paying $10K in federal taxes out of the same $25K stipend (about 40%)**.

I think a 40% tax rate for a PhD student is a *bit* unreasonable; the remaining $15K a year is not too far from the poverty line.

(The relevant sentence is page 96, line 20 of the GOP tax bill.)

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**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).

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(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.)

**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.

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**.

**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 .

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.

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 .

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.

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.

First, we prove that

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.

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 .

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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.

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.

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.)

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:

- Direct angle chasing
- Power of a point / radical axis
- Homothety, similar triangles, ratios
- Recognizing some standard configuration (see Yufei for a list)
- Doing some length calculations
- Complex numbers
- Barycentric coordinates
- Inversion
- Harmonic bundles or pole/polar and homography
- 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.

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.

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.

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.

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Spoiler warnings: USAMO 2014/1, and hints for Putnam 2014 A4 and B2. You may want to work on these problems yourself before reading this post.

At last year’s USA IMO training camp, I prepared a handout on writing/style for the students at MOP. One of the things I talked about was the “ocean-crossing point”, which for our purposes you can think of as the discrete jump from a problem being “essentially not solved” () to “essentially solved” (). The name comes from a Scott Aaronson post:

Suppose your friend in Boston blindfolded you, drove you around for twenty minutes, then took the blindfold off and claimed you were now in Beijing. Yes, you do see Chinese signs and pagoda roofs, and no, you can’t immediately disprove him — but based on your knowledge of both cars and geography, isn’t it more likely you’re just in Chinatown? . . .

We start in Boston, we end up in Beijing, and at no point is anything resembling an ocean ever crossed.

I then gave two examples of how to write a solution to the following example problem.

**Problem 1** **(USAMO 2014)**

Let , , , be real numbers such that and all zeros , , , and of the polynomial are real. Find the smallest value the product

can take.

*Proof:* (Not-so-good write-up) Since for every (where ), we get which equals to . If this is and . Also, , this is .

*Proof:* (Better write-up) The answer is . This can be achieved by taking , whence the product is , and .

Now, we prove this is a lower bound. Let . The key observation is that

Consequently, we have

This proves the lower bound.

You’ll notice that it’s much easier to see the key idea in the second solution: namely,

which allows you use the enigmatic condition .

Unfortunately I have the following confession to make:

In practice, most solutions are written more like the first one than the second one.

The truth is that writing up solutions is sort of a chore that people never really want to do but have to — much like washing dishes. So must solutions won’t be written in a way that helps you learn from them. This means that when you read solutions, you should assume that the thing you really want (i.e., the ocean-crossing point) is buried somewhere amidst a haystack of other unimportant details.

But in practice even the “better write-up” I mentioned above still has too much information in it.

Suppose you were explaining how to solve this problem to a friend. You would probably not start your explanation by saying that the minimum is , achieved by — even though this is indeed a logically necessary part of the solution. Instead, the first thing you would probably tell them is to notice that

In fact, if your friend has been working on the problem for more than ten minutes, this is probably the *only* thing you need to tell them. They probably already figured out by themselves that there was a good chance the answer would be , just based on the condition . This “one-liner” is all that they need to finish the problem. You don’t need to spell out to them the rest of the details.

When you explain a problem to a friend in this way, you’re communicating just the difference: the one or two sentences such that your friend could work out the rest of the details themselves with these directions. When reading the solution yourself, you should try to extract the main idea in the same way. Olympiad problems generally have only a few main ideas in them, from which the rest of the details can be derived. So reading the solution should feel much like searching for a needle in a haystack.

In particular: you should rarely read most of the words in the solution, and you should almost never read every word of the solution.

Whenever I read solutions to problems I didn’t solve, I often read less than 10% of the words in the solution. Instead I search aggressively for the one or two sentences which tell me the key step that I couldn’t find myself. (Functional equations are the glaring exception to this rule, since in these problems there sometimes isn’t any main idea other than “stumble around randomly”, and the steps really are all about equally important. But this is rarer than you might guess.)

I think a common mistake students make is to treat the solution as a sequence of logical steps: that is, reading the solution line by line, and then verifying that each line follows from the previous ones. This seems to entirely miss the point, because not all lines are created equal, and most lines can be easily *derived* once you figure out the main idea.

If you find that the only way that you can understand the solution is reading it step by step, then the problem may simply be too hard for you. This is because what counts as “details” and “main ideas” are relative to the absolute difficulty of the problem. Here’s an example of what I mean: the solution to a USAMO 3/6 level geometry problem, call it , might look as follows.

*Proof:* First, we prove lemma . (Proof of , which is USAMO 1/4 level.)

Then, we prove lemma . (Proof of , which is USAMO 1/4 level.)

Finally, we remark that putting together and solves the problem.

Likely the main difficulty of is actually *finding* and . So a very experienced student might think of the sub-proofs as “easy details”. But younger students might find challenging in their own right, and be unable to solve the problem even after being told what the lemmas are: which is why it is hard for them to tell that were the main ideas to begin with. In that case, the problem is probably way over their head.

This is also why it doesn’t make sense to read solutions to problems which you have not worked on at all — there are often details, natural steps and notation, et cetera which are obvious to you if and only if you have actually tried the problem for a little while yourself.

The earlier sections describe how to extract the main idea of an olympiad solution. This is neat because instead of having to remember an entire solution, you only need to remember a few sentences now, and it gives you a good understanding of the solution at hand.

But this still isn’t achieving your ultimate goal in learning: you are trying to maximize your scores on *future* problems. Unless you are extremely fortunate, you will probably never see the exact same problem on an exam again.

So one question you should often ask is:

“How could I have thought of that?”

(Or in my case, “how could I train a student to think of this?”.)

There are probably some surface-level skills that you can pick out of this. The lowest hanging fruit is things that are technical. A small number of examples, with varying amounts of depth:

- This problem is “purely projective”, so we can take a projective transformation!
- This problem had a segment with midpoint , and a line parallel to , so I should consider projecting through a point on .
- Drawing a grid of primes is the only real idea in this problem, and the rest of it is just calculations.
- This main claim is easy to guess since in some small cases, the frogs have “violating points” in a large circle.
- In this problem there are numbers on a circle, odd. The counterexamples for even alternate up and down, which motivates proving that no three consecutive numbers are in sorted order.
- This is a juggling problem!

(Brownie points if any contest enthusiasts can figure out which problems I’m talking about in this list!)

But now I want to point out that the best answers to the above question are often *not formalizable*. Lists of triggers and actions are “cheap forms of understanding”, because going through a list of methods will only get so far.

On the other hand, the un-formalizable philosophy that you can extract from reading a question, is part of that legendary “intuition” that people are always talking about: you can’t describe it in words, but it’s certainly there. Maybe I would even be better if I reframed the question as:

“What does this problem feel like?”

So let’s talk about our feelings. Here is David Yang’s take on it:

Whenever you see a problem you really like, store it (and the solution) in your mind like a cherished memory . . . The point of this is that

you will see problems which will remind you of that problem despite having no obvious relation.You will not be able to say concretely what the relation is, but think a lot about it and give a name to the common aspect of the two problems. Eventually, you will see new problems for which you feel like could also be described by that name.Do this enough, and you will have a very powerful intuition that cannot be described easily concretely (and in particular, that nobody else will have).

This itself doesn’t make sense without an example, so here is an example of one philosophy I’ve developed. Here are two problems on Putnam 2014:

**Problem 2** **(Putnam 2014 A4)**

Suppose is a random variable that takes on only nonnegative integer values, with , , and . Determine the smallest possible value of the probability of the event .

**Problem 3** **(Putnam 2014 B2)**

Suppose that is a function on the interval such that for all and

How large can be?

At a glance there seems to be nearly no connection between these problems. One of them is a combinatorics/algebra question, and the other is an integral. Moreover, if you read the official solutions or even my own write-ups, you will find very little in common joining them.

Yet it turns out that these two problems do have something in common to me, which I’ll try to describe below. My thought process in solving either question went as follows:

In both problems, I was able to quickly make a good guess as to what the optimal / was, and then come up with a heuristic explanation (not a proof) why that guess had to be correct, namely, “by smoothing, you should put all the weight on the left”. Let me call this optimal argument .

That conjectured gave a numerical answer to the actual problem: but for both of these problems, it turns out that numerical answer is completely uninteresting, as are the exact details of . It should be philosophically be interpreted as “this is the number that happens to pop out when you plug in the optimal choice”. And indeed that’s what both solutions feel like. These solutions don’t actually care what the exact values of are, they only care about the properties that made me think they were optimal in the first place.

I gave this philosophy the name **Equality**, with poster description “problems where looking at the equality case is important”. This text description feels more or less useless to me; I suppose it’s the thought that counts. But ever since I came up with this name, it has helped me solve new problems that come up, because they would give me the same feeling that these two problems did.

Two more examples of these themes that I’ve come up with are **Global** and **Rigid**, which will be described in a future post on how I design training materials.

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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 ?

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?

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?

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 .

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 .

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.

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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.

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.

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

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.)

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 .

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

- (Lie II) is fully faithful, and
- (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 ).

]]>

One of the most fundamental problems in graph theory is that of a *graph coloring*, in which one assigns a color to every vertex of a graph so that no two adjacent vertices have the same color. The most basic invariant related to the graph coloring is the chromatic number:

**Definition 1**

A simple graph is **-colorable** if it’s possible to properly color its vertices with colors. The smallest such is the **chromatic number** .

In this exposition we study a more general notion in which the set of permitted colors is different for each vertex, as long as at least colors are listed at each vertex. This leads to the notion of a so-called choice number, which was introduced by Erdös, Rubin, and Taylor.

**Definition 2**

A simple graph is **-choosable** if its possible to properly color its vertices given a list of colors at each vertex. The smallest such is the **choice number** .

**Example 3**

We have for any integer (here is the cycle graph on vertices). To see this, we only have to show that given a list of two colors at each vertex of , we can select one of them.

- If the list of colors is the same at each vertex, then since is bipartite, we are done.
- Otherwise, suppose adjacent vertices , are such that some color at is not in the list at . Select at , and then greedily color in , \dots, in that order.

We are thus naturally interested in how the choice number and the chromatic number are related. Of course we always have

Näively one might expect that we in fact have an equality, since allowing the colors at vertices to be different seems like it should make the graph easier to color. However, the following example shows that this is not the case.

**Example 4** **(Erdös)**

We claim that for any integer we have

The latter equality follows from being partite.

Now to see the first inequality, let have vertex set , where is the set of functions and . Then consider colors for . On a vertex , we list colors , , \dots, . On a vertex , we list colors , , \dots, . By construction it is impossible to properly color with these colors.

The case is illustrated in the figure below (image in public domain).

This surprising behavior is the subject of much research: how can we bound the choice number of a graph as a function of its chromatic number and other properties of the graph? We see that the above example requires exponentially many vertices in .

**Theorem 5** **(Noel, West, Wu, Zhu)**

If is a graph with vertices then

In particular, if then .

One of the most major open problems in this direction is the following.

**Definition 6**

A **claw-free** graph is a graph with no induced . For example, the line graph (also called edge graph) of any simple graph is claw-free.

If is a claw-free graph, then . In particular, this conjecture implies that for *edge* coloring, the notions of “chromatic number” and “choice number” coincide.

In this exposition, we prove the following result of Alon.

**Theorem 7** **(Alon)**

A bipartite graph is choosable, where

is half the maximum of the average degree of subgraphs .

In particular, recall that a *planar* bipartite graph with vertices contains at most edges. Thus for such graphs we have and deduce:

**Corollary 8**

A planar bipartite graph is -choosable.

This corollary is sharp, as it applies to which we have seen in Example 4 has .

The rest of the paper is divided as follows. First, we begin in §2 by stating Theorem 9, the famous combinatorial nullstellensatz of Alon. Then in §3 and §4, we provide descriptions of the so-called *graph polynomial*, to which we then apply combinatorial nullstellensatz to deduce Theorem 18. Finally in §5, we show how to use Theorem 18 to prove Theorem 7.

The main tool we use is the Combinatorial Nullestellensatz of Alon.

**Theorem 9** **(Combinatorial Nullstellensatz)**

Let be a field, and let be a polynomial of degree . Let such that for all .

Assume the coefficient of of is not zero. Then we can pick , \dots, such that

**Example 10**

Let us give a second proof that

for every positive integer . Our proof will be an application of the Nullstellensatz.

Regard the colors as real numbers, and let be the set of colors at vertex (hence , and ). Consider the polynomial

The coefficient of is . Therefore, one can select a color from each so that does not vanish.

Motivated by Example 10, we wish to apply a similar technique to general graphs . So in what follows, let be a (simple) graph with vertex set .

**Definition 11**

The **graph polynomial** of is defined by

We observe that coefficients of correspond to differences in directed orientations. To be precise, we introduce the notation:

**Definition 12**

Consider **orientations** on the graph with vertex set , meaning we assign a direction to every edge of to make it into a directed graph . An oriented edge is called **ascending** if and , i.e. the edge points from the smaller number to the larger one.

Then we say that an orientation is

**even**if there are an even number of ascending edges, and**odd**if there are an odd number of ascending edges.

Finally, we define

- to the be set of all even orientations of in which vertex has indegree .
- to the be set of all odd orientations of in which vertex has indegree .

Set .

**Example 13**

Consider the following orientation:

There are exactly two ascending edges, namely and . The indegrees of are , and . Therefore, this particular orientation is an element of . In terms of , this corresponds to the choice of terms

which is a term.

*Proof:* Consider expanding . Then each expanded term corresponds to a choice of or from each , as in Example 13. The term has coefficient is the orientation is even, and if the orientation is odd, as desired.

Thus we have an explicit combinatorial description of the coefficients in the graph polynomial .

We now give a second description of the coefficients of .

**Definition 15**

Let , viewed as a directed graph. An **Eulerian suborientation** of is a subgraph of (not necessarily induced) in which every vertex has equal indegree and outdegree. We say that such a suborientation is

**even**if it has an even number of edges, and**odd**if it has an odd number of edges.

Note that the empty suborientation is allowed. We denote the even and odd Eulerian suborientations of by and , respectively.

Eulerian suborientations are brought into the picture by the following lemma.

*Proof:* Consider any orientation , Then we define a suborietation of , denoted , by including exactly the edges of whose orientation in is in the opposite direction. It’s easy to see that this induces a bijection

Moreover, remark that

- is even if and are either both even or both odd, and
- is odd otherwise.

The lemma follows from this.

*Proof:* Combine Lemma 14 and Lemma 16.

We now arrive at the main result:

**Theorem 18**

Let be a graph on , and let be an orientation of . If , then given a list of colors at each vertex of , there exists a proper coloring of the vertices of .

In particular, is -choosable.

*Proof:* Combine Corollary 17 with Theorem 9.

Armed with Theorem 18, we are almost ready to prove Theorem 7. The last ingredient is that we need to find an orientation on in which the maximal degree is not too large. This is accomplished by the following.

*Proof:* This is an application of Hall’s marriage theorem.

Let . Construct a bipartite graph

Connect and if is an endpoint of . Since we satisfy Hall’s condition (as is a condition for all subgraphs ) and can match each edge in to a (copy of some) vertex in . Since there are exactly copies of each vertex in , the conclusion follows.

Now we can prove Theorem 7. *Proof:* According to Lemma 19, pick where . Since is bipartite, we obviously have , since cannot have any odd cycles. So Theorem 18 applies and we are done.

]]>

`loadkeys`

and `keyfuzz`

that are the first search entries don’t work for me, so some more sophisticated black magic was necessary.
This step is technically optional, but I did it because the function keys are a pain anyways. Normally on Apple keyboards one needs to use the `Fn`

key to get the normal Fn keys to behave as a `F<n>`

keystroke. I prefer to reverse this behavior, so that the SysRq combinations is `Alt+F13+F`

rather than `Fn+Alt+F13+F`

, say.

For this, the advice on the Arch Wiki worked, although it is not thorough on some points that I think should’ve been said. On newer kernels, one does this by creating the file `/etc/modprobe.d/hid_apple.conf`

and writing

```
options hid_apple fnmode=2
```

Then I edited the file `/etc/mkinitcpio.conf`

to include the new file:

```
...
BINARIES=""
# FILES
# This setting is similar to BINARIES above, however, files are added
# as-is and are not parsed in any way. This is useful for config files.
FILES="/etc/modprobe.d/hid_apple.conf"
# HOOKS
...
```

Finally, recompile the kernel for this change to take effect. On Arch Linux one can just do this by issuing the command

```
$ sudo pacman -S linux
```

which will reinstall the entire kernel.

Next, I needed to get the scancode of the key I wanted to turn into the SysRQ key. For me attempting `showkey -s`

did not work so I instead had to use evtest, as described in this Arch Wiki.

```
$ sudo pacman -S evtest
$ sudo evtest
No device specified, trying to scan all of /dev/input/event*
Available devices:
/dev/input/event0: Logitech USB Receiver
/dev/input/event1: Logitech USB Receiver
/dev/input/event2: Apple, Inc Apple Keyboard
/dev/input/event3: Apple, Inc Apple Keyboard
/dev/input/event4: Apple Computer, Inc. IR Receiver
/dev/input/event5: HDA NVidia Headphone
/dev/input/event6: HDA NVidia HDMI/DP,pcm=3
/dev/input/event7: Power Button
/dev/input/event8: Sleep Button
/dev/input/event9: Power Button
/dev/input/event10: Video Bus
/dev/input/event11: PC Speaker
/dev/input/event12: HDA NVidia HDMI/DP,pcm=7
/dev/input/event13: HDA NVidia HDMI/DP,pcm=8
Select the device event number [0-13]: 2
Input driver version is 1.0.1
Input device ID: bus 0x3 vendor 0x5ac product 0x220 version 0x111
Input device name: "Apple, Inc Apple Keyboard"
```

This is on my Mac Mini; the list of devices looks different on my laptop. After this pressing the desired key yields something which looked like

```
Event: time 1456870457.844237, -------------- SYN_REPORT ------------
Event: time 1456870457.924097, type 4 (EV_MSC), code 4 (MSC_SCAN), value 70068
Event: time 1456870457.924097, type 1 (EV_KEY), code 183 (KEY_F13), value 1
```

This is the F13 key which I want to map into a SysRq — the keycode 70068 above (which is in fact a hex code) is the one I wanted.

Now that I had the scancode, I cd’ed to `/etc/udev/hwdb.d`

and added a file

`90-keyboard-sysrq.hwdb`

with the content

```
evdev:input:b0003*
KEYBOARDKEY_70068=sysrq
```

One then updates `hwdb.bin`

by running the command

```
$ sudo udevadm hwdb --update
$ sudo udevadm trigger
```

The latter command makes the changes take effect immediately. You should be able to test this by running `sudo evtest`

again; `evtest`

should now report the new keycode (but the same scancode).

One can test the SysRQ key by running Alt+SysRq+H, and then checking the `dmesg`

output to see if anything happened:

```
$ dmesg | tail -n 1
[ 283.001240] sysrq: SysRq : HELP : loglevel(0-9) reboot(b) crash(c) ...
```

It remains to actually enable SysRQ, according to the bitmask described here. My system default was apparently 16:

```
$ sysctl kernel.sysrq
kernel.sysrq = 16
```

For my purposes, I then edited `/etc/sysctl.d/99-sysctl.conf`

and added the line

```
kernel.sysrq=254
```

This gave me everything except the nicing of real-time tasks. Of course the choice of value here is just personal preference.

Personally, my main use for this is killing Chromium, which has a bad habit of freezing up my computer (especially if Firefox is open too). I remedy the situation by repeatedly running Alt+SysRq+F to kill off the memory hogs. If this doesn’t work, just Alt+SysRq+K kills off all the processes in the current TTY.

]]>