## On Detiling Polynomials: A Generalization of the Euler MacLaurin Formula

Today, we’ll be talking about a relation between the discrete and continuous.

A friend of mine, Aaron Slipper, explained to me a simple and striking perspective of the Euler-MacLaurin formula — and his work extending it.

He came upon this through the proof of Jacobi, relayed to him by our beloved teacher Laurens Gunnarsen. The proof of Jacobi uses the lattice group $$\mathbb{Z} \subset \mathbb{R}$$, so Aaron sought and found analogous relationships for other lattice groups.

## The translation group, 1-d case

One mental model for Aaron’s construction is as follows. We are given a discrete group (a pattern motif and a set of allowed transformations), and wish to examine how we can build a lattice with this group over time (each time evolution step is an application of one of the allowed translations. We might do so in the following fashion.

Each element in the discrete group can be written as a sequence of moves from the identity.

MISSING IMAGE

Here we have

$$F(0) = f(0)$$
$$F(1) = f(0) + f(1)$$
$$F(n) = f(0) + f(1) + …. + f(n)$$

That is:

\$$F(x) = \sum_{x} f(x)\$$

We have a group action of time for any $$t \in \mathbb{R}$$, a successor function, which maps $$F(X) \to F(X+t)$$. That is,

\$$e^{tD}F(X) \mapsto F(X+t)\$$

where $$D = \frac{d}{dX}$$.

This is the operator version of Taylor’s theorem. This is saying that $$d/dx$$ is an infinitesimal transformation, as the exponential “flows” $$d/dx$$ along, it generates the real line.

Sidenote (a physical incarnation): The equation \$$e^{tD}F(x) = F(x+t)\$$ reminds me of a theorem about a wave function for an electron in a crystal with no defects (such wavefunctions are of the form $$\phi(r) = e^{ik\cdot r}u(r)$$, and are said to be “Bloch,” after the phyiscist).

The Bloch theorem states that a wave function in a crystal (with no defects) can change under translation only by a phase factor \$$e^{ik \cdot T} \phi(r) = \phi(r + T)\$$ This formal resemblence isn’t so surprising: the defining property of a crystal is translational symmetry.

Usually $$t$$ ranges over $$R$$, but let’s restrict our exponential, $$e^{tD}$$, by removing all $$t$$ that aren’t supported by the type $$F$$. For example, if $$F: \mathbb{Z} \to \mathbb{Z}^2$$, then restrict to $$t \in \mathbb{Z}$$.

Aaron explained a simple example of such a system. Start from the origin, $$0$$, of $$R^1$$:

\$$f(0) + f(1) + … + f(n) = F(n)\$$

\$$1 + 1 + … + 1 = n\$$

So then we can express f(n) in terms of the previous F(n).

\$$f(n) = F(n) – F(n-1)\$$

\$$1 = n – (n-1)\$$

Using the group action of time, $$e^{tD}F(x) = F(x+t)$$, we rexpress:

\$$f(n) = e^{0 \cdot D} F(n) – e^{-1 \cdot D} F(n)\$$

\$$f = (1-e^{-D})F\$$

\$$frac{1}{(1-e^{-D})} f = F\$$

We call $$1-e^{-D}$$, or, equally, $$\frac{1}{(1-e^{-D})}$$ the “detiling polynomial.”

Note that the latter is the Bernoulli numbers, replacing $$D$$ with a natural number $$n$$.

## The translation group, n-d case

Another example, found by Aaron, is the square lattice (in the upper half complex plane), i.e., $$Z^2$$ \in $$R^2 \simeq C$$:

Recall the set theoretic fact: $$A \cup B – A – B + A \cap B = 0$$

\$$f(x) = F(x) – F(x-1) – F(x-i) + F(x-i-1)\$$

\$$f(n) = e^{0 \cdot D} F(n) – e^{-1 \cdot D} F(n) – e^{-i \cdot D} F(n) + e^{-1-i \cdot D} F(n)\$$

\$$f = (1-e^{-D})(1-e^{-iD})F\$$

It’s important to note that the choices of i and 1 as generators of the square lattice are arbitrary, we could have also chosen something like (i+1) and 1.

Also note that the detiling polynomial is not invariant under such a generator change.

In summary, we can express the “tiling” of lattices starting from a point as discrete dynamical systems, s.t. our successor function marches us across a configuration space (allowed states as you build the lattice) fibered over the discrete group $$G$$ (aka, indexed by $$G$$).

If we think about this from the point of view of Stokes theorem, it is then natural to assign positive and negative weights to the lattice such that we build a connected structure by tesselating, all but the “corners” will cancel.

If we assign a positive and negative weight to each vertex, that is:
– +
+ –

This is simple to see if you take the assignments of weights to be the incidence matrix of a graph, that is,

$\left[ {\begin{array}{cc} -1 & 1 \ 1 & -1 \end{array} } \right]$

## Background on the Möbius tiling

Due to a misunderstanding, I thought Aaron had stopped at the 2-d square lattice, and this presentation of the concept is bursting with generalizations — so I went home and drew a lot of lattices, derived the detiling polynomial for n dimensional square lattice case, for the truncated icosahedron (C60), and for the n-dimensional hexagonal tiling (which reduces to the n-dimensional square case).

It turns out he’d already derived these cases, and he’d asked how to derive the hyperbolic detiling polynomial, but playing with them got me very interested. They’ve been on my mind during my whole cross country road trip.

I’ve been reading Poincare’s Analysis Situs, and became interested in Fuchsian (a German name, pronounced Fook-see-in) groups, as these motivated Poincare to define what we now call the fundamental group (or Poincare group) of a space.

Since I was reading about them, I realized a simple case of what Aaron was saying about non-Euclidean lattices is the case of a Fuchsian group, specifically, the PSL(2,C), otherwise known as the Mobius transformation group:

What is the detiling polynomial for the Mobius group?

That is, what are the generators that define a fundmental domain which we can translate to build the hyperbolic tiling of the complex plane, like the one we saw above?

Wait a moment, those are squares… but I’m asking about a translation group of hyperbolic triangles, how did we go from one to the other?

## Picturing SL(2,R) and $$\mathfrak{sl}(2, R)$$

Let’s ignore the projectivization for the moment, and just look at $$SL(2, Z)$$. In fact, let’s look first at $$SL(2, R)$$, to make things even simpler.

Now, SL(2, R) is a 3-dimensional Lie group. It’s basically that portion of R^4 consisting of 4-tuples (a, b, c, d) for which $$ad – bc = 1$$.

So that’s one equation constraining four unknowns. So SL(2, R) is 3-dimensional.

By the way, $$ad – bc = 1$$ iff $$4ad – 4bc = 4$$ and this immediately translates into

\$$(a + d)^2 – (a – d)^2 – (b + c)^2 + (b – c)^2 = 4\$$

so that, if we change variables in a rather simple way, we conclude that SL(2, R) is basically just the level set of the function

\$$u^2 + v^2 – x^2 – y^2\$$

In other words, $$SL(2, R)$$ can be presented as the set of all $$(u, v, x, y)$$ for which that equation holds.

Now, let’s look at the Lie algebra of this Lie group. That is, let’s suppose that $$Id + tA$$ has determinant $$1$$, where Id is the identity matrix, and $$t$$ is infinitesimal — in the sense that we can discard outright any term that includes $$t^2$$ or any higher power.

We want to see what this means for the matrix $$A$$.

What’s important to remember is that the determinant of a 2×2 matrix is, as we have just finished noticing, a quadratic function of its entries. But quadratic things in $$t$$ are to be discarded — so the contribution of the off-diagonal terms in $$Id + tA$$ can be ignored. Because $$Id$$ is diagonal, it follows that the off-diagonal terms are each going to be proportional to $$t$$. So their product, which is how they enter into the determinant, is going to be order $$t^2$$, and therefore is set to $$0$$.

$$Id + tA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + t \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

Due to this, only the diagonal elements of $$Id + tA$$ contribute to its determinant, if we are only going to retain first-order terms in $$t$$.

$$\text{det} \begin{pmatrix} 1 + ta & tb \\ tc & 1 + td \end{pmatrix}$$

And indeed, the determinant to first order in $$t$$ is just $$(1 + ta)(1 + td) = 1 + t(a + d)$$, which is only equal to 1 (to first order in t) when $$a + d = 0$$. That is, the Lie algebra of $$SL(2, R)$$ is just the set of 2×2 matrices with trace (i.e., the sum of their diagonal elements) equal to zero.

The general element of this Lie algebra looks like this:

$$\begin{pmatrix} a & b + c \\ b – c & -a \end{pmatrix}$$

because we can always write the off-diagonal ones as the sum and the difference of two other numbers.

Look at what becomes of the determinant here.

For a matrix of this form, the determinant is \$$c^2 – a^2 – b^2\$$ Keep that in mind, because it is the Casimir element of this Lie algebra.

Now, the group acts on its Lie algebra by conjugation, giving rise to what is called the adjoint representation. Given A in the Lie algebra, and U in the Lie group, we put \$$[Ad_U](A) = UAU^{-1}\$$

The crucial thing to appreciate here is that the trace of $$A$$ and the trace of $$UAU^{-1}$$ are guaranteed to be the same, and the determinants are also going to match. That is, trace and determinants are both functions of the conjugacy class of A, rather than of A itself.

How does a non-Euclidean plane arise from SL(2, R)?

We first have to know where the plane is, before we can start to tile it!

The plane is actually in the Lie algebra, which is why we’ve been droning on and on about it.

In fact, the non-Euclidean plane we’ll be tiling is exactly the set of points $$(a,b,c) \in \mathbb{R}^3$$ corresponding to the trace-free matrices:

$$\begin{pmatrix} a & b + c \\ b – c & -a \end{pmatrix}$$

for which

\$$c^2 – a^2 = b^2 = 1\$$

This set is just a paraboloid of revolution in $$R^3$$.

The point is that the natural (i.e., group-invariant) notion of distance between a pair of points in this paraboloid is not the obvious Euclidean distance in $$R^3$$. Instead, we have the Hyperboloid model (thanks Minkowski):

Now, an action of a discrete subgroup of the non-Euclidean isometry group is a tiling. That is, it is characterized by a “fundamental domain.”

So there’s our translation group of hyperbolic triangles! It’s the plain old triangular tiling on the non-Euclidean plane.

## Perspectives on the detiling polynomial

After talking over each other for a bit, Aaron pointed out that I was speaking about the 1st perspective, and he was speaking about the second:

1. A sum over a discrete subgroup of a Lie Group, compared to an integral over a corresponding region. This is actually done ON the Lie Group itself.
2. Groups acting via automoprhisms on a corresponding space. The PSL(2,R) acts as a group of isometries, and the discrete group acts on a tiling. (That is, the discrete group is a Fuschian group.)

## Afternote: Unfinished speculation wrt the Lattice of Integers on a Hyperbolic Line

From here on, the post is scraps and thoughts in progress, feel free to stop here.

So we can write down the detiling polynomial with the generators presented as matrices, and so if we wanted to, say, as what the integers of a hyperbolic line are, we just need to solve for d/dt=: D in our shift operator.

Then once we can write down a detiling polynomial for a line with a non Euclidean metric, that is, write down d/dt such that
The operator d/dt generates the hyperbolic line by translation by the exponential (as a flow on the hyperbola), the issue that I have is figuring out if I want the integers of a hyperbolic to be

Of a hyperbola
1) projection
Or
2) Arc length based
Or
3) The subset of the hyperbola that consists of Points P such that |P-A| is an integer, and |P-B| is an integer

So then I guess really we want our integers to be a discrete subgroup of wrt this group operation.

I tried to derive the particular D we need to have e^{nD}F(t) = F(t+n) make sense by writing a hyperbola parametrically and trying to solve for dt in terms of dx and dy, but it felt more complicated than necessary.

Let’s reparameterize the hyperbola in terms of one variable, $$t$$, to get a viable $$d/dt$$ to stick into our shift operator $$e^n\frac{d}{dt}$$. To get a hyperbolic shift operator, we look at the line $$(a \sec t, b \tan t)$$, for an arbitrary choice of $$a$$ and $$b$$.

We can define the distance between two points to be the infimum distance of all curves between them.

One way to make a hyperbola is to pick two points on your y-axis, let them be c units apart, and look at all points p such that $$|p-A| – |p-B| = c/2$$. Let’s call this hyperbola $$R_h$$.

What are the integers of a hyperbola? Maybe they are the subset of the points $$p$$ with the property that $$|p-A|$$ AND $$|p-B|$$ are whole numbers, let’s call this subset $$Z_h$$. Perhaps this is not quite what we want.

There’s something called operator calculus or umbral calculus, and it’s suited for our goal of writing down a detiling polynomial for a discrete subgroup of the hyperbola $$Z_h \subset R_h$$.

I went from subset to subgroup with no justification, so let me give some. A hyperbola is a group, in fact, any conic section is a group.

For example, the group structure on a line is pretty simple. Take $$y = ax + c$$ with origin $$(0,c)$$, then for $$(x_1,y_1),(x_2,y_2)$$ lying on the line, their sum is the point $$(x_1 +x_2,y_1 +y_2 −c)$$.

In greater generality, given a conic section and a choice of origin, say $$O$$. Then for points $$A, B$$ lying on the conic, take the straight line joining them (or tangent if the points are the same) and draw the parallel line through $$O$$. This intersects the conic at a third point, $$C$$, and define the sum $$A + B$$ to be $$C$$.

You may object because I’m using the parellel postulate, but we’re looking at a hyperbola embedded in $$R^2$$, with the usual Euclidean metric, for the moment.

## A bit of background on Fuschian functions

“So three parameters specify a Fuchsian function; Three integers, the sum of whose reciprocals is less than 1. These then correspond to the angles of a hyperbolic triangle, And Poincaré showed how you can tesselate the Poincaré disk with them in such a way that a discrete group of isometries of the Poincaré disk (a discrete subgroup of PSL(2, R) ) acts transitively on them.” – Aaron

This section is incomplete, it’s a trainwreck from here on out.

We know it’s bigraded, like the torus group, but how do we count such a bigrading. Do we use a trigrading, that is, number of copies of J, K, and $$(JK)^{-1}$$?

If we then take our generators $$i$$ and $$1$$ for the square lattice, which generate translation “up” and “over” respectively, these guys live on the complex plane with the usual metric. What if we say that they live on this new metric? What if we simply replace each generator with:

(abcd) (i1) = ai+b/ci+d

Here we see that there are actually 4 transformations being done in succession,

1. a translation
2. an inversion,
3. a homothety/rotation,
4. and another translation.

If, like me, you ask “how the hell did they come up with az+b/cz+d” I offer two pieces of thought $\left[ {\begin{array}{cc} a & b \end{array} } \right] \left[ {\begin{array}{c} z \ 1 \end{array} } \right] = \left[ {\begin{array}{c} az+b \ cz+d \end{array} } \right]$

% farrey diagram, multipliying a/c*z/z is still az/cz

For the translations, we know that $$e^{tD} \cdot f(x) = f(x+t)$$. What is the analogue of $$exp$$ for inversions, that is, for what function f and predecessor $$?$$ does the following hold?

\$$? \cdot f(x) = f(\frac{1}{x})\$$

An example of such an $$f$$ and $$?$$ is

\$$\frac{1}{\sqrt{x}} \cdot \Theta(x) = \Theta\big(\frac{1}{x}\big)\$$

(Also, yes, we could be boring and just use good old translation, $$e^{\frac{1-x}{x}D} f(x) = f(\frac{1}{x}$$ This is actually pretty weird, as reflection and translation are not always so nice (1234 example).)

## Some thoughts on detiling polynomials wrt lie algebras of finite groups

(Finite Lie groups Lie algebras)

This week, I asked Alan Weinstein about the concept of a Lie algebra associated to a finite group. He explained that he’d thought about the dual Lie algebra of a discrete group, and felt that looking for the cotangent bundle, not the tangent bundle, was the most natural construction.

Since $$F(x) := \int_0^x f(t) dt$$ is the configuration space of our lattice, then $$d/dx F(x) = f(x) = (1-e^{-D}) F(x)$$

These detiling polynomials, e.g., $$(1-e^{-D})$$, are taking us from our configuration space F to the real numbers – – so the detiling polynomials live in the cotangent bundle (aka Lie coalgebra) NOT the Lie algebra.

%….the hyperbolic lattice’s lie coalgebra using aaron’s thing, my thing with an altered filter….

An interesting thing that Ken Ribet mentioned offhandedly: forget the group structure, can you predict which point will be the next point added? He mentioned this while repeating back what I’d explained to see if he’d understood me correctly.

## A Quick Note on a Geometric Definition of $$v_n$$

This post assumes knowledge of the definition of the oriented cobordism ring, as well as the equivalence $$\pi_*MU \simeq MU^*(pt) =: MU^*$$, and familiarity with the Landweber exact-functor theorem.

A quick post on a nice thing. I was reading Quillen and stumbled across what seems to be the first nod toward the importance of the coefficients $$p, v_1, v_2, … \in MU^*$$.

I have complained about my confusion wrt these coefficients and the concept of complex orientation in a few past blog posts. I’ve read about it so many times in many different equivalent forms, but finally, this one stuck.

They are defined as normal bundles which correspond to a choice of weakly complex structure. Thanks to Tyler Lawson for confirming and clearing up my suspicions on this connection.

For those who haven’t encountered homotopy theory, I’d like to show you the following excerpt of Whitehead so you may realize that Quillen is observing a creature in its native habitat.

“A complex orientation of a map of manifolds $$f: Z \to X$$ is a generalization of a weakly complex structure on $$Z$$ when $$X$$ is a point. By a complex orientation of $$f$$, we mean an equivalence class of factorizations of $$f$$,

\$$Z \xrightarrow{i} E \xrightarrow{\pi} X\$$

where $$p: E \to X$$ is a complex vector bundle and $$i$$ is an embedding endowed with a complex structure on its normal bundle $$v_i$$.”

Does the normal bundle $$v_i$$ have to do with the $$v_i$$ in the sequence $$(p, v_1, v_2, …)$$ in the coefficient ring of MU? Are these just the same letters being used?

Since an equivalence class of factorizations of $$f$$ is an equivalence of choices of the embedding $$i$$, then this is also an equivalence class of the complex structures of the normal bundles $$v_i$$.

Next, if we take the equivalence class of factorizations Z -i-> E -p-> X to be up to cobordism, then each $$v_i$$ is represented by an element of $$MU^*(X)$$. We have to choose $$X$$ to be $$pt$$ for the $$v_i$$ to be in the coefficient ring as we’d expect.

Sidenote: Quillen also mentions that if the dimension of E is “sufficiently large” then one obtains each complex orientation of $$f$$ from exactly one homotopy class of complex structures on $$v_i$$. I am not sure why $$E$$ must be large for this to be true, but I have been told it is for the following two reasons. One is that $$E$$ has to be sufficiently large for $$Z$$ to be able to embed. The second is that it also has to be large for some ambiguities in the process (the isomorphism class of the normal bundle, for example) to be eliminated.

Another sidenote: here’s the excerpt from Whitehead which is not as directly relevant:

It seems that Dan Quillen’s guiding conviction was to understand a mathematical phenomena by seeking out its very simplest concrete manifestation. Due to this, I doubly appreciated the following quote from this seminal paper of his that we’ve been talking about:

I have been strongly influenced by Grothendieck’s theory of motives and like to think of a cobordism theory as a universal contravariant functor on the category of $$C^\infty$$ manifolds endowed with Gysin homomorphism for a class of proper “oriented” maps, instead of as the generalized cohomology theory given by a specific Thom spectrum.

—-Dan Quillen, [Elementary Proofs of Results in Cobordism Theory]

He introduced formal groups as a tool in algebraic topology due to his interest to understand from first principles the result from cobordism theory that the coefficient ring of $$MU$$ is a polynomial ring with an infinite number of even generators.

This caught on. If you’re curious wrt these $$v_n$$, I wrote this little Toda-Smith article on nlab which brushes by how they come up as periodic maps.

An awareness of the classification results on formal group laws gives us some computational tools to try and wrassle the impossible beast of the homotopy groups of spheres. We usually do this a prime at a time, that is, study the homotopy groups of the p-local sphere, because it is way easier. Still impossibly hard, but easier.

For example, his methods led to our finding that $$v_{n}^{-1} BP_*/(p^\infty, …, v_n^\infty$$ is like $$H^*_{gp}(\mathbb{S}_n, E_*)$$. People (who actually know how to ram their heads against these things!) compute some of the higher ones (?) by playing a super-hard game along these lines:

1. $$Ext_{BP_*BP}^{*,*}(v_{n}^{-1} BP_*/(p^\infty, …, v_n^\infty)$$
2. apply the chromatic spectral sequence to get $$Ext^{*,*}BP_*$$ ,
3. then apply the Adams-Novikov to get $$\pi_*\mathbb{S}_{(p)}$$

## Some comments on math communication

I read Bill Thurston’s On Proof and Progress this morning. This led me to consider a few things I’ve learned this summer about the sociology and psychology of being a part of the mathematical community, which I figured I’d share on the off chance that you might find it encouraging or helpful.

0. Communicating in a pedagogically correct manner

I have spent a large part of the summer learning how to speak to other mathematicians, and the standards generally enforced, and I have much more to learn. It is somewhat complicated to keep in mind what is commonly thought of in various subfields as “easy to understand” or “basic,” and what is “complex” or “impossible to grasp.” You musn’t allow it to affect what you view as natural. Just recognize that what you see as simple is not necessarily so in the eyes of others, and vice versa.

Keep track of what causes people to shut off, what causes people to feel like they are being talked down to; probe to find out what their mental models are and what they think of as important. Teaching in a way that feels collaborative involves a large amount of empathy and a change of language, for example:

• No: I was explaining this, and your question confused me.
• Yes: We were doing this, and I got confused.

1. Articulating vague thoughts

It is very important to realize that there are many ways to make precise a vague question (e.g., an undeveloped feeling of connection between two things usually not thought of as connected, an uncertainty about a concept which you are not able to pinpoint).

Spending a large amount of time alone to develop things in your own mental sandbox until they are ready to be translated into words is healthy and natural. But do translate them into words, even if they are not fully formed, s.t. you don’t end up with a theory so removed and technical from the outside that it remains unabsorbed.

Trying to communicate the way you think about an idea in its purest form is not always useful or interesting to the other party, but sometimes it is.

2. Giving a lecture

Giving a good talk and controlling the room are, unfortunately, entirely different skills.

It seems that in a lecture setting, one must abandon hope of communicating formal information, and instead try to communicate key insights and mental models — usually just one or two — and back up philosophical statements with numerous simple examples!

(I gave a talk where each sentence I said aimed to convey things that had originally taken me months to even begin to appreciate. I also let a few people interrupt the flow of the talk in order to heckle wrt small vocabulary differences and technical details. Someone stood up in the middle of my talk and talked for 10 minutes. This is not the way to go.)

3. Building your own mental models

You are going to reinvent things, lots of things. This is good: it is important to practice making original discoveries! If you hold the belief that understanding = you could have invented it yourself, then reinvention is especially encouraging!

Most of my current mental models come from sitting alone and drawing in my notebook what concepts mean to me, the questions they lead to and flowed from, what images they invoke…

It is an intimate act, personally understanding a concept. For me, it involves a lot of doodling and staring out into space.

It also involves a lot of chunking (e.g., a CW-space is an indexed array with attaching maps; a functor is a generalized manifold; approximation of continuous processes via power series; reducing complexity by looking for the base objects and laws which generate the objects you care about).

I started reading histories of mathematics and found that some of the connections and motivations I’d come to myself, and many I hadn’t seen, were the historical reasons for their invention (H-spaces are more general versions of Lie groups, homotopy theory came from complex analysis so what comes from the calculus of variations, etc).

For this reason, there is incredible joy in reading original papers, or in historical and careful recounts of such papers (e.g., Dirichlet’s lectures on Gauss’s Disquisitiones Arithmeticae), as the life of concepts often seems to be lost through a game of paper telephone (the citations of old papers usually don’t convey the interests of the old author).

4. Vocab hunting

This is an incredibly fun and superficially rewarding game: coming across a technical term (e.g., pre-mouse) in a language in which you are not conversant (e.g., model theory), and then “chasing” (via wikipedia links and journal articles) the concept until you find/reformulate the definition into a language you speak. This is best done when you need a pick me up or can’t sleep.