An Entropy Challenge

If you like computer calculations, here’s a little challenge for you. Oscar Dahlsten may have solved it, but we’d love for you to check his work. It’s pretty important for the foundations of thermodynamics, but you don’t need to know any physics or even anything beyond a little algebra to tackle it! First I’ll explain it in really simple terms, then I’ll remind you a bit of why it matters.

We’re looking for two lists of nonnegative numbers, of the same length, listed in decreasing order:

that sum to 1:

and that obey this inequality:

for all (ignoring ), yet do not obey these inequalities:

for all

Oscar’s proposed solution is this:

Can you see if this works? Is there a simpler example, like one with lists of just 3 numbers?

This question came up near the end of my post More Second Laws of Thermodynamics. I phrased the question with a bit more jargon, and said a lot more about its significance. Suppose we have two probability distributions on a finite set, say and We say majorizes if

for all when we write both lists of numbers in decreasing order. This means is ‘less flat’ than , so it should have less entropy. And indeed it does: not just for ordinary entropy, but also for Rényi entropy! The Rényi entropy of is defined by

where or . We can also define Rényi entropy for by taking a limit, and at we get the ordinary entropy

The question is whether majorization is more powerful than Rényi entropy as a tool to to tell when one probability distribution is less flat than another. I know that if majorizes its Rényi entropy is less than than that of for all Your mission, should you choose to accept it, is to show the converse is not true.

More Second Laws of Thermodynamics

Oscar Dahlsten is visiting the Centre for Quantum Technologies, so we’re continuing some conversations about entropy that we started last year, back when the Entropy Club was active. But now Jamie Vicary and Brendan Fong are involved in the conversations.

I was surprised when Oscar told me that for a large class of random processes, the usual second law of thermodynamics is just one of infinitely many laws saying that various kinds of disorder increase. I’m annoyed that nobody ever told me about this before! It’s as if they told me about conservation of energy but not conservation of schmenergy, and phlenergy, and zenergy…

So I need to tell you about this. You may not understand it, but at least I can say I tried. I don’t want you blaming me for concealing all these extra second laws of thermodynamics!

Here’s the basic idea. Not all random processes are guaranteed to make entropy increase. But a bunch of them always make probability distributions flatter in a certain precise sense. This makes the entropy of the probability distribution increase. But when you make a probability distribution flatter in this sense, a bunch of other quantities increase too! For example, besides the usual entropy, there are infinitely many other kinds of entropy, called ‘Rényi entropies’, one for each number between 0 and ∞. And a doubly stochastic operator makes all the Rényi entropies increase! This fact is a special case of Theorem 10 here:

• Tim van Erven and Peter Harremoës, Rényi divergence and majorization.

Let me state this fact precisely, and then say a word about how this is related to quantum theory and ‘the collapse of the wavefunction’.

To keep things simple let’s talk about probability distributions on a finite set, though Erven and Harremoës generalize it all to a measure space.

How do we make precise the concept that one probability distribution is flatter than another? You know it when you see it, at least some of the time. For example, suppose I have some system in thermal equilibrium at some temperature, and the probabilities of it being in various states look like this:

Then say I triple the temperature. The probabilities flatten out:

But how can we make this concept precise in a completely general way? We can do it using the concept of ‘majorization’. If one probability distribution is less flat than another, people say it ‘majorizes’ that other one.

Here’s the definition. Say we have two probability distributions and on the same set. For each one, list the probabilities in decreasing order:

Then we say majorizes if

for all So, the idea is that the biggest probabilities in the distribution add up to more than the corresponding biggest ones in

In 1960, Alfred Rényi defined a generalization of the usual Shannon entropy that depends on a parameter If is a probability distribution on a finite set, its Rényi entropy of order is defined to be

where Well, to be honest: if is 0, 1, or we have to define this by taking a limit where we let creep up to that value. But the limit exists, and when we get the usual Shannon entropy

As I explained a while ago, Rényi entropies are important ways of measuring biodiversity. But here’s what I learned just now, from the paper by Erven and Harremoës:

Theorem 1. If a probability distribution majorizes a probability distribution its Rényi entropies are smaller:

for all

And here’s what makes this fact so nice. If you do something to a classical system in a way that might involve some randomness, we can describe your action using a stochastic matrix. An matrix is called stochastic if whenever is a probability distribution, so is This is equivalent to saying:

• the matrix entries of are all and

• each column of sums to 1.

If is stochastic, it’s not necessarily true that the entropy of is greater than or equal to that of not even for the Shannon entropy.

Puzzle 1. Find a counterexample.

However, entropy does increase if we use specially nice stochastic matrices called ‘doubly stochastic’ matrices. People say a matrix doubly stochastic if it’s stochastic and it maps the probability distribution

to itself. This is the most spread-out probability distribution of all: every other probability distribution majorizes this one.

Why do they call such matrices ‘doubly’ stochastic? Well, if you’ve got a stochastic matrix, each column sums to one. But a stochastic operator is doubly stochastic if and only if each row sums to 1 as well.

Here’s a really cool fact:

Theorem 2. If is doubly stochastic, majorizes for any probability distribution Conversely, if a probability distribution majorizes a probability distribution then for some doubly stochastic matrix .

Taken together, Theorems 1 and 2 say that doubly stochastic transformations increase entropy… but not just Shannon entropy! They increase all the different Rényi entropies, as well. So if time evolution is described by a doubly stochastic matrix, we get lots of ‘second laws of thermodynamics’, saying that all these different kinds of entropy increase!

Finally, what does all this have to do with quantum mechanics, and collapsing the wavefunction? There are different things to say, but this is the simplest:

Theorem 3. Given two probability distributions , then majorizes if and only there exists a self-adjoint matrix with eigenvalues and diagonal entries

The matrix will be a density matrix: a self-adjoint matrix with positive eigenvalues and trace equal to 1. We use such matrices to describe mixed states in quantum mechanics.

Theorem 3 gives a precise sense in which preparing a quantum system in some state, letting time evolve, and then measuring it ‘increases randomness’.

How? Well, suppose we have a quantum system whose Hilbert space is If we prepare the system in a mixture of the standard basis states with probabilities we can describe it with a diagonal density matrix Then suppose we wait a while and some unitary time evolution occurs. The system is now described by a new density matrix

where is some unitary operator. If we then do a measurement to see which of the standard basis states our system now lies in, we’ll get the different possible results with probabilities the diagonal entries of But the eigenvalues of will still be the numbers So, by the theorem, majorizes !

So, not only Shannon entropy but also all the Rényi entropies will increase!

Of course, there are some big physics questions lurking here. Like: what about the real world? In the real world, do lots of different kinds of entropy tend to increase, or just some?

Of course, there’s a huge famous old problem about how reversible time evolution can be compatible with any sort of law saying that entropy must always increase! Still, there are some arguments, going back to Boltzmann’s H-theorem, which show entropy increases under some extra conditions. So then we can ask if other kinds of entropy, like Rényi entropy, increase as well. This will be true whenever we can argue that time evolution is described by doubly stochastic matrices. Theorem 3 gives a partial answer, but there’s probably much more to say.

I don’t have much more to say right now, though. I’ll just point out that while doubly stochastic matrices map the ‘maximally smeared-out’ probability distribution

to itself, a lot of this theory generalizes to stochastic matrices that map exactly one other probability distribution to itself. We need to work with relative Rényi entropy instead of Rényi entropy, and so on, but I don’t think these adjustments are really a big deal. And there are nice theorems that let you know when a stochastic matrix maps exactly one probability distribution to itself, based on the Perron–Frobenius theorem.

References

I already gave you a reference for Theorem 1, namely the paper by Erven and Harremoës, though I don’t think they were the first to prove this particular result: they generalize it quite a lot.

What about Theorem 2? It goes back at least to here:

• Barry C. Arnold, Majorization and the Lorenz Order: A Brief Introduction, Springer Lecture Notes in Statistics 43, Springer, Berlin, 1987.

The partial order on probability distributions given by majorization is also called the ‘Lorenz order’, but mainly when we consider probability distributions on infinite sets. This name presumably comes from the Lorenz curve, a measure of income inequality. This curve shows for the bottom x% of households, what percentage y% of the total income they have:

Puzzle 2. If you’ve got two different probability distributions of incomes, and one majorizes the other, how are their Lorenz curves related?

When we generalize majorization by letting some other probability distribution take the place of

it seems people call it the ‘Markov order’. Here’s a really fascinating paper on that, which I’m just barely beginning to understand:

• A. N. Gorban, P. A. Gorban and G. Judge, Entropy: the Markov ordering approach, Entropy 12 (2010), 1145–1193.

What about Theorem 3? Apparently it goes back to here:

• A. Uhlmann, Wiss. Z. Karl-Marx-Univ. Leipzig 20 (1971), 633.

though I only know this thanks to a more recent paper:

• Michael A. Nielsen, Conditions for a class of entanglement transformations, Phys. Rev. Lett. 83 (1999), 436–439.

By the way, Nielsen’s paper contains another very nice result about majorization! Suppose you have states and of a 2-part quantum system. You can trace out one part and get density matrices describing mixed states of the other part, say and . Then Nielsen shows you can get from to using ‘local operations and classical communication’ if and only if majorizes . Note that things are going backwards here compared to how they’ve been going in the rest of this post: if we can get from to , then all forms of entropy go down when we go from to ! This ‘anti-second-law’ behavior is confusing at first, but familiar to me by now.

When I first learned all this stuff, I naturally thought of the following question—maybe you did too, just now. If are probability distributions and

for all , is it true that majorizes ?

Apparently the answer must be no, because Klimesh has gone to quite a bit of work to obtain a weaker conclusion: not that majorizes , but that majorizes for some probability distribution He calls this catalytic majorization, with serving as a ‘catalyst’:

• Matthew Klimesh, Inequalities that collectively completely characterizes the catalytic majorization relation.

I thank Vlatko Vedral here at the CQT for pointing this out!

Finally, here is a good general introduction to majorization, pointed out by Vasileios Anagnostopoulos:

• T. Ando, Majorization, doubly stochastic matrices, and comparison of eigenvalues, Linear Algebra and its Applications 118 (1989), 163-–248.

Network Theory (Part 24)

Now we’ve reached the climax of our story so far: we’re ready to prove the deficiency zero theorem. First let’s talk about it informally a bit. Then we’ll review the notation, and then—hang on to your seat!—we’ll give the proof.

The crucial trick is to relate a bunch of chemical reactions, described by a ‘reaction network’ like this:

to a simpler problem where a system randomly hops between states arranged in the same pattern:

This is sort of amazing, because we’ve thrown out lots of detail. It’s also amazing because this simpler problem is linear. In the original problem, the chance that a reaction turns a B + E into a D is proportional to the number of B’s times the number of E’s. That’s nonlinear! But in the simplified problem, the chance that your system hops from state 4 to state 3 is just proportional to the probability that it’s in state 4 to begin with. That’s linear.

The wonderful thing is that, at least under some conditions, we can find equilibrium solutions of our original problem starting from equilibrium solutions of the simpler problem.

Let’s roughly sketch how it works, and where we are so far. Our simplified problem is described by an equation like this:

where is a function that the probability of being in each state, and describes the probability per time of hopping from one state to another. We can easily understand quite a lot about the equilibrium solutions, where doesn’t change at all:

because this is a linear equation. We did this in Part 23. Of course, when I say ‘easily’, that’s a relative thing: we needed to use the Perron–Frobenius theorem, which Jacob introduced in Part 20. But that’s a well-known theorem in linear algebra, and it’s easy to apply here.

In Part 22, we saw that the original problem was described by an equation like this, called the ‘rate equation’:

Here is a vector whose entries describe the amount of each kind of chemical: the amount of A’s, the amount of B’s, and so on. The matrix is the same as in the simplified problem, but is a matrix that says how many times each chemical shows up in each spot in our reaction network:

The key thing to notice is where we take a vector and raise it to the power of a matrix. We explained this operation back in Part 22. It’s this operation that says how many B + E pairs we have, for example, given the number of B’s and the number of E’s. It’s this that makes the rate equation nonlinear.

Now, we’re looking for equilibrium solutions of the rate equation, where the rate of change is zero:

But in fact we’ll do even better! We’ll find solutions of this:

And we’ll get these by taking our solutions of this:

and adjusting them so that

while remains a solution of

But: how do we do this ‘adjusting’? That’s the crux of the whole business! That’s what we’ll do today.

Remember, is a function that gives a probability for each ‘state’, or numbered box here:

The picture here consists of two pieces, called ‘connected components’: the piece containing boxes 0 and 1, and the piece containing boxes 2, 3 and 4. It turns out that we can multiply by a function that’s constant on each connected component, and if we had to begin with, that will still be true afterward. The reason is that there’s no way for to ‘leak across’ from one component to another. It’s like having water in separate buckets. You can increase the amount of water in one bucket, and decrease it another, and as long as the water’s surface remains flat in each bucket, the whole situation remains in equilibrium.

That’s sort of obvious. What’s not obvious is that we can adjust this way so as to ensure

for some

And indeed, it’s not always true! It’s only true if our reaction network obeys a special condition. It needs to have ‘deficiency zero’. We defined this concept back in Part 21, but now we’ll finally use it for something. It turns out to be precisely the right condition to guarantee we can tweak any function on our set of states, multiplying it by a function that’s constant on each connected component, and get a new function with

When all is said and done, that is the key to the deficiency zero theorem.

Review

The battle is almost upon us—we’ve got one last chance to review our notation. We start with a stochastic reaction network:

This consists of:

• finite sets of transitions complexes and species

• a map giving a rate constant for each transition,

• source and target maps saying where each transition starts and ends,

• a one-to-one map saying how each complex is made of species.

Then we extend and to linear maps:

Then we put inner products on these vector spaces as described last time, which lets us ‘turn around’ the maps and by taking their adjoints:

More surprisingly, we can ‘turn around’ and get a nonlinear map using ‘matrix exponentiation’:

This is most easily understood by thinking of as a row vector and as a matrix:

Remember, complexes are made out of species. The matrix entry says how many things of the th species there are in a complex of the th kind. If says how many complexes there are of each kind, says how many things there are of each species. Conversely, if says how many things there are of each species, says how many ways we can build each kind of complex from them.

So, we get these maps:

Next, the boundary operator

describes how each transition causes a change in complexes:

As we saw last time, there is a Hamiltonian

describing a Markov processes on the set of complexes, given by

But the star of the show is the rate equation. This describes how the number of things of each species changes with time. We write these numbers in a list and get a vector with nonnegative components. The rate equation says:

We can read this as follows:

• says how many things of each species we have now.

• says how many complexes of each kind we can build from these species.

• says how many transitions of each kind can originate starting from these complexes, with each transition weighted by its rate.

• is the rate of change of the number of complexes of each kind, due to these transitions.

• is the rate of change of the number of things of each species.

The zero deficiency theorem

We are looking for equilibrium solutions of the rate equation, where the number of things of each species doesn’t change at all:

In fact we will find complex balanced equilibrium solutions, where even the number of complexes of each kind doesn’t change:

More precisely, we have:

Deficiency Zero Theorem (Child’s Version). Suppose we have a reaction network obeying these two conditions:

1. It is weakly reversible, meaning that whenever there’s a transition from one complex to another there’s a directed path of transitions going back from to

2. It has deficiency zero, meaning .

Then for any choice of rate constants there exists a complex balanced equilibrium solution of the rate equation where all species are present in nonzero amounts. In other words, there exists with all components positive and such that:

Proof. Because our reaction network is weakly reversible, the theorems in Part 23 show there exists with

This may not be of the form but we shall adjust so that it becomes of this form, while still remaining a solution of latex . To do this, we need a couple of lemmas:

Lemma 1.

Proof. We need to use a few facts from linear algebra. If is a finite-dimensional vector space with inner product, the orthogonal complement of a subspace consists of vectors that are orthogonal to everything in :

We have

where and are subspaces of and denotes the sum of two subspaces: that is, the smallest subspace containing both. Also, if is a linear map between finite-dimensional vector spaces with inner product, we have

and

Now, because our reaction network has deficiency zero, we know that

Taking the orthogonal complement of both sides, we get

and using the rules we mentioned, we obtain

as desired.   █

Now, given a vector in or with all positive components, we can define the logarithm of such a vector, component-wise:

Similarly, for any vector in either of these spaces, we can define its exponential in a component-wise way:

These operations are inverse to each other. Moreover:

Lemma 2. The nonlinear operator

is related to the linear operator

by the formula

which holds for all

Proof. A straightforward calculation. By the way, this formula would look a bit nicer if we treated as a row vector and multiplied it on the right by : then we would have

The problem is that we are following the usual convention of multiplying vectors by matrices on the left, yet writing the matrix on the right in Taking the transpose of the matrix serves to compensate for this.   █

Now, given our vector with we can take its logarithm and get Lemma 1 says that

so we can write

where and Moreover, we can write

for some so that

Exponentiating both sides component-wise, we get

where at right we are taking the component-wise product of vectors. Thanks to Lemma 2, we conclude that

So, we have taken and almost written it in the form —but not quite! We can adjust to make it be of this form:

Clearly all the components of are positive, since the same is true for both and So, the only remaining task is to check that

We do this using two lemmas:

Lemma 3. If and then

Proof. It is enough to check that multiplication by commutes with the Hamiltonian since then

Recall from Part 23 that is the Hamiltonian of a Markov process associated to this ‘graph with rates’:

As noted here:

• John Baez and Brendan Fong, A Noether theorem for Markov processes.

multiplication by some function on commutes with if and only if that function is constant on each connected component of this graph. Such functions are called conserved quantities.

So, it suffices to show that is constant on each connected component. For this, it is enough to show that itself is constant on each connected component. But this will follow from the next lemma, since   █

Lemma 4. A function is a conserved quantity iff In other words, is constant on each connected component of the graph iff .

Proof. Suppose or in other words, or in still other words, To show that is constant on each connected component, it suffices to show that whenever we have two complexes connected by a transition, like this:

then takes the same value at both these complexes:

To see this, note

and since we conclude

But calculating this inner product, we see

as desired.

For the converse, we simply turn the argument around: if is constant on each connected component, we see whenever there is a transition It follows that for every transition so

And thus concludes the proof of the lemma!   █

And thus concludes the proof of the theorem!   █

And thus concludes this post!

Network Theory (Part 23)

We’ve been looking at reaction networks, and we’re getting ready to find equilibrium solutions of the equations they give. To do this, we’ll need to connect them to another kind of network we’ve studied. A reaction network is something like this:

It’s a bunch of complexes, which are sums of basic building-blocks called species, together with arrows called transitions going between the complexes. If we know a number for each transition describing the rate at which it occurs, we get an equation called the ‘rate equation’. This describes how the amount of each species changes with time. We’ve been talking about this equation ever since the start of this series! Last time, we wrote it down in a new very compact form:

Here is a vector whose components are the amounts of each species, while and are certain matrices.

But now suppose we forget how each complex is made of species! Suppose we just think of them as abstract things in their own right, like numbered boxes:

We can use these boxes to describe states of some system. The arrows still describe transitions, but now we think of these as ways for the system to hop from one state to another. Say we know a number for each transition describing the probability per time at which it occurs:

Then we get a ‘Markov process’—or in other words, a random walk where our system hops from one state to another. If is the probability distribution saying how likely the system is to be in each state, this Markov process is described by this equation:

This is simpler than the rate equation, because it’s linear. But the matrix is the same—we’ll see that explicitly later on today.

What’s the point? Well, our ultimate goal is to prove the deficiency zero theorem, which gives equilibrium solutions of the rate equation. That means finding with

Today we’ll find all equilibria for the Markov process, meaning all with

Then next time we’ll show some of these have the form

So, we’ll get

and thus

as desired!

So, let’s get to to work.

The Markov process of a graph with rates

We’ve been looking at stochastic reaction networks, which are things like this:

However, we can build a Markov process starting from just part of this information:

Let’s call this thing a ‘graph with rates’, for lack of a better name. We’ve been calling the things in ‘complexes’, but now we’ll think of them as ‘states’. So:

Definition. A graph with rates consists of:

• a finite set of states

• a finite set of transitions

• a map giving a rate constant for each transition,

• source and target maps saying where each transition starts and ends.

Starting from this, we can get a Markov process describing how a probability distribution on our set of states will change with time. As usual, this Markov process is described by a master equation:

for some Hamiltonian:

What is this Hamiltonian, exactly? Let’s think of it as a matrix where is the probability per time for our system to hop from the state to the state This looks backwards, but don’t blame me—blame the guys who invented the usual conventions for matrix algebra. Clearly if this probability per time should be the sum of the rate constants of all transitions going from to :

where we write when is a transition with source and target

Now, we saw in Part 11 that for a probability distribution to remain a probability distribution as it evolves in time according to the master equation, we need to be infinitesimal stochastic: its off-diagonal entries must be nonnegative, and the sum of the entries in each column must be zero.

The first condition holds already, and the second one tells us what the diagonal entries must be. So, we’re basically done describing But we can summarize it this way:

Puzzle 1. Think of as the vector space consisting of finite linear combinations of elements Then show

Equilibrium solutions of the master equation

Now we’ll classify equilibrium solutions of the master equation, meaning with

We’ll do only do this when our graph with rates is ‘weakly reversible’. This concept doesn’t actually depend on the rates, so let’s be general and say:

Definition. A graph is weakly reversible if for every edge there is directed path going back from to meaning that we have edges

This graph with rates is not weakly reversible:

but this one is:

The good thing about the weakly reversible case is that we get one equilibrium solution of the master equation for each component of our graph, and all equilibrium solutions are linear combinations of these. This is not true in general! For example, this guy is not weakly reversible:

It has only one component, but the master equation has two linearly independent equilibrium solutions: one that vanishes except at the state 0, and one that vanishes except at the state 2.

The idea of a ‘component’ is supposed to be fairly intuitive—our graph falls apart into pieces called components—but we should make it precise. As explained in Part 21, the graphs we’re using here are directed multigraphs, meaning things like

where is the set of edges (our transitions) and is the set of vertices (our states). There are actually two famous concepts of ‘component’ for graphs of this sort: ‘strongly connected’ components and ‘connected’ components. We only need connected components, but let me explain both concepts, in a futile attempt to slake your insatiable thirst for knowledge.

Two vertices and of a graph lie in the same strongly connected component iff you can find a directed path of edges from to and also one from back to

Remember, a directed path from to looks like this:

Here’s a path from to that is not directed:

and I hope you can write down the obvious but tedious definition of an ‘undirected path’, meaning a path made of edges that don’t necessarily point in the correct direction. Given that, we say two vertices and lie in the same connected component iff you can find an undirected path going from to In this case, there will automatically also be an undirected path going from to

For example, and lie in the same connected component here, but not the same strongly connected component:

Here’s a graph with one connected component and 3 strongly connected components, which are marked in blue:

For the theory we’re looking at now, we only care about connected components, not strongly connected components! However:

Puzzle 2. Show that for weakly reversible graph, the connected components are the same as the strongly connected components.

With these definitions out of way, we can state today’s big theorem:

Theorem. Suppose is the Hamiltonian of a weakly reversible graph with rates:

Then for each connected component there exists a unique probability distribution that is positive on that component, zero elsewhere, and is an equilibrium solution of the master equation:

Moreover, these probability distributions form a basis for the space of equilibrium solutions of the master equation. So, the dimension of this space is the number of components of

Proof. We start by assuming our graph has one connected component. We use the Perron–Frobenius theorem, as explained in Part 20. This applies to ‘nonnegative’ matrices, meaning those whose entries are all nonnegative. That is not true of itself, but only its diagonal entries can be negative, so if we choose a large enough number will be nonnegative.

Since our graph is weakly reversible and has one connected component, it follows straight from the definitions that the operator will also be ‘irreducible’ in the sense of Part 20. The Perron–Frobenius theorem then swings into action, and we instantly conclude several things.

First, has a positive real eigenvalue such that any other eigenvalue, possibly complex, has absolute value Second, there is an eigenvector with eigenvalue and all positive components. Third, any other eigenvector with eigenvalue is a scalar multiple of

Subtracting it follows that is the eigenvalue of with the largest real part. We have and any other vector with this property is a scalar multiple of

We can show that in fact To do this we copy an argument from Part 20. First, since is positive we can normalize it to be a probability distribution:

Since is infinitesimal stochastic, sends probability distributions to probability distributions:

for all On the other hand,

so we must have

We conclude that when our graph has one connected component, there is a probability distribution that is positive everywhere and has Moreover, any with is a scalar multiple of

When has several components, the matrix is block diagonal, with one block for each component. So, we can run the above argument on each component and get a probability distribution that is positive on We can then check that and that every with can be expressed as a linear combination of these probability distributions in a unique way.   █

This result must be absurdly familiar to people who study Markov processes, but I haven’t bothered to look up a reference yet. Do you happen to know a good one? I’d like to see one that generalizes this theorem to graphs that aren’t weakly reversible. I think I see how it goes. We don’t need that generalization right now, but it would be good to have around.

The Hamiltonian, revisited

One last small piece of business: last time I showed you a very slick formula for the Hamiltonian I’d like to prove it agrees with the formula I gave this time.

We start with any graph with rates:

We extend and to linear maps between vector spaces:

We define the boundary operator just as we did last time:

Then we put an inner product on the vector spaces and So, for we let the elements of be an orthonormal basis, but for we define the inner product in a more clever way involving the rate constants:

where This lets us define adjoints of the maps and via formulas like this:

Then:

Theorem. The Hamiltonian for a graph with rates is given by

Proof. It suffices to check that this formula agrees with the formula for given in Puzzle 1:

Here we are using the complex as a name for one of the standard basis vectors of Similarly shall we write things like or for basis vectors of

First, we claim that

To prove this it’s enough to check that taking the inner products of either sides with any basis vector we get results that agree. On the one hand:

On the other hand:

where the factor of in the inner product on cancels the visible factor of So indeed the results match.

Using this formula for we now see that

which is precisely what we want.   █

I hope you see through the formulas to their intuitive meaning. As usual, the formulas are just a way of precisely saying something that makes plenty of sense. If is some state of our Markov process, is the sum of all transitions starting at this state, weighted by their rates. Applying to a transition tells us what change in state it causes. So tells us the rate at which things change when we start in the state That’s why is the Hamiltonian for our Markov process. After all, the Hamiltonian tells us how things change:

Okay, we’ve got all the machinery in place. Next time we’ll prove the deficiency zero theorem!

Network Theory (Part 22)

Okay, now let’s dig deeper into the proof of the deficiency zero theorem. We’re only going to prove a baby version, at first. Later we can enhance it:

Deficiency Zero Theorem (Baby Version). Suppose we have a weakly reversible reaction network with deficiency zero. Then for any choice of rate constants there exists an equilibrium solution of the rate equation where all species are present in nonzero amounts.

The first step is to write down the rate equation in a new, more conceptual way. It’s incredibly cool. You’ve probably seen Schrödinger’s equation, which describes the motion of a quantum particle:

If you’ve been following this series, you’ve probably also seen the master equation, which describes the motion of a ‘stochastic’ particle:

A ‘stochastic’ particle is one that’s carrying out a random walk, and now describes its probability to be somewhere, instead of its amplitude.

Today we’ll see that the rate equation for a reaction network looks somewhat similar:

where is some matrix, and is defined using a new thing called ‘matrix exponentiation’, which makes the equation nonlinear!

If you’re reading this you probably know how to multiply a vector by a matrix. But if you’re like me, you’ve never seen anyone take a vector and raise to the power of some matrix! I’ll explain it, don’t worry… right now I’m just trying to get you intrigued. It’s not complicated, but it’s exciting how this unusual operation shows up naturally in chemistry. That’s just what I’m looking for these days: new math ideas that show up in practical subjects like chemistry, and new ways that math can help people in these subjects.

Since we’re looking for an equilibrium solution of the rate equation, we actually want to solve

or in other words

In fact we will do better: we will find a solution of

And we’ll do this in two stages:

• First we’ll find all solutions of

This equation is linear, so it’s easy to understand.

• Then, among these solutions we’ll find one that also obeys

This is a nonlinear problem involving matrix exponentiation, but still, we can do it, using a clever trick called ‘logarithms’.

Putting the pieces together, we get our solution of

and thus our equilibrium solution of the rate equation:

That’s a rough outline of the plan. But now let’s get started, because the details are actually fascinating. Today I’ll just show you how to rewrite the rate equation in this new way.

The rate equation

Remember how the rate equation goes. We start with a stochastic reaction network, meaning a little diagram like this:

This contains quite a bit of information:

• a finite set of transitions,

• a finite set of complexes,

• a finite set of species,

• a map giving a rate constant for each transition,

• source and target maps saying where each transition starts and ends,

• a one-to-one map saying how each complex is made of species.

Given all this, the rate equation says how the amount of each species changes with time. We describe these amounts with a vector So, we want a differential equation filling in the question marks here:

Now last time, we started by thinking of as a subset of and thus of the vector space Back then, we wrote the rate equation as follows:

where vector exponentiation is defined by

when and are vectors in

However, we’ve now switched to thinking of our set of complexes as a set in its own right that is mapped into by This is good for lots of reasons, like defining the concept of ‘deficiency’, which we did last time. But it means the rate equation above doesn’t quite parse anymore! Things like and live in ; we need to explicitly convert them into elements of using for our equation to make sense!

So now we have to write the rate equation like this:

This looks more ugly, but if you’ve got even one mathematical bone in your body, you can already see vague glimmerings of how we’ll rewrite this the way we want:

Here’s how.

First, we extend our maps and to linear maps between vector spaces:

Then, we put an inner product on the vector spaces and For we do this in the most obvious way, by letting the complexes be an orthonormal basis. So, given two complexes we define their inner product by

We do the same for But for we define the inner product in a more clever way involving the rate constants. If are two transitions, we define their inner product by:

This will seem perfectly natural when we continue our study of circuits made of electrical resistors, and if you’re very clever you can already see it lurking in Part 16. But never mind.

Having put inner products on these three vector spaces, we can take the adjoints of the linear maps between them, to get linear maps going back the other way:

These are defined in the usual way—though we’re using daggers here they way physicists do, where mathematicians would prefer to see stars! For example, is defined by the relation

and so on.

Next, we set up a random walk on the set of complexes. Remember, our reaction network is a graph with complexes as vertices and transitions as edges, like this:

Each transition has a number attached to it: the rate constant So, we can randomly hop from complex to complex along these transitions, with probabilities per unit time described by these numbers. The probability of being at some particular complex will then be described by a function

which also depends on time, and changes according to the equation

for some Hamiltonian

I defined this Hamiltonian back in Part 15, but now I see a slicker way to write it:

I’ll justify this next time. For now, the main point is that with this Hamiltonian, the rate equation is equivalent to this:

The only thing I haven’t defined yet is the funny exponential That’s what makes the equation nonlinear. We’re taking a vector to the power of a matrix and getting a vector. This sounds weird—but it actually makes sense!

It only makes sense because we have chosen bases for our vector spaces. To understand it, let’s number our species as we’ve been doing all along, and number our complexes Our linear map then becomes a matrix of natural numbers. Its entries say how many times each species shows up in each complex:

The entry says how many times the th species shows up in the th complex.

Now, let’s be a bit daring and think of the vector as a row vector with entries:

Then we can multiply on the right by the matrix and get a vector in :

So far, no big deal. But now you’re ready to see the definition of which is very similar:

It’s exactly the same, but with multiplication replacing addition, and exponentiation replacing multiplication! Apparently my class on matrices stopped too soon: we learned about matrix multiplication, but matrix exponentiation is also worthwhile.

What’s the point of it? Well, suppose you have a certain number of hydrogen molecules, a certain number of oxygen molecules, a certain number of water molecules, and so on—a certain number of things of each species. You can list these numbers and get a vector Then the components of describe how many ways you can build up each complex from the things you have. For example,

say roughly how many ways you can build complex 1 by picking things of species 1, things of species 2, and so on.

Why ‘roughly’? Well, we’re pretending we can pick the same thing twice. So if we have 4 water molecules and we need to pick 3, this formula gives The right answer is To get this answer we’d need to use the ‘falling power’ as explained in Part 4. But the rate equation describes chemistry in the limit where we have lots of things of each species. In this limit, the ordinary power becomes a good approximation.

Puzzle. In this post we’ve seen a vector raised to a matrix power, which is a vector, and also a vector raised to a vector power, which is a number. How are they related?

There’s more to say about this, which I’d be glad to explain if you’re interested. But let’s get to the punchline:

Theorem. The rate equation:

is equivalent to this equation:

or in other words:

Proof. It’s enough to show

So, we’ll compute and think about the meaning of each quantity we get en route.

We start with This is a list of numbers saying how many things of each species we have: our raw ingredients, as it were. Then we compute

This is a vector in It’s a list of numbers saying how many ways we can build each complex starting from our raw ingredients.

Alternatively, we can write this vector as a sum over basis vectors:

Next let’s apply to this. We claim that

In other words, we claim is the sum of all the transitions having as their source, weighted by their rate constants! To prove this claim, it’s enough to take the inner product of each side with any transition and check that we get the same answer. For the left side we get

To compute the right side, we need to use the cleverly chosen inner product on Here we get

In the first step here, the factor of in the cleverly chosen inner product canceled the visible factor of For the second step, you just need to think for half a minute—or ten, depending on how much coffee you’ve had.

Either way, we we conclude that indeed

Next let’s combine this with our formula for :

We get this:

In other words, is a linear combination of transitions, where each one is weighted both by the rate it happens and how many ways it can happen starting with our raw ingredients.

Our goal is to compute We’re almost there. Remember, says which complex is the input of a given transition, and says which complex is the output. So, says the total rate at which complexes are created and/or destroyed starting with the species in as our raw ingredients.

That sounds good. But let’s just pedantically check that everything works. Applying to both sides of our last equation, we get

Remember, our goal was to prove that this equals

But if you stare at these a while and think, you’ll see they’re equal.   █

It took me a couple of weeks to really understand this, so I’ll be happy if it takes you just a few days. It seems peculiar at first but ultimately it all makes sense. The interesting subtlety is that we use the linear map called ‘multiplying by ’:

to take a bunch of complexes and work out the species they contain, while we use the nonlinear map called ‘raising to the th power’:

to take a bunch of species and work out how many ways we can build each complex from them. There is much more to say about this: for example, these maps arise from a pair of what category theorists call ‘adjoint functors’. But I’m worn out and you probably are too, if you’re still here at all.

References

I found this thesis to be the most helpful reference when I was trying to understand the proof of the deficiency zero theorem:

• Jonathan M. Guberman, Mass Action Reaction Networks and the Deficiency Zero Theorem, B.A. thesis, Department of Mathematics, Harvard University, 2003.

I urge you to check it out. In particular, Section 3 and Appendix A discuss matrix exponentiation. Has anyone discussed this before?

Here’s another good modern treatment of the deficiency zero theorem:

• Jeremy Gunawardena, Chemical reaction network theory for in silico biologists, 2003.

The theorem was first proved here:

• Martin Feinberg, Chemical reaction network structure and the stability of complex isothermal reactors: I. The deficiency zero and deficiency one theorems, Chemical Engineering Science 42 (1987), 2229-2268.

However, Feinberg’s treatment here relies heavily on this paper:

• F. Horn and R. Jackson, General mass action kinetics, Archive for Rational Mechanics and Analysis 47 (1972), 81-116.

(Does anyone work on ‘irrational mechanics’?) These lectures are also very helpful:

• Martin Feinberg, Lectures on reaction networks, 1979.

If you’ve seen other proofs, let us know.

Network Theory (Part 21)

Recently we’ve been talking about ‘reaction networks’, like this:

The letters are names of chemicals, and the arrows are chemical reactions. If we know how fast each reaction goes, we can write down a ‘rate equation’ describing how the amount of each chemical changes with time.

In Part 17 we met the deficiency zero theorem. This is a powerful tool for finding equilibrium solutions of the rate equation: in other words, solutions where the amounts of the various chemicals don’t change at all. To apply this theorem, two conditions must hold. Both are quite easy to check:

• Your reaction network needs to be ‘weakly reversible’: if you have a reaction that takes one bunch of chemicals to another, there’s a series of reactions that takes that other bunch back to the one you started with.

• A number called the ‘deficiency’ that you can compute from your reaction network needs to be zero.

The first condition makes a lot of sense, intuitively: you won’t get an equilibrium with all the different chemicals present if some chemicals can turn into others but not the other way around. But the second condition, and indeed the concept of ‘deficiency’, seems mysterious.

Luckily, when you work through the proof of the deficiency zero theorem, the mystery evaporates. It turns out that there are two equivalent ways to define the deficiency of a reaction network. One makes it easy to compute, and that’s the definition people usually give. But the other explains why it’s important.

In fact the whole proof of the deficiency zero theorem is full of great insights, so I want to show it to you. This will be the most intense part of the network theory series so far, but also the climax of its first phase. After a little wrapping up, we will then turn to another kind of network: electrical circuits!

Today I’ll just unfold the concept of ‘deficiency’ so we see what it means. Next time I’ll show you a slick way to write the rate equation, which is crucial to proving the deficiency zero theorem. Then we’ll start the proof.

Reaction networks

Let’s recall what a reaction network is, and set up our notation. In chemistry we consider a finite set of ‘species’ like C, O₂, H₂O and so on… and then consider reactions like

CH₄ + 2O₂ CO₂ + 2H₂O

On each side of this reaction we have a finite linear combination of species, with natural numbers as coefficients. Chemists call such a thing a complex.

So, given any finite collection of species, say let’s write to mean the set of finite linear combinations of elements of with natural numbers as coefficients. The complexes appearing in our reactions will form a subset of this, say

We’ll also consider a finite collection of reactions—or in the language of Petri nets, ‘transitions’. Let’s call this . Each transition goes from some complex to some other complex: if we want a reaction to be reversible we’ll explicitly include another reaction going the other way. So, given a transition it will always go from some complex called its source to some complex called its target

All this data, put together, is a reaction network:

Definition. A reaction network consists of:

• a finite set of species,

• a finite set of transitions,

• a finite set of complexes,

• source and target maps

We can draw a reaction network as a graph with complexes as vertices and transitions as edges:

The set of species here is and the set of complexes is

But to write down the ‘rate equation’ describing our chemical reactions, we need a bit more information: constants saying the rate at which each transition occurs. So, we define:

Definition. A stochastic reaction network is a reaction network together with a map assigning a rate constant to each transition.

Let me remind you how the rate equation works. At any time we have some amount of each species These numbers are the components of a vector which of course depends on time. The rate equation says how this vector changes:

Here I’m writing for the th component of the vector and similarly for I should remind you what means, since here we are raising a vector to a vector power, which is a bit unusual. So, suppose we have any vector

and we raise it to the power of

Then by definition we get

Given this, I hope the rate equation makes intuitive sense! There’s one term for each transition The factor of shows up because our transition destroys things of the th species and creates of them. The big product

shows up because our transition occurs at a rate proportional to the product of the numbers of things it takes as inputs. The constant of proportionality is the reaction rate

The deficiency zero theorem says lots of things, but in the next few episodes we’ll prove a weak version, like this:

Deficiency Zero Theorem (Baby Version). Suppose we have a weakly reversible reaction network with deficiency zero. Then for any choice of rate constants there exists an equilibrium solution of the rate equation. In other words:

An important feature of this result is that all the components of the vector are positive. In other words, there’s actually some chemical of each species present!

But what do the hypotheses in this theorem mean?

A reaction network is weakly reversible if for any transition going from a complex to a complex there is a sequence of transitions going from back to But what about ‘deficiency zero’? As I mentioned, this requires more work to understand.

So, let’s dive in!

Deficiency

In modern math, we like to take all the stuff we’re thinking about and compress it into a diagram with a few objects and maps going between these objects. So, to understand the deficiency zero theorem, I wanted to isolate the crucial maps. For starters, there’s an obvious map

sending each complex to the linear combination of species that it is.

Indeed, we can change viewpoints a bit and think of as an abstract set equipped with this map saying how each complex is made of species. From now on this is what we’ll do. Then all the information in a stochastic reaction network sits in this diagram:

This is fundamental to everything we’ll do for the next few episodes, so take a minute to lock it into your brain.

We’ll do lots of different things with this diagram. For example, we often want to use ideas from linear algebra, and then we want our maps to be linear. For example, extends uniquely to a linear map

sending real linear combinations of complexes to real linear combinations of species. Reusing the name here won’t cause confusion. We can also extend and to linear maps in a unique way, getting a little diagram like this:

Linear algebra lets us talk about differences of complexes. Each transition gives a vector

saying the change in complexes that it causes. And we can extend uniquely to a linear map

defined on linear combinations of transitions. Mathematicians would call a boundary operator.

So, we have a little sequence of linear maps

This turns a transition into a change in complexes, and then a change in species.

If you know fancy math you’ll be wondering if this sequence is a ‘chain complex’, which is a fancy way of saying that The answer is no. This equation means that every linear combination of reactions leaves the amount of all species unchanged. Or equivalently: every reaction leaves the amount of all species unchanged. This only happens in very silly examples.

Nonetheless, it’s possible for a linear combination of reactions to leave the amount of all species unchanged.

For example, this will happen if we have a linear combination of reactions that leaves the amount of all complexes unchanged. But this sufficient condition is not necessary. And this leads us to the concept of ‘deficiency zero’:

Definition. A reaction network has deficiency zero if any linear combination of reactions that leaves the amount of every species unchanged also leaves the amount of every complex unchanged.

In short, a reaction network has deficiency zero iff

for every In other words—using some basic concepts from linear algebra—a reaction network has deficiency zero iff is one-to-one when restricted to the image of Remember, the image of is

Roughly speaking, this consists of all changes in complexes that can occur due to reactions.

In still other words, a reaction network has deficiency zero if 0 is the only vector in both the image of and the kernel of

Remember, the kernel of is

Roughly speaking, this consists of all changes in complexes that don’t cause changes in species. So, ‘deficiency zero’ roughly says that if a reaction causes a change in complexes, it causes a change in species.

(All this ‘roughly speaking’ stuff is because in reality I should be talking about linear combinations of transitions, complexes and species. But it’s a bit distracting to keep saying that when I’m trying to get across the basic ideas!)

Now we’re ready to understand deficiencies other than zero, at least a little. They’re defined like this:

Definition. The deficiency of a reaction network is the dimension of

How to compute the deficiency

You can compute the deficiency of a reaction network just by looking at it. However, it takes a little training. First, remember that a reaction network can be drawn as a graph with complexes as vertices and transitions as edges, like this:

There are three important numbers we can extract from this graph:

• We can count the number of vertices in this graph; let’s call that since it’s just the number of complexes.

• We can count the number of pieces or ‘components’ of this graph; let’s call that for now.

• We can also count the dimension of the image of This space, is called the stochiometric subspace: vectors in here are changes in species that can be accomplished by transitions in our reaction network, or linear combinations of transitions.

These three numbers, all rather easy to compute, let us calculate the deficiency:

Theorem. The deficiency of a reaction network equals

Proof. By definition we have

but another way to say this is

where we are restricting to the subspace and taking the dimension of the kernel of that.

The rank-nullity theorem says that whenever you have a linear map between finite-dimensional vector spaces, then

where is the domain of namely the vector space It follows that

The domain of is just while its image equals so

The theorem then follows from this:

Lemma.

Proof. In fact this holds whenever we have a finite set of complexes and a finite set of transitions going between them. We get a diagram

We can extend the source and target functions to linear maps as usual:

and then we can define We claim that

where is the number of connected components of the graph with as vertices and as edges.

This is easiest to see using an inductive argument where we start by throwing out all the edges of our graph and then put them back in one at a time. When our graph has no edges, and the number of components is so our claim holds: both sides of the equation above are zero. Then, each time we put in an edge, there are two choices: either it goes between two different components of the graph we have built so far, or it doesn’t. If goes between two different components, it increases by 1 and decreases the number of components by 1, so our equation continues to hold. If it doesn’t, neither of these quantities change, so our equation again continues to hold.   █

Examples

This reaction network is not weakly reversible, since we can get from and to but not back. It becomes weakly reversible if we throw in another transition:

Taking a reaction network and throwing in the reverse of an existing transition never changes the number of complexes, the number of components, or the dimension of the stoichiometric subspace. So, the deficiency of the reaction network remains unchanged. We computed it back in Part 17. For either reaction network above:

• the number of complexes is 5:

• the number of components is 2:

• the dimension of the stoichiometric subspace is 3. For this we go through each transition, see what change in species it causes, and take the vector space spanned by these changes. Then we find a basis for this space and count it:

so

As a result, the deficiency is zero:

Now let’s throw in another complex and some more transitions:

Now:

• the number of complexes increases by 1:

• the number of components is unchanged:

• the dimension of the stoichiometric subspace increases by 1. We never need to include reverses of transitions when computing this:

so

As a result, the deficiency is still zero:

Do all reaction networks have deficiency zero? That would be nice. Let’s try one more example:

• the number of complexes is the same as in our last example:

• the number of components is also the same:

• the dimension of the stoichiometric subspace is also the same:

so

So the deficiency is still zero:

It’s sure easy to find examples with deficiency zero!

Puzzle 1. Can you find an example where the deficiency is not zero?

Puzzle 2. If you can’t find an example, prove the deficiency is always zero. If you can, find 1) the smallest example and 2) the smallest example that actually arises in chemistry.

Note that not all reaction networks can actually arise in chemistry. For example, the transition $A \to A + A$ would violate conservation of mass. Nonetheless a reaction network like this might be useful in a very simple model of amoeba reproduction, one that doesn’t take limited resources into account.

Different kinds of graphs

I’ll conclude with some technical remarks that only a mathematician could love; feel free to skip them if you’re not in the mood. As you’ve already seen, it’s important that a reaction network can be drawn as a graph:

But there are many kinds of graph. What kind of graph is this, exactly?

As I mentioned in Part 17, it’s a directed multigraph, meaning that the edges have arrows on them, more than one edge can go from one vertex to another, and we can have any number of edges going from a vertex to itself. Not all those features are present in this example, but they’re certainly allowed by our definition of reaction network!

After all, we’ve seen that a stochastic reaction network amounts to a little diagram like this:

If we throw out the rate constants, we’re left with a reaction network. So, a reaction network is just a diagram like this:

If we throw out the information about how complexes are made of species, we’re left with an even smaller diagram:

And this precisely matches the slick definition of a directed multigraph: it’s a set of edges, a set of vertices, and functions saying where each edge starts and where it ends: its source and target.

Since this series is about networks, we should expect to run into many kinds of graphs. While their diversity is a bit annoying at first, we must learn to love it, at least if we’re mathematicians and want everything to be completely precise.

There are at least 2³ = 8 kinds of graphs, depending on our answers to three questions:

• Do the edges have arrows on them?

• Can more than one edge can go between a specific pair of vertices?

and

• Can an edge can go from a vertex to itself?

We get directed multigraphs if we answer YES, YES and YES to these questions. Since they allow all three features, directed multigraphs are very important. For example, a category is a directed multigraph equipped with some extra structure. Also, what mathematicians call a quiver is just another name for a directed multigraph.

We’ve met two other kinds of graph so far:

• In Part 15 and Part 16 we described circuits made of resistors—or, in other words, Dirichlet operators—using ‘simple graphs’. We get simple graphs when we answer NO, NO and NO to the three questions above. The slick definition of a simple graph is that it’s a set of vertices together with a subset of the collection of 2-element subsets of .

• In Part 20 we described Markov processes on finite sets—or, in other words, infinitesimal stochastic operators—using ‘directed graphs’. We get directed graphs when we answer YES, NO and YES to the three questions. The slick definition of a directed graph is that it’s a set of vertices together with a subset of the ordered pairs of : .

There is a lot to say about this business, but for now I’ll just note that you can use directed multigraphs with edges labelled by positive numbers to describe Markov processes, just as we used directed graphs. You don’t get anything more general, though! After all, if we have multiple edges from one vertex to another, we can replace them by a single edge as long as we add up their rate constants. And an edge from a vertex to itself has no effect at all.

In short, both for Markov processes and reaction networks, we can take ‘graph’ to mean either ‘directed graph’ or ‘directed multigraph’, as long as we make some minor adjustments. In what follows, we’ll use directed multigraphs for both Markov processes and reaction networks. It will work a bit better if we ever get around to explaining how category theory fits into this story.

Symmetry and the Fourth Dimension (Part 6)

Last time I showed what happened if you took a cube and chopped off its corners more and more until you reached its dual: the octahedron. Today let’s do the same thing starting with a dodecahedron!

Just as a cube has 3 squares meeting at each vertex, a dodecahedron has 3 pentagons meeting at each vertex. So, instead of the Coxeter diagram for the cube:

V—4—E—3—F

everything today will be based on the Coxeter diagram for the dodecahedron:

V—5—E—3—F

The number 5 is much cooler than the number 4, which is, frankly, a bit square. So the shapes we get today look much more sophisticated, at least to my eyes. But the underlying math is very similar: we can use diagrams to keep track of these shapes as we did before.

Dodecahedron: •—5—o—3—o

First we have the dodecahedron, with all pentagons as faces:

I like this shape so much I gave a lecture about it, and you can see the slides here:

• John Baez, Tales of the dodecahedron: from Pythagoras through Plato to Poincaré.

Truncated dodecahedron: •—5—•—3—o

Then we get the truncated dodecahedron, with decagons (10-sided shapes) and triangles as faces:

Icosidodecahedron: o—3—•—3—o

Then, halfway through, we get the aptly named icosidodecahedron, with pentagons and triangles as faces:

Like that other ‘halfway through’ shape the cuboctahedron, every edge of the icosidodecahedron lies on a great circle’s worth of edges.

Truncated icosahedron: o—5—•—3—•

Then we get the truncated icosahedron, with pentagons and hexagons as faces:

This one is so beautiful that a whole sport was developed in its honor!

Icosahedron: o—5—o—3—•

And then finally we get the icosahedron, with triangles as faces:

Again, I like this one so much I gave a talk about it:

• John Baez, Who discovered the icosahedron?

I almost feel like telling you all the stuff that’s in these talks of mine… and if I turn these blog posts into a book, I’ll definitely want to include it all! But there’s a lot of it, and I’m feeling a bit lazy—so why not just go check it out?

Puzzle. Why did Thomas Heath, the great scholar of Greek mathematics, think that Geminus of Rhodes is responsible for the remark in Euclid’s Elements crediting Theatetus with discovering the icosahedron?