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
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.
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:
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 :
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
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
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!