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This page is a blog article in progress, written by John Baez . To see discussions of this article while it is being written, visit go to the Azimuth Forum.Azimuth Forum.
The storm is brewing for the final climax, so let’s remember where we are. We start with a stochastic reaction network:
This consists of:
• finite sets of transitions $T$, complexes $K$ and species $S$,
• a map $r: T \to (0,\infty)$ giving a rate constant for each transition,
• source and target maps $s,t : T \to K$ saying where each transition starts and ends,
• a map $Y : K \to \mathbb{N}^S$ saying how each complex is made of species.
Then we extend $s, t$ and $Y$ to linear maps:
Then we put inner products on these vector spaces as described last time, which lets us ‘turn around’ the maps $s$ and $t$ by taking their adjoints:
More surprisingly, we can ‘turn around’ $Y$ and get a nonlinear map using ‘matrix exponentiation’:
This is most easily understood by thinking of $x$ as a row vector and $Y$ as a matrix:
So, we get these maps:
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
The 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$x \in \mathbb{R}^S$ with nonnegative components. The rate equation says:
We are looking for equilibrium solutions of the rate equation, where the number of things of each species doesn’t change with time at all:
In fact we will find complex balanced equilibrium solutions, where even the number of complexes of each kind don’t change:
More precisely, we have:
Deficiency Zero Theorem (Child’s Version). Suppose we have a reaction network obeying these two conditions:
It is weakly reversible, meaning that whenever there’s a transition from one complex $\kappa$ to another $\kappa'$, there’s a sequence of transitions going back from $\kappa'$ to $\kappa$.
It has deficiency zero, meaning $\mathrm{im} d \cap \mathrm{ker} Y = \{ 0 \}$.
Then for any choice of rate constants there exists an complex balanced equilibrium solution of the rate equation where all species are present in nonzero amounts. In other words, there exists $x \in (0,\infty)^S$ with$H x^Y = 0$.