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Stream: deprecated: chemistry

Topic: a puzzle from today's meeting

Sophie Libkind (May 03 2022 at 20:04):

Given a polynomial dynamical system, is it the law of mass action applied to some Petri net with rates?

Sam Tenka (May 03 2022 at 20:11):

My guess is yes (contrary to the claim! So my guess is probably wrong, but as an educational exercise I'll try to prove my guess tp see why...)

Sophie Libkind (May 03 2022 at 20:29):

@Sam Tenka, it would help me parse this if you used Latex! $1 + 1 = 2$

Sophie Libkind (May 03 2022 at 20:29):

This is a very cool feature of Zulip imo :smiley:

Sam Tenka (May 03 2022 at 20:30):

@Sophie Libkind oh ah cool! I'll rewrite in $\LaTeX$ shortly!

Sam Tenka (May 03 2022 at 20:52):

NOTE :
my original proof from tens of minutes ago had an (easy-to-fix) arithmetic error, but instead of fixing it in a minimal way, I chose to replace it by an essentially different and somewhat simpler construction.

SETUP:
Let $X$ be a finite set of formal variables. The latter generate the ring $R = \mathbb{R}[X]$. A formal diffeq is an $|X|$-tuplet $(f_x : x \in X) \in R^X$ of members of $R$; each formal diffeq determines an actual diffeq (first order with $|X|$ equations): $(D x = f_x(y : y\in X) : x\in X)$. Here, $D$ denotes differentiation-wrt-time.

Another structure we consider is that of a chemistry homework problem (CHP): a CHP is a member of $M(\mathbb{R}_{\geq 0} \times M(X) \times M(X))$. Here, $M:\textsf{Set}\to\textsf{Set}$ is the "finite multiset constructor". $M$ sends a set to the set of finite multisets whose every member is in that set. Though $M$ is a functor with great properties, we're introducing it merely to save ink. Of course, the terminal object in $\textsf{Set}$ induces for each $X$ a map $\text{weight}_X:M(X)\to M(1)=\mathbb{N}$. Using this and pullbacks, it's not hard to make the good ol' "tell me the multiplicity in a given multiset of an element" function $\text{count}_X:M(X)\times X \to \mathbb{N}$. Terminals and pullbacks are both limits, so it seems that counting is more "rightish/limity" than "leftish/colimity".

The point is that each CHP $c$ induces an formal diffEQ as follows: for each $x\in X$ we define $f_x = \sum_{(r,l_{in},l_{out}) \in c} (\prod_{y \in l_{in}} y) \cdot (- \text{count}_X(l_{in}, x) + \text{count}_X(l_{out}, x))$.

Sam's CLAIM:
every formal diffEQ arises as above from some CHP.

PROOF (with a gap discussed at end):
Let's just construct $c$ from $(f_x : x\in X)$.
Observing that the construction from CHPs to formal diffEQs sends disjoint unions to component-wise sums, it suffices to consider the case where one $f_x$ is a monomial and the other $f_y$'s are zero. Say that $f_x = s \cdot \prod_{y \in L} y$, where $s \in \mathbb{R}$ and $L \in M(X)$.

Well, if $s \in \mathbb{R}_{\geq 0}$ is non-negative, then we make the CHP $c = [(s, L, \sum_{y\in X} (\delta(y,x)+\text{count}_X(L,y)) \cdot [y])]$. Here, summing multisets just sums their multiplicities; multiplying a multiset by a natural number just multiplies all the multiplicities accordingly; and $\delta(y,x)$ is one on the diagonal and zero elsewhere. One verifies by staring hard that $c$ is a CHP and that it induces the desired formal diffEQ.

What if $s=-|s|$ is negative? Well, we make a CHP $c^\prime = [(|s|/\text{count}_X(L,x), L, [])]$. This gives us the right behavior for $D x$, but it decreases the concentrations of the other species too much; to fix this, we just apply the non-negative case to add more elements into our CHP to cancel out those decreases. One verifies by staring hard that making $c$ from $c^\prime$ in this way induces the desired formal diffEQ --- EXCEPT that in the process of staring hard, one finds that $c^\prime$ ain't always a well-defined CHP, since we might divide by zero!!!!!

Thus, QEND --- quod erat non demonstrandum.

REMARK:
the above proof works so long as the "diagonal counts" $(\text{count}_X(f_x, x) : x\in X)$ are all non-zero (i.e. positive integers). Call a formal diffEQ for which this holds, everywhere loopy. Then

PROVEN CLAIM:
every everywhere loopy formal diffEQ arises from some chemistry homework problem!

(converse untrue)

Sophie Libkind (May 03 2022 at 20:54):

Another useful Zulip tip: You can use the button that looks like :eye: to preview your message before you send it. :stuck_out_tongue:

Sam Tenka (May 03 2022 at 20:58):

The above considerations suggest this system:

$D x = -y, D y = 0$
or a cousin
$D x = -y, D y = -x$

Can some invariant argument can show that one of these cannot come from a petri net with nonnegative rates?
I'm not sure!

John Baez (May 03 2022 at 21:16):

Sam Tenka said:

The above considerations suggest this system:

$D x = -y, D y = 0$

or this cousin [...]

Can some invariant argument can show that one of these cannot come from a petri net with nonnegative rates?
I'm not sure!

This looks good. If a Petri net with rates gives the equation

$\frac{d}{dt} x = - y$

the law of mass action says there must be a transition (at least one) whose input is just one y. In theory such a transition could have any number of x's and y's as output, but no such transition can make the concentration of x decrease, as this equation does.

John Baez (May 03 2022 at 21:18):

This is not a rigorous proof but it's good enough for me right now.