Vertices of binding polyhedra as minimal representations

Chapter II: Polyhedral constraints enable holistic analysis of bioregulation

3.7 Vertices of binding polyhedra as minimal representations

From the previous section’s thorough analysis of one binding reaction, we observed that one central feature of the range of values that the reaction orders can take is that it forms a polyhedral set. The polyhedral set has vertices that correspond to constant reaction orders.

Before studying how the polyhedral shape arise in the next section, in this section we focus on just the vertices. We show that vertices in a species’ reaction order polyhedra correspond to minimal expressions of a species in terms of others. We then investigate how these minimal expressions can be found in a systematic way, using zonotopes from the theory of convex polytopes and oriented matroids (Chapter 6 and 7 of [123]).

Vertices and minimal representations

For reaction orders, we are interested in the log derivative of one species𝑥_𝑗^*considered to be catalytically active, for some𝑗^* ∈ {1, . . . , 𝑛}, with respect to the totals𝑡 ∈ R^𝑑_>0. In the one binding reaction example𝐸+𝑆 ⇌𝐶, if we take the active species𝑥_𝑗^* to be𝐶, then the reaction order of interest is ^𝜕log𝐶

𝜕log(𝑡𝑆,𝑡𝐸). From our investigation of this example, we see that vertices correspond to extreme cases where one term dominates the totals. The vertex with reaction order (1,1) corresponds to approximate expression𝐶 ≈ ^𝑡𝐸_𝐾^𝑡𝑆 is achieved when 𝐾 ≫ 𝐸, 𝑆 so that most of enzyme and substrate are in free form. In other words, free enzyme𝐸 dominates total enzyme so that𝑡𝐸 ≈𝐸, and free substrate dominates total substrate 𝑡_𝑆 ≈ 𝑆. In this case, we have approximate expression 𝐶 = ^𝐸𝑆_𝐾 ≈ ^𝑡𝐸_𝐾^𝑡𝑆. Then the log derivative of ^{𝑡𝐸𝑡𝑆}

𝐾 to(𝑡_𝑆, 𝑡_𝐸)is just their exponents, namely(1,1). Let us inspect another vertex. The vertex(1,0)corresponds to𝐶 ≈ 𝑡_𝑆 with condition𝑡_𝐸 ≫ 𝑡_𝑆, 𝐾. This is a scenario where enzymes are overabundant so most substrates are bound. So bound substrate𝐶dominates𝑡_𝑆, i.e. 𝑡_𝑆 ≈𝐶. Then𝐶 ≈𝑡_𝑆, and of course the log derivative of𝑡_𝑆 to(𝑡_𝑆, 𝑡_𝐸)is(1,0).

This provides a reasoning for what the vertices are. The totals𝑡=𝐿𝑥come from summing the concentration of several species in 𝑥. When some of the totals𝑡 are dominated by one of the species in the sum and the active species𝑥_𝑗^*can be expressed as a monomial in these species, then the reaction orders are just the exponents of this monomial. Let us write down exactly what this means. We assume𝑥^*_𝑗 can be written as a monomial in terms of variables in 𝑥and 𝑘. This corresponds to 𝑥_𝑗^* = 𝑥^𝑎𝑘^𝑏 for some vector 𝑎 ∈ R^𝑛, 𝑏∈R^𝑟. Here the notation𝑥^𝑎means𝑥^𝑎 =𝑥^𝑎₁¹. . . 𝑥^𝑎_𝑛^𝑛 =𝑒^{𝑎⊺log𝑥}. Since the totals𝑡has just𝑑 variables, if the variables in this monomial expression are all from dominant species of each total, then there can be at most𝑑different variables of𝑥involved. This corresponds to the exponent vector𝑎has at most𝑑nonzero entries, i.e. its support size is no larger than 𝑑. Let𝒥 ⊂ {1, . . . , 𝑛}denote the support of𝑎, i.e. 𝒥 ={𝑗 :𝑎_𝑗 ̸= 0}. The support of𝑎has no more than𝑑nonzero entries means|𝒥 | ≤𝑑. Then if there exists𝑡_𝑖𝑗 for each𝑗 ∈ 𝒥 such that𝑡_𝑖𝑗 ≈𝑥_𝑗, then𝑥_𝑗^* =𝑥^𝑎𝑘^𝑏 =𝑘^𝑏∏︀_𝑗∈𝒥 𝑥^𝑎_𝑗^𝑗 ≈𝑘^𝑏∏︀_𝑗∈𝒥(𝑡_𝑖𝑗)^𝑎𝑗. So the log derivative of𝑥_𝑗^* with respect to(𝑡,𝑘)is just(𝑎,𝑏).

From this observation, we see finding vertices comes down to finding monomial expressions of𝑥_𝑗^* in terms of𝑥and𝑘of the form𝑥_𝑗^* =𝑥^𝑎𝑘^𝑏, such that𝑎has no more than𝑑nonzero entries. Taking log, the monomial expression becomeslog𝑥_𝑗^* = 𝑎^⊺log𝑥+𝑏^⊺log𝑘. The space of such expressions for a binding network is naturally given by the steady state equations defining the equilibrium manifold𝑁log𝑥= log𝑘(see Eq (3.12)). The condition on the support of𝑎motivates us to find vectors with minimal support yield a monomial representation. The desire for minimal support is also motivated by the mathematical reason that all other monomials expressions can be written as linear sums of minimal support ones (see Lemma 6.7 in [123] for example). Therefore, we would like to find the minimal representations of𝑥_𝑗^* in terms of variables𝑥for a given binding network. This is what we study next.

Minimal representations in binding networks

To begin with, we define precisely what we mean by minimal. In words, a vector𝑣∈𝑉 ⊂R^𝑛 is minimal in a set of vectors𝑉 if there is no nonzero vector in𝑉 with smaller support vectors. Support vector is defined as𝑢:= supp𝑣has𝑢_𝑗 = 1if𝑣_𝑗 ̸= 0, and𝑢_𝑗 = 0if𝑣_𝑗 = 0. The partial order on support vectors is the canonical one: 0<1applied component-wise.

For convenience, we slightly abuse notation and usesupp𝑣to also denote the set of support indices𝒥 = {𝑗 = 1, . . . , 𝑛:𝑣_𝑗 ̸= 0}. This is identifying the space of subsets of{1, . . . , 𝑛}

with space{0,1}^𝑛.

Definition 3.7.1. Given set of vectors𝑉 ⊂R^𝑛. 𝑣^* ∈𝑉 is avector of minimal supportin𝑉

if it is nonzero and there does not exist nonzero vector𝑣 ∈𝑉, such thatsupp𝑣 <supp𝑣^*. In other words, for any nonzero𝑣 ∈𝑉, eithersupp𝑣^* ≤supp𝑣or they are not comparable.

We denote the set of minimal vectors of a set𝑉 byMINSUPP(𝑉).

We note that if𝑉 is a vector space, then 𝑣^* has minimal support in 𝑉 implies 𝛼𝑣^* has minimal support in𝑉 for all nonzero scalar𝛼, since they all have the same support.

Now we consider the minimal representations of𝑥_𝑗^* in terms of variables𝑥for detailed balanced solutions of an isomer-atomic network, so 𝑥 is in equilibrium manifold (see Eq (3.12)). A (monomial) representation of 𝑥_𝑗^* in terms of 𝑥 and 𝑘 can be written as log𝑥_𝑗^* =𝑎^⊺log𝑥+𝑏^⊺log𝑘for some vector𝑎∈R^𝑛and𝑏∈R^𝑟. This can be re-written as

(𝑒_𝑗^*−𝑎)^⊺log𝑥=𝑏^⊺log𝑘.

Since 𝑥and𝑘are in equilibrium manifoldℳ, it satisfies𝑁log𝑥 = log𝑘. So the above condition can be satisfied for some𝑏∈R^𝑟if and only if(𝑒_𝑗^*−𝑎)∈rowspan𝑁. Indeed, this condition implies there exists some𝑏 ∈R^𝑟such that(𝑒_𝑗^*−𝑎) = 𝑏^⊺𝑁, so we can calculate that(𝑒_𝑗^* −𝑎)^⊺log𝑥=𝑏^⊺𝑁log𝑥=𝑏^⊺log𝑘. In terms of set equivalence,

{representations of𝑥_𝑗^* in𝑥}

={𝑎∈R^𝑛: there exists𝑏 ∈R^𝑟, such that(𝑒_𝑗^*−𝑎)^⊺log𝑥=𝑏^⊺log𝑘}

={𝑎: (𝑒𝑗^* −𝑎)∈rowspan𝑁}=𝑒𝑗^*−rowspan𝑁 =𝑒𝑗^*+𝒮.

The last step we used thatrowspan𝑁 =𝒮 is a linear subspace, so−𝒮 =𝒮. The minimal representations of𝑥_𝑗^* in terms of all species then comes down to finding the vectors of minimal support in the affine subspace𝑒_𝑗^*+𝒮. In other words, we define

{minimal representations of𝑥_𝑗^* in𝑥}:= MINSUPP(𝑒_𝑗^*+𝒮). (3.46) We would like to express minimal support vectors on the affine subspace𝑒_𝑗^*+𝒮in terms of minimal support vectors of linear subspaces. This requires taking𝑒_𝑗^* and𝒮out of the minimal support operator. We do so below by inspecting how the minimal support vectors in𝑒𝑗^*+𝒮relate to those in𝒮.

Proposition 3.7.2.

MINSUPP(𝑒_𝑗^*+𝒮) =𝑒_𝑗^*+{0} ∪MINSUPP_𝑗^*(𝒮), (3.47) where we defineMINSUPP_𝑗^*(𝒮) := {𝑢∈MINSUPP(𝒮) :𝑢𝑗^* =−1}.

Proof. We first note that𝑒_𝑗^* has minimal support in this affine subspace. This is simply expressing𝑥_𝑗^* as itself, corresponding to the zero vector0in𝒮.

Now consider any vector of minimal support𝑣 ∈𝑒_𝑗^*+𝒮that is not𝑎𝑒_𝑗^* for some nonzero constant𝑎. 𝑣 should have𝑣_𝑗^* = 0. This is because if𝑣_𝑗^* ̸= 0, then we can always divide out 𝑥_𝑗^* to get a vector with smaller support, contradicting that 𝑣 is minimal. Seen in another way, if𝑣𝑗^* ̸= 0, thensupp𝑒𝑗^* <supp𝑣. Therefore, 𝑣 satisfies𝑣 = 𝑒𝑗^* +𝑢,𝑢 ∈ 𝒮 and𝑢_𝑗^* =−1. In other words, we have shown

MINSUPP(𝑒_𝑗^* +𝒮)⊂ {𝑒_𝑗^*}∪(𝑒˙ _𝑗^*+{𝑢 ∈ 𝒮 :𝑢_𝑗^* =−1}),

where∪˙ denote disjoint union, or union of disjoint sets. If this𝑢with𝑢_𝑗^* =−1has minimal support in𝒮, then the corresponding𝑣=𝑒_𝑗^*+𝑢has minimal support in𝑒_𝑗^*+𝒮. This is becausesupp𝑣 = supp𝑢∖{𝑗^*}.

We then show that this is also necessary. If𝑢is not of minimal support in𝒮, then there exists nonzero vector𝑢^′ ∈ 𝒮 with smaller support than𝑣. Consider the case𝑢^′_𝑗* = 0. Since 𝑢^′ is nonzero, there exists another𝑗^′ ̸= 0so that𝑢^′_𝑗′ ̸= 0. Since𝑢^′has smaller support than 𝑢,𝑢_𝑗^′ ̸= 0as well. Now consider

𝑢^′′ :=𝑢−𝑢_𝑗^′ 𝑢^′_𝑗′

𝑢^′.

This is linear combination of𝑢and𝑢^′, so𝑢^′′ ∈ 𝒮. By construction,𝑢^′′_𝑗′ = 0, sosupp𝑢^′′ ⊂ supp𝑢∖{𝑗^′}. Also, since𝑢^′_𝑗^* = 0, we have𝑢^′′_𝑗^* =𝑢_𝑗^* = −1. Therefore,𝑣^′′ :=𝑒_𝑗^*+𝑢^′′has smaller support than𝑣.

Now consider the case 𝑢^′_𝑗* ̸= 0. Since 𝑢^′ has smaller support, there exists 𝑗 so that 𝑢^′_𝑗 = 0 while 𝑢_𝑗 ̸= 0. Define 𝑢^′′ = −(𝑢^′_𝑗*)⁻¹𝑢^′ ∈ 𝒮. Then 𝑢^′′_𝑗* = −1, while 𝑢^′′_𝑗 = 0. So supp𝑢^′′ ⊂supp𝑢∖{𝑗}, and𝑣^′′ :=𝑢^′′+𝑒_𝑗^*has smaller support than𝑣.

With the above result, we have characterized the vertices of reaction order polyhedra in terms of minimal vectors of linear subspace. When the binding network gets large, we would like a way to compute these vertices in a systematic fashion. Below, we use the tool of zonotopes to compute the minimal vectors of linear subspaces, and therefore vertices of reaction order polyhedra.

Computation of minimal representations through zonotopes

Vectors of minimal support in𝒮 can be explicitly computed through zonotopes. For this, we utilize the theory of polytopes and oriented matroids (see Chapter 6 and 7 of [123]).

Some of the basic tools below may be well known in the fields of polytopes, oriented matroids, and matroid theory. But for completeness, exposition, and accessibility to readers not familiar with those fields, I provide a streamlined development of necessary tools nonetheless. To begin with, instead of the partial order on vectors’ support, we consider the partial order on sign vectors, where0<+and0<−is applied component-wise. Note that smaller in sign implies smaller in support, but not vice versa in general. For example, (++)and(+-)are not comparable in the sign order, but their support are both(11), and therefore equal in support order.

Since we consider vectors in a linear subspace, the correspondence between vectors of minimal sign and vectors of minimal support is simple.

Lemma 3.7.3. For a vector𝑣in linear subspace𝒱 ⊂R^𝑛, it is of minimal sign iff it is of minimal support.

Proof. ( ⇐=). Contrapositive is not minimal sign implies not minimal support. This is obvious because smaller in sign order implies smaller in support order.

(=⇒). Contrapositive is not minimal support implies not minimal sign. 𝑣is not minimal support implies there exists𝑢∈ 𝒱 so thatsupp𝑢 <supp𝑣, so there exists𝑗₀,𝑢_𝑗0 = 0while 𝑣_𝑗0 ̸= 0. Let u and v denote𝑢 and𝑣’s sign vectors. Let 𝑆(u,v) = {𝑗 :u_𝑗 =−v_𝑗 ̸= 0}be the separation set of u and v. If 𝑆(u,v) = ∅, then u < v. If𝑆(u,v) ̸= ∅, let𝑞 =|𝑆(u,v)|. Take𝑗₁ ∈𝑆(u,v). We can eliminate𝑗₁ between u and v to obtain w¹ ∈sgn𝒱. By definition, w¹

𝑗1 = 0, and w¹

𝑗0 =v_𝑗

0. Since u has smaller support ahtn v, we also havesuppw¹ <suppv, but now|𝑆(w¹,v)| ≤ 𝑞−1, since they have the same sign at𝑗1 now. If|𝑆(w¹,v)| >0, we take its element𝑗₂and perform the same process. Then after𝑞^′steps,𝑞^′ ≤𝑞, we reach a sign vector w^𝑞

′ ∈sgn𝒱 withsuppw^𝑞

′ <suppv while𝑆(w^𝑞

′,v) = ∅. This implies w^𝑞

′ <v.

In fact, the correspondence between vectors of minimal support and minimal sign in a linear subspace goes even further to have a natural bijection between pairs of minimal sign vectors{v,−v}and their support vectorsuppv.

Lemma 3.7.4. For a linear subspace𝒱 ⊂ R^𝑛, consider the support map on sign vectors supp :{+,−,0}^𝑛→ {0,1}^𝑛.

The pre-image of any minimal support vector insupp𝒱 ⊂ {0,1}^𝑛through the support map is a pair{v,−v}for some minimal sign vector v insgn𝒱 ⊂ {+,−,0}^𝑛.

Proof. Let supp𝑣 denote a minimal support vector in supp𝒱. Let v = sgn𝑣, and then suppv= supp(−v) = supp𝑣. Now we are left to show there are no other sign vectors in sgn𝒱 with the same support. Let u be any sign vector insgn𝒱 with the same support, i.e. suppu = suppv. Because they have the same support, u ∈ {v,−v}is equivalent to 𝑆(u,v) = ∅or𝑆(u,−v) = ∅. Assume not, then there exists𝑗₊ ∈𝑆(u,v)and𝑗− ∈𝑆(u,−v), 𝑗₊ ̸=𝑗−. This implies u_𝑗

+ =−v_𝑗

+ ̸= 0, while u_𝑗

− =v_𝑗

− ̸= 0. Then we can form w ∈sgn𝒱 that eliminates 𝑗₊ between u and v. This meanssuppw ≤ suppu = suppv, while at the same time, w_𝑗

+ = 0, so it has strictly smaller support. At the same time, it is not zero, because w_𝑗

− =u_𝑗

− =v_𝑗

− ̸= 0. This contradicts that u and v are of minimal support.

Now we show how to explicitly construct vectors with minimal sign in the stoichiometry subspace 𝒮 = rowspan𝑁. Consider 𝑁 ∈ R^𝑟×𝑛 as a vector configuration of 𝑛 column vectors{𝑦₁, . . . ,𝑦_𝑛}inR^𝑟. Then the (centered) zonotope of𝑁 can be equivalently defined in the following ways: as combinations of vectors, Minkowski sum of line segments, or linear image of a cube.

𝑍(𝑁) :=

{︃

𝑧 ∈R^𝑟 :𝑧=

𝑛

∑︁

𝑖=1

𝜆𝑖𝑦𝑖,−1≤𝜆𝑖 ≤1

}︃

=⊕^𝑛_𝑖=1[−𝑦𝑖,𝑦𝑖] =𝑁𝐶𝑛,

where⊕denote Minkowski sum,[−𝑦_𝑖,𝑦_𝑖]denote the line segment between−𝑦_𝑖 and𝑦_𝑖, and𝐶_𝑛denote the𝑛-cube.

As shown in Corollary 7.17 of [123], there is a bijection between the facets of𝑍(𝑁)and the minimal sign vectors (called cocircuits) in rowspan𝑁 (also see end of Section 3.3).

Explicitly, any vector𝑢inrowspan𝑁 has unique coefficient vector𝑐∈R^𝑟so that𝑁^⊺𝑐=𝑢. We associate with𝑐, therefore𝑢, a face of𝑍(𝑁)defined by

𝑍(𝑁)^𝑐:=

{︃

𝑧 ∈𝑍(𝑁) :𝑐^⊺𝑧 = max

𝑧^′∈𝑍(𝑁)𝑐^⊺𝑧^′

}︃

.

The vector𝑢has a minimal sign vector, therefore minimal support, if and only if the face 𝑍(𝑁)^𝑐is a facet of𝑍(𝑁). This comes from the bijection between the face lattice of𝑍(𝑁) with the partial order of set inclusion, and the sign vectors ofrowspan𝑁 with the partial order on sign vectors. See Chapter 7.3 of [123] for details.

We can then obtain the minimal vectors as the vertices of the polar dual of the zonotope:

vert^{︁𝑍(𝑁)^Δ}︁, where Δ denote polar dual, because polar dual maps facets of𝑍(𝑁)to vertices of𝑍(𝑁)^Δ. Indeed, the polar dual of the face𝑍(𝑁)^𝑐is defined as

(𝑍(𝑁)^𝑐)^◇ :={𝑦∈R^𝑟:𝑦^⊺𝑧 ≤1,∀𝑧 ∈𝑍(𝑁); and𝑦^⊺𝑧= 1,∀𝑧∈𝑍(𝑁)^𝑐}.

We see that

𝑎⁻¹𝑐∈(𝑍(𝑁)^𝑐)^◇, 𝑎:= max

𝑧^′∈𝑍(𝑁)𝑐^⊺𝑧^′.

For the case where𝑍(𝑁)^𝑐is a facet, its polar dual(𝑍(𝑁)^𝑐)^◇ is a vertex. So it is a singleton set{𝑎⁻¹𝑐}. Since the vector𝑎⁻¹𝑐is proportional to the coefficient vector of the minimal vector𝑢 ∈rowspan𝑁, we can explicitly calculate the minimals vectors by

MINSUPP(𝒮) = MINSUPP(rowspan𝑁) = 𝑁^⊺vert^{︁𝑍(𝑁)^Δ}︁. (3.48)

Now we combine this result with (3.47) to find minimal representations of𝑥_𝑗^* in terms of 𝑥:

MINSUPP(𝑒_𝑗^*+𝒮) =𝑒_𝑗^*+{0} ∪MINSUPP_𝑗^*(𝒮)

=𝑒_𝑗^*+{0} ∪^{︁𝑢:𝑢∈𝑁^⊺vert^{︁𝑍(𝑁)^Δ}︁, 𝑢𝑗^* =−1^}︁.

(3.49)

The above constitute the proof of the following theorem.

Theorem 3.7.5. The minimal representations of 𝑥_𝑗^* in terms of𝑥 in the equilibrium manifold ℳof a binding network with transpose-reduce stoichiometry matrix𝑁 ∈R^𝑑×𝑛can be written as 𝑥_𝑗^* =𝑥^𝑎𝑘^𝑏, where𝑎is contained in the set defined in(3.49), and for a given𝑎vector,𝑏∈R^𝑟can be computed from(𝑒_𝑗^*−𝑎) =𝑁^⊺𝑏.

With this result, we can compute minimal representations of a given species simply by computing the vertices of a zonotope constructed from the transpose-reduced stoichiometry matrix𝑁. This can be implemented via linear programming for example, or using off-the- shelf packages such as SageMath [94] and the multi-parameteric toolbox [61]. In terms of computational complexity, we are asking for the vertices of a polytope given its facets, which has a computational complexity that is exponential in the number of facets (see page 80 of [20] and [14]). Since the polytope of concern here is the zonotope of the stoichiometry matrix𝑁, the number of facets is roughly𝑛the number of species, therefore the complexity is exponential in𝑛. This is intractable when𝑛is large, but is a significant improvement over numerically scanning all solutions of polynomial systems of equations. A rough generic computational complexity for solving polynomial systems of equations is𝑂(degree^num.eqn.), wheredegreeis the degree of each equation andnum.eqn.is the number of equations [62].

If there are𝑟binding reactions, each corresponding to a degree2polynomial equation, then this is𝑂(2^𝑟). It is often the case that𝑟scales proportionally with𝑛, so each numerical solution has exponential in 𝑛 complexity. Let 𝑁 denote the number of points to scan in the solution space, which satisfy 𝑁 ≫ 2^𝑛 always so as to obtain the full regulatory

profile numerically. Then obtaining the full regulatory profile via numerical solutions of polynomial equations has complexity 𝑂(𝑁2^𝑛). This is costly compared to 𝑂(𝑒^𝑛) of solving for the vertices of the reaction order polyhedra directly using zonotopes. Recall that inverting a matrix of𝑛 dimensions has a cost of roughly𝑂(𝑛^2.3). Then numerically sampling the reaction order polyhedra using the log derivative formula incur a cost of 𝑂(𝑁 𝑛^2.3). This also improves over numerically solving the polynomial equations.

Minimal representations yield vertices

Recall that for a given (monomial) representation of𝑥_𝑗^*in terms of𝑥of the form𝑥_𝑗^* =𝑥^𝑎𝑘^𝑏 to become a vertex, we also need𝑎to have no more than𝑑nonzero entries. Is this satisfied for all minimal representations we defined in previous subsections? The answer is yes by the following lemma for isomer-atomic binding networks.

Lemma 3.7.6. Given an isomer-atomic binding network with𝑑atomic species. Vectors of minimal support in𝒮 have at most𝑑+ 1nonzero entries. The minimal representations of𝑥_𝑗^* in terms of𝑥 have at most𝑑nonzero entries.

Proof. Argument via stoichiometrically atomic and then dimensionality counting. The second statement is implied by the first statement, because if a vector𝑢inMINSUPP_𝑗^*(𝒮) has𝑞nonzero entries, then𝑒_𝑗^* +𝑢has𝑞−1nonzero entries, since𝑢_𝑗^* =−1. So suffice to showMINSUPP(𝒮)has at most𝑑+ 1nonzero entries.

𝑣 ∈ 𝒮if and only if𝐿𝑣= 0, i.e. 𝑣¹+𝐿₂𝑣² = 0, where we split vector𝑣 ∈R^𝑛into its first𝑑 entries𝑣¹ ∈R^𝑑and its last𝑟entries𝑣² ∈R^𝑟. So we have bijection betweenR^𝑟 and𝒮, which maps𝑐∈R^𝑟to𝑣 ∈ 𝒮 defined by𝑣¹ =−𝐿₂𝑐and𝑣² =𝑐. In other words,

𝒮 =

⎧

⎨

⎩

⎡

⎣

−𝐿₂𝑐 𝑐

⎤

⎦:𝑐∈R^𝑟

⎫

⎬

⎭

.

We specify the zero pattern of a vector𝑣 ∈ 𝒮by indices of zeros in𝑣¹,ℐ ={𝑖= 1, . . . , 𝑑:𝑣_𝑖¹ = 0}, and indices of nonzeros in 𝑣², 𝒥 = ^{︁𝑗 = 1, . . . , 𝑟 :𝑣_𝑗² ̸= 0^}︁. Enforcing 𝑣¹_𝑖 = 0for 𝑖 ∈ ℐ then corresponds to 𝑐 satisfying the system of linear equations 0 = 𝐿^ℐ₂𝑐, where 𝐿^ℐ₂ is the submatrix withℐ rows from𝐿₂. Since𝑣² =𝑐, enforcing the zero pattern𝑣_𝑗² = 0for 𝑗 /∈ 𝒥 then corresponds to restricting𝑐to subspaces, so that the linear system of equations becomes0=𝐿^ℐ_2,𝒥𝑐𝒥, where𝐿^ℐ_2,𝒥 denote the submatrix of𝐿₂ withℐ rows and𝒥 columns, and𝑐𝒥 denote the subvector of𝑐with𝒥 entries. This problem has unique solution (𝑐𝒥 =0) if𝐿^ℐ_2,𝒥 is full column rank, i.e. |𝒥 |. It has infinitely many solutions, and therefore a nonzero solution, if𝐿^ℐ_2,𝒥 is not full column rank.

The above can be summarized into the following statement. There exists nonzero𝑣 ∈ 𝒮 such that𝑣¹_𝑖 = 0for𝑖∈ ℐ and𝑣_𝑗² = 0for𝑗 /∈ 𝒥 if and only if𝐿^ℐ_2,𝒥 is not full column rank.

Note that for anyℐ,𝒥 such that|𝒥 | =|ℐ|+ 1, we have more columns than rows, which guarantees𝐿^ℐ_2,𝒥 is not full column rank, so there exists a nonzero vector𝑣 ∈ 𝒮 with the zero patterns specified byℐand𝒥. Hence, vector𝑣 ∈ 𝒮 has minimal support implies itsℐ and𝒥 satisfy|𝒥 | ≤ |ℐ|+ 1.

For any vector𝑣∈ 𝒮 with zero pattern described byℐ and𝒥, the number of zeros in𝑣¹ is|ℐ|, and the number of zeros in𝑣² =𝑐is𝑟− |𝒥 |. Together, the number of zeros of𝑣is 𝑟+|ℐ| − |𝒥 |. The number of nonzeros of𝑣 isnnz_𝑣 =𝑑+|𝒥 | − |ℐ|, by𝑛=𝑟+𝑑. If𝑣has minimal support, then|𝒥 | ≤ |ℐ|+ 1, sonnz_𝑣 ≤𝑑+ 1.

Note that𝑣is minimal also implies|𝒥 |=𝑞+ 1, where𝑞is rank of𝐿^ℐ_2,𝒥. Because if not then 𝑣is nonzero, which implies|𝒥 | ≥𝑞+ 1, so|𝒥 | ≥𝑞+ 2. Eliminating a linearly dependent column yields𝒥^′, which together withℐcorresponds to a vector𝑣^′that has smaller support than𝑣.

Examples

We illustrate our way of finding vertices developed in this section via one example.

Example 6(two paths).

𝐴+𝐵 ⇌𝐶_AB, 𝐴+𝐶 ⇌𝐶_AC, 𝐶_AB+𝐶_AC ⇌𝐶_2ABC (3.50) Here𝐴participates in the formation of𝐶_2ABCthrough two paths, one through𝐶_AB, and another through𝐶_AC.

This network is stoichiometrically atomic. With the following species order, we have the stoichiometry and conservation law matrices as the following:

𝑥= (𝐴, 𝐵, 𝐶, 𝐶_AB, 𝐶_AC, 𝐶_2ABC),

⎡

⎣

𝐿 𝑁

⎤

⎦=

⎡

⎢

⎢

⎢

⎢

⎢

⎢

⎢

⎢

⎢

⎢

⎢

⎢

⎣

1 0 0 1 1 2 0 1 0 1 0 1 0 0 1 0 1 1

−1 −1 0 1 0 0

−1 0 −1 0 1 0 0 0 0 −1 −1 1

⎤

⎥

⎥

⎥

⎥

⎥

⎥

⎥

⎥

⎥

⎥

⎥

⎥

⎦

. (3.51)

Consider species𝐴for example. By inspection, we have𝐴∝𝐴,^𝐶_𝐵^AB,^𝐶_𝐶^AC,^(︁𝐶2ABC_𝐵𝐶 ^)︁1/2, ^𝐶2ABC

𝐶AB𝐶

and ^𝐶2ABC

𝐶AC𝐵. We can find these minimal expressions from zonotope computation. The following is a implementation of the zonotope computation using SageMath in python.

import numpy as np from sage.all import *

n_mat=np.array([[-1,-1,0,1,0,0], [-1,0,-1,0,1,0], [0,0,0,-1,-1,1]]) d,n=n_mat.shape

P1 = polytopes.parallelotope(n_mat.T) P1dual = P1.polar()

cs=np.array(P1dual.Vrepresentation()) vs=cs.dot(n_mat)

idx=0 # index of A ei=np.zeros(n) ei[idx]=1

vs_idx=[i for i in range(len(vs)) if vs[i,idx]>0]

[ei-vs[i]/vs[i,idx] for i in vs_idx]

This code outputs the following vectors.

[array([ 0. , -0.5, -0.5, 0. , 0. , 0.5]), array([ 0., -1., 0., 0., -1., 1.]),

array([ 0., -1., 0., 1., 0., 0.]), array([ 0., 0., -1., -1., 0., 1.]), array([ 0., 0., -1., 0., 1., 0.])]

We can check that indeed these correspond to the minimal representations we obtained by

inspection. △

Vertices for the matrix of reaction orders

As we see in the analysis for one binding reaction in Section3.6, especially Eq (3.42), that instead of studying the reaction order polyhedra of one species of interest for ^{𝜕log𝑥𝑗}

*

𝜕log(𝑡,𝑘), we can also obtain the reaction order polyhedra on the matrix for all species together for

𝜕log𝑥

𝜕log(𝑡,𝑘). We provide a preliminary investigation into this using a similar approach as before.

This problem is worth further investigation.

Similar to the case of reaction order vector of one species, a vertex happen for the reaction order matrix when𝑑species of𝑥dominates the totals𝑡, and there is a monomial relation of𝑥in terms of the dominant𝑥and𝑘. We can represent the condition that𝑑species of𝑥 are dominant in𝑡by the expression

𝑃_𝑡log𝑥≈log𝑡, (3.52)

where𝑃_𝑡 ∈R^𝑑×𝑛is a projection matrix, with the𝑖th row a basis vector𝑒_𝑗𝑖 ∈R^𝑛for some 𝑗_𝑖 ∈ {1, . . . , 𝑛} and 𝑖 = 1, . . . , 𝑑. The condition for the monomial relation then can be written as

log𝑥=𝐴^⊺𝑃_𝑡log𝑥+𝐵^⊺log𝑘, (3.53) where𝐴∈R^𝑑×𝑛, and𝐵∈R^𝑟×𝑛. We see that with these two conditions, we have

𝜕log𝑥

𝜕log(𝑡,𝑘) = 𝜕log𝐴^⊺𝑃_𝑡log𝑥+𝐵log𝑘

𝜕log(𝑡,𝑘) ≈ 𝜕𝐴^⊺log𝑡+𝐵^⊺log𝑘

𝜕(log𝑡,log𝑘) =^[︁𝐴^⊺ 𝐵^⊺]︁. So if these two conditions are satisfied, then we obtain a constant log derivative.

So to find vertices for reaction order matrix comes down to restricting what𝐴and𝑃_𝑡can be. We note that Eq (3.7) can be written as

(I_𝑛−𝐴^⊺𝑃𝑡) log𝑥=𝐵^⊺log𝑘=𝐵^⊺𝑁log𝑥,

where the last step usedlog𝑘=𝑁log𝑥from conditions of equilibrium manifold. So we have

I−𝐴^⊺𝑃_𝑡 =𝐵^⊺𝑁.

If we denote𝑉 =𝐴^⊺𝑃𝑡, then there exists some𝐵 ∈R^𝑟×𝑛such that𝑉 =𝐵^⊺𝑁 corresponds to each row ofI−𝑉 is inrowspan𝑁 =𝒮. In other words,

{︁𝑉 ∈R^𝑛×𝑛: there exists𝐵 ∈R^𝑟×𝑛such thatI−𝑉 =𝐵^⊺𝑁^}︁

=^{︁𝑉 ∈R^𝑛×𝑛:𝑣_𝑗 ∈𝑒_𝑗+𝒮, 𝑗 = 1, . . . , 𝑛^}︁, where𝑣_𝑗 is the𝑗th row of𝑉.

Now, what is the restriction on𝑉 from the fact that𝑉 actually𝑉 =𝐴^⊺𝑃_𝑡for some choice of𝐴and projection matrix𝑃_𝑡^⊺. Since𝑃_𝑡is a projection matrix with𝑖th row taking value𝑒_𝑗𝑖, we have𝐴^⊺𝑃_𝑡is just taking the row vectors of𝐴and place them as column vectors, namely the𝑖th row vector of𝐴denoted𝑎𝑖 is mapped to the 𝑗𝑖th column of𝐴^⊺𝑃𝑡. So there are