Total binomial decomposition (TBD) Thomas Kahle Otto-von-Guericke Universität Magdeburg

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1 Total binomial decomposition (TBD) Thomas Kahle Otto-von-Guericke Universität Magdeburg

2 Setup Let k be a field. For computations we use k = Q. k[p] := k[p 1,..., p n ] the polynomial ring in n indeterminates For each u N n there is a monomial p u = n i=j pu j j. For u, v N n, λ k there is a binomial p u λp v. Definition A binomial ideal I k[p 1,..., p n ] is an ideal that can be generated by binomials.

3 Binomial ideals Monomial ideals have boring varieties Binomial ideals: tractable and flexible For many purposes a trinomial ideal is a general ideal.

4 Binomial prime ideals can be characterized. Up to scaling p j they are: Definition Let A Z d n. The toric ideal for A is the prime ideal I A := p u p v : u, v N n, u v ker A

5 Binomial prime ideals can be characterized. Up to scaling p j they are: Definition Let A Z d n. The toric ideal for A is the prime ideal I A := p u p v : u, v N n, u v ker A Primary ideals can be characterized too, but depends on char(k).

6 Monomial maps Let k[t ± ] = k[t ± 1,..., t± d ]. Consider the k-algebra homomorphism φ A : k[p] k[t ± ], p j t A j = t A 1j 1 t A dj d where A j is the j-th column of A.

7 Monomial maps Let k[t ± ] = k[t ± 1,..., t± d ]. Consider the k-algebra homomorphism φ A : k[p] k[t ± ], p j t A j = t A 1j 1 t A dj d where A j is the j-th column of A. Claim I A = ker φ A. : p u?? : Exercise 1

8 Monomial maps Let k[t ± ] = k[t ± 1,..., t± d ]. Consider the k-algebra homomorphism φ A : k[p] k[t ± ], p j t A j = t A 1j 1 t A dj d where A j is the j-th column of A. Claim I A = ker φ A. : p u?? : Exercise 1 This proves that I A is prime The toric variety V (I A ) has a monomial parametrization.

9 Toric ideals in application: Log-linear models One discrete random variable with values in [n]. A distribution is an element of the probability simplex n 1 = {p R n : p j 0, j p j = 1}. A model is a subset M n 1.

10 Log-linear models A log-linear model is specified by linear constraints on logs of p j log p = Mθ, θ R d. for a fixed model matrix M R n d.

11 Log-linear models A log-linear model is specified by linear constraints on logs of p j log p = Mθ, θ R d. for a fixed model matrix M R n d. Let s write M = A T and assume A Z d n. Then log p j = θa j where A j is the j-th column of A.

12 The log-linear constraint encodes a monomial parametrization: log p j = θa j p j = e θa j p j = t A j if we put t j = e θ j and let t j > 0, j = 1,..., d be the parameters.

13 Observation Each log-linear model is the intersection of a toric variety with n 1. The independence model = P 1 P 1

14 Some consequences Testing if a given distribution is in the model is checking binomial equations. Nearest point methods, Kullback Leibler geometry Binomial equations can have meaning in terms of (conditional) independence Graphical models. The boundary of a log-linear model looks like the boundary of the polytope conv{a i, i = 1,..., n} Existence of the MLE.

15 Computational problems Given A, how to find a finite generating set of I A? Let B ker Z A be a lattice basis. Decompose b = b + b with b ± i = max{±b i, 0} Then p b+ p b I A.

16 Computational problems Given A, how to find a finite generating set of I A? Let B ker Z A be a lattice basis. Decompose b = b + b with b ± i = max{±b i, 0} Then p b+ p b I A. Equality does not hold, but p b+ p b : j p j = I A

17 Generators of toric ideals The most efficient computational way to find them is 4ti2 (FourTiTwo package in Macaulay2). The exponents appearing in a finite generating set are sometimes called a Markov basis Database Exercise: Given a toric ideal, how to find A?

18 Some combinatorial commutative algebra An abstract reason why binomial ideals are good are monoid gradings. Define a Z d -valued grading on k[p] via deg p j = A j.

19 Some combinatorial commutative algebra An abstract reason why binomial ideals are good are monoid gradings. Define a Z d -valued grading on k[p] via deg p j = A j. I A is homogeneous

20 Some combinatorial commutative algebra An abstract reason why binomial ideals are good are monoid gradings. Define a Z d -valued grading on k[p] via deg p j = A j. I A is homogeneous The Hilbert function of k[p]/i A takes values only 0 and 1. 1 for all b NA = {Au : u N n } the monoid generated by A 0 for all other b Z d \ NA

21 Let Q be a commutative Noetherian monoid.

22 Let Q be a commutative Noetherian monoid. Monoid Algebras The monoid algebra over Q is the k-vector space k[q] := k {x q } with x q x u := x q+u. q Q A binomial ideal is an ideal generated by binomials x q λx u, q, u Q, λ k. Examples k[n n ] = k[p 1,..., p n ] k[na] = k[p 1,..., p n ]/I A

23 This generalizes to Eisenbud Sturmfels An ideal I k[p] is binomial if and only if k[p]/i is finely graded by a commutative Noetherian monoid. Combinatorial commutative algebra This leads to a very nice theory of binomial ideals based on the separation of combinatorics (the monoid) and arithmetics (the coefficients)

24 Not every ideal is prime or toric! Every ideal I k[p 1,..., p n ] is a finite intersection of primary ideals I = Q i, Qi = P i is prime i (Q is primary, if in k[p]/q every element is regular or nilpotent.) If k is algebraically closed, every binomial ideal is an intersection of primary binomial ideals (Eisenbud/Sturmfels). Independent of k, decompositions of congruences point the way! mesoprimary decomposition

25 Combinatorial versions of binomial ideals A congruence on Q is an equivalence relation such that a b a + q b + q q Q Congruences are the kernels of monoid homomorphisms Quotients Q := Q/ are monoids again. Congruences from binomial ideals Each binomial ideal I k[q] induces a congruence I on Q: a I b λ 0 : x a λx b I

26 y 3, y 2 (x 1), y(x 2 1) x 2 xy, xy y 2

27 Decompositions of binomial ideals in action Consider distributions of 3 binary random variables: (p 000, p 001,..., p 111 ). Assume we want to study the following conditional independencies: C = {X 1 X 2 X 3, X 1 X 3 X 2 } As you will see, this leads to binomial conditions: p 000 p 010 p 100 p 110 = 0, p 001 p 011 p 101 p 111 = 0 X 1 X 2 X 3 p 000 p 001 p 100 p 101 = 0, p 010 p 011 p 110 p 111 = 0 X 1 X 3 X 2

28 The prime decomposition of the corresponding ideal I C is ( ) p000 p I C = rk 001 p 010 p 011 = 1 p 100 p 101 p 110 p 111 p 000, p 100, p 011, p 111 p 001, p 010, p 101, p 110 The model (inside 7 ) consists of three (toric) components An independence model (d = 4) X 1 {X 2, X 3 }. conv A i = 1 1 is a prism over a 3d-simplex. 2 copies of 3 embedded in faces of 7.

29 Theorem If for the distribution of 3 binary random variables both X 1 X 2 X 3 and X 1 X 3 X 2 hold, then either X 1 {X 2, X 3 } ( the intersection axiom holds ), or p 000 = p 100 = p 011 = p 111 = 0 ( X 2 = 1 X 3 ), or p 001 = p 010 = p 101 = p 110 = 0 ( X 2 = X 3 ).