Ehrhart Positivity
Federico Castillo
University of California, Davis
Joint work with Fu Liu

January 16, 2017

Lattice points of a polytope

A (convex) polytope is a bounded solution set of a finite system of
linear inequalities, or is the convex hull of a finite set of points.

Lattice points of a polytope

A (convex) polytope is a bounded solution set of a finite system of
linear inequalities, or is the convex hull of a finite set of points.
An integral polytope is a polytope whose vertices are all lattice points.
i.e., points with integer coordinates.

Lattice points of a polytope

A (convex) polytope is a bounded solution set of a finite system of
linear inequalities, or is the convex hull of a finite set of points.
An integral polytope is a polytope whose vertices are all lattice points.
i.e., points with integer coordinates.

Definition
For any polytope P ⊂ Rd and positive integer m ∈ N, the mth dilationof P is mP = {mx : x ∈ P}. We definei(P, m) = |mP ∩ Zd |to be the number of lattice points in the mP.

ExampleP

3P

ExampleP

3P

In this example we can see that i(P, m) = (m + 1)2

Theorem of Ehrhart (on integral polytopes)

Figure: Eugene Ehrhart.

Theorem of Ehrhart (on integral polytopes)

Figure: Eugene Ehrhart.

Theorem (Ehrhart)
Let P be a d-dimensional integral polytope. Then i(P, m) is apolynomial in m of degree d.

Theorem of Ehrhart (on integral polytopes)

Figure: Eugene Ehrhart.

Theorem (Ehrhart)
Let P be a d-dimensional integral polytope. Then i(P, m) is apolynomial in m of degree d.

Different bases.

Therefore, we call i(P, m) the Ehrhart polynomial of P.

Different bases.

Therefore, we call i(P, m) the Ehrhart polynomial of P.We study its
coefficients.

Different bases.

Therefore, we call i(P, m) the Ehrhart polynomial of P.We study its
coefficients. ... however, there is another popular point of view.

Different bases.

Therefore, we call i(P, m) the Ehrhart polynomial of P.We study its
coefficients. ... however, there is another popular point of view.
For example, it is not hard to see that the Ehrhart polynomial of theunit 3d-cube [0, 1]3 is equal to:Different bases.

Therefore, we call i(P, m) the Ehrhart polynomial of P.We study its
coefficients. ... however, there is another popular point of view.
For example, it is not hard to see that the Ehrhart polynomial of theunit 3d-cube [0, 1]3 is equal to:i([0, 1]3 , m) = (m + 1)3 ,i([0, 1]3 , m) = m3 + 3m2 + 3m + 1,






 
m+3
m+2
m+1
m
3
i([0, 1] , m) = 1
+4
+1
+0
3
3
3
3

The h∗ or δ vector.

An alternative basis
We can write:




 
m+d
m+d −1mi(P, m) = h0∗ (P)+ h1∗ (P)+ · · · + hd∗ (P).
ddd
The vector (h0∗ , h1∗ , · · · , hd∗ ) has many good properties.

The h∗ or δ vector.

An alternative basis
We can write:




 
m+d
m+d −1mi(P, m) = h0∗ (P)+ h1∗ (P)+ · · · + hd∗ (P).
ddd
The vector (h0∗ , h1∗ , · · · , hd∗ ) has many good properties.

Theorem(Stanley)
For any lattice polytope P, hi∗ (P) is nonnegative integer.

The h∗ or δ vector.

An alternative basis
We can write:




 
m+d
m+d −1mi(P, m) = h0∗ (P)+ h1∗ (P)+ · · · + hd∗ (P).
ddd
The vector (h0∗ , h1∗ , · · · , hd∗ ) has many good properties.

Theorem(Stanley)
For any lattice polytope P, hi∗ (P) is nonnegative integer.
Additionally it has an algebraic meaning.

Back to coefficients of Ehrhart polynomials

Back to coefficients of Ehrhart polynomials
What is known?

Back to coefficients of Ehrhart polynomials
What is known?
1. The leading coefficient of i(P, m) is the volume vol(P) of P.

Back to coefficients of Ehrhart polynomials
What is known?
1. The leading coefficient of i(P, m) is the volume vol(P) of P.
2. The second coefficient equals 1/2 of the sum of the normalized
volumes of each facet.

Back to coefficients of Ehrhart polynomials
What is known?
1. The leading coefficient of i(P, m) is the volume vol(P) of P.
2. The second coefficient equals 1/2 of the sum of the normalized
volumes of each facet.
3. The constant term of i(P, m) is always 1.

Back to coefficients of Ehrhart polynomials
What is known?
1. The leading coefficient of i(P, m) is the volume vol(P) of P.
2. The second coefficient equals 1/2 of the sum of the normalized
volumes of each facet.
3. The constant term of i(P, m) is always 1.
No simple forms known for other coefficients for general polytopes.

Back to coefficients of Ehrhart polynomials
What is known?
1. The leading coefficient of i(P, m) is the volume vol(P) of P.
2. The second coefficient equals 1/2 of the sum of the normalized
volumes of each facet.
3. The constant term of i(P, m) is always 1.
No simple forms known for other coefficients for general polytopes.

Warning
It is NOT even true that all the coefficients are positive.
For example, for the polytope P with vertices(0, 0, 0), (1, 0, 0), (0, 1, 0) and (1, 1, 13), its Ehrhart polynomial isi(P, n) =13 3
1
n + n2 − n + 1.
6
6

By the way...

Theorem (A.Tsuchiya)
Other than the leading, second, and constant being positive, any signpattern can occur in the other coefficients.

General philosophy.

They are related to volumes.

Ehrhart Positivity

Main Definition.
We say an integral polytope is Ehrhart positive (or just positive for thistalk) if it has positive coefficients in its Ehrhart polynomial.

Ehrhart Positivity

Main Definition.
We say an integral polytope is Ehrhart positive (or just positive for thistalk) if it has positive coefficients in its Ehrhart polynomial.
In the literature, different techniques have been used to proved
positivity.

Example I

Polytope: Standard
simplex.

Example I

Polytope: Standard
simplex.
Reason: Explicit
verification.

Standard simplex.
In the case of∆d = {x ∈ Rd+1 : x1 + x2 + · · · + xd+1 = 1, xi ≥ 0},

Standard simplex.
In the case of∆d = {x ∈ Rd+1 : x1 + x2 + · · · + xd+1 = 1, xi ≥ 0},
It can be computed that its Ehrhart polynomial is


m+d
.
d
(Notice how simple this h∗ vector is).

Standard simplex.
In the case of∆d = {x ∈ Rd+1 : x1 + x2 + · · · + xd+1 = 1, xi ≥ 0},
It can be computed that its Ehrhart polynomial is


m+d
.
d
(Notice how simple this h∗ vector is). More explicitly we havem+d
(m + d)(m + d − 1) · · · (m + 1)=dd!

Standard simplex.
In the case of∆d = {x ∈ Rd+1 : x1 + x2 + · · · + xd+1 = 1, xi ≥ 0},
It can be computed that its Ehrhart polynomial is


m+d
.
d
(Notice how simple this h∗ vector is). More explicitly we havem+d
(m + d)(m + d − 1) · · · (m + 1)=dd!
which expands positively in powers of m.

Hypersimplices.

In the case of∆d+1,k = conv{x ∈ {0, 1}d+1 : x1 + x2 + · · · + xd+1 = k },

Hypersimplices.

In the case of∆d+1,k = conv{x ∈ {0, 1}d+1 : x1 + x2 + · · · + xd+1 = k },it can be computed that its Ehrhart polynomial isHypersimplices.

In the case of∆d+1,k = conv{x ∈ {0, 1}d+1 : x1 + x2 + · · · + xd+1 = k },it can be computed that its Ehrhart polynomial is


d+1 
X
d + 1 d + 1 + mk − (m + 1)i − 1
(−1)i
i
d
i=0

Hypersimplices.

In the case of∆d+1,k = conv{x ∈ {0, 1}d+1 : x1 + x2 + · · · + xd+1 = k },it can be computed that its Ehrhart polynomial is


d+1 
X
d + 1 d + 1 + mk − (m + 1)i − 1
(−1)i
i
d
i=0

Not clear if the coefficients are positive.

Example II

Polytope: Crosspolytope

Example II

Polytope: Crosspolytope
Reason: Roots havenegative real part.

Crosspolytope.

In the case of the crosspolytope:♦d = conv{±ei : 1 ≤ i ≤ d},Crosspolytope.

In the case of the crosspolytope:♦d = conv{±ei : 1 ≤ i ≤ d},
It can be computed that its Ehrhart polynomial is
d
X
k=0

2k

  
dm,kk

Crosspolytope.

In the case of the crosspolytope:♦d = conv{±ei : 1 ≤ i ≤ d},
It can be computed that its Ehrhart polynomial is
d
X
k=0

2k

  dm,kkwhich is not clear if it expands positively in powers of m.

Crosspolytope.

HoweverCrosspolytope.

However, according to EC1, Exercise 4.61(b), every zero of theEhrhart polynomial has real part −1/2.

Crosspolytope.

However, according to EC1, Exercise 4.61(b), every zero of theEhrhart polynomial has real part −1/2. Thus it is a product of factors(n + 1/2) or(n + 1/2 + ia)(n + 1/2 − ia) = n2 + n + 1/4 + a2 ,where a is real, so positivity follows.

Crosspolytope.

However, according to EC1, Exercise 4.61(b), every zero of theEhrhart polynomial has real part −1/2. Thus it is a product of factors(n + 1/2) or(n + 1/2 + ia)(n + 1/2 − ia) = n2 + n + 1/4 + a2 ,where a is real, so positivity follows.

What are the roots about?
This opens more questions.

Birkhoff Poytope
The following is the graph (Beck-DeLoera-Pfeifle-Stanley) of zeros for
the Birkhoff polytope of 8 × 8 doubly stochastic matrices.

Example III

Polytope: Zonotopes.

Example III

Polytope: Zonotopes.
Reason: Formula for them.

Example III

Polytope: Zonotopes.
Reason: Formula for them.
One of the few examples in which the formula is explicit on the coefficients.

Zonotopes.

Definition
The Minkowski sum of vectors
Z(v1 , · · · , vk ) = v1 + v2 + · · · + vk .
The Ehrhart polynomiali(Z(v1 , · · · , vk ), m) = ad md + ad−1 md−1 + · · · a0 m0 ,has a coefficient by coefficient interpretation.

Zonotopes.

Theorem(Stanley)
In the above expression, ai is equal to (absolute value of) the greatest
common divisor (g.c.d.) of all i × i minors of the matrix···M =  v1 v2 · · · vk ···

Zonotopes.

This includes the unit cube [0, 1]d which has Ehrhart polynomiali(d , m) = (m + 1)d .

Zonotopes.

This includes the unit cube [0, 1]d which has Ehrhart polynomiali(d , m) = (m + 1)d .
And also the regular permutohedronXΠn =[ei , ej ],1≤i<j≤n+1= conv{ σ(1), σ(2), · · · , σ(n + 1) ∈ Rn+1 : σ ∈ Sn+1 }.

Permutohedron.

Figure: A permutohedron in dimension 3.

The Ehrhart polynomial is 1 + 6m + 15m2 + 16m3 .

Example IV

Polytope: Cyclic polytopes.

Example IV

Polytope: Cyclic polytopes.
Reason: Higher integrality
conditions.

Cyclic polytopes.

Consider the moment map m : R → Rd that sends
x 7→ (x, x 2 , · · · , x d ).
The convex hull of any(!) n points on that curve is what is called acyclic polytope C(n, d).

Cyclic polytopes.

Consider the moment map m : R → Rd that sends
x 7→ (x, x 2 , · · · , x d ).
The convex hull of any(!) n points on that curve is what is called acyclic polytope C(n, d).

Ehrhart Polynomial.
Fu Liu proved that under certain integrality conditions, the coefficientof t k in the Ehrhart polynomal of P is given by the volume of the
projection that forgets the last k coordinates.

Not a combinatorial property

Theorem (Liu)For any polytope P there is a polytope P 0 with the same face latticeand Ehrhart positivity.

Plus many unknowns.

Other polytopes have been observed to be positive.
I

CRY (Chan-Robbins-Yuen).

Tesler matrices (Mezaros-Morales-Rhoades).
I Birkhoff polytopes (Beck-DeLoera-Pfeifle-Stanley).
I

I

Matroid polytopes (De Loera - Haws- Koeppe).

Plus many unknowns.

Other polytopes have been observed to be positive.
I

CRY (Chan-Robbins-Yuen).

Tesler matrices (Mezaros-Morales-Rhoades).
I Birkhoff polytopes (Beck-DeLoera-Pfeifle-Stanley).
I

I

Matroid polytopes (De Loera - Haws- Koeppe).

Also:

Littlewood Richardson
Ronald King conjecture that the stretch littlewood richardsontν
coefficients ctλ,tµare polynomials in N[t]. This polynomials are known
to be Ehrhart polynomials.

General approach?

General approach?

“Though it be madness, yet there’s
method in’t...” Hamlet, Act II.

Method in the madness.

Coming from the theory of toric varieties, we haveMethod in the madness.

Coming from the theory of toric varieties, we have

Definition
A McMullen formula is a function α such thatX
|P ∩ Zd | =α(F , P)nvol(F ).
F ⊆P

where the sum is over all faces and α depends locally on F and P.
More precisely, it is defined on the normal cone of F in P.

Method in the madness.

Coming from the theory of toric varieties, we have

Definition
A McMullen formula is a function α such thatX
|P ∩ Zd | =α(F , P)nvol(F ).
F ⊆P

where the sum is over all faces and α depends locally on F and P.
More precisely, it is defined on the normal cone of F in P.
McMullen proved the existence of such α in a nonconstructive andnonunique way.

Constructions

There are at least three different constructions
1. Pommersheim-Thomas. Need to choose a flag of subspaces.
2. Berline-Vergne. No choices, invariant under On (Z). This is what
we use.
3. Schurmann-Ring. Need to choose a fundamental cell.

Example

1
6
5
12

5
12

McMullen Formula:3
10|P ∩ Z| = (Area of P) + 21 (Perimeter of P) + 1.

1
4

The way one gets the +1 is different.

Pommersheim-Thomas

920
Berline-Vergne

1
4
1
2

1
4
Schurmann-RingRefinement of positivity.

This gives expressions for the coefficients.
X|nP ∩ Zd | =α(F , nP)nvol(F )
F ⊂nP

=

X
F ⊂P

α(F , P)nvol(F )ndim(F )Refinement of positivity.

This gives expressions for the coefficients.
X|nP ∩ Zd | =α(F , nP)nvol(F )
F ⊂nP

=

X

α(F , P)nvol(F )ndim(F )F ⊂P

We see that

Coefficient
The coefficient of nk is

X
F :dim(F )=kα(P, F )vol(F ).

Refinement of positivity.

This gives expressions for the coefficients.
X|nP ∩ Zd | =α(F , nP)nvol(F )
F ⊂nP

=

X

α(F , P)nvol(F )ndim(F )F ⊂P

We see that

Coefficient
The coefficient of nk isX

α(P, F )vol(F ).

F :dim(F )=k

As long as all α are positive, then the coefficients will be positive.

Main properties.

The important facts about the Berline-Vergne construction areI

It exists.

I

Symmetric under rearranging coordinates.

I

It is a valuation.

We exploit these.

A refined conjecture.

We pose the following.

Conjecture.
The regular permutohedron is (Berline-Vergne) α positive.

A refined conjecture.

We pose the following.

Conjecture.
The regular permutohedron is (Berline-Vergne) α positive.
We care because this imply Ehrhart positivity for a family of polytopes.

A refined conjecture.

We pose the following.

Conjecture.
The regular permutohedron is (Berline-Vergne) α positive.
We care because this imply Ehrhart positivity for a family of polytopes.

Proposition.
The above conjecture implies that Generalized Permutohedra arepositive.

A refined conjecture.

We pose the following.

Conjecture.
The regular permutohedron is (Berline-Vergne) α positive.
We care because this imply Ehrhart positivity for a family of polytopes.

Proposition.
The above conjecture implies that Generalized Permutohedra arepositive.
This would expand on previous results from Postnikov, and a
conjecture of De Loera-Haws-Koeppe stating that matroid polytopes
are positive.

Partial results.

We’ve checked the conjecture in the cases:
1. The linear term (corresponding to edges) in dimensions up to100.
2. The third and fourth coefficients.
3. Up to dimension 6.

Regular permutohedra revisited.

Figure: A permutohedron in dimension 3.

Regular permutohedra revisited.

Figure: A permutohedron in dimension 3.
1
For example, α(v , Π3 ) = 24
for any vertex. Since they are all
symmetric and they add up to 1.

A deformation.

Figure: Truncated octahedronA deformation.

Figure: Truncated octahedron

Computing with the properties.
Note that we have just two types of edges (with normalized volume1). From the permutohedron we get
24α1 + 12α2 = 6.
Now looking at the octhaedron, the alpha values are the same, since
the normal cones didn’t change. In this case we get
12α2 = 7/3

Computing with the properties.
Note that we have just two types of edges (with normalized volume1). From the permutohedron we get
24α1 + 12α2 = 6.
Now looking at the octhaedron, the alpha values are the same, since
the normal cones didn’t change. In this case we get
12α2 = 7/3
This information is enough to conclude thatα1 = 11/72
α2 = 14/72

Computing with the properties.
Note that we have just two types of edges (with normalized volume1). From the permutohedron we get
24α1 + 12α2 = 6.
Now looking at the octhaedron, the alpha values are the same, since
the normal cones didn’t change. In this case we get
12α2 = 7/3
This information is enough to conclude thatα1 = 11/72
α2 = 14/72

Remark.
We did not use the explicit construction at all, just existence andproperties. This line of thought is the one we generalize.

A computation
There is just one 3 dimensional face, with α = 1 and volume44−2 = 16 which contributes 16n3
I There are six 2 dimensional faces with volume 1 and eight withvolume 3. The α value is 1/2 for all these faces so we get a11contribution of 6 1 · n2 + 8 3 · n2 = 15n22
2
I Two types of edges, both with volume 1. There are 24 shortedges with value 11/72 and 12 long edges with value 14/72, for
11
14
a contribution of 24 n + 12 n = 6n72
72
I There are 24 vertices, all with value 1/24 with a contribution of 1
I

Putting all together we get the Ehrhart polynomial
1 + 6n + 15n2 + 16n3

Main result.

We have a combinatorial formula for the α values of faces of regularpermutohedra. This formula involves mixed Ehrhart coefficients ofhypersimplices. The takeaway from this is

Uniqueness theorem.
Any McMullen formula that is symmetric under the coordinates is
uniquely determined on the faces of permutohedra.

Main result.

We have a combinatorial formula for the α values of faces of regularpermutohedra. This formula involves mixed Ehrhart coefficients ofhypersimplices. The takeaway from this is

Uniqueness theorem.
Any McMullen formula that is symmetric under the coordinates is
uniquely determined on the faces of permutohedra.
Which leads to the question.

Question.
Is Berline and Vergne the only construction that satisfies additivity
and symmetry?

Warning

We want to remark that it is not true that zonotopes are BV α
positive, even though they are Ehrhart positive.

A bit about the formula

Let P1 , · · · , Pm be a list of polytopes of dimension n, then

Mixed Valuations
The expression Lat(w1 P1 + · · · + wm Pm ) is a polynomial on the wi
variables. The coefficients are called mixed Ehrhart coefficients.

A bit about the formula

Let P1 , · · · , Pm be a list of polytopes of dimension n, then

Mixed Valuations
The expression Lat(w1 P1 + · · · + wm Pm ) is a polynomial on the wi
variables. The coefficients are called mixed Ehrhart coefficients.
On the top degree we have the mixed volumes.

A bit about the formula

Let P1 , · · · , Pm be a list of polytopes of dimension n, then

Mixed Valuations
The expression Lat(w1 P1 + · · · + wm Pm ) is a polynomial on the wi
variables. The coefficients are called mixed Ehrhart coefficients.
On the top degree we have the mixed volumes. Volumes are always
positive and mixed volumes are too, although this is not clear from theabove definition.

Permutohedra
We define a permutohedron for any vectorx) = (x1 , · · · , xn+1 ) ∈ Rn+1 . Let’s assume x1 ≤ · · · ≤ xn+1 .

Permutohedra
We define a permutohedron for any vectorx) = (x1 , · · · , xn+1 ) ∈ Rn+1 . Let’s assume x1 ≤ · · · ≤ xn+1 .

Perm(x) := conv{ xσ(1) , xσ(2) , · · · , xσ(n+1) ∈ Rn+1 : σ ∈ Sn+1 }.

Permutohedra
We define a permutohedron for any vectorx) = (x1 , · · · , xn+1 ) ∈ Rn+1 . Let’s assume x1 ≤ · · · ≤ xn+1 .

Perm(x) := conv{ xσ(1) , xσ(2) , · · · , xσ(n+1) ∈ Rn+1 : σ ∈ Sn+1 }.
If we define wi := xi+1 − xi , for i = 1, · · · , n, thenPerm(x) = w1 ∆1,n+1 + w2 ∆2,n+1 + · · · + wn ∆n,n+1 .

Permutohedra
We define a permutohedron for any vectorx) = (x1 , · · · , xn+1 ) ∈ Rn+1 . Let’s assume x1 ≤ · · · ≤ xn+1 .

Perm(x) := conv{ xσ(1) , xσ(2) , · · · , xσ(n+1) ∈ Rn+1 : σ ∈ Sn+1 }.
If we define wi := xi+1 − xi , for i = 1, · · · , n, thenPerm(x) = w1 ∆1,n+1 + w2 ∆2,n+1 + · · · + wn ∆n,n+1 .
So the number of integer points depends polynomially on theparameters wi .

Permutohedra
We define a permutohedron for any vectorx) = (x1 , · · · , xn+1 ) ∈ Rn+1 . Let’s assume x1 ≤ · · · ≤ xn+1 .

Perm(x) := conv{ xσ(1) , xσ(2) , · · · , xσ(n+1) ∈ Rn+1 : σ ∈ Sn+1 }.
If we define wi := xi+1 − xi , for i = 1, · · · , n, thenPerm(x) = w1 ∆1,n+1 + w2 ∆2,n+1 + · · · + wn ∆n,n+1 .
So the number of integer points depends polynomially on theparameters wi .These parameters are the lenghts of the edges inPerm(x).

Permutohedra
We define a permutohedron for any vectorx) = (x1 , · · · , xn+1 ) ∈ Rn+1 . Let’s assume x1 ≤ · · · ≤ xn+1 .

Perm(x) := conv{ xσ(1) , xσ(2) , · · · , xσ(n+1) ∈ Rn+1 : σ ∈ Sn+1 }.
If we define wi := xi+1 − xi , for i = 1, · · · , n, thenPerm(x) = w1 ∆1,n+1 + w2 ∆2,n+1 + · · · + wn ∆n,n+1 .
So the number of integer points depends polynomially on theparameters wi .These parameters are the lenghts of the edges inPerm(x).
For instance, the coefficient of w1 w2 is, by definition,
2!MLat2 (∆1,n+1 , ∆2,n+1 )Formula

Roughly
What we have looks likeα(F , P) = A × B.
Where A is some combinatorial expression, evidently positive.
And B is one (depending of F ) mixed Ehrhart coefficient ofhypersimplices.

Formula

Roughly
What we have looks likeα(F , P) = A × B.
Where A is some combinatorial expression, evidently positive.
And B is one (depending of F ) mixed Ehrhart coefficient ofhypersimplices.
In particular, our conjecture is equivalent to the positivity of such
coefficients.

Formula

Roughly
What we have looks likeα(F , P) = A × B.
Where A is some combinatorial expression, evidently positive.
And B is one (depending of F ) mixed Ehrhart coefficient ofhypersimplices.
In particular, our conjecture is equivalent to the positivity of such
coefficients. It is not even clear if hypersimplices themselves (without
any mixing) are Ehrhart positive.

Example

An instance of the formula looks like:A facet in Π3
Formula would say it is equal to
2·2
2!MLat2 (∆1,4 , ∆3,4 ).
24
where M stands for mixed and Lat 2 is the quadratic coefficient of
Ehrhart polynomial.

Example

An instance of the formula looks like:A facet in Π3
Formula would say it is equal to
2·2
2!MLat2 (∆1,4 , ∆3,4 ).
24
where M stands for mixed and Lat 2 is the quadratic coefficient of
Ehrhart polynomial.
Remark: The value at facets is always 12 .

Example

An instance of the formula looks like:A facet in Π3
Formula would say it is equal to
2·2
2!MLat2 (∆1,4 , ∆3,4 ).
24
where M stands for mixed and Lat 2 is the quadratic coefficient of
Ehrhart polynomial.
Remark: The value at facets is always 12 .
This mixed valuations can be evaluated in the usual alternating form.
We can check if the above expression is right. Let’s do it!

Example

What is 2!MLat2 (∆1,4 , ∆3,4 )?

Example

What is 2!MLat2 (∆1,4 , ∆3,4 )?
Lat(w1 ∆1,4 + w2 ∆2,4 ) = 1 + 1!MLat(∆1,4 )w1 + 1!MLat(∆3,4 )w2+ 2!MLat(∆1,4 , ∆3,4 )w1 w2 + 2!MLat(∆1,4 , ∆1,4 )w12
+ 2!MLat(∆3,4 , ∆3,4 )w22
+ 3!MLat(∆1,4 , ∆3,4 , ∆3,4 )w1 w22+ 3!MLat(∆1,4 , ∆1,4 , ∆3,4 )w12 w2+ 3!MLat(∆1,4 , ∆1,4 , ∆1,4 )w13
+ 3!MLat(∆3,4 , ∆3,4 , ∆3,4 )w23

Example

What is 2!MLat2 (∆1,4 , ∆3,4 )?
Lat(w1 ∆1,4 + w2 ∆2,4 ) = 1 + 1!MLat(∆1,4 )w1 + 1!MLat(∆3,4 )w2+ 2!MLat(∆1,4 , ∆3,4 )w1 w2 + 2!MLat(∆1,4 , ∆1,4 )w12
+ 2!MLat(∆3,4 , ∆3,4 )w22
+ 3!MLat(∆1,4 , ∆3,4 , ∆3,4 )w1 w22+ 3!MLat(∆1,4 , ∆1,4 , ∆3,4 )w12 w2+ 3!MLat(∆1,4 , ∆1,4 , ∆1,4 )w13
+ 3!MLat(∆3,4 , ∆3,4 , ∆3,4 )w23
How to find that coefficient?

Example

What is 2!MLat2 (∆1,4 , ∆3,4 )?
Lat(w1 ∆1,4 + w2 ∆2,4 ) = 1 + 1!MLat(∆1,4 )w1 + 1!MLat(∆3,4 )w2+ 2!MLat(∆1,4 , ∆3,4 )w1 w2 + 2!MLat(∆1,4 , ∆1,4 )w12
+ 2!MLat(∆3,4 , ∆3,4 )w22
+ 3!MLat(∆1,4 , ∆3,4 , ∆3,4 )w1 w22+ 3!MLat(∆1,4 , ∆1,4 , ∆3,4 )w12 w2+ 3!MLat(∆1,4 , ∆1,4 , ∆1,4 )w13
+ 3!MLat(∆3,4 , ∆3,4 , ∆3,4 )w23
How to find that coefficient? Inclusion-Exclusion!

Examplei (∆14 + ∆34 , t)=i (∆14 , t)=i (∆34 , t)

=

10 3
11
t + 5t 2 +t + 1,
3
3
11
1 3
t + t2 +t + 1,
6
6
1 3
11
t + t2 +t + 1.
6
6

Therefore,
2!MLat2 (∆1,4 , ∆3,4 ) =5 − 1 − 1 = 3
So we get
4
1
2·2
2!MLat2 (∆1,4 , ∆3,4 ) =
·3=
24
24
2

Further direction.

Some observations lead to the very natural question:

Sum of positives.
If P and Q are positive, is it true that P + Q is positive?