Main properties

Definition

Let $Σ$ be a balanced unimodular fan. We use fan conventions.

Definition 1 (Ehrhart fan)

A fan $Σ$ is an Ehrhart fan if:

1. for any nonzero face $σ$, $Σ^σ$ is Ehrhart,
2. there is a unique function $f: A^1(Σ; ℤ) → ℤ$, called the Ehrhart polynomial of $Σ$, such that:
   1. $f(0) = 1$
   2. $f(z + x_ρ) = f(z) + f^ρ(i^*_ρ(z + x_ρ))$ for any $z∈ℤ^{Σ_1}$ and any ray $ρ$, where $f^ρ$ is the Ehrhart polynomial of $Σ^ρ$.

Note that $f$ is indeed a polynomial by Proposition 2 (Polynomiality). We denote it by $\mathrm{Ehr}(Σ)$.

Pre-Ehrhart fans and properties of the Ehrhart polynomial

Definition 2 (Pre-Ehrhart fan)

A fan $Σ$ is pre-Ehrhart if it only verifies the point 1. of the above definition.

Example 1 (One dimensional Ehrhart fans)

One dimensional balanced fans $Σ$ are Ehrhart of Ehrhart polynomial $$\mathrm{Ehr}(Σ)(∑_{ρ ∈ Σ_1} a_ρ x_ρ) = 1 + ∑_ρ a_ρ.$$

Proposition 1

If $Σ$ is pre-Ehrhart, then there exists a unique function $f: ℤ^{Σ_1} → ℤ$ which verifies points 2.a. and the following variant of 2.b: $$∀z,ρ, \quad f(z + x_ρ) = f(z) + f^ρ(i^*_ρ(Ψ(z + x_ρ)))$$ where $Ψ: ℤ^{Σ_1} → A^1(Σ)$ is the projection.

We call this unique function the Ehrhart polynomial of $Σ$ and denote it $\mathrm{Ehr}(Σ)$.

Proof

The uniqueness is clear. Denote by $K = ⟨x_ρ,-x_ρ⟩_{ρ ∈ Σ_1}$ the free monoid generated by the $x_ρ$ and there opposite. Let $Φ: K → ℤ^{Σ_1}$ be the projection. We denote by $;$ the binary operation of the group. Let $f: K → ℤ$ by :

• $f(0) = 1$,
• $f(z ; x_ρ) = f(z) + f^ρ(i^*_ρ(Φ(z + x_ρ)))$ and
• $f(z ; -x_ρ) = f(z) - f^ρ(i^*_ρ(Φ(z)))$.

It suffices to prove that, for any $y, z ∈ K$, any $ρ, ρ'$ and any $ε, ε' ∈ \{±1\}$,

• $f(y ; εx_ρ ; -εx_ρ ; z) = f(y ; z)$,
• $f(y ; εx_ρ ; ε'x_{ρ'} ; z) = f(y ; ε'x_{ρ'} ; εx_ρ ; z)$

Using $Ψ(Φ(y ; x_ρ ; -x_ρ)) = Ψ(Φ(y))$, the first point reduces to the case $z=0$, and then it is trivial from the definition of $f$. Then we can reduce the second case to $ε = ε' = +1$; indeed, with this case, we obtain the other with algebraic manipulations similar to the following one: $$f(-x_ρ ; x_{ρ'}) = f(-x_ρ ; x_{ρ'} ; x_ρ ; -x_ρ) = f(-x_ρ ; x_ρ ; x_{ρ'} ; -x_ρ) = f(x_{ρ'} ; -x_ρ).$$

Both cases $ρ=ρ'$ and $ρ∧ρ'= 𝟏$ are trivial. It remains the case $ρ ∧ ρ' = τ$ with $|τ| = 2$. As before, we can assume that $z=0$. Then we get $$\begin{align} f(y ; x_ρ ; x_{ρ'}) &= f(y) + f^ρ(i^*_ρ(Ψ(Φ(z ; x_ρ)))) + f^{ρ'}(i^*_{ρ'}(Ψ(Φ(z ; x_ρ ; x_{ρ'})))) \\ &= f(y) + f^ρ(i^*_ρ(Ψ(Φ(z ; x_ρ)))) + f^{ρ'}(i^*_{ρ'}(Ψ(Φ(z ; x_{ρ'}))) + x_{π^ρ'(ρ)}) \\ &= f(y) + f^ρ(i^*_ρ(Ψ(Φ(z ; x_ρ)))) + f^{ρ'}(i^*_{ρ'}(Ψ(Φ(z ; x_{ρ'})))) + f^τ(i^*_{ρ'≺τ}(i^*_{ρ'}(Ψ(Φ(z ; x_{ρ'}))) + x_{π^ρ'(ρ)})) \\ &= f(y) + f^ρ(i^*_ρ(Ψ(Φ(z ; x_ρ)))) + f^{ρ'}(i^*_{ρ'}(Ψ(Φ(z ; x_{ρ'})))) + f^τ(i^*_{τ}(Ψ(Φ(z ; x_{ρ'} ; x_ρ)))). \end{align}$$ The last formula is symmetric in $ρ$ and $ρ'$, which concludes the proof.

Any element $z$ of $ℤ^{Σ_1}$ can be written in a unique way under the form $∑_ρ a_ρ x_ρ$. In the following, we will say that we work by induction on $𝐳$ to mean that we work by induction on $∑_{ρ} |a_ρ|$. Moreover, we often omit to check the step case where we remove $x_ρ$ from $z$, since it directly follows from the case where we add it. We also omit regularly the map $i^*$.

Lemma 1 (Pre-Ehrhart to Ehrhart fans)

Let $Σ$ be a pre-Ehrhart fan, and let $ℓ ∈ M$. Then, for any $z$, $$\mathrm{Ehr}(Σ)(x_ℓ + z) = \mathrm{Ehr}(Σ)(x_ℓ)-1 + \mathrm{Ehr}(Σ)(z).$$ In particular, a pre-Ehrhart fan is Ehrhart if and only if its Ehrhart polynomial values one on any $x_ℓ$, $ℓ ∈ M$.

Proof

Working by induction on $z$, for $z = 0$, this is trivial, and otherwise $$\begin{align} \mathrm{Ehr}(Σ)(x_ℓ + z + x_ρ) &= \mathrm{Ehr}(Σ)(x_ℓ + z) + \mathrm{Ehr}(Σ^ρ)(x_ℓ + z + x_ρ) \\ &= \mathrm{Ehr}(Σ)(x_ℓ)-1 + \mathrm{Ehr}(Σ)(z) + \mathrm{Ehr}(Σ^ρ)(x_ℓ + z + x_ρ) \\ &= \mathrm{Ehr}(Σ)(x_ℓ)-1 + \mathrm{Ehr}(Σ)(z) + \mathrm{Ehr}(Σ^ρ)(z + x_ρ) \\ &= \mathrm{Ehr}(Σ)(x_ℓ)-1 + \mathrm{Ehr}(Σ)(z + x_ρ). \end{align}$$

Proposition 2 (Polynomiality)

Assume $Σ$ is pre-Ehrhart. Then the function $f := \mathrm{Ehr}(Σ)$ is polynomial, that is $f(∑_{ρ ∈ Σ_1}) a_ρ x_ρ ∈ ℚ[(a_ρ)_{ρ ∈ Σ_1}]$.

Proof

The proof goes by induction. Assume the proposition is true for any fans of dimension $d-1$. Let $ρ_1, …, ρ_k$ be the rays of $Σ$. Then, simplifying the notations with $a_k = a_{ρ_k}$ and $x_k = x_{ρ_k}$ and omitting the $i^*$, $$f(∑_{i=1}^k a_k x_k) = ∑_{j=1}^{a_1}f^{ρ_1}(j x_1) \ + \ ∑_{j=1}^{a_2}f^{ρ_2}(a_1 x_1 + j x_2) \ + \ ⋯ \ + \ ∑_{j=1}^{a_k}f^{ρ_k}(a_1 x_1 + ⋯ + a_{k-1} x_{k-1} + j x_k).$$ We claim that each term is a polynomial. Indeed, let $1 ≤ i ≤ k$, and let $ℓ$ be a integral linear form which is 1 on $e_{ρ_i}$. Then, for any $j$ $$f^{ρ_i} (a_1 x_1 + ⋯ + a_{i-1} x_{i-1} + j x_i) = ∑_{l, l < i, ρ_l ∧ ρ_1 ≠ 𝟏} f^{ρ_i}((a_l-jℓ(e_{ρ_l}))x_l).$$ By induction, this is a polynomial in $ℤ[(a_l)_l, j]$. Since sum of $r$-th powers is a polynomial for any $r$, $∑_{j=1}^{a_i} f^{ρ_i} (a_1 x_1 + ⋯ + a_{i-1} x_{i-1} + j x_i)$ is a polynomial in $ℤ[(a_l)_l, a_i]$. Summing all together, we get the conclusion follows.

Stability properties

Lemma 2 (Ehrhart on $-x_\rho$)

Assume $Σ$ verifies point 1. Then the function $f: ℤ^{Σ_1} → ℤ$ introduced in Proposition 1 verifies $f(-x_ρ) = 0$ for all $ρ$.

Proof

Clear from the definition.

Proposition 3 (Stability by product)

Let $Σ$ and $Σ'$ be two Ehrhart fans in $N_ℝ$ and $N'_ℝ$ respectively, both verifying point 1, and let $ι_1: Σ → Σ×Σ'$ and $ι_2: Σ' → Σ×Σ'$ be the two inclusions. Then, for any $z$, $$\mathrm{Ehr}(Σ×Σ')(z) = \mathrm{Ehr}(Σ)(ι_1^*(z)) ⋅ \mathrm{Ehr}(Σ')(ι_2^*(z)).$$ In particular, the product of two Ehrhart fans is Ehrhart.

Proof

We get the proof by induction, both on the dimension of the product, and on the sum $∑_ρ|a_ρ|$. Clearly $\mathrm{Ehr}(Σ×Σ')(0) = \mathrm{Ehr}(Σ)(0)\mathrm{Ehr}(Σ')(0)$. Let $ρ$ be a ray of $Σ$. We also denote by $ρ$ the corresponding ray in $Σ×Σ'$. We have, for any $z$ and omitting the $i^*$, $$\begin{align*} \mathrm{Ehr}(Σ×Σ')(z + x_ρ) &= \mathrm{Ehr}(Σ×Σ')(z) + \mathrm{Ehr}((Σ×Σ')^ρ)(z + x_ρ) \\ &= \mathrm{Ehr}(Σ)(ι_1^*(z))⋅\mathrm{Ehr}(Σ')(ι_2^*(z)) + \mathrm{Ehr}(Σ^ρ×Σ')(z + x_ρ) \\ &= \mathrm{Ehr}(Σ)(ι_1^*(z))⋅\mathrm{Ehr}(Σ')(ι_2^*(z)) + \mathrm{Ehr}(Σ^ρ)(ι_1^*(z + x_ρ))⋅\mathrm{Ehr}(Σ')(ι_2^*(z + x_ρ)) \\ &= \mathrm{Ehr}(Σ)(ι_1^*(z))⋅\mathrm{Ehr}(Σ')(ι_2^*(z + x_ρ)) + \mathrm{Ehr}(Σ^ρ)(ι_1^*(z + x_ρ))⋅\mathrm{Ehr}(Σ')(ι_2^*(z + x_ρ)) \\ &= (\mathrm{Ehr}(Σ)(ι_1^*(z)) + \mathrm{Ehr}(Σ^ρ)(ι_1^*(z + x_ρ))) \ ⋅ \ \mathrm{Ehr}(Σ')(ι_2^*(z + x_ρ)) \\ &= \mathrm{Ehr}(Σ)(ι_1^*(z + x_ρ))⋅\mathrm{Ehr}(Σ')(ι_2^*(z + x_ρ)). \end{align*}$$ The proof of the first part easily follows. For the stability of being Ehrhart, simply note that if $ℓ$ is a linear form on $N_ℝ × N'_ℝ$, then $ι_1^*(x_ℓ) = x_{ℓ|_Σ}$.

Proposition 4 (Stability by blow-up)

Let $Σ$ be a fan and let $Σ'$ be the unimodular blow-up of $Σ$ on some face $σ$, and let $ρ_0$ be the new ray. Then $Σ$ is pre-Ehrhart if and only if $Σ'$ is. Moreover, let $ι: Σ' → Σ$ be the inclusion. Then, for any $z$, $$\mathrm{Ehr}(Σ')(ι^*(z)) = \mathrm{Ehr}(Σ).$$ In particular, Ehrhart fans are stable by unimodular blow-ups and blow-downs.

We recall that $ι^*(x_ρ) = x_ρ + x_ρ_0$ if $ρ ∈ σ$.

Proof

We work by induction on both the dimension of $Σ$, and on $z$. The only nontrivial case to check is the following one. Let $ρ$ be a ray of $σ$. Then, for any $z$ $$\begin{align*} \mathrm{Ehr}(Σ')(ι^*(z + x_ρ)) &= \mathrm{Ehr}(Σ')(ι^*(z) + x_ρ_0 + x_ρ) \\ &= \mathrm{Ehr}(Σ')(ι^*(z)) + \mathrm{Ehr}(Σ'^{ρ_0})(ι^*(z) + x_ρ_0) + \mathrm{Ehr}(Σ'^ρ)(ι^*(z) + x_ρ_0 + x_ρ) \\ &= \mathrm{Ehr}(Σ)(z) + \mathrm{Ehr}(Σ'^{ρ_0})(ι^*(z) + x_ρ_0) + \mathrm{Ehr}(Σ'^ρ)(ι^*(z + x_ρ)) \\ &= \mathrm{Ehr}(Σ)(z + x_ρ) + \mathrm{Ehr}(Σ'^{ρ_0})(ι^*(z) + x_ρ_0). \end{align*}$$ Hence we just have to prove that the second term is zero. Note that $Σ'^{ρ_0}$ can be canonically identified with $Σℙ^{|σ|-1} × Σ^σ$, where $Σℙ^{|σ|-1}$ is the projective fan of rays corresponding to the rays of $σ$. Let $𝚥_1: Σℙ^{|σ|-1} → Σ'^{ρ_0}$ be the inclusion. Then, $𝚥_1^*(ι^*(z))$ is linear of $Σℙ^{|σ|-1}$. Since $σ$ is unimodular, there exists a linear form $ℓ$ which is 1 on both $ρ_0$ and $ρ$, and that is 0 on other rays of $σ$. Hence, using Lemma 2 (Ehrhart on $-x_\rho$), $$\begin{align*} \mathrm{Ehr}(Σℙ^{|σ|-1})(𝚥_1^*(ι^*(z) + x_ρ_0)) &= \mathrm{Ehr}(Σℙ^{|σ|-1})(𝚥_1^*(ι^*(z) + x_ρ_0 - x_ℓ)) \\ &= \mathrm{Ehr}(Σℙ^{|σ|-1})(𝚥_1^*(ι^*(z)) - x_ρ) \\ &= \mathrm{Ehr}(Σℙ^{|σ|-1})(-x_ρ) \\ &= 0. \end{align*}$$ With Proposition 3 (Stability by product), we get that the second term is zero, which concludes the proof.

Proposition 5 (Tropical modifications and Ehrhart fans)

Let $Σ$ be an Ehrhart fan. Let $f$ be a integral piecewise linear function of $Σ$ whose divisor is trivial or reduced. Set $x_f := ∑_ρ f(e_ρ)x_ρ$. Set $Σ' := \mathrm{Trop}_f(Σ)$ and $π: Σ' → Σ$ the projection. Then $Σ'$ is Ehrhart if and only if:

• either $\mathrm{div}(f)$ is trivial and $\mathrm{Ehr}(Σ)(-x_f) = 1$,
• or $\mathrm{div}(f)$ is a nontrivial Ehrhart fan and $\mathrm{Ehr}(Σ)(-x_f) = 0$.

Moreover, in both cases, $\mathrm{Ehr}(Σ')(π^*(z)) = \mathrm{Ehr}(Σ)(z)$ for any $z$.

Proof

We prove it by induction on the dimension of $Σ$. Recall the [description of the star fan of a tropical modification 🏗️]. Assume that $\mathrm{div}(f)$ is either trivial or Ehrhart. In both case, we get that $Σ'$ is pre-Ehrhart by induction. The last equality now follows by a direct induction on $z$. In particular, for all linear forms of the form $π^*(x_ℓ)$, $ℓ ∈ M$, $\mathrm{Ehr}(Σ')(x_ℓ) = 1$. It remains to check that “vertical” linear form $l$, which is the new coordinate, verifies $\mathrm{Ehr}(Σ')(x_l) = 1$. Two cases.

• If $\mathrm{div}(f)$ is trivial, then $x_l = π^*(-x_f)$ and we get the result,
• otherwise, $Σ'$ has a new ray $ρ$, and $x_l = x_ρ + π^*(-x_f)$. Hence $$\begin{align} \mathrm{Ehr}(Σ')(x_l) &= \mathrm{Ehr}(Σ')(π^*(-x_f) + x_ρ) \\ &= \mathrm{Ehr}(Σ')(π^*(-x_f)) + \mathrm{Ehr}(Σ'^ρ)(x_l) \\ &= \mathrm{Ehr}(Σ)(-x_f) + 1. \end{align}$$

which concludes the proof.

Bergman fans are Ehrhart

Theorem 1 (Complete unimodular fans are Ehrhart)

Complete unimodular fans are Ehrhart.

Proof

The point is Ehrhart. By Example 1 (One dimensional Ehrhart fans), the line is Ehrhart. The results follows from Proposition 3 (Stability by product), Proposition 4 (Stability by blow-up) and the weak factorization theorem.

Theorem 6 (Bergman fans are Ehrhart)

Any unimodular generalized Bergman fan is Ehrhart.

Proof

Using Theorem 1 (Complete unimodular fans are Ehrhart), Proposition 4 (Stability by blow-up), and Proposition 5 (Tropical modifications and Ehrhart fans), we can reduce the theorem to the following lemma.

Lemma 3

Let $𝔐$ be a matroid without loop which is not free. Take an element $a$ of $𝔐$ which is not free, and another element $o$. Recall that $Σ_𝔐$ is the tropical modification of $Σ_{𝔐-a}$ along the map $f$ such that, for any flat $F$ of $𝔐-a$: $$f(e_F) = \begin{cases} 1 & \text{if $o ∈ \mathrm{cl}_𝔐(F)$ and $a ∉ \mathrm{cl}_𝔐(F)$,} \\ -1 & \text{if $a ∈ \mathrm{cl}_𝔐(F)$ and $o ∉ \mathrm{cl}_𝔐(F)$,} \\ 0 & \text{otherwise.} \end{cases}$$ Then $\mathrm{Ehr}(Σ_{𝔐-a})(-x_f) = 0$.

Proof

We work by induction on the rank of the matroid. Put an order on proper nontrivial flats of $𝔐-a$ which is non-decreasing for the rank : $F_1, \dots, F_r$, with $F_1 = \{o\}$. Assume by induction that we have proven that for some $m ≥ 2$, $$\mathrm{Ehr}(Σ_{𝔐-a})(-∑_{i=1}^{m-1}f(e_F_k)x_F_k) = 0$$ This is clear for $m=2$. Denote the argument by $z$. Let us prove that $\mathrm{Ehr}(Σ_{𝔐-a})(z - f(e_F_m)x_F_m)=0$.

• If $f(e_F_m) = 0$, this is trivial.
• If $o ∈ \mathrm{cl}_𝔐(F)$ and $a ∉ \mathrm{cl}_𝔐(F)$, then $$\begin{align*} \mathrm{Ehr}(Σ_{𝔐-a})(z - x_F_m) &= -\mathrm{Ehr}(Σ^{ρ_F_m}_{𝔐-a})(z) \\ &= -\mathrm{Ehr}(Σ_{(𝔐-a)|_F} × Σ_{(𝔐-a)/F})(z) \\ &= -\mathrm{Ehr}(Σ_{(𝔐-a)|_F})(π_1(z)) ⋅ \mathrm{Ehr}(Σ_{(𝔐-a)/F})(π_2(z)). \end{align*}$$ By induction, the first factor is zero. Indeed, it correspond to the case of the matroid $𝔐|_F = 𝔐'-b$ where $𝔐'$ of ground set $F+\{b\}$ is such that the only flat of $𝔐|_F$ whose closure in $𝔐'$ contains $b$ is $F$ itself.
• The last case is $a ∈ \mathrm{cl}_𝔐(F)$ and $o ∉ \mathrm{cl}_𝔐(F)$. Let $o'$ be another element of $F$ and let $ℓ$ be the linear form which is 1 on $e_F_{\{o'\}}$, -1 on $e_F_{\{a\}}$ and 0 on $e_F_{\{b\}}$ for other elements $b$ of $𝔐-a$. Then, $$\begin{align*} \mathrm{Ehr}(Σ_{𝔐-a})(z + x_F_m) &= \mathrm{Ehr}(Σ^{ρ_F_m}_{𝔐-a})(z + x_F_m) \\ &= \mathrm{Ehr}(Σ^{ρ_F_m}_{𝔐-a})(z + x_F_m + x_ℓ) \\ &= \mathrm{Ehr}(Σ_{(𝔐-a)|_F})(π_1(z + x_F_m + x_ℓ)) ⋅ \mathrm{Ehr}(Σ_{(𝔐-a)/F})(π_2(z + x_F_m + x_ℓ)). \end{align*}$$ Once again, the first factor is zero, since it corresponds to the current lemma applied to $𝔐|_{F + a}$ with elements $a$ and $o'$ instead of $a$ and $o$.

This concludes the proof.

Counter-examples

A counter-example

Take $Σ$ the pseudo fan which is the double cover of $Σℙ^2$ with six rays $ρ_1,ρ_2, ρ_0=-ρ_1-ρ_2$ and their copies $ρ_4,ρ_5,ρ_3$ and six 2-cones $ρ_i + ρ_{i+1}$ for $i ∈ ℤ/6ℤ$. It is Ehrhart. Take the piecewise-linear function $f$ with $$f(ρ_0) = -1, \qquad f(ρ_5) = f(ρ_1) = 1, \qquad f(ρ_2) = f(ρ_3) = f(ρ_4) = 0.$$ Then $\mathrm{div}(f)$ is just the tropical line and is Ehrhart, but $\mathrm{Ehr}(Σ)(-x_f) = -1 ≠ 0$.

Hence, $\mathrm{Trop}_f(Σ)$ is just pre-Ehrhart. It is not a fan but still a pseudo-fan, but one can preform some blow-ups and a new tropical modification along a tropical line to get a classical fan. This fan cannot be Ehrhart because of the last equality of Proposition 5 (Tropical modifications and Ehrhart fans). This fan is the counter-example studied by Babaee and Huh.

Simpler counter-examples

It should be easy to find simpler counter-example. We just have to find a Ehrhart fan, for instance $Σℙ^2$, and a reduced divisor whose associated function $f$ verifies $\mathrm{Ehr}(Σ)(-x_f) ≠ 0$.

And indeed, take $Σ = Σ_{U_{3,3}}$ with the conewise-linear function $f$ which is 1 on each unit vector. Then its divisor is the one-skeleton of $Σ$ and $\mathrm{Ehr}(Σ)(f) = -1$. Hence, its tropical modification is trivial.

Actually, take any integral polytope which has a nonzero number of interior points and whose facets are all dimension one for the lattice volume (a unimodular simplex has volume 1). Take its normal fan, and subdivide it to get a unimodular fan. Take the height function $f$ of the polytope. Take the tropical modification of your fan along $f$. Then you get a balanced fan which is not Ehrhart.

Extensions

Weighted fans

For weighted fans, and in order to preserve Example 1 (One dimensional Ehrhart fans), we could adapt Definition 1 (Ehrhart fan) asking for $f(0)$ on a point to be the weight of the point. Not sure what to put for higher dimensional fans.

Non unimodular

For rational fans that are not unimodular, the main point which does not obviously extends is Proposition 4 (Stability by blow-up). Actually, the proof should work since one can perform unimodular blow-ups until reaching a unimodular fan.