1. Preliminaries

The paper is set in $\mathbb{R}^d$ with the standard Euclidean topology. Particularly, we denote the Euclidean norm of a vector $v\in \mathbb{R}^d$ by

\begin{equation*}\|v\|:=\sqrt {v\cdot v},\end{equation*}

where $\cdot :\mathbb{R}^d\times \mathbb{R}^d\rightarrow \mathbb{R}$ denotes the dot product in $\mathbb{R}^d$ . By a convex body $K$ in the Euclidean space $\mathbb{R}^d$ , we refer to a non-empty compact convex set. Particularly, the convex bodies we consider do not need to have non-empty interior and thus can be less than full-dimensional. When we talk about the dimension of a convex body, we refer to the dimension of the smallest affine space containing the convex body.

We fix some notation and terminology concerning convex bodies and their volume polynomials. As a general reference, we suggest the monograph of Schneider [Reference Schneider18, Chapter 1]. Let $n$ and $d$ be positive integers. We write $[n]:=\{1,\ldots ,n\}$ and

\begin{equation*}\Delta _n^d:=\{\alpha \in \mathbb{N}_0^n\,\mid \,\sum _{i=1}^n\alpha _i=d\}\end{equation*}

for the $d$ -th discrete simplex. The space of homogeneous polynomials of degree $d$ in $n$ variables over $\mathbb{R}$ is denoted by $H_n^d$ . For a polynomial $f\in H_n^d$ , we denote its degree in the variable $x_i$ by $\deg _i(f)$ .

Let $\mathcal{K}:=(K_1,\ldots ,K_n)$ be convex bodies in $\mathbb{R}^d$ . Their volume polynomial is the homogeneous polynomial

\begin{equation*}\mathrm{vol}_{\mathcal{K}}(x):=\mathrm{vol}(x_1K_1+\cdots +x_nK_n):=\sum _{\alpha \in \Delta _n^d}\frac {d!}{\alpha !}V_\alpha (\mathcal{K})x^\alpha \end{equation*}

for non-negative $x_1,\ldots ,x_n$ and $x:=(x_1,\ldots ,x_n)$ and we call $V_\alpha (\mathcal{K})$ for $\alpha \in \Delta _n^d$ the mixed volumes of $\mathcal{K}$ . For a multi-index $\alpha \in \Delta _n^d$ we write

\begin{equation*}\alpha !:=\alpha _1!\cdot \ldots \cdot \alpha _n! \hspace {20px} \text{and} \hspace {20px} x^\alpha := x_1^{\alpha _1}\cdot \ldots \cdot x_n^{\alpha _n}.\end{equation*}

We generally assume $\mathcal{K}$ to be full-dimensional, meaning that the affine dimension of the convex body

\begin{equation*}\sum \mathcal{K}:=\sum _{i=1}^nK_i\end{equation*}

equals $d$ . This guarantees that the volume polynomial is non-zero. We further use the notation

\begin{equation*}V_\alpha (\mathcal{K})=V(\mathcal{K}^\alpha )=V(K_1^{\alpha _1},\ldots ,K_n^{\alpha _n})\end{equation*}

to refer to the mixed volume of the convex bodies $\mathcal{K}=(K_1,\ldots ,K_n)$ . The set of all volume polynomials is denoted by $V_n^d$ .

The mixed volumes satisfy several useful properties of which we will only list a few here. For a more thorough understanding, we refer to the monograph of Schneider [Reference Schneider18, Chapter 5].

Due to part (b) of the above Proposition1.1, we may always assume that the considered convex bodies contain the origin, which often times simplifies our settings and calculation. For a $k$ -dimensional linear subspace $E\subset \mathbb{R}^d$ , we denote by $\mathrm{vol}_E$ (resp. $V_E$ ) the volume (resp. the mixed volume) in the space $E$ . We omit the subspace in the notation if it can be deduced from the context. By $K|E$ , we denote the orthogonal projection of a convex body $K\subset \mathbb{R}^d$ onto the space $E$ . If some of the convex bodies we are considering lie in a common linear subspace, we can use this to represent the mixed volume in $\mathbb{R}^d$ as a product of the mixed volumes in the smaller subspace and its orthogonal complement.

Proposition 1.2. [Reference Schneider18, Theorem 5.3.1] Let $E$ be a $k$ -dimensional linear subspace of $\mathbb{R}^d$ and let $L_1,\ldots ,L_k\subset E$ as well as $K_1,\ldots ,K_{d-k}\subset \mathbb{R}^d$ be convex bodies. We have

\begin{equation*}\binom {d}{k}V(L_1,\ldots ,L_k,K_1,\ldots ,K_{d-k})=V_E(L_1,\ldots ,L_k)V_{E^\perp }(K_1| E^\perp ,\ldots ,K_{d-k}| E^\perp ),\end{equation*}

where $E^\perp$ refers to the orthogonal space of $E$ .

The mixed volumes satisfy several useful inequalities, the most famous being the Alexandrov–Fenchel inequality (see [Reference Alexandrov1, Reference Fenchel10])

\begin{equation*}V(K_1,\ldots ,K_n)^2\geq V(K_1^2,K_3,\ldots ,K_n)V(K_2^2,K_3,\ldots ,K_n).\end{equation*}

A generalisation of polynomials with coefficients satisfying this inequality leads us to the set of Lorentzian polynomials.

Definition 1.3. (see [Reference Brändén and Huh7]) A subset $J\subseteq \mathbb{N}^n$ is called M-convex if for any $\alpha , \beta \in J$ and any index $i\in [n]$ with $\alpha _i\gt \beta _i$ , there exists an index $j\in J$ with $\alpha _j\lt \beta _j$ and $\alpha -e_i+e_j,\,\beta -e_j+e_i\in J$ . We denote by $M_n^d$ the set of all polynomials in $H_n^d$ with non-negative coefficients and M-convex support. We further define the set of Lorentzian polynomials as $L_n^1:= M_n^1$ and for $d\geq 2$ as

\begin{equation*}L_n^d:=\{f\in M_n^d\,\mid \,\text{for all }\alpha \in \Delta _n^{d-2}:\,\mathcal{H}_{\partial ^\alpha f}\,\text{has at most one positive eigenvalue}\},\end{equation*}

where $\mathcal{H}_f$ refers to the Hessian of a polynomial $f\in H_n^d$ .

The conditions for the Hessian matrices lead to the fact that the coefficients of Lorentzian polynomials always satisfy the Alexandrov-Fenchel inequality. In fact, we have the inclusion $V_n^d\\ \subseteq L_n^d$ .

As Lorentzian polynomials have been thoroughly studied, especially concerning operations that preserve their properties, we will focus on some basic notions here and refer the reader to the work of Brändén and Huh [Reference Brändén and Huh7] for a broader understanding.

Both of these properties can be transferred to volume polynomials. To do so in the case of products of volume polynomials, it is enough to first consider two polynomials in distinct variables in $V_{n_1}^{d_1}$ (resp. in $V_{n_2}^{d_2}$ ) and then embed the associated convex bodies in $\mathbb{R}^{d_1}$ (resp. $\mathbb{R}^{d_2}$ ) in the Euclidean space $\mathbb{R}^{d_1+d_2}$ . The volume polynomial of the resulting convex bodies is exactly the product of the two polynomials. The general case follows immediately from an appropriate substitution of the variables.

Generally, the set of Lorentzian polynomials allows more operations that preserve it than the set of volume polynomials. One such operation is the derivative: Whereas the definition of Lorentzian polynomials immediately shows that the derivative of any polynomial $f\in L_n^d$ is again Lorentzian, the same is not true for volume polynomials. This can be seen by considering the Lorentzian polynomial

\begin{equation*}f:=\frac {1}{2}x_1^2x_2+\frac {1}{2}x_1^2x_3+\frac {1}{2}x_1^2x_4+2x_1x_2x_3+2x_1x_2x_4+\frac {1}{2}x_1x_3x_4+x_2x_3x_4\in L_4^3,\end{equation*}

which is the volume polynomial of the four convex bodies

\begin{align*} K_1&:={\text{conv}}(0,e_1,e_2),\quad K_2:={\text{conv}}(0,e_3), \\ K_3&:={\text{conv}}(0,2e_1+e_3) \quad \text{and}\quad K_4:={\text{conv}}\left (0,\left (\begin{matrix} \dfrac{3}{2}\\[8pt] \dfrac{1}{2}\\[1pt] 1 \end{matrix}\right )\right ) \end{align*}

in $\mathbb{R}^3$ . The derivative

\begin{equation*}\partial _1f=x_1x_2+x_1x_3+x_1x_4+2x_2x_3+2x_2x_4+\frac {1}{2}x_3x_4\end{equation*}

on the other hand cannot be a volume polynomial due to the findings of Heine (see [Reference Heine13, p. 119]). In Section 3, we will go into more depth as to why the polynomial $\partial _1 f$ cannot be represented as the volume polynomial of any four convex bodies in $\mathbb{R}^2$ .

The derivative is just one example of such an operation. Another one is provided by the next Proposition1.7.

Proposition 1.7. [Reference Brändén, Leake and Pak8, Lemma 4.4] Let $f\in L_n^d$ be a Lorentzian polynomial and let us write

\begin{equation*}f(x_1,\ldots ,x_n)=\sum _{i=0}^dx_n^{d-i}f_i(x_1,\ldots ,x_{n-1}).\end{equation*}

Then $f_i$ is a Lorentzian polynomial of degree $i$ for every $i\in [d]$ .

In contrast to Proposition1.5 and Remark1.6, an easy example shows that Proposition1.7 is not necessarily true for volume polynomials. Let us consider the polynomial

\begin{align*} f:=\,&x_5^3+x_5^2(x_1+x_2+x_3+\textstyle \frac {3}{2}x_4)+x_5(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4) \\ &+x_1x_2x_3+\textstyle \frac {1}{2}x_1x_2x_4+\frac {1}{2}x_1x_3x_4+\frac {1}{2}x_2x_3x_4, \end{align*}

which is the volume polynomial of the convex bodies

\begin{align*} K_1&:={\text{conv}}(0,e_1), \\ K_2&:={\text{conv}}(0,e_2), \\ K_3&:={\text{conv}}(0,e_3), \\ K_4&:={\text{conv}}(0,\frac {1}{2}(e_1+e_2+e_3)), \end{align*}

and the unit cube

\begin{align*} K_5&:={\text{conv}}(0,e_1,e_2,e_3,e_1+e_2,e_1+e_3,e_2+e_3,e_1+e_2+e_3). \end{align*}

Just as above, according to Heine (see [Reference Heine13, p. 119]), the elementary symmetric polynomial in four variables of degree two, and thus $f_2$ , cannot be represented as a volume polynomial.

2. Operations

We have noted that the product of Lorentzian (resp. volume) polynomials is again a Lorentzian (resp. volume) polynomial. Generally, the factors of Lorentzian or volume polynomials do not have to be either. For example, the polynomial

\begin{equation*}f:= x^3+3x^2y+3xy^2=x(x^2+3xy+3y^2)\end{equation*}

is Lorentzian as the Hessian matrices of $\partial _x f$ and $\partial _y f$ have exactly one positive eigenvalue. Because of the fact that $f$ is bivariate, it is due to Shephard [Reference Shephard19, Theorem 4] that $f$ is a volume polynomial. On the other hand, the factor $x^2+3xy+3y^2$ is not Lorentzian due to its Hessian matrix having two positive eigenvalues and thus, it cannot be a volume polynomial (Theorem1.4).

Nevertheless, there are certain cases where the factors of Lorentzian (resp. volume) polynomials are Lorentzian (resp. volume) polynomials.

Proof. Let $f:= gh\in L_{n_1+n_2}^{d_1+d_2}$ be a Lorentzian polynomial with $g$ and $h$ having distinct variables. We define the $(n_1+n_2)\times (n_2+1)$ -matrix

\begin{equation*}A:=\left(\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} \frac {1}{|g|_1} & 0 &\ldots & \ldots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ \frac {1}{|g|_1} & 0 &\ldots & \ldots & 0 \\ 0 & 1 & 0 & \ldots & 0 \\ \vdots & 0 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & 1 & 0 \\ 0 & 0& \ldots & 0 & 1 \end{array}\right),\end{equation*}

where $|g|_1$ defines the sum of the (non-negative) coefficients of $g$ . Taking this matrix we know that the polynomial

\begin{equation*}f\left (A\begin{pmatrix} x \\ y_1 \\ \vdots \\ y_{n_2} \end{pmatrix}\right )=x^{d_1}h\end{equation*}

is Lorentzian by Proposition1.5. Essentially, we plug in the variable $x$ for any of the $n_1$ variables that appear in the polynomial $g$ , while simultaneously scaling the polynomial. Now, as the derivative of any Lorentzian polynomial is again Lorentzian, the polynomial $h$ , which equals exactly the derivative

\begin{equation*}\partial _x^{d_1} x^{d_1}h=d_1!h\end{equation*}

of the above polynomial up to scaling, is a Lorentzian polynomial and analogously, the same is true for $g$ .

The next proposition shows that we can skip the restriction of distinct variables if we restrict the degree of the (only) common variable of the two factors.

Proof. As the derivative of any Lorentzian polynomial is again Lorentzian, we can gather that the polynomial

\begin{equation*}\partial _1^{d_1}f=d_1!(1+d_1x_1\partial _1)g=d_1!g((1+d_1)x_1,x_2,\ldots ,x_n)\end{equation*}

is a Lorentzian polynomial. Proposition1.5 then allows us to conclude $g\in L_n^{d_2}$ by scaling the variable $x_1$ .

Using the same technique of transforming the variables as in the proof of Proposition2.1, we can deduce the following corollary.

Proof. Let the polynomial $g$ be in the variables $x_1,\ldots ,x_{n_1}$ and $h$ be in the variables $x_1,y_2,\ldots ,y_{n_2}$ . We set $x_1=\ldots =x_{n_1}$ in $f$ and scale the resulting polynomial by $\frac {1}{|g|_1}$ which preserves the Lorentzian property by Proposition1.5. The resulting polynomial is of the form

\begin{equation*}x_1^{d_1}h\end{equation*}

with $\deg _1(h)\leq 1$ . Thus, it satisfies the conditions of Proposition2.2 and $h$ is Lorentzian.

In order to conclude similar results for volume polynomials, we need to use different techniques as the derivative of a volume polynomial is generally not a volume polynomial anymore, which renders most of the previously used techniques useless for volume polynomials. Instead, we can use the geometric aspects of the given convex bodies to transfer the results to volume polynomials.

Proof. We proceed similar to the the case of Lorentzian polynomials and first assume that $f:= x_{n+1}^{d_1}g\in V^{d_1+d_2}_{n+1}$ is a volume polynomial with $g\in H_n^{d_2}$ being a polynomial in the variables $x_1,\ldots ,x_n$ . Let $\mathcal{K}:= (K_1,\ldots ,K_{n+1})$ be the convex bodies in $\mathbb{R}^{d_1+d_2}$ associated to the corresponding variables $x_1,\ldots ,x_{n+1}$ . As we have $\deg _{n+1}(f)=d_1$ , the convex body $K_{n+1}$ must lie in a $d_1$ -dimensional linear subspace $E\subseteq \mathbb{R}^{d_1+d_2}$ due to Proposition1.1(d). Due to the factorisation of $f$ , it must be

\begin{equation*}V(K_1^{\alpha _1},\ldots ,K_n^{\alpha _n},K_{n+1}^{d_1-1})=0\end{equation*}

for all $\alpha \in \Delta _n^{d_2+1}$ . With Proposition1.1(c) this allows us to assume $K_1,\ldots ,K_n\subseteq E^\perp$ as it means, that we cannot expand any system of $d_1-1$ linearly independent vectors from $E$ with $d_2+1$ linearly independent vectors coming from line segments in $K_1,\ldots ,K_n$ to get a basis of $\mathbb{R}^{d_1+d_2}$ . Now we take $\alpha \in \Delta _n^{d_2}$ and we have

\begin{align*} V_{(\alpha ,d_1)}(K)&=\binom {d_1+d_2}{d_1}^{-1}V(K_{n+1}^{d_1})V(K_1^{\alpha _1},\ldots ,K_n^{\alpha _n}) \\ &=\frac {d_1!d_2!}{(d_1+d_2)!}V(K_{n+1}^{d_1})V(K_1^{\alpha _1},\ldots ,K_n^{\alpha _n}) \end{align*}

by Proposition1.2. For the volume polynomial $f$ , this leads to

\begin{align*} f&=\sum _{\alpha \in \Delta _n^{d_2}}\frac {(d_1+d_2)!}{\alpha !d_1!}V_{(\alpha ,d_1)}(\mathcal{K})x^\alpha x_{n+1}^{d_1} \\ &=x_{n+1}^{d_1}\sum _{\alpha \in \Delta _n^{d_2}}\frac {d_2!}{\alpha !} \mathrm{vol}_{d_1}(K_{n+1})V(K_1^{\alpha _1},\ldots ,K_n^{\alpha _n})x^\alpha \\ &=x_{n+1}^{d_1} \mathrm{vol}_{d_1}(K_{n+1})\mathrm{vol}(x_1K_1+\cdots +x_nK_n), \end{align*}

where $\mathrm{vol}_{d_1}$ refers to the $d_1$ -dimensional volume in the subspace $E$ of $\mathbb{R}^{d_1+d_2}$ . Hence, $g$ is a volume polynomial.

For the general case, we use a transformation of the variables as before and obtain the result.

Proof. Let $f$ be the volume polynomial of the convex bodies $\mathcal{K}:=(K_1,\ldots ,K_n)$ in $\mathbb{R}^{d_1+d_2}$ , so that we have

\begin{align*} f=&\sum _{\alpha \in \Delta _n^{d_1+d_2}}\frac {(d_1+d_2)!}{\alpha !}V_\alpha (\mathcal{K})x^\alpha \\ =&\sum _{\substack {\alpha \in \Delta _n^{d_1+d_2}\\\alpha _1=d_1}}\frac {(d_1+d_2)!}{\alpha !}V_\alpha (\mathcal{K})x^\alpha +\sum _{\substack {\alpha \in \Delta _n^{d_1+d_2}\\\alpha _1=d_1+1}}\frac {(d_1+d_2)!}{\alpha !}V_\alpha (\mathcal{K})x^\alpha \\ =&x_1^{d_1}\left (\sum _{\substack {\alpha \in \Delta _n^{d_2}\\\alpha _1=0}}\frac {(d_1+d_2)!}{(\alpha +d_1e_1)!}V_{\alpha +d_1e_1}(\mathcal{K})x^\alpha +\sum _{\substack {\alpha \in \Delta _n^{d_2}\\\alpha _1=1}}\frac {(d_1+d_2)!}{(\alpha +d_1e_1)!}V_{\alpha +d_1e_1}(\mathcal{K})x^\alpha \right ) \end{align*}

If we have $\deg _1(g)=0$ , the statement follows immediately from Proposition2.6. Thus, we only consider the case that $\deg _1(g)=1$ . With Proposition1.1(d), we gather $\dim (K_1)=d_1+1$ and due to Proposition1.1(c), we can assume $K_2,\ldots ,K_n\subseteq V$ for a $d_2$ -dimensional linear subspace $V\subseteq \mathbb{R}^{d_1+d_2}$ . We denote by $U_1$ the $(d_1+1)$ -dimensional linear subspace containing $K_1$ and get $U_1\cap V=\mathbb{R}v$ for a vector $v\in \mathbb{R}^{d_1+d_2}$ . We can now write $U_1=U+\mathbb{R}v$ for a $d_1$ -dimensional subspace $U\subseteq \mathbb{R}^{d_1+d_2}$ and by a change of basis and Proposition1.1(b), we assume $U=V^\perp$ without loss of generality, particularly $v\in U^\perp$ . We write $C_1:= K_1|U$ and chose the length of $v$ such that we get

\begin{equation*}\mathrm{vol}_{d_1+1}(K_1)=\mathrm{vol}_{d_1+1}(C_1+{\text{conv}}(0,v))=\|v\|\mathrm{vol}_{d_1}(C_1).\end{equation*}

For an $\alpha \in \Delta _n^{d_2}$ with $\alpha _1=0$ , we have $\alpha +d_1e_1\in \Delta _n^{d_1+d_2}$ and with Proposition1.2, we can rewrite the mixed volume

\begin{equation*}\frac {(d_1+d_2)!}{d_1!\alpha !}V(K_1^{d_1},K_2^{\alpha _2},\ldots ,K_n^{\alpha _n})=\frac {d_2!}{\alpha !}V_U(C_1)V_V(K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}). \end{equation*}

On the other hand, for an $\alpha \in \Delta _n^{d_2}$ with $\alpha _1=1$ , we get with Proposition1.2

\begin{align*} &\frac {(d_1+d_2)!}{(d_1+1)!\hat {\alpha }!}V(K_1^{d_1+1},K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}) \\ =&\frac {(d_2-1)!}{\hat {\alpha }!}\mathrm{vol}_{d_1+1}(K_1)V_{U_1^\perp }\left ((K_2\mid U_1^\perp )^{\alpha _2},\ldots ,(K_n\mid U_1^\perp )^{\alpha _n}\right ), \end{align*}

where $\hat {\alpha }$ refers to $(\alpha _2,\ldots ,\alpha _n)$ . Our choice of $v$ allows us to further rewrite the above mixed volume, so that we gather

\begin{align*} &\frac {(d_2-1)!}{\hat {\alpha }!}\mathrm{vol}_{d_1+1}(K_1)V_{U_1^\perp }\left ((K_2\mid U_1^\perp )^{\alpha _2},\ldots ,(K_n\mid U_1^\perp )^{\alpha _n}\right ) \\ =&\frac {(d_2-1)!}{\hat {\alpha }!}\mathrm{vol}_{d_1+1}(C_1+{\text{conv}}(0,v))V_{U_1^\perp }\left ((K_2\mid U_1^\perp )^{\alpha _2},\ldots ,(K_n\mid U_1^\perp )^{\alpha _n}\right ) \\ =&\frac {(d_1+d_2)!}{(d_1+1)!\hat {\alpha }!}V((C_1+{\text{conv}}(0,v))^{d_1+1},K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}). \end{align*}

By multilinearity in the first argument due to Proposition1.1(a), we can split the convex body $C_1+{\text{conv}}(0,v)$ , which allows us to then use Proposition1.1(c) and finally Proposition1.2 to conclude

\begin{align*} &\frac {(d_1+d_2)!}{(d_1+1)!\hat {\alpha }!}V((C_1+{\text{conv}}(0,v))^{d_1+1},K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}) \\ =&\frac {(d_1+d_2)!}{(d_1+1)!\hat {\alpha }!}\sum _{i=0}^{d_1+1}\binom {d_1+1}{i}V(C_1^{d_1+1-i},{\text{conv}}(0,v)^i,K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}) \\ =&\frac {d_2!}{\hat {\alpha }!}\binom {d_1+d_2}{d_1}V(C_1^{d_1},{\text{conv}}(0,v),K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}) \\ = &\frac {d_2!}{\alpha !}V_U(C_1)V_V({\text{conv}}(0,v),K_2^{\alpha _2},\ldots ,K_n^{\alpha _n}). \end{align*}

Inserting both cases into our polynomial $f$ , we get

\begin{equation*}f=x_1^{d_1}\mathrm{vol}_{d_1}(C_1)\mathrm{vol}_{d_2}(x_1{\text{conv}}(0,v)+x_2K_2+\ldots +x_nK_n)\end{equation*}

and thus, $g$ is a volume polynomial.

As in the case of Corollary2.3 for Lorentzian polynomials, we can now deduce the following Corollary by referring to Remark1.6 and then using the above Proposition2.6.

The above results illustrate how we can use our knowledge of Lorentzian polynomials to obtain new information on volume polynomials as we have first studied the factors of Lorentzian polynomials and then transferred this knowledge to volume polynomials appropriately. But as mentioned before, the vast majority of results and operations for Lorentzian polynomials are not transferable to volume polynomials. Instead, we often need further restrictions or some adjusting of the results to be able to transfer the operations preserving the Lorentzian property to volume polynomials. We have seen one such example of a non-transferable result in Proposition1.7. As we explicitly do not require the convex bodies to have non-empty interior, we can transfer at least parts of Proposition1.7 to volume polynomials.

Proposition 2.9. Let $f\in V_n^d$ be the volume polynomial of $n$ convex bodies $\mathcal{K}:=(K_1,\ldots ,K_n)$ in $\mathbb{R}^d$ and let us write

\begin{equation*}f(x_1,\ldots ,x_n)=\sum _{i=0}^dx_n^{d-i}f_i(x_1,\ldots ,x_{n-1}).\end{equation*}

Then $f_d$ is a volume polynomial of degree $d$ and $f_{d-m}$ for $m:=\dim (K_n)$ is a volume polynomial of degree $d-m$ .

Proof. We have $f_d=f(x_1,\ldots ,x_{n-1},0)=\mathrm{vol}(x_1K_1+\ldots +x_{n-1}K_{n-1})$ . Let $U\subseteq \mathbb{R}^d$ be the $m$ -dimensional linear subspace with $K_n\subseteq U$ . We have

\begin{align*} f_{d-m}&=\sum _{\alpha \in \Delta _{n-1}^{d-m}}\frac {d!}{m!\alpha !}V(\mathcal{K}^\alpha ,K_n^m)x^\alpha \\ &=\sum _{\alpha \in \Delta _{n-1}^{d-m}}\frac {(d-m)!}{\alpha !}V_U(K_n^m)V_{U^\perp }((K|U^\perp )^\alpha )x^\alpha \\ &=\mathrm{vol}_m(K_1)\mathrm{vol}(x_1(K_1| U^\perp )+\ldots +x_{n-1}(K_{n-1}| U^\perp )). \end{align*}

3. Volume polnomials as a subset of Lorentzian polynomials

The Alexandrov-Fenchel inequality (see [Reference Alexandrov1, Reference Fenchel10]), being the first major restriction for sequences that can be realised as a sequence of coefficients of a volume polynomial, started a long line of further inequalities that can be deduced from it. The set of homogeneous polynomials with coefficients satisfying these inequalities contains the set of Lorentzian polynomials ([Reference Gurvits11, Example 1.2(3)] and [Reference Brändén and Huh7, Proposition 4.4]) which allows us to solely focus on this smaller set as Brändén and Huh found that every volume polynomial is Lorentzian [Reference Brändén and Huh7, Theorem 4.1].

We denote by $AF_n^d$ the set of homogeneous polynomials in $n$ variables of degree $d$ with non-negative coefficients satisfying the Alexandrov-Fenchel inequality as well as the resulting inequalities [Reference Shephard19, p. 132]

\begin{equation*}V(K^\alpha ,K_i^{r-1},K_j)V(K^\alpha ,K_i,K_j^{r-1})\geq V(K^\alpha ,K_i^r)V(K^\alpha ,K_j^r)\end{equation*}

for $\alpha \in \Delta _n^{d-r}$ and

\begin{equation*}(-1)^r\det \left (\left (V(K^\beta ,K_i,K_j)\right )_{i,j\in [r]}\right )\leq 0\end{equation*}

for $\beta \in \Delta _n^{d-2}$ and $r\leq n$ . Considering the polynomial

\begin{equation*}g:= c_{111}x_1^3+3c_{223}x_2^2x_3+3c_{233}x_2x_3^2\end{equation*}

with $c_{111},c_{223},c_{233}\gt 0$ which lies in $AF_3^3$ but not in $L_3^3$ , one sees that focusing on the set $L_n^d$ instead of $AF_n^d$ already reduces the number of polynomials due to the additional condition of the $M$ -convexity of the support of polynomials in $L_n^d$ . Thus going forward, we regard the set $V_n^d$ as a subset of $L_n^d$ instead of a subset of the bigger set $AF_n^d$ .

Shephard proved [Reference Shephard19, Theorem 4] that for any degree $d\in \mathbb{N}$ , we have

\begin{equation*}V_2^d=L_2^d\end{equation*}

and he further proved [Reference Shephard19, Theorem 5] that for $(d+2)$ -many variables, the inclusion

\begin{equation*}V_{d+2}^d\subsetneq L_{d+2}^d\end{equation*}

is strict. This generalised a result of Heine [Reference Heine13, p. 119] for polynomials in four variables and of degree two. To illustrate the idea behind the proof, we will mention Heine’s example here.

Example 3.1. [Reference Heine13, p. 119] The elementary symmetric polynomial in four variables of degree two

\begin{equation*}f:= x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\end{equation*}

is Lorentzian as can be seen straight forwardly by computing the Hessian matrix. If it were the volume polynomial of convex bodies $K_1,\ldots ,K_4\subseteq \mathbb{R}^2$ , these would have to be line segments by Remark 1.1. Without loss of generality, we assume $K_i={\text{conv}}(0,e_i)$ for $i=1,2$ and $K_3={\text{conv}}(0,a)$ , $K_4={\text{conv}}(0,b)$ for $a,b\in \mathbb{R}^2$ . Computing the mixed volumes of these convex bodies leads to

\begin{equation*} 1=\pm a_i = \pm b_i=\pm (a_1b_2-a_2b_1)\end{equation*}

for $i=1,2$ and thus to a contradiction.

This example also illustrates why it is often useful to first refer to multiaffine polynomials as they allow an easy computation of the mixed volumes, which would otherwise be more difficult (see [Reference Bárány and Füredi6, Reference Dyer, Gritzmann and Hufnagel9]).

In the case of three variables, Heine [Reference Heine13, p. 118] proved

\begin{equation*}V_3^2=L_3^2.\end{equation*}

Later, Gurvits [Reference Gurvits11, Conjecture 5.1] conjectured that this might be true for all degrees. This was disproved by Brändén and Huh ( [Reference Brändén and Huh7, Footnote 15] and [Reference Huh14, Example 14]), who constructed the Lorentzian polynomial

\begin{equation*}f:= 14x_1^3+6x_1^2x_2+24x_1^2x_3+12x_1x_2x_3+6x_1x_3^2+3x_2x_3^2,\end{equation*}

which cannot be a volume polynomial as the coefficients do not satisfy the reverse Khovanskii-Teissier inequality [Reference Lehmann and Xiao15, Theorem 5.7]. This inequality states that for three convex bodies $K_1,K_2,K_3$ in $\mathbb{R}^d$ , the mixed volumes satisfy

\begin{equation*}\binom {d}{k}V(K_1^{d-k},K_2^k)V(K_1^k,K_3^{d-k})\geq V(K_1^d)V(K_2^k,K_3^{d-k})\end{equation*}

for all non-negative integers $k\leq d$ .

To give some geometric motivation for the inequality, we assume that we have three convex bodies $K_1,K_2,K_3\subseteq \mathbb{R}^d$ with $\dim (K_2)=k\leq d$ and $\dim (K_3)=d-k$ . Let $U\subseteq \mathbb{R}^d$ be the $k$ -dimensional linear subspace with $K_2\subseteq U$ . As the inequality is trivial when the right hand side equals zero, we can assume $K_3\subseteq U^\perp$ due to Proposition1.1(c). With this in mind, Proposition1.2 leads us to

\begin{align*} \binom {d}{k}V(K_2^k,K_3^{d-k})&=\mathrm{vol}_k(K_2)\mathrm{vol}_{d-k}(K_3), \\ \binom {d}{k}V(K_1^{d-k},K_2^k)&=\mathrm{vol}_{d-k}(K_1| U^\perp )\mathrm{vol}_k(K_2), \\ \binom {d}{k}V(K_1^k,K_3^{d-k})&=\mathrm{vol}_k(K_1| U)\mathrm{vol}_{d-k}(K_3). \end{align*}

By approximating the volume of $K_1$ , we get

\begin{align*} V(K_1^d)&\leq \mathrm{vol}_k(K_1| U)\mathrm{vol}_{d-k}(K_1|U^\perp ) \\ &\leq \binom {d}{k}\frac {V(K_1^k,K_3^{d-k})V(K_1^{d-k},K_2^k)}{V(K_2^k,K_3^{d-k})}. \end{align*}

As was communicated by Ivan Soprunov, this shows that the above example of a polynomial in $L_3^3\setminus V_3^3$ by Brändén and Huh cannot be a volume polynomial (without using Hodge theory, as in their proof). In the general case, when the convex bodies $K_1,K_2,K_3\subseteq \mathbb{R}^d$ have dimension greater than $k$ or $d-k$ , one cannot use the above technique to see that the mixed volumes satisfy the reverse Khovanskii–Teissier inequality.

Using the above polynomials and our prior results, we are now in the position to prove our main theorem and thus to fully classify when the inclusion $V_n^d\subseteq L_n^d$ is strict. First, the case $n=1$ obviously leads to $V_1^d\subseteq L_1^d$ for all $d \in \mathbb{N}$ . Second, the case $d=1$ obviously leads to $V_n^1\subseteq L_n^1$ for all $n \in \mathbb{N}$ . The remaining cases are solved in the following.

Proof. Shephard [Reference Shephard19, Theorem 4] proved that the sets are equal for $n=2$ and Heine [Reference Heine13, p. 118] proved the same for $(d,n)=(2,3)$ . We define the polynomial

\begin{equation*}f_k:= x_2^k(14x_1^3+6x_1^2x_2+24x_1^2x_3+12x_1x_2x_3+6x_1x_3^2+3x_2x_3^2)\end{equation*}

for $k\in \mathbb{N}_0$ which is a Lorentzian polynomial in $L_3^{3+k}$ by Proposition1.5 as it is the product of two Lorentzian polynomials. By the results of Brändén and Huh ([Reference Brändén and Huh7, Footnote 15] and [Reference Huh14, Example 14]), the second factor cannot be realised as a volume polynomial. By Proposition2.6, the polynomial $f_k$ cannot lie in $V_3^{3+k}$ either. Hence, we have $V_3^{3+k}\subsetneq L_3^{3+k}$ for all $k\in \mathbb{N}_0$ . The polynomial

\begin{equation*}f:= x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\end{equation*}

leads to the strict inclusion $V_4^2\subsetneq L_4^2$ (Shephard [Reference Shephard19, Theorem 5] and Heine [Reference Heine13, p. 119]). Given $n$ with $V_n^d\subsetneq L_n^d$ , we can deduce $V_{n+1}^d\subsetneq L_{n+1}^d$ by taking a polynomial $g\in L_n^d\setminus V_n^d$ . By Proposition1.5, the polynomial

\begin{equation*}g(x_1,\ldots ,x_{n-1},x_n+x_{n+1})\end{equation*}

is a Lorentzian polynomial in $L_{n+1}^d$ . If the new polynomial was a volume polynomial, the same would be true for $g$ as setting $x_{n+1}=0$ preserves volume polynomials due to Remark1.6.