# Viro’s patchworking and the signed reduced A-discriminant

Weixun Deng  Department of Mathematics, Texas A&M University, College Station,
Texas, USA<EMAIL_ADDRESS>,  J. Maurice Rojas  Department of
Mathematics, Texas A&M University, College Station, Texas, USA
<EMAIL_ADDRESS>and  Máté L. Telek  Department of Mathematical Sciences,
University of Copenhagen, Universitetsparken 5, 2100 Copenhagen, Denmark
<EMAIL_ADDRESS>
###### Abstract.

Computing the isotopy type of a hypersurface, defined as the positive real
zero set of a multivariate polynomial, is a challenging problem in real
algebraic geometry. We focus on the case where the defining polynomial has
combinatorially restricted exponent vectors and fixed coefficient signs,
enabling faster computation of the isotopy type. In particular, Viro’s
patchworking provides a polyhedral complex that has the same isotopy type as
the hypersurface, for certain choices of the coefficients. So we present
properties of the signed support, focussing mainly on the case of n-variate
(n+3)-nomials, that ensure all possible isotopy types can be obtained via
patchworking. To prove this, we study the signed reduced A-discriminant and
show that it has a simple structure if the signed support satisfies some
combinatorial conditions.

_Keywords: exponential sum, signed support, Viro’s patchworking, isotopy type_

##  1\. Introduction

Tropical geometry bridges the worlds of algebraic and polyhedral geometry. The
key idea is to transform algebraic varieties into polyhedral objects, turning
their algebraic structure functorially into a combinatorial structure. This
approach has been fruitful in the case of varieties over algebraically closed
fields [22, 17]. Recently, the tropicalization of semialgebraic sets over the
real numbers (or more generally over real closed fields) has received
increasing attention [29, 31, 18, 4, 5].

In the early 1980s, Oleg Viro showed that it is possible to associate a
polyhedral complex with a parametrized polynomial in such a way that there
exists some choice of coefficients such that the polyhedral complex and the
positive real zero set of the polynomial with that choice of coefficients have
the same isotopy type [33]. This result is known as _Viro’s patchworking_ in
the literature and was one of the first examples of tropical geometry.

Classifying the isotopy type of real hypersurfaces is a challenging question
in real algebraic geometry, tracing its origins to Hilbert’s 16th problem
[34]. In his original formulation, Hilbert asked for a classification of
isotopy types of plane real algebraic curves. Based on his patchworking
method, Viro provided such a classification for curves of degree $7$ [32].

Using Viro’s method, under certain conditions, one can determine the possible
isotopy types for hypersurfaces defined by the positive roots of polynomials
of the form $f(x)\\!=\\!\sum_{a\in\mathcal{A}}c_{a}x^{a}$, with
$\mathcal{A}\subseteq\mathbb{Z}^{n}$ a finite set and $c_{a}$ real and nonzero
for all $a\\!\in\\!\mathcal{A}$. We call $\mathcal{A}$ the support of $f$, let
$\varepsilon\\!=\\!(\varepsilon_{a}\;|\;a\\!\in\\!\mathcal{A})\\!\in\\!\\{\pm
1\\}^{\mathcal{A}}$ denote the vector of signs of the coefficients $c_{a}$,
and call the pair $(\mathcal{A},\varepsilon)$ the signed support of $f$. To
know whether one can build all possible isotopy types for a given
$(\mathcal{A},\varepsilon)$ via Viro’s patchworking one must first understand
how the isotopy type of a hypersurface changes while varying the coefficient
vector $c$. We will abuse notation slightly by calling $\varepsilon_{a}$ the
sign of the exponent vector $a$ when the underlying polynomial and support are
clear. Hence we will sometimes speak of positive or negative exponents in this
sense.

It is known that using discriminant varieties, one can decompose the
coefficient space into a disjoint union of open connected sets — called
chambers — such that in each chamber the topological type of any corresponding
hypersurface is constant [14, 3]. More specifically, such a chamber
decomposition is given by connected components of the complement of the union
of the signed $A$-discriminants $\nabla_{A_{F},\varepsilon_{F}}$ for the faces
$F$ of $\operatorname{Conv}(\mathcal{A})$. In [26], a reduced version of the
signed $A$-discriminant, denoted $\Gamma_{\varepsilon}(A,B)$, was introduced.
The complement of the union of the signed reduced $A$-discriminants has the
same property as its non-reduced counterpart, but has the advantage of
reducing the number of parameters. We will review these constructions in
Section 2.3.

When considering chambers (reduced or non-reduced), there will sometimes be
chambers where the isotopy type of the hypersurfaces can not be obtained by
Viro’s patchworking (see, e.g., [11, Example 2.9] and [3, Theorem 7.8]). These
chambers are called inner chambers, and reduced inner chambers are usually
bounded sets.

Our main goal is to give conditions on the signed support
$(\mathcal{A},\varepsilon)$ enabling Viro’s patchworking to find all isotopy
types, i.e., conditions that obstruct the existence of inner chambers. For
instance, we show that the signed $A$-discriminant is empty if and only if the
exponent vectors in $\mathcal{A}$ with positive signs can be separated from
those with negative signs by an affine hyperplane (Theorem 3.5). In that case,
all hypersurfaces with signed support $(\mathcal{A},\varepsilon)$ have the
same isotopy type, and the isotopy type can be obtained by Viro’s
patchworking. Moreover, for a given set of exponent vectors $\mathcal{A}$, we
give an upper bound on the number of sign distributions $\varepsilon\in\\{\pm
1\\}^{\mathcal{A}}$ such that the signed support $(\mathcal{A},\varepsilon)$
does not have a separating hyperplane (Proposition 3.4).

As we will see, it will be convenient (and natural) to assume
$\operatorname{Conv}(\mathcal{A})$ has dimension $n$ and cardinality $n+k+1$,
and then consider $k$ as a measure of the complexity of the resulting family
of isotopy types. The special cases $k\\!\in\\!\\{0,1\\}$ are addressed in [2,
8], so we will contribute to the case $k\\!=\\!2$: We show that the complement
of the signed reduced $A$-discriminant has exactly two connected components if
$\mathcal{A}$ contains exactly one negative exponent vector (Theorem 4.11) or
if the positive and the negative exponent vectors of $\mathcal{A}$ are
separated by an $n$-simplex in a certain way (Theorem 4.12). Under the
additional assumption that the signed $A$-discriminants associated to proper
faces of $\operatorname{Conv}(\mathcal{A})$ are empty, one can use Viro’s
patchworking to find all possible isotopy types (Corollary 4.13).

Signed supports with a separating hyperplane or a separating simplex have
previously been studied in [9]. In that work, the authors used these
conditions to show that the set of points where the polynomial takes negative
values has, at most, one connected component.

For $n=2$ and $\mathcal{A}$ consisting $5$ points, we show that if the
negative and positive exponent vectors are separated by two pairs of affine
lines (Theorem 4.10), then the complement of $\Gamma_{\varepsilon}(A,B)$ has
at most two connected components, which are unbounded. In Section 5, we study
further bivariate $5$-nomials and show that for a bounded chamber to exist in
the complement of $\Gamma_{\varepsilon}(A,B)$, the set of exponent vectors
must satisfy very restrictive inequalities, see Theorem 5.5 and Remark 5.6.

We will work in the more general context of exponential sums on
$\mathbb{R}^{n}$, instead of polynomials on $\mathbb{R}^{n}_{>0}$, and this
will in fact simplify some of our arguments. Note that any real polynomial can
be transformed into a real exponential sum while preserving topological
properties of the corresponding zero sets: Any polynomial
$f\colon\mathbb{R}^{n}_{>0}\to\mathbb{R}$ give rise to an exponential sum
$\mathbb{R}^{n}\to\mathbb{R},\,(x_{1},\dots x_{n})\mapsto
f(e^{x_{1}},\dots,e^{x_{n}})$. Since the map
$\operatorname{Exp}\colon\mathbb{R}^{n}\to\mathbb{R}^{n}_{>0},\,(x_{1},\dots
x_{n})\mapsto(e^{x_{1}},\dots,e^{x_{n}})$ is a homeomorphism, two subsets of
$\mathbb{R}^{n}_{>0}$ have the same isotopy type if and only if their images
under $\operatorname{Exp}$ have the same isotopy type.

### Notation

For two vectors $v,w\in\mathbb{R}^{n}$, $v\cdot w$ denotes the Euclidean
scalar product, and $v\ast w$ denotes the coordinate-wise product of $v$ and
$w$. The transpose of a matrix $M$ will be denoted by $M^{\top}$. We denote
the interior of a set $X\subseteq\mathbb{R}^{n}$ by $\operatorname{int}(X)$.
If $X\subseteq\mathbb{R}^{n}$ is a polyhedron, $\operatorname{relint}(X)$
denotes the relative interior of $X$. By $\\#S$ we denote the cardinality of a
finite set $S$. For a differentiable function
$f\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$, the Jacobian matrix at
$x\in\mathbb{R}^{n}$ is denoted by $J_{f}(x)$.

##  2\. Preliminaries

###  2.1. Signed support of an exponential sum

Let $\mathcal{A}=\\{\alpha_{1},\dots,\alpha_{n+k+1}\\}\subseteq\mathbb{R}^{n}$
be a finite set. We think about the elements of $\mathcal{A}$ as the exponent
vectors of an _exponential sum_ :

(1) $\displaystyle f_{c}\colon\mathbb{R}^{n}\to\mathbb{R},\quad x\mapsto
f_{c}(x)=\sum_{i=1}^{n+k+1}c_{i}e^{\alpha_{i}\cdot x},$

where $c\in(\mathbb{R}\setminus\\{0\\})^{n+k+1}$. We call
$\varepsilon=\operatorname{sign}(c)\in\\{\pm 1\\}^{n+k+1}$ a _sign
distribution_ , and $(\mathcal{A},\varepsilon)$ a _signed support_. For a
fixed order of the exponent vectors $\alpha_{1},\dots,\alpha_{n+k+1}$, there
is an isomorphism between the vector spaces
$\mathbb{R}^{\mathcal{A}}\cong\mathbb{R}^{n+k+1}$. We might use these two
notations interchangebly. For a fixed sign distribution $\varepsilon\in\\{\pm
1\\}^{n+k+1}$, we write

$\mathbb{R}^{\mathcal{A}}_{\varepsilon}=\\{c\in\mathbb{R}^{\mathcal{A}}\mid\operatorname{sign}(c)=\varepsilon\\}.$

for the orthant in $\mathbb{R}^{\mathcal{A}}$ containing the coefficients
matching the signs given by $\varepsilon$.

We split the support set $\mathcal{A}$ into the sets of _positive_ and
_negative exponent vectors_ , that is, we define

$\mathcal{A}_{+}:=\\{\alpha_{i}\in\mathcal{A}\mid\varepsilon_{i}=1\\},\qquad\mathcal{A}_{-}:=\\{\alpha_{i}\in\mathcal{A}\mid\varepsilon_{i}=-1\\}.$

We call $\mathcal{A}$ _full-dimensional_ if $\operatorname{Conv}(\mathcal{A})$
has dimension $n$. For a set $S\subseteq\mathbb{R}^{n}$, we denote the
restriction of $f_{c}$ to $S$ by

$f_{c|S}\colon\mathbb{R}^{n}\to\mathbb{R},\quad x\mapsto
f_{c}(x)=\sum_{\alpha_{i}\in\mathcal{A}\cap S}c_{i}e^{\alpha_{i}\cdot x}.$

Furthermore, we set $\mathcal{A}_{S}:=\mathcal{A}\cap S$, and define
$\varepsilon_{S}$ to be the sign distribution containing the signs
corresponding to the elements in $\mathcal{A}_{S}$.

The real zero set of $f_{c}$ is denoted by

$Z(f_{c}):=\\{x\in\mathbb{R}^{n}\mid f_{c}(x)=0\\}.$

We are interested in the possible _isotopy types_ of $Z(f_{c})$ when
$c\in\mathbb{R}^{n+k+1}$ varies over all coefficients such that
$\operatorname{sign}(c)=\varepsilon$. Two subsets
$Z_{0},Z_{1}\subseteq\mathbb{R}^{n}$ are _isotopic_ (ambient in
$\mathbb{R}^{n}$) if there exists a continuous map
$H\colon[0,1]\times\mathbb{R}^{n}\to\mathbb{R}^{n}$, called an _isotopy_ ,
such that

  * •

$H(t,\cdot)$ is a homeomorphism for all $t\in[0,1]$,

  * •

$H(0,\cdot)$ is the identity on $\mathbb{R}^{n}$,

  * •

$H(1,Z_{0})=Z_{1}$.

Being isotopic gives an equivalence relation on subsets of $\mathbb{R}^{n}$
[20, Chapter 10.1], which allows us to talk about their isotopy types.

###  2.2. Viro’s patchworking

Viro’s patchworking method provides possible isotopy types of
$Z(f_{c}),\,c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$ for a fixed signed
support $(\mathcal{A},\varepsilon)$. In this section, we recall this method.
We follow the notation used in [7].

Let $\mathcal{A}=\\{\alpha_{1},\dots,\alpha_{n+k+1}\\}\subseteq\mathbb{Z}^{n}$
be a finite set and $h\in\mathbb{R}^{n+k+1}$. We consider the _lifted_ points

$\mathcal{A}^{h}:=\big{\\{}(\alpha_{i},h_{i})\in\mathbb{R}^{n+1}\mid\alpha_{i}\in\mathcal{A}\big{\\}}.$

A face $F\subseteq\operatorname{Conv}(\mathcal{A}^{h})$ is called an _upper
face_ if there exists a vector $(v_{F},1)\in\mathbb{R}^{n+1}$ such that

$F=\big{\\{}x\in\operatorname{Conv}(\mathcal{A}^{h})\mid(v_{F},1)\cdot
x\geq(v_{F},1)\cdot y\text{ for all
}y\in\operatorname{Conv}(\mathcal{A}^{h})\big{\\}}.$

For a generic choice of $h\in\mathbb{R}^{n+k+1}$, each upper face
$F\subseteq\operatorname{Conv}(\mathcal{A}^{h})$ contains exactly $n+1$ points
of $\mathcal{A}^{h}$. The projection of upper faces of
$\operatorname{Conv}(\mathcal{A}^{h})$ onto $\mathbb{R}^{n}$ gives a
polyhedral subdivision $\mathcal{P}$ of $\operatorname{Conv}(\mathcal{A})$. If
$h$ is generic, each polyhedron in $\mathcal{P}$ is a simplex.

The _tropical hypersurface_ associated to $\mathcal{A}$ and
$h\in\mathbb{R}^{\mathcal{A}}$ is defined as

$\operatorname{Trop}(\mathcal{A},h):=\big{\\{}v\in\mathbb{R}^{n}\mid\max_{i=1,\dots,n+k+1}(v\cdot\alpha_{i}+h_{i})\text{
is attained at least twice}.\big{\\}}.$

It is dual to the $(n-1)$-skeleton of the subdivision $\mathcal{P}$ induced by
$h$. For a sign distribution $\varepsilon\in\\{\pm 1\\}^{\mathcal{A}}$, we
define the _signed tropical hypersurface_

$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h):=\Big{\\{}v\in\mathbb{R}^{n}\mid\max_{i=1,\dots,n+k+1}(v\cdot\alpha_{i}+h_{i})\;\text{
is attained for some }\alpha\in\mathcal{A}_{+}\text{ and
}\alpha^{\prime}\in\mathcal{A}_{-}\Big{\\}}.$

######  Example 2.1.

Consider the following set of exponent vectors

(2) $\displaystyle\mathcal{A}=\Big{\\{}\alpha_{1}=\begin{bmatrix}0\\\
0\end{bmatrix},\quad\alpha_{2}=\begin{bmatrix}1\\\
0\end{bmatrix},\quad\alpha_{3}=\begin{bmatrix}0\\\
1\end{bmatrix},\quad\alpha_{4}=\begin{bmatrix}4\\\
1\end{bmatrix},\quad\alpha_{5}=\begin{bmatrix}1\\\ 4\end{bmatrix}\Big{\\}},$

the sign distribution $\varepsilon=(1,1,1,-1,-1)$ and $h=(5,4,4,5,5)$. The
upper faces of the convex hull of the lifted points

$\displaystyle\mathcal{A}^{h}=\Big{\\{}\begin{bmatrix}5\\\ 0\\\
0\end{bmatrix},\quad\begin{bmatrix}4\\\ 1\\\
0\end{bmatrix},\quad\begin{bmatrix}4\\\ 0\\\
1\end{bmatrix},\quad\begin{bmatrix}5\\\ 4\\\
1\end{bmatrix},\quad\begin{bmatrix}5\\\ 1\\\ 4\end{bmatrix}\Big{\\}},$

are shown in Figure 1(a). The induced subdivision of
$\operatorname{Conv}(\mathcal{A})$ contains $3$ triangles, $7$ edges, and $5$
vertices. The tropical hypersurface $\operatorname{Trop}(\mathcal{A},h)$,
which is dual to the $1$ skeleton of the subdivision, contains $3$ vertices
and $7$ edges. We depicted $\operatorname{Trop}(\mathcal{A},h)$ and the signed
tropical hypersurface $\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$ in
Figure 1(b).

(a)

(b)

Figure 1. (a) Upper convex hull of the lifted points from Example 2.1 and the
induced polyhedral subdivision. (b) Signed tropical hypersurface associated to
the points in (a).

Viro’s patchworking is usually stated for polynomials and their zero sets in
the positive orthant $\mathbb{R}_{>0}^{n}$, however using the coordinate-wise
exponential map $\operatorname{Exp}\colon\mathbb{R}^{n}\to\mathbb{R}_{>0}^{n}$
it possible to translate Viro’s theorem to exponential sums.

######  Theorem 2.2.

[33][14, Ch.11 Theorem 5.6][17, Theorem 2.19] Let $(\mathcal{A},\varepsilon)$
be a signed support such that $\mathcal{A}\subseteq\mathbb{Z}^{n}$ and let
$h\in\mathbb{R}^{\mathcal{A}}$ be generic. For $t\in\mathbb{R}$, consider the
exponential sum

$g_{t}\colon\mathbb{R}^{n}\to\mathbb{R},\quad
x\mapsto\sum_{i=1}^{n+k+1}\varepsilon_{i}e^{h_{i}t}e^{\alpha_{i}\cdot x}.$

Then the signed tropical hypersurface
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$ is isotopic to $Z(g_{t})$
for $t\gg 0$ sufficiently large.

###  2.3. Signed $A$-discriminant

The goal of this subsection is to recall the notion of the _A-discriminant_
from [14, 26, 3]. Let $f_{c}$ be an exponential sum as in (1). A point
$x\in\mathbb{R}^{n}$ is a _singular zero_ of $f_{c}$ if and only if

(3) $\displaystyle f_{c}(x)=\frac{\partial f_{c}(x)}{\partial
x_{1}}=\dots=\frac{\partial f_{c}(x)}{\partial x_{n}}=0.$

We denote the set of singular zeros of $f_{c}$ by
$\operatorname{Sing}(f_{c})$. For a fixed signed support
$(\mathcal{A},\varepsilon)$, we define the _signed $A$-discriminant_ as

$\displaystyle\nabla_{\mathcal{A},\varepsilon}:=\big{\\{}c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}\mid\operatorname{Sing}(f_{c})\neq\emptyset\big{\\}}.$

Thus, $\nabla_{\mathcal{A},\varepsilon}$ contains all coefficients
$c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ such that the exponential sum
$f_{c}$ has a singular zero in $\mathbb{R}^{n}$.

For the sake of completeness, we recall that the signed $A$-discriminant does
not change under affine transformations of the exponent vectors.

######  Proposition 2.3.

Let $f_{c}$ be an exponential sum with support
$\mathcal{A}=\\{\alpha_{1},\dots,\alpha_{n+k+1}\\}\subseteq\mathbb{R}^{n}$.
For an invertible matrix $M\in\mathbb{R}^{n\times n}$ and $v\in\mathbb{R}^{n}$
consider the exponential sum

$g_{c}\colon\mathbb{R}^{n}\to\mathbb{R},\quad x\mapsto
g_{c}(x)=\sum_{i=1}^{n+k+1}c_{i}e^{(M\alpha_{i}+v)\cdot x}.$

Then we have:

  * (i)

If $\det(M)>0$, then the hypersurfaces $Z(f_{c})$ and $Z(g_{c})$ are isotopic.

  * (ii)

$\operatorname{Sing}(f_{c})=M^{\top}\operatorname{Sing}(g_{c})$.

  * (iii)

For all $x\in\operatorname{Sing}(g_{c})$ the Hessian matrices
$\operatorname{Hess}_{f_{c}}(M^{\top}x)$ and $\operatorname{Hess}_{g_{c}}(x)$
have the same number of positive, negative and zero eigenvalues.

###### Proof.

Note that $g_{c}(x)=e^{v\cdot x}f_{c}(M^{\top}x)$ for all
$x\in\mathbb{R}^{n}$. Since $e^{v\cdot x}\neq 0$, we have

(4) $\displaystyle Z(g_{c})$
$\displaystyle=Z(f_{c}(M^{\top}x))=\\{x\in\mathbb{R}^{n}\mid\sum\limits_{i=1}^{n+k+1}c_{i}e^{\alpha_{i}\cdot(M^{\top}x)}=0\\}$
$\displaystyle=\\{(M^{\top})^{-1}y\in\mathbb{R}^{n}\mid\sum\limits_{i=1}^{n+k+1}c_{i}e^{\alpha_{i}\cdot
y}=0\\}=(M^{\top})^{-1}Z(f_{c})$

Since the group of invertible real $n\times n$ matrices with positive
determinant is path-connected (see, e.g. [36, Theorem 3.68]), there exists a
continuous path from the identity matrix to $(M^{\top})^{-1}$, which induces
an isotopy. This shows (i).

Applying the product and the chain rule from calculus, we have

$J_{g_{c}}(x)=v^{\top}f_{c}(M^{\top}x)+e^{v\cdot
x}J_{f_{c}}(M^{\top}x)M^{\top}.$

Using (4) and that $M^{\top}$ is invertible, for $x\in Z(g_{c})$ it follows
that $J_{g_{c}}(x)=0$ if and only if $J_{f_{c}}(M^{\top}x)=0$, which implies
(ii).

For the rest of the proof, we assume that $x\in\operatorname{Sing}(g_{c})$.
From [28, Corollary 1] it follows that

$\operatorname{Hess}_{g_{c}}(x)=e^{v\cdot
x}M\operatorname{Hess}_{f_{c}}(M^{\top}x)M^{\top}.$

Thus, $\operatorname{Hess}_{g_{c}}(x)$ and
$\operatorname{Hess}_{f_{c}}(M^{\top}x)$ have the same number of positive,
negative and zero eigenvalues by Sylvester’s law of inertia [25, Chapter 7].

∎

######  Corollary 2.4.

Let $(\mathcal{A},\varepsilon)$ be a signed support, $M\in\mathbb{R}^{n\times
n}$ an invertible matrix and $v\in\mathbb{R}^{n}$. For
$M\mathcal{A}+v=\\{M\alpha+v\mid\alpha\in\mathcal{A}\\}$, we have

$\nabla_{\mathcal{A},\varepsilon}=\nabla_{M\mathcal{A}+v,\varepsilon}.$

###### Proof.

The statement follows directly from Proposition 2.3(ii). ∎

######  Remark 2.5.

Using Proposition 2.3, one might transform any full-dimensional support
$\mathcal{A}\subseteq\mathbb{R}^{n}$ to a support containing the standard
basis vectors $e_{1},\dots,e_{n}\in\mathbb{R}^{n}$ and the zero vector without
changing the isotopy types of the corresponding hypersurfaces.

To be more precise, from $\dim\operatorname{Conv}(\mathcal{A})=n$ it follows
that $\mathcal{A}$ contains $n+1$ affinely independent vectors
$\alpha_{1},\dots,\alpha_{n+1}$. Thus, there exists an invertible matrix
$M\in\mathbb{R}^{n\times n}$ such that
$M(\alpha_{1}-\alpha_{n+1})=e_{1},\dots,M(\alpha_{n}-\alpha_{n+1})=e_{n}$. If
$\det(M)<0$, we change the order of $\alpha_{1},\dots,\alpha_{n}$ such that
the corresponding matrix $M$ has positive determinant. For
$v:=-M\alpha_{n+1}$, the affine linear map
$L\colon\mathbb{R}^{n}\to\mathbb{R}^{n},\,\alpha\mapsto M\alpha+v$ satisfies
$\\{0,e_{1},\dots,e_{n}\\}\subseteq L(\mathcal{A})$.

For each face $F\subseteq\operatorname{Conv}(\mathcal{A})$, we define
$\nabla_{\mathcal{A}_{F},\varepsilon_{F}}$ in a similar way, and set

$\tilde{\nabla}_{\mathcal{A}_{F},\varepsilon_{F}}:=\big{\\{}(c_{\alpha_{i}})_{i=1,\dots,n+k+1}\in\mathbb{R}^{n+k+1}_{\varepsilon}\mid(c_{\alpha_{i}})_{\alpha_{i}\in\mathcal{A}_{F}}\in\nabla_{\mathcal{A}_{F},\varepsilon_{F}}\big{\\}}.$

In [3], the authors proved the following statement regarding the topology of
hypersurfaces corresponding to different connected components of the
complement of the union of the signed $A$-discriminants
$\tilde{\nabla}_{\mathcal{A}_{F},\varepsilon_{F}}$.

######  Proposition 2.6.

[3, Proposition 2.9] Let $(\mathcal{A},\varepsilon)$ be a full-dimensional
signed support such that $\mathcal{A}\subseteq\mathbb{Z}^{n}$. If $c$ and
$c^{\prime}$ are in the same connected component of

$\mathbb{R}^{\mathcal{A}}_{\varepsilon}\setminus\big{(}\bigcup_{F\subseteq\operatorname{Conv}(\mathcal{A})\text{
a face}}\tilde{\nabla}_{\mathcal{A}_{F},\varepsilon_{F}}\big{)},$

then the zero sets $Z(f_{c})$ and $Z(f_{c^{\prime}})$ are homeomorphic.

Finding the defining equalities of the signed $A$-discriminant is challenging,
but an explicit parametrization is much simpler to find. First, let
$\operatorname{diag}(c)$ denote the $\\#\mathcal{A}\times\\#\mathcal{A}$
diagonal matrix with $(a,a)$-entry $c_{a}$, and let us rewrite the equalities
in (3) as:

(5) $\displaystyle\hat{A}\operatorname{diag}(c)\Big{(}e^{\alpha_{i}\cdot
x}\Big{)}_{i=1,\dots,n+k+1}=\hat{A}_{\varepsilon}\operatorname{diag}(\lvert
c\rvert)\Big{(}e^{\alpha_{i}\cdot x}\Big{)}_{i=1,\dots,n+k+1}=0,$

where the matrix $\hat{A}$ is given by

(8) $\displaystyle\hat{A}=\left[\begin{array}[]{ccc}1&\ldots&1\\\
\alpha_{1}&\ldots&\alpha_{n+k+1}\end{array}\right]\in\mathbb{R}^{(n+1)\times(n+k+1)},$

$\varepsilon=\operatorname{sign}(c)\in\\{\pm 1\\}^{n+k+1}$ and
$\hat{A}_{\varepsilon}=\hat{A}\operatorname{diag}(\varepsilon)$. We refer to
the equation system (5) as the _critical system_ of
$(\mathcal{A},\varepsilon)$. Note that the assumption
$\dim\operatorname{Conv}(\mathcal{A})=n$ is equivalent to
$\operatorname{rk}(\hat{A})=n+1$ [38]. If $\operatorname{rk}(\hat{A})=n+1$,
the kernel of $\hat{A}$ has dimension $k$, which is usually called the
_codimension_ of $\mathcal{A}$.

If $x\in\mathbb{R}^{n}$ is a singular zero of $f_{c}$, then
$\operatorname{diag}(c)\big{(}e^{\alpha_{i}\cdot
x}\big{)}_{i=1,\dots,n+k+1}\in\ker(\hat{A})$. Choose a basis of
$\ker(\hat{A})$ and write these vectors as columns of a matrix
$B\in\mathbb{R}^{(n+k+1)\times k}$. Such a choice of $B$ is called a _Gale
dual matrix_ of $\hat{A}$. With slight abuse of notation, we might call $B$ a
Gale dual matrix of $\mathcal{A}$. Since $\operatorname{im}(B)=\ker(\hat{A})$
for each $x\in\operatorname{Sing}(f_{c})$ there exists
$\lambda\in\mathbb{R}^{k}$ such that

(9) $\displaystyle B\lambda=\operatorname{diag}(c)\big{(}e^{\alpha_{i}\cdot
x}\big{)}_{i=1,\dots,n+k+1}.$

Since $e^{\alpha_{i}\cdot x}$ is positive for all $x\in\mathbb{R}^{n}$ and all
$\alpha_{1},\dots,\alpha_{n+k+1}$, the signs of the vector in (9) are given by
$\operatorname{sign}(c)=\varepsilon$. Therefore, it is enough to look at the
set

$\mathcal{C}_{B,\varepsilon}:=\Big{\\{}\lambda\in\mathbb{R}^{k}\mid\operatorname{sign}(B\lambda)=\varepsilon\Big{\\}}.$

We define the _signed Horn-Kapranov Uniformization_ map as

(10)
$\displaystyle\psi\colon\mathcal{C}_{B,\varepsilon}\times\mathbb{R}^{n}\to\mathbb{R}^{n+k+1}_{\varepsilon},\quad(\lambda,x)\mapsto
B\lambda\ast\big{(}e^{\alpha_{i}\cdot(-x)}\big{)}_{i=1,\dots,n+k+1},$

where $\ast$ denotes the point-wise multiplication of two vectors.

######  Proposition 2.7.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support with Gale
dual matrix $B$. For the signed Horn-Kapranov Uniformization map (10), it
holds that

$\operatorname{im}(\psi)=\nabla_{\mathcal{A},\varepsilon}.$

###### Proof.

If $\psi(\lambda,x)=c$, then

$\hat{A}\operatorname{diag}(c)\big{(}e^{\alpha_{i}\cdot
x}\big{)}_{i=1,\dots,n+k+1}=\hat{A}\Big{(}B\lambda\ast\big{(}e^{\alpha_{i}\cdot-x}\big{)}_{i=1,\dots,n+k+1}\ast\big{(}e^{\alpha_{i}\cdot
x}\big{)}_{i=1,\dots,n+k+1}\Big{)}=\hat{A}B\lambda=0,$

which implies that $x\in\operatorname{Sing}(f_{c})$ and therefore
$c\in\nabla_{\mathcal{A},\varepsilon}$.

On the contrary, if $c\in\nabla_{\mathcal{A},\varepsilon}$, then there exists
a point $x\in\operatorname{Sing}(f_{c})$. From (9) follows that there exists
$\lambda\in\mathcal{C}_{B,\varepsilon}$ such that $\psi(\lambda,x)=c$. ∎

The signed $A$-discriminant lives in an ambient space of dimension $n+k+1$.
Following [26], we reduce the dimension of the ambient space to $k$ by
quotienting out some homogeneities without losing essential information as
follows. We define the _signed reduced A-discriminant_
$\Gamma_{\varepsilon}(A,B)$ [26, Definition 2.5] to be

$\Gamma_{\varepsilon}(A,B):=B^{\top}\operatorname{Log}\lvert\nabla_{\mathcal{A},\varepsilon}\rvert,$

where $\operatorname{Log}$ is the coordinate-wise natural logarithm map and
$\lvert\cdot\rvert$ denotes the coordinate-wise absolute value map. In [26],
bounded (resp. unbounded) connected components of
$\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$ have been called _signed
reduced inner_ (resp. _outer_) _chambers_.

######  Theorem 2.8.

[26, Theorem 3.8.] Let $(\mathcal{A},\varepsilon)$ be a full-dimensional
signed support with Gale dual matrix $B$ and let
$c,c^{\prime}\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$. If
$B^{\top}\operatorname{Log}\lvert c\rvert$ and
$B^{\top}\operatorname{Log}\lvert c^{\prime}\rvert$ are in the same connected
component of

$\mathbb{R}^{k}\setminus\big{(}\bigcup_{F\subseteq\operatorname{Conv}(\mathcal{A})\text{
a
face}}B^{\top}\operatorname{Log}\lvert\tilde{\nabla}_{\mathcal{A}_{F},\varepsilon_{F}}\rvert\big{)},$

then the zero sets $Z(f_{c})$ and $Z(f_{c^{\prime}})$ are ambiently isotopic
in $\mathbb{R}^{n}$.

The signed reduced $A$-discriminant admits a parametrization as well.

######  Proposition 2.9.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support with Gale
dual matrix $B$. The image of the map

(11)
$\displaystyle\xi_{B,\varepsilon}\colon\mathcal{C}_{B,\varepsilon}\to\mathbb{R}^{k},\quad\lambda\mapsto
B^{\top}\operatorname{Log}\lvert B\lambda\rvert$

is the signed reduced $A$-discriminant $\Gamma_{\varepsilon}(A,B)$.

###### Proof.

Let
$\operatorname{pr}_{k}\colon\mathcal{C}_{B,\varepsilon}\times\mathbb{R}^{n}\to\mathcal{C}_{B,\varepsilon}$
be the natural coordinate projection and $\psi$ be the signed Horn-Kapranov
Uniformization (10). Furthermore, denote $\tilde{A}$ the matrix obtained from
$\hat{A}$ in (8) by removing its top row, that is, the columns of $\tilde{A}$
are given by the vectors in $\mathcal{A}$. For every
$(\lambda,x)\in\mathcal{C}_{B,\varepsilon}\times\mathbb{R}^{n}$, we have

$\psi(\lambda,x)=B^{\top}\operatorname{Log}\lvert
B\lambda\ast(e^{\alpha_{i}\cdot(-x)})_{i=1,\dots,n+k+1}\rvert=B^{\top}\operatorname{Log}\lvert
B\lambda\rvert-B^{\top}\tilde{A}^{\top}x=\xi_{B,\varepsilon}(\lambda),$

where the last equality holds, since $\tilde{A}B=0$. Thus, the following
diagram commutes

(12)
${\mathcal{C}_{B,\varepsilon}\times\mathbb{R}^{n}}$${\mathbb{R}^{n+k+1}_{\varepsilon}}$${\mathbb{R}^{n+k+1}}$${\mathcal{C}_{B,\varepsilon}}$${\mathbb{R}^{k}}$$\scriptstyle{\operatorname{pr}_{k}}$$\scriptstyle{\psi}$$\scriptstyle{\operatorname{Log}\lvert\cdot\rvert}$$\scriptstyle{B^{\top}}$$\scriptstyle{\xi_{B,\varepsilon}}$

which gives that

$\operatorname{im}(\xi_{B,\varepsilon})=\operatorname{im}(\xi_{B,\varepsilon}\circ\operatorname{pr}_{k})=\operatorname{im}(B^{\top}\circ\operatorname{Log}\lvert\cdot\rvert\circ\psi)=B^{\top}\operatorname{Log}\lvert\nabla_{\mathcal{A},\varepsilon}\rvert=\Gamma_{\varepsilon}(A,B),$

where the first equality holds since $\operatorname{pr}_{k}$ is surjective,
and the second-to-last equality follows from Proposition 2.7. ∎

Since the first row of the matrix $\hat{A}$ is given by the all one vector
$\mathds{1}\in\mathbb{R}^{n+k+1}$, we have $B^{\top}\mathds{1}=0$, which
implies that the map $\xi_{B,\varepsilon}$ is homogeneous, i.e. for all
$a\in\mathbb{R}$:

$\xi_{B,\varepsilon}(a\lambda)=B^{\top}\operatorname{Log}\lvert
B(a\lambda)\rvert=\log(\lvert
a\rvert)B^{\top}\mathds{1}+B^{\top}\operatorname{Log}\lvert
B\lambda\rvert=B^{\top}\operatorname{Log}\lvert
B\lambda\rvert=\xi_{B,\varepsilon}(\lambda).$

Thus, one could projectivize the domain
$\mathcal{C}_{B,\varepsilon}\subseteq\mathbb{R}^{k}$ of $\xi_{B,\varepsilon}$.

######  Remark 2.10.

Modifying a Gale dual matrix using elementary column operations gives another
choice of Gale dual matrix. Thus, one can assume without any restriction that
the last row of the Gale dual matrix $B$ has the form
$B_{n+k+1}=(0,\dots,0,-1)$. Such a choice of $B$ fixes the sign
$\varepsilon_{n+k+1}=-1$. Since $Z(f_{c})=Z(f_{-c})$, we can always fix one of
the signs of the coefficients.

Since $\xi_{B,\varepsilon}$ is homogeneous, one can replace $\mathbb{R}^{k}$
(assuming $B_{n+k+1}=(0,\dots,0,-1)$) by the upper half of the $(k-1)-$sphere

$C_{k-1}=\\{\lambda\in\mathbb{R}^{k}\mid\lVert\lambda\rVert=1,\,\lambda_{k}>0\\},$

or by the $(k-1)-$dimensional affine subspace

$\\{\lambda\in\mathbb{R}^{k}\mid\lambda_{k}=1\\}.$

In Section 4, we will prefer this latter choice and work with the map

(13)
$\displaystyle\bar{\xi}_{B,\varepsilon}\colon\\{\mu\in\mathbb{R}^{k-1}\mid\operatorname{sign}(B\begin{bmatrix}\mu\\\
1\end{bmatrix})=\varepsilon\\}\to\mathbb{R}^{k},\quad\lambda\mapsto
B^{\top}\operatorname{Log}\lvert B\begin{bmatrix}\mu\\\ 1\end{bmatrix}\rvert.$

######  Example 2.11.

To give an illustration of the signed reduced $A$-discriminant, we recall [11,
Example 2.9]. Consider the same set of exponent vectors as in Example 2.1

(14) $\displaystyle\mathcal{A}=\Big{\\{}\alpha_{1}=\begin{bmatrix}0\\\
0\end{bmatrix},\quad\alpha_{2}=\begin{bmatrix}1\\\
0\end{bmatrix},\quad\alpha_{3}=\begin{bmatrix}0\\\
1\end{bmatrix},\quad\alpha_{4}=\begin{bmatrix}4\\\
1\end{bmatrix},\quad\alpha_{5}=\begin{bmatrix}1\\\ 4\end{bmatrix}\Big{\\}}.$

Unlike in Example 2.1, here we consider the sign distribution
$\varepsilon=(-1,1,1,-1,-1)$. We depicted the signed support
$(\mathcal{A},\varepsilon)$ in Figure 2(a). Since $\mathcal{A}$ has
codimension $2$, the signed reduced $A$-discriminant
$\Gamma_{\varepsilon}(A,B)$ is in the plane $\mathbb{R}^{2}$. For an
illustration we refer to Figure 2(b). The complement of
$\Gamma_{\varepsilon}(A,B)$ has $3$ connected components. For the coefficient
vectors $c=(-1,6,3,-1,-1)$,$(-1,1,1,-1,-1)$,$(-1,0.5,1,-1,-1)$, their
projection $B^{\top}\operatorname{Log}\lvert c\rvert$ lies in different
connected components of $\mathbb{R}^{2}\setminus\Gamma_{\varepsilon}(A,B)$.
The corresponding hypersurfaces $Z(f_{c})$ are shown in Figure 2(c),(d),(e)
respectively.

One remarkable property of this particular signed support is that the isotopy
type of the hypersurface in Figure 2(d), corresponding to the coefficient
$c=(-1,1,1-1,-1)$, cannot be obtained by Viro’s patchworking (cf. Theorem
2.2). All possible signed tropical hypersurfaces
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$, with
$h\in\mathbb{R}^{\mathcal{A}}$ generic, consist of $2$ unbounded connected
components, but the hypersurface $Z(f_{c}),\,c=(-1,1,1-1,-1)$ has $3$
connected components, one bounded and two unbounded.

(a)

(b)

(c)

(d)

(e)

Figure 2. (a) Signed support from Example 2.11. The positive and negative
exponent vectors are depicted by red circles and blue dots respectively. (b)
Signed reduced $A$-discriminant of the signed support in (a). (c)(d)(e)
Hypersurfaces $Z(f_{c})$ corresponding to different connected components of
the complement of the signed reduced $A$-discriminant.

###  2.4. Some useful results from topology

In the proof of Proposition 4.6, we need some classical results from topology.
Let us also introduce these briefly here (see, for example, Chapter I.11 &
Chapter IV.19 in [6]).

######  Lemma 2.12.

Suppose that $X$ and $Y$ are locally compact, Hausdorff spaces and that
$f\colon X\to Y$ is continuous. Let $X^{+}$ be the one-point compactification
space of $X$. Then $f$ is proper (i.e., the preimage of any compact subset is
compact) $\Longleftrightarrow$ $f$ extends to a continuous map $f^{+}\colon
X^{+}\to Y^{+}$ by setting $f^{+}(\infty_{X})=\infty_{Y}$.

######  Lemma 2.13.

(Jordan-Brouwer Separation Theorem) If
$f\colon\mathbf{S}^{n-1}\to\mathbf{S}^{n}$ (where $\mathbf{S}^{n}$ denotes
$n$-sphere) is an injective continuous map, then $\mathbf{S}^{n}\setminus
f(\mathbf{S}^{n-1})$ consists of exactly two connected components. Moreover,
$f(\mathbf{S}^{n-1})$ is the topological boundary of each of these components.

######  Corollary 2.14.

If $f\colon\mathbb{R}^{n}\to\mathbb{R}^{n+1}$ is injective, continuous and
proper, then $\mathbb{R}^{n+1}\setminus f(\mathbb{R}^{n})$ consists of exactly
two unbounded connected components.

###### Proof.

Note that the one point compactification of $\mathbb{R}^{n}$ is
$\mathbf{S}^{n}$. By Lemma 2.12, $f$ can be extended to
$f^{+}\colon\mathbf{S}^{n}\to\mathbf{S}^{n+1}$ with $f^{+}(\infty)=\infty$.
Then $f^{+}$ is also injective. By Lemma 2.13, $\mathbf{S}^{n+1}\setminus
f^{+}(\mathbf{S}^{n})$ consists of two connected components and the point
$\infty$ is in the boundary. Since the point $\infty$ is in the boundary of
each component, we can always find a sequence
$\\{x_{l}\\}\subseteq\mathbb{R}^{n+1}=\mathbf{S}^{n+1}\setminus\\{\infty\\}$
in each of the components such that $x_{l}\to\infty$. Therefore, these two
components are unbounded in $\mathbb{R}^{n+1}$. ∎

Finally, we need the following version of the mean value theorem:

######  Lemma 2.15.

(Cauchy’s Mean Value Theorem) If the functions $f,g\colon[a,b]\to\mathbb{R}$
are both continuous and differentiable on the open interval $(a,b)$, then
there exists some $c\in(a,b)$, such that

$(f(b)-f(a))g^{\prime}(c)=(g(b)-g(a))f^{\prime}(c).$

##  3\. Signed supports without singular zeros

In this section, we give a necessary and sufficient condition on the signed
support $(\mathcal{A},\varepsilon)$ such that the signed reduced
$A$-discriminant $\nabla_{\mathcal{A},\varepsilon}$ is empty. Building on this
result and Theorem 2.8, we give conditions on $(\mathcal{A},\varepsilon)$ such
that for all $c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$ the hypersurfaces
$Z(f_{c})$ have the same isotopy type (Theorem 3.5). First, we start with a
simple observation.

######  Proposition 3.1.

Let $\mathcal{A}=\\{\alpha_{1},\dots,\alpha_{n+k+1}\\}\subseteq\mathbb{R}^{n}$
be a set of exponent vectors and $\varepsilon\in\\{\pm 1\\}^{n+k+1}$ be a
fixed sign distribution. Let $\hat{A}$ the matrix defined in (8).

  * (a)

If
$\ker(\hat{A})\cap\mathbb{R}_{\varepsilon}^{n+k+1}=\ker(\hat{A}_{\varepsilon})\cap\mathbb{R}_{>0}^{n+k+1}=\emptyset$,
then for all $c\in\mathbb{R}_{\varepsilon}^{n+k+1}$ the critical system (5)
does not have any solution $x\in\mathbb{R}^{n}$.

  * (b)

If $\ker(\hat{A}_{\varepsilon})\cap\mathbb{R}_{>0}^{n+k+1}\neq\emptyset$, then
there exists $c\in\mathbb{R}_{\varepsilon}^{n+k+1}$ such that the critical
system has a solution $x\in\mathbb{R}^{n}$.

###### Proof.

Part (a) follows directly, since for any
$c\in\mathbb{R}^{n+k+1}_{\varepsilon}$ and any solution $x\in\mathbb{R}^{n}$
of (5), we have $\operatorname{diag}(c)\big{(}e^{\alpha_{i}\cdot
x}\big{)}_{i=1,\dots,n+k+1}\in\ker(\hat{A})\cap\mathbb{R}_{\varepsilon}^{n+k+1}$.

If $v\in\ker(\hat{A}_{\varepsilon})\cap\mathbb{R}_{>0}^{n+k+1}$, then for
$c=\operatorname{diag}(\varepsilon)v$, the point $x=(0,\dots,0)$ is a solution
of (5). ∎

In the following, we interpret the conditions in Proposition 3.1 in terms of
the geometry of the support $\mathcal{A}$ and the sign distribution
$\varepsilon$. An _affine hyperplane_ is a set of the form

$\mathcal{H}_{v,a}:=\\{\mu\in\mathbb{R}^{n}\mid v\cdot\mu=a\\},$

for some $v\in\mathbb{R}^{n}\setminus\\{0\\}$ and $a\in\mathbb{R}$. Each
affine hyperplane defines two half-spaces

$\mathcal{H}_{v,a}^{+}:=\\{\mu\in\mathbb{R}^{n}\mid v\cdot\mu\geq
a\\},\quad\mathcal{H}_{v,a}^{-}:=\\{\mu\in\mathbb{R}^{n}\mid v\cdot\mu\leq
a\\}.$

Following [9], we call $\mathcal{H}_{v,a}$ a _separating hyperplane_ of
$(\mathcal{A},\varepsilon)$ if

(15)
$\displaystyle\mathcal{A}_{+}\subseteq\mathcal{H}_{v,a}^{+},\quad\text{and}\quad\mathcal{A}_{-}\subseteq\mathcal{H}_{v,a}^{-}.$

A separating hyperplane $\mathcal{H}_{v,a}$ is called _non-trivial_ , if at
least one of the open half-spaces
$\operatorname{int}(\mathcal{H}_{v,a}^{+}),\,\operatorname{int}(\mathcal{H}_{v,a}^{-})$
contains a point of $\mathcal{A}$. A non-trivial separating hyperplane is
called _very strict_ if $\mathcal{H}_{v,a}$ does not contain any point in
$\mathcal{A}$. For an illustration of separating hyperplanes, we refer to
Figure 3.

Figure 3. The hyperplane $\mathcal{H}_{v,0}$ with $v=(0,0,1)$ is a non-trivial
separating hyperplane of
$\mathcal{A}_{+}=\big{\\{}(1,0,0)^{\top},(2,2,0)^{\top},(0,2,0)^{\top},(1,1,1)^{\top}\big{\\}}$
(depicted as red circles) and
$\mathcal{A}_{-}=\big{\\{}(0,0,0)^{\top},(2,0,0)^{\top},(1,1,-1)^{\top}\big{\\}}$
(blue dots). For the face
$F=\operatorname{Conv}\big{(}(0,0,0)^{\top},(0,0,2)^{\top}\big{)}$ (marked by
thick line segment), the restricted signed support
$\mathcal{A}_{F,+}=\big{\\{}(1,0,0)^{\top}\big{\\}},\,\mathcal{A}_{F,-}=\big{\\{}(0,0,0)^{\top},(2,0,0)^{\top}\big{\\}}$
does not have any non-trivial separating hyperplane.

######  Proposition 3.2.

A signed support $(\mathcal{A},\varepsilon)$ has a non-trivial separating
hyperplane if and only if
$\ker(\hat{A}_{\varepsilon})\cap\mathbb{R}_{>0}^{n+k+1}=\emptyset$, where
$\hat{A}\in\mathbb{R}^{(n+1)\times(n+k+1)}$ denotes the matrix from (8).

###### Proof.

By Stiemke’s Theorem [30] (which is a version of Farkas’ Lemma, see also [38,
Section 6.2]), exactly one of the following holds. Either there exists
$w\in\mathbb{R}^{n+1}$ such that

(16) $\displaystyle\hat{A}_{\varepsilon}^{\top}w\geq 0,$

and at least one of the inequalities is strict, or there exists
$u\in\mathbb{R}_{>0}^{n+k+1}$ such that

(17) $\displaystyle\hat{A}_{\varepsilon}u=0,$

but not both. Condition (17) is equivalent to
$\ker(\hat{A}_{\varepsilon})\cap\mathbb{R}_{>0}^{n+k+1}\neq\emptyset$. Note
that one can rewrite (16) as

$\varepsilon_{i}((w_{2},\dots,w_{n+1})\cdot\alpha_{i})\geq\varepsilon_{i}(-w_{1}),$

for all $\alpha_{i}\in\mathcal{A}$. Thus, if such a $w$ exists, then
$\mathcal{H}_{v,a}$ with $v=(w_{2},\dots,w_{n+1}),\,a=-w_{1}$ is a non-trivial
separating hyperplane of $(\mathcal{A},\varepsilon)$. On the other hand, if
$\mathcal{H}_{v,a}$ is a non-trivial hyperplane of
$(\mathcal{A},\varepsilon)$, then $w=(-a,v)$ satisfies (16). ∎

######  Theorem 3.3.

Let $(\mathcal{A},\varepsilon)$ be a signed support with Gale dual matrix $B$.
Then the following are equivalent:

  * (i)

$\nabla_{\mathcal{A},\varepsilon}=\emptyset$

  * (ii)

$\Gamma_{\varepsilon}(A,B)=\emptyset$

  * (iii)

$(\mathcal{A},\varepsilon)$ has a non-trivial separating hyperplane.

###### Proof.

The equivalence between (i) and (ii) follows directly from the definition of
the signed reduced $A$-discriminant, since
$\Gamma_{\varepsilon}(A,B)=B^{\top}\operatorname{Log}\lvert\nabla_{\mathcal{A},\varepsilon}\rvert$.

From Proposition 3.1 it follows that
$\nabla_{\mathcal{A},\varepsilon}=\emptyset$ if and only if
$\ker(\hat{A}_{\varepsilon})\cap\mathbb{R}_{>0}^{n+k+1}=\emptyset$, which is
equivalent to the existence of a non-trivial separating hyperplane of
$(\mathcal{A},\varepsilon)$ by Propositon 3.2. This shows that (i) and (ii)
are equivalent. ∎

For fixed set of exponent vectors $\mathcal{A}\subseteq\mathbb{R}^{n}$, using
the correspondence between hyperplane arrangements and zonotopes, one derives
a bound on the number of sign distributions for which
$(\mathcal{A},\varepsilon)$ does not have a non-trivial separating hyperplane.

######  Proposition 3.4.

Let $\mathcal{A}=\\{\alpha_{1},\dots,\alpha_{n+k+1}\\}\subseteq\mathbb{R}^{n}$
be a finite set such that $\dim\operatorname{Conv}(\mathcal{A})=n$. The number
of sign distributions $\varepsilon\in\\{\pm 1\\}^{n+k+1}$ for which
$(\mathcal{A},\varepsilon)$ does not have a non-trivial separating hyperplane
is bounded above by:

$2\sum_{i=0}^{k-1}\binom{n+k}{i}.$

###### Proof.

Let $\hat{A}$ be the matrix defined in (8). By Proposition 3.2, the signed
support $(\mathcal{A},\varepsilon)$ does not have a non-trivial separating
hyperplane if and only if
$\ker(\hat{A})\cap\mathbb{R}_{\varepsilon}^{n+k+1}\neq\emptyset$. So all we
have to do is to count how many orthants $\mathbb{R}_{\varepsilon}^{n+k+1}$
$\ker(\hat{A})$ intersects.

The assumption $\dim\operatorname{Conv}(\mathcal{A})=n$ implies that
$\dim\ker(\hat{A})=k$. Let $B\in\mathbb{R}^{(n+k+1)\times k}$ be Gale dual to
$\hat{A}$ and denote by $B_{1},\dots,B_{n+k+1}$ the rows of $B$. By [12, Lemma
0.16] (see also [38, Corollary 7.17]), the orthants
$\mathbb{R}_{\varepsilon}^{n+k+1}$ that $\operatorname{im}(B)=\ker(\hat{A})$
intersects, correspond one-to-one to the vertices of the zonotope

$[-B_{1},B_{1}]+\dots+[-B_{n+k+1},B_{n+k+1}]\subseteq\mathbb{R}^{k}.$

By [13, Table 2.1] (see also [37]) such a zonotope can have at most

$2\sum_{i=0}^{k-1}\binom{n+k}{i}$

many vertices. ∎

We finish the section by characterizing signed supports for which all
corresponding hypersurfaces have the same isotopy type.

######  Theorem 3.5.

Let $(\mathcal{A},\varepsilon)$ be a signed support. If for all faces
$F\subseteq\operatorname{Conv}(\mathcal{A})$ the signed support
$(\mathcal{A}_{F},\varepsilon_{F})$ has a non-trivial separating hyperplane,
then for all $c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ the hypersurfaces
$Z(f_{c})$ have the same isotopy type.

###### Proof.

From Theorem 3.3 follows that the signed reduced $A$-discriminants associated
to the faces $F\subseteq\operatorname{Conv}(\mathcal{A})$ are empty. Thus, all
the hypersurfaces $Z(f_{c}),\,c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$ have
the same isotopy type by Theorem 2.8. ∎

######  Corollary 3.6.

If a signed support $(\mathcal{A},\varepsilon)$ has a very strict separating
hyperplane, then the hypersurfaces $Z(f_{c})$ have the same isotopy type for
all $c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$.

###### Proof.

If $\mathcal{H}_{v,a}$ is a very strict separating hyperplane of
$(\mathcal{A},\varepsilon)$, then it is also a very strict separating
hyperplane of $(\mathcal{A}_{F},\varepsilon_{F})$ for all faces
$F\subseteq\operatorname{Conv}(\mathcal{A})$. Now, the statement follows from
Theorem 3.5. ∎

######  Example 3.7.

The signed support $(\mathcal{A},\varepsilon)$ from Example 2.1 has a very
strict separating hyperplane. Thus, by Corollary 3.6, all hypersurfaces
$Z(f_{c}),\,c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ have the same isotopy
type. From Theorem 2.2 follows that this isotopy type agrees with the isotopy
type of the signed tropical hypersurface
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$ for every generic
$h\in\mathbb{R}^{\mathcal{A}}$. We refer to Figure 1(b) for an illustration of
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$ with $h=(5,4,4,5,5)$.

######  Remark 3.8.

If a signed support $(\mathcal{A},\varepsilon)$ has a non-trivial separating
hyperplane, it might happen that for one of the faces
$F\subseteq\operatorname{Conv}(\mathcal{A})$ the restricted signed support
$(\mathcal{A}_{F},\varepsilon_{F})$ does not have a non-trivial separating
hyperplane. For such an example, we revisit the signed support from Figure 3.
The face $F=\operatorname{Conv}((0,0,0)^{\top},(2,0,0)^{\top})$, contains two
negative exponent vectors
$\alpha_{1}=(0,0,0)^{\top},\alpha_{2}=(2,0,0)^{\top}$ and one positive
exponent vector $\alpha_{3}=(1,0,0)^{\top}$. Since $\alpha_{3}$ lies in the
relative interior of $\operatorname{Conv}(\alpha_{1},\alpha_{2})$, it follows
that $\\{\alpha_{1},\alpha_{2}\\}$ and $\\{\alpha_{3}\\}$ cannot be separated
by an affine hyperplane.

##  4\. A-discriminants with two signed reduced outer chambers

The goal of this section is to describe conditions on the signed support
$(\mathcal{A},\varepsilon)$ that ensure
$\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$ has at most two connected
components, which are unbounded. In Section 4.1, we focus on the case where
$\mathcal{A}$ has exactly $n+3$ exponent vectors. We show that the complement
of $\Gamma_{\varepsilon}(A,B)$ has at most two chambers if the parametrization
map $\overline{\xi}_{B,\varepsilon}$ has at most one critical point
(Proposition 4.6). It is known that $\overline{\xi}_{B,\varepsilon}$ can have
at most $n$ critical points [27], however there did not exist any known
example in the literature where this bound is attained. In Example 4.3, we
describe a family of signed supports such that
$\overline{\xi}_{B,\varepsilon}$ has $n$ critical points for every
$n\in\mathbb{N}$.

In Section 4.2, we investigate the relation between critical points of
$\overline{\xi}_{B,\varepsilon}$ and degenerate singular points of $Z(f_{c})$
, and show that if $Z(f_{c})$ has no degenerate singular point for all
$c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ and the codimension of
$\mathcal{A}$ is $2$, then $\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$
has at most two connected components (Theorem 4.9). In Section 4.3, we give
several conditions on the geometry of the signed support
$(\mathcal{A},\varepsilon)$ precluding the existence of degenerate singular
points in $Z(f_{c}),\,c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$.

During the whole section, we assume that the Gale dual matrix
$B\in\mathbb{R}^{(n+k+1)\times k}$ is chosen in a way such that its last row
has the form $(0,\dots,0,-1)$ (cf. Remark 2.10).

###  4.1. Critical points of the signed reduced A-discriminant

Let $\mathcal{A}=\\{\alpha_{1},\dots,\alpha_{n+k+1}\\}\subseteq\mathbb{R}^{n}$
be a set of exponent vectors such that
$\dim\operatorname{Conv}(\mathcal{A})=n$ and fix a sign distribution
$\varepsilon\in\\{\pm 1\\}^{n+k+1}$. Let
$\hat{A}\in\mathbb{R}^{(n+1)\times(n+k+1)}$ be as given in (8) and choose a
Gale dual matrix $B\in\mathbb{R}^{(n+k+1)\times k}$ with rows
$B_{1},\dots,B_{n+k+1}$ such that its last row has the form
$B_{n+k+1}=(0,\dots,0,-1)$.

Let $\bar{\xi}_{B,\varepsilon}$ be the parametrization map of
$\Gamma_{\varepsilon}(A,B)$ as defined in (13). Following [1, Section 1.2], we
call a point $\mu\in\mathbb{R}^{k-1}$ a _critical point_ of
$\bar{\xi}_{B,\varepsilon}$ if $\bar{\xi}_{B,\varepsilon}(\mu)$ is well-
defined and the Jacobian matrix $J_{\bar{\xi}_{B,\varepsilon}}(\mu)$ does not
have full rank, that is, it has rank strictly less than $k-1$.

######  Lemma 4.1.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support with Gale
dual matrix $B$, and let $\bar{\xi}_{B,\varepsilon}$ be as defined in (13).
Then for each $\mu\in\mathbb{R}^{k-1}$ where $\bar{\xi}_{B,\varepsilon}$ is
defined, we have the following equality for the Jacobian matrix

(18) $\displaystyle
J_{\bar{\xi}_{B,\varepsilon}}(\mu)=B^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}\tilde{B},$

where $\tilde{B}$ denotes the matrix obtained from $B$ by deleting its last
column.

###### Proof.

By definition (cf. (13)), the $j-$th coordinate of $\bar{\xi}_{B,\varepsilon}$
is given by

$\big{(}\bar{\xi}_{B,\varepsilon}(\mu)\big{)}_{j}=\sum_{i=1}^{n+k+1}\log\lvert
B_{i}\cdot\hat{\mu}\rvert B_{i,j},$

where $\hat{\mu}=\begin{bmatrix}\mu\\\ 1\end{bmatrix}$. Thus, the partial
derivatives of $\bar{\xi}_{B,\varepsilon}$ have the form

(19)
$\displaystyle\frac{\partial\big{(}\bar{\xi}_{B,\varepsilon}(\mu)\big{)}_{j}}{\partial\mu_{\ell}}=\sum_{i=1}^{n+k+1}\frac{B_{i,j}B_{i,\ell}}{B_{i}\cdot\hat{\mu}}$

for all $j=1,\dots,k$, and $\ell=1,\dots,k-1$. Comparing (19) with the entries
of the right-hand side of (18) the result follows. ∎

Using Lemma 4.1, an easy computation shows that

(20)
$\displaystyle\hat{\mu}^{\top}J_{\bar{\xi}_{B,\varepsilon}}(\mu)=\Big{(}B\hat{\mu}\Big{)}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}\tilde{B}=\mathds{1}^{\top}\tilde{B}=0.$

Therefore, if $\bar{\xi}_{B,\varepsilon}$ is differentiable at $\mu$, then a
normal vector at $\bar{\xi}_{B,\varepsilon}(\mu)$ is given by
$\hat{\mu}=\begin{bmatrix}\mu\\\ 1\end{bmatrix}$. This statement was proven by
Kapranov [21, Theorem 2.1].

In the remaining of the section, we focus on the case $k=2$. Under this
assumption, we have

$J_{\bar{\xi}_{B,\varepsilon}}(\mu)=\begin{bmatrix}b_{1}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}\\\
b_{2}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}\end{bmatrix},$

where $b_{1},b_{2}$ denote the first and second column of the Gale dual matrix
$B$.

######  Lemma 4.2.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support of
codimension $2$ with Gale dual $B\in\mathbb{R}^{(n+3)\times 2}$, and let
$\bar{\xi}_{B,\varepsilon}$ be as defined in (13). For
$\mu\in\mathbb{R}\setminus\\{0\\}$ the following are equivalent.

  * (i)

$\mu$ is a critical point of $\bar{\xi}_{B,\varepsilon}$.

  * (ii)

$\operatorname{sign}(B\hat{\mu})=\varepsilon$ and $\mu$ is a zero of the
univariate polynomial

(21) $\displaystyle
q_{B}(\mu):=\Big{(}\prod_{i=1}^{n+3}(B\hat{\mu})_{i}\Big{)}b_{1}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}.$

  * (iii)

$\operatorname{sign}(B\hat{\mu})=\varepsilon$ and $\mu$ is a zero of the
univariate polynomial

(22)
$\displaystyle\tilde{q}_{B}(\mu):=\Big{(}\prod_{i=1}^{n+3}(B\hat{\mu})_{i}\Big{)}b_{2}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}.$

Moreover, $q_{B}$ has degree at most $n$ and $\bar{\xi}_{B,\varepsilon}$ has
at most $n$ critical points.

###### Proof.

Note that $\bar{\xi}_{B,\varepsilon}(\mu)$ is defined only if
$\operatorname{sign}(B\hat{\mu})=\varepsilon$. Furthermore, the factor
$\prod_{i=1}^{n+3}(B\hat{\mu})_{i}$ clears the denominator of
$b_{1}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}$ and
of $b_{2}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}$.
Moreover, $\prod_{i=1}^{n+3}(B\hat{\mu})_{i}\neq 0$ if
$\operatorname{sign}(B\hat{\mu})=\varepsilon$. From (20) follows that

$\mu
b_{1}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}=-b_{2}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}.$

Thus, $J_{\bar{\xi}_{B,\varepsilon}}(\mu)=0$ if and only if $q_{B}(\mu)=0$,
which is also equivalent to $\tilde{q}_{B}(\mu)=0$.

The polynomial $q_{B}$ has been studied previously in [27], where it has been
shown that its degree is at most $n$. ∎

In [27, Theorem 3.10], the author constructed several matrices
$B\in\mathbb{R}^{(n+3)\times 2}$ such that $q_{B}(\mu)$ has exactly $n$ real
roots. However, these roots correspond to different sign distributions
$\varepsilon$. To show that the bound on the critical points of
$\bar{\xi}_{B,\varepsilon}$ in Lemma 4.2 is attained, one needs to construct
$B\in\mathbb{R}^{(n+3)\times 2}$ such that $q_{B}$ (or $\tilde{q}_{B}$) has
$n$ real roots $\mu_{1},\dots,\mu_{n}$ such that
$\operatorname{sign}(B\hat{\mu}_{1})=\dots=\operatorname{sign}(B\hat{\mu}_{n})=\varepsilon$
for some fixed $\varepsilon\in\\{\pm 1\\}^{\mathcal{A}}$. We provide such a
construction in the following example.

######  Example 4.3.

Let $n\in\mathbb{N}$ and let $\mu_{1},\ldots,\mu_{n}$ be distinct positive
numbers different from $1$. Consider the univariate polynomials
$f(\mu)=(\mu-\mu_{1})(\mu-\mu_{2})\cdots(\mu-\mu_{n})$,
$g(\mu)=(\mu+\mu_{1})(\mu+\mu_{2})\cdots(\mu+\mu_{n})(\mu+1)\in\mathbb{R}[\mu]$.
Since $\deg(f)=n<\deg(g)=n+1$, the fraction $\tfrac{f(\mu)}{g(\mu)}$ admits a
partial fraction decomposition:

(23)
$\displaystyle\frac{(\mu-\mu_{1})(\mu-\mu_{2})\cdots(\mu-\mu_{n})}{(\mu+\mu_{1})(\mu+\mu_{2})\cdots(\mu+\mu_{n})(\mu+1)}=\frac{a_{1}}{\mu+\mu_{1}}+\frac{a_{2}}{\mu+\mu_{2}}+\cdots+\frac{a_{n}}{\mu+\mu_{n}}+\frac{a_{n+1}}{\mu+1},$

where $a_{1},\dots,a_{n+1}\in\mathbb{R}$. The $a_{i}$’s satisfy the following
properties:

  * (1)

${\displaystyle\frac{a_{1}}{\mu_{1}}+\frac{a_{2}}{\mu_{2}}+\frac{a_{3}}{\mu_{3}}+\cdots+\frac{a_{n}}{\mu_{n}}+a_{n+1}=(-1)^{n}}$,

  * (2)

$a_{1}+a_{2}+\cdots+a_{n+1}=1$.

  * (3)

$a_{i}\neq 0$, $i=1,\dots,n+1$.

Property (1) follows by plugging in $\mu=0$, (2) follows by comparing the
leading coefficients of numerators on both sides, (3) follows by comparing the
degree of denominators on both sides.

We use the $a_{i}$’s to build the matrix:

$B=\begin{bmatrix}\frac{a_{1}}{\mu_{1}}&\frac{a_{2}}{\mu_{2}}&\frac{a_{3}}{\mu_{3}}&\cdots&\frac{a_{n}}{\mu_{n}}&a_{n+1}&(-1)^{n+1}&0\\\
a_{1}&a_{2}&a_{3}&\cdots&a_{n}&a_{n+1}&0&-1\end{bmatrix}^{\top}.$

Properties (1) and (2) imply that $\mathds{1}^{\top}B=0$, thus it is possible
to choose a matrix $\hat{A}\in\mathbb{R}^{(n+1)\times(n+3)}$ as in (8) such
that $B$ is its Gale dual. Denoting by $b_{1},b_{2}$ the columns of $B$, we
have:

$b_{2}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}=\frac{\frac{a_{1}^{2}}{\mu_{1}}}{\frac{a_{1}}{\mu_{1}}\mu+a_{1}}+\frac{\frac{a_{2}^{2}}{\mu_{2}}}{\frac{a_{2}}{\mu_{2}}\mu+a_{2}}+\frac{\frac{a_{3}^{2}}{\mu_{3}}}{\frac{a_{3}}{\mu_{3}}\mu+a_{3}}+\cdots+\frac{\frac{a_{n}^{2}}{\mu_{n}}}{\frac{a_{n}}{\mu_{n}}\mu+a_{n}}+\frac{a_{n+1}^{2}}{a_{n+1}\mu+a_{n+1}}.$

The right-hand side of this equality agrees with the right-hand side of (23),
as the $a_{i}$’s are nonzero. Therefore, the zeros of
$\tilde{q}_{B}(\mu)=b_{2}^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}}\Big{)}b_{1}$
are $\mu_{1},\dots,\mu_{n}$. Since $\mu_{1},\dots,\mu_{n}$ are positive, it
follows for all $i=1,\dots,n$ that

$\operatorname{sign}(B\hat{\mu}_{i})=(\operatorname{sign}(a_{1}),\operatorname{sign}(a_{2}),\ldots,\operatorname{sign}(a_{n+1}),(-1)^{n+1},-1)=:\varepsilon.$

We conclude that $\bar{\xi}_{B,\varepsilon}$ has $n$ critical points using
Lemma 4.2.

Let $n=4$ and pick $\mu_{1}=5,\mu_{2}=6,\mu_{3}=7,\mu_{4}=8$. We have

$\displaystyle\frac{(\mu-5)(\mu-6)(\mu-7)(\mu-8)}{(\mu+5)(\mu+6)(\mu+7)(\mu+8)(\mu+1)}$
$\displaystyle=$
$\displaystyle\frac{-715}{\mu+5}+\frac{\frac{12012}{5}}{\mu+6}+\frac{-2730}{\mu+7}+\frac{1040}{\mu+8}+\frac{\frac{18}{5}}{\mu+1}$

and

$B=\begin{bmatrix}-143&\frac{2002}{5}&-390&130&\frac{18}{5}&-1&0\\\
-715&\frac{12012}{5}&-2730&1040&\frac{18}{5}&0&-1\end{bmatrix}^{\top}.$

By the above, the map $\bar{\xi}_{B,\varepsilon}$ has $4$ critical points for
$\varepsilon=(-1,1,-1,1,1,-1,-1)$. The signed reduced $A$-discriminant is
drawn to the right and its critical points are highlighted by red circles.

In Example 2.11 (cf. Figure 2(b)), we saw that the complement of
$\Gamma_{\varepsilon}(A,B)$ has three connected components if
$\bar{\xi}_{B,\varepsilon}$ has $2$ critical points. In the following, we show
that if $\bar{\xi}_{B,\varepsilon}$ has at most one critical point, then
$\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$ cannot have more then two
connected components. Before we get to this result, let us prove the following
lemma for self-intersections of curves.

######  Lemma 4.4.

Let $\varphi\colon\mathbb{R}\to\mathbb{R}^{2}$ be a smooth map such that the
Jacobian matrix $J_{\varphi}(\mu)$ has full rank for all $\mu\in\mathbb{R}$
except for at most one point. Let $S\subseteq\mathbb{R}^{2}$ be the curve
parametrized by $\varphi$. If there exist two distinct points
$a,b\in\mathbb{R}$ such that $\varphi(a)=\varphi(b)$, then there exist two
distinct points $\mu_{1},\mu_{2}\in\mathbb{R}$ such that the tangent lines of
$S$ at $\varphi(\mu_{1})$ and at $\varphi(\mu_{2})$ are parallel.

###### Proof.

Denote $\varphi_{1},\varphi_{2}$ the first and the second coordinate of
$\varphi$. Suppose $a<b$, and assume there exists $t\in\mathbb{R}$ such that
$\varphi^{\prime}_{1}(t)=\varphi^{\prime}_{2}(t)=0$, that is, $J_{\varphi}(t)$
does not have full rank. We start by choosing $c\in\mathbb{R}$ such that
$a<c<b$, $\varphi(c)\neq\varphi(a)$ and $\varphi$ is smooth on both intervals
$(a,c)$ and $(c,b)$. If $t\leq a$ or $t\geq b$, such $c$ exists since the
Jacobian matrix $J_{\varphi}$ has full rank on $(a,b)$. If $a<t<b$ and
$\varphi(t)=\varphi(a)$, then the curve is smooth on the interval $(a,t)$, and
we pick a $c$ as before. Finally, if $a<t<b$ and $\varphi(t)\neq\varphi(a)$,
then we choose $c=t$. If $\varphi$ does not have any singular point, then we
pick $c$ as in the case $t\leq a$ or $t\geq b$.

By Lemma 2.15, on the interval $(a,c)$, there exists some $\mu_{1}\in(a,c)$
such that

$(\varphi_{1}(c)-\varphi_{1}(a))\varphi^{\prime}_{2}(\mu_{1})=(\varphi_{2}(c)-\varphi_{2}(a))\varphi^{\prime}_{1}(\mu_{1}).$

Similarly, on the interval $(c,b)$, there exists some $\mu_{2}\in(c,b)$ such
that

$(\varphi_{1}(c)-\varphi_{1}(b))\varphi^{\prime}_{2}(\mu_{2})=(\varphi_{2}(c)-\varphi_{2}(b))\varphi^{\prime}_{1}(\mu_{2}).$

Thus, since $\varphi(a)=\varphi(b)$ and $\varphi(c)\neq\varphi(a)$, we have

$\varphi^{\prime}_{1}(\mu_{1})\varphi^{\prime}_{2}(\mu_{2})=\varphi^{\prime}_{1}(\mu_{2})\varphi^{\prime}_{2}(\mu_{1})$

and hence the tangent lines at $\mu_{1},\mu_{2}$ are parallel. ∎

######  Remark 4.5.

Lemma 4.4 is not true for hypersurfaces in $\mathbb{R}^{n}$ when $n\geq 3$.
The surface given by
$\varphi\colon(t,s)\mapsto(e^{-s}(t^{2}-1),e^{-s}t(t^{2}-1),s)$ is a
counterexample for $n=3$. The map $\varphi$ is not injective but there are no
pairs of points with parallel tangent planes. The image of $\varphi$ is shown
on the right.

Now we are able to prove the following result bounding the number of connected
components of $\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$.

######  Proposition 4.6.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support of
codimension $2$ with Gale dual matrix $B\in\mathbb{R}^{(n+3)\times 2}$. If
$\bar{\xi}_{B,\varepsilon}$ has at most one critical point, then the
complement of the signed reduced A-discriminant $\Gamma_{\varepsilon}(A,B)$
has at most two connected components, which are unbounded.

###### Proof.

Recall that for $\mu\in\mathbb{R}$, we used the notation
$\hat{\mu}=\begin{bmatrix}\mu\\\ 1\end{bmatrix}$. If
$(\mathcal{A},\varepsilon)$ has a non-trivial separating hyperplane, then
$\Gamma_{\varepsilon}(A,B)=\emptyset$ by Theorem 3.3. Thus,
$\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$ has one connected
component. If $(\mathcal{A},\varepsilon)$ does not have a non-trivial
separating hyperplane, then from Proposition 3.2 follows that there exist
$\mu_{1},\mu_{2}\in\mathbb{R}$ such that
$\operatorname{sign}(B\hat{\mu}_{1})=\operatorname{sign}(B\hat{\mu}_{2})=\varepsilon$.
By (20), $\hat{\mu}_{1}$ and $\hat{\mu}_{2}$ are normal vectors at
$\bar{\xi}_{B,\varepsilon}(\mu_{1})$ and at
$\bar{\xi}_{B,\varepsilon}(\mu_{2})$ respectively. If the tangent lines at
$\bar{\xi}_{B,\varepsilon}(\mu_{1})$ and at
$\bar{\xi}_{B,\varepsilon}(\mu_{2})$ are parallel, then
$\hat{\mu}_{1}=\lambda\hat{\mu}_{2}$ for some
$\lambda\in\mathbb{R}\setminus\\{0\\}$, which implies that $\mu_{1}=\mu_{2}$.
This shows that there is no pair of points in $\Gamma_{\varepsilon}(A,B)$ with
parallel tangent lines.

Lemma 4.4 implies that $\bar{\xi}_{B,\varepsilon}$ is injective. Also,
$\bar{\xi}_{B,\varepsilon}$ maps an open interval of $\mathbb{R}$ to
$\mathbb{R}^{2}$, and $\bar{\xi}_{B,\varepsilon}(\mu)\to\infty$ as $\mu$
approaches the endpoints of the interval. Therefore,
$\bar{\xi}_{B,\varepsilon}$ is proper by Lemma 2.12, which implies that the
complement of $\Gamma_{\varepsilon}(A,B)$ has exactly two unbounded connected
components by Corollary 2.14. ∎

###  4.2. Critical points and degenerate singularities

Let $f_{c}$ be an exponential sum as in (1). A singular point
$x\in\operatorname{Sing}(f_{c})$ is called _degenerate_ if the Hessian matrix
$\operatorname{Hess}_{f_{c}}(x)$ is not invertible. We have the following
relationship between critical points of the signed reduced $A$-discriminant
and degenerate singular points in the corresponding hypersurface.

######  Lemma 4.7.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support with Gale
dual matrix $B\in\mathbb{R}^{(n+k+1)\times k}$. If
$\mu^{*}\in\mathbb{R}^{k-1}$ is a critical point of
$\bar{\xi}_{B,\varepsilon}$, then for $c^{*}=B\begin{bmatrix}\mu^{*}\\\
1\end{bmatrix}$, the point $x^{*}=(0,\dots,0)\in\mathbb{R}^{n}$ is a
degenerate singular point of $f_{c^{*}}$.

###### Proof.

Since $\operatorname{diag}(c^{*})\Big{(}e^{\alpha_{i}\cdot
x^{*}}\Big{)}_{i=1,\dots,n+k+1}=B\begin{bmatrix}\mu^{*}\\\
1\end{bmatrix}\in\ker(\hat{A})$, we have that $x^{*}$ is a singular point of
$f_{c^{*}}$ (cf.(5)). Thus, we only have to show that it is a degenerate
singular point. Let $\psi$ denote the Horn-Kapranov Uniformization map (10).
From [10, Theorem 3.4, Theorem 3.5], it follows that $x^{*}$ is a degenerate
singular point if

(24) $\displaystyle\operatorname{rk}J_{\psi}(\hat{\mu}^{*},x^{*})\leq n+k-1,$

where $\hat{\mu}^{*}=\begin{bmatrix}\mu^{*}\\\ 1\end{bmatrix}$.

We prove (24) in two steps. First we show that

(25)
$\displaystyle\operatorname{rk}J_{\xi_{B,\varepsilon}}(\hat{\mu}^{*})=\operatorname{rk}J_{\bar{\xi}_{B,\varepsilon}}(\mu^{*}).$

To see this, a similar computation as in (18) shows

$J_{\xi_{B,\varepsilon}}(\hat{\mu}^{*})=B^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}^{*}}\Big{)}B.$

Thus, the first $k-1$ columns of $J_{\xi_{B,\varepsilon}}(\hat{\mu}^{*})$ and
$J_{\bar{\xi}_{B,\varepsilon}}(\mu^{*})$ are the same. To show that the two
matrices have the same rank, it is enough to show that the last column of
$J_{\xi_{B,\varepsilon}}(\hat{\mu}^{*})$ is contained in the linear space
spanned by the columns of $J_{\bar{\xi}_{B,\varepsilon}}(\mu^{*})$, which
holds since

$B^{\top}\operatorname{diag}\Big{(}\tfrac{1}{B\hat{\mu}^{*}}\Big{)}B\hat{\mu}^{*}=B^{\top}\mathds{1}=0,$

where the last equality holds since $B^{\top}\hat{A}^{\top}=0$ and the first
column of $\hat{A}^{\top}$ equals $\mathds{1}$. This shows (25).

In the second part of the proof, we show (24). Using that the diagram (12)
commutes and the chain rule, we have

$J_{B^{\top}\circ\operatorname{Log}\lvert\cdot\rvert\circ\psi}(\hat{\mu}^{*},x^{*})=J_{\xi_{B,\varepsilon}\circ\operatorname{pr}_{k}}(\hat{\mu}^{*},x^{*})=J_{\xi_{B,\varepsilon}}(\mu^{*})J_{\operatorname{pr}_{k}}(\hat{\mu}^{*},x^{*}).$

Using (25) and that
$\operatorname{rk}J_{\operatorname{pr}_{k}}(\hat{\mu}^{*},x^{*})=k$, it
follows that

$\operatorname{rk}J_{B^{\top}\circ\operatorname{Log}\lvert\cdot\rvert\circ\psi}(\hat{\mu}^{*},x^{*})=\operatorname{rk}J_{\xi_{B,\varepsilon}}(\hat{\mu}^{*})=\operatorname{rk}J_{\bar{\xi}_{B,\varepsilon}}(\mu^{*})\leq
k-2,$

where the last inequality holds since $\mu^{*}$ is a critical point of
$\bar{\xi}_{B,\varepsilon}$.

Using again the chain rule

$J_{B^{\top}\circ\operatorname{Log}\lvert\cdot\rvert\circ\psi}(\hat{\mu}^{*},x^{*})=B^{\top}J_{\operatorname{Log}\lvert\cdot\rvert}(\psi(\hat{\mu}^{*},x^{*}))J_{\psi}(\hat{\mu}^{*},x^{*}).$

Note that $B^{\top}$ has rank $k$ and
$J_{\operatorname{Log}\lvert\cdot\rvert}(\psi(\hat{\mu}^{*},x^{*}))$ is a
diagonal matrix with nonzero diagonal entries. Thus,
$\operatorname{rk}B^{\top}J_{\operatorname{Log}\lvert\cdot\rvert}(\psi(\hat{\mu}^{*},x^{*}))=k$.
From Sylvester’s rank inequality follows that

$\operatorname{rk}B^{\top}J_{\operatorname{Log}\lvert\cdot\rvert}(\psi(\hat{\mu}^{*},x^{*}))+\operatorname{rk}J_{\psi}(\hat{\mu}^{*},x^{*})-(n+k+1)\leq\operatorname{rk}J_{B^{\top}\circ\operatorname{Log}\lvert\cdot\rvert\circ\psi}(\hat{\mu}^{*},x^{*})\leq
k-2,$

which imples (24). ∎

######  Proposition 4.8.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support with Gale
dual matrix $B$. If for all $c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$, all
singular points of $Z(f_{c})$ are non-degenerate, then
$\bar{\xi}_{B,\varepsilon}$ does not have any critical point.

###### Proof.

The statement is a direct consequence of Lemma 4.7. ∎

######  Theorem 4.9.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support of
codimension $2$ with Gale dual matrix $B$. If for all
$c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$, all singular points of
$Z(f_{c})$ are non-degenerate, then the complement of the signed reduced
$A$-discriminant $\Gamma_{\varepsilon}(A,B)$ has at most two connected
components, which are unbounded.

###### Proof.

The statement follows directly from Proposition 4.6 and Proposition 4.8. ∎

###  4.3. Signed supports without degenerate singular points

Now, we show that for certain signed supports $(\mathcal{A},\varepsilon)$, the
singular points of the hypersurfaces $Z(f_{c})$ are non-degenerate singular
for all $c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$ .

We call a pair of parallel affine hyperplanes
$\mathcal{H}_{v,a},\mathcal{H}_{v,b}\subseteq\mathbb{R}^{n}$ ($a\geq b$)
_enclosing hyperplanes of the positive exponents_ $\mathcal{A}_{+}$ if

$\mathcal{A}_{+}\subseteq\mathcal{H}^{-}_{v,a}\cap\mathcal{H}^{+}_{v,b}\quad\text{and}\quad\mathcal{A}_{-}\subseteq\mathbb{R}^{n}\setminus\big{(}\operatorname{int}(\mathcal{H}^{-}_{v,a})\cap\operatorname{int}(\mathcal{H}^{+}_{v,b})\big{)}.$

Enclosing hyperplanes $\mathcal{H}_{v,a},\mathcal{H}_{v,b}$ are _strict
enclosing hyperplanes_ of $\mathcal{A}_{+}$ if additionally
$\operatorname{int}(\mathcal{H}^{+}_{v,a})\cap\mathcal{A}_{-}\neq\emptyset$
and
$\operatorname{int}(\mathcal{H}^{-}_{v,b})\cap\mathcal{A}_{-}\neq\emptyset$.
We define _strict enclosing hyperplanes of the negative exponents_
$\mathcal{A}_{-}$ in a similar way. For an illustration, we refer to Figure 4.

Our first statement concerns exponential sums in two variables.

######  Theorem 4.10.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support in
$\mathbb{R}^{2}$ and assume that both $\mathcal{A}_{+}$ and $\mathcal{A}_{-}$
have a pair of strict enclosing hyperplanes. Then

  * (i)

for every $c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ and
$x\in\operatorname{Sing}(f_{c})$, the Hessian matrix
$\operatorname{Hess}_{f_{c}}(x)$ has a positive and a negative eigenvalue.

  * (ii)

If $\mathcal{A}$ consists of $5$ exponent vectors, then the complement of the
signed reduced $A$-discriminant $\Gamma_{\varepsilon}(A,B)$ consists of at
most two connected components.

###### Proof.

Let $\mathcal{H}_{v,a},\mathcal{H}_{v,b}$ (resp.
$\mathcal{H}_{w,a^{\prime}},\mathcal{H}_{w,b^{\prime}}$) enclosing hyperplanes
of $\mathcal{A}_{+}$ (resp. $\mathcal{A}_{-}$). Using an affine change of
coordinates as in Proposition 2.3, we assume without loss of generality that
$v=(1,0)^{\top}$, $w=(0,1)^{\top}$.

For $x^{*}\in\operatorname{Sing}(f_{c})$ consider the univariate exponential
sums:

$\displaystyle f_{c,x^{*},v}\colon\mathbb{R}\mapsto\mathbb{R},\quad s\mapsto
f_{c,x^{*},v}(s):=\sum_{i=1}^{2+k+1}c_{i}e^{\alpha_{i}\cdot(x^{*}+sv)}$
$\displaystyle f_{c,x^{*},w}\colon\mathbb{R}\mapsto\mathbb{R},\quad s\mapsto
f_{c,x^{*},w}(s):=\sum_{i=1}^{2+k+1}c_{i}e^{\alpha_{i}\cdot(x^{*}+sw)}.$

By construction it holds:

(26) $\displaystyle 0=f_{c}(x^{*})=f_{c,x^{*},v}(0)=f_{c,x^{*},w}(0).$

By denoting $\alpha_{1,i},\alpha_{2,i}$ the first and the second coordinate of
the vector $\alpha_{i}$, it is easy to check that

$\displaystyle\frac{\partial f_{c,x^{*},v}}{\partial
s}(s)=\sum_{i=1}^{2+k+1}c_{i}\alpha_{1,i}e^{\alpha_{i}\cdot(x^{*}+sv)},\qquad\frac{\partial
f_{c,x^{*},w}}{\partial
s}(s)=\sum_{i=1}^{2+k+1}c_{i}\alpha_{2,i}e^{\alpha_{i}\cdot(x^{*}+sw)},$

It follows that

(27) $\displaystyle\frac{\partial f_{c,x^{*},v}}{\partial s}(0)=\frac{\partial
f_{c}}{\partial x_{1}}(x^{*})=0,\qquad\frac{\partial f_{c,x^{*},w}}{\partial
s}(0)=\frac{\partial f_{c}}{\partial x_{2}}(x^{*})=0,$

since $x^{*}\in Z(f_{c})$ is a singular point. Combining (26),(27), we have
that $0$ is a root of $f_{c,x^{*},v}$ (resp. $f_{c,x^{*},w}$) of multiplicity
at least two.

The condition that $\mathcal{H}_{v,a},\mathcal{H}_{v,b}$ (resp.
$\mathcal{H}_{w,a^{\prime}},\mathcal{H}_{w,b^{\prime}}$) are strict enclosing
hyperplanes of $\mathcal{A}_{+}$ (resp. $\mathcal{A}_{-}$) implies that both
exponential sums have at most two sign changes in their coefficient sequence.
Since Descartes’ rule of signs is valid for polynomials with real exponents
[35], one can extend the result to exponential sums. Using Descartes’ rule of
signs, it follows that the multiplicity of $0$ is exactly two for both
$f_{c,x^{*},v}$ and $f_{c,x^{*},w}$ . Furthermore,

$f_{c,x^{*},v}(s)<0\quad\text{and}\quad f_{c,x^{*},w}(s)>0\qquad\text{for all
}s\neq 0.$

So $0$ is a local maximum of $f_{c,x^{*},v}$ and a local minimum of
$f_{c,x^{*},w}$. Therefore

$\displaystyle\frac{\partial^{2}f_{c}}{\partial
x_{1}^{2}}(x)=\frac{\partial^{2}f_{c,x^{*},v}}{\partial
s^{2}}(0)<0,\qquad\frac{\partial^{2}f_{c}}{\partial
x_{2}^{2}}(x)=\frac{\partial^{2}f_{c,x^{*},w}}{\partial s^{2}}(0)>0,$

which implies that

$\det(\operatorname{Hess}_{f_{c}}(x^{*}))=\frac{\partial^{2}f_{c}}{\partial
x_{1}^{2}}(x^{*})\frac{\partial^{2}f_{c}}{\partial
x_{2}^{2}}(x^{*})-\Big{(}\frac{\partial^{2}f_{c}}{\partial x_{1}\partial
x_{2}}(x^{*})\Big{)}^{2}<0.$

Thus, $\operatorname{Hess}_{f_{c}}(x^{*})$ is invertible and must have a
negative and a positive eigenvalue.

If $\mathcal{A}$ contains $5$ exponent vectors, then the codimension of
$\mathcal{A}$ is $2$. Since all singular points of $Z(f_{c})$ are non-
degenerate for all $c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ by (i), part
(ii) follows from Theorem 4.9. ∎

Figure 4. An illustration of strict enclosing hyperplanes. Consider the signed
exponent vectors $\mathcal{A}_{+}=\big{\\{}\begin{bmatrix}0\\\
0\end{bmatrix},\begin{bmatrix}3\\\ 0\end{bmatrix},\begin{bmatrix}0\\\
3\end{bmatrix}\big{\\}}$ (depicted by red circles) and
$\mathcal{A}_{-}=\big{\\{}\begin{bmatrix}-1\\\
2\end{bmatrix},\begin{bmatrix}4\\\ -2\end{bmatrix}\big{\\}}$ (depicted by blue
dots). The hyperplanes $\mathcal{H}_{v,3},\mathcal{H}_{v,0}$ with $v=(1,0)$
are strict enclosing hyperplanes of $\mathcal{A}_{+}$. The negative exponent
vectors $\mathcal{A}_{-}$ also have a pair of strict enclosing hyperplanes
given by $\mathcal{H}_{w,1.2},\mathcal{H}_{w,1.2}$ with $w=(0.8,1)$.

The next condition on the signed support that precludes the existence of
degenerate singular points is valid for every number of variables $n$.
Specifically, we require that there is only one exponent vector with negative
sign.

######  Theorem 4.11.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support with Gale
dual $B$ such that $\\#\mathcal{A}_{-}=1$. Then we have

  * (i)

for all $c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$ and
$x\in\operatorname{Sing}(f_{c})$ the Hessian matrix
$\operatorname{Hess}_{f_{c}}(x)$ has only positive eigenvalues.

  * (ii)

If $(\mathcal{A},\varepsilon)$ has codimension $2$, then the complement of the
signed reduced $A$-discriminant $\Gamma_{\varepsilon}(A,B)$ consists of at
most two connected components.

###### Proof.

Write $\mathcal{A}_{+}=\\{\alpha_{1},\dots,\alpha_{n+k}\\}$ and
$\mathcal{A}_{-}=\\{\alpha_{n+k+1}\\}$. Using Proposition 2.3, we assume
without loss of generality that $\alpha_{n+k+1}=0$. Under this assumption, the
Hessian of $f_{c}$ at $x\in\operatorname{Sing}(f_{c})$ is given by

(28)
$\displaystyle\operatorname{Hess}_{f_{c}}(x)=\sum_{i=1}^{n+k}(e^{\alpha_{i}\cdot
x}c_{i})\alpha_{i}\cdot\alpha_{i}^{\top}=\tilde{A}\operatorname{diag}\big{(}(e^{\alpha_{i}\cdot
x}c_{i})_{i=1,\dots,n+k})\big{)}\tilde{A}^{\top},$

where

$\tilde{A}=\begin{bmatrix}\alpha_{1}&\dots&\alpha_{n+k}\end{bmatrix}\in\mathbb{R}^{n\times(n+k)}.$

Since the affine hull of $\alpha_{1},\dots,\alpha_{n+k+1}$ has dimension $n$
and $\alpha_{n+k+1}=0$, it follows that $\operatorname{rk}(\tilde{A})=n$.

Since $e^{\alpha_{i}\cdot x}c_{i}$ is positive for $i=1,\dots,n+k$, their
square root is a real number. This gives

$\operatorname{Hess}_{f_{c}}(x)=\big{(}\tilde{A}\operatorname{diag}\big{(}(\sqrt{e^{\alpha_{i}\cdot
x}c_{i}})_{i=1,\dots,n+k})\big{)}\big{(}\tilde{A}\operatorname{diag}\big{(}(\sqrt{e^{\alpha_{i}\cdot
x}c_{i}})_{i=1,\dots,n+k})\big{)}\big{)}^{\top}$

and as $\tilde{A}$ has full rank

$\operatorname{rk}\big{(}\tilde{A}\operatorname{diag}\big{(}(e^{\alpha_{i}\cdot
x}c_{i})_{i=1,\dots,n+k})\big{)}\tilde{A}^{\top}\big{)}=\operatorname{rk}(\tilde{A}\operatorname{diag}\big{(}(\sqrt{e^{\alpha_{i}\cdot
x}c_{i}})_{i=1,\dots,n+k})\big{)})=\operatorname{rk}\tilde{A}=n.$

Thus, $\operatorname{Hess}_{f_{c}}(x)$ is positive semi-definite and of full
rank, which implies that all of its eigenvalues are positive. This shows (i).

Since all singular points of $Z(f_{c})$ are non-degenerate for all
$c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$, part (ii) follows from Theorem
4.9. ∎

Our final condition on the signed support, precluding the existence of
degenerate singular points in the hypersurfaces $Z(f_{c})$, requires the
positive and negative exponent vectors to be separated by a simplex, as
follows. We recall the definition of the negative vertex cone of a simplex
from [9, Section 4]. For an $n$-simplex $P\subseteq\mathbb{R}^{n}$ with
vertices $\mu_{0},\dots,\mu_{n}$, the _negative vertex cone_ at vertex
$\mu_{k}$ equals

$P^{-,k}:=\mu_{k}+\operatorname{Cone}\big{(}\mu_{k}-\mu_{0},\dots,\mu_{k}-\mu_{n}\big{)}.$

We write $P^{-}$ for the union of $P^{-,0},\dots,P^{-n}$. We refer to Figure 5
for such a simplex and its negative vertex cones in the plane.

######  Theorem 4.12.

Let $P\subseteq\mathbb{R}^{n}$ be an $n$-simplex and let
$(\mathcal{A},\varepsilon)$ be a full-dimensional signed support in
$\mathbb{R}^{n}$ with Gale dual matrix $B$ such that $\mathcal{A}_{+}\subseteq
P$, $\mathcal{A}_{-}\subseteq P^{-}$, and
$\mathcal{A}\cap\operatorname{int}\big{(}P\cup P^{-}\big{)}\neq\emptyset$.
Then

  * (i)

for all $c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ and all singular points
$x\in\operatorname{Sing}(f_{c})$ the eigenvalues of
$\operatorname{Hess}_{f_{c}}(x)$ are negative.

  * (ii)

If $\mathcal{A}$ has codimension $2$, then the complement of the signed
reduced A-discriminant $\Gamma_{\varepsilon}(A,B)$ consists of at most two
connected components.

###### Proof.

Note that the negative vertex cones are preserved under affine transformation
of $P$, see e.g. [9, Lemma 4.5]. Thus by Proposition 2.3, we can assume
without loss of generality that $P=\operatorname{Conv}(0,e_{1},\dots,e_{n})$.

Denote $\operatorname{Exp}\colon\mathbb{R}^{n}\to\mathbb{R}^{n}_{>0}$ and
$\operatorname{Log}\colon\mathbb{R}^{n}_{>0}\to\mathbb{R}^{n}$ the coordinate-
wise exponential and logarithm maps. From [24, Theorem 7], it follows that the
Hessian of the function

$f_{c}\circ\operatorname{Log}\colon\mathbb{R}^{n}_{>0}\to\mathbb{R},\quad
y\mapsto\sum_{i=1}^{n+k+1}c_{i}y^{\alpha_{i}}$

is negative definite for all $y\in\mathbb{R}^{n}_{>0}$. If
$x\in\mathbb{R}^{n}$ is a singular point of $f_{c}$, then from [28, Corollary
1] it follows that

$\operatorname{Hess}_{f_{c}}(x)=\operatorname{Hess}_{f_{c}\circ\operatorname{Log}\circ\operatorname{Exp}}(x)=\big{(}J_{\operatorname{Exp}}(x)\big{)}^{\top}\operatorname{Hess}_{f_{c}\circ\operatorname{Log}}(\operatorname{Exp}(x))J_{\operatorname{Exp}}(x).$

Thus, all the eigenvalues of $\operatorname{Hess}_{f_{c}}(x)$ are negative by
Sylvester’s law of inertia [25, Chapter 7]. In particular, all singular points
of $Z(f_{c})$ are non-degenerate for all
$c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$. Part (ii) follows from Theorem
4.9. ∎

Figure 5. A simplex $P=\operatorname{Conv}((0,0),(3,0),(0,3))$ separating the
signed exponent vectors $\mathcal{A}_{+}=\big{\\{}\begin{bmatrix}0\\\
0\end{bmatrix},\begin{bmatrix}0\\\ 3\end{bmatrix},\begin{bmatrix}1\\\
1\end{bmatrix}\big{\\}}$ (marked by red circles) and
$\mathcal{A}_{-}=\big{\\{}\begin{bmatrix}-1\\\
5\end{bmatrix},\begin{bmatrix}5\\\ -1\end{bmatrix},\begin{bmatrix}-1\\\
-1\end{bmatrix},\begin{bmatrix}-2\\\ 0\end{bmatrix}\big{\\}}$ (marked by blue
dots).

Using Theorem 4.11 and Theorem 4.12, we give conditions on the signed support
$(\mathcal{A},\varepsilon)$ such that all possible isotopy types of
$Z(f_{c}),\,c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ are given by some
signed tropical hypersurface (cf. Theorem 2.2).

######  Corollary 4.13.

Let $(\mathcal{A},\varepsilon)$ be a full-dimensional signed support of
codimension $2$ with Gale dual $B$ such that either $\\#\mathcal{A}_{-}=1$ or
$\mathcal{A}_{+}$ and $\mathcal{A}_{-}$ are separated by a simplex as in
Theorem 4.12. If for each proper face
$F\subsetneq\operatorname{Conv}(\mathcal{A})$ the restricted signed support
$(\mathcal{A}_{F},\varepsilon_{F})$ has a non-trivial separating hyperplane,
then for each smooth hypersurface $Z(f_{c})$ with
$c\in\mathbb{R}_{\varepsilon}^{\mathcal{A}}$ there exists
$h\in\mathbb{R}^{\mathcal{A}}$ such that the signed tropical hypersurface
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$ and $Z(f_{c})$ have the
same isotopy type.

###### Proof.

In both cases, the complement of $\Gamma_{\varepsilon}(A,B)$ has at most two
connected components by Theorem 4.11 or Theorem 4.12. Since
$(\mathcal{A}_{F},\varepsilon_{F})$ has a non-trivial separating hyperplane
for every proper face $F\subsetneq\operatorname{Conv}(\mathcal{A})$, we have
$\nabla_{\mathcal{A}_{F},\varepsilon_{F}}=\emptyset$ by Theorem 3.3. From
Theorem 2.8 follows that the hypersurfaces $Z(f_{c})$ with
$c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ have at most two different
isotopy types.

First, we focus on the case $\\#\mathcal{A}_{-}=1$. Assume with out loss of
generality that $\mathcal{A}_{-}=\\{\alpha_{n+3}\\}$. If $\alpha_{n+3}$ is
contained in the boundary of $\operatorname{Conv}(\mathcal{A})$, then there
exist a hyperplane $\mathcal{H}_{v,a}\subseteq\mathbb{R}^{n}$ such that
$\alpha_{n+3}\in\mathcal{H}_{v,a}$ and
$\mathcal{A}\subseteq\mathcal{H}_{v,a}^{+}$ (cf. [19, Corollary 2.5]). Thus
$(\mathcal{A},\varepsilon)$ has a non-trivial separating hyperplane, which
implies that all $Z(f_{c})$ with $c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$
have the same isotopy type by Theorem 3.5.

If $\alpha_{n+3}\in\operatorname{int}(\operatorname{Conv}(\mathcal{A}))$, then
choose a generic $h\in\mathbb{R}^{\mathcal{A}}\cong\mathbb{R}^{n+3}$ such that
$h_{n+3}>h_{i}$ for $i=1,\dots,n+2$. By construction, we have
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)\neq\emptyset$ and
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},-h)=\emptyset$. By Theorem 2.2,
there exist $c_{1},c_{2}\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ such that
$Z(f_{c_{1}})$ and $Z(f_{c_{2}})$ are isotopic to
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$ and to
$\operatorname{Trop}_{\varepsilon}(\mathcal{A},-h)$ respectively. Since the
number of possible isotopy types is at most two, it follows that the possible
isotopy types are given by $\operatorname{Trop}_{\varepsilon}(\mathcal{A},h)$
and $\operatorname{Trop}_{\varepsilon}(\mathcal{A},-h)$.

Next, we consider the case when $\mathcal{A}_{+}$ and $\mathcal{A}_{-}$ are
separated by a simplex $P\subseteq\mathbb{R}^{n}$. If one of the negative
simplex cones $P^{-,0},\dots,P^{-,n}$ does not contain any positive exponent
vector, then $(\mathcal{A},\varepsilon)$ has a non-trivial separating
hyperplane and all $Z(f_{c})$ with
$c\in\mathbb{R}^{\mathcal{A}}_{\varepsilon}$ have the same isotopy type by
Theorem 3.5. If $P^{-,i}\cap\mathcal{A}_{+}\neq\emptyset$ for each
$i=0,\dots,n$, then a similar argument as above shows that there exists two
signed tropical hypersurfaces wich are not isotopic to each other. This
concludes the proof. ∎

##  5\. Bivariate $5$-nomials

For bivariate $5$-nomials, the signed reduced $A$-discriminant has at most $2$
critical points by Lemma 4.2. If there is only one critical point, then
$\mathbb{R}^{k}\setminus\Gamma_{\varepsilon}(A,B)$ has a simple structure, it
has at most two connected components, which are unbounded (cf. Proposition
4.6). In this section, we give a complete description of the geometry of the
signed support of a bivariate $5$-nomial whose signed reduced $A$-discriminant
has two critical points. In our experiments, if the signed reduced
$A$-discriminant had two critical points, then its complement had a bounded
chamber. We conjecture that this is always true, however we do not have a
proof of this statement nor a counter example.

######  Conjecture 5.1.

Let $(\mathcal{A},\varepsilon)$ be the signed support of a bivariate
$5$-nomial and let

$\bar{\xi}_{B,\varepsilon}\colon\\{\mu\in\mathbb{R}\mid\operatorname{sign}(B\begin{bmatrix}\mu\\\
1\end{bmatrix})=\varepsilon\\}\to\mathbb{R}^{2}$

be the parametrization map of $\Gamma_{\varepsilon}(A,B)$ as defined in (13).
If $\bar{\xi}_{B,\varepsilon}$ has two critical points, then the complement of
$\Gamma_{\varepsilon}(A,B)$ has a bounded connected component.

Given a $2$-simplex $P=\operatorname{Conv}(\mu_{0},\mu_{1},\mu_{2})$, denote
by
$\mathcal{H}_{v_{0},d_{0}},\mathcal{H}_{v_{1},d_{1}},\mathcal{H}_{v_{2},d_{2}}$
the supporting hyperplanes of the facets of $P$. We choose these hyperplanes
such that

$P=\mathcal{H}^{+}_{v_{0},d_{0}}\cap\mathcal{H}^{+}_{v_{1},d_{1}}\cap\mathcal{H}^{+}_{v_{2},d_{2}}$

and $\mu_{i}\notin\mathcal{H}_{v_{i},d_{i}}$ for each $i=0,1,2$. The
complement of the union of the hyperplanes
$\mathcal{H}_{v_{0},d_{0}},\mathcal{H}_{v_{1},d_{1}},\mathcal{H}_{v_{2},d_{2}}$
has $7$ chambers. One of these chambers is the simplex $P$. Three other
chambers are the negative vertex cones $P^{-,0},P^{-,1},P^{-,2}$, as
introduced in Section 4.3. For these chambers we have

(29) $\displaystyle P^{-,i}=\bigcap_{j=0,j\neq
i}^{2}\mathcal{H}^{-}_{v_{j},d_{j}}\cap\mathcal{H}^{+}_{v_{i},d_{i}},\quad\text{
for }i=0,1,2.$

The three other chambers in the hyperplane arrangment can be written as

(30) $\displaystyle P^{+,i}=\bigcap_{j=0,j\neq
i}^{2}\mathcal{H}^{+}_{v_{j},d_{j}}\cap\mathcal{H}^{-}_{v_{i},d_{i}},\quad\text{
for }i=0,1,2.$

For $i\neq j\in\\{0,1,2\\}$, we define the subset of $P^{+,i}$

(31) $\displaystyle P^{+,i,j}:=\bigcap_{j=0,j\neq
i}^{2}\mathcal{H}^{+}_{v_{j},d_{j}}\cap\mathcal{H}^{-}_{v_{i},d_{i}}\cap\mathcal{H}^{-}_{v_{j},D_{j}},$

where $D_{j}:=v_{j}\cdot\mu_{j}>d_{j}$. For an illustration of these chambers,
we refer to Figure 6.

In the following lemmata, we focus on the special case when $P$ is the
standard $2$-simplex
$\Delta_{2}=\operatorname{Conv}((0,0)^{\top},(1,0)^{\top},(0,1)^{\top})$ and

(32) $\displaystyle\mathcal{A}_{+}=\Big{\\{}\begin{bmatrix}0\\\
0\end{bmatrix},\begin{bmatrix}1\\\ 0\end{bmatrix},\begin{bmatrix}0\\\
1\end{bmatrix}\Big{\\}},\quad\mathcal{A}_{-}=\Big{\\{}\begin{bmatrix}x_{1}\\\
y_{1}\end{bmatrix},\begin{bmatrix}x_{2}\\\ y_{2}\end{bmatrix}\Big{\\}}.$

Afterward we extend the results to the general case in Theorem 5.5. Choose a
Gale dual matrix corresponding to the exponent vectors in (32) as

(33) $\displaystyle B=\begin{bmatrix}1-x_{1}-y_{1}&1-x_{2}-y_{2}\\\
x_{1}&x_{2}\\\ y_{1}&y_{2}\\\ -1&0\\\ 0&-1\end{bmatrix}.$

With that choice of the Gale dual matrix, the polynomial from (21) is a
quadratic polynomial $q_{B}(t):=at^{2}+bt+c$, where

(34) $\displaystyle a$ $\displaystyle:=-x_{1}y_{1}\left(1-x_{1}-y_{1}\right),$
$\displaystyle b$
$\displaystyle:=-x_{1}^{2}y_{2}^{2}+2x_{1}x_{2}y_{1}y_{2}-x_{2}^{2}y_{1}^{2}+x_{1}^{2}y_{2}+y_{2}^{2}x_{1}+x_{2}^{2}y_{1}+y_{1}^{2}x_{2}-x_{1}y_{2}-x_{2}y_{1},$
$\displaystyle c$ $\displaystyle:=-x_{2}y_{2}\left(1-x_{2}-y_{2}\right).$

A point $t\in\mathbb{R}$ is a critical point of $\bar{\xi}_{B,\varepsilon}$,
$\varepsilon=(1,1,1,-1-1)$ if and only if $q_{B}(t)=0$ and it satisfies the
inequalities:

(35) $\displaystyle(1-x_{1}-y_{1})t+1-x_{2}-y_{2}>0,\quad x_{1}t+x_{2}>0,\quad
y_{1}t+y_{2}>0,\quad t>0.$

######  Lemma 5.2.

Let $\mathcal{A}_{+},\mathcal{A}_{-}$ be the set of exponent vectors as
defined in (32) and let $B$ the corresponding Gale dual matrix from (33). If

  * (i)

$\alpha_{4}\in\Delta_{2}$ and $\alpha_{5}\in\Delta_{2}^{+,i}$ for some
$i\in\\{0,1,2\\}$, or

  * (ii)

$\alpha_{4}\in\Delta_{2}^{-,i}$ and $\alpha_{5}\in\Delta_{2}^{+,i}$ for some
$i\in\\{0,1,2\\}$,

then $\bar{\xi}_{B,\varepsilon}$ has at most one critical point.

###### Proof.

Let $a,c$ denote the coefficients of $q_{B}$ as in (34). Both in case (i) and
(ii), we have $a\leq 0,\,c\geq 0$, which implies that $q_{B}$ has at most one
sign change in its coefficient sequence. From Descartes’ rule of signs, it
follows that $q_{B}$ has at most $1$ positive real root. By (35), every
critical point of $\bar{\xi}_{B,\varepsilon}$ is a positive root of $q_{B}$.
Therefore, $\bar{\xi}_{B,\varepsilon}$ has at most one critical point. ∎

######  Lemma 5.3.

Let $\mathcal{A}_{+},\mathcal{A}_{-}$ be the set of exponent vectors as
defined in (32) and let $B$ the corresponding Gale dual matrix from (33). If
$\alpha_{4}\in\operatorname{int}(\Delta_{2})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{-,0})$, then
$\bar{\xi}_{B,\varepsilon}$ has one critical point.

###### Proof.

The inequalities in (35) is equivalent to
$M:=\max\\{\tfrac{-x_{2}}{x_{1}},\tfrac{-y_{2}}{y_{1}}\\}<t$. Note that $M>0$,
The number of critical points of $\bar{\xi}_{B,\varepsilon}$ is the same as
the number of roots of $q_{B}$ in the interval $(M,\infty)$.

Let $a,c$ denote the coefficients of $q_{B}$ as in (34). Under the assumption
of the lemma, we have $a<0$ and $c<0$. Thus, $q_{B}$ has $0$ or $2$ sign
changes in its coefficient sequence. By Descartes’ rule of signs $q_{B}$ has
at most two positive roots. Moreover, if $q_{B}(M)>0$, then $q_{B}$ has
exactly one root in the interval $(M,\infty)$.

Evaluating $q_{B}$ at $\tfrac{-x_{2}}{x_{1}}$ or at $\tfrac{-y_{2}}{y_{1}}$,
depending which one is larger, we used the Maple function IsEmpty [23] and the
Mathematica function Reduce [16], to verify that $q_{B}(M)>0$. Thus, $q_{B}$
has exactly one root in the interval $(M,\infty)$, which concludes the proof.
∎

######  Lemma 5.4.

Let $\mathcal{A}_{+},\mathcal{A}_{-}$ be the set of exponent vectors as
defined in (32), let $B$ the corresponding Gale dual matrix from (33) and let
$a,b,c$ defined in (34). Assume
$\alpha_{4}\in\operatorname{int}(\Delta_{n}^{+,1})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{n}^{+,2})$. The map
$\bar{\xi}_{B,\varepsilon}$ has two critical points if and only if
$\alpha_{4}\in\operatorname{int}(P^{+,1,2})$ and
$\alpha_{5}\in\operatorname{int}(P^{+,2,1})$ and the coordinates of
$\alpha_{4}$ and $\alpha_{5}$ satisfy the following inequalities:

(36) $\displaystyle b^{2}-4ac$ $\displaystyle>0$ $\displaystyle b^{2}-4ac$
$\displaystyle<(2x_{2}y_{1}(1-x_{1}-y_{1})+b)^{2},\quad
0<2x_{2}y_{1}(1-x_{1}-y_{1})+b$ $\displaystyle b^{2}-4ac$
$\displaystyle<(2x_{1}y_{2}(1-x_{1}-y_{1})+b)^{2},\quad
0>2x_{1}y_{2}(1-x_{1}-y_{1})+b.$

###### Proof.

If
$\operatorname{relint}(\operatorname{Conv}(\\{\alpha_{4},\alpha_{5}\\})\cap\operatorname{relint}(\Delta_{2})=\emptyset$,
then $\mathcal{A}_{+}$ and $\mathcal{A}_{-}$ can be separated by an affine
hyperplane [15, Section 2.2, Theorem 2] and therefore
$\Gamma_{\varepsilon}(A,B)=\emptyset$ by Theorem 3.3. In particular,
$\Gamma_{\varepsilon}(A,B)$ does not have any critical point.

If $\operatorname{relint}(P)\cap\operatorname{relint}(Q)\neq\emptyset$, then
there exists a pair of strict enclosing hyperplanes of $\mathcal{A}_{-}$,which
are parallel to the affine hull of $\alpha_{4}$ and $\alpha_{5}$. If
additionally
$\alpha_{5}\in\operatorname{int}(\Delta_{n}^{+,2})\setminus\operatorname{int}(\Delta_{n}^{+,2,1})$,
then by perturbing the hyperplanes
$\mathcal{H}_{e_{1},0},\mathcal{H}_{e_{1},1}$ we get strict enclosing
hyperplanes of $\mathcal{A}_{+}$. Thus, from Proposition 4.8 and Theorem 4.10
it follows that $\bar{\xi}_{B,\varepsilon}$ does not have any critical point.

A similar argument shows that $\bar{\xi}_{B,\varepsilon}$ does not have
critical points if
$\alpha_{5}\in\operatorname{int}(\Delta_{n}^{+,1})\setminus\operatorname{int}(\Delta_{n}^{+,1,2})$.
This shows that to get a critical point of $\bar{\xi}_{B,\varepsilon}$ the
negative exponent vectors must satisfy
$\alpha_{4}\in\operatorname{int}(\Delta_{n}^{+,1,2})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{n}^{+,2,1})$, which is equivalent to

$\displaystyle x_{1}<0,\quad 0<y_{1}<1,\quad 1-x_{1}-y_{1}>0,$ $\displaystyle
0<x_{2}<1,\quad y_{2}<0,\quad 1-x_{2}-y_{2}>0.$

Assuming these inequalities are satisfied, the inequality $y_{1}t+y_{2}>0$ in
(35) implies $t>\tfrac{-y_{2}}{y_{1}}>0$, and $t>0$ implies
$(1-x_{1}-y_{1})t+1-x_{2}-y_{2}>0$. Thus, the first and the last inequalities
in (35) are redundant.

The roots of $q_{B}$ are given by

$t_{1}=\tfrac{-b+\sqrt{b^{2}-4ac}}{2a},\qquad
t_{2}=\tfrac{-b-\sqrt{b^{2}-4ac}}{2a}.$

An easy computation shows that $t_{1}\neq t_{2}$ and both satisfy the second
and the third inequality in (35) if and only if

$\displaystyle b^{2}-4ac$ $\displaystyle>0$ $\displaystyle b^{2}-4ac$
$\displaystyle<(2x_{2}y_{1}(1-x_{1}-y_{1})+b)^{2},\quad
0<2x_{2}y_{1}(1-x_{1}-y_{1})+b$ $\displaystyle b^{2}-4ac$
$\displaystyle<(2x_{1}y_{2}(1-x_{1}-y_{1})+b)^{2},\quad
0>2x_{1}y_{2}(1-x_{1}-y_{1})+b.$

This concludes the proof. ∎

Using Lemma 5.2, Lemma 5.3, and Lemma 5.4, we characterize the signed support
of a bivariate $5$-nomial such that signed reduced $A$-discriminant has two
critical points.

######  Theorem 5.5.

Let $(\mathcal{A},\varepsilon)$ be the full-dimensional signed support of a
bivariate $5$-nomial with Gale dual matrix $B\in\mathbb{R}^{5\times 2}$. The
map $\bar{\xi}_{B,\varepsilon}$ has two critical points only if
$\\#\mathcal{A}_{+}=3,\,\\#\mathcal{A}_{-}=2$ and
$\dim\operatorname{Conv}(\mathcal{A}_{+})=2$, or
$\\#\mathcal{A}_{+}=2,\,\\#\mathcal{A}_{-}=3$ and
$\dim\operatorname{Conv}(\mathcal{A}_{-})=2$.

Assume that
$\mathcal{A}_{+}=\\{\alpha_{1},\alpha_{2},\alpha_{3}\\},\,\mathcal{A}_{-}=\\{\alpha_{4},\alpha_{5}\\}$
and $P=\operatorname{Conv}(\mathcal{A}_{+})$ has dimension $2$. Let
$M\in\mathbb{R}^{2\times 2}$ be an invertible matrix such that
$M(\alpha_{2}-\alpha_{1})=e_{1},\,M(\alpha_{3}-\alpha_{1})=e_{2}$ and
$v=-M\alpha_{1}$. Denote $(x_{1},y_{1})^{\top}=M\alpha_{4}+v$,
$(x_{2},y_{2})^{\top}=M\alpha_{5}+v$ and $a,b,c$ the expressions in
$x_{1},y_{1},x_{2},y_{2}$ from (34). The map $\bar{\xi}_{B,\varepsilon}$ has
two critical points if and only if
$\alpha_{4}\in\operatorname{int}(P^{+,j,i})$ and
$\alpha_{5}\in\operatorname{int}(P^{+,i,j})$ for $i\neq j\in\\{0,1,2\\}$ and
the coordinates of $\alpha_{4}$ and $\alpha_{5}$ satisfy the following
inequalities:

(37) $\displaystyle b^{2}-4ac$ $\displaystyle>0$ $\displaystyle b^{2}-4ac$
$\displaystyle<(2x_{2}y_{1}(1-x_{1}-y_{1})+b)^{2},\quad
0<2x_{2}y_{1}(1-x_{1}-y_{1})+b$ $\displaystyle b^{2}-4ac$
$\displaystyle<(2x_{1}y_{2}(1-x_{1}-y_{1})+b)^{2},\quad
0>2x_{1}y_{2}(1-x_{1}-y_{1})+b.$

###### Proof.

If all the exponent vectors are positive (resp. negative), then
$\Gamma_{\varepsilon}(A,B)=\emptyset$. If $\\#\mathcal{A}_{+}=1$ or
$\\#\mathcal{A}_{-}=1$, then $\bar{\xi}_{B,\varepsilon}$ does not have any
critical point by Theorem 4.11 and Proposition 4.8. Thus, in order to have a
critical point of $\bar{\xi}_{B,\varepsilon}$ one needs
$\\#\mathcal{A}_{+}\geq 2,\,\\#\mathcal{A}_{-}\geq 2$.

Write $P=\operatorname{Conv}(\mathcal{A}_{+})$ and
$Q=\operatorname{Conv}(\mathcal{A}_{-})$ and assume $\dim P=\dim Q=1$. Since
$\dim\operatorname{Conv}(P\cup Q)=\dim\operatorname{Conv}(\mathcal{A})=2$, the
intersection $P\cap Q$ is either empty or a point. If
$\operatorname{relint}(P)\cap\operatorname{relint}(Q)=\emptyset$, then $P$ and
$Q$ can be separated by an affine hyperplane [15, Section 2.2, Theorem 2] and
therefore $\Gamma_{\varepsilon}(A,B)=\emptyset$ by Theorem 3.3. If
$\operatorname{relint}(P)\cap\operatorname{relint}(Q)\neq\emptyset$, then the
affine hull of $P$ (resp. $Q$) is a strict enclosing hyperplane of
$\mathcal{A}_{+}$ (resp. $\mathcal{A}_{-}$). In that case, we use Theorem 4.10
and Proposition 4.8 to conclude that $\bar{\xi}_{B,\varepsilon}$ does not have
any critical point. This shows the first part of the theorem.

In the rest of the proof, we assume
$\mathcal{A}_{+}=\\{\alpha_{1},\alpha_{2},\alpha_{3}\\},\,\mathcal{A}_{-}=\\{\alpha_{4},\alpha_{5}\\}$
and $\dim P=2$. Choose the order of $\alpha_{1},\alpha_{2},\alpha_{3}$ so that
${\displaystyle\det\begin{bmatrix}1&1&1\\\
\alpha_{1}&\alpha_{2}&\alpha_{3}\end{bmatrix}>0}$. Then there exists an
invertible matrix $M\in\mathbb{R}^{2\times 2}$ with positive determinant such
that $M(\alpha_{2}-\alpha_{1})=e_{1}$ and $M(\alpha_{3}-\alpha_{1})=e_{2}$.
Let $v:=-M\alpha_{1}$. By construction, the affine linear map

$L\colon\,\mathbb{R}^{2}\to\mathbb{R}^{2},\quad a\mapsto Ma+v,$

satisfies $L(\alpha_{1})=0,\,L(\alpha_{2})=e_{1},\,L(\alpha_{3})=e_{2}$ and
$L(P)=\Delta_{2}$. By Proposition 2.3, we assume without loss of generality
that

$\displaystyle\alpha_{1}=\begin{bmatrix}0\\\
0\end{bmatrix},\alpha_{2}=\begin{bmatrix}1\\\
0\end{bmatrix},\alpha_{3}=\begin{bmatrix}0\\\
1\end{bmatrix},\alpha_{4}=\begin{bmatrix}x_{1}\\\
y_{1}\end{bmatrix},\alpha_{5}=\begin{bmatrix}x_{2}\\\ y_{2}\end{bmatrix},$

and choose the Gale dual matrix $B$ as in (33).

If $\alpha_{4},\alpha_{5}$ are separated from $\Delta_{2}$ by an affine
hyperplane, then $\Gamma_{\varepsilon}(A,B)=\emptyset$ by Theorem 3.3. If
$\alpha_{4},\alpha_{5}\in\operatorname{int}(\Delta_{2})$, then
$\bar{\xi}_{B,\varepsilon}$ does not have any critical point by Theorem 4.12
and Proposition 4.8. If $\alpha_{4}$ (resp. $\alpha_{5}$) lies on a supporting
hyperplane of a facet of $P$, then $a=0$ (resp. $c=0$). From this follows that
$q_{B}$ has at most two monomial terms, so $q_{B}$ has at most one positive
root. Thus, $\bar{\xi}_{B,\varepsilon}$ has at most one critical point.

In the following, we investigate the remaining cases:

  * (I)

$\alpha_{4}\in\operatorname{int}(\Delta_{2})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{+,i})$ for some
$i\in\\{0,1,2\\}$.

  * (II)

$\alpha_{4}\in\operatorname{int}(\Delta_{2}^{-,i})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{+,i})$ for some
$i\in\\{0,1,2\\}$.

  * (III)

$\alpha_{4}\in\operatorname{int}(\Delta_{2})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{-,i})$ for some
$i\in\\{0,1,2\\}$.

  * (IV)

$\alpha_{4}\in\operatorname{int}(\Delta_{2}^{+,i})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{+,j})$ for some $i\neq
j\in\\{0,1,2\\}$.

For (I) and (II), Lemma 5.2 implies that $\bar{\xi}_{B,\varepsilon}$ has at
most one critical point. In case (III), by rotating the exponent vectors we
assume without loss of generality that
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{-,0})$. Now, it follows from
Lemma 5.3 that $\bar{\xi}_{B,\varepsilon}$ has one critical point. Under the
assumption in (IV), we rotate again the exponent vectors to achieve
$\alpha_{4}\in\operatorname{int}(\Delta_{2}^{+,1})$ and
$\alpha_{5}\in\operatorname{int}(\Delta_{2}^{+,2})$. This rotation does not
change the number of critical points of $\bar{\xi}_{B,\varepsilon}$ by
Corollary 2.4. We use Lemma 5.4 to conclude that $\bar{\xi}_{B,\varepsilon}$
has two critical points if and only if the inequalities in (37) are satisfied.
∎

######  Remark 5.6.

Consider the signed support

$\displaystyle\mathcal{A}_{+}=\Big{\\{}\begin{bmatrix}0\\\
0\end{bmatrix},\begin{bmatrix}1\\\ 0\end{bmatrix},\begin{bmatrix}0\\\
1\end{bmatrix}\Big{\\}},\quad\mathcal{A}_{-}=\Big{\\{}\begin{bmatrix}x_{1}\\\
y_{1}\end{bmatrix},\begin{bmatrix}x_{2}\\\ y_{2}\end{bmatrix}\Big{\\}}.$

Using the Mathematica function Reduce [16], we verified that for every fixed
$(x_{1},y_{1})\in\operatorname{int}(\Delta_{2}^{+,1,2})$ and $0<x_{2}<1$,
there always exists $y_{2}<0$ satisfying the inequalities in (37). With other
words, for given $(x_{1},y_{1})\in\operatorname{int}(\Delta_{2}^{+,1,2})$
there exists a $(x_{2},y_{2})\in\operatorname{int}(\Delta_{2}^{+,2,1})$ such
that $\bar{\xi}_{B,\varepsilon}$ has two critical points.

In Figure 6, we depicted the region of $(x_{2},y_{2})$’s satisfying the
inequalities in (37) for $(x_{1},y_{1})=(-0.1,0.3)$.

Figure 6. An illustration of the chambers as defined in (29),(30),(31) for
$P=\operatorname{Conv}((0,0),(1,0),(0,1))$. The red circles denote positive
exponent vectors $\mathcal{A}_{+}=\big{\\{}\begin{bmatrix}0\\\
0\end{bmatrix},\begin{bmatrix}1\\\ 0\end{bmatrix},\begin{bmatrix}0\\\
1\end{bmatrix}\big{\\}}$, the blue dot denote a negative exponent vector
$\begin{bmatrix}-0.1\\\ 0.3\end{bmatrix}$. The blue region contains all
negative exponent vectors $\begin{bmatrix}x_{2}\\\ y_{2}\end{bmatrix}\in
P^{+,2,1}$ such that $\bar{\xi}_{B,\varepsilon}$ has two critical points.

#### Acknowledgements.

WD and JMR were partially supported by NSF grant CCF-1900881. MLT thanks
Elisenda Feliu for useful discussions and comments on the manuscript. MLT was
supported by the European Union under the Grant Agreement no. 101044561,
POSALG. Views and opinions expressed are those of the author(s) only and do
not necessarily reflect those of the European Union or European Research
Council (ERC). Neither the European Union nor ERC can be held responsible for
them.

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