A $\tau$-Conjecture for Newton Polygons

Abstract

One can associate to any bivariate polynomial $P(X,Y)$ its Newton polygon. This is the convex hull of the points $(i,j)$ such that the monomial $X^i Y^j$ appears in $P$ with a nonzero coefficient. We conjecture that when $P$ is expressed as a sum of products of sparse polynomials, the number of edges of its Newton polygon is polynomially bounded in the size of such an expression. We show that this so-called $\tau$-conjecture for Newton polygons, even in a weak form, implies that the permanent polynomial is not computable by polynomial-size arithmetic circuits. We make the same observation for a weak version of an earlier real $\tau$-conjecture. Finally, we make some progress toward the $\tau$-conjecture for Newton polygons using recent results from combinatorial geometry.

Appendix: Newton polygon of \(fg+1\)

In this section we denote by $0$ the point in the plane with coordinates $(0,0)$.

We give here (in Theorem 7) a linear upper bound assuming the following two properties:

• (i) The polynomials $f$ and $g$ have the same support, i.e., ${\mathrm {Mon}}(f)={\mathrm {Mon}}(g)$. We denote by $\{p_0,\ldots ,p_{t-1}\}$ this common support.

• (ii) If $f$ and $g$ have a constant term, we assume without loss of generality that $p_0=0$ and we add the following requirement: if $p_j$ is an extremal point of ${\mathrm {conv}}(p_1,p_2,\ldots ,p_{t-1})$, then $2p_j$ is not in the support of $f$ and $g$.

We do not know how to prove a linear upper bound assuming only (i). Condition (ii) is satisfied in particular when the points in ${\mathrm {Mon}}(f)={\mathrm {Mon}}(g)$ are convexly independent.

The interesting case, which we consider first, is when $f$ and $g$ have a constant term but $fg+1$ has no constant term. As explained previously, we assume that $p_0$ corresponds to the constant terms of $f$ and $g$, i.e., $p_0=0$. Under these hypotheses, we have the following result.

Proposition 2

Under assumptions (i) and (ii),

$$\begin{aligned} {\mathrm {Newt}}(fg+1)={\mathrm {conv}}(2p_1,\ldots ,2p_{t-1},(p_i)_{i \in I}), \end{aligned}$$

where $(p_i)_{i \in I}$ is the subset of those monomials in ${\mathrm {Mon}}(f)$ that appear in $fg+1$ with a nonzero coefficient.

Proof

We first prove the inclusion from left to right. Since $fg+1$ has no constant term, all monomials of $fg+1$ are of the form $p_i+p_j$, where $i \ge 1$ or $j \ge 1$. Consider first the case where $i$ and $j$ are both nonzero. If $i=j$, then this monomial appears on the right-hand side, and if $i \ne j$, then it is the middle point of two points (namely, $2p_i$ and $2p_j$) appearing on the right-hand side. The remaining case is when $i=0$ or $j=0$. If, for example, $j=0$, then we have $p_i+p_j=p_i$, and we see from the definition of $I$ that this monomial also appears on the right-hand side.

Now we prove the inclusion from right to left. Again by definition of $I$, all the $p_i$ with $i \in I$ are monomials of $fg+1$. Hence, it remains to show that

$$\begin{aligned} {\mathrm {conv}}(2p_1,\ldots ,2p_{t-1}) \subseteq {\mathrm {Newt}}(fg+1). \end{aligned}$$

The left-hand side can be written as ${\mathrm {conv}}((2p_j)_{j \in J})$, where the $(p_j)_{j\in J}$ form a convexly independent subset of $\{p_1,\ldots ,p_{t-1}\}$. Any monomial of the form $2p_j$ with $j \in J$ appears in $fg+1$ with a nonzero coefficient because it can be obtained in a unique way by expansion of the product $fg$. Assume indeed that $2p_j=p_i+p_k$, with $i \ne k$. Then $p_j$ is the middle point of $p_i$ and $p_k$. If $i \ge 1$ and $k \ge 1$, this is impossible by construction of $J$. If $i=0$ or $k=0$, this is also impossible by hypothesis (ii). We thus have ${\mathrm {conv}}((2p_j)_{j \in J}) \subseteq {\mathrm {Newt}}(fg+1)$, and the proof is complete. $\square$

We note that this proposition does not hold without assumption (ii), as shown by the following example. Take $f=1+X^2Y+XY^2+(1/2)X^2Y^4+(1/2)X^4Y^2$ and $g=-1+X^2Y+XY^2-(1/2)X^2Y^4-(1/2)X^4Y^2$. Then $fg+1=2X^3Y^3-(1/2)X^6Y^6-(1/4)X^4Y^8-(1/4)X^8Y^4$. The monomial $X^3Y^3$ is a vertex of ${\mathrm {Newt}}(fg+1)$ but is not of the form $p_i$ or $2p_j$ prescribed by Proposition 2.

Theorem 7

Under the same assumptions (i) and (ii) as previously, ${\mathrm {Newt}}(fg+1)$ has at most $t+1$ edges, where $t$ denotes the number of monomials of $f$ and $g$.

Proof

We continue to denote the common support of $f$ and $g$ by $\{p_0,\ldots ,p_{t-1}\}$. If $0$ does not belong to this support, then ${\mathrm {Newt}}(fg+1)$ is the convex hull of $\{0\}$ and ${\mathrm {Newt}}(fg)$. Moreover, ${\mathrm {Newt}}(fg)={\mathrm {Newt}}(f)+{\mathrm {Newt}}(g)={\mathrm {conv}}(2p_0,\ldots ,2p_{t-1})$.

If $0$ is in the support and $fg+1$ has a constant term, then ${\mathrm {Newt}}(fg+1)={\mathrm {Newt}}(fg)$ has at most $t$ edges ($t$ and not $2t$ since $f$ and $g$ have the same support).

In the remaining case ($0$ is in the support but $fg+1$ has no constant term), we need to use hypothesis (ii). This case is treated in Proposition 2. At first sight, it seems that ${\mathrm {Newt}}(fg+1)$ can have up to $2(t-1)$ vertices, but the list of possible vertices can be shortened by picking a convexly independent subsequence. More precisely, write ${\mathrm {conv}}(2p_1,\ldots ,2p_{t-1},(p_i)_{i \in I})={\mathrm {conv}}((2p_j)_{j \in J},(p_k)_{k \in K})$, where $J \subseteq \{1,\ldots ,t-1\}$ and $K \subseteq I$ are chosen such that the points in this sequence are convexly independent. By the lemma that follows, $| J \cap K| \le 2$. As a result, the number of points in the sequence is $|J|+|K| = |J \cup K| + | J \cap K| \le (t-1)+2=t+1$. $\square$

Lemma 3

If $p,\, q,\, r$ are three distinct nonzero points in the plane, then the six points $p,\, q,\, r,\, 2p,\, 2q,\, 2r$ are not convexly independent.

This is clear from a picture and can be proved, for instance, by considering the four points $0,\, p,\, q,\, r$. There are two cases:

• 1. If these four points are convexly independent, assume, for instance, that $pq$ is a diagonal of the quadrangle $0prq$. Then the line $pq$ separates $0$ from $r$. As a result, $r \in {\mathrm {conv}}(p,q,2r)$.

• 2. If the four points are not convexly independent, assume, for instance, that $r \in {\mathrm {conv}}(0,p,q)$. In this case, $2r \in {\mathrm {conv}}(2p,2q,r)$. $\square$