Multijoints and Factorisation

Abstract

We solve the dual multijoint problem and prove the existence of so-called factorisations for arbitrary fields and multijoints of $k_j$-planes. More generally, we deduce a discrete analogue of a theorem due in essence to Bourgain and Guth. Our result is a universal statement which describes a property of the discrete wedge product without any explicit reference to multijoints and is stated as follows: Suppose that $k_1 + \ldots + k_d = n$. There is a constant $C=C(n)$ so that for any field $\mathbb {F}$ and for any finitely supported function $S : \mathbb {F}^n \rightarrow \mathbb {R}_{\ge 0}$, there are factorising functions $s_{k_j} : \mathbb {F}^n\times {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)\rightarrow \mathbb {R}_{\ge 0}$ such that

$$\begin{aligned} \left( V_1 \wedge \cdots \wedge V_d\right) S\left( p\right) ^d \le \prod _{j=1}^d s_{k_j}\left( p, V_j\right) , \end{aligned}$$

for every $p\in \mathbb {F}^n$ and every tuple of planes $V_j\in {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)$, and

$$\begin{aligned} \sum _{p\in \pi _j} s(p, e(\pi _j)) =C \left| \left| S\right| \right| _d, \end{aligned}$$

for every $k_j$-plane $\pi _j\subset \mathbb {F}^n$, where $e(\pi _j)\in {{\,\mathrm{{Gr}}\,}}(k_j,\mathbb {F}^n)$, is the translate of $\pi _j$ that contains the origin and $\wedge$ denotes the discrete wedge product.

1 Introduction

The multijoint problem is the discrete analogue of the multilinear Kakeya problem. This discrete problem is cast in affine space over an arbitrary field $\mathbb {F}$, finite or otherwise. Consider the projective space of lines through the origin in $\mathbb {F}^d$,¹ which we denote by $\mathbb {S}^{d-1}$, and its elements, which we call directions. If $l\subset \mathbb {F}^d$ is a line, let $e(l) \in \mathbb {S}^{d-1}$, the direction of l, be the translate of l which contains the origin. Let $\omega _j\in \mathbb {S}^{d-1}$ for $1\le j \le d$. We define the discrete wedge product $\wedge _{j=1}^d \omega _j = \omega _1 \wedge \cdots \wedge \omega _d = 1$ if the directions $\omega _1, \ldots , \omega _d$ span $\mathbb {F}^d$, and 0 otherwise. Then, we can define the multijoint kernel $\delta$ by

$$\begin{aligned} \delta (p,l_1, \ldots , l_d):= \left( \prod _{j=1}^d \chi _{l_j}(p)\right) e(l_1)\wedge \cdots \wedge e(l_d), \end{aligned}$$

for lines $l_j \subset \mathbb {F}^d$ and points $p\in \mathbb {F}^d$. For each $1\le j \le d$, let $\mathcal {L}_j$ be a set of lines in $\mathbb {F}^d$ and we define set of their multijoints by

$$\begin{aligned} J:= \left\{ p\in \mathbb {F}^d: \exists (l_j)_j\in \mathcal {L}_1\times \cdots \times \mathcal {L}_d \text { s.t. } \delta (p,l_1, \ldots , l_d)=1\right\} . \end{aligned}$$

Then the multijoint problem consists of establishing the inequality

$$\begin{aligned} \sum _{p\in \mathbb {F}^d} \left( \sum _{l_1\in \mathcal {L}_1}\cdots \sum _{l_d\in \mathcal {L}_d} \delta (p,l_1, \ldots , l_d)f_1(l_1) \cdots f_d(l_d)\right) ^{\frac{1}{d-1}} \lesssim \prod _{j=1}^d \left( \sum _{l_j\in \mathcal {L}_j}f_j(l_j)\right) ^{\frac{1}{d-1}} \end{aligned}$$

(1)

for arbitrary $f_j: \mathcal {L}_j \rightarrow \mathbb {R}_{\ge 0}$, with implicit constants independent of $\mathcal {L}_j$ and $f_j$. This was first proved by Zhang [1].

1.1 Multijoint Inequalities as Boundedness of Operators

The multijoint inequality describes the boundedness of a multilinear operator and we give a proof of this boundedness by establishing two assertions.

We define the multijoint operator by

$$\begin{aligned} T[f_1, \ldots , f_d](p)= \sum _{l_1\in \mathcal {L}_1}\cdots \sum _{l_d\in \mathcal {L}_d} \delta (p,l_1, \ldots , l_d)f_1(l_1)\cdots f_d(l_d), \end{aligned}$$

for $p \in \mathbb {F}^d$ and arbitrary $f_j : \mathcal {L}_j\rightarrow \mathbb {R}_{\ge 0}$. Then we can express the multijoint inequality (1) as the boundedness of the multilinear operator T,

$$\begin{aligned} \left| \left| T[f_1, \ldots , f_d]^{\frac{1}{d}}\right| \right| _{L^{\frac{d}{d-1}}(J)} \le C \prod _{j=1}^d \left| \left| f_j\right| \right| _{L^1(\mathcal {L}_j)}^{\frac{1}{d}}, \end{aligned}$$

(2)

for arbitrary functions $f_j: \mathcal {L}_j \rightarrow \mathbb {R}_{\ge 0}$, where we use $\left| \left| f\right| \right| _{L^p(A)}$ to denote $\left( \sum _{a\in A}|{f(a)}|^p\right) ^\frac{1}{p}$ for any discrete set A.

Now, consider an arbitrary non-negative test function $S : J \rightarrow \mathbb {R}_{\ge 0}$. Suppose that we can find a “factorising” function $s: J\times \mathbb {S}^{d-1} \rightarrow \mathbb {R}_{\ge 0}$ so that

$$\begin{aligned} T[f_1, \ldots , f_d](p)S(p)^d \le \sum _{l_1\in \mathcal {L}_1}\cdots \sum _{l_d\in \mathcal {L}_d} \prod _{j=1}^d s(p, e(l_j))f_j(l_j) \end{aligned}$$

uniformly over all $f_j:\mathcal {L}_j \rightarrow \mathbb {R}_{\ge 0}$ - this is the first assertion mentioned above. Defining positive linear operators $T_j; L^1(\mathcal {L}_j) \rightarrow L^1(J)$ by

$$\begin{aligned} T_j[f_j](p) = \sum _{l_j\in \mathcal {L}_j}s(p,e(l_j))f_j(l_j), \end{aligned}$$

for any $p\in J$, let us further suppose that s can be chosen so that the operators $T_j$ each satisfy

$$\begin{aligned} \left| \left| T_j[f_j]\right| \right| _{L^1(J)}\le C\left| \left| S\right| \right| _d\left| \left| f_j\right| \right| _{L^1(\mathcal {L}_j)} \end{aligned}$$

for all $f_j: \mathcal {L}_j\rightarrow \mathbb {R}_{\ge 0}$ and $C = C(d)$ - this is the second assertion. Then

$$\begin{aligned} \sum _{p\in J}S(p)T[f_1, \ldots , f_d](p)^{\frac{1}{d}}\le & {} \sum _{p\in J} \prod _{j=1}^d T_j [f_j](p)^{\frac{1}{d}}\\\le & {} \prod _{j=1}^d \left| \left| T_j[f_j]\right| \right| _1^\frac{1}{d} \le C\left| \left| S\right| \right| _d\prod _{j=1}^d \left| \left| f_j\right| \right| _1^\frac{1}{d}, \end{aligned}$$

where we have used Hölder’s inequality followed by boundedness of each $T_j$. Therefore, to prove (2), and hence (1), it suffices, for arbitrary S, to find bounded linear operators $T_j$, as above, so that

$$\begin{aligned} S(p)T[f_1, \ldots , f_d](p)^{1/d} \le T_1[f_1](p)^{1/d}\cdots T_d[f_d](p)^{1/d} \end{aligned}$$

for all $f_j: \mathcal {L}_j \rightarrow \mathbb {R}_{\ge 0}$ and $p\in J$.

This flavour of analysis is closely related to so-called geometric multilinear duality [2]. Such methods were used by Carbery and Valdimarsson in their proof of Guth’s endpoint multilinear Kakeya theorem [3]. Importantly, to analyse the multijoint problem in terms of functional operators, it is crucial we bound $ST^{1/d}$ by the geometric mean of the operators $T_j$, which contains precisely d factors and weights that depend on d only. That we bound T from above by a geometric mean is what motivates the description of the results described in this article as factorisation theorems.

This analysis was motivated by the following theorem, essentially due to Bourgain and Guth [4, Sect.7], which formed the cornerstone of their proof of the general multilinear Kakeya theorem. The precise formulation of this result was not stated by Bourgain and Guth, but was given in [3] by Carbery and Valdimarsson.

To state these factorisation results, we momentarily use $\mathbb {S}^{d-1}$ to denote the classical unit sphere in $\mathbb {R}^d$. For any non-zero vectors $\omega _1, \ldots , \omega _d\in \mathbb {S}^{d-1}$, we define the Euclidean wedge product $\omega _1 \wedge \cdots \wedge \omega _d$ to be the unsigned volume of the parallelepiped with edges $\omega _1, \ldots , \omega _d$.

Theorem

(Multilinear Kakeya Factorisation Theorem [3,4,5]) Let $\mathcal {Q}$ be the lattice of unit cubes in $\mathbb {R}^d$ and let $S: \mathcal {Q}\rightarrow [1, \infty )$ be finitely supported. Then there exists a function $s: \mathcal {Q}\times \mathbb {S}^{d-1} \rightarrow \mathbb {R}_{\ge 0}$ so that

$$\begin{aligned} (\omega _1 \wedge \cdots \wedge \omega _d) S(Q)^d\le \prod _{j=1}^ds(Q, \omega _j) \end{aligned}$$

for all $Q\in \mathcal {Q}$ and $\omega _j\in \mathbb {S}^{d-1}$, where $\wedge$ denotes the Euclidean wedge product, and so that

$$\begin{aligned} \sum _{Q: T\cap Q \ne \emptyset } s(Q,e(T)) \lesssim \left| \left| S\right| \right| _d \end{aligned}$$

for any tube $T\subset \mathbb {R}^d$ with unit cross-sectional area and direction e(T).

The function s that appears in the multilinear Kakeya factorisation theorem is constructed in terms of the so-called visibility and directional surface area of a suitable polynomial hypersurface associated to the given configuration of tubes.

1.2 Results

The multijoint problem is a discrete analogue of the multilinear Kakeya problem. This can be seen from (1) by taking $\mathbb {F}=\mathbb {R}$, replacing the counting measure with the Lebesgue measure, the discrete wedge product (in the definition of $\delta$) with the absolute value of the Euclidean wedge product, and lines with 1-tubes. It was observed in [3] (and implicity in [4, 5]) that the multilinear Kakeya theorem follows from the multilinear Kakeya factorisation theorem, above. Although the multijoint problem was proved by Zhang [1] and more recently, higher-dimensional generalisations were proved by Tidor, Yu and Zhao [6], a discrete analogue to the multilinear Kakeya factorisation theorem has remained unproven, until now.

From now on, we will fix our notation so that n denotes the underlying spacial dimension and d denotes the degree of multilinearity.

Suppose $V_1, \ldots , V_d$ are $k_1$-, $\ldots , k_d$-dimensional vector spaces in $\mathbb {F}^n$, respectively, where $k_1 + \ldots + k_d = n$. We define the discrete wedge product on these $k_j$-planes by

$$\begin{aligned} \wedge _{j=1}^dV_j:= \wedge _{j=1}^d\wedge _{k=1}^{k_j} \omega _{j, k}, \end{aligned}$$

where $\omega _{j, 1}, \ldots , \omega _{j, k_j} \in \mathbb {S}^{n-1}$ is a choice of $k_j$ linearly independent directions contained in $V_j$, for each $1\le j \le d$. We can now define the ($k_j$-plane) multijoint kernel by

$$\begin{aligned} \delta (p,\pi _1, \ldots , \pi _d) = \left( \prod _{j=1}^d \chi _{\pi _j}(p)\right) e(\pi _1)\wedge \cdots \wedge e(\pi _d), \end{aligned}$$

for all $p\in \mathbb {F}^n$ and all $k_j$-planes $\pi _j$. Given sets of $k_j$-planes, $\Pi _j$, we say that p is a $k_j$-multijoint, or multijoint in short, if there are planes $\pi _j\in \Pi _j$ so that $\delta (p,\pi _1, \ldots , \pi _d)=1$, and we say that the planes $\pi _j$ form a multijoint at p.

Let $1 \le k \le n$. Recall that the Grassmannian with respect to k and $\mathbb {F}^n$ is the set of all k-subspaces of $\mathbb {F}^n$, which we denote by ${{\,\mathrm{{Gr}}\,}}(k, \mathbb {F}^n)$. Given any affine k-plane $\pi$, let $e(\pi )\in {{\,\mathrm{{Gr}}\,}}(k, \mathbb {F}^n)$ denote the translate of $\pi$ that contains the origin.

Theorem 1

(Discrete Factorisation Theorem) Let $k_1 + \ldots + k_d = n$. For all finitely supported $S:\mathbb {F}^n\rightarrow \mathbb {R}_{\ge 0}$ with $\left| \left| S\right| \right| _{d}=1$, for each $1\le j \le d$, there exists a function $s_{k_j}: \mathbb {F}^n\times {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n) \rightarrow \mathbb {R}_{\ge 0}$ for each $1\le j \le d$ so that

$$\begin{aligned} (V_1\wedge \cdots \wedge V_d )S(p)^d \le \prod _{j=1}^ds_{k_j}(p,V_j), \end{aligned}$$

(3)

for all $p\in \mathbb {F}^n$ and $V_j\in {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)$ for $1\le j \le d$, and so that for any $1\le j \le d$,

$$\begin{aligned} \sum _{p\in \pi _j}s_{k_j}(p,e(\pi _j)) \lesssim _n 1, \end{aligned}$$

(4)

for all affine $k_j$-subspaces $\pi _j\subset \mathbb {F}^n$.

Theorem 1 is precisely an analogue of the Multilinear Kakeya Factorisation Theorem due to Bourgain and Guth. While dual to the multijoint problem, this result does not refer to any specific sets of lines or planes in $\mathbb {F}^n$. In this sense, it is a universal result, which describes a property of the pair $(\mathbb {F}^n, \wedge )$. In particular, since this is a discrete analogue of the factorisation result of Bourgain and Guth, we have thus established a perspective from which we may understand the geometry of affine space and Euclidean space to be the same. For example, it is intuitive that transversality in affine and Euclidean space are closely related and our result offers a new formalisation of this heuristic.

The function s described by this result is precisely the one alluded to in Sect. 1.1 where the first and second assertions are described by the first and second displays, respectively. Hence, Theorem 1 implies that (1) is true. On the other hand, in closely related work in collaboration with Carbery [7], we show in the case where $k_1, \ldots , k_d = 1$ (multijoints of lines) that Theorem 1 follows if (1) is true. Hence, these two statements are equivalent and it is this equivalence that we describe as geometric multilinear duality.

1.3 Geometric Interpretation

We take a moment to describe a geometrical interpretation of Theorem 1. Loosely, the discrete wedge product is well approximated by a geometric mean. Alternatively, to understand the geometry determined by d-tuples of directions at finitely many points in affine space, it suffices to consider each component of the tuple independently. To make these heuristics precise, we re-examine the arguments of Sect. 1.1 in the case for multijoints with each $k_j=1$.

Fix d finite sets of lines $\mathcal {L}_j\subset \mathbb {F}^d$ and let $f_j : \mathcal {L}_j\rightarrow \mathbb {R}_{\ge 0}$. To evaluate the $d/(d-1)$-norm of $T[f_1, \ldots , f_d]^\frac{1}{d}$, given an arbitrary test function $S : \mathbb {F}^d\rightarrow \mathbb {R}_{\ge 0}$, we consider

$$\begin{aligned} \sum _{p\in \mathbb {F}^d}S(p)\left( \sum _{l_1, \ldots , l_d}\left( \prod _j\chi _{l_j}(p)f_j(l_j)\right) \left( \wedge _j e(l_j)\right) \right) ^\frac{1}{d}. \end{aligned}$$

Observing that to evaluate this expression, at every point in p, we must consider the wedge product of a d-tuple of lines. By applying the first inequality from Theorem 1, we bound this expression from above by

$$\begin{aligned} \sum _{p\in \mathbb {F}^d}\left( \sum _{l_1, \ldots , l_d} \prod _j \chi _{l_j}(p)f_j(l_j)s(p,e(l_j))\right) ^\frac{1}{d}. \end{aligned}$$

The innermost summand is now a product of independent functions, so we may apply Hölder’s inequality to consider each set of lines separately, as

$$\begin{aligned} \prod _j \left( \sum _{p\in \mathbb {F}^d}\sum _{l_j}\chi _{l_j}(p)f_j(l_j)s(p, (l_j))\right) ^\frac{1}{d}. \end{aligned}$$

Finally, changing the order of summation and applying the second display of Theorem 1, this is bounded above by

$$\begin{aligned} \prod _j \left( \sum _{l_j}f_j(l_j) \sum _{p\in l_j} s(p,e(l_j))\right) ^\frac{1}{d}\lesssim \prod _j \left| \left| f_j\right| \right| _{L^1(\mathcal {L}_j)}^\frac{1}{d}, \end{aligned}$$

as desired.

We expect this universal geometric theorem, describing the properties of the pair $(\mathbb {F}^d, \wedge )$ to be useful in the study of non-transversal problems such as non-transversal multilinear restriction. Removing the assumption of transversality would require the inclusion of a wedge product as a geometric weight, for example in a multilinear Fourier extension operator. Thereafter, if one can further refine Theorem 1 so that the first display is an exact equality, then using arguments similar to those found in [8], one can reduce the non-transversal multilinear extension problem to a generalised Mizohata–Takeuchi conjecture. It is important in this application that our statement of Theorem 1 describes a geometric mean of precisely d factors and is not an application-dependent mean.

1.4 Overview of the Article

In Sect. 2, we recall the notion of “handicaps”, as introduced by Yu and Zhao in [9], and further developed in [6]. Handicaps allow us to choose polynomial vanishing conditions so that we may apply the polynomial method while respecting the geometry of multijoints.

In Sect. 3, we diverge from the approach in [6, 9] and adapt the recent novel developments introduced therein to establish our factorisation theorems.

2 Notation

We write $A\lesssim B$ to mean that there is a non-negative constant C, depending only on dimension, so that $A\le CB$. We write $A\lesssim _{n} B$ to mean that there is a constant C, depending only on n and the dimension, so that $A\le CB$. We write $B\gtrsim A$ and $B\gtrsim _{n} A$ to mean $A\lesssim B$ and $A \lesssim _{n}B$, respectively. Moreover, by $A\sim B$, we mean $A\lesssim B$ and $B \lesssim A$, and finally, by $A\sim _{n} B$, we mean $A\lesssim _{n}B$ and $B\lesssim _{n} A$.

3 Polynomials, Handicaps and Vanishing Conditions

3.1 The Polynomial Method

This section introduces the main tools we will employ and motivates the arguments that follow. We will work in an arbitrary field, $\mathbb {F}$, and any derivative will be the Hasse derivative. All arguments remain valid when $\mathbb {F}=\mathbb {R}$ with the usual derivative operator.

Let $f\in \mathbb {F}[x_1, \ldots , x_n]$ and $\alpha \in \mathbb {N}^n$ be a multi-index. The $\alpha$-th Hasse derivative of f is the coefficient of the monomial $z^\alpha = z_1^{\alpha _1}\cdots z_n^{\alpha _n}$ in the polynomial $p(x+z)\in \left( \mathbb {F}[x_1, \ldots , x_n]\right) [z_1, \ldots ,z_n]$. This is denoted by $D^\alpha f$. For further details, see [10] or [11].

Let $\mathbb {F}_\lambda [x_1, \ldots , x_n] = \{f\in \mathbb {F}[x_1, \ldots , x_n] : \textrm{deg}\, f \le \lambda \}$. Recall that for any field $\mathbb {F}$ and $\lambda \in \mathbb {N}$,

$$\begin{aligned} \dim \mathbb {F}_\lambda [x_1, \ldots , x_n] = {\lambda + n \atopwithdelims ()n}, \end{aligned}$$

(5)

and hence, $\dim \left( \left( \mathbb {F}_\lambda [x_1, \ldots , x_n]\right) ^*\right) = {\lambda + n \atopwithdelims ()n}$. From this fact, we can deduce a commonly used consequence, known as the parameter counting lemma. Indeed, checking whether a (Hasse) derivative of a low degree polynomial at a point is zero is equivalent to checking whether $\langle \phi , f\rangle =0$ for an appropriately chosen $\phi \in \left( \mathbb {F}_\lambda [x_1, \ldots , x_n]\right) ^*$. Therefore, the following so-called parameter counting Lemma follows from elementary linear algebra.

Lemma 1

(Parameter Counting) Let $\phi _1, \ldots , \phi _m$ be homogeneous linear functionals which act on $\mathbb {F}_\lambda [x_1, \ldots , x_n]$. If $m < {n + \lambda \atopwithdelims ()n}$, then there is a non-zero $f\in \mathbb {F}_\lambda [x_1, \ldots , x_n]$ so that $\langle \phi _i, f\rangle =0$, for each $1\le i \le m$.

We organise our arguments in such a way that the only dimensional constants that appear in our stated results arise directly from the implied constants in the inequalities described by ${n + \lambda \atopwithdelims ()n} \sim \lambda ^n$ or for any positive $k_1, \ldots , k_d$,

$$\begin{aligned} {k_1 + \lambda \atopwithdelims ()k_1} \cdots {k_d + \lambda \atopwithdelims ()k_d} \sim _n \lambda ^{k_1 + \ldots + k_d}. \end{aligned}$$

3.2 Handicaps

For any finite subset $J\subseteq \mathbb {F}^n$, a handicap is a function $\alpha : J \rightarrow \mathbb {Z}$. We will use $\alpha$ to equip $J\times \mathbb {Z}_{\ge 0}$ with a linear order so that it has a least element. We will then accumulate vanishing conditions as we increment along this linear order, starting from the least element.

Remark 1

The handicap keeps track of degrees of freedom in choosing vanishing conditions. More precisely, given a fixed $p_0\in J$, the degrees of freedom are the $(|{J}|-1)$ entries of $(\alpha _{p_0}-\alpha _p)_{p\in J\setminus \{p\}}$.   $\blacktriangleleft$

Since J is finite, we may equip it with a total order so that for $p, q\in J$, at most one of $p < q$, $p=q$ or $p>q$ holds. We define a total order (called the priority order in [6]), denoted by $\prec$, on $J\times \mathbb {Z}_{\ge 0}$ as follows. We say $(p,r) \prec (p^\prime , r^\prime )$ if

• $r-\alpha _p < r^\prime - \alpha _{p^\prime }$, or

• $r-\alpha _p = r^\prime - \alpha _{p^\prime }$ and $p<p^\prime$.

Moreover, $(p,r)=(p^\prime , r^\prime )$ if and only if $p=p^\prime$ and $r=r^\prime$. We write $\preceq$ to allow for this equality case.

3.3 Vanishing Conditions

Let $\Pi _1, \ldots , \Pi _d$ be finite sets of $k_1$-$, \ldots , k_d$-dimensional planes in $\mathbb {F}^n$, respectively, where $k_1 + \ldots + k_d= n$. Let $J = \{p : \exists (\pi _j)_j \text { so that }\delta (p, \pi _1 \ldots , \pi _d)=1\}$ be the associated set of multijoints. Let $\lambda \in \mathbb {N}$ be a parameter and let $\alpha : J \rightarrow \mathbb {Z}$ be a handicap. Choose an ordering of the set J and thereafter, define the total order $\prec = \prec _\alpha$ on $J\times \mathbb {Z}_{\ge 0}$, as in Sect. 2.2.

Let $\pi \subset \mathbb {F}^n$ be a k-plane which we fix for the remainder of Sect. 2.3. Without loss of generality, we assume that $\pi$ is spanned by the coordinate vectors $e_1, \ldots , e_k$, so we make the identification

$$\begin{aligned} \left\{ f|_\pi : f\in \mathbb {F}[x_1, \ldots , x_n]\right\} = \mathbb {F}[x_1, \ldots , x_k]. \end{aligned}$$

(6)

For each (p, r), we define $\mathbb {B}_r(p,\pi ,\lambda )$ to be the vector space of linear functionals which act on $\mathbb {F}_\lambda [x_1, \ldots , x_k]$ and are of the form $f\mapsto Df(p)$ for some differential operator D of order r acting on k-variate polynomials.

With $\alpha$, $\lambda$ and $\pi$ fixed, we now define sets $B(p,\pi , \alpha , \lambda )$ for $p\in \pi \cap J$. To do so, we perform an iterative procedure starting with the $\prec$-least element of $(\pi \cap J)\times \mathbb {Z}_{\ge 0}$ and proceeding to the $\prec$-next element on each iteration. Starting with (p, 0), the least element of $(\pi \cap J)\times \mathbb {Z}_{\ge 0}$, we choose a set $B_0(p,\pi ,\alpha ,\lambda )\subset \mathbb {B}_0(p,\pi ,\lambda )$ that is a basis for $\mathbb {B}_0(p,\pi ,\lambda )$.

Assume for each $(p^\prime ,r^\prime )\prec (p,r)$, that we have chosen sets $B_{r^\prime }(p^\prime ,\pi ,\alpha ,\lambda )$ so that the disjoint union, $\cup _{(p^\prime , r^\prime )\prec (p,r)} B_{r^\prime }(p^\prime , \pi , \alpha , \lambda )$ is a basis for ${{\,\textrm{span}\,}}_{(p^\prime , r^\prime )\prec (p,r)} \mathbb {B}_{r^\prime }(p^\prime , \pi , \lambda )$. Thereafter, choose $B_r(p,\pi ,\alpha ,\lambda ) \subset \mathbb {B}_r(p,\pi ,\lambda )$ so that the disjoint union

$$\begin{aligned} \bigcup _{(p^\prime , r^\prime )\preceq (p,r)} B_{r^\prime }(p^\prime , \pi , \alpha , \lambda ) \end{aligned}$$

is a basis for ${{\,\textrm{span}\,}}_{(p^\prime , r^\prime )\preceq (p,r)} \mathbb {B}_{r^\prime }(p^\prime , \pi , \lambda )$.

Remark 2

Writing these unions out with more complete notation,

$$\begin{aligned} \bigcup _{(p^\prime , r^\prime )\preceq (p,r)} B_{r^\prime }(p^\prime , \pi , \alpha , \lambda ) = \bigcup _{\begin{array}{c} (p^\prime , r^\prime ) \in (\pi \cap J) \times \mathbb {Z}_{\ge 0}:(p^\prime , r^\prime )\preceq (p,r) \end{array}} B_{r^\prime }(p^\prime , \pi , \alpha , \lambda ). \end{aligned}$$

Observe that we can write these unions as

$$\begin{aligned} \bigcup _{\begin{array}{c} p^\prime \in (p + e(\pi ))\cap J, \, r^\prime \in \mathbb {Z}_{\ge 0}:(p^\prime , r^\prime )\preceq (p,r) \end{array}} B_{r^\prime }(p^\prime , \pi , \alpha , \lambda ), \end{aligned}$$

since $\pi = p+e(\pi )$ for any k-plane $\pi$. Since the first argument of B is p, the only information this construction requires from the second argument of B is which vector space is parallel to $\pi$, namely, $e(\pi )\in {{\,\mathrm{{Gr}}\,}}(k,\mathbb {F}^n)$. It follows that for this k-plane construction, we may also reduce the second argument to $e(\pi )$, although for notational convenience, we again continue to simply write $B(p,\pi ,\alpha ,\lambda )$.

Further observe that for any other k-plane $\pi ^\prime$ so that $\pi \cap J = \pi ^\prime \cap J$, we have that $B(p,\pi ,\alpha ,\lambda ) = B(p,\pi ^\prime , \alpha ,\lambda )$ for all $p\in \pi \cap J$.   $\blacktriangleleft$

Let $B(p,\pi ,\alpha ,\lambda ) = \cup _{r\ge 0} B_r(p,\pi ,\alpha ,\lambda )$. Since any functional D acting on elements of $\mathbb {F}_\lambda [x_1, \ldots , x_k]$ can be expressed as a finite sum, $D=\sum _{p,r} D_{p,r}$, where $D_{p,r} \in \mathbb {B}_r(p,\pi , \lambda )$, the disjoint union $\bigcup _{p\in \pi \cap J} B(p,\pi ,\alpha ,\lambda )$ is a basis for the space of linear functionals on $\mathbb {F}_\lambda [x_1, \ldots , x_k]$ by construction. Hence, by (5),

$$\begin{aligned} \sum _{p\in \pi \cap J}|{B(p,\pi ,\alpha ,\lambda )}| = \dim \mathbb {F}_{\lambda }[x_1, \ldots , x_k] = {\lambda +k \atopwithdelims ()k}. \end{aligned}$$

(7)

Equation (7) replaces the use of the Fundamental Theorem of Algebra to bound the number of zeros of a polynomial, which we often see in polynomial method arguments such as in [12, 13].

Let

$$\begin{aligned} {\tilde{S}}_k(p,\pi , \alpha ,\lambda ) = |{B(p,\pi , \alpha , \lambda )}| \end{aligned}$$

for $\pi \in \cup _{j: k_j = k}\Pi _j$ and $p\in \pi \cap J$. Our analysis will only consider pairs $(p,\pi )$ so that $p\in \pi$. Therefore, it is not necessary to define ${\tilde{S}}_k$ for pairs such that $p\not \in \pi$. These quantities remain central to our analysis. Moreover, it follows immediately from (7) that

$$\begin{aligned} \sum _{p\in \pi \cap J}{\tilde{S}}_k(p,\pi ,\alpha ,\lambda ) = \dim \mathbb {F}_{\lambda }[x_1, \ldots , x_k] = {\lambda +k \atopwithdelims ()k}. \end{aligned}$$

(8)

Given a polynomial $f\in \mathbb {F}_\lambda [x_1,\ldots x_k]$, each set $B(p,\pi ,\alpha ,\lambda )$ indexes the vanishing conditions, $\langle \phi , f\rangle =0$ for $\phi \in B(p,\pi ,\alpha ,\lambda )$. A vanishing condition is a condition of the form $\langle \phi , f\rangle =0$ where $\phi$ is a linear form on $\mathbb {F}[x_1, \ldots , x_k]$. Such conditions include evaluation at a point, or evaluation of derivatives at a point.

Note that the sets B are translation-invariant with respect to $\alpha$.

Proposition 1

(Translation Invariance) Let $\pi$ be a k-plane. Let $\alpha$ be a handicap and $c\in \mathbb {Z}$. Let $(\alpha + c)_p:= \alpha _p + c$ for all $p\in J$. Then

$$\begin{aligned} B(p,\pi ,\alpha ,\lambda ) = B(p,\pi ,\alpha + c, \lambda ) \end{aligned}$$

for all $p\in (l\cap J)$. Furthermore, $\tilde{S}(p,\pi ,\alpha ,\lambda ) = {\tilde{S}}(p,\pi ,\alpha +c, \lambda )$.

3.4 The Vanishing Lemma

The implied constant $C = C(k_1, \ldots , k_d)$ in our Theorem 1 is derived from the polynomial method. A traditional application of the polynomial method would find a non-zero polynomial $f\in \mathbb {F}[x_1, \ldots , x_d]$ of low degree which vanishes at every point of J. More generally, we may ask that f satisfies vanishing conditions, $\langle \phi , f\rangle =0$ for some $\phi \in (\mathbb {F}[x_1, \ldots , x_d])^*$.

It was observed in [6] that choosing vanishing conditions in the traditional way is naïve and unwittingly imposes more vanishing conditions than necessary. Following [6], we will begin by choosing vanishing conditions, with some degrees of freedom, in an optimal way such that a vanishing lemma remains valid. Once the vanishing lemma is established, the use of polynomials is concluded. We then show that these vanishing conditions were chosen in a way that satisfies desirable uniform boundedness, monotonicity and continuity properties with respect to these degrees of freedom. This allows for a heuristically simple perturbation argument and we conclude the argument by choosing vanishing conditions that respect the geometry of the particular multijoint configuration with which we are working. In Sect. 3.1, our results diverge from [6]. We make a different choice of handicap, and with that, we ultimately establish the discrete Bourgain–Guth theorem.

The sets $B(p,\pi ,\alpha ,\lambda )$, as constructed in Sect. 2.3 above, are comprised of dual maps of the form Df(p) for some derivative operators which act on polynomials restricted to $\pi$. Let us abuse notation and write $D\in B(p,\pi ,\alpha ,\lambda )$ to mean a derivative operator D, so that $D f(p) = \langle \phi , f\rangle$, for some $\phi \in B(p,\pi ,\alpha ,\lambda )$. Then, it is a cornerstone of the Tidor–Yu–Zhao handicap construction that for any fixed $\alpha$ and $\lambda$, the vanishing conditions indexed by $B_j(p,\pi _j,\alpha ,\lambda )$, for $p\in J$ and $\pi _j\in \Pi _j$, satisfy a vanishing lemma.

Lemma 2

(Vanishing Lemma [6, Lemma 5.8]) Fix $\alpha$ and $\lambda$. For each $1\le j \le d$, $\pi _j\in \Pi _j$ and $p\in J$, build the sets $B(p,\pi _j) = B(p,\pi _j, \alpha ,\lambda )$ and $B_r(p,\pi _j) = B_r(p,\pi _j, \alpha ,\lambda )$ as above, in Sect. 2.3. For each $p\in J$, choose planes $\pi _1(p)\in \Pi _1, \ldots , \pi _d(p)\in \Pi _d$ so that $\delta (p,\varvec{\pi }(p))=1$. If $f\in \mathbb {F}_\lambda [x_1, \ldots , x_n]$ is non-zero then there exist $p\in J$ and $D_1\in B(p,\pi _1(p)), \ldots , D_d\in B(p,\pi _d(p))$ so that

$$\begin{aligned} D_1\cdots D_d f(p) \ne 0. \end{aligned}$$

Corollary 1

Fix $\alpha$ and $\lambda$. For each $1\le j \le d$, $\pi _j\in \Pi _j$ and $p\in J$, build the sets $B(p,\pi _j, \alpha , \lambda )$ as above, in Sect. 2.3. For each $p\in J$, choose planes $\pi _1(p) \in \Pi _1, \ldots , \pi _d(p) \in \Pi _d$ so that $\delta (p,\varvec{\pi }(p)) =1$. Let ${\tilde{S}}_{k_j}(p,\pi _j,\alpha , \lambda ) := |{B(p,\pi _j, \alpha , \lambda )}|$. Then

$$\begin{aligned} \sum _{p\in J} \prod _{j=1}^d{\tilde{S}}_{k_j}(p,\pi _j(p),\alpha ,\lambda ) \ge {\lambda + n\atopwithdelims ()n}. \end{aligned}$$

Proof

For a contradiction, suppose that the conclusion is false. By parameter counting, there exists a non-zero $f\in \mathbb {F}_\lambda [x_1, \ldots , x_n]$ so that

$$\begin{aligned} D_1 \cdots D_d f(p)=0 \end{aligned}$$

for all $D_j\in B(p,\pi _j(p),\alpha ,\lambda )$, and all $1\le j \le d$, contrary to Lemma 2. $\square$

This application of the vanishing lemma concludes the use of polynomials in our argument and it remains to prove that there exists a “good” handicap.

3.5 Handicap Properties

In this subsection, we recall and remark upon the uniform boundedness, monotonicity and continuity of the numbers $\tilde{S}(p,l,\alpha ,\lambda )$ with respect to $\alpha$, as stated below. We use this case to highlight the relation between the handicap argument and our discrete Bourgain–Guth theorem.

In the case of lines l, it is possible to deduce the aforementioned properties of ${\tilde{S}}(p,l,\alpha ,\lambda )$ from the definition of $\prec$, alone, without reference to polynomials, or vector spaces thereof.

The advantages of using the new Tidor–Yu–Zhao approach are twofold. Firstly, the choice of vanishing conditions varies nicely with respect to the handicap. Secondly, interpreting vanishing conditions as elements of a dual space allows us to establish the uniform equality (8), which counts the number of points “with multiplicity” on a plane.

The good properties of handicaps are as follows and are proved in [6, Lemma 5.4, Lemma 5.5, Lemma 5.6].

Lemma 3

(k-Plane Uniform Boundedness) Let $\lambda \in \mathbb {N}$ and $\alpha : J \rightarrow \mathbb {Z}$ be a handicap. If $\alpha$ is such that $\alpha _p < \alpha _q - \lambda$ for some $p,q\in \pi \cap J$, then ${\tilde{S}}(p,\pi , \alpha ,\lambda )=0$.

Lemma 4

(k-Plane Monotonicity) Let $\lambda \in \mathbb {N}$, and $\alpha ^{(1)}, \alpha ^{(2)}\in \mathbb {Z}^J$ be two handicaps. Suppose $\exists p\in \pi \cap J$ so that $\alpha _p^{(1)} - \alpha _{p^\prime }^{(1)} \le \alpha _p^{(2)} - \alpha _{p^\prime }^{(2)}$ for all $p^\prime \in \pi \cap J$. Then

$$\begin{aligned} {\tilde{S}}_k(p,\pi ,\alpha ^{(1)}, \lambda ) \le {\tilde{S}}_k(p, \pi , \alpha ^{(2)}, \lambda ). \end{aligned}$$

Lemma 5

(k-Plane Continuity) Let $p\in \pi \cap J$, let $\alpha ^{(1)}, \alpha ^{(2)}\in \mathbb {Z}^J$ be handicaps and $\lambda \in \mathbb {N}$. Then

$$\begin{aligned}{} & {} \bigg |{{\tilde{S}}(p,\pi ,\alpha ^{(1)}, \lambda ) - {\tilde{S}}(p,\pi , \alpha ^{(2)}, \lambda )}\bigg |\\{} & {} \hspace{4em}\le {\lambda + k-1 \atopwithdelims ()k-1}\sum _{p^\prime \in J}\bigg |{(\alpha _p^{(1)} - \alpha ^{(1)}_{p^\prime }) - (\alpha ^{(2)}_p-\alpha ^{(2)}_{p^\prime })}\bigg |. \end{aligned}$$

By translation invariance, Proposition 1, we may choose $\alpha _p^{(i)}=0$ for $i=1,2$. The resulting inequality reads as

$$\begin{aligned} \bigg |{{\tilde{S}}(p,\pi ,\alpha ^{(1)}, \lambda ) - {\tilde{S}}(p,\pi , \alpha ^{(2)}, \lambda )}\bigg |\le {\lambda + k-1 \atopwithdelims ()k-1}\sum _{p^\prime \in J}\bigg |{\alpha ^{(1)}_{p^\prime }-\alpha ^{(2)}_{p^\prime }}\bigg |. \end{aligned}$$

This is more easily recognisable as a Lipschitz continuity result. In fact, if desired, we could redefine handicaps and identify $\alpha$ by its equivalence class, $\alpha + \mathbb {Z}^{|{J}|}$, under equivalence by translation invariance.

4 The Discrete Bourgain–Guth Theorem

We now prove our discrete Bourgain–Guth theorem by making a material modification of the Tidor–Yu–Zhao perturbation argument [6].

4.1 There Exists a Good Handicap

We prove that there exists a handicap with properties which are good for our purposes. The main lemma is stated as follows.

Lemma 6

(k-Plane Handicap) Let $\lambda \in \mathbb {N}$, and $\Pi _1, \ldots , \Pi _d$ be finite sets of $k_1$-$, \ldots , k_d$- planes in $\mathbb {F}^n$ where $k_1+\ldots + k_d=n$, with associated multijoints J and let $S : J\rightarrow \mathbb {R}_{\ge 0}$ be finitely supported. Then there is a handicap $\alpha : J \rightarrow \mathbb {Z}$ so that for all $p\in {{\,\textrm{supp}\,}}S$ and $\pi _j\in \Pi _j$,

$$\begin{aligned} \frac{1}{S(p)^d}\left( \prod _{j=1}^d \frac{{\tilde{S}}_{k_j}(p,\pi _j,\alpha , \lambda )}{{\lambda +k_j\atopwithdelims ()k_j}}\right) \end{aligned}$$

(9)

lies in a common interval with length $\le h^\prime /\lambda$ for some $h^\prime =h^\prime (S,J,n)$, which does not depend on $\lambda$. Furthermore, we may choose $\alpha$ so that $\tilde{S}_{k_j}(p,\cdot ,\alpha ,\lambda )=0$ for all $p\not \in {{\,\textrm{supp}\,}}S$.

Remark 3

This lemma is different from its precursor [6, Lemma 5.10]. Specifically, the number of factors in the geometric mean (9) is now precisely d rather than

$$\begin{aligned} \sum _{\pi _1, \ldots , \pi _d} \delta (p,\pi _1, \ldots , \pi _d)f_1(\pi _1) \cdots f_d(\pi _d) \end{aligned}$$

terms, i.e. the multijoint multiplicity, at each $p\in J$ in [6]. Moreover, the weights for our geometric mean (9) are uniform and not problem-dependent as they were in [6].   $\blacktriangleleft$

We now prepare for our proof of Lemma 6. Let $\alpha$ be a handicap. For any $p\in J$ and $(\pi _j)_j\in \Pi _1\times \cdots \times \Pi _d$ so that $p\in \cap _j \pi _j$, let

$$\begin{aligned} W((\pi _j)_j, \alpha ):= \frac{1}{S(p)^d}\prod _{j=1}^d {\tilde{S}}_{k_j}(p,\pi _j,\alpha ,\lambda ). \end{aligned}$$

Note that W depends additionally on p and $\lambda$. However, $\lambda$ is fixed for this lemma, and the dependence on p is implicit because $p\in \cap _j \pi _j$. Therefore, we suppress p and $\lambda$. Now, for each $p\in J$, we define

$$\begin{aligned} w_p(\alpha ):= \min _{(\pi _j)_j: \delta (p, (\pi _j)_j)=1} W((\pi _j)_j, \alpha ). \end{aligned}$$

To prove this result, we introduce the notion of “connectedness” of the multijoint configuration. We say $\pi \in \Pi _1\cup \cdots \cup \Pi _d$ contributes to a multijoint p if there is some tuple $(\pi _j)_j\in \Pi _1\times \cdots \times \Pi _d$ so that $\pi = \pi _j$ for some j and $\delta (p,(\pi _j)_j)=1$. We say that $p,q\in J$ are adjacent if there is some $\pi \in \Pi _1\cup \cdots \cup \Pi _d$ that contributes to p and contributes to q. We say a set $E\subseteq J$ is connected if given any $p,q\in E$, there is a sequence of points $p=p^{(1)}, \ldots , p^{(N)} = q\in E$ so that $p^{(i)}$ and $p^{(i+1)}$ are adjacent for all $1\le i < N$. This defines an equivalence relation on any $E\subseteq J$.

Suppose that p and q are not contained in the same connected component of ${{\,\textrm{supp}\,}}S$. Then the value of $W((\pi _j)_j, \alpha )$, for any tuple $(\pi _j)_j$ forming a multijoint at some p, is independent of ${\tilde{S}}_{k_j}(q^\prime ,\pi _j, \alpha , \lambda )$ for all $\pi _j\in \Pi _j$ and all $q^\prime$ in the same connected component as q. Hence, in the construction of the functions $s_{k_j}$, we can restrict our attention to each connected component of ${{\,\textrm{supp}\,}}S$ separately. Thus, it suffices to assume that any two multijoints in ${{\,\textrm{supp}\,}}S$ are connected by multijoints in ${{\,\textrm{supp}\,}}S$.

Let $w: \{\alpha :J\rightarrow \mathbb {Z}\}\rightarrow \mathbb {R}_{\ge 0}^{|{J}|}$ be the map such that $w(\alpha ) = (w_p(\alpha ))_p$. Define $A=A(\lambda )\subset \{\alpha : J \rightarrow \mathbb {Z}\}$ to be the set of $\alpha$ such that ${\tilde{S}}(p,\cdot ,\alpha ,\lambda )=0$ for all $p\not \in {{\,\textrm{supp}\,}}S$. By uniform boundedness, Lemma 3, the image w(A) is finite and non-empty.

Label each $p\in J$ so that $J = \{p_1, \ldots , p_{|{J}|}\}$. For any $\alpha$, there exists a permutation $\sigma = \sigma _\alpha \in S_{|{J}|}$ so that $w_{p_{\sigma (1)}}(\alpha ) \ge \cdots \ge w_{p_{\sigma (|{J}|)}}(\alpha )$. Since the set w(A) is finite, of all $w(\alpha )\in w(A)$, we can choose one so that $(w_{p_{\sigma (i)}}(\alpha ))_{1\le i \le |{J}|} \in \mathbb {R}_{\ge 0}^{|{J}|}$ is minimal with respect to lexicographical order on $\mathbb {R}^{|{J}|}$.

Let

$$\begin{aligned} w(\alpha ) = (w_{p_{\sigma (i)}}(\alpha ))_{1\le i \le |{J}|} \end{aligned}$$

(10)

be such a minimum and let $\alpha$ be a minimiser. By relabelling the indices of each $p\in J$, we may assume that $\sigma$ is the identity permutation so that

$$\begin{aligned} w_{p_1}(\alpha ) \ge \cdots \ge w_{p_{|{J}|}}(\alpha )\ge 0. \end{aligned}$$

For ease of notation, let $w_i :=w_{p_i}(\alpha )$ for each $1\le i \le |{J}|$.

Proposition 2

(Continuity of Perturbations) Let $1\le t \le |{J}|$. Let

$$\begin{aligned} v = \sum _{\begin{array}{c} 1\le i \le t:\\ p_i \in {{\,\textrm{supp}\,}}S \end{array}}e_i + \sum _{\begin{array}{c} i:\\ p_i\not \in {{\,\textrm{supp}\,}}S \end{array}}e_i \end{aligned}$$

and $\alpha ^\prime = \alpha - v$, where $e_i: J\rightarrow \{0,1\}$ is such that $e_i(p) = 1$ if $p=p_i$, and zero otherwise. There is a constant h which depends on S, n and $|{J}|$, but not on $\lambda$, so that

$$\begin{aligned} |{w_i(\alpha )-w_i(\alpha ^\prime )}| \le \frac{h}{2\lambda } \end{aligned}$$

for all $1\le i \le N$, where $N = |{{{\,\textrm{supp}\,}}S}|$.

Proof

We construct h directly. Fix $1\le i \le N$, fix a d-tuple of planes $\varvec{\pi }$ which realises $w_i(\alpha )$, and fix $(\pi ^\prime _j)_j$ which realises $w_i(\alpha ^\prime )$. Consider

$$\begin{aligned} |{w_i(\alpha ) - w_i(\alpha ^\prime )}| = |{W((\pi _j)_j,\alpha ) - W((\pi ^\prime _j)_j, \alpha ^\prime )}|. \end{aligned}$$

If $(\pi _j)_j = (\pi ^\prime _j)_j$ then $|{w_i(\alpha ) - w_i(\alpha ^\prime )}| = |{W((\pi _j)_j, \alpha ) - W((\pi _j)_j,\alpha ^\prime )}|$. Otherwise, $(\pi _j)_j \ne (\pi ^\prime _j)_j$, in which case

$$\begin{aligned} |{w_i(\alpha ) - w_i(\alpha )}| \le \max _{{(\tilde{\pi }_j)_j} } |{W({ (\tilde{\pi }_j)_j}, \alpha ) - W({(\tilde{\pi }_j)_j}, \alpha ^\prime )}|, \end{aligned}$$

where the maximum is over all ${(\tilde{\pi }_j)_j} \in ({{\,\mathrm{arg\, min}\,}}W(\cdot , \alpha ) \cup {{\,\mathrm{arg\, min}\,}}W(\cdot , \alpha ^\prime ))$. Hence, we may assume

$$\begin{aligned} |{w_i(\alpha ) - w_i(\alpha ^\prime )}| \le |{W({(\tilde{\pi }_j)_j},\alpha ) - W({ (\tilde{\pi }_j)_j}, \alpha ^\prime )}|, \end{aligned}$$

for some ${ (\tilde{\pi }_j)_j}$, which minimises either $W(\cdot , \alpha )$ or $W(\cdot , \alpha ^\prime )$. However, for any $(\pi _j)_j\in \Pi _1\times \cdots \times \Pi _d$ which forms a multijoint at $p=p_i$,

$$\begin{aligned} \begin{aligned}&S(p)^d\left( \prod _{j=1}^d {\lambda + k_j\atopwithdelims ()k_j}\right) |{W((\pi _j)_j,\alpha )-W((\pi _j)_j, \alpha ^\prime )}| \\&= \bigg |{\prod _{j=1}^d \left( {\tilde{S}}(p,\pi _j,\alpha ^\prime , \lambda ) - ({\tilde{S}}(p,\pi _j,\alpha ^\prime , \lambda ) - {\tilde{S}}(p,\pi _j,\alpha , \lambda ))\right) - \prod _{j=1}^d {\tilde{S}}(p,\pi _j,\alpha ^\prime , \lambda )}\bigg |. \end{aligned} \end{aligned}$$

(11)

Taking (11), we expand the first product so that we can cancel both occurrences of $\prod _j {\tilde{S}}(p,\pi _j, \alpha ^\prime , \lambda )$. We are then left with terms of the following form:

$$\begin{aligned} \left( \prod _{j\in A} {\tilde{S}}(p,\pi _j,\alpha ^\prime ,\lambda )\right) \left( \prod _{j^\prime \in B} ({\tilde{S}}(p,\pi _j,\alpha ^\prime , \lambda ) - {\tilde{S}}(p,\pi _j,\alpha , \lambda ))\right) , \end{aligned}$$

(12)

where $A\sqcup B = \{1, \ldots , d\}$ and $B\ne \emptyset$. To establish an upper bound on (11), by the triangle inequality, it suffices to bound each such term of the form (12) separately. By continuity (Lemma 5), for any $\pi _j \in \Pi _j$ and $p\in \pi _j\cap J$,

$$\begin{aligned} |{{\tilde{S}}(p, \pi _j,\alpha , \lambda ) -{\tilde{S}}(p, \pi _j,\alpha ^\prime , \lambda )}| \le {\lambda + k_j-1\atopwithdelims ()k_j-1}|{J}|, \end{aligned}$$

since $\left| \left| \alpha - \alpha ^\prime \right| \right| _{L^1(J)} =\left| \left| v\right| \right| _{L^1(J)}\le |{J}|$. Combining this with the fact that each ${\tilde{S}}(p,\pi , \alpha ^\prime , \lambda ) = {\lambda + k_j\atopwithdelims ()k_j}$ by construction, each term (12) is bounded by

$$\begin{aligned} \prod _{j\in A} {\lambda + k_j\atopwithdelims ()k_j}\prod _{j\in B}{\lambda + k_j-1\atopwithdelims ()k_j-1}|{J}| \sim _{n, J} \lambda ^{\sum _{j\in A}k_j+\sum _{j^\prime \in B} (k_{j^\prime }-1)}, \end{aligned}$$

for sets A, B so that $A\sqcup B = \{1, \ldots , d\}$ and $B \ne \emptyset$. Hence, (11) is dominated by

$$\begin{aligned} S(p)^d\left( \prod _{j=1}^d {\lambda + k_j\atopwithdelims ()k_j}\right) |{W((\pi _j)_j,\alpha )-W((\pi _j)_j, \alpha ^\prime )}| \le h_{n-1}\lambda ^{n-1} + \ldots + h_1 \lambda + h_0 \end{aligned}$$

for $h_{n-1},\ldots , h_0$ sufficiently large, depending on n and $|{J}|$. This in turn is bounded above by $(h/2)\lambda ^{n-1}$ for sufficiently large h, depending only on $h_{n-1}, \ldots , h_1$ and not depending on $\lambda$. Dividing by

$$\begin{aligned} \left( \prod _{j=1}^d {\lambda + k_j\atopwithdelims ()k_j}\right) \sim _{n, J} \lambda ^n \end{aligned}$$

and updating h to additionally depend on $S(p)^{-d}$, we deduce

$$\begin{aligned} |{w_i(\alpha )-w_i(\alpha ^\prime )}| \le |{W((\pi _j)_j,\alpha )-W((\pi _j)_j,\alpha ^\prime )}| \le \frac{h}{2\lambda } \end{aligned}$$

for some h which can be expressed as a function in $|{J}|, n$ and the quantities $\{S(p)^{-d}: p\in J\}$, but does not depend on $\lambda$. This defines h. $\square$

The remainder of this subsection is dedicated to proving Lemma 6. Before we embark on the proof itself, we give a brief outline of the argument.

Since $w_1(\alpha ) \ge \ldots \ge w_{|{J}|}(\alpha )\ge 0$, it suffices to show that all the differences ${w_i(\alpha )-w_{i+1}(\alpha )}$ are small. We will prove this by contradiction. To begin, we assume there is some index $1\le t <N$ so that ${w_{t}(\alpha )-w_{t+1}(\alpha )} > h/\lambda$, with h as given by Proposition 2. We construct a perturbation of the handicap $\alpha$. Let $\alpha ^\prime$ be the perturbed handicap. The following three claims are established for the perturbation:

• 1. The large entries of the tuple $w(\alpha )$ remain large, and the small entries remain small.

• 2. The perturbed handicap $\alpha ^\prime$ is such that $w(\alpha ^\prime )\in w(A)$, so $w(\alpha ) \le w(\alpha ^\prime )$.

• 3. If $w(\alpha )\ne w(\alpha ^\prime )$ then $w(\alpha ^\prime ) < w(\alpha )$. Hence, $w(\alpha ^\prime )=w(\alpha )$.

Thereafter, we realise that the perturbation can be applied to $\alpha ^\prime$, the already perturbed handicap. Moreover, by the connectedness of ${{\,\textrm{supp}\,}}S$, there is a pair $p,q\in {{\,\textrm{supp}\,}}S$ so that $w_p(\alpha )$ is large, $w_q(\alpha )$ is small and there is a plane which contributes to both p and q. If we perturb sufficiently many times, what results is a perturbed handicap $\alpha ^\prime$ so that $w(\alpha )=w(\alpha ^{\prime })$, and $\alpha _p^\prime < \alpha _q^\prime - \lambda$. By uniform boundedness (Lemma 3), $\tilde{S}_{k_j}(p,\pi ,\alpha ^\prime ,\lambda )=0$, and hence $w_q(\alpha ) =0$. So $0 = w_p(\alpha )\ge w_q(\alpha ) \ge 0$, and hence $w_q(\alpha )=0$. However, we assumed for a contradiction that $w_p(\alpha )$ was large and that $w_q(\alpha )$ was small so that $w_p(\alpha ) > w_q(\alpha )$, which is a contradiction.

(Proof of Lemma 6)

Let ${{\,\textrm{supp}\,}}S = \{p_1, \ldots , p_N\}$ for some $N\le |{J}|$ be connected, and let $h^\prime = |{J}|h$, where h is given by Proposition 2. Since the tuple $(w_i)_i$ has at most $|{{J}}|$ non-zero entries, it suffices to show that

$$\begin{aligned} w_i - w_{i+1} \le \frac{h}{\lambda }\end{aligned}$$

(13)

for all $1\le i < N$. Indeed, summing this inequality over all $1\le i \le N$, we deduce $\max _i w_i - \min _i w_i \le |{J}|h/\lambda$. That is, the interval containing all $(w_i)_{1\le i\le N}$ has width $\sim _{k_j, J, S}1/\lambda$, where the implied constant does not depend on $\lambda$.

We now prove that (13) holds for all $1\le i < N$. Suppose for a contradiction that there is some index $1\le i< N$ so that (13) does not hold. Let t be the least such index, and for this choice of t, define

$$\begin{aligned} v = \sum _{\begin{array}{c} 1\le i \le t:\\ p_i \in {{\,\textrm{supp}\,}}S \end{array}}e_i + \sum _{\begin{array}{c} i:\\ p_i\not \in {{\,\textrm{supp}\,}}S \end{array}}e_i \end{aligned}$$

and let $\alpha ^\prime = \alpha - v$.

We say that $w_i$ is large if $i \le t$, and $w_i$ is small otherwise.

Let $\pi \in \Pi _j$ for some $1\le j \le d$. Recall that $\sum _{p\in \pi } {\tilde{S}}_{k_j}(p,\pi ,\alpha ,\lambda ) = {\lambda + k_j \atopwithdelims ()k_j}$ for any $\alpha$. It follows that, if $\tilde{S}_{k_j}(p,\pi ,\alpha ^\prime ,\lambda ) > \tilde{S}_{k_j}(p,\pi ,\alpha ,\lambda )$ for some $p\in \pi \cap J$, then there exists some $p^\prime \in \pi \cap J$ so that $p^\prime \ne p$ and ${\tilde{S}}_{k_j}(p^\prime , \pi , \alpha ^\prime , \lambda ) < \tilde{S}_{k_j}(p^\prime , \pi , \alpha , \lambda )$. For those i such that $p_i\not \in {{\,\textrm{supp}\,}}S$, we have that $w_i(\alpha ) = 0 = w_i(\alpha ^\prime )$. Moreover, by monotonicity (Lemma 4), for i such that $p_i\in {{\,\textrm{supp}\,}}S$, we have that if $i\le t$ then $w_i(\alpha ^\prime ) \le w_i(\alpha )$ (where the handicap is decreased) and if $i > t$ then $w_i(\alpha ^\prime ) \ge w_i(\alpha )$ (where the handicap is unchanged). One inequality is strict if and only if the other is too. By Proposition 2, the difference $|{w_i(\alpha ) - w_i(\alpha ^\prime )}|$ is bounded above by $h/(2\lambda )$ for all $1\le i \le N$, and by the definition of t,

$$\begin{aligned} w_t(\alpha ) -w_{t+1}(\alpha )> \frac{h}{\lambda }. \end{aligned}$$

(14)

Let $\sigma ^\prime \in S_{|{J}|}$ be a permutation such that

$$\begin{aligned} w_{\sigma ^\prime (1)}(\alpha ^\prime ) \ge \cdots \ge w_{{\sigma ^\prime (|{J}|)}}(\alpha ^\prime ), \end{aligned}$$

(15)

where $\sigma (i)=i$ for $N<i\le |{J}|$.

Claim 1

If $1\le i \le t$, then $1\le \sigma ^\prime (i) \le t$, and if $t < i^\prime \le N$, then $t < \sigma ^\prime (i^\prime ) \le |{J}|$.

Proof

Suppose that $p_i, p_{i^\prime } \in {{\,\textrm{supp}\,}}S$ are such that $i\le t < i^\prime$. By Proposition 2,

$$\begin{aligned} w_i(\alpha ^\prime ) - w_{i^\prime }(\alpha ^\prime )= & {} \left( w_i(\alpha ^\prime ) - w_i(\alpha )\right) + \left( w_i(\alpha )- w_{i^\prime }(\alpha )\right) + \left( w_{i^\prime }(\alpha ) - w_{i^\prime }(\alpha ^\prime )\right) \\\ge & {} -\frac{h}{2\lambda } + \left( w_i(\alpha )- w_{i^\prime }(\alpha )\right) - \frac{h}{2\lambda }. \end{aligned}$$

Using inequalities (14) and (15), this is at least

$$\begin{aligned} \left( w_{t}(\alpha )- w_{t+1}(\alpha )\right) -\frac{h}{\lambda } > -\frac{h}{\lambda }+ \frac{h}{\lambda }=0. \end{aligned}$$

Hence, if $1\le i \le t$, then $w_{i}(\alpha ^\prime ) \ge w_{i^\prime }(\alpha ^\prime )$ for all $i^\prime >t$. Therefore $1\le \sigma ^\prime (i) \le t$. Similarly, if $t < i \le N$, then $w_{i^\prime }(\alpha ^\prime ) \ge w_{i}(\alpha ^\prime )$ for all $1\le i^\prime \le t$. Hence, $t < \sigma ^\prime (i) \le N$, as desired. $\square$

Claim 2

With $\alpha ^\prime$ as above, $w(\alpha ^\prime )\in w(A)$. Hence $w(\alpha ^\prime )$ was among those considered when the minimiser $w(\alpha )$ was chosen.

Proof

Since $w(\alpha ) \in w(A)$, we have that $\tilde{S}_{k_j}(p,\cdot ,\alpha ,\lambda )=0$ for all $1\le j \le d$ and $p\not \in {{\,\textrm{supp}\,}}S$. We may assume that $\alpha$ is such that for all $q\not \in {{\,\textrm{supp}\,}}S$, and all $\pi \in \Pi _1,\cup \cdots \cup \Pi _d$ so that $\pi \ni q$, if $p\in \pi \cap {{\,\textrm{supp}\,}}S$, then

$$\begin{aligned} \alpha _q < \alpha _p - \lambda . \end{aligned}$$

We will show that ${\tilde{S}}_{k_j}(q,\cdot ,\alpha ,\lambda )=0$ for all $1\le j \le d$ and $q\not \in {{\,\textrm{supp}\,}}S$. Let $q\not \in {{\,\textrm{supp}\,}}S$ and $1\le j \le d$. If $\pi \in \Pi _1\cup \cdots \cup \Pi _d$ is such that $\pi \ni q$, and $p\in \pi \cap {{\,\textrm{supp}\,}}S$, then

$$\begin{aligned} \alpha _q^\prime = \alpha _q-1 < (\alpha _p-\lambda ) -1 = (\alpha _p -1)- \lambda \le \alpha _p^\prime -\lambda . \end{aligned}$$

We may assume that every plane in $\Pi _1\cup \cdots \cup \Pi _d$ intersects ${{\,\textrm{supp}\,}}S$, so such p exists. Hence, by uniform boundedness (Lemma 3), $\tilde{S}_{k_j}(q,\pi ,\alpha ^\prime ,\lambda )=0$. Hence $w(\alpha ^\prime )\in w(A)$ and so $(w_p(\alpha ^\prime ))_p$ was among those tuples considered when $(w_p(\alpha ))_p$ was chosen. $\square$

Claim 3

If $w(\alpha ^\prime )\ne w(\alpha )$ then $w(\alpha ^\prime )< w(\alpha )$. Hence $w(\alpha ) = w(\alpha ^\prime )$.

Proof

Suppose that there is some $1\le i \le |{J}|$ so that $w_i(\alpha ^\prime ) \ne w_i(\alpha )$. Then there is an $i \le t$ so that $w_i(\alpha ^\prime ) < w_i(\alpha )$ and $1\le \sigma ^\prime (i) \le t$. That is, $w_i(\alpha ^\prime )$ is strictly smaller than $w_i(\alpha )$ and $w_{\sigma ^{\prime }(i)}(\alpha ^\prime )$ is among the t largest values of $(w_i(\alpha ^\prime ))_{1\le i \le |{J}|}$ by Claim 1. Hence

$$\begin{aligned} (w_{\sigma ^\prime (i)}(\alpha ^\prime ))_{1\le i \le t} <_\text {lex} (w_i(\alpha ))_{1\le i \le t}, \end{aligned}$$

and therefore, $w(\alpha ^\prime )$ is of strictly lower lexicographical order than $w(\alpha )$. Moreover, by Claim 2, $w(\alpha ^\prime )\in w(A)$. This contradicts that $w(\alpha )\in w(A)$ was chosen to be minimising. $\square$

Thus, $w_i(\alpha ^\prime ) = w_i(\alpha - v)= w_i(\alpha )$ for all $1\le i \le |{J}|$. We have not yet contradicted our assumption that (13) is false; however, we have deduced that the perturbation $\alpha \mapsto (\alpha - v)$ must leave $(w_i)_i$ unchanged. To conclude, we observe that we may return to (10), use $\alpha ^\prime = \alpha -v$, and repeat the application of Claim 1, Claim 2 and Claim 3. We may repeat this process c times to deduce that

$$\begin{aligned} w_i(\alpha )= w_i(\alpha -v) = w_i(\alpha - 2v) = \cdots = w_i(\alpha -cv) \end{aligned}$$

for all $1\le i \le |{J}|$. By connectedness of ${{\,\textrm{supp}\,}}S$, we can find a plane $\pi \in \Pi _j$ for some $1\le j \le d$ contributing to distinct multijoints $p_i,p_j\in {{\,\textrm{supp}\,}}S$ for some i and j satisfying $i\le t < j\le N$. Taking c sufficiently large (depending on $\lambda$) forces $w_i(\alpha ^\prime ) = 0$ by uniform boundedness (Lemma 3) and hence $w_i(\alpha )=0$. That is, one of the large entries of the tuple $(w_i(\alpha ))_i$ is zero. Since $(w_i)_i$ is decreasing in i and each $w_i$ is non-negative, $w_{i^\prime }(\alpha )=0$ for all $i\le i^\prime \le |{J}|$. This contradicts our assumption that $w_t(\alpha ) - w_{t+1}(\alpha ) > h/\lambda$, and hence (13) holds for all $1\le i < N$. $\square$

4.2 Good Vanishing Conditions Yield A Factorisation

We now deduce Theorem 1 in the special case where the functions $s_{k_j}$ have their second arguments supported on the finite sets $\Pi _1, \ldots , \Pi _d$, respectively.

Let $S:J\rightarrow \mathbb {R}_{\ge 0}$ be non-negative with $\left| \left| S\right| \right| _{d} = 1$ and let $\lambda$ be sufficiently large. By Lemma 6, we may choose a handicap $\alpha =\alpha (\lambda )$, so that the numbers $\tilde{S}_{k_j}(p,\pi _j,\alpha ,\lambda )$, as constructed in Sect. 2.3, are such that there is an interval containing

$$\begin{aligned} w_p(\alpha ) = \min _{\delta (p,(\pi _j)_j)=1} \frac{1}{S(p)^d}\left( \prod _{j=1}^d \frac{{\tilde{S}}(p,\pi _j,\alpha , \lambda )}{{\lambda +k_j\atopwithdelims ()k_j}}\right) \end{aligned}$$

for all $p\in {{\,\textrm{supp}\,}}S$ with length at most $h^\prime /\lambda$. Moreover, ${\tilde{S}}(p,\cdot , \alpha ,\lambda )=0$ for all $p\not \in {{\,\textrm{supp}\,}}S$. That is, we can find $w\in \mathbb {R}$ so that

$$\begin{aligned} w-\varepsilon \le w_p(\alpha ) \le w \end{aligned}$$

for all ${p\in {{\,\textrm{supp}\,}}S}$ and so that $\tilde{S}(p,\cdot ,\alpha ,\lambda ) =0$ if $S(p)=0$, where $\varepsilon = h^\prime /\lambda$, and $h^\prime$ is the constant given in Lemma 6. For each $p\in J$, let $(\pi _j(p))_j$ be a choice of $k_j$-planes which minimises $\prod _{j=1}^d \tilde{S}(p,\pi _j(p),\alpha ,\lambda )$ over all tuples $(\pi _j)_j$ so that $\delta (p,(\pi _j)_j)=1$. Corollary 1 applies, so,

$$\begin{aligned} w=\sum _{p\in J}S(p)^{d}w \ge \sum _{p\in J}\prod _{j=1}^d \frac{{\tilde{S}}(p, \pi _j(p),\alpha , \lambda )}{{\lambda +k_j\atopwithdelims ()k_j}} \ge \frac{1}{\prod _{j=1}^d{\lambda +k_j\atopwithdelims ()k_j}}{\lambda + n \atopwithdelims ()n} \gtrsim _{n} 1, \end{aligned}$$

(16)

where we have used that $w_p(\alpha ) = 0$ if $p\not \in {{\,\textrm{supp}\,}}S$, and $k_1+\ldots + k_d = n$. Hence, $w\gtrsim _{n}1$.

If $\lambda$ is large enough, then the length of the interval $\varepsilon =h^\prime /\lambda < w$ so that the left endpoint $w-\varepsilon$ will be strictly positive. Hence, each $B(p,\pi ,\alpha , \lambda )\ne \emptyset$ for all $p\in J$ and every $\pi$ that contributes to p. Define

$$\begin{aligned} s_{k_j, \lambda }(p, \pi _j):= \frac{{\tilde{S}}(p,\pi _j,\alpha , \lambda )}{{\lambda +k_j\atopwithdelims ()k_j}}. \end{aligned}$$

For each $p\in J$, if $\pi _j\in \Pi _j$ is such that $p\not \in \pi _j$, then set $s_{k_j, \varepsilon }(p,\pi _j)=0$. Thus,

$$\begin{aligned} 1\lesssim _{n} w \le \frac{h^\prime }{\lambda }+ \frac{1}{S(p)^d} \prod _{j=1}^d s_{k_j, \lambda }(p,\pi _j(p)). \end{aligned}$$

for any $p\in {{\,\textrm{supp}\,}}S$. Since $\left| \left| S\right| \right| _d =1$, $S(p) \le 1$ for all $p\in J$. Hence

$$\begin{aligned} S^{d}(p) \lesssim _{n} \frac{h^\prime }{\lambda }+ \prod _{j=1}^d s_{k_j, \lambda }(p,\pi _j(p)). \end{aligned}$$

Since the tuple of planes $(\pi _j(p))_j$ minimises the right-hand side over all tuples such that $\delta (p,(\pi _j)_j)=1$,

$$\begin{aligned} \delta (p, (\pi _j)_j)S^{d}(p) \lesssim _{n}\frac{h^\prime }{\lambda }+ \prod _{j=1}^d s_{k_j,\lambda }(p,\pi _j) \end{aligned}$$

(17)

for all tuples of planes $(\pi _j)_j$. By construction, for all $1\le j \le d$, $\pi _j\in \Pi _j$

$$\begin{aligned} \sum _{p\in \pi _j\cap J} s_{k_j, \lambda }(p, \pi _j) = \frac{1}{{\lambda +k_j\atopwithdelims ()k_j}}\sum _{p\in \pi _j\cap J}\tilde{S}_{k_j}(p,\pi _j,\alpha , \lambda ) =1. \end{aligned}$$

(18)

Both inequality (17) and equation (18) are uniform in $\varepsilon$ and each function

$$\begin{aligned} \frac{1}{S(p)}s_{k_j, \lambda }(p,\pi _j) = \frac{1}{S(p)}\frac{{\tilde{S}}_{k_j}(p,\pi _j,\alpha , \lambda )}{{\lambda +k_j\atopwithdelims ()k_j}} \end{aligned}$$

can be realised as an $\mathbb {R}$-valued vector in $[0,1]^{|{J}|\times |{\Pi _j}|}$. Hence, passing to a subsequence if necessary, we may let $s_{k_j} = \lim _{\lambda \rightarrow \infty } s_{k_j, \lambda }$ for each $1\le j \le d$. Letting $\lambda \rightarrow \infty$ in (17) and (18) concludes our proof of Theorem 1 under the assumption that the factorising functions have their second argument supported on the finite sets $\Pi _j$.

We now make our concluding arguments. If the field $\mathbb {F}$ is finite, then we can apply our result thus far with each $\pi _j = {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)$. Hence, we may assume that the field is infinite.

Let $S: \mathbb {F}^n \rightarrow \mathbb {R}_{\ge 0}$ be finitely supported. Consider

$$\begin{aligned} \mathcal {O}:=\left\{ \left( \pi _j \cap {{\,\textrm{supp}\,}}S\right) _j: \delta (p,\pi _1, \ldots , \pi _d)=1, \, p\in \mathbb {F}^n,\, \pi _j \subset \mathbb {F}^n \right\} , \end{aligned}$$

where in the definition of $\mathcal {O}$, we consider any tuple of $k_j$-planes (of which there may be infinitely many) and any $p\in {{\,\textrm{supp}\,}}S$. However, since ${{\,\textrm{supp}\,}}S$ is finite, so too is $\mathcal {O}$. For every $(E_j)_j\in \mathcal {O}$, choose a tuple of $k_j$-planes so that $\delta (p,\pi _1, \ldots , \pi _d)=1$ and $\pi _j\cap {{\,\textrm{supp}\,}}S = E_{j}$ for every $1\le j \le d$. We hence define the finite sets $\Pi _j$ to consist of all such planes $\pi _j$. In particular, since the field is infinite, for each $p\in {{\,\textrm{supp}\,}}S$, the tuple $(\{p\})_j$ belongs to $\mathcal {O}$ and hence ${{\,\textrm{supp}\,}}S$ is a subset of the multijoints formed by the finite sets $\Pi _1, \ldots , \Pi _d$. Hence, there exist factorising functions ${\tilde{s}}_{k_j}$ that satisfy the displays described in Theorem 1, but for only finitely many tuples of $k_j$-planes.

Recall from Remark 2 that if two $k_j$-planes $\pi$ and $\pi ^\prime$ are distinct and satisfy $\pi \cap {{\,\textrm{supp}\,}}S = \pi ^\prime \cap {{\,\textrm{supp}\,}}S$, then $B(\cdot , \pi ,\alpha , \lambda ) = B(\cdot , \pi ^\prime , \alpha ,\lambda )$. Hence, the functions $\tilde{s}_{k_j}$ satisfy ${\tilde{s}}_{k_j}(\cdot , e(\pi )) = \tilde{s}_{k_j}(\cdot , e(\pi ^\prime ))$. Now, consider the pairs $(p,V_j)\in {{\,\textrm{supp}\,}}S\times {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)$. By construction, there exists $\pi _j \in \Pi _j$ so that $\pi _j \cap {{\,\textrm{supp}\,}}S = (p+V_j)\cap {{\,\textrm{supp}\,}}S$ and it is well-defined to set $s_{k_j}(p, V_j) := {\tilde{s}}_{k_j}(p, e(\pi _j))$. We additionally set $s_{k_j}(p,\cdot )=0$ for any $p\not \in {{\,\textrm{supp}\,}}S$, whereby each $s_{k_j}$ is finitely supported and defined on $\mathbb {F}^n \times {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)$. Each $s_{k_j}$ automatically satisfies

$$\begin{aligned} \sum _{p \in \pi _j} s_{k_j}(p, e(\pi _j)) =\sum _{p \in \pi _j\cap J} s_{k_j}(p, e(\pi _j)) = 1 \end{aligned}$$

for any $k_j$-plane $\pi _j\subset \mathbb {F}^n$, establishing the second display of Theorem 1. Turning to the first display, let $p\in {{\,\textrm{supp}\,}}S$ and let $V_j \in {{\,\mathrm{{Gr}}\,}}(k_j, \mathbb {F}^n)$ be such that $V_1\wedge \cdots \wedge V_d = 1$. For each j, there exists $\pi _j \in \Pi _j$ so that $(p+V_j)\cap {{\,\textrm{supp}\,}}S = \pi _j\cap {{\,\textrm{supp}\,}}S$ and $\delta (p,\pi _1, \ldots , \pi _d) = 1$ by construction. Hence

$$\begin{aligned} (V_1\wedge \cdots \wedge V_d)S(p)^d= & {} \delta (p, \pi _1, \ldots , \pi _d)S(p)^d \\{} & {} \lesssim _{n} \prod _{j=1}^d {\tilde{s}}_{k_j}(p, e(\pi _j)) = \prod _{j=1}^d s_{k_j}(p,V_j). \end{aligned}$$

This concludes the proof of Theorem 1.

An alternative presentation of this argument in the case where each $k_j=1$ can also be found in [7, Section 4].

4.3 Multijoints of Varieties

Given $k_j$-dimensional varieties $\gamma _j$, respectively, we extend the definition of $\delta$ so that for any $p\in \mathbb {F}^n$ which is a regular point of each $\gamma _j$,

$$\begin{aligned} \delta (p,\gamma _1, \ldots , \gamma _d):= \left( \prod _{j=1}^d \chi _{\gamma _j}(p)\right) e(T_p\gamma _1)\wedge \cdots \wedge e(T_p \gamma _d), \end{aligned}$$

where $T_p\gamma _j$ denotes the tangent plane to $\gamma _j$ at p. Although we will not include a proof, a suitable modification to the argument above establishes Theorem 2:

Theorem 2

(Multijoint Factorisation for Varieties) Let $\Gamma _1, \ldots , \Gamma _d$ be sets of $k_1$-$, \ldots , k_d$-dimensional varieties in $\mathbb {F}^n$, respectively, where $k_1 + \ldots + k_d=n$, and let $J = \{p : \exists (\gamma _j)_j \text { so that } \delta (p,\gamma _1, \ldots , \gamma _d)=1\}$.

For all finitely supported $S:J\rightarrow \mathbb {R}_{\ge 0}$ with $\left| \left| S\right| \right| _{d}=1$, there exist functions $s_{k}: J\times \left( \cup _{j : k_j = k}\Gamma _{j}\right) \rightarrow \mathbb {R}_{\ge 0}$ for each $k\in \{k_1, \ldots , k_d\}$, so that

$$\begin{aligned} \delta (p,\gamma _1, \ldots , \gamma _d)S(p)^d \lesssim _{n} \prod _{j=1}^ds_{k_j}(p,\gamma _j), \end{aligned}$$

for all $p\in J$ and $(\gamma _1, \ldots , \gamma _d)\in \Gamma _1\times \cdots \times \Gamma _d$, and so that

$$\begin{aligned} \sum _{p\in \gamma _j\cap J}s_{k_j}(p, \gamma _j) = \deg \gamma _j \end{aligned}$$

for all $\gamma _j\in \Gamma _j$ and all $1\le j \le d$.

Indeed, the inputs to our new Lemma 6, above, were numbers ${\tilde{S}}_{k_j}$ which satisfy equation (7); Corollary 1; Lemma 3; Lemma 4 and Lemma 5. To deduce Theorem 2, our proof of Lemma 6 remains valid, with appropriately generalised numerology of the aforementioned inputs, and these are all given in [6].