2 Mathematical preliminaries

Here, we review some mathematical background for this work that was not covered in the introduction. For vector spaces $\mathcal {V}_1, \dots , \mathcal {V}_m$ over a field $\mathbb {F}$ , we use $\mathrm {Prod}\left (\mathcal {V}_1 : \dots : \mathcal {V}_m \right )$ to denote the set of (nonzero) product tensors in $\mathcal {V}_1\otimes \dots \otimes \mathcal {V}_m$ . This set forms an algebraic variety given by the affine cone over the Segre variety $\mathrm {Seg}(\mathbb {P} \mathcal {V}_1 \times \dots \times \mathbb {P} \mathcal {V}_m)$ , with the point $0$ removed. We use symbols like $a, b$ to index tensors and symbols like $i, j$ to index subsystems. For vector spaces $\mathcal {V}$ and $\mathcal {W}$ , let $\mathrm {Hom}(\mathcal {V},\mathcal {W})$ denote the space of linear maps from $\mathcal {V}$ to $\mathcal {W}$ , and let $\mathrm {End}(\mathcal {V})=\mathrm {Hom}(\mathcal {V},\mathcal {V})$ . For a vector space $\mathcal {V}$ of dimension d, let $\{e_1,\dots , e_d\}$ be a standard basis for $\mathcal {V}$ .

For a product tensor $z \in \mathrm {Prod}\left (\mathcal {V}_1 : \dots : \mathcal {V}_m \right )$ , the vectors ${z_{j} \in \mathcal {V}_j}$ for which ${z=z_1 \otimes \dots \otimes z_m}$ are uniquely defined up to scalar multiples $\alpha _{1} z_1, \dots , \alpha _m z_m$ such that $\alpha _1\cdots \alpha _m=1$ . For positive integers n and m, we frequently define multisets of product tensors

$$ \begin{align*} \{ x_a : a \in [n]\} \subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \dots : \mathcal{V}_m \right) \end{align*} $$

without explicitly defining corresponding vectors $\{x_{a,j}\}$ such that

$$ \begin{align*} x_a=x_{a,1}\otimes \dots \otimes x_{a,m} \end{align*} $$

for all $a \in [n]$ . In this case, we implicitly fix some such vectors and refer to them without further introduction.

We use the notation

$$ \begin{align*} x_{a, \hat{j}}&= x_{a,1} \otimes \dots \otimes x_{a,j-1}\otimes x_{a, j+1} \otimes \dots \otimes x_{a,m},\\ \mathcal{V}_{\hat{j}}&=\mathcal{V}_1 \otimes \dots \otimes \mathcal{V}_{j-1}\otimes \mathcal{V}_{j+1}\otimes\dots \otimes \mathcal{V}_{m}, \end{align*} $$

so $x_{a,\hat {j}}\in \mathcal {V}_{\hat {j}}$ . Note that $\mathcal {V}_1\otimes \dots \otimes \mathcal {V}_m$ is naturally isomorphic to $\mathrm {Hom}(\mathcal {V}_j^*,\mathcal {V}_{\hat {j}})$ for any $j \in [m]$ , where $\mathcal {V}_j^*$ is the dual vector space to $\mathcal {V}_j$ . The rank of a tensor in $\mathcal {V}_1 \otimes \mathcal {V}_2$ is equal to the rank of the corresponding linear operator in $\mathrm {Hom}(\mathcal {V}_1^*, \mathcal {V}_2)$ . We denote the rank of a tensor $v\in \mathcal {V}$ , viewed as an element of $\mathrm {Hom}(\mathcal {V}_j^*,\mathcal {V}_{\hat {j}})$ , by $\operatorname {rank}_j(v)$ . The flattening rank of v is defined as $\max \{\operatorname {rank}_1(v),\dots , \operatorname {rank}_m(v)\}$ . Note that the tensor rank of v is lower bounded by the flattening rank of v.

We write $S \cup T$ to denote the union of two sets S and T. If S and T happen to be disjoint, we often write $S \sqcup T$ instead to remind the reader of this fact. For a positive integer t, we say that a collection of subsets $S_1,\dots , S_t \subseteq T$ partitions T if $S_p \cap S_q =\{\}$ for all $p\neq q \in [t]$ , and $S_1 \sqcup \dots \sqcup S_t=T$ .

For a multiset of nonzero vectors $E=\{v_1,\dots , v_n\}\subseteq \mathcal {V}$ , a connected component of E is an inclusion-maximal connected subset of E. Any multiset of nonzero vectors E can be (uniquely, up to reordering) partitioned into disjoint connected components ${T_1 \sqcup \dots \sqcup T_t=E}$ [Reference OxleyOxl06, Proposition 4.1.2]. Observe that

$$ \begin{align*} \operatorname{\mathrm{span}}(E)=\bigoplus_{i\in [t]} \operatorname{\mathrm{span}}(T_i), \end{align*} $$

and note that $S \subseteq E$ separates E if and only if

$$ \begin{align*} \dim \operatorname{\mathrm{span}} \{ v_1,\dots, v_n\} = \dim \operatorname{\mathrm{span}} \{v_a : a \in S\} + \dim\operatorname{\mathrm{span}}\{v_a : a \in S^c\} \end{align*} $$

if and only if

$$ \begin{align*} \operatorname{\mathrm{span}} \{v_a : a \in S\} \cap \operatorname{\mathrm{span}}\{v_a : a \in S^c\}= \{0\} \end{align*} $$

(see [Reference OxleyOxl06, Proposition 4.2.1]).

In the remainder of this section, we formally introduce symmetric tensors and symmetric tensor decompositions, which are natural analogues of tensors and tensor decompositions. For a positive integer $m \geq 2$ and a vector space $\mathcal {W}$ over a field $\mathbb {F}$ with $\mathrm {Char}(\mathbb {F})>m$ or ${\mathrm {Char}}(\mathbb {F})=0$ , we say that a tensor $v \in \mathcal {W}^{\otimes m}$ is symmetric if it is invariant under permutations of the subsystems. The Waring rank of a symmetric tensor v, denoted by $\mathrm {WaringRank}(v)$ , is the minimum number n for which v is equal to a linear combination of n symmetric product tensors. A decomposition of v into a linear combination of $\mathrm {WaringRank}(v)$ symmetric product tensors is called a Waring rank decomposition of v. A decomposition of v into a linear combination of symmetric product tensors (not necessarily of minimum number) is known simply as a symmetric decomposition of v.

A symmetric decomposition of v

(4) $$ \begin{align} v = \sum_{a \in [n]} \alpha_a v_a^{\otimes m} \end{align} $$

is said to be the unique Waring rank decomposition of v if for any nonnegative integer $r \leq n$ , multiset of nonzero vectors $\{u_a : a \in [r]\} \subseteq \mathcal {W} \setminus \{0\}$ and nonzero scalars $\{\beta _a : a \in [r]\} \subseteq \mathbb {F}^{\times }$ for which

(5) $$ \begin{align} v = \sum_{a \in [r]} \beta_a u_a^{\otimes m}, \end{align} $$

it holds that $r=n$ and

$$ \begin{align*} \{\alpha_a v_a^{\otimes m} : a \in [n]\} = \{\beta_a u_a^{\otimes m} : a \in [n]\}. \end{align*} $$

More generally, for a positive integer $\tilde {n} \geq n$ , we say that the symmetric decomposition (4) is the unique symmetric decomposition of vinto at most $\tilde {n}$ terms if for any $r \leq \tilde {n}$ and symmetric decomposition (5), either

$$ \begin{align*} {\operatorname{k-rank}(u_a : a \in [r])=1}, \end{align*} $$

or $r=n$ and

$$ \begin{align*} \{\alpha_a v_a^{\otimes m} : a \in [n]\} = \{\beta_a u_a^{\otimes m} : a \in [n]\}. \end{align*} $$

Note that equation (4) is the unique Waring rank decomposition of v if and only if it is the unique symmetric decomposition of v into at most n terms. Note also that if equation (4) is the unique symmetric decomposition of v into at most $\tilde {n}>n$ terms, then it is the unique Waring rank decomposition of v. We refer to results that certify uniqueness of a symmetric decomposition into at most $\tilde {n}>n$ terms as uniqueness results for non-Waring rank decompositions. We present such results in Section 8.

Our assumption that $\mathrm {Char}(\mathbb {F})>m$ or ${\mathrm {Char}}(\mathbb {F})=0$ in the symmetric case ensures that the symmetric subspace is isomorphic to the space of homogeneous polynomials over $\mathbb {F}$ of degree m in $\dim (\mathcal {W})$ variables, and that every symmetric tensor has finite Waring rank (see, e.g., [Reference Iarrobino and KanevIK99, Appendix A] and [Reference LandsbergLan12, Section 2.6.4]).

3 Proving the splitting theorem

In this section, we prove Theorem 4. We first observe the following basic fact.

Proposition 5. Let $n \geq 2$ be an integer, let $\mathcal {V}=\mathcal {V}_1\otimes \mathcal {V}_2$ be a vector space over a field $\mathbb {F}$ and let

$$ \begin{align*} E=\{x_a \otimes y_a: a \in[n] \} \subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \mathcal{V}_2 \right) \end{align*} $$

be a multiset of product tensors. If E is connected, then $\{x_a : a \in [n]\}$ and $\{y_a : a \in [n]\}$ are both connected.

Proof. Suppose toward contradiction that E is connected and $\{x_a : a \in [n]\}$ splits, that is,

(6) $$ \begin{align} \dim\operatorname{\mathrm{span}}\{x_a: a \in [n]\}=\operatorname{\mathrm{span}}\{x_a: a \in S\} \oplus \operatorname{\mathrm{span}}\{x_a:a \in S^c\} \end{align} $$

for some nonempty proper subset $S \subseteq [n]$ . Since E is connected, there exists a nonzero vector

$$ \begin{align*} v \in \operatorname{\mathrm{span}}\{x_a \otimes y_a: a \in S\} \cap \operatorname{\mathrm{span}}\{x_a \otimes y_a: a \in S^c\}. \end{align*} $$

Let $f \in \mathcal {V}_2^*$ be any linear functional such that Then is a nonzero element of

$$ \begin{align*} \operatorname{\mathrm{span}}\{x_a: a \in S\} \cap \operatorname{\mathrm{span}}\{x_a:a \in S^c\}, \end{align*} $$

contradicting equation (6). The result is obviously symmetric under permutation of $\mathcal {V}_1$ and $\mathcal {V}_2$ .

It is not difficult to see that Theorem 4 follows directly from the $m=2$ case of Theorem 4, Proposition 5 and induction on m: Let $\mathcal {V}_1,\dots , \mathcal {V}_m$ be $\mathbb {F}$ -vector spaces, and let

$$ \begin{align*} E=\{x_a : a \in [n]\} \subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \cdots : \mathcal{V}_m \right) \end{align*} $$

be a multiset of product tensors for which $\dim \operatorname {\mathrm {span}}(E) \leq \sum _{j=1}^m (d_j-1)$ . If

$$ \begin{align*} E_{\hat{m}}:=\{x_{a,\hat{m}}: a \in [n]\} \end{align*} $$

splits, then E splits by Proposition 5 (recall that $x_{a,\hat {m}}=x_{a,1}\otimes \dots \otimes x_{a,m-1}$ ). Otherwise, by the induction hypothesis $\dim \operatorname {\mathrm {span}}(E_{\hat {m}})>\sum _{j=1}^{m-1}(d_j-1)$ . But this implies ${\dim \operatorname {\mathrm {span}}(E) \leq \dim \operatorname {\mathrm {span}}(E_{\hat {m}})+d_m-2}$ , so E splits by the $m=2$ case of Theorem 4. Hence, we need only prove the $m=2$ case of Theorem 4, which we now explicitly state for clarity.

Theorem 6 ( $m=2$ case of Theorem 4).

Let $n \geq 2$ be an integer, let $\mathcal {V}=\mathcal {V}_1\otimes \mathcal {V}_2$ be a vector space over a field $\mathbb {F}$ , and let

$$ \begin{align*} E=\{x_a \otimes y_a: a \in[n] \} \subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \mathcal{V}_2 \right) \end{align*} $$

be a multiset of product tensors. Let

$$ \begin{align*} d_1=\dim\operatorname{\mathrm{span}} \{ x_a: a \in [n]\} \end{align*} $$

and

$$ \begin{align*} d_2=\dim\operatorname{\mathrm{span}} \{ y_a: a \in [n]\}. \end{align*} $$

If E is connected, then $\dim \operatorname {\mathrm {span}}(E)\geq d_1+d_2-1$ .

To prove Theorem 6, we require a matroid-theoretic construction called the ear decomposition of a connected matroid (see, e.g., [Reference Coullard and HellersteinCH96]).

Lemma 7 (Ear decomposition (see, e.g., [Reference Coullard and HellersteinCH96])).

Let $n \geq 2$ be an integer, let $\mathcal {V}$ be a vector space over a field $\mathbb {F}$ and let $E=\{v_1,\dots , v_n\} \subseteq \mathcal {V}\setminus \{0\}$ be a multiset of nonzero vectors. If E is connected, then there exists a collection of circuits $C_1, \dots , C_t \subseteq E$ such that

$$ \begin{align*} E=C_1 \cup C_2 \cup \dots \cup C_t, \end{align*} $$

and for each $p \in [t]$ , the multisets $C_p$ and $E_p := C_1 \cup \dots \cup C_p$ satisfy the following two properties:

1. 1. $C_p \cap E_{p-1} \neq \{\}$

2. 2. $\dim \operatorname {\mathrm {span}}(E_p)-\dim \operatorname {\mathrm {span}}(E_{p-1})=|E_p \setminus E_{p-1}|-1$

Now, we prove Theorem 6.

Proof of Theorem 6.

For a subset $S \subseteq [n]$ , let

$$ \begin{align*} d^S&=\dim\operatorname{\mathrm{span}}\{x_a \otimes y_a : a \in S\},\\ d_1^S&=\dim\operatorname{\mathrm{span}}\{x_a : a \in S\},\\ d_2^S&=\dim\operatorname{\mathrm{span}}\{y_a: a \in S\}. \end{align*} $$

In a slight change of notation from Lemma 7, let $C_1,\dots , C_t \subseteq [n]$ be the index sets corresponding to an ear decomposition of E, and let $E_p=C_1 \cup \dots \cup C_p \subseteq [n]$ for each $p \in [t]$ . The theorem follows from the following two claims

Claim 8. $d^{E_1} \geq d^{E_1}_1+d^{E_1}_2-1.$

Claim 9. For each $p \in \{2,\dots , t\}$ ,

$$ \begin{align*} \lvert E_p \setminus E_{p-1} \rvert-1 \geq d_1^{E_p}-d_1^{E_{p-1}}+d_2^{E_p}-d_2^{E_{p-1}}. \end{align*} $$

Before proving these claims, let us first use them to complete the proof. Note that

$$ \begin{align*} d^{E_2}&=d^{E_1}+\lvert E_2 \setminus E_1 \rvert-1\\ &\geq d^{E_1}_1+d^{E_1}_2-1+\lvert E_2 \setminus E_1 \rvert-1\\ &\geq d^{E_2}_1+d^{E_2}_2-1. \end{align*} $$

The first line is a property of the ear decomposition, the second line follows from Claim 8, and the third line follows from Claim 9. So Claim 8 holds with $E_1$ replaced with $E_2$ . Repeating this process inductively gives $d^{[n]} \geq d^{[n]}_1+d^{[n]}_2-1$ , which is what we wanted to prove. This completes the proof, modulo proving the claims.

Proof of Claim 8.

By permuting $[n]$ , we may assume that $C_1=[q]$ for some $q \in [n]$ and that $\{ x_a : a \in [d_1^{[q]}]\}$ is a basis for $\operatorname {\mathrm {span}}\{x_a : a \in [q]\}$ . Let $s=d_1^{[q]}$ .

Suppose that there exists $b \in [s]$ such that $y_b \notin \operatorname {\mathrm {span}} \{y_a : a \in [q]\setminus [s]\}$ . Let $f \in \mathcal {V}_1^*$ , $g \in \mathcal {V}_2^*$ be linear functionals such that $f(x_b)=g(y_b)=1$ , $f(x_a)=0$ for all $a \in [s] \setminus \{b\}$ , and $g(y_a)=0$ for all $a \in [q]\setminus [s]$ . So

$$ \begin{align*} (f \otimes g)(x_a \otimes y_a)= \begin{cases} 1, & a=b\\ 0, & a \neq b \end{cases}. \end{align*} $$

It follows that $x_b \otimes y_b \notin \operatorname {\mathrm {span}} \{x_a \otimes y_a : a \in [q] \setminus \{b\}\}$ , contradicting the fact that $C_1$ indexes a circuit. So $\{y_a : a \in [s]\} \subseteq \operatorname {\mathrm {span}}\{y_a : a\in [q]\setminus [s]\}$ , which implies

$$ \begin{align*} d_2^{C_1} &\leq q-s\\ &=d^{C_1}+1-d_1^{C_1}, \end{align*} $$

completing the proof. $\triangle $

Now, we prove Claim 9.

Proof of Claim 9.

Let $B\subseteq E_{p-1}$ be such that $\{x_a : a \in B\}$ is a basis for ${\operatorname {\mathrm {span}} \{x_a : a \in E_{p-1}\}}$ . By permuting $[n]$ , we may assume that $E_p \setminus E_{p-1}=[q]$ for some $q \in [n]$ and that ${B \cup \{x_a : a \in [s]\}}$ is a basis for $\operatorname {\mathrm {span}}\{x_a : a \in E_p\}$ , where $s=d^{E_p}-d^{E_{p-1}}$ . If there exists $b \in [s]$ for which $y_b \notin \operatorname {\mathrm {span}} \{y_a : a \in [q]\setminus [s]\}$ , then, as in the proof of Claim 8,

$$ \begin{align*} x_b \otimes y_b \notin \operatorname{\mathrm{span}} \{x_a \otimes y_a : a \in E_p \setminus \{b\}\}. \end{align*} $$

But this contradicts connectedness of E. It follows that $d_2^{E_p \setminus E_{p-1}}\leq q-s$ , so

$$ \begin{align*} d_2^{E_p}-d_2^{E_{p-1}}&\leq d_2^{C_p}-d_2^{C_p \cap E_{p-1}}\\ &\leq d_2^{C_p \setminus (C_p \cap E_{p-1})}-1\\ &= d_2^{E_p \setminus E_{p-1}}-1\\ &\leq q-s-1\\ &=\lvert E_p \setminus E_{p-1} \rvert-(d_1^{E_p}-d_1^{E_{p-1}})-1. \end{align*} $$

The first line is easy to verify (in matroid-theoretic terms, this is submodularity of the rank function). The second line follows from the fact that $\{y_a : a \in C_p\}$ is connected. The third line is obvious, the fourth line follows from $d_2^{E_p \setminus E_{p-1}}\leq q-s$ , and the fifth line follows from our definitions. This completes the proof.

The proofs of Claims 8 and 9 complete the proof of the theorem.

4 Using our splitting theorem to generalize Kruskal’s theorem

In this section, we use our splitting theorem (Theorem 4) to prove our generalization of Kruskal’s theorem (Theorem 2). We then introduce a reshaped version of Theorem 2, which has many more degrees of freedom than the standard reshaping of Kruskal’s theorem.

To prove Theorem 2, we first observe the following useful corollary to our splitting theorem.

Corollary 10. Let $n \geq 2$ and $m\geq 2$ be integers, let $\mathcal {V}=\mathcal {V}_1\otimes \dots \otimes \mathcal {V}_m$ be a vector space over a field $\mathbb {F}$ , let

$$ \begin{align*} E=\{x_a: a \in[n] \} \subseteq \mathrm{Prod}\left(\mathcal{V}_1:\dots : \mathcal{V}_m \right) \end{align*} $$

be a multiset of product tensors, and for each $j \in [m]$ , let

$$ \begin{align*} d_j=\dim\operatorname{\mathrm{span}} \{ x_{a,j}: a \in [n]\}. \end{align*} $$

If $n \leq \sum _{j=1}^m (d_j-1)+1$ , then E splits.

Proof. If E is linearly independent, then it obviously splits. Otherwise,

$$ \begin{align*} {\dim \operatorname{\mathrm{span}} (E) \leq n-1}, \end{align*} $$

and the result follows immediately from our splitting theorem.

Now, we use this corollary to prove our generalization of Kruskal’s theorem.

Proof of Theorem 2.

Let $x_a=x_{a,1}\otimes \dots \otimes x_{a,m}$ for each $a \in [n]$ , and suppose that ${\sum _{a \in [n]} x_a = \sum _{a \in [r]} y_a}$ for some nonnegative integer $r \leq n$ and multiset of product tensors $\{y_a: a \in [r]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1:\dots : \mathcal {V}_m \right )$ . For notational convenience, for each $a \in [r]$ let $x_{n+a}=-y_a$ , so that $\sum _{a\in [n+r]} x_a =0$ . Let $T_1 \sqcup \dots \sqcup T_t=[n+r]$ be the index sets of the connected components of $\{x_a : a \in [n+r]\}$ . Since $\sum _{a\in [n+r]} x_a =0$ , it follows that $\sum _{a \in T_p} x_a =0$ for all $p \in [t]$ , so $\lvert T_p \rvert \geq 2$ for all $p \in [t]$ .

For each $p \in [t]$ , if

(7) $$ \begin{align} \bigl\lvert T_p \cap [n] \bigr\rvert \geq \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert, \end{align} $$

then it must hold that

(8) $$ \begin{align} \bigl\lvert T_p \cap [n] \bigr\rvert = \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert=1, \end{align} $$

otherwise $\{x_a : a \in T_p\}$ would split by Corollary 10, a contradiction. Since $r \leq n$ and the inequality (7) can never be strict, it follows that $r=n$ and equation (8) holds for all $p \in [t]$ . This completes the proof.

For $m \geq 4$ , both Kruskal’s theorem and our Theorem 2 can be ‘reshaped’ by regarding multiple subsystems as a single subsystem, to give potentially stronger uniqueness criteria. It is worth noting that the reshaped version of Theorem 2 has quite a different flavour from the reshaped version of Kruskal’s theorem; in particular, there are many more degrees of freedom to choose from. We omit the proof of the following reshaped version of Theorem 2 because it is similar to the proof of Theorem 2.

Theorem 11 (Reshaped generalization of Kruskal’s theorem).

Let $n \geq 2$ and $m \geq 3$ be integers, let $\mathcal {V}=\mathcal {V}_1\otimes \dots \otimes \mathcal {V}_m$ be a vector space over a field $\mathbb {F}$ and let

$$ \begin{align*} \{x_a: a \in [n]\} \subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \dots : \mathcal{V}_m \right) \end{align*} $$

be a multiset of product tensors. For each $S \subseteq [n]$ and $J \subseteq [m]$ , let

$$ \begin{align*} d_J^S=\dim\operatorname{\mathrm{span}}\Big\{\bigotimes_{j \in J} x_{a,j} : a \in S\Big\}. \end{align*} $$

If for every subset $S \subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ , there exists a partition $J_1 \sqcup \dots \sqcup J_t =[m]$ (which may depend on S) such that $2 \lvert S \rvert \leq \sum _{i \in [t]} (d_{J_i}^S-1)+1$ , then $\sum _{a \in [n]} x_a$ constitutes a unique tensor rank decomposition.

It is instructive to compare Theorem 11 to the standard reshaping of Kruskal’s theorem:

Theorem 12 (Reshaped Kruskal’s theorem).

Let $n \geq 2$ and $m \geq 3$ be integers, let ${\mathcal {V}=\mathcal {V}_1\otimes \cdots \otimes \mathcal {V}_m}$ be a vector space over a field $\mathbb {F}$ and let

$$ \begin{align*} \{x_a: a \in [n] \}\subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \dots : \mathcal{V}_m \right) \end{align*} $$

be a multiset of product tensors. For each $J \subseteq [m]$ , let

$$ \begin{align*} k_J = \operatorname{k-rank}\left( \bigotimes_{j \in J} x_{a,j} : a \in [n]\right). \end{align*} $$

If there exists a partition of $[m]$ into three disjoint subsets $J \sqcup K \sqcup L=[m]$ such that ${2n \leq k_J+k_K+k_L-2},$ then $\sum _{a \in [n]} x_a$ constitutes a unique tensor rank decomposition.

Theorem 12 clearly follows from our Theorem 11. In Theorem 12, one could of course consider more general partitions of $[m]$ into more than three subsets, but since the k-rank satisfies $k_{J \cup K} \geq \min \{n,k_{J}+k_K-1\}$ for any disjoint subsets $J,K \subseteq [m]$ (See Lemma 1 in [Reference Sidiropoulos and BroSB00]), it suffices to consider tripartitions $J \sqcup K \sqcup L=[m]$ . In contrast, it is not clear that one can restrict to tripartitions in Theorem 11. There is another major difference between these two theorems: In Theorem 12, one chooses a single partition of $[m]$ , whereas in Theorem 11, one is free to choose a different partition of $[m]$ for every S.

We remark that many other statements in this work (for example, the splitting theorem itself) can be reshaped similarly to Theorem 11. We do not explicitly state these reshapings.

5 The inequality appearing in our splitting theorem cannot be weakened

In this section, we find a connected multiset of product tensors $E=\{x_a : a \in [n]\}$ that satisfies ${\dim \operatorname {\mathrm {span}}(E) = \sum _{j=1}^m (d_j-1)+1}$ . In fact, we prove that this multiset of product tensors forms a circuit, which is stronger than being connected. This proves that the bound in Corollary 21, and the inequality $\dim \operatorname {\mathrm {span}}(E) \leq \sum _{j=1}^m (d_j-1)$ appearing in Theorem 4, cannot be weakened. The example we use is Derksen’s [Reference DerksenDer13], which he used to prove that the inequality appearing in Kruskal’s theorem cannot be weakened.

Fact 13. For any field $\mathbb {F}$ with $\mathrm {Char}(\mathbb {F})=0$ and positive integers $d_1,\dots , d_m$ with ${n-1=\sum _{j=1}^m (d_j-1)+1}$ , there exist vector spaces $\mathcal {V}_1,\dots ,\mathcal {V}_m$ over $\mathbb {F}$ and a multiset of product tensors $\{x_a:a \in [n]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1 : \cdots : \mathcal {V}_m \right )$ that forms a circuit and satisfies

$$ \begin{align*} \dim\operatorname{\mathrm{span}}\{x_{a,j} : a \in [n]\}\geq d_j \end{align*} $$

for all $a \in [n]$ .

We note that if $d_1=\dots = d_m$ , then the multiset of product tensors $\{x_a : a \in [n]\}$ can be taken to be symmetric in the sense introduced in Section 2 (this is obvious from Derksen’s construction [Reference DerksenDer13]). As a result, our splitting theorem is also sharp for symmetric product tensors. We use this fact in Sections 7 and 8 to prove optimality of our results on symmetric decompositions. We remark that the assumption $\mathrm {Char}(\mathbb {F})=0$ can be weakened, see [Reference DerksenDer13].

6 Between splitting and uniqueness: intermediate consequences of the splitting theorem

For the entirety of this section, we fix nonnegative integers $n\geq 2$ and $m\geq 2$ , a vector space $\mathcal {V}=\mathcal {V}_1\otimes \dots \otimes \mathcal {V}_m$ over a field $\mathbb {F}$ and a multiset of product tensors ${\{x_a: a \in [n] \} \subseteq \mathrm {Prod}\left (\mathcal {V}_1:\dots : \mathcal {V}_m \right )}$ . For each subset $S\subseteq [n]$ and index $j \in [m]$ , we define

$$ \begin{align*} d_j^S=\dim\operatorname{\mathrm{span}} \{ x_{a,j}: a \in S\} \end{align*} $$

and use the shorthand $d_j= d_j^{[n]}$ for all $j \in [m]$ .

As a consequence of our splitting theorem, if $n \leq \sum _{j=1}^m (d_j -1)+1$ , then ${\{x_a: a \in [n]\}}$ splits (Corollary 10). We used Corollary 10 to prove our generalization of Kruskal’s theorem: If $2 \lvert S \rvert \leq \sum _{j=1}^m(d_j^S-1)+1$ for every subset $S \subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ , then $\sum _{a \in [n]} x_a$ constitutes a unique tensor rank decomposition. It is natural to ask what happens when other, similar inequalities hold. In particular, suppose that

(9) $$ \begin{align} {\lvert S \rvert+\mathcal{R}(\lvert S \rvert) \leq \sum_{j=1}^m (d_j^S -1)+1} \end{align} $$

for all $S \subseteq [n]$ with $s+1 \leq \lvert S \rvert \leq n$ , for some $s \in [n-1]$ and function $\mathcal {R}: [n]\setminus [s] \rightarrow \mathbb {Z}$ . What can be said about the tensors $v \in \operatorname {\mathrm {span}} \{x_a : a \in [n]\}$ ?

In this section, we use Corollary 10 to answer this question for choices of s and $\mathcal {R}$ that produce useful results on sets of product tensors. In Section 6.1, we prove uniqueness results for low-rank tensors in $\operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ . These results can be viewed as a family of statements that connects the choice of parameters in Corollary 10, where $s=n-1$ and $\mathcal {R}(n)=n$ , to the choice of parameters in our generalization of Kruskal’s theorem, where $s=1$ and . We use this family of statements to prove a new lower bound on tensor rank, to improve (and sharpen) results of Ballico on the number of nontrivial subsystems in a circuit of product tensors [Reference Ballico, Bernardi, Chiantini and GuardoBBCG18, Reference BallicoBal20], to strengthen a known sufficient condition for a set of product tensors to be linearly independent and to strengthen a known sufficient condition for a set of product tensors to contain no other product tensors in their span (aside from scalar multiples) [Reference Ha and KyeHK15]. In Section 6.2, we prove uniqueness results for nonrank decompositions of $\sum _{a \in [n]} x_a$ (i.e., decompositions into a nonminimal number of product tensors), which appear to be the first known results of this kind. We prove all of these results using the language of subpartitions of pairs of decompositions, which we now introduce.

Definition 14. For positive integers n and r, multisets of product tensors

$$ \begin{align*} \{x_a : a \in [n]\},\{y_a : a \in [r]\} \subseteq \mathrm{Prod}\left(\mathcal{V}_1 : \dots : \mathcal{V}_m \right), \end{align*} $$

and nonzero scalars

$$ \begin{align*} \{\alpha_a : a\in [n]\}, \{\beta_a : a \in [r]\} \subseteq \mathbb{F}^\times, \end{align*} $$

for which

$$ \begin{align*} \sum_{a \in [n]} \alpha_a x_a=\sum_{a \in [r]} \beta_a y_a, \end{align*} $$

we say that the (ordered) pair of decompositions $(\sum _{a \in [n]} \alpha _a x_a, \sum _{a \in [r]} \beta _a y_a)$ has an $(s,l)$ -subpartition for some positive integers s and l if there exist pairwise disjoint subsets $Q_1,\dots , Q_l \subseteq [n]$ and pairwise disjoint subsets ${R_1,\dots , R_l \subseteq [r]}$ for which

$$ \begin{align*} \max\{1, \lvert R_p \rvert\}\leq \lvert Q_p \rvert\leq s \end{align*} $$

and $\sum _{a \in Q_p} \alpha _a x_a = \sum _{a \in R_p} \beta _a y_a$ for all $p \in [l]$ . We say that the pair $(\sum _{a \in [n]} \alpha _a x_a, \sum _{a \in [r]} \beta _a y_a)$ has an $(s,l)$ -partition if the sets $Q_1,\dots , Q_l \subseteq [n]$ and $R_1,\dots ,R_l \subseteq [r]$ can be chosen to partition $[n]$ and $[r]$ , respectively.

We say that the pair $(\sum _{a \in [n]} \alpha _a x_a, \sum _{a \in [r]} \beta _a y_a)$ is reducible if there exist subsets $Q \subseteq [n]$ and $R \subseteq [r]$ for which $\lvert Q \rvert> \lvert R \rvert $ and $\sum _{a \in Q} \alpha _a x_a = \sum _{a \in R} \beta _a y_a$ . We say that the pair is irreducible if it is not reducible.

(Technically, the linear combinations appearing in the pair $(\sum _{a \in [n]} \alpha _a x_a, \sum _{a \in [r]} \beta _a y_a)$ should be regarded formally so that they contain the data of the decompositions, and the linear combinations appearing elsewhere should be regarded as standard linear combinations in $\mathcal {V}$ .)

For brevity, we will often abuse notation and say that $\sum _{a \in [n]} \alpha _a x_a=\sum _{a \in [r]} \beta _a y_a$ has an $(s,l)$ -subpartition (or is reducible) to mean that $(\sum _{a \in [n]} \alpha _a x_a,\sum _{a \in [r]} \beta _a y_a)$ has an $(s,l)$ -subpartition (or is reducible). Note that the properties of $(s,l)$ -subpartitions and reducibility are not symmetric with respect to permutation of the first and second decompositions. Typically, the first decomposition $\sum _{a \in [n]} \alpha _a x_a$ will be known, and the second decomposition $\sum _{a \in [r]} \beta _a y_a$ will be some unknown decomposition that we want to control.

An immediate consequence of Corollary 10 is that if $\sum _{a \in [n]} x_a = \sum _{a \in [r]} y_a$ for some $r \leq n$ , and the inequality (9) holds for $s=n-1$ and $\mathcal {R}(n)=r$ , then this pair of decompositions has an $(n-1, 1)$ -subpartition (see Corollary 20 for a slight extension of this statement). By comparison, our generalization of Kruskal’s theorem states that if $r \leq n$ , and equation (9) holds for $s=1$ and , then $r=n$ and this pair of decompositions has a $(1,n)$ -subpartition. In Section 6.1, we prove statements on the existence of $(s,l)$ -subpartitions for $r\leq n$ , which interpolate between these two statements by trading stronger assumptions for stronger notions of uniqueness. In Section 6.2, we prove a similar family of statements for $r\geq n+1$ , obtaining novel uniqueness results for nonrank decompositions.

We conclude the introduction to this section by making a few notes about our definitions of $(s,l)$ -subpartitions and reducibility. It may seem a bit strange at first that the inequality $\lvert R_p \rvert \leq \lvert Q_p \rvert $ appears in our definition of an $(s,l)$ -subpartition. We have chosen to include this inequality because we typically want to reduce the number of product tensors that appear a decomposition. Our definition of reducibility captures a similar idea: If $n \leq r$ and $(\sum _{a \in [n]} \alpha _a x_a,\sum _{a \in [r]} \beta _a y_a)$ is reducible, then these decompositions can easily be combined to produce a decomposition into fewer than n product tensors. (When $r \leq n$ , reducibility of $(\sum _{a \in [r]} \beta _a y_a,\sum _{a \in [n]} \alpha _a x_a)$ captures a similar idea.) Assuming irreducibility will allow us to avoid certain pathological cases. Note that if $\sum _{a \in [n]} \alpha _a x_a$ is a tensor rank decomposition, then $(\sum _{a \in [n]} \alpha _a x_a,\sum _{a \in [r]} \beta _a y_a)$ is automatically irreducible.

Note that when $(\sum _{a \in [n]} \alpha _a x_a, \sum _{a \in [r]} \beta _a y_a)$ is irreducible, the existence of an $(s,l)$ -subpartition is equivalent to the existence of pairwise disjoint subsets $Q_1,\dots , Q_l \subseteq [n]$ and pairwise disjoint subsets $R_1,\dots , R_l \subseteq [r]$ for which

$$ \begin{align*} 1 \leq \lvert R_p \rvert = \lvert Q_p \rvert\leq s \end{align*} $$

and $\sum _{a \in Q_p} \alpha _a x_a = \sum _{a \in R_p} \beta _a y_a$ for all $p \in [l]$ . When $s=1$ , these statements are equivalent even without the irreducibility assumption.

6.1 Low-rank tensors in the span of a set of product tensors

In this subsection, we prove statements about low-rank tensors in $\operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ . Most of our results in this section are consequences of Theorem 15, which is a somewhat complicated statement on the existence of $(s,l)$ -partitions. For $s=1$ , and any ${r \in \{0,1,\dots , n\}}$ , we obtain a condition on $\{x_a : a \in [n]\}$ for which the only rank $\leq r$ tensors in $\operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ are those that can be written (uniquely) as a linear combination of $\leq r$ elements of $\{x_a : a \in [n]\}$ . For $s=1,r=0$ , we obtain a sufficient condition for linear independence of $\{x_a : a \in [n]\}$ . For $s=1,r=1$ , we obtain a sufficient condition for the only product tensors in $\operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ to be scalar multiples of $x_1,\dots , x_n$ . These generalize Proposition 3.1 and Theorem 3.2 in [Reference Ha and KyeHK15], respectively. The case $s=1, r=n$ reproduces our generalization of Kruskal’s theorem. For $s=n-1$ , we strengthen recent results in [Reference Ballico, Bernardi, Chiantini and GuardoBBCG18, Reference BallicoBal20] on circuits of product tensors.

Most of the statements in this subsection are consequences of the following theorem, which is complicated to state but easy to prove with our splitting theorem.

Theorem 15. Let $s \in [n-1]$ , and $r\in \{0,1,\dots ,n\}$ be integers. Suppose that for every subset $S \subseteq [n]$ with $s+1\leq \lvert S \rvert \leq n,$ it holds that

(10) $$ \begin{align} \min\{2 \lvert S \rvert,\lvert S \rvert+r\} \leq\sum_{j=1}^m (d_j^S-1)+1. \end{align} $$

Then for any $v \in \operatorname {\mathrm {span}} \{x_a : a \in [n]\}$ with $\operatorname {rank}(v)\leq r$ , and any decomposition ${v=\sum _{a \in [\tilde {r}]} y_a}$ of v into $\tilde {r} \leq r$ product tensors $\{y_a : a \in [\tilde {r}]\} \subseteq \mathrm {Prod}\left (\mathcal {V}_1:\dots : \mathcal {V}_m \right )$ , the following holds: For any subset $S \subseteq [n]$ for which $\lvert S \rvert \geq s+1$ , and nonzero scalars $\{\alpha _a : a \in S\} \subseteq \mathbb {F}^\times $ for which it holds that

$$ \begin{align*} \sum_{a \in S} \alpha_a x_a = \sum_{a \in [\tilde{r}]} y_a \end{align*} $$

and $( \sum _{a \in [\tilde {r}]} y_a,\sum _{a \in S} \alpha _a x_a)$ is irreducible, the pair of decompositions $(\sum _{a \in [n]} \alpha _a x_a, \sum _{a \in [\tilde {r}]} y_a)$ has an $(s,l)$ -partition, for $l=\lceil \lvert S \rvert /s \rceil $ .

Proof. For each $a \in [\tilde {r}]$ , let $x_{n+a}=-y_a$ , and let $E = S \cup ([n+\tilde {r}]\setminus [n]) \subseteq [n+\tilde {r}]$ . Let ${T_1\sqcup \dots \sqcup T_t = E}$ be a partition of E into index sets corresponding to the connected components of $\{x_a : a \in E\}$ . Since $( \sum _{a \in [\tilde {r}]} y_a,\sum _{a \in S} \alpha _a x_a)$ is irreducible, it must hold that

$$ \begin{align*} \bigl\lvert T_p \cap S \bigr\rvert\geq \bigl\lvert T_p \cap (E \setminus S) \bigr\rvert \end{align*} $$

for all $p \in [t]$ , and hence

$$ \begin{align*} \lvert T_p \rvert \leq \min\big\{ 2 \bigl\lvert T_p \cap S \bigr\rvert, \bigl\lvert T_p \cap S \bigr\rvert +r\big\}. \end{align*} $$

If $\lvert T_p \cap S \rvert \geq s+1,$ then $\{x_a : a \in T_p\}$ splits by equation (10) and Corollary 10, a contradiction. So it must hold that ${\lvert T_p \cap S \rvert \leq s}$ for all $p \in [t]$ . It follows that $t \geq \lceil \lvert S \rvert /s \rceil $ by the pigeonhole principle, and one can take $Q_p=T_p \cap S$ and

$$ \begin{align*} R_p = \{a \in [\tilde{r}] : n+a \in T_p \cap (E \setminus S)\} \end{align*} $$

for all $p \in [t]$ to conclude.

6.1.1 $s=1$ case of Theorem 15

The $s=1$ case of Theorem 15 gives a sufficient condition for which the only tensor rank $\leq r$ elements of $\operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ are those which can be written (uniquely) as a linear combination of $\leq r$ elements of $\operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ . In this subsection, we state this case explicitly and observe several consequences of this case. In particular, we observe a lower bound on tensor rank and a sufficient condition for a set of product tensors to be linearly independent.

Corollary 16 ( $s=1$ case of Theorem 15).

Let $r\in \{0,1,\dots ,n\}$ be an integer. Suppose that for every subset $S \subseteq [n]$ such that $2\leq \lvert S \rvert \leq n$ , it holds that

(11) $$ \begin{align} \lvert S \rvert+\min\{\lvert S \rvert,r\} \leq\sum_{j=1}^m (d_j^S-1)+1. \end{align} $$

Then any nonzero linear combination of more than r elements of $\{x_a : a \in [n]\}$ has tensor rank greater than r, and every tensor $v \in \operatorname {\mathrm {span}}\{x_a : a \in [n]\}$ of tensor rank at most r has a unique tensor rank decomposition into a linear combination of elements of $\{x_a : a \in [n]\}$ .

Note that a sufficient condition for the inequality (11) to hold is that

$$ \begin{align*} n+r \leq \sum_{j=1}^m (k_j-1)+1, \end{align*} $$

where $k_j=\operatorname {k-rank}(x_{1,j},\dots , x_{n,j})$ for all $j \in [m]$ . This recovers Proposition 3.1 and Theorem 3.2 in [Reference Ha and KyeHK15] in the $r=0$ and $r=1$ cases, respectively, and interpolates between Kruskal’s theorem and these results. For clarity, we will explicitly state the $r=0$ and $r=1$ cases of Corollary 16 at the end of this subsection.

Proof of Corollary 16.

Let $S \subseteq [n]$ be a subset, let $\{\alpha _a : a \in S\} \subseteq \mathbb {F}^{\times }$ be a multiset of nonzero scalars, let $\tilde {r}=\operatorname {rank}[\sum _{a \in S} \alpha _a x_a]$ and let $\{y_a : a \in [\tilde {r}]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1 : \dots : \mathcal {V}_m \right )$ be such that $\sum _{a \in S} \alpha _a x_a=\sum _{a \in [\tilde {r}]} y_a$ . If $\tilde {r}\leq r$ , then by the $s=1$ case of Theorem 15, this pair of decompositions has a $(1,\lvert S \rvert )$ -partition. It follows that $\lvert S \rvert =\tilde {r}$ . Hence, every linear combination of more than r elements of $\{x_a : a \in [n]\}$ has tensor rank greater than r.

Let $v \in \operatorname {\mathrm {span}} \{x_a : a \in [n]\}$ have tensor rank $\tilde {r} \leq r$ . Then ${v= \sum _{a \in Q} \alpha _a x_a}$ for some set $Q \subseteq [n]$ of size $\lvert Q \rvert =\tilde {r}$ and nonzero scalars $\{ \alpha _a : a \in Q\}$ . It follows from equation (11) and Theorem 2 that this is the unique tensor rank decomposition of v.

Corollary 16 immediately implies the following lower bound on $\operatorname {rank}[\sum _{a \in [n]} x_a]$ .

Corollary 17. If for every subset $S \subseteq [n]$ for which $2\leq \lvert S \rvert \leq n,$ it holds that

(12) $$ \begin{align} \lvert S \rvert+\min\{\lvert S \rvert,r\} \leq\sum_{j=1}^m (d_j^S-1)+1, \end{align} $$

then $\operatorname {rank}[\sum _{a \in [n]} x_a]\geq r+1$ .

In particular, Corollary 17 implies that

(13) $$ \begin{align} \operatorname{rank}\bigg[\sum_{a \in [n]} x_a\bigg]\geq \min\bigg\{n, \sum_{j=1}^m (k_j-1)+2-n\bigg\}. \end{align} $$

In Section 7, we prove that when the Kruskal ranks are sufficiently balanced, two of the k-ranks $k_i,k_j$ appearing in the bound (13) can be replaced with standard ranks $d_i,d_j$ (Theorem 28). Our Theorem 28 is independent of the bound in Corollary 17 (see Example 29).

We close this subsection by stating the $r=0$ and $r=1$ cases of Corollary 16, which generalize Proposition 3.1 and Theorem 3.2 in [Reference Ha and KyeHK15], respectively. We remark that the $m=2$ subcase of Corollary 18 was proven by Pierpaola Santarsiero in unpublished work, using a different proof technique.

Corollary 18 ( $s=1$ , $r=0$ case of Theorem 15).

If for every subset $S \subseteq [n]$ for which $2\leq \lvert S \rvert \leq n,$ it holds that

$$ \begin{align*} \lvert S \rvert \leq\sum_{j=1}^m (d_j^S-1)+1, \end{align*} $$

then $\{x_a : a \in [n]\}$ is linearly independent.

Corollary 19 ( $s=1$ , $r=1$ case of Theorem 15).

If for every subset $S \subseteq [n]$ for which $2\leq \lvert S \rvert \leq n,$ it holds that

$$ \begin{align*} \lvert S \rvert \leq\sum_{j=1}^m (d_j^S-1), \end{align*} $$

then

$$ \begin{align*} \operatorname{\mathrm{span}}\{x_a : a \in [n]\} \cap \mathrm{Prod}\left(\mathcal{V}_1 : \dots : \mathcal{V}_m \right)=\mathbb{C}^\times x_1 \sqcup \dots \sqcup \mathbb{C}^\times x_n. \end{align*} $$

6.1.2 $s=n-1$ case of Theorem 15

In this subsection, we state a slight adaptation of the $s=n-1$ case of Theorem 15, which gives sufficient conditions for a pair of decompositions to have an $(n-1,1)$ -subpartition. After stating this case, we observe that the subcase $r=1$ improves recent results in [Reference Ballico, Bernardi, Chiantini and GuardoBBCG18, Reference BallicoBal20] concerning circuits of product tensors. We then remark on applications of this special case in quantum information theory.

Corollary 20 ( $s=n-1$ case of Theorem 15).

Let $r \in \{0,1,\dots , n\}$ be an integer. If ${n+r \leq \sum _{j=1}^m (d_j-1)+1}$ , then for any nonnegative integer $\tilde {r} \leq r$ and multiset of product tensors $\{y_a : a \in [\tilde {r}]\}$ for which $\sum _{a \in [n]} x_a = \sum _{a \in [\tilde {r}]} y_a$ , the pair of decompositions $(\sum _{a \in [n]} x_a,\sum _{a \in [\tilde {r}]} y_a)$ has an $(n-1,1)$ -subpartition.

Moreover, if ${n+r \leq \sum _{j=1}^m (d_j-1)+1}$ , $\tilde {r}=\operatorname {rank}[\sum _{a \in [n]} x_a]$ , and $1\leq \tilde {r} \leq \min \{r,n-1\}$ , then there exists a subset $S \subseteq [n]$ with $\tilde {r} \leq \lvert S \rvert \leq n-1$ for which

$$ \begin{align*} \operatorname{rank}\bigg[\sum_{a \in S} x_a\bigg] < \tilde{r}. \end{align*} $$

Proof. The statement of the first paragraph is slightly different from the $s=n-1$ case of Theorem 15, and it follows easily from Corollary 10. To prove the statement of the second paragraph, let $\{z_a : a \in [\tilde {r}]\}\in \mathrm {Prod}\left (\mathcal {V}_1: \dots : \mathcal {V}_m \right )$ be any multiset of product tensors for which $\sum _{a \in [n]} x_a = \sum _{a \in [\tilde {r}]} z_a$ , and let $Q \subseteq [n]$ , $R \subseteq [\tilde {r}]$ be subsets for which

$$ \begin{align*} \max\{\lvert R \rvert,1\} \leq \lvert Q \rvert \leq n-1 \end{align*} $$

and $\sum _{a \in Q} x_a = \sum _{a \in R} z_a$ . If $\lvert R \rvert <\lvert Q \rvert $ and $\lvert Q \rvert \geq \tilde {r}$ , then we can take $S=Q$ . If $\lvert R \rvert <\lvert Q \rvert $ and $\lvert Q \rvert \leq \tilde {r}-1$ , then we can take $S\subseteq [n]$ to be any subset for which $S \supseteq Q$ and $\lvert S \rvert =\tilde {r}$ . It remains to consider the case $\lvert R \rvert =\lvert Q \rvert $ . In this case, it must hold that $\bigl \lvert [\tilde {r}]\setminus R \bigr \rvert < \bigl \lvert [n] \setminus Q \bigr \rvert ,$ so we can find S using the same arguments as in the case $\lvert R \rvert <\lvert Q \rvert $ .

A special case of the $r=1$ case of Corollary 20 gives an upper bound of $n-2$ on the number of subsystems $j\in [m]$ for which a circuit of product tensors can have $d_j>1$ . This bound improves those obtained in [Reference BallicoBal20, Theorem 1.1] and [Reference Ballico, Bernardi, Chiantini and GuardoBBCG18, Lemma 4.5], and is sharp (see Section 5).

Corollary 21. If $\{{x_a}: a \in [n] \}$ forms a circuit, then ${d_j>1}$ for at most $n-2$ indices $j \in [m]$ .

Proof. This follows immediately from Corollary 10 since circuits are connected. Alternatively, this follows from the second paragraph in the statement of Corollary 20 since for any circuit it holds that $\sum _{a \in S} x_a \neq 0$ for all $S \subseteq [n]$ with $1 \leq \lvert S \rvert \leq n-1$ .

As an immediate consequence of Corollary 21, a sum of two product tensors is again a product tensor if and only if $d_j>1$ for at most a single subsystem index $j \in [m]$ (see Corollary 15 in [Reference LovitzLov20]). This statement is well known. In particular, it was used in [Reference WestwickWes67, Reference JohnstonJoh11] to characterize the invertible linear operators that preserve the set of product tensors. In [Reference LovitzLov21, Reference LovitzLov20], the first author used this statement to study decomposable correlation matrices and observed that it directly provides an elementary proof of a recent result in quantum information theory [Reference Boyer, Liss and MorBLM17] (see Corollary 16 in [Reference LovitzLov20]).

6.2 Uniqueness results for nonrank decompositions

In this subsection, we prove uniqueness results for decompositions of $\sum _{a \in [n]} x_a$ into ${r\geq n+1}$ product tensors. Namely, we provide conditions on $\{x_a : a \in [n]\}$ for which whenever $\sum _{a \in [n]} x_a = \sum _{a \in [r]} y_a$ for some multiset of product tensors $\{y_a : a \in [r]\}$ , this pair of decompositions has an $(s,l)$ -subpartition. In particular, for $s=1$ we obtain sufficient conditions for the existence of subsets $Q \subseteq [n]$ , $R\subseteq [r]$ of size $\lvert Q \rvert =\lvert R \rvert =l$ for which $\{x_a : a\in Q\}=\{y_a : a \in R\}$ . We refer the reader also to Section 8, in which we prove uniqueness results on non-Waring rank decompositions of symmetric tensors and identify applications of our nonrank uniqueness results.

In Theorem 22, we give sufficient conditions for which whenever $(\sum _{a \in [n]} x_a, \sum _{a \in [r]} y_a)$ is irreducible, it has an $(s,l)$ -subpartition. We then observe that for $s=1$ we can drop the irreducibility assumption and obtain the result described in the previous paragraph. We then prove a modified version of Theorem 22, which drops the irreducibility assumption for arbitrary $s\in [n-1]$ . At the end of this subsection, we review these statements in the $s=n-1$ case.

Theorem 22. Let $n \geq 2$ , $q \in [n-1]$ , $s \in [q]$ and r be positive integers for which

(14) $$ \begin{align} n+1 \leq r \leq n+\Bigl\lceil \frac{n-q}{s} \Bigr\rceil, \end{align} $$

and let $l=\lfloor q/s \rfloor $ . If for every subset $S \subseteq [n]$ for which $s+1\leq \lvert S \rvert \leq n,$ it holds that

(15) $$ \begin{align} 2\lvert S \rvert+\max\left\{0,(r-n)-\biggl\lceil \frac{n-q+s}{\lvert S \rvert} \biggr\rceil+1\right\} \leq \sum_{j=1}^m (d_j^S -1)+1, \end{align} $$

then for any multiset of product tensors $\{y_a : a \in [r]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1 : \dots : \mathcal {V}_m \right )$ for which ${\sum _{a \in [n]} x_a= \sum _{a \in [r]} y_a}$ and $(\sum _{a \in [n]} x_a,\sum _{a \in [r]} y_a)$ is irreducible, this pair of decompositions has an $(s, l)$ -subpartition.

One may be concerned about whether the complicated collection of inequalities (15) can ever be satisfied. The answer is yes, simply because the right-hand side can depend on m, whereas the left-hand side does not. So for m large enough, one can always find $\{x_a : a \in [n]\}$ that satisfies these inequalities. In fact, they can even be satisfied nontrivially for $m=3$ , as we observe in Example 26.

Proof of Theorem 22.

For each $a \in [r]$ , let $x_{n+a}=-y_a$ , and let $T_1 \sqcup \dots \sqcup T_t=[n+r]$ be the index sets of the decomposition of $\{x_a : a \in [n+r]\}$ into connected components. Note that for each $p \in [t]$ , it must hold that

$$ \begin{align*} \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\geq \bigl\lvert T_p \cap [n] \bigr\rvert, \end{align*} $$

otherwise we would contradict irreducibility. For each $p \in [t]$ , if

$$ \begin{align*} \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert= \bigl\lvert T_p \cap [n] \bigr\rvert, \end{align*} $$

then $\bigl \lvert T_p \cap [n] \bigr \rvert \leq s$ , otherwise $\{ x_a : a \in T_p\}$ would split by equation (15) and Corollary 10. Assume without loss of generality that

$$ \begin{align*} \bigl\lvert T_1 \cap [n] \bigr\rvert-\bigl\lvert T_1 \cap [n+r]\setminus [n] \bigr\rvert &\geq \bigl\lvert T_2 \cap [n] \bigr\rvert-\bigl\lvert T_2 \cap [n+r]\setminus [n] \bigr\rvert\\ &\;\; \vdots \\ & \geq \bigl\lvert T_{{t}} \cap [n] \bigr\rvert-\bigl\lvert T_{{t}} \cap [n+r]\setminus [n] \bigr\rvert. \end{align*} $$

If

$$ \begin{align*} \bigl\lvert T_{1}\cap [n] \bigr\rvert = \bigl\lvert T_{1} \cap [n+r]\setminus [n] \bigr\rvert, \end{align*} $$

then let $\tilde {l}\in [t]$ be the largest integer for which

(16) $$ \begin{align} \bigl\lvert T_{\tilde{l}}\cap [n] \bigr\rvert = \bigl\lvert T_{\tilde{l}} \cap [n+r]\setminus [n] \bigr\rvert. \end{align} $$

Otherwise, let $\tilde {l}=0$ . Then for all $p \in [t]\setminus [\tilde {l}]$ it holds that

(17) $$ \begin{align} \lvert T_p \cap [n] \rvert < \lvert T_p \cap [n+r]\setminus [n] \rvert \end{align} $$

(recall that we define $[0]=\{\}$ ). To complete the proof, we will show that $\tilde {l} \geq l$ , for then we can take $Q_p=T_p \cap [n]$ and $R_p=T_p \cap [n+r]\setminus [n]$ for all $p \in [l]$ to conclude.

Suppose toward contradiction that $\tilde {l}< l$ . We require the following two claims.

Claim 23. It holds that $\tilde {l}<t$ , $\bigl \lceil \frac {n- s \tilde {l} }{t-\tilde {l}} \bigr \rceil \geq s+1$ , and there exists $p \in [t]\setminus [\tilde {l}]$ for which

(18) $$ \begin{align} \bigl\lvert T_p \cap [n] \bigr\rvert \geq \biggl\lceil \frac{n- s \tilde{l} }{t-\tilde{l}} \biggr\rceil. \end{align} $$

Claim 24. For all $p \in [t] \setminus [\tilde {l}]$ , it holds that

(19) $$ \begin{align} \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert \leq \bigl\lvert T_p \cap [n] \bigr\rvert +r-n+\tilde{l} -t+1 \end{align} $$

Before proving these claims, we first use them to complete the proof of the theorem. Let $p \in [t]\setminus [\tilde {l}]$ be as in Claim 23. Then,

$$ \begin{align*} \lvert T_p \rvert &= \bigl\lvert T_p \cap [n] \bigr\rvert+\bigl\lvert T_p \cap [n+r] \setminus [n] \bigr\rvert\\ &\leq 2 \bigl\lvert T_p \cap [n] \bigr\rvert+ r-n+\tilde{l} -t+1\\ &\leq 2 \bigl\lvert T_p \cap [n] \bigr\rvert+ r-n - \biggl\lceil \frac{n-s \tilde{l}}{\bigl\lvert T_p \cap [n] \bigr\rvert} \biggr\rceil+1\\ &\leq 2 \bigl\lvert T_p \cap [n] \bigr\rvert+ r-n- \biggl\lceil \frac{n-q+s}{\bigl\lvert T_p \cap [n] \bigr\rvert} \biggr\rceil+1\\ & \leq \sum_{j=1}^m (d_j^{T_p \cap [n]}-1)+1, \end{align*} $$

where the first line is obvious, the second follows from Claim 24, the third follows from Claim 23, the fourth follows from $\tilde {l}< l$ and the fifth follows from equation (15) and the fact that $\bigl \lvert T_p \cap [n] \bigr \rvert \geq s+1$ . So $\{x_a : a \in T_p\}$ splits, a contradiction. This completes the proof, modulo proving the claims.

Proof of Claim 23.

To prove the claim, we first observe that $n>st$ . Indeed, if $n\leq st$ , then

$$ \begin{align*} r & = \sum_{p=1}^t \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\\ &\geq n+t-\tilde{l} \\ &\geq n+\frac{n-q}{s}+1, \end{align*} $$

where the first line is obvious, the second follows from equations (16) and (17) and the third follows from $n\leq st$ and $\tilde {l}<l$ . This contradicts equation (14), so it must hold that $n>st$ .

Note that $\tilde {l}<t$ , for otherwise we would have $n \leq st$ by the fact that $\bigl \lvert T_p \cap [n] \bigr \rvert \leq s$ for all $p \in [\tilde {l}]$ . To verify that $\bigl \lceil \frac {n- s \tilde {l} }{t-\tilde {l}} \bigr \rceil \geq s+1$ , it suffices to prove $\frac {n- s \tilde {l} }{t-\tilde {l}}> s,$ which follows from $n>st$ . To verify equation (39), since $\bigl \lvert T_p \cap [n] \bigr \rvert \leq s$ for all $p \in [\tilde {l}]$ , by the pigeonhole principle there exists $p \in [t]\setminus [\tilde {l}]$ for which

$$ \begin{align*} \bigl\lvert T_p \cap [n] \bigr\rvert \geq \biggl\lceil \frac{n- s \tilde{l} }{t-\tilde{l}} \biggr\rceil. \end{align*} $$

This proves the claim.

Proof of Claim 24.

Suppose toward contradiction that the inequality (19) does not hold for some $\tilde {p}\in [t] \setminus [\tilde {l}]$ . Then

$$ \begin{align*} r &= \sum_{p=1}^t \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\\ & \geq \sum_{p\neq \tilde{p}} \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert + \bigl\lvert T_{\tilde{p}} \cap[n] \bigr\rvert + (r-n)+\tilde{l}-t+2\\ &\geq r+1, \end{align*} $$

where the first two lines are obvious, and the last line follows from equations (16) and (17), a contradiction. $\triangle $

The proofs of Claims 23 and 24 complete the proof of the theorem.

6.2.1 $s=1$ case of Theorem 22

In the $s=1$ case of Theorem 22, we can drop the assumption that the pair of decompositions is irreducible. This is because the other assumptions already imply that $\sum _{a \in [n]} x_a$ constitutes a (unique) tensor rank decomposition by Theorem 2, so $\sum _{a \in [n]} x_a=\sum _{a \in [r]} y_a$ will automatically be irreducible (see the discussion at the beginning of Section 6).

Corollary 25 ( $s=1$ case of Theorem 22).

Let $q \in [n-1]$ and r be positive integers for which $n+1 \leq r \leq 2n-q$ . If for every subset $S \subseteq [n]$ with $2\leq \lvert S \rvert \leq n$ it holds that

(20) $$ \begin{align} 2\lvert S \rvert+\max\left\{0,(r-n)-\biggl\lceil \frac{n-q+1}{\lvert S \rvert} \biggr\rceil+1\right\} \leq \sum_{j=1}^m (d_j^S -1)+1, \end{align} $$

then for any multiset of product tensors $\{y_a : a \in [r]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1 : \dots : \mathcal {V}_m \right )$ for which $\sum _{a \in [n]} x_a= \sum _{a \in [r]} y_a$ , there exist subsets $Q \subseteq [n]$ and $R \subseteq [r]$ of size $\lvert Q \rvert =\lvert R \rvert =q$ for which ${\{x_a : a \in Q\}=\{y_a : a \in R\}}$ (in other words, this pair of decompositions has a $(1,q)$ -subpartition).

It is worth noting that although the assumptions of Corollary 25 require $\sum _{a \in [n]} x_a$ to constitute a unique tensor rank decomposition, this result can also be applied to arbitrary decompositions $\sum _{a \in [n]} x_a$ , provided that $\sum _{a\in S} x_a$ constitutes a unique tensor rank decomposition for some subset $S\subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ , as one can simply apply Corollary 25 to the pair of decompositions $(\sum _{a \in S} x_a , \sum _{a\in [r]} y_a- \sum _{a\in [n]\setminus S} x_a)$ . It is not difficult to produce explicit examples in which Corollary 25 can be applied in this way (for instance, by modifying Example 26).

As an example, we now use Corollary 25 to prove uniqueness of nonrank decompositions of the identity tensor $\sum _{a \in [n]} e_a^{\otimes 3}$ .

Example 26. Let $n \geq 2$ , $q \in [n-1],$ and r be positive integers for which $n+1 \leq r \leq 2n-q$ and

$$ \begin{align*} q \leq n+1-\frac{1}{4}\left((r-n+2)^2+1\right). \end{align*} $$

If

$$ \begin{align*} \sum_{a \in [n]} e_a ^{\otimes 3}=\sum_{a \in [r]} y_a \end{align*} $$

for some multiset of product tensors $\{y_a : a \in [r]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1 : \mathcal {V}_2 : \mathcal {V}_3 \right )$ , then there exist subsets $Q \subseteq [n]$ and $R \subseteq [n+r]$ of sizes $\lvert Q \rvert =\lvert R \rvert =q$ such that ${\{x_a : a \in Q\} = {\{y_a : a \in R\}}}$ . For example, if $r=n+1$ , then we can take $q= n-2$ for any $n \geq 3$ .

To verify Example 26, it suffices to show that the inequality (20) holds for all $S \subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ . This reduces to proving that

$$ \begin{align*} \lvert S \rvert(r-n+2-\lvert S \rvert) - (n-q+1)<0, \end{align*} $$

which occurs whenever the polynomial in $\lvert S \rvert $ on the lefthand side has no real roots, that is, whenever

$$ \begin{align*} (r-n+2)^2 \leq 4(n-q+1)-1. \end{align*} $$

6.2.2 Modifying Theorem 22 to apply to reducible pairs of decompositions

A drawback to Theorem 22 is that it only applies to irreducible pairs of decompositions. We now present a modification of this result, which can certify the existence of an $(s,l)$ -subpartition even for reducible decompositions, at the cost of stricter assumptions. We defer this proof to the appendix, as it is very similar to that of Theorem 22.

Theorem 27. Let $q \in [n-1]$ , $s \in [q]$ and r be positive integers for which

$$ \begin{align*} n+1 \leq r \leq \biggl\lceil \left(\frac{s+1}{s}\right)(n-q+s) \biggr\rceil-1, \end{align*} $$

and let $l=\lfloor q/s \rfloor $ . If for every subset $S \subseteq [n]$ for which $s+1\leq \lvert S \rvert \leq n,$ it holds that

$$ \begin{align*} 2\lvert S \rvert+\max\left\{0,(r-n+q-s)-\biggl\lceil \frac{n-q+s}{\lvert S \rvert} \biggr\rceil+1\right\} \leq \sum_{j=1}^m (d_j^S -1)+1, \end{align*} $$

then for any multiset of product tensors $\{y_a : a \in [r]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1 : \dots : \mathcal {V}_m \right )$ for which $\sum _{a \in [n]} x_a= \sum _{a \in [r]} y_a$ , this pair of decompositions has an $(s,l)$ -subpartition.

6.2.3 $s=n-1$ case of Theorem 27

When $s=n-1$ , then it necessarily holds that $r=n+1$ and $q=n-1$ and Theorem 27 simply says that if $2n+1 \leq \sum _{j=1}^m (d_j-1)+1$ , then $\sum _{a \in [n]} x_a= \sum _{a \in [n+1]} y_a$ has an $(n-1,1)$ -subpartition. Theorem 22 yields a weaker statement.

7 A lower bound on tensor rank

In Section 6.1.1, we saw that for a multiset of product tensors $\{x_a : a \in [n]\}$ with k-ranks $k_j=\operatorname {k-rank}(x_{a,j} : a \in [n])$ , it holds that

(21) $$ \begin{align} \operatorname{rank}\bigg[\sum_{a \in [n]} x_a\bigg]\geq \min\bigg\{n, \sum_{j=1}^m (k_j-1)+2-n\bigg\}. \end{align} $$

In this section, we prove that when the k-ranks are sufficiently balanced, two of the k-ranks $k_i,k_j$ appearing in this bound can be replaced with standard ranks $d_i,d_j$ , which improves this bound when the k-ranks and ranks are not equal and specializes to Sylvester’s matrix rank inequality when $m=2$ . We prove that this improved bound is independent of a different lower bound on tensor rank that we observed in Corollary 17. We furthermore observe that this improved bound is sharp in a wide parameter regime. As a consequence, we obtain a lower bound on Waring rank, which we also prove is sharp.

Theorem 28 (Tensor rank lower bound).

Let $n \geq 2$ and $m\geq 2$ be integers, let ${\mathcal {V}=\mathcal {V}_1\otimes \dots \otimes \mathcal {V}_m}$ be a vector space over a field $\mathbb {F}$ and let

$$ \begin{align*} {E=\{x_a: a \in[n] \} \subseteq \mathrm{Prod}\left(\mathcal{V}_1:\dots : \mathcal{V}_m \right)} \end{align*} $$

be a multiset of product tensors. For each index $j \in [m]$ , let $k_j=\operatorname {k-rank}(x_{a,j}: a \in [n])$ and ${d_j=\dim \operatorname {\mathrm {span}}\{x_{a,j} : a \in [n]\}}$ . Define

(22) $$ \begin{align} \mu=\max_{\substack{i, j \in [m]\\i \neq j}}\{d_i-k_i+d_j-k_j\}. \end{align} $$

If for every index $i \in [m]$ it holds that

(23) $$ \begin{align} k_i \leq \sum_{\substack{j \in [m]\\ j \neq i}} (k_j-1)+1, \end{align} $$

then

(24) $$ \begin{align} \operatorname{rank}\bigg[\sum_{a\in [n]} x_a\bigg] \geq \min\bigg\{n,\mu+\sum_{j=1}^m (k_j-1)+2-n\bigg\}. \end{align} $$

Intuitively, the condition (23) ensures that the k-ranks are sufficiently balanced. This inequality is satisfied, for example, when the product tensors are symmetric. While we are unaware whether the precise inequality (23) is necessary for the lower bound (24) to hold, the following example illustrates that some inequality of this form must hold.

Example 29. The set of product tensors

$$ \begin{align*} E=\{e_1^{\otimes 3}, e_2^{\otimes 3}, e_3^{\otimes 3}, e_4^{\otimes 3}, e_5 \otimes (e_1+e_2)^{\otimes 2}, e_6\otimes (e_1-e_2)^{\otimes 2}\} \end{align*} $$

does not satisfy equation (24). Indeed,

$$ \begin{align*} \operatorname{rank}[\Sigma(E)]&=5\\ &< q+k_1+k_2+k_3-1-n\\ &=d_2+d_3-1\\ &=7. \end{align*} $$

This example illustrates that in order for the bound (24) to hold, the k-ranks must be sufficiently ‘balanced’ in order to avoid cases such as this. In particular, some inequality resembling equation (23) is necessary. We remark that this example can be extended to further parameter regimes using Derksen’s example [Reference DerksenDer13], and similar arguments as in Sections 7.1 and 8.1.

Note that when $m=2$ , Theorem 28 states that

$$ \begin{align*} \operatorname{rank}\bigg[\sum_{a \in [n]} x_a\bigg] \geq d_1+d_2-n, \end{align*} $$

provided that $k_1=k_2$ . This is Sylvester’s matrix rank inequality (although Sylvester’s result holds also when $k_1 \neq k_2$ ) [Reference Horn and JohnsonHJ13].

The following example demonstrates that our two lower bounds on tensor rank in Theorem 28 and Corollary 17 are independent.

Example 30. By Theorem 28, the sum of the set of product tensors

$$ \begin{align*} \{e_1^{\otimes 3},e_2^{\otimes 3}, (e_1+e_2)^{\otimes 2}\otimes e_3,e_3^{\otimes 2} \otimes (e_1+e_2+e_3)\} \end{align*} $$

has tensor rank $4$ . Note that this bound cannot be achieved with the flattening rank lower bound, nor with Corollary 17, as the first three vectors do not satisfy equation (12). Many more such examples can be obtained using the construction in Section 7.1.

Conversely, the sum of the set of product tensors

$$ \begin{align*}\{& e_1^{\otimes 3}, e_2^{\otimes 3}, e_3^{\otimes 3}, e_4^{\otimes 3},(e_2+e_3)\otimes (e_2+e_4)\otimes (e_1+e_4)\} \end{align*} $$

has tensor rank 5 by Corollary 17, while Theorem 28 only certifies that this sum has tensor rank at least $4$ .

Now, we prove Theorem 28.

Proof of Theorem 28.

Let $r =\operatorname {rank}[\sum _{a \in [n]} x_a]$ , and let $\{y_a : a \in [r]\}\subseteq \mathrm {Prod}\left (\mathcal {V}_1: \dots : \mathcal {V}_m \right )$ be a multiset of product tensors for which $\sum _{a \in [n]} x_a = \sum _{a \in [r]} y_a$ is a tensor rank decomposition. We need to prove that r satisfies the inequality (24). For each $a \in [r]$ , let $x_{n+a}=-y_a$ , and let ${T_1\sqcup \dots \sqcup T_t =[n+r]}$ be the index sets of the connected components of ${\{x_a : a \in [n+r]\}}$ , so that $\sum _{a \in T_j} x_a=0$ for all $j\in [t]$ . For each subset ${S \subseteq [n]}$ and index $j \in [m]$ , let

$$ \begin{align*} d_j^S=\dim\operatorname{\mathrm{span}}\{x_{a,j} : a \in S\}. \end{align*} $$

We first consider the case $t=1$ , that is, $\{x_a : a \in [n+r]\}$ is connected. By the splitting theorem, it holds that

$$ \begin{align*} n+r &\geq \sum_{j=1}^m (d_j-1)+2\\ & \geq \mu+\sum_{j=1}^m (k_j-1)+2, \end{align*} $$

completing the proof in this case.

We proceed by induction on t. Suppose the theorem holds whenever the number of connected components is less than t. Assume without loss of generality that

$$ \begin{align*} \bigl\lvert T_1 \cap [n] \bigr\rvert-\bigl\lvert T_1 \cap [n+r]\setminus [n] \bigr\rvert &\geq \bigl\lvert T_2 \cap [n] \bigr\rvert-\bigl\lvert T_2 \cap [n+r]\setminus [n] \bigr\rvert\\ &\;\; \vdots \\ & \geq \bigl\lvert T_{{t}} \cap [n] \bigr\rvert-\bigl\lvert T_{{t}} \cap [n+r]\setminus [n] \bigr\rvert\\ &\geq 0, \end{align*} $$

where the last line follows from $\sum _{a \in T_t} x_a=0$ and the fact that $\sum _{a \in [r]} y_a$ is a tensor rank decomposition (so in particular, $\sum _{a \in T_t \cap [n+r]\setminus [n]} x_a$ is a tensor rank decomposition). If

$$ \begin{align*} \bigl\lvert T_1 \cap [n] \bigr\rvert = \bigl\lvert T_1 \cap [n+r]\setminus [n] \bigr\rvert, \end{align*} $$

then $r=n$ and we are done. Otherwise,

$$ \begin{align*} \bigl\lvert T_{[t-1]} \cap [n] \bigr\rvert> \bigl\lvert T_{[t-1]} \cap [n+r]\setminus [n] \bigr\rvert, \end{align*} $$

where $T_{[t-1]}=T_1 \sqcup \dots \sqcup T_{t-1}$ .

Observe that $k_j < \bigl \lvert T_{[t-1]}\cap [n] \bigr \rvert $ for all $j \in [m]$ . Indeed, since

$$ \begin{align*} \operatorname{rank}\bigg[\sum_{a \in T_{[t-1]} \cap [n]} x_a\bigg] < \bigl\lvert T_{[t-1]} \cap [n] \bigr\rvert, \end{align*} $$

it must hold that

$$ \begin{align*} 2 \bigl\lvert T_{[t-1]} \cap [n] \bigr\rvert-1 \geq \sum_{j=1}^m \bigg(\min\big\{\bigl\lvert T_{[t-1]}\cap [n] \bigr\rvert,k_j\big\}-1\bigg)+2, \end{align*} $$

by equation (21). If $k_i \geq \bigl \lvert T_{{[t-1]}}\cap [n] \bigr \rvert $ for some $i \in [m]$ , then this inequality implies that ${k_j < \bigl \lvert T_{{[t-1]}}\cap [n] \bigr \rvert }$ for all $j \neq i$ , and hence

$$ \begin{align*} k_i \geq \bigl\lvert T_{{[t-1]}}\cap [n] \bigr\rvert \geq \sum_{\substack{j \in [m]\\ j \neq i}} (k_j-1)+2, \end{align*} $$

contradicting equation (23). So $k_j < \bigl \lvert T_{{[t-1]}}\cap [n] \bigr \rvert $ for all $j \in [m]$ .

Since $k_j< \bigl \lvert T_{{[t-1]}}\cap [n] \bigr \rvert $ for all $j \in [m]$ , the k-ranks of $\{x_a : a \in T_{[t-1]}\cap [n]\}$ satisfy equation (23), so by the induction hypothesis,

(25) $$ \begin{align} \bigl\lvert T_{[t-1]} \bigr\rvert \geq \mu^{T_{[t-1]}\cap[n]}+\sum_{j=1}^m(k_j-1)+2, \end{align} $$

where

$$ \begin{align*} \mu^{T_{[t-1]}\cap[n]}=\max_{\substack{i,j \in [m]\\i\neq j}}\bigg\{d_i^{T_{[t-1]}\cap[n]}-k_i+d_j^{T_{[t-1]}\cap[n]}-k_j\bigg\}. \end{align*} $$

To complete the proof, we will show that

$$ \begin{align*} \bigl\lvert T_{[t-1]} \bigr\rvert+\lvert T_t \rvert \geq \mu+\sum_{j=1}^m (k_j-1)+2. \end{align*} $$

Let $i,i' \in [m]$ be such that $\mu =d_i-k_i+d_{i'}-k_{i'}$ . Then

$$ \begin{align*} \bigl\lvert T_{[t-1]} \bigr\rvert+\lvert T_t \rvert&\geq d_i^{T_{[t-1]}\cap[n]}-k_i+d_{i'}^{T_{[t-1]}\cap[n]}-k_{i'}+\sum_{j=1}^m(k_j-1)+\sum_{j=1}^m(d_j^{T_t\cap [n]}-1)+4\\ &\geq d_i^{T_{[t-1]}\cap[n]}-k_i+d_{i'}^{T_{[t-1]}\cap[n]}-k_{i'}+\sum_{j=1}^m(k_j-1)+d_{i}^{T_t\cap [n]}+d_{i'}^{T_t\cap [n]}+2\\ &\geq d_i-k_i+d_{i'}-k_{i'}+\sum_{j=1}^m(k_j-1)+2\\ &= \mu+\sum_{j=1}^m(k_j-1)+2, \end{align*} $$

where the first line follows from equation (25) and the fact that $\{x_a : a \in T_t\}$ is connected, the second is obvious, the third is easy to verify (in matroid-theoretic terms, this is submodularity of the rank function), and the fourth is by definition. This completes the proof.

As an immediate corollary to Theorem 28, we obtain the following lower bound on the Waring rank of a symmetric tensor in terms of a known symmetric decomposition.

Corollary 31 (Waring rank lower bound).

Let $n\geq 2$ , and $m \geq 2$ be integers, let $\mathcal {W}$ be a vector space over a field $\mathbb {F}$ with ${\mathrm {Char}}(\mathbb {F})=0$ or ${\mathrm {Char}}(\mathbb {F})>m$ and let ${\{v_a : a \in [n]\} \subseteq \mathcal {W}\setminus \{0\}}$ be a multiset of nonzero vectors. Let

$$ \begin{align*}k={\operatorname{k-rank}(v_a : a \in [n])} \end{align*} $$

and

$$ \begin{align*} {d=\dim\operatorname{\mathrm{span}}\{v_a : a \in [n]\}}. \end{align*} $$

Then for any multiset of nonzero scalars

$$ \begin{align*} \{\alpha_a : a \in [n]\} \subseteq \mathbb{F}^\times, \end{align*} $$

it holds that

(26) $$ \begin{align} \mathrm{WaringRank}\bigg[ \sum_{a \in [n]} \alpha_a v_a^{\otimes m}\bigg] \geq \min\{n,2d+(m-2)(k-1)-n\}. \end{align} $$

7.1 Our tensor rank lower bound is sharp

In this subsection, we observe that, in a wide parameter regime, the inequalities (24) and (26) appearing in Theorem 28 and Corollary 31 cannot be improved.

Let $\mathbb {F}$ be a field with $\mathrm {Char}(\mathbb {F})=0$ , let $n \geq 2$ , ${m \geq 2}$ ,

$$ \begin{align*} 2 \leq d_1,\dots, d_m \leq n, \end{align*} $$

and

$$ \begin{align*} {k_1\leq d_1,\dots, k_m \leq d_m} \end{align*} $$

be positive integers and let

$$ \begin{align*} \lambda=\sum_{j=1}^m (k_j-1)+2. \end{align*} $$

Suppose that the following conditions hold:

1. 1. ${\mu =2(d_{i}-k_i)}$ for some index $i \in [m]$ , where $\mu $ is defined as in equation (22).

2. 2. $\max \{k_j: j \in [m]\}+d_i-k_i+1 \leq n \leq d_i-k_i+\lambda $ .

3. 3. The inequality (23) is satisfied.

Then there exists a multiset of product tensors E corresponding to these choices of parameters that satisfies equation (24) with equality. Indeed, the bound $\operatorname {rank}[\Sigma (E)] \geq n$ is trivial to attain with equality, and the bound

(27) $$ \begin{align} \operatorname{rank}[\Sigma(E)] \geq 2(d_i-k_i)+\lambda-n \end{align} $$

can be attained with equality as follows. Let

$$ \begin{align*} \{x_a : a \in [\lambda]\} \subseteq \mathrm{Prod}\left(\mathbb{F}^{d_1}: \dots : \mathbb{F}^{d_m} \right) \end{align*} $$

be a multiset of product tensors that forms a circuit and satisfies

(28) $$ \begin{align} \dim\operatorname{\mathrm{span}}\{x_{a,j}: a \in [\lambda]\}=\operatorname{k-rank}(x_{a,j} : a \in [\lambda])=k_j \end{align} $$

for all $j \in [m]$ . An example of such a circuit is presented in [Reference DerksenDer13] and reviewed in Section 5. Now, let

$$ \begin{align*} \{x_{a} : a\in [\lambda+d_i-k_i]\setminus [\lambda]\}\subseteq \mathrm{Prod}\left(\mathbb{F}^{d_1}: \dots : \mathbb{F}^{d_m} \right) \end{align*} $$

be any multiset of product tensors for which

(29) $$ \begin{align} \dim\operatorname{\mathrm{span}}\{x_{a,j} : a \in [\lambda+d_i-k_i]\}=d_j \end{align} $$

and

$$ \begin{align*} \operatorname{k-rank}(x_{a,j} : a\in [\lambda+d_i-k_i]) = k_j \end{align*} $$

for all $j \in [m]$ , which is guaranteed to exist since $\mathbb {F}$ is infinite. Let

$$ \begin{align*} E=\{x_a : a \in [n-d_i+k_i]\} \sqcup \{x_a : a \in [\lambda+d_i-k_i]\setminus [\lambda]\} \end{align*} $$

and

$$ \begin{align*} F=\{x_a : a \in [\lambda] \setminus [n-d_i+k_i]\} \sqcup \{x_a : a \in [\lambda+d_i-k_i]\setminus [\lambda]\}. \end{align*} $$

Recall that $n \leq d_i-k_i+\lambda $ by assumption, so the set $[\lambda ] \setminus [n-d_i+k_i]$ that appears in the definition of F is well defined. Since $n -d_i+k_i \geq k_j+1$ for all $j \in [m]$ , E has k-ranks $k_1,\dots , k_m,$ as desired. It is also clear that E has ranks $d_1,\dots ,d_m$ , by equations (28) and (29). Since $\{x_a: a \in [\lambda ]\}$ forms a circuit, some nonzero linear combination of E is equal to a nonzero linear combination of F. Since $\lvert F \rvert $ is equal to the right-hand side of equation (27), this completes the proof.

Out of the three conditions required for our construction, $\mu =2(d_i-k_i)$ seems the most restrictive. Unfortunately, our methods appear to require this condition. A nearly identical construction shows that the inequality (26) appearing in Corollary 31 cannot be improved (and our restrictive condition on $\mu $ is automatically satisfied in this case). The only difference in the construction is to choose the product tensors $\{x_a : a \in [\lambda +d_i-k_i]\}$ to be symmetric in this case, which can always be done (in particular, the product tensors appearing in Derksen’s example can be taken to be symmetric).

8 A uniqueness result for non-Waring rank decompositions

In this section, we prove a sufficient condition on a symmetric decomposition

$$ \begin{align*} v=\sum_{a \in [n]} \alpha_a v_a^{\otimes m} \end{align*} $$

under which any distinct decomposition $v=\sum _{a \in [r]} \beta _a u_a^{\otimes m}$ must have r lower bounded by some quantity, which we call $r_{\min }$ for now. When $r_{\min } \leq n$ , this yields a lower bound on $\mathrm {WaringRank}(v)$ that is contained in Corollary 31. When ${r_{\min } =n+1}$ , this yields a uniqueness criterion for symmetric decompositions that is contained in Theorem 2 but improves Kruskal’s theorem in a wide parameter regime. The main result in this section is the case ${r_{\min }> n+1}$ , where we obtain an even stronger statement than uniqueness: Every symmetric decomposition of v into less than $r_{\min }$ terms must be equal to $\sum _{a \in [n]} \alpha _a v_a^{\otimes m}$ (in the language introduced in Section 2, $\sum _{a \in [n]} \alpha _a v_a^{\otimes m}$ is the unique symmetric decomposition of v into less than $r_{\min }$ terms). In Section 8.1, we prove that our bound $r_{\min }$ cannot be improved. In Section 8.2, we identify potential applications of our nonrank uniqueness results.

Our results in this section were inspired by, and generalize, Theorem 6.8 and Remark 6.14 in [Reference ChiantiniChi19]. Our results in this section should be compared with those of Section 6.2 on uniqueness of nonrank decompositions of tensors that are not necessarily symmetric.

Theorem 32. Let $n\geq 2$ and $m \geq 2$ be integers, let $\mathcal {W}$ be a vector space over a field $\mathbb {F}$ with ${\mathrm {Char}}(\mathbb {F})=0$ or ${\mathrm {Char}}(\mathbb {F})>m$ , let $E={\{v_a : a \in [n]\} \subseteq \mathcal {W}\setminus \{0\}}$ be a multiset of nonzero vectors with ${\operatorname {k-rank}(v_a : a \in [n]) \geq 2}$ and let

$$ \begin{align*} {d=\dim\operatorname{\mathrm{span}}\{v_a : a \in [n]\}}. \end{align*} $$

Then for any nonnegative integer $r \geq 0$ , multiset of nonzero vectors ${F={\{u_a : a \in [r]\} \subseteq \mathcal {W}\setminus \{0\}}}$ with ${\operatorname {k-rank}(u_a : a \in [r]) \geq \min \{2,r\}}$ , and multisets of nonzero scalars

$$ \begin{align*} \{\alpha_a : a \in [n]\}, \{\beta_a: a \in [r]\} \subseteq \mathbb{F}^\times \end{align*} $$

for which

(30) $$ \begin{align} \{\alpha_a v_a^{\otimes m} : a \in [n]\} \neq \{\beta_a u_a^{\otimes m} : a \in [r]\} \end{align} $$

and

(31) $$ \begin{align} \sum_{a \in [n]} \alpha_a v_a^{\otimes m} = \sum_{a \in [r]} \beta_a u_a^{\otimes m}, \end{align} $$

it holds that

(32) $$ \begin{align} n+ r \geq m+2d-2. \end{align} $$

In the language of the introduction to this section, $r_{\min } = m+2d-2-n$ . For comparison, the result we have referred to in [Reference ChiantiniChi19] asserts that, under the condition $n \leq m$ , it holds that $n+r \geq m+d$ , which is weaker than our bound (32).

Proof of Theorem 32.

By subtracting terms from both sides of equation (31), and combining parallel product tensors into single terms (or to zero), it is clear that it suffices to prove the statement when E is linearly independent (so $d=n$ ).

Note that $r \geq n$ by Kruskal’s theorem. For each $a \in [r]$ , let $v_{n+a}=u_a$ , and let ${T_1\sqcup \dots \sqcup T_t =[n+r]}$ be the index sets of the connected components of ${\{v_a^{\otimes m} : a \in [n+r]\}}$ . Assume without loss of generality that $\bigl \lvert T_1 \cap [n] \bigr \rvert \geq \dots \geq \bigl \lvert T_t \cap [n] \bigr \rvert $ , and let $\tilde {t}\in [t]$ be the largest integer for which $\bigl \lvert T_{\tilde {t}} \cap [n] \bigr \rvert \geq 1$ . By equation (30), there must exist $\tilde {p} \in [\tilde {t}]$ for which $\lvert T_{\tilde {p}} \rvert \geq 3$ . Note that

$$ \begin{align*} \dim\operatorname{\mathrm{span}}\{v_a : a \in T_{\tilde{p}}\} \geq \max\left\{2, \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right\}. \end{align*} $$

Since $\{v_a^{\otimes m} : a \in T_{\tilde {p}}\}$ is connected, it follows from our splitting theorem that

(33) $$ \begin{align} \lvert T_{\tilde{p}} \rvert \geq m(\max\left\{2, \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right\}-1)+2. \end{align} $$

Now,

$$ \begin{align*} n+r &\geq \sum_{p \in [\tilde{t}]} \lvert T_p \rvert\\ & \geq \sum_{p \neq \tilde{p}} \big[m\left(\bigl\lvert T_p\cap [n] \bigr\rvert -1\right)+2\big]+ m\left(\max\left\{2, \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right\}-1\right)+2\\ & = m\left(n-\lvert T_{\tilde{p}}\cap [n] \rvert\right)-(m-2)\left(\tilde{t}-1\right)+m\left(\max\left\{2, \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right\}-1\right)+2\\ & \geq m\left(n-\lvert T_{\tilde{p}}\cap [n] \rvert\right)-(m-2)\left(n- \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right)+m\left(\max\left\{2, \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right\}-1\right)+2\\ &=2n-2\bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert+m\left(\max\left\{2, \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert\right\}-1\right)+2\\ &\geq 2n+m-2. \end{align*} $$

The first line is obvious, the second follows from equation (33) and the fact that every multiset $\{v_a^{\otimes m} : a \in T_p\}$ is connected, the third is algebra, the fourth uses the fact that $\bigl \lvert T_p \cap [n] \bigr \rvert \geq 1$ for all $p \in [\tilde {t}]$ and the rest is algebra. This completes the proof.

Theorem 32 immediately implies the following uniqueness result for non-Waring rank decompositions.

Corollary 33 (Uniqueness result for non-Waring rank decompositions).

Let $n\geq 2$ and $m \geq 2$ be integers, let $\mathcal {W}$ be a vector space over a field $\mathbb {F}$ with ${\mathrm {Char}}(\mathbb {F})=0$ or $\mathrm {Char}(\mathbb {F})>m$ , let ${\{v_a : a \in [n]\} \subseteq \mathcal {W}\setminus \{0\}}$ be a multiset of nonzero vectors with ${\operatorname {k-rank}(v_a : a \in [n]) \geq 2}$ , let $\{\alpha _a : a \in [n]\} \subseteq \mathbb {F}^\times $ be a multiset of nonzero scalars and let ${d=\dim \operatorname {\mathrm {span}}\{v_a : a \in [n]\}}$ . If

$$ \begin{align*} 2n+1 \leq m+2d-2, \end{align*} $$

then $\sum _{a \in [n]} \alpha _a v_a^{\otimes m}$ constitutes a unique Waring rank decomposition. More generally, if

$$ \begin{align*} n +r +1\leq m+2d-2, \end{align*} $$

for some $r \geq n$ , then $\sum _{a \in [n]} \alpha _a v_a^{\otimes m}$ is the unique symmetric decomposition of this tensor into at most r terms.

Note that the $r=n$ case of Corollary 33 improves Kruskal’s theorem for symmetric decompositions as soon as $2d> m(k-2)+4$ , where $k=\operatorname {k-rank}(v_a : a \in [n])$ . This case of Corollary 33 is in fact contained in our generalization of Kruskal’s theorem (Theorem 2), since for every subset $S\subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ , it holds that

$$ \begin{align*} 2 \lvert S \rvert&=2n-2\bigl\lvert [n]\setminus S \bigr\rvert\\ &\leq m+2d-2\bigl\lvert [n]\setminus S \bigr\rvert-3\\ &\leq m+2d^{S}-3\\ &\leq m(d^S-1)+1, \end{align*} $$

where $d^S=\dim \operatorname {\mathrm {span}}\{v_a : a \in S\}$ . This demonstrates that our generalization of Kruskal’s theorem is stronger than Kruskal’s theorem, even for symmetric tensor decompositions.

Our main result in this section is the $r>n$ case of Corollary 33, which yields uniqueness results for non-Waring rank decompositions of $\sum _{a \in [n]} \alpha _a v_a^{\otimes n}$ . The following example illustrates this case in practice.

It is natural to ask if Corollary 33 can be improved under further restrictions on $\operatorname {k-rank}(v_a : a \in [n])$ . At the end of Section 8.1, we prove that this cannot be done, at least in a particular parameter regime.

8.1 The inequality appearing in our uniqueness result is sharp

In this subsection, we prove that the inequality (32) that appears in Theorem 32 cannot be improved, by constructing explicit multisets of symmetric product tensors that satisfy this bound with equality.

Let $\mathbb {F}$ be a field with ${\mathrm {Char}}(\mathbb {F})=0$ . We will prove that, for any choice of positive integers $m \geq 2$ , $d \geq 2$ , $r \geq d-2$ , and $n\geq d$ for which $n+r=m+2d-2$ , there exist multisets of nonzero vectors E and F that satisfy the assumptions of Theorem 32. Note that the inequality ${r \geq d-2}$ automatically holds when $r \geq n$ , so this assumption does not restrict the parameter regime in which the inequality appearing in our uniqueness result (Corollary 33) is sharp as a consequence.

We first consider the case $d=2$ . Let $\{v_a^{\otimes m} : a \in [m+2]\} \subseteq \mathrm {Prod}\left (\mathbb {F}^{2} : \dots : \mathbb {F}^2 \right )$ be a circuit of symmetric product tensors for which

$$ \begin{align*} \operatorname{k-rank}(v_a: a \in [m+2])= 2. \end{align*} $$

An example of such a circuit is given in [Reference DerksenDer13] and reviewed in Section 5. So there exist nonzero scalars $\{\alpha _a : a \in [m+2]\}\subseteq \mathbb {F}^\times $ for which $\sum _{a \in [m+2]} \alpha _a v_a^{\otimes m}=0$ , and we can take the multisets $E=\{v_a : a \in [n]\}$ and $F=\{v_a : a \in [m+2]\setminus [n]\}$ to conclude.

For $d \geq 3$ , let $\{v_a^{\otimes m} : a \in [m+2]\} \subseteq \mathrm {Prod}\left (\mathbb {F}^{d} : \dots : \mathbb {F}^d \right )$ be the same multiset of symmetric product tensors as above, embedded in a larger space. Let

$$ \begin{align*} \{v_a: a \in [d+m]\setminus [m+2]\} \subseteq \mathbb{F}^d\setminus \{0\} \end{align*} $$

be any multiset of nonzero vectors for which

$$ \begin{align*} \dim\operatorname{\mathrm{span}}\{v_a : a \in [d+m]\}=d \end{align*} $$

and

$$ \begin{align*} \operatorname{k-rank}\{v_a : a \in [d+m]\}\geq 2, \end{align*} $$

which is guaranteed to exist since $\mathbb {F}$ is infinite. Since $r \geq d-2$ , we can take the multisets

$$ \begin{align*} E=\{v_a: a \in [n-d+2]\}\sqcup\{v_a: a \in [d+m]\setminus [m+2]\} \end{align*} $$

and

$$ \begin{align*} F=\{v_a : a \in [m+2]\setminus[n-d+2]\} \sqcup \{v_a: a \in [d+m]\setminus [m+2]\} \end{align*} $$

to conclude.

Somewhat surprisingly, the inequality (32) is very nearly sharp even when the k-rank condition is tightened to ${\operatorname {k-rank}(v_a : a \in [n]) \geq k}$ for some $k \geq 3$ under certain parameter constraints. More specifically, for any $k \in \{3,4,\dots , d-1\}$ , it is almost sharp under the choice $n=d+1$ and $r=m+d-1$ . Let

$$ \begin{align*} E= \{v_a : a \in [d+m]\setminus[m]\} \sqcup \bigg\{\sum_{a\in [k]} v_a\bigg\}, \end{align*} $$

and

$$ \begin{align*} F=\{v_a: a \in [m]\} \sqcup \{v_a : a \in [d+m]\setminus [m+2] \}\sqcup \bigg\{\sum_{a\in [k]} v_a\bigg\}. \end{align*} $$

Here, $\lvert E \rvert +\lvert F \rvert =2d+m,$ exceeding our lower bound by 2. When $k=d$ , take the same multisets E and F, with $\sum _{a\in [d]} v_a$ removed, to observe that our bound is sharp under the choice $n=d$ and $r=m+d-2$ . Note that the k-rank is brought down to k because of a single vector in the multiset. This is a concrete demonstration of the fact that the k-rank is a very crude measure of genericity. We emphasize that this construction relies on the particular choice of parameters $n=d+1$ , and $r=m+d-1$ . It is possible that the inequality (32) could be significantly strengthened for other choices of n and r. Indeed, we have exhibited such an improvement for $r \leq n$ in Corollary 31.

9 Comparing our generalization of Kruskal’s theorem to the uniqueness criteria of Domanov, De Lathauwer and Sørensen

In this section, we compare our generalization of Kruskal’s theorem to uniqueness criteria obtained by Domanov, De Lathauwer and Sørensen (DLS) in the case of three subsystems [Reference Domanov and De LathauwerDL13a, Reference Domanov and De LathauwerDL13b, Reference Domanov and De LathauwerDL14, Reference Sørensen and De LathauwerSL15, Reference Sørensen and De De LathauwerSDL15], which are the only previously known extensions of Kruskal’s theorem that we are aware of. A drawback to the uniqueness criteria of DLS is that, similarly to Kruskal’s theorem, they require the k-ranks to be above a certain threshold. In Section 9.1, we make this statement precise and show by example that our generalization of Kruskal’s theorem can certify uniqueness below this threshold. Moreover, in Section 9.2 we observe that our generalization of Kruskal’s theorem contains many of the uniqueness criteria of DLS. The uniqueness criteria of DLS are spread across five papers and can be difficult to keep track of. For clarity and future reference, in Theorem 37 we combine all of these criteria into a single statement. In Section 9.3, we use insight gained from this synthesization and our Theorem 2 as evidence to support a conjectural uniqueness criterion that would contain and unify every uniqueness criteria of DLS into a single, elegant statement.

For the remainder of this section, we fix a vector space $\mathcal {V}=\mathcal {V}_1 \otimes \mathcal {V}_2 \otimes \mathcal {V}_3$ over a field $\mathbb {F}$ and a multiset of product tensors

$$ \begin{align*} \{x_a : a \in [n]\} \subseteq \mathrm{Prod}\left(\mathcal{V}_1:\mathcal{V}_2 : \mathcal{V}_3 \right) \end{align*} $$

with k-ranks $k_j=\operatorname {k-rank}(x_{a,j} : a \in [n])$ for each $j \in [3]$ . For each subset $S \subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ and index $j \in [3]$ , we let

$$ \begin{align*} d_j^S=\dim\operatorname{\mathrm{span}}\{x_{a,j} : a \in [n]\}. \end{align*} $$

We also let $d_j=d_j^{[n]}$ for all $j \in [3]$ .

9.1 Uniqueness below the k-rank threshold of DLS

All of the uniqueness criteria of DLS require the k-ranks to be above a certain threshold. In this subsection, we show by example that our generalization of Kruskal’s theorem can certify uniqueness below this threshold.

Making this threshold precise, the uniqueness criteria of DLS cannot be applied whenever

(34) $$ \begin{align} &\min\{k_2,k_3\} \leq n-d_1+1,\nonumber\\ \text{ and } &\min\{k_1,k_3\} \leq n-d_2 + 1,\nonumber\\ \text{ and } & \min\{k_1,k_2\} \leq n-d_3 +1. \end{align} $$

For example, if $k_2= k_3=2$ , then the uniqueness criteria of DLS can only certify uniqueness if $d_1=n$ . The following example shows that our generalization of Kruskal’s theorem (Theorem 2) can certify uniqueness even if equation (34) holds.

Example 35. Consider the multiset of product tensors

$$ \begin{align*}\{&\alpha_1 e_1^{\otimes 3},\alpha_2 e_2^{\otimes 3}, \alpha_3 e_3^{\otimes 3},\alpha_4 e_4^{\otimes 3},\alpha_5(e_2+e_3)\otimes (e_2+e_4)\otimes (e_1+e_4)\} \quad \text{for} \quad \alpha_1,\dots, \alpha_5 \in \mathbb{F}^{\times}. \end{align*} $$

In this example, $k_1=k_2=k_3=2$ , $d_1=d_2=d_3=4$ , and $n-d_j+2=3$ for all $j \in [3]$ , so equation (34) holds. Nevertheless, for arbitrary $\alpha _1,\dots , \alpha _5 \in \mathbb {F}^{\times }$ , our generalization of Kruskal’s theorem certifies that the sum of these product tensors constitutes a unique tensor rank decomposition. We note that uniqueness for $\alpha _2=\dots =\alpha _5=1$ was proven in [Reference Domanov and De LathauwerDL13b, Example 5.2], using a proof specific to this case, in order to demonstrate that their uniqueness criteria are not also necessary for uniqueness.

Example 35 shows that Theorem 2 is strictly stronger than Kruskal’s theorem and is independent of the uniqueness criteria of DLS. It is natural to ask if Theorem 2 is stronger than Kruskal’s theorem even for symmetric tensor decompositions. We have observed in Section 8 that this is indeed the case.

9.2 Extending several uniqueness criteria of DLS

In this subsection, we observe that several of the uniqueness criteria of DLS are contained in our generalization of Kruskal’s theorem and prove a further, independent uniqueness criterion. The uniqueness criteria of DLS are numerous and can be difficult to keep track of. To more easily analyze these criteria, in Theorem 37 we combine them all into a single statement.

9.2.1 Conditions U, H, C and S

Here, we introduce several different conditions on multisets of product tensors, which will make the uniqueness criteria of DLS easier to state and also make them easier to relate to our generalization of Kruskal’s theorem. We first recall Conditions U, H and C from [Reference Domanov and De LathauwerDL13a, Reference Domanov and De LathauwerDL13b]. For notational convenience, we have changed these definitions slightly from [Reference Domanov and De LathauwerDL13a, Reference Domanov and De LathauwerDL13b]. For example, our Condition U is their Condition $U_{n-d_1+2}$ , with the added condition that $k_1 \geq 2$ . After reviewing Conditions U, H and C, we introduce Condition S, which captures the conditions of our generalization of Kruskal’s theorem in the case ${m=3}$ . Unlike Conditions U, H and C, our Condition S does not appear in [Reference Domanov and De LathauwerDL13a, Reference Domanov and De LathauwerDL13b] nor anywhere else that we are aware of.

For a vector $\alpha \in \mathbb {F}^n$ , we let $\omega (\alpha )$ denote the number of nonzero entries in $\alpha $ .

Condition U. It holds that $k_{1} \geq 2$ , and for all $\alpha \in \mathbb {F}^n$ ,

(35) $$ \begin{align} \operatorname{rank}\Big[\sum_{a \in [n]} \alpha_a x_{a, 2}\otimes x_{a,3} \Big]\geq \min\{\omega(\alpha),n-d_1+2\}. \end{align} $$

Condition H. It holds that $k_{1} \geq 2$ , and

$$ \begin{align*} d_{2}^S\!+d_{3}^S\!-\lvert S \rvert \geq \min\{\lvert S \rvert, n-d_{1}\!+2\} \end{align*} $$

for all $S \subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ .

Condition C takes a bit more work to describe. We use coordinates for this condition in order to avoid having to introduce further multilinear algebra notation. For positive integers $q,r,$ and t and matrices

$$ \begin{align*} Y=(y_1,\dots, y_t) \in \mathrm{Hom}(\mathbb{F}^t, \mathbb{F}^q)\\ Z=(z_1,\dots, z_t)\in \mathrm{Hom}(\mathbb{F}^t, \mathbb{F}^r), \end{align*} $$

let

$$ \begin{align*} Y \odot Z=(y_1\otimes z_1, \dots, y_t \otimes z_t) \in \mathrm{Hom}(\mathbb{F}^t,\mathbb{F}^{qr}) \end{align*} $$

denote the Khatri–Rao product of Y and Z. Suppose $\mathcal {V}_j=\mathbb {F}^{d_j}$ for each $j \in [3]$ , and consider the matrices

$$ \begin{align*} X_j=(x_{1,j},\dots, x_{n,j})\in \mathrm{Hom}(\mathbb{F}^n,\mathbb{F}^{d_j}) \end{align*} $$

for $j \in [3]$ . For a positive integer $s\leq d_j$ , let $\mathcal {C}_s(X_j)$ be the $\binom {d_j}{s} \times \binom {n}{s}$ matrix of $s \times s$ minors of $X_j$ , with rows and columns arranged according to the lexicographic order on the size s subsets of $[d_j]$ and $[n]$ , respectively. Define the matrix

$$ \begin{align*} C_s=\mathcal{C}_s(X_{2})\odot \mathcal{C}_s(X_{3}) \in \mathrm{Hom}(\mathbb{F}^{\binom{n}{s}}, \mathbb{F}^{q}), \end{align*} $$

where $q=\big (\substack {d_{2}\\s}\big )\big (\substack {d_{3}\\s}\big )$ . Now, we can state Condition C.

Condition C. It holds that $k_{1} \geq 2$ , $\min \{d_2,d_3\} \geq n-d_1+2$ , and

$$ \begin{align*} \operatorname{rank}(C_{n-d_{1}+2})=\left(\substack{ n \\ n-d_{1}+2} \right). \end{align*} $$

To more easily compare our generalization of Kruskal’s theorem to the uniqueness criteria of DLS, we give a name (Condition S) to the condition of our Theorem 2 in the case $m=3$ .

Condition S. It holds that

$$ \begin{align*} 2 \lvert S \rvert \leq d_1^S+d_2^S+d_3^S-2 \end{align*} $$

for all $S \subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ .

These conditions are related to each other as follows:

(36)

All of the implications in equation (36) except (Condition H $\Rightarrow $ Condition S) were proven in [Reference Domanov and De LathauwerDL13a]. To see that Condition H $\Rightarrow $ Condition S, note that, for any subset $S\subseteq [n]$ with $2 \leq \lvert S \rvert \leq n$ , the condition $k_1 \geq 2$ implies

$$ \begin{align*} d_{1}^S \geq \max\{2,d_{1}-(n-\lvert S \rvert)\}, \end{align*} $$

so by Condition H,

$$ \begin{align*} d_{1}^S+d_{2}^S+d_{3}^S &\geq \max\{2,d_{1}-(n-\lvert S \rvert)\} + \lvert S \rvert+\min\{\lvert S \rvert,n-d_{1}+2\}\\ &\geq 2 \lvert S \rvert+2, \end{align*} $$

and Condition S holds. The following example shows that Condition C $\not \Rightarrow $ Condition S.

Example 36. Let $\{x_{a,1}\otimes x_{a,2}\otimes x_{a,3} : a \in [9]\} \subseteq \mathbb {Q}^7 \otimes \mathbb {Q}^6 \otimes \mathbb {Q}^6$ be the set of product tensors defined by

$$ \begin{align*} X_1&=(x_{1,1},\dots, x_{9,1})=\begin{bmatrix} 1& 2& 1& 1& 2& -1& 1& 1& -2\\ 0& 2& 0& -1& 2& 0& 1& 2& -1\\ -2& -2& -1& -1& 0& -1& -2& 2& 1\\ 0& 2& 2& 1& -2& 2& 1& 0& 2\\ 0& -1& 0& -1& -2& -2& 2& -2&-2\\ 1& 0& -1& 0& 0& -2& 2& 0& 2\\ 0& 2& -2& 1& 2& -1& -2& -1& 1 \end{bmatrix},\\ \end{align*} $$

$$ \begin{align*} X_2&=(x_{1,2},\dots,x_{9,2})=\begin{bmatrix} 0& -2& 0& -1& 2& -1& 0& -1& 1\\ 2& -1& 0& 0& 2& 0& -2& 0& 1\\ 1& 0& -2& 2& 1& 1& 2& 1& 0\\ 2& 0& 2& -2& 0& 2& -1& 2& -1\\ 1& 1& 1& 0& 2& -1& 1& 1& 0\\ 1& -2& 2& -1& -2& 2& -1& 0& 2 \end{bmatrix},\\ \end{align*} $$

$$ \begin{align*} X_3&=(x_{1,3},\dots,x_{9,3})=\begin{bmatrix} 2& -1& 1& 1& 1& 1& -1& -2& -1\\ -2& 1& 1& 0& 0& 1& 1& 2& -2\\ 2& 1& 1& 1& -1& 2& 1& -2& -1\\ 1& 2& 1& 0& -2& -1& -1& 1& -2\\ 0& 2& 1& 2& 2& 2& 1& 1& 0\\ 2& -1& 2& 1& 0& -1& 2& -1& -2 \end{bmatrix}. \end{align*} $$

Condition C holds for this set of product tensors, but Condition S does not. In fact, it also holds that $k_1=7$ and $k_2=k_3=6$ , so $\sum _{a\in [9]} x_{a,1}\otimes x_{a,2} \otimes x_{a,3}$ is a unique tensor rank decomposition by Theorem 37.1 below. Together, this example and Example 35 demonstrate that our generalization of Kruskal’s theorem is independent of the uniqueness criteria of DLS (it is neither weaker nor stronger). This example was generated by randomly picking three matrices $X_1,X_2,X_3$ with entries in $\{-2,-1,0,1,2\}$ .

By Example 35, Condition S $\not \Rightarrow $ Condition U. In [Reference Domanov and De LathauwerDL13a], it is asked whether Condition H $\Rightarrow $ Condition C. Condition U is theoretically computable, as it can be phrased as an ideal membership problem; however, we are unaware of an efficient implementation. By comparison, Conditions C, H, and S are easy to check.

In the case of three subsystems, our Theorem 2 states that Condition S implies uniqueness. Since Condition H $\Rightarrow $ Condition S, then a corollary to Theorem 2 is that Condition H implies uniqueness. Similarly, Theorem 37 below states that Condition U + extra assumptions implies uniqueness. By equation (36), this implies that Condition H + the same extra assumptions implies uniqueness, and similarly, Condition C + the same extra assumptions implies uniqueness. Since we have proven that Condition H alone implies uniqueness, it is natural to ask whether Conditions C or U alone imply uniqueness. We reiterate this line of reasoning in Section 9.3 and pose this question formally.

9.2.2 Synthesizing the uniqueness criteria of DLS

The following theorem contains every uniqueness criterion of DLS for which we are aware of an efficient implementation. This theorem is stated in terms of Condition U to maintain generality; however, only the implied statements in which Condition U is replaced by Conditions H or C have an efficient implementation. Note that our Theorem 2 generalizes the Condition H version of this theorem to the statement that Condition S alone implies uniqueness (so, in particular, Condition H alone implies uniqueness).

Theorem 37. Suppose that Condition U holds and any one of the following conditions holds:

1. 1. $k_{1}+\min \{k_{2},k_{3}-1\}\geq n+1.$

2. 2. It holds that $k_2 \geq 2$ and for all $\alpha \in \mathbb {F}^n$ ,

   $$ \begin{align*} \mathrm{rank}\left[\sum_{a \in [n]} \alpha_a x_{a,1}\otimes x_{a,3}\right]\geq \min\{\omega(\alpha),n-d_2+2\}. \end{align*} $$
   (Note that this is just Condition U with the first subsystem replaced by the second.)

3. 3. There exists a subset $S\subseteq [n]$ with $0 \leq |S| \leq d_{1}$ such that the following three conditions hold:

   1. (a) $d_{1}^S=|S|.$

   2. (b) $d_{2}^{[n]\setminus S}=n-|S|.$

   3. (c) For any linear map $\Pi \in \mathrm {End}(\mathcal {V}_{1})$ with ${\ker (\Pi ) = \operatorname {\mathrm {span}} \{x_{a ,1}: a \in S\}}$ , scalars $\alpha _1,\dots , \alpha _n \in \mathbb {F}$ , and index $b \in [n]\setminus S$ such that

      $$ \begin{align*} \sum_{a \in [n]\setminus S} \alpha_a & \Pi x_{a, 1} \otimes x_{a, 3} = \Pi x_{b, 1} \otimes z \end{align*} $$
      for some $z \in \mathcal {V}_{\sigma (3)}$ , it holds that $\omega (\alpha )\leq 1$ .

4. 4. There exists a permutation $\tau \in S_n$ for which the matrix

   $$ \begin{align*} X_{1}^\tau = (x_{\tau(1),1},\dots, x_{\tau(n),n}) \end{align*} $$
   has reduced row echelon form
   $$ \begin{align*} Y=\left[\begin{array}{@{}c | c@{}} \begin{matrix} 1 & & \\ & \ddots &\\ && 1 \end{matrix} & \begin{matrix} & & \\ & {\Huge Z} &\\ && \end{matrix} \end{array}\right], \end{align*} $$
   where $Z \in \mathrm {Hom}(\mathbb {F}^{n-d_{1}},\mathbb {F}^{d_{1}})$ and the blank entries are zero. Furthermore, for each $a \in [d_{1}-1]$ , the columns of the submatrix of Y with row index $\{a,a+1,\dots , d_{1}\}$ and column index $\{a, a+1,\dots , n\}$ have k-rank at least two.

5. 5. $k_{1}=d_{1}.$

6. 6. For all $\alpha \in \mathbb {F}^n$ ,

   $$ \begin{align*} \mathrm{rank}\left[\sum_{a \in [n]} \alpha_a x_{a,2}\otimes x_{a,3}\right]\geq \min\{\omega(\alpha),n-k_1+2\}. \end{align*} $$
   (Note that this is a stronger statement than Condition U, as it replaces the quantity ${n-d_1+2}$ with the possibly larger quantity $n-k_1+2$ .)

Then $\sum _{a\in [n]} x_a$ constitutes a unique tensor rank decomposition.

For each $i \in [5]$ , we will refer to Theorem 37.i as the statement that Condition U and the i-th condition appearing in Theorem 37 imply uniqueness. Theorems 37.1 and 37.2 are Corollary 1.23 and Proposition 1.26 in [Reference Domanov and De LathauwerDL13b, Reference Domanov and De LathauwerDL14]. The Condition C version of Theorem 37.3 is stated in Theorem 2.2 in [Reference Sørensen and De De LathauwerSDL15], although the proof is contained in [Reference Domanov and De LathauwerDL13a, Reference Domanov and De LathauwerDL13b, Reference Sørensen and De LathauwerSL15]. Condition 3b in Theorem 37 can be formulated as checking the rank of a certain matrix (see [Reference Sørensen and De De LathauwerSDL15]). Theorem 37.4 is a new result that we will prove (see Proposition 38 for a coordinate-free statement). The Condition C version of Theorems 37.5 and 37.6 are Theorems 1.6 and 1.7 in [Reference Domanov and De LathauwerDL14]. It is easy to see that our Theorem 37.4 contains Theorem 37.5, which in turn contains Theorem 37.6, by the arguments used in [Reference Domanov and De LathauwerDL14].

Most of these statements have previously only been formulated for $\mathbb {F}=\mathbb {R}$ or $\mathbb {F}=\mathbb {C}$ ; however, in all of these cases the proof can be adapted to hold over an arbitrary field. The first step in proving all of these statements is to show that Condition U implies uniqueness in the first subsystem. This is Proposition 4.3 in [Reference Domanov and De LathauwerDL13a], and it is proven using Kruskal’s permutation lemma [Reference KruskalKru77] (the proof of the permutation lemma in [Reference LandsbergLan12] holds word-for-word over an arbitrary field). In fact, uniqueness in the first subsystem holds even with the assumption $k_1 \geq 2$ removed from Condition U [Reference Domanov and De LathauwerDL13a].

A less-restrictive condition than Condition U, which we would call Condition W, also appears in [Reference Domanov and De LathauwerDL13a, Reference Domanov and De LathauwerDL13b] and is the same as Condition U except that it only requires equation (35) to hold when ${\alpha = (f(x_{1,1}),\dots , f(x_{n,1}))}$ for some linear functional $f \in \mathcal {V}_1^*$ . We note that Theorem 37 also holds with Condition U replaced by Condition W. Although the Condition W version of Theorem 37 is slightly stronger than the Condition U version, we are not aware of an efficient algorithm to check either Condition U or Condition W, and the existence of such an algorithm seems unlikely.

We conclude this subsection by proving Theorem 37.4. For this, we require the following proposition, which restates Condition 4 in a coordinate-free manner.

Proposition 38. Condition 4 in Theorem 37 holds if and only if there exists a permutation $\tau \in S_n$ such that for each $a \in [d_1-1]$ there is a linear operator $\Pi _a \in \mathrm {End}(\mathcal {V}_1)$ for which

$$ \begin{align*} \Pi_a (x_{\tau(b),1})=0 \end{align*} $$

for all $b \in [a-1]$ , and

(37) $$ \begin{align} \operatorname{k-rank}( \Pi_a x_{\tau(a),1},\dots,\Pi_a x_{\tau(n),1})\geq 2. \end{align} $$

Proof. Assume without loss of generality that $\mathcal {V}_1=\mathbb {F}^{d_1}$ . To see that the first statement implies the second, for each $a\in [d_1-1]$ let $\Pi _a=D_a P$ , where $P\in \mathrm {End}(\mathbb {F}^{d_1})$ is the invertible matrix for which $PX_1^\tau =Y$ , and $D_a \in \mathrm {End}(\mathbb {F}^{d_1})$ is the diagonal matrix with the first $a-1$ entries zero and the remaining entries $1$ . It is easy to verify that equation (37) holds.

Conversely, suppose that the reduced row echelon form of $X_1^\tau $ , given by $P X_1^\tau $ for some invertible matrix $P \in \mathrm {End}(\mathbb {F}^{d_1})$ , does not have the specified form. Then there exists ${a\in [d_1-1]}$ for which the columns of $D_a P X_1^\tau $ have k-rank at most one. Any matrix $\Pi _a\in \mathrm {End}(\mathbb {F}^{d_1})$ for which $\Pi _a (x_{\tau (b),1})=0$ for all $b \in [a-1]$ satisfies

$$ \begin{align*} \Pi_a =\Pi_a P^{-1} D_a P. \end{align*} $$

Since the k-rank is nonincreasing under matrix multiplication from the left, equation (37) does not hold.

With Proposition 38 in hand, we can now prove Theorem 37.4.

Proof of Theorem 37.4.

The question of whether or not the decomposition $\sum _{a\in [n]} x_a$ constitutes a unique tensor rank decomposition is invariant under permutations $\tau \in S_n$ of the tensors, so it suffices to prove the statement under the assumption that the permutation $\tau $ appearing in Condition 4 is trivial. We prove the statement by induction on $d_1$ . If $d_1=2$ , then Condition U implies $k_2=k_3=n$ , so uniqueness follows from Kruskal’s theorem. For $d_1>2$ , suppose $\sum _{a\in [n]} x_a =\sum _{a\in [r]} y_a$ for some nonnegative integer $r \leq n$ and multiset of product tensors

$$ \begin{align*} \{y_a : a \in [r]\} \subseteq \mathrm{Prod}\left(\mathcal{V}_1: \mathcal{V}_2 : \mathcal{V}_3 \right). \end{align*} $$

By Proposition 4.3 in [Reference Domanov and De LathauwerDL13a] (or rather, the extension of this result to an arbitrary field), $r=n$ , and there exists a permutation $\sigma \in S_n$ and nonnegative scalars $\alpha _1,\dots , \alpha _n \in \mathbb {F}^\times $ such that $\alpha _a x_{a,1}=y_{\sigma (a),1}$ for all $a \in [n]$ . Let $\Pi _1 \in \mathrm {End}(\mathcal {V}_1)$ be any operator for which ${\ker (\Pi _1)=\operatorname {\mathrm {span}}\{x_{a,1}\}}$ and equation (37) holds (recall that $\tau $ is trivial). Then

$$ \begin{align*} \sum_{a \in [n] \setminus \{1\}} (\Pi_1 x_{a,1})\otimes x_{a,2} \otimes x_{a,3}=\sum_{a \in [n] \setminus \{1\}} (\alpha_{a} \Pi_1 x_{a,1})\otimes y_{\sigma(a),2} \otimes y_{\sigma(a),3}. \end{align*} $$

Now, $\dim \operatorname {\mathrm {span}}\{\Pi _1 x_{a,1}: a \in [n]\setminus \{1\}\}=d_1-1$ , and Condition U again holds for the multiset of product tensors

$$ \begin{align*} \{(\Pi_1 x_{a,1}) \otimes x_{a,2} \otimes x_{a,3} : a \in [n] \setminus \{1\}\}. \end{align*} $$

Furthermore, these product tensors again satisfy Condition 4 of Theorem 37, so by the induction hypothesis

$$ \begin{align*} (\Pi_1 x_{a,1}) \otimes x_{a,2} \otimes x_{a,3}=(\alpha_a \Pi_1 x_{a,1}) \otimes y_{\sigma(a),2} \otimes y_{\sigma(a),3}\quad \text{for all}\quad a \in [n]\setminus \{1\}. \end{align*} $$

It follows that $x_a=y_{\sigma (a)}$ for all $a \in [n] \setminus \{1\}$ , so $x_1=y_{\sigma (1)}$ . This completes the proof.

10 Appendix

In this appendix, we prove Theorem 27. The proof is very similar to that of Theorem 22.

Proof of Theorem 27.

For each $a \in [r]$ , let $x_{n+a}=-y_a$ , and let $T_1 \sqcup \dots \sqcup T_t=[n+r]$ be the index sets of the decomposition of $\{x_a : a \in [n+r]\}$ into connected components. Note that for each $p \in [t]$ , if

$$ \begin{align*} \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\leq \bigl\lvert T_p \cap [n] \bigr\rvert, \end{align*} $$

then $\bigl \lvert T_p \cap [n] \bigr \rvert \leq s$ , otherwise $\{x_a : a \in T_p\}$ would split. Assume without loss of generality that

$$ \begin{align*} \bigl\lvert T_1 \cap [n] \bigr\rvert-\bigl\lvert T_1 \cap [n+r]\setminus [n] \bigr\rvert &\geq \bigl\lvert T_2 \cap [n] \bigr\rvert-\bigl\lvert T_2 \cap [n+r]\setminus [n] \bigr\rvert\\ &\;\; \vdots \\ & \geq \bigl\lvert T_{{t}} \cap [n] \bigr\rvert-\bigl\lvert T_{{t}} \cap [n+r]\setminus [n] \bigr\rvert, \end{align*} $$

If

$$ \begin{align*} \bigl\lvert T_{1}\cap [n] \bigr\rvert \geq \bigl\lvert T_{1} \cap [n+r]\setminus [n] \bigr\rvert, \end{align*} $$

then let $\tilde {l}\in [t]$ be the largest integer for which

(38) $$ \begin{align} \bigl\lvert T_{\tilde{l}}\cap [n] \bigr\rvert \geq \bigl\lvert T_{\tilde{l}} \cap [n+r]\setminus [n] \bigr\rvert. \end{align} $$

Otherwise, let $\tilde {l}=0$ . Then for all $p \in [t]\setminus [\tilde {l}]$ it holds that

$$ \begin{align*} \bigl\lvert T_p \cap [n] \bigr\rvert < \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert. \end{align*} $$

To complete the proof, we will show that $\tilde {l} \geq l$ , for then we can take $Q_p=T_p \cap [n]$ and $R_p=T_p \cap [n+r]\setminus [n]$ for all $p \in [l]$ to conclude.

Suppose toward contradiction that $\tilde {l}< l$ . We will require the following two claims.

Claim 40. It holds that $\tilde {l}<t$ , $\bigl \lceil \frac {n- s \tilde {l} }{t-\tilde {l}} \bigr \rceil \geq s+1$ , and there exists $p \in [t]\setminus [\tilde {l}]$ for which

(39) $$ \begin{align} \bigl\lvert T_p \cap [n] \bigr\rvert \geq \biggl\lceil \frac{n- s \tilde{l} }{t-\tilde{l}} \biggr\rceil. \end{align} $$

Claim 41. For all $p \in [t] \setminus [\tilde {l}]$ , it holds that

(40) $$ \begin{align} \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert \leq \bigl\lvert T_p \cap [n] \bigr\rvert +(r-n)+(s+1)\tilde{l} -t+1. \end{align} $$

Before proving these claims, we first use them to complete the proof of the theorem. Let $p \in [t]\setminus [\tilde {l}]$ be as in Claim 40. Then,

$$ \begin{align*} \lvert T_p \rvert &= \bigl\lvert T_p \cap [n] \bigr\rvert+\bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\\ &\leq 2 \bigl\lvert T_p \cap [n] \bigr\rvert+ r-n+(s+1)\tilde{l} -t+1\\ &\leq 2 \bigl\lvert T_p \cap [n] \bigr\rvert+ r-n + s \tilde{l} - \biggl\lceil \frac{n-s \tilde{l}}{\bigl\lvert T_p \cap [n] \bigr\rvert} \biggr\rceil+1\\ &\leq 2 \bigl\lvert T_p \cap [n] \bigr\rvert+ (r-n+q-s)- \biggl\lceil \frac{n-q+s}{\bigl\lvert T_p \cap [n] \bigr\rvert} \biggr\rceil+1\\ & \leq \sum_{j=1}^m (d_j^{T_p \cap [n]}-1)+1, \end{align*} $$

where the first line is obvious, the second follows from Claim 41, the third follows from Claim 40, the fourth follows from $\tilde {l}< l$ and the fifth follows from the assumptions of the theorem and the fact that $\lvert T_p \cap [n] \rvert \geq s+1$ . So $\{x_a : a \in T_p\}$ splits, a contradiction. This completes the proof, modulo proving the claims.

Proof of Claim 23.

To prove the claim, we first observe that $n>st$ . Indeed, if $n \leq st$ , then

$$ \begin{align*} r& \geq \sum_{p=\tilde{l}+1}^t \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\\ &\geq\sum_{p=\tilde{l}+1}^t \left(\bigl\lvert T_p \cap [n] \bigr\rvert+1\right)\\ &=n-\bigl\lvert (T_1 \sqcup \dots \sqcup T_{\tilde{l}})\cap [n] \bigr\rvert+t-\tilde{l}\\ &\geq n+t - (s+1)\tilde{l}\\ & \geq n+\biggl\lceil \frac{n}{s}-(s+1)(q/s-1) \biggr\rceil\\ &= \biggl\lceil \left(\frac{s+1}{s}\right)(n-q+s) \biggr\rceil, \end{align*} $$

where the first line is obvious, the second follows from equation (17), the third is obvious, the fourth follows from $\lvert T_p \cap [n] \rvert \leq s$ for all $p \in [\tilde {l}]$ , the fifth follows from $n\leq st$ and $\tilde {l}<l$ and the sixth is algebra. This contradicts the assumptions of the theorem, so it must hold that $n>st$ .

Note that $\tilde {l}<t$ , for otherwise we would have $n \leq st$ by the fact that $\bigl \lvert T_p \cap [n] \bigr \rvert \leq s$ for all $p \in [\tilde {l}]$ . To verify that $\bigl \lceil \frac {n- s \tilde {l} }{t-\tilde {l}} \bigr \rceil \geq s+1$ , it suffices to prove $\frac {n- s \tilde {l} }{t-\tilde {l}}> s,$ which follows from $n>st$ . To verify equation (39), since $\lvert T_p \cap [n] \rvert \leq s$ for all $p \in [\tilde {l}]$ , by the pigeonhole principle there exists $p \in [t]\setminus [\tilde {l}]$ for which

$$ \begin{align*} \bigl\lvert T_p \cap [n] \bigr\rvert \geq \biggl\lceil \frac{n- s \tilde{l} }{t-\tilde{l}} \biggr\rceil. \end{align*} $$

This proves the claim.

Proof of Claim 41.

Suppose toward contradiction that the inequality (40) does not hold for some $\tilde {p}\in [t] \setminus [\tilde {l}]$ . Then

$$ \begin{align*} r &\geq \sum_{p=\tilde{l}+1}^t \bigl\lvert T_p \cap [n+r]\setminus [n] \bigr\rvert\\ & \geq \sum_{p \neq \tilde{p}} \big(\bigl\lvert T_p \cap [n] \bigr\rvert+1\big)+ \bigl\lvert T_{\tilde{p}} \cap [n] \bigr\rvert+ (r-n)+(s+1)\tilde{l} -t+2\\ &= \sum_{p=\tilde{l}+1}^t \bigl\lvert T_p \cap [n] \bigr\rvert+ (r-n)+s\tilde{l}+1\\ &\geq r+1, \end{align*} $$

where the first three lines are obvious, and the fourth follows from equation (38), a contradiction. $\triangle $

The proofs of Claims 40 and 41 complete the proof of the theorem.