2.1 Multilinear algebra

We assume that the reader is familiar with tensors. A main feature in this paper is that we deal with infinite-dimensional spaces and their duals; and this infinite dimensionality causes some technical complications. For example, multilinear functions on a direct product $\prod_{i = 1}^n V_i$ of vector spaces can be naturally identified with the dual space $(\bigotimes_{i = 1}^n V_i)^*$ of linear functions on $\bigotimes_{i = 1}^n V_i$ . Moreover, $\bigotimes_{i = 1}^n V_i^*$ canonically embeds into $(\bigotimes_{i = 1}^n V_i)^*$ via $(\otimes_{i = 1}^n f_i) (\otimes_{i = 1}^n v_i) = \prod_{i = 1}^n f_i(v_i)$ . A special case is that $(V^*)^{\otimes n}$ embeds into $(V^{\otimes n})^*$ . If all $V_i$ ’s are finite dimensional then this embedding is an isomorphism. However, for infinite dimensional $V_i$ , it is well known that this embedding is not surjective. We describe this in Subsection A.1 of the appendix.

For any symmetric tensor $A \in {\text{Sym}}^n(V)$ we define the symmetric rank of A to be the least $r \ge 0$ for which A can be expressed as

\begin{equation*}A = \sum_{i = 1}^r \lambda_i \textbf{v}_i^{\otimes n}, \quad \lambda_i \in \mathbb{F}, \textbf{v}_i \in V,\end{equation*}

and we denote it by $\text{rk}_{\textsf{S}}(A)$ . If there is no such decomposition we define $\text{rk}_{\textsf{S}}(A) = \infty$ . If $\text{rk}_{\textsf{S}}(A) < \infty$ then in any such expression of A as a sum of $\text{rk}_{\textsf{S}}(A)$ terms all $\lambda_i \ne 0$ , all $\textbf{v}_i \ne 0$ and are pairwise linearly independent. In Subsection A.2 we show that if $\mathbb{F}$ is infinite, then $\text{rk}_{\textsf{S}}(A) < \infty$ for all $A \in {\text{Sym}}^n(V)$ . We prove all needed technical multilinear algebra statements in Section A.

2.2 Weighted graph homomorphisms

We recap the notion of weighted graph homomorphisms [Reference Freedman, Lovász and Schrijver27], but state it for an arbitrary field $\mathbb{F}$ .

An ( $\mathbb{F}$ -)weighted graph H is a graph with a weight $\alpha_H(i) \in \mathbb{F} \setminus \{0\}$ associated with each node i and a weight $\beta_H(i, j) \in \mathbb{F}$ associated with each edge i j. For undirected GH, we assume $\beta_H(i, j)= \beta_H(j, i)$ .

Let G be an unweighted graph (with possible multiple edges, but no loops) and H a weighted graph (with possible loops, but no multiple edges). A map $\phi \colon V(G) \to V(H)$ is a homomorphism if every edge of G goes to an edge or loop of H. In this paper, it is convenient to assume that H is a complete graph with a loop at all nodes by adding all missing edges and loops with weight 0. Then the weighted graph H is described by an integer $q = |V(H)| \ge 0$ (H can be the empty graph), a nowhere zero vector $a = (\alpha_1, \ldots, \alpha_q) \in \mathbb{F}^q$ and a symmetric matrix $B = (\beta_{i j}) \in \mathbb{F}^{q \times q}$ . In this setting every map $\phi \colon V(G) \to V(H)$ is a homomorphism. We assign the weights

(2) \begin{equation}\alpha_\phi = \prod_{u \in V(G)} \alpha_H(\phi(u)),\quad \quad \quad\hom_\phi(G, H) = \prod_{u v \in E(G)} \beta_H(\phi(u), \phi(v)),\end{equation}

and define

(3) \begin{equation}\hom(G, H) = \sum_{\phi \colon V(G) \to V(H)} \alpha_\phi \hom_\phi(G, H).\end{equation}

When G is the empty graph, i.e. $V(G) = \emptyset$ , the only map $\phi \colon \emptyset \to V(H)$ is the empty map $\phi = \emptyset$ ; in that case we have the empty products $\alpha_\emptyset = 1$ , $\hom_\emptyset = 1$ and $\hom(G, H) = 1$ .

If all node weights and edge weights in H are 1, then this is the number of homomorphisms from G into H. Without loss of generality we require all vertex weights $\alpha_H(i) \ne 0$ since any vertex i with $\alpha_H(i) = 0$ can be deleted together with all incident edges i j and loops at i.

Note that when H is the empty graph, then $\hom(G, H) = 0$ if G is not the empty graph (because there is no map $\phi \colon V(G) \to V(H)$ in this case), and $\hom(G, H) = 1$ if G is the empty graph (because there is precisely one empty map $\phi = \emptyset$ in this case.) The function $f_H = \hom(\cdot, H)$ is a graph parameter, a concept to be formally defined shortly.

3.1 Basic definitions

An $\mathbb{F}$ -valued graph parameter is a function from finite graph isomorphism classes to $\mathbb{F}$ .Footnote 2 For convenience, we think of a graph parameter as a function defined on finite graphs and invariant under graph isomorphism. We allow multiple edges in our graphs, but no loops, as input to a graph parameter. A graph parameter f is called multiplicative, if for any disjoint union $G_1 \sqcup G_2$ of graphs $G_1$ and $G_2$ we have $f(G_1 \sqcup G_2) = f(G_1) f(G_2)$ .

A k-labelled graph ( $k \ge 0$ ) is a finite graph in which k nodes are labelled by $1, 2, \ldots, k$ (the graph can have any number of unlabelled nodes). Two k-labelled graphs are isomorphic if there is a label-preserving isomorphism between them. We identify a (k-labelled) graph with its (k-labelled) graph isomorphism class. We denote by $K_k$ the k-labelled complete graph on k nodes, and by $U_k$ , the k-labelled graph on k nodes with no edges. In particular, $K_0 = U_0$ is the empty graph with no nodes and no edges. A graph parameter on a labelled graph ignores its labels.

It is easy to see that for a multiplicative graph parameter f, either f is identically 0 or $f(K_0) = 1$ . Every weighted graph homomorphism $f_H = \hom(\cdot, H)$ is a multiplicative graph parameter.

The product of two k-labelled graphs $G_1$ and $G_2$ is defined as follows: we take their disjoint union, and then identify nodes with the same label. Hence for two 0-labelled graphs, $G_1 G_2 = G_1 \sqcup G_2$ (disjoint union). Clearly, the graph product is associative and commutative with the identity $U_k$ , so the set of all isomorphism classes of finite k-labelled graphs together with the product operation forms a commutative monoid which we denote by ${\mathcal{PLG}}[k]$ .

Let $\mathcal{G}[k]$ denote the monoid algebra $\mathbb{F} {\mathcal{PLG}}[k]$ consisting of all finite formal linear combinations in ${\mathcal{PLG}}[k]$ with coefficients from $\mathbb{F}$ ; they are called (k-labelled, $\mathbb{F}$ -)quantum graphs. This is a commutative algebra with $U_k$ being the multiplicative identity, and the empty sum as the additive identity. Later, in Section 5 we will expand these definitions to allow label sets to be arbitrary finite subsets of ${\mathbb{Z}_{>0}}$ .

3.2 Connection tensors

Now we come to the central concept for our treatment. Let f be any graph parameter. For all integers $k, n \ge 0$ , we define the following n-dimensional array $T(f, k, n) \in \mathbb{F}^{({\mathcal{PLG}}[k])^n}$ , which can be identified with $(V^{\otimes n})^*$ , where V is the infinite-dimensional vector space with coordinates indexed by ${{\mathcal{PLG}}[k]}$ , i.e. $V = \bigoplus_{{\mathcal{PLG}}[k]} \mathbb{F}$ . The entry of T(f, k, n) at coordinate $(G_1, \ldots, G_n)$ is $f(G_1 \cdots G_n)$ ; when $n = 0$ , we define T(f, k, n) to be the scalar $f(U_k)$ . Furthermore, by the commutativity of the product the arrays T(f, k, n) are symmetric with respect to coordinates, i.e. $T(f, k, n) \in {\text{Sym}}(\mathbb{F}^{({\mathcal{PLG}}[k])^n})$ . Fix f, k and n, we call the n-dimensional array T(f, k, n) the (kth, n-dimensional) connection tensor of the graph parameter f. When $n = 2$ , a connection tensor is exactly a connection matrix of the graph parameter f studied in [Reference Freedman, Lovász and Schrijver27], i.e. $T(f, k, 2) = M(f, k)$ .

In contrast to [Reference Freedman, Lovász and Schrijver27], we will be concerned with only one property of connection tensors, namely their symmetric rank. The symmetric rank $\text{rk}_{\textsf{S}}(f, k, n) = \text{rk}_{\textsf{S}}(T(f, k, n))$ , as a function of k, n, will be called the symmetric rank connectivity function of the parameter f. This may be infinite, but for many interesting parameters it is finite, and its growth rate will be important for us.

Proof. Suppose $f \ne 0$ is multiplicative. Then $f(K_0)^2 = f(K_0)$ , showing that $f(K_0) \in \{ 0, 1 \}$ . If $f(K_0) = 0$ , then the relation $f(G) = f(G) f(K_0)$ implies that $f(G) = 0$ for every G, which is excluded. So $f(K_0) = 1$ . Trivially $\text{rk}_{\textsf{S}} T(f, 0, n) = 1$ for $n = 0, 1$ . Fix any $n \ge 2$ . Then $f(G_1 \cdots G_n) = f(G_1) \cdots f(G_n)$ for any 0-labelled graphs $G_1, \ldots, G_n$ , which implies that $\text{rk}_{\textsf{S}} T(f, 0, n) = 1$ .

Now suppose $f(K_0) = 1$ and for some $n \ge 2$ , $\text{rk}_{\textsf{S}}(f, 0, n) = 1$ . This implies that there is a graph parameter $\phi$ and a constant $c_n$ such that $f(G_1 \cdots G_n) = c_n \phi(G_1) \cdots \phi(G_n)$ . Putting all $G_i = K_0$ , we get $c_n \phi(K_0)^n = f(K_0) = 1$ so $\phi(K_0) \ne 0$ and $c_n = 1 / \phi(K_0)^n$ . Dividing $\phi$ by $\phi(K_0)$ we can assume that $\phi$ is normalized so that $f(G_1 \cdots G_n) = \phi(G_1) \cdots \phi(G_n)$ and $\phi(K_0) = 1$ . Next, taking $G_1 = G$ and $G_i = K_0$ for $2 \le i \le n$ we see that $f(G) = \phi(G)$ for every G and therefore $f(G_1 \cdots G_n) = f(G_1) \cdots f(G_n)$ . Finally, substituting $G_i = K_0$ for $2 < i \le n$ , we get $f(G_1 G_2) = f(G_1) f(G_2)$ so f is multiplicative.

3.3 Connection tensors of homomorphisms

Fix a weighted graph $H = (\alpha, B)$ . Recall that in the definition of $\hom(\cdot, H)$ we assume H to be a complete graph with possible 0 weighted edges and loops, but no 0 weighted vertices. For any k-labelled graph G and mapping $\phi \colon [k] \to V(H)$ , let

(4) \begin{equation}\hom_\phi(G, H)= \sum_{\text{ $\substack{\psi \colon V(G) \to V(H) \\ \psi \text{ extends } \phi}$}} \frac{\alpha_\psi}{\alpha_\phi} \hom_\psi(G, H),\end{equation}

where $\alpha_\phi = \prod_{i \in [k]} \alpha_H(\phi(i))$ , and $\alpha_\psi$ and $\hom_\psi$ are defined by (2). Here $\psi \text{ extends } \phi$ means that if $u_i \in V(G)$ is labelled by $i \in [k]$ then $\psi(u_i) = \phi(i)$ , so $\frac{\alpha_\psi}{\alpha_\phi}$ is the product of vertex weights of $\alpha_\psi$ not in $\alpha_\phi$ . Then

(5) \begin{equation}\hom(G, H) = \sum_{\phi \colon [k] \to V(H)} \alpha_\phi \hom_\phi(G, H).\end{equation}

Our main contribution in this paper is that a simple exponential bound in k on the symmetric rank of the connection tensor of a graph parameter characterizes it being expressible as $\hom(\cdot, H)$ . This holds over all fields $\mathbb{F}$ . In the following theorems, the rank function $\text{rk}_{\textsf{S}}(f_H, k, n)$ is defined over the field $\mathbb{F}$ .

Proof. The first claim is obvious, as an empty product is 1, and the sum in (3) is over the unique empty map $\emptyset$ which is the only possible map from the empty set $V(K_0)$ . For the second claim notice that for any k-labelled graphs $G_1, \ldots, G_n$ and $\phi \colon [k] \to V(H)$ ,

(6) \begin{equation}\hom_\phi(G_1 \cdots G_n, H) = \hom_\phi(G_1, H) \cdots \hom_\phi(G_n, H).\end{equation}

When $n = 0$ , this equality is $\hom_\phi(U_k, H) = 1$ according to (2), as an empty product is 1.

By (5) and (6), for the connection tensor $T(f_H, k, n)$ we have the following decomposition:

\begin{equation*}T(f_H, k, n) = \sum_{\phi \colon [k] \to V(H)} \alpha_\phi (\hom_\phi(\cdot, H))^{\otimes n},\end{equation*}

where each $\hom_\phi(\cdot, H) \in \mathbb{F}^{{\mathcal{PLG}}[k]}$ and $k, n \ge 0$ . Let $q = |V(H)|$ . Then each $T(f_H, k, n)$ is a linear combination of $q^k$ tensor n-powers and therefore $\text{rk}_{\textsf{S}} T(f_H, k, n) \le q^k$ for $k, n \ge 0$ .

The main results of this paper are Theorems 3.2 and 3.3, a converse to Theorem 3.1.

More generally, we have the following stronger theorem.

Theorem 3.3 implies Theorem 3.2 by choosing a large n. Indeed if $\text{rk}_{\textsf{S}}(f, k, n) \le q^k$ , we may choose any $n \ge \max(q^k + 1, 2)$ , which is $q^k + 1$ unless $q = 0$ and $k > 0$ .

In Section 5 we will prove Theorem 3.3, then Theorem 3.2 also follows.

4.1 Tensor rank lower bound of certain tensors

We first prove a lemma about the rank of the connection tensor for graph matchings. Let $\mathcal{M}_{a, b} = \mathcal{M}_{n; a, b} \in {\text{Sym}}^n(\mathbb{F}^2)$ denote the function $\{0, 1\}^n \rightarrow \mathbb{F}$ ( $n \ge 0$ ), such that on the all-0 input $\textbf{0}$ it takes value a, on all inputs of Hamming weight one it takes value b, and on all other inputs it takes value 0. This function is denoted by $[a, b, 0, \ldots, 0]$ in the Holant literature. ( $\mathcal{M}_{0; a, b}$ is just a constant a.) We have the following lemma; the proof is adapted from the proof of Lemma 5.1 in [Reference Comon, Golub, Lim and Mourrain20].

Proof. For $n = 0$ the lemma is trivial. Let $n \ge 1$ . Clearly $\mathcal{M}_{n; a, b} \ne 0$ , and so $r = \text{rk}_{\textsf{S}} \mathcal{M}_{n; a, b} \ge 1$ . Suppose $r < n$ for a contradiction. Then we can write $\mathcal{M}_{n; a, b} = \sum_{i = 1}^r \lambda_i v_i^{\otimes n}$ where $\lambda_i \in \mathbb{F}$ , and $v_i = (\alpha_i, \beta_i) \in \mathbb{F}^2$ are non-zero and pairwise linearly independent. The decomposition implies that the linear system $A x = \vec b$ with the extended matrix

(7) \begin{equation}\widehat A = [A \mid \vec b] =\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c|c}\alpha_1^n & \alpha_2^n & \ldots & \alpha_r^n & a \\ \\[-15pt]\alpha_1^{n - 1} \beta_1 & \alpha_2^{n - 1} \beta_2 & \ldots & \alpha_r^{n - 1} \beta_r & b \\ \\[-15pt]\alpha_1^{n - 2} \beta_1^2 & \alpha_2^{n - 2} \beta_2^2 & \ldots & \alpha_r^{n - 2} \beta_r^2 & 0 \\ \\[-15pt]\vdots & \vdots & \ddots & \vdots & \vdots \\ \\[-15pt]\beta_1^n & \beta_2^n & \ldots & \beta_r^n & 0\end{array}\right]\end{equation}

has a solution $x_i = \lambda_i,\, 1 \le i \le r$ . Note that $\widehat A$ is $(n + 1) \times (r + 1)$ and A has only r columns. We show that ${\text{rank}} \, \widehat A = r + 1 > {\text{rank}} \, A$ . This is a contradiction. We consider the following two cases.

1. (1). All $\beta_i \ne 0$ . Then, by the pairwise linear independence of $v_i$ , the ratios $\alpha_i / \beta_i$ are pairwise distinct. Then the last r rows of A, i.e. rows $n - r + 2$ to $n + 1$ form an $r \times r$ Vandermonde matrix of rank r. Note that $n+1 \ge n - r + 2 > 2$ . By $b \ne 0$ , we get an $(r + 1) \times (r + 1)$ submatrix of $\widehat A$ of rank $r + 1$ by taking row 2 and the last r rows.

2. (2). Some $\beta_i = 0$ . Without loss of generality we can assume it is $\beta_1$ . Again by the pairwise linear independence of $v_i$ , all other $\beta_i \ne 0$ , and all $\alpha_i / \beta_i$ are pairwise distinct for $2 \le i \le r$ (vacuously true if $r = 1$ ). Then since $b \ne 0$ , the submatrix of $\widehat A$ formed by taking rows 1, 2 and the last $r - 1 \ge 0$ rows have rank $r + 1$ .

We now show that for an infinite field $\mathbb{F}$ , we can give a tight upper bound for $\text{rk}_{\textsf{S}} \mathcal{M}_{n; a, b}$ where $b \ne 0$ and $n \ge 1$ . The existence of a decomposition $\mathcal{M}_{n; a, b} = \sum_{i = 1}^r \lambda_i v_i^{\otimes n}$ where $r \ge 1$ , $\lambda_i \in \mathbb{F}$ and $v_i = (\alpha_i, \beta_i) \in \mathbb{F}^2$ ( $1 \le i \le r$ ) is equivalent to the statement that system (7) has a solution $x_i = \lambda_i\, (1 \le i \le r)$ . Note that by Lemma 4.1, we must have $r \ge n$ .

Assume $\mathbb{F}$ is infinite. We show how to achieve $r = n$ with one exceptional case. First, we set all $\beta_i = 1$ . By comparing with the Vandermonde determinant $\det([A \mid \vec t\,])$ as a polynomial in t, where the last column is $\vec t = (t^n, t^{n - 1}, \ldots, t, 1)^T$ , we have

\begin{equation*}\det \widehat A = \det([A \mid \vec b]) = (-1)^n \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j) \cdot \Big(a - b \sum_{i = 1}^n \alpha_i\Big).\end{equation*}

To see this equation, note that as a polynomial in t the Vandermonde determinant $\det([A \mid \vec t\,]) = \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j) \prod_{i = 1}^n (\alpha_i - t) = \sum_{i = 0}^n c_i t^i$ for some $c_i \in \mathbb{F}$ ( $0 \le i \le n$ ). Then $\det \widehat A = a c_n + b c_{n - 1}$ .

If we set $\alpha_1, \ldots, \alpha_n \in \mathbb{F}$ to be pairwise distinct, and $\sum_{i = 1}^n \alpha_i = a / b$ , then ${\text{rank}}\ A = {\text{rank}}\ \hat A = n$ . The (affine) hyperplane $\Pi: \sum_{i = 1}^n \alpha_i = a / b$ has points away from its intersections with finitely many hyperplanes $x_i = x_j$ ( $i \ne j$ ), as long as each of these hyperplanes is distinct from $\Pi$ . This is trivially true if $n = 1$ . Let $n \ge 2$ . Under an affine linear transformation we may assume $\Pi$ is the hyperplane $x_n = 0$ in $\mathbb{F}^n$ and we only need to show $\mathbb{F}^{n - 1}$ is not the union of finitely many, say k, affine hyperplanes. Consider the cube $S^{n - 1}$ for a large subset $S \subseteq \mathbb{F}$ . The union of these k affine hyperplanes intersecting $S^{n - 1}$ has cardinality at most $k |S|^{n - 2} < |S|^{n - 1}$ , for a large S.

Each hyperplane $x_i = x_j$ ( $i \ne j$ ) is distinct from $\Pi$ , except in one case

(8) \begin{equation}a = 0,\, n = 2, \text{ and } {\text{char}}\ \mathbb{F} = 2.\end{equation}

In this exceptional case, we can easily prove that indeed $\text{rk}_{\textsf{S}} \mathcal{M}_{2; 0, b} =3$ .

We have proved the following.

We remark that for any infinite $\mathbb{F}$ not in case (8), for $n \ge 2$ we can achieve ${\text{rank}}\ A = {\text{rank}}\ \widehat A = n$ in (7) by further requiring that all $\alpha_i \ne 0$ (in addition to being pairwise distinct, and all $\beta_i=1$ ). This is simply to avoid the intersections of $\Pi$ with another finitely many hyperplanes distinct from $\Pi$ . Setting $a_i = 1 / \alpha_i$ , we can set $a_i \in \mathbb{F}$ such that ${\text{rank}}\ A = {\text{rank}}\ \widehat A = n$ for the following $\widehat A$ .

(9) \begin{equation}\widehat A = [A \mid \vec b] =\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c|c}1 & 1 & \ldots & 1 & a\\ \\[-15pt]a_1 & a_2 & \ldots & a_n & 1 \\ \\[-15pt]\vdots & \vdots & \ddots & \vdots & \vdots \\ \\[-15pt]a_1^{n - 1} & a_2^{n - 1} & \ldots & a_n^{n - 1} & 0 \\ \\[-15pt]a_1^n & a_2^n & \ldots & a_n^n & 0\end{array}\right].\end{equation}

The linear system $A x = \vec b$ has a solution $(\lambda_1, \ldots, \lambda_n)$ implies that $\mathcal{M}_{m; a, b} = \sum_{i = 1}^n \lambda_i v_i^{\otimes m}$ , where $v_i = (1, a_i)$ , for all $0 \le m \le n$ . (For $m = 0$ , $\mathcal{M}_{0; a, 1} = a$ is a constant, and $v_i^{\otimes 0} = 1$ .)

4.2 Perfect matchings

Let $\mathcal{F}$ be any set of $\mathbb{F}$ -valued constraint functions, a.k.a. signatures, from some finite set [q]. For example, the binary Equality $(=_2)$ signature on (x, y) outputs value 1 if $x = y$ , and 0 otherwise. Similarly one can define All-Distinct on [q], and Exact-One and Exact-Two on the Boolean domain $(q = 2)$ . An input to a Holant problem ${\text{Holant}}(\mathcal{F})$ is $\Omega = (G, \pi)$ where $G = (V, E)$ is a graph (with possible multiple edges and loops), and $\pi$ assigns to each $v \in V$ some $f_v \in \mathcal{F}$ of arity $\deg(v)$ , and associate its incident edges as input variables to $f_v$ . The output is ${\text{Holant}}(G; \mathcal{F}) = \sum_{\sigma} \prod_{v \in V} f_v(\sigma \mid_{E(v)})$ , where the sum is over all edge assignments $\sigma \colon E \to [q]$ , E(v) denotes the incident edges of v and $\sigma \mid_{E(v)}$ denotes the restriction of $\sigma$ . Bipartite ${\text{Holant}}(G; \mathcal{F} \mid \mathcal{G})$ are defined on bipartite graphs $G = (U, V, E)$ where vertices in U and V are assigned signatures from $\mathcal{F}$ and $\mathcal{G}$ , respectively.

The graph parameter that counts the number of perfect matchings in a graph, denoted by #Perfect-Matching (or ${\text{pm}}$ ), is a quintessential Holant problem, corresponding to the Exact-One function. In this subsection we show it is not expressible as a GH function over any field. This was proved in [Reference Freedman, Lovász and Schrijver27] for GH functions with real edge and positive vertex weights. However that proof does not work for arbitrary fields, e.g. for the field of complex numbers $\mathbb{C}$ , or even for real numbers with arbitrary (not necessarily positive) vertex weights. A crucial condition in [Reference Freedman, Lovász and Schrijver27] is positive semidefiniteness. Our main result (Theorems 3.1, 3.2 and 3.3) indicates that the property of being expressible as a GH function is completely characterized by tensor rank.

Let ${\text{pm}}(G) = m \cdot 1 \in \mathbb{F}$ (the sum of m copies of $1 \in \mathbb{F}$ ) where m is the number of perfect matchings in G. Obviously, ${\text{pm}}$ is a multiplicative graph parameter with ${\text{pm}}(K_0) = 1$ . Next, let G be a k-labelled graph, let $X \subseteq [k]$ , and let ${\text{pm}}(G, X)$ denote the number of matchings in G (expressed in $\mathbb{F}$ ) that match all the unlabelled nodes and, for labelled nodes, exactly the nodes in X. Then for any k-labelled graphs $G_1, \ldots, G_n$ ,

\begin{equation*}{\text{pm}}(G_1 \cdots G_n)= \sum_{X_1 \sqcup \ldots \sqcup X_n = [k]} {\text{pm}}(G_1, X_1) \cdots {\text{pm}}(G_n, X_n).\end{equation*}

This means that $T({\text{pm}}, k, n)$ is the product $N_k^{\otimes n} W_{k, n}$ where $N_k$ has infinitely many rows indexed by all k-labelled graphs G, but only $2^k$ columns indexed by the subsets X of [k], with the entry at (G, X)

\begin{equation*}N_{k; G, X} = {\text{pm}}(G, X),\end{equation*}

and $W_{k, n}$ is a symmetric $\underbrace{2^k \times \ldots \times 2^k}_{n \text{ times}}$ tensor (from ${\text{Sym}}^n(\mathbb{F}^{2^k})$ ), where

\begin{equation*}W_{k, n; X_1, \ldots, X_n} =\begin{cases}& 1 \quad \text{if } X_1 \sqcup \ldots \sqcup X_n = [k], \\[3pt] & 0 \quad \text{otherwise.}\end{cases}\end{equation*}

For any k, if $W_{k, n} = \sum_{i = 1}^r a_i v_i^{\otimes n}$ , then $T({\text{pm}}, k, n) = \sum_{i = 1}^r a_i (N_k v_i)^{\otimes n}$ . Hence $\text{rk}_{\textsf{S}} T({\text{pm}}, k, n) \le \text{rk}_{\textsf{S}} W_{k, n}$ . We show that in fact equality holds. Consider the family of k-labelled graphs $\{ P_X \}_{X \subseteq [k]}$ of cardinality $2^k$ indexed by the subsets of [k] and is defined as follows: each $P_X$ has $|X|$ unlabelled vertices $\{ x_i \}_{i \in X}$ and k-labelled vertices $\{ y_i \}_{i = 1}^k$ labelled 1 to k, with an edge between $x_i$ and $y_j$ iff $i = j$ . It is easy so see that for $X, Y \subseteq [k]$ , $N_{k; P_X, Y} = 1$ if $X = Y$ and 0 otherwise. Then if we consider the subset of rows in $N_k$ corresponding to $\{ P_X \}_{X \subseteq [k]}$ we see that they form the identity matrix $I_{2^k}$ with a suitable order of rows. Therefore $\text{rk}_{\textsf{S}} W_{k, n} = \text{rk}_{\textsf{S}} \left( I_{2^k}^{\otimes n} W_{k, n} \right) \le \text{rk}_{\textsf{S}} T({\text{pm}}, k, n)$ and so $\text{rk}_{\textsf{S}} T({\text{pm}}, k, n) = \text{rk}_{\textsf{S}} W_{k, n}$ .

Note that for $k = 1$ , $W_{1, n}$ is just the perfect matching tensor (or the Exact-One function on n inputs) $\mathcal{M}_{0, 1} \in {\text{Sym}}^n(\mathbb{F}^2)$ where $n \ge 1$ . Applying Lemma 4.1 with $a = 0, b = 1$ , we get $\text{rk}_{\textsf{S}} W_{1, n} \ge n$ and therefore $\text{rk}_{\textsf{S}} T({\text{pm}}, 1, n) \ge n$ for $n \ge 1$ . Now if ${\text{pm}}$ were expressible as $\hom(\cdot, H)$ for some weighted graph H with $q = |V(H)|$ , then by Theorem 3.1, $\text{rk}_{\textsf{S}} T({\text{pm}}, k, n) \le q^k$ for $k, n \ge 0$ so that $\text{rk}_{\textsf{S}} T({\text{pm}}, 1, n) \le q$ for $n \ge 0$ . However, as we have just shown $\text{rk}_{\textsf{S}} T({\text{pm}}, 1, n) \ge n$ for $n \ge 1$ which contradicts the upper bound when $n > q$ . Hence ${\text{pm}}$ is not expressible as a graph homomorphism function over any field. We state it as a theorem.

In this proof we have only used simple k-labelled graphs that do not have edges between the k-labelled vertices. The graphs $\{ P_X \}_{X \subseteq [k]}$ clearly have this property, and this property is preserved under product of k-labelled graphs. It follows that Theorem 4.1 holds even when ${\text{pm}}$ is restricted to simple graphs.

We can prove the same inexpressibility results for other Holant problems, such as weighted matchings, proper edge colourings, and vertex disjoint cycle covers. We will also discuss bounded degree cases of weighted matchings.

4.3 Weighted matchings

We prove that the more general problem #Weighted-Matching $_a$ defined by ${\text{wm}}_a(G) = {\text{Holant}}(G; \{ \mathcal{M}_{n; a, 1} \}_{n \ge 0})$ is not expressible as a GH function over any field, for any a.

Clearly, ${\text{wm}}_a$ is a multiplicative graph parameter with ${\text{wm}}_a(K_0) = 1$ . If $a = 0$ then ${\text{wm}}_a$ is just #Perfect-Matching ( ${\text{pm}}$ ). Counting all matchings is ${\text{wm}}_a$ for $a = 1$ .

Let G be a k-labelled graph, $X \subseteq [k]$ , and let ${\text{wm}}_a(G, X)$ denote the partial ${\text{Holant}}$ sum in ${\text{wm}}_a(G)$ over all $\{ 0, 1 \}$ -edge assignments of G such that within [k], those in X have exactly one incident edge assigned 1 and all nodes in $[k] \setminus X$ have no incident edges assigned 1. Then we have for any k-labelled graphs $G_1, \ldots, G_n$ ,

\begin{equation*}{\text{wm}}_a(G_1 \cdots G_n)= \sum_{X_1 \sqcup \ldots \sqcup X_n \subseteq [k]} a^{k - |X_1 \sqcup \ldots \sqcup X_n|} {\text{wm}}_a(G_1, X_1) \cdots {\text{wm}}_a(G_n, X_n).\end{equation*}

This means that $T({\text{wm}}_a, k, n)$ is the product $N_{k; a}^{\otimes n} W_{k, n; a}$ , where $N_{k; a}$ has infinitely many rows indexed by all k-labelled graphs G, and $2^k$ columns indexed by $X \subseteq [k]$ , with the entry at (G, X)

\begin{equation*}N_{k; a; G, X} = {\text{wm}}_a(G, X),\end{equation*}

and $W_{k, n; a}$ is a symmetric $\underbrace{2^k \times \ldots \times 2^k}_{n \text{ times}}$ tensor (from ${\text{Sym}}^n(\mathbb{F}^{2^k})$ ), where

\begin{equation*}W_{k, n; a; X_1, \ldots, X_n} =\begin{cases}& a^{k - |X_1 \sqcup \ldots \sqcup X_n|} \quad \text{if } X_1 \sqcup \ldots \sqcup X_n \subseteq [k], \\[3pt] & 0 \quad \text{otherwise.}\end{cases}\end{equation*}

Hence $\text{rk}_{\textsf{S}} T({\text{wm}}_a, k, n) \le \text{rk}_{\textsf{S}} W_{k, n; a}$ . We show that in fact equality holds. Consider the same family of k-labelled graphs $\{ P_X \}_{X \subseteq [k]}$ defined in the previous subsection. It is easy to see that for $X, Y \subseteq [k]$ , $N_{k; a; P_X, Y} = a^{|X| - |Y|}$ if $Y \subseteq X$ and 0 otherwise. Here by convention, $a^0 = 1$ even if $a = 0$ . Consider the rows in $N_{k; a}$ corresponding to $\{ P_X \}_{X \subseteq [k]}$ . They form the non-singular matrix ${{\Big[\begin{array}{ll} {1} & {0} \\[-8pt] {a} & {1}^{\otimes n} \end{array}\Big]}}$ if the rows and columns are ordered lexicographically for $X, Y \subseteq[k]$ . Therefore $\text{rk}_{\textsf{S}} W_{k, n; a} = \text{rk}_{\textsf{S}} \big({{\Big[\begin{array}{ll} {1} & {0} \\[-8pt] {a} & {1} \end{array}\Big]^{\otimes n}}} W_{k, n; a} \Big) \le \text{rk}_{\textsf{S}} T({\text{wm}}_a, k, n)$ and so $\text{rk}_{\textsf{S}} T({\text{wm}}_a, k, n) = \text{rk}_{\textsf{S}} W_{k, n; a}$ .

Note that for $k = 1$ , $W_{1, n; a} = \mathcal{M}_{n; a, 1} = [a, 1, 0, \ldots 0] \in {\text{Sym}}^n(\mathbb{F}^2)$ where $n \ge 1$ . Applying Lemma 4.1 with $b = 1$ , we get $\text{rk}_{\textsf{S}} W_{1, n; a} \ge n$ and therefore $\text{rk}_{\textsf{S}} T({\text{wm}}_a, 1, n) \ge n$ for $n \ge 1$ . Now if ${\text{wm}}_a$ were expressible as $\hom(\cdot, H)$ for some weighted graph H with $q = |V(H)|$ , then by Theorem 3.1, $\text{rk}_{\textsf{S}} T({\text{wm}}_a, k, n) \le q^k$ for $k, n \ge 0$ so that $\text{rk}_{\textsf{S}} T({\text{wm}}_a, 1, n) \le q$ for $n \ge 0$ . This contradicts $\text{rk}_{\textsf{S}} T({\text{wm}}_a, 1, n) \ge n$ when $n > q$ . Hence ${\text{wm}}_a$ is not expressible as a graph homomorphism function over any field.

By the same remark for Theorem 4.1, the proof for Theorem 4.2 carries over to simple graphs.

4.4 Bounded degree graphs

Fix any $d \ge 2$ . A degree-d bounded graph is a graph with maximum degree at most d. In this subsection, we investigate the expressibility of the graph parameter #Weighted-Matching $_a$ ( ${\text{wm}}_a$ ) as a GH function on bounded degree graphs. More precisely, we are interested when ${\text{wm}}_a$ is expressible as a $\hom(\cdot, H)$ with $|V(H)| = q$ on degree-d bounded graphs. For convenience, we temporarily allow vertex weights to be 0, and it will be addressed later.

Given a graph G (possibly with multiple edges but no loops), let $G^{\prime}$ be the vertex-edge incidence graph of G. The vertex set V( $G^{\prime}$ ) consists of the original vertices from V(G) on the LHS and the edges E(G) on the RHS. Let H be the weighted graph specified by vertex weights $(\alpha_1, \ldots, \alpha_q) \in \mathbb{F}^q$ and a symmetric matrix $B = (\beta_{i j}) \in \mathbb{F}^{q \times q}$ for edge weights, then for any G

(10) \begin{equation}\hom(G, H) = {\text{Holant}}(G^{\prime}; \{ \sum_{i = 1}^q \alpha_i e_{q, i}^{\otimes n} \}_{n \ge 0} \mid B),\end{equation}

where $\{ e_{q, i} \}_{i = 1}^q \in \mathbb{F}^q$ has a single 1 at the ith position and 0 elsewhere. Here $\hom(G, H)$ is expressed in (10) as a domain-q Holant sum on $G^{\prime}$ : any LHS vertex of $G^{\prime}$ of degree n is assigned the signature $\sum_{i = 1}^q \alpha_i e_{q, i}^{\otimes n}$ (which takes value $\alpha_i$ if all incident edges have value $i \in [q]$ , and 0 otherwise), and any RHS vertex of $G^{\prime}$ (an edge of G) is assigned the symmetric binary signature specified by B.

First, we show that for any $a \in \mathbb{F}$ , if ${\text{wm}}_a$ is expressible as $\hom(\cdot, H)$ with $|V(H)| = q$ on degree-d bounded simple graphs over any field $\mathbb{F}$ , then $q \ge d$ . Recall the proof from Section 4.3. This time we restrict the connection tensor $T({\text{wm}}_a, 1, d)$ (for $k = 1$ ) to the 1-labelled graphs $P_\emptyset$ and $P_{[1]}$ , which are just $K_1$ and $K_2$ without the label. The product of d-labelled graphs from $\{P_\emptyset, P_{[1]}\}$ having $\ell$ copies of $P_{[1]}$ is the star graph $S_\ell$ with one internal node labelled by 1 and $\ell$ external unlabelled nodes. All these are degree-d bounded simple graphs. Note that $T({\text{wm}}_a, 1, d)_{|\{ P_\emptyset, P_{[1]} \}^d} = {{\Big[\begin{array}{ll} {1} & {0} \\[-8pt] {a} & {1} \end{array}\Big]^{\otimes d}}} W_{1, d; a}$ . Therefore $\text{rk}_{\textsf{S}} \left( T({\text{wm}}_a, 1, d)_{|\{ P_\emptyset, P_{[1]} \}^d} \right)= \text{rk}_{\textsf{S}} \left( {{\Big[\begin{array}{ll} {1} & {0} \\[-8pt] {a} & {1} \end{array}\Big]^{\otimes d}}} W_{1, d; a} \right) = \text{rk}_{\textsf{S}} W_{1, d; a} \ge d$ , the last step is by Lemma 4.1. On the other hand, if ${\text{wm}}_a$ is expressible as $\hom(\cdot, H)$ with $|V(H)| = q$ on degree-d bounded simple graphs, then arguing similarly to the proof of Theorem 3.1 but restricting the domain of the arguments $G_i, \, 1 \le i \le d$ in (6) to $\{P_\emptyset, P_{[1]} \}$ , we have $\text{rk}_{\textsf{S}} \left( T({\text{wm}}_a, 1, d)_{|\{ P_\emptyset, P_{[1]} \}^d} \right) \le q$ so $d \le q$ . Then clearly this bound also holds if we do not allow 0-weighted vertices.

Now let $\mathbb{F}$ be infinite. By the remark after Lemma 4.2, if we are not in the exceptional case (8) (we put $n = d$ ), then for some $a_i, \alpha_i \in \mathbb{F}$ we have $\mathcal{M}_{m; a, 1} = \sum_{i = 1}^d \alpha_i (1, a_i)^{\otimes m}$ , for every $0 \le m \le d$ . Let $T \in \mathbb{F}^{d \times 2}$ be the matrix whose 1st and 2nd columns are $(1, 1, \ldots, 1)^\texttt{T}$ and $(a_1, a_2, \ldots, a_d)^\texttt{T}$ respectively. Then $e_{d, i} T = (1, a_i)$ . Define the symmetric matrix $B = (\beta_{i j}) = T T^\texttt{T} \in \mathbb{F}^{d \times d}$ . Now let H be a weighted graph on d vertices specified by $(\alpha_1, \ldots, \alpha_d) \in \mathbb{F}^d$ and $B = (\beta_{i j}) \in \mathbb{F}^{d \times d}$ . Then for any degree-d bounded graph G we have the following equality chain:

\begin{align*}{\text{wm}}_a(G) &= {\text{Holant}}(G;\! \left\{\! \mathcal{M}_{m; a, 1} \!\right\}_{0 \le m \le d}) = {\text{Holant}}(G^{\prime};\! \left\{\! \mathcal{M}_{m; a, 1} \!\right\}_{0 \le m \le d} \!\mid\! (\!=_2) ) \\[3pt] &= {\text{Holant}}(G^{\prime};\! \left\{\! \sum_{i = 1}^d \!\alpha_i (1, a_i)^{\otimes m} \!\right\}_{0 \le m \le d} \!\mid\! (\!=_2) ) = {\text{Holant}}(G^{\prime};\! \left\{\! \sum_{i = 1}^d \!\alpha_i (e_{d, i} T )^{\otimes m} \!\right\}_{0 \le m \le d} \!\mid\! (\!=_2\!) ) \\[3pt] &= {\text{Holant}}(G^{\prime};\! \left\{\! \sum_{i = 1}^d \!\alpha_i e_{d, i}^{\otimes m} T^{\otimes m} \!\right\}_{0 \le m \le d} \!\mid\! (\!=_2\!) ) = {\text{Holant}}(G^{\prime};\! \left\{\! \sum_{i = 1}^d\! \alpha_i e_{d, i}^{\otimes m} \!\right\}_{0 \le m \le d} \!\mid\! T^{\otimes 2} (=_2\!) ).\end{align*}

where the last equation moving $T^{\otimes n}$ from the LHS of the Holant problem to $T^{\otimes 2}$ in the RHS, is called a holographic transformation [Reference Cai and Chen11, Reference Valiant55] (the argument works for arbitrary fields). This follows from the associativity of the operation of tensor contraction. The Equality function $(=_2)$ is transformed to $T^{\otimes 2} (=_2)$ , which has the matrix form $T T^\texttt{T} = B$ . Hence this is precisely the function $\hom(G, H)$ . We have temporarily allowed 0-weighted vertices; but in fact by the lower bound $q \ge d$ no 0-weighted vertex exists, since otherwise by removing 0-weighted vertices we would have ${\text{wm}}_a(\cdot) = \hom(\cdot, H^{\prime})$ with fewer vertices.

The inexpressibility with $|V(H)| = 2$ for the exceptional case (8) holds even for simple graphs, by considering paths of 0, 1 or 2 edges. Also, in this case it can be easily shown that ${\text{wm}}_a = {\text{pm}}$ is expressible as $\hom(\cdot, H)$ where $|V(H)| = 3$ : this can be done similarly to the expressibility proof above via a holographic transformation (then H cannot have 0-weighted vertices).

This proves Theorem 4.3.

Note that #Perfect-Matching ( ${\text{pm}}$ ) is just the special case $a = 0$ . Hence Theorem 4.3 also holds for ${\text{pm}}$ .

4.5 Proper edge d-colourings

Next we show that the graph parameter #d-Edge-Coloring ( ${\text{ec}}_d$ ) is not expressible as a GH function over any field of characteristic 0. Given a graph G, ${\text{ec}}_d(G)$ counts the number of proper edge d-colourings in a graph, where $d \ge 1$ is the number of colours available.

Clearly, ${\text{ec}}_d$ is a multiplicative graph parameter with ${\text{ec}}(K_0) = 1$ . Since ${\text{char}}\ \mathbb{F} = 0$ , $\mathbb{F}$ is infinite. Consider $K_1$ as a 1-labelled graph and the star graph $S_d$ with one internal node labelled by 1 and d unlabelled external nodes all connected to node 1, where $d \ge 1$ . Consider the connection tensor $T({\text{ec}}_d, k, n)$ restricted to $\{ K_1, S_d \}^n$ . For $(G_1, \ldots, G_n) \in \{ K_1, S_d \}^n$ , if more than one $G_i = S_d$ then the product $G_1 \cdots G_n$ has no proper edge d-colouring because the labelled vertex has degree $\ge 2d > d$ . Then it is easy to see that the connection tensor $T({\text{ec}}_d, k, n)$ restricted to $\{ K_1, S_d \}^n$ has the form $\mathcal{M}_{1, d!} = [1, d!, 0, \ldots, 0] \in {\text{Sym}}^n(\mathbb{F}^2)$ , and $d! \ne 0$ in $\mathbb{F}$ as ${\text{char}}\ \mathbb{F} = 0$ . Therefore by Lemma 4.1, $\text{rk}_{\textsf{S}} \mathcal{M}_{1, d!} \ge n$ for $n \ge 1$ . Hence $\text{rk}_{\textsf{S}} T({\text{ec}}_d, 1, n) \ge n$ for $n \ge 1$ .

Now if ${\text{ec}}_d$ were expressible as $\hom(\cdot, H)$ for some weighted graph H with $q = |V(H)|$ , then by Theorem 3.1, $\text{rk}_{\textsf{S}} T({\text{ec}}_d, 1, n) \le q$ for $n \ge 0$ . This contradicts the upper bound when $n > q$ . Hence ${\text{ec}}_d$ is not expressible as a graph homomorphism function over any field $\mathbb{F}$ of ${\text{char}}\ \mathbb{F} = 0$ .

By the same remark for Theorem 4.1, the proof for Theorem 4.4 carries over to simple graphs.

4.6 Vertex-disjoint cycle covers

We show that the graph parameter ${\text{vdcc}}$ (#Vertex-Disjoint-Cycle-Cover) which counts the number of vertex disjoint cycle covers in a graph is not expressible as a GH function over an arbitrary field. In a multigraph without loops a cycle is a vertex disjoint closed path of length at least 2. The graph parameter ${\text{vdcc}}(G) = m \cdot 1 \in \mathbb{F}$ , where m is the number of edge subsets $E^{\prime}$ that form a vertex disjoint set of cycles that cover all vertices.

Clearly, ${\text{vdcc}}$ is a multiplicative graph parameter with ${\text{vdcc}}(K_0) = 1$ . Next, consider $K_1$ and $K_3$ as 1-labelled graphs. Note that $K_3$ is a cycle of 3 vertices. It is easy to see that the connection tensor $T({\text{vdcc}}, k, n)$ restricted to $\{ K_1, K_3 \}^n$ has the form $\mathcal{M}_{0, 1} = [0, 1, 0, \ldots, 0] \in {\text{Sym}}^n(\mathbb{F}^2)$ and therefore by Lemma 4.1, $\text{rk}_{\textsf{S}} \mathcal{M}_{0, 1} \ge n$ for $n \ge 1$ . Hence $\text{rk}_{\textsf{S}} T({\text{vdcc}}, 1, n) \ge n$ for $n \ge 1$ .

Now if ${\text{vdcc}}$ were expressible as $\hom(\cdot, H)$ for some weighted graph H with $q = |V(H)|$ , then by Theorem 3.1, $\text{rk}_{\textsf{S}} T({\text{vdcc}}, 1, n) \le q$ for $n \ge 0$ . This contradicts the upper bound when $n > q$ . Hence ${\text{vdcc}}$ is not expressible as a GH function over any field.

By the same remark for Theorem 4.1, the proof for Theorem 4.5 carries over to simple graphs.

4.7 Some previous work and model distinction

Schrijver [46] gave a beautiful characterization of a graph property $G \mapsto f(G)$ to be expressible as $Z_H(\cdot)$ in (1) with complex $\beta_{i j}$ but all $\alpha_i = 1$ . However, there are some subtle differences in the definition. In particular, somewhat deviating from the standard definition (as in [Reference Freedman, Lovász and Schrijver27] and in this paper) the graphs G are allowed to have loops (indeed multiloops and multiedges) in [46]. The criterion is expressed in terms of a sum $\sum_{P \in \Pi_{V(G)}} \mu_P f(G / P) = 0$ for all G with $|V(G)| > f(K_1)$ . Here $\Pi_{V(G)}$ is the partition lattice on V(G), $\mu_P$ is the Möbius inversion function on partitions, and the graph $G / P$ is obtained from G by condensing all vertices in P into one vertex. This condensation naturally creates loops. In [46] Schrijver carefully makes the distinction between the results in that paper with that of [Reference Freedman, Lovász and Schrijver27], and states that ‘[A]n interesting question is how these results relate’.

In [48] Schrijver gave another characterization of expressibility as $Z_H(\cdot)$ in (1) with complex $\beta_{i j}$ but all $\alpha_i = 1$ . Again, the graphs G may have multiloops and multiedges. The criterion is expressed in terms of a rank bound of the kth ‘connection matrix’ for every k. However, here the definition of the ‘connection matrix’ differs from that of [Reference Freedman, Lovász and Schrijver27]. They are defined over ‘k-marked graphs’ where the k marked vertices are not necessarily distinct. This is in contrast to ‘k-labeled graphs’ in [Reference Freedman, Lovász and Schrijver27], as well as in this paper. Thus the kth ‘connection matrix’ in [48] is a super matrix of the kth ‘connection matrix’ in [Reference Freedman, Lovász and Schrijver27], and the rank bound is a stronger requirement.

At the end of this subsection we will show that this distinction is material, by showing that the well-known hardcore gas model in statistical physics, $\sum_{\text{\scriptsize ind } I \subseteq V(G)} \lambda^{|I|}$ , where the sum is over all independent sets of G, cannot be expressed in the model discussed in [46, 48], i.e. without vertex weight. On the other hand, obviously the hardcore gas model is defined as a partition function of GH with vertex weight. In particular the rank bound for the ‘connection matrix’ in [48] must fail, while it must hold for that of [Reference Freedman, Lovász and Schrijver27], as well as for our connection tensor.

The terminology in the literature on this subject is unfortunately not uniform. In Lovász’s book [Reference Lovász39], separate from the partition function of GH as in (1), the model studied by Szegedy in [Reference Szegedy50] is called the ‘edge coloring model’. This is essentially what we called Holant problems, or edge models. The slight difference is that Holant problems allow different constraint functions from a set assigned at different vertices, while the edge colouring model studied in [Reference Szegedy50] is a special case of Holant problems where for each arity d a single symmetric vertex function $f_d$ is given and placed at all vertices of degree d.

In short, in edge models edges play the role of variables, and constraint functions are at vertices, and in vertex models vertices play the role of variables, and edges have binary constraint functions as well as vertices have unary constraint functions. Counting matchings, or perfect matchings, or valid edge colourings, or cycle covers, etc., are naturally expressible as edge models. It turns out that many orientation problems can also be expressed as edge models after a holographic transformation (by $Z = \frac{1}{\sqrt{2}} \Big[\begin{array}{ll} 1 & 1 \\[-8pt] \textit{i} & -\textit{i} \end{array}\Big]$ ). However, confusingly, the edge colouring model had also been called a vertex model in [Reference Draisma, Gijswijt, Lovász, Regts and Schrijver22]. We also note that a preliminary version of [49] appeared as [47] which used the terminology ‘vertex model’, and that was changed to ‘edge coloring model’ in the final version [49]. In particular, these papers [Reference Draisma, Gijswijt, Lovász, Regts and Schrijver22, 49, Reference Szegedy50] do not address the expressibility of counting perfect matchings in the vertex model as in (1) with arbitrary vertex and edge weights, which is the subject of the present paper. However, Szegedy [Reference Szegedy51] proved that counting perfect matchings can be expressed as the limit of the partition function of a parametrized vertex model (see below).

Now we prove that the hardcore gas model fails the expressibility criterion of Schrijver [48] as a partition function of GH without vertex weight. We observe that if we take $A = \Big[\begin{array}{ll} 1 & 1 \\[-8pt] 1 & 0 \end{array}\Big]$ indexed by $\{0, 1\}$ , which is the binary Nand function representing the independent set constraint, and the vertex weight $\alpha_0 = 1, \alpha_1 = \lambda$ , then $\sum_{\text{\scriptsize ind } I \subseteq V(G)} \lambda^{|I|}$ is clearly an instance of the expression (1). We now show that such an expression is impossible without vertex weight.

To state Schrijver’s criterion we need a few definitions. Let f be a graph property. A k-marked graph is a pair $(G, \mu)$ where $G = (V, E)$ is graph, and $\mu \colon [k] \rightarrow V$ is a function marking k (not necessarily distinct) vertices. So a vertex may have several marks; this is the key distinction with k-labelled graphs. For k-marked graphs $(G_1, \mu)$ and $(G_2, \nu)$ , the product $(G_1, \mu) (G_2, \nu)$ is the k-marked graph obtained by first taking a disjoint union and then merging all vertices of $G_1$ and $G_2$ that share a common mark. (Note that this merging process may produce multiedges and multiloops.) The kth connection matrix $C_{k, f}$ in the sense of [48] is the infinite matrix whose rows and columns are indexed by all finite k-marked graphs and the entry at $((G_1, \mu), (G_2, \nu))$ is $f((G_1, \mu) (G_2, \nu))$ . The expressibility criterion by Schrijver in [48] (Theorem 1) is that $f(\emptyset) = 1$ and for some c, ${\text{rank}}(C_{k, f}) \le c^k$ for all k.

We show that for the hardcore gas model this rank grows super exponentially. We consider the following submatrix. Let $\Pi_k$ be the set of all partitions of [k]. The rows and columns of the submatrix are indexed by the following k-marked graphs. For every $P = \{C_1, \ldots, C_s\} \in \Pi_k$ we have a k-marked graph $G_P$ with s vertices denoted by $v_1, \ldots, v_s$ , with no edges. For every $j \in [k]$ , let $C_t$ be the unique set in P that contains j, then we mark $v_t$ with j. Now suppose $P, Q \in \Pi_k$ are the indices of a row and column respectively, then the entry at (P, Q) is the hardcore gas function evaluated at the product graph $G_P G_Q$ . We observe that this product graph is just $G_{P \vee Q}$ where $P \vee Q$ is the least-upper-bound of P and Q in the lattice order of refinement: $P \le P^{\prime}$ iff P refines P $^{\prime}$ . It follows that the entry at (P, Q) is

\begin{equation*}\prod_{D \in P \vee Q} (1 + \lambda) = (1 + \lambda)^{|P \vee Q|}.\end{equation*}

It is proved in [48] (Proposition 1) for $\lambda \ne -1, 0, 1, 2, \ldots, k - 2$ , this matrix is nondegenerate. As its dimension $|\Pi_k|$ grows superexponentially in k, we see that the criterion in [48] is not satisfied. (However, it is easy to see that on d-regular graphs ( $d \ge 1$ ) the hardcore gas model is expressible as a partition function of GH without vertex weight, with the edge weight matrix $\Big[\begin{array}{ll} 1 & \lambda^{1 / d} \\[-8pt] \lambda^{1 / d} & 0 \end{array}\Big]$ , where $\lambda^{1 / d}$ can be taken to be any dth root from $\lambda$ . Basically, the entry 0 ensures that only independent sets can contribute non-zero weight to the partition function, and for each vertex assigned to be in an independent set each of the d incident edges contributes a factor $\lambda^{1 / d}$ .)

In contrast to Theorem 4.1, B. Szegedy in [Reference Szegedy51] shows that graph parameters arising from finite rank edge colouring models (defined in [Reference Szegedy51]) are in fact partition functions of singular vertex colouring models. Roughly speaking, for any fixed size (complete) graph H we consider vertex and edge weights of H that depend on a parameter t, and denoting the weighted graph by $H_t$ , we say a graph parameter is the partition function of a singular vertex colouring model if it is the limit of the graph homomorphism function $\hom(G, H_t)$ when $t \to 0$ . In particular, their connection matrices are exponentially rank bounded, a consequence by taking limits. The graph parameter #Perfect-Matching ( ${\text{pm}}$ ) is easily seen to be the partition function of a finite rank edge colouring model, so it can be expressed as a singular vertex colouring model. In fact, in [Reference Szegedy51] it is shown that ${\text{pm}}(G) = \lim_{t \to 0} \hom(G, H_t)$ , where $H_t$ has the vertex weights $(1 / t, -1 / t)$ , and edge weights $\Big[\begin{array}{ll} 1 + t^2 & 1 \\[-8pt] 1 & 1 \end{array}\Big]$ . Now we give a simple proof of this fact using holographic transformations.

We have the factorization $\Big[\begin{array}{ll} 1 + t^2 & 1 \\[-8pt] 1 & 1 \end{array}\Big] = M M^{\tt T}$ , where $M = \Big[\begin{array}{ll} 1 & t \\[-8pt] 1 & 0 \end{array}\Big]$ . A holographic transformation by M transforms the vertex weight function of a vertex of degree d to

\begin{equation*}\frac{1}{t} \left[(1,\, 0)^{\otimes d} - (0,\, 1)^{\otimes d}\right] M^{\otimes d}= \frac{1}{t} \left[(1,\, t)^{\otimes d} - (1,\, 0)^{\otimes d}\right]\end{equation*}

which is $[0, 1, t, \ldots, t^{d - 1}]$ in the symmetric signature notation, i.e. it takes values 0 and 1 on inputs of Hamming weight 0 and 1, respectively, and $t^{i - 1}$ on inputs of Hamming weight i, where $2 \le i \le d$ . Meanwhile, the edge weight function is transformed to the binary Equality (represented by the identity matrix $I_2 = \Big[\begin{array}{ll} 1 & 0 \\[-8pt] 0 & 1 \end{array}\Big]$ ). Taking limit as $t \to 0$ we get the graph parameter ${\text{pm}}$ .

5.1 The monoid and algebra of graphs

For every finite set $S \subseteq {\mathbb{Z}_{>0}}$ , we denote by $U_S$ the graph with $|S|$ nodes labelled by S and no edges. Note that $U_\emptyset = K_0$ is the empty graph.

We put all k-labelled graphs into a single structure as follows. By a partially labelled graph we mean a finite graph in which some of the nodes are labelled by distinct positive integers. (All label sets are finite.) Two partially labelled graphs are isomorphic if there is an isomorphism between them preserving all labels. For two partially labelled graphs $G_1$ and $G_2$ , let $G_1 G_2$ denote the partially labelled graph obtained by taking the disjoint union of $G_1$ and $G_2$ , and identifying the nodes with the same label; the union of the label sets becomes the labels of $G_1 G_2$ . This way we obtain a commutative monoid ${\mathcal{PLG}}$ consisting of all isomorphism classes of finite partially labelled graphs with the empty graph $U_\emptyset$ being the identity.Footnote 3 For every finite set $S \subseteq {\mathbb{Z}_{>0}}$ , we call a partially labelled graph S-labelled, if its labels form the set S. We call a partially labelled graph ${\subseteq} S$ -labelled, if its labels form a subset of S. We define ${\mathcal{PLG}}(S)$ and ${\mathcal{PLG}}_{\subseteq}(S)$ to be the subsets of ${\mathcal{PLG}}$ consisting of all isomorphism classes of S-labelled and ${\subseteq} S$ -labelled graphs, respectively. Clearly ${\mathcal{PLG}}(S) \subseteq {\mathcal{PLG}}_{\subseteq}(S)$ . Then both ${\mathcal{PLG}}(S)$ and ${\mathcal{PLG}}_{\subseteq}(S)$ are commutative monoids in ${\mathcal{PLG}}$ . ${\mathcal{PLG}}_{\subseteq}(S)$ is a submonoid of ${\mathcal{PLG}}$ with the same identity $U_\emptyset$ , while ${\mathcal{PLG}}(S)$ is a submonoid of ${\mathcal{PLG}}$ iff $S = \emptyset$ , as the identity in ${\mathcal{PLG}}(S)$ is $U_S$ .

Let $\mathcal G$ denote the monoid algebra $\mathbb{F} {\mathcal{PLG}}$ consisting of all finite formal linear combinations in ${\mathcal{PLG}}$ with coefficients from $\mathbb{F}$ ; they are called (partially labelled, $\mathbb{F}$ -)quantum graphs. Restricting the labels to precisely S or to subsets of S, we have $\mathbb{F} {\mathcal{PLG}}(S)$ or $\mathbb{F} {\mathcal{PLG}}_{\subseteq}(S)$ , the algebras of S-labelled or ${\subseteq} S$ -labelled quantum graphs, denoted by $\mathcal{G}(S)$ or $\mathcal{G}_{\subseteq}(S)$ , respectively. $\mathcal{G}(S)$ is an algebra inside $\mathcal{G}$ with $U_S$ being the multiplicative identity, and $\mathcal{G}_{\subseteq}(S)$ is a subalgebra of $\mathcal{G}$ . The empty sum is the additive identity in all.

Because many definitions, notations and statements for ${\mathcal{PLG}}(S), \mathcal{G}(S)$ and ${\mathcal{PLG}}_{\subseteq}(S), \mathcal{G}_{\subseteq}(S)$ appear similar, we will often commingle them to minimize repetitions, e.g. we use $\mathcal{G}_{(\subseteq)}(S)$ to denote either $\mathcal{G}(S)$ or $\mathcal{G}_{\subseteq}(S)$ (and the statements are asserted for both).

We can extend f to a linear map on $\mathcal{G}$ , and define an n-fold multilinear form, where $n \ge 1$ ,

\begin{equation*}\langle x_1, \ldots, x_n \rangle_{(n)} = f(x_1 \cdots x_n),\quad \text{for } x_1, \ldots, x_n \in \mathcal{G}.\end{equation*}

It is symmetric because $\mathcal{G}$ is commutative. Note that if we restrict each argument to $\mathcal{G}[k]$ and then write it with respect to the basis ${\mathcal{PLG}}[k]$ of $\mathcal{G}[k]$ , we get precisely the connection tensor (array) T(f, k, n).

Let

\begin{equation*}\mathcal{K} = \{ x \in \mathcal{G} \mid \forall y \in \mathcal{G},\, f(x y) = \langle x, y \rangle = 0 \}\end{equation*}

be the annihilator of $\mathcal{G}$ . Clearly, $\mathcal{K}$ is an ideal in $\mathcal{G}$ , so we can form the quotient algebra $\hat{\mathcal{G}} = \mathcal{G} / \mathcal{K}$ which is commutative as well. We denote its identity by $u_\emptyset = U_\emptyset + \mathcal{K}$ . More generally, we let $u_S = U_S + \mathcal{K}$ for any finite subset $S \subseteq {\mathbb{Z}_{>0}}$ . If $x \in \mathcal{K}$ , then $f(x) = f(x U_\emptyset) = 0$ and so f can also be considered as a linear map on $\hat{\mathcal{G}}$ by $f(x + \mathcal{K}) = f(x)$ for $x \in \mathcal{G}$ . For a partially labelled graph G we denote by $\hat G = G + \mathcal{K}$ the corresponding element of $\hat{\mathcal{G}}$ . More generally, we write $\hat x = x + \mathcal{K}$ for any $x \in \mathcal{G}$ . Since $\mathcal{K}$ is an ideal in $\mathcal{G}$ , the form $\langle \cdot, \ldots, \cdot \rangle_{(n)}$ on $\mathcal{G}$ induces an n-fold multilinear symmetric form on $\hat{\mathcal{G}}$ , where $n \ge 1$ ,

(11) \begin{equation}\langle x_1, \ldots, x_n \rangle_{(n)} = f(x_1 \cdots x_n),\quad \text{for } x_1, \ldots, x_n \in \hat{\mathcal{G}}.\end{equation}

We can also define

\begin{equation*}\hat{\mathcal{G}}_{(\subseteq)}(S) = (\mathcal{G}_{(\subseteq)}(S) + \mathcal{K}) / \mathcal{K} = \{ x + \mathcal{K} \mid x \in \mathcal{G}_{(\subseteq)}(S) + \mathcal{K} \}= \{ x + \mathcal{K} \mid x \in \mathcal{G}_{(\subseteq)}(S)\}.\end{equation*}

It is easy to see that $\hat{\mathcal{G}}_{\subseteq}(S)$ is a subalgebra of $\hat{\mathcal{G}}$ with the same identity $u_\emptyset = U_\emptyset + \mathcal{K}$ , and $\hat{\mathcal{G}}(S)$ is an algebra inside $\hat{\mathcal{G}}$ with the identity $u_S = U_S + \mathcal{K}$ .Footnote 4

If $S, T \subseteq {\mathbb{Z}_{>0}}$ are finite subsets, then ${\mathcal{PLG}}_{(\subseteq)}(S) \cdot {\mathcal{PLG}}_{(\subseteq)}(T) \subseteq {\mathcal{PLG}}_{(\subseteq)}(S \cup T)$ so by linearity we get $\mathcal{G}_{(\subseteq)}(S) \, \mathcal{G}_{(\subseteq)}(T) \subseteq \mathcal{G}_{(\subseteq)}(S \cup T)$ and so, going to the quotients, we have $\hat{\mathcal{G}}_{(\subseteq)}(S) \hat{\mathcal{G}}_{(\subseteq)}(T) \subseteq \hat{\mathcal{G}}_{(\subseteq)}(S \cup T)$ . Also note that for a finite $S \subseteq {\mathbb{Z}_{>0}}$ , we have ${\mathcal{PLG}}(S) \subseteq {\mathcal{PLG}}_{\subseteq}(S)$ so by linearity $\mathcal{G}(S) \subseteq \mathcal{G}_{\subseteq}(S)$ and then by going to the quotients we obtain $\hat{\mathcal{G}}(S) \subseteq \hat{\mathcal{G}}_{\subseteq}(S)$ .

Since $\mathcal{G}_{(\subseteq)}(S) \cap \mathcal{K}$ is an ideal in $\mathcal{G}_{(\subseteq)}(S)$ , we can also form another quotient algebra

\begin{equation*}\tilde{\mathcal{G}}_{(\subseteq)}(S) = \mathcal{G}_{(\subseteq)}(S) / (\mathcal{G}_{(\subseteq)}(S) {\mathcal{K}}).\end{equation*}

We have the following canonical isomorphisms between $\tilde{\mathcal{G}}_{(\subseteq)}(S)$ and $\hat{\mathcal{G}}_{(\subseteq)}(S)$ .

Proof. It follows from the Second Isomorphism Theorem for algebraic structures (see [Reference Cohn19] p. 8).

For finite $S \subseteq {\mathbb{Z}_{>0}}$ , it is convenient to treat the algebras $\tilde{\mathcal{G}}_{(\subseteq)}(S)$ and $\hat{\mathcal{G}}_{(\subseteq)}(S)$ as separate objects despite this isomorphism in Claim 1. As it will be seen later, the algebras $\tilde{\mathcal{G}}(S)$ with $S = [k]$ , where $k \ge 0$ , are naturally associated with the kth connection tensors $T(f, k, n), n \ge 0$ . Later we will need to work with various finite $S \subseteq {\mathbb{Z}_{>0}}$ simultaneously and need an ambient algebra in which dependencies between elements can be established. The algebras $\tilde{\mathcal{G}}(S)$ do not naturally possess this property as they are the quotients of the algebras $\mathcal{G}(S)$ which have no common element except 0. However, the fact that $\hat{\mathcal{G}}(S) \subseteq \hat{\mathcal{G}}$ for any finite $S \subseteq {\mathbb{Z}_{>0}}$ will allow us to establish dependencies between their elements. In other words, $\hat{\mathcal{G}}$ will serve as the ambient algebra in which further derivations will take place. Next, the $\subseteq$ -definitions will be needed to define a projection $\hat \pi_S \colon \mathcal{G} \to \mathcal{G}_{\subseteq}(S)$ (see Claims 6 and 7). This projection will be used later in the proof.

We say that elements $x, y \in \mathcal{G}\, (\text{or } \hat{\mathcal{G}})$ are orthogonal (with respect to f), if $f(x y) = 0$ and denote it by $x \perp y$ . For a subset $A \subseteq \mathcal{G} \,(\text{or } \hat{\mathcal{G}})$ , denote by $A^\perp = \{ x \in \mathcal{G} \,(\text{or } \hat{\mathcal{G}}) \mid \forall y \in A, \, x \perp y\}$ the set of those elements in $\mathcal{G} \,(\text{or } \hat{\mathcal{G}})$ orthogonal to all elements in A. Next, we say that subsets $A, B \subseteq \mathcal{G} \,(\text{or } \hat{\mathcal{G}})$ are orthogonal (with respect to f), if $x \perp y$ for all $x \in A$ and $y \in B$ . Similarly, we can talk about an element of $\mathcal{G} \,(\text{or } \hat{\mathcal{G}})$ being orthogonal to a subset of $\mathcal{G} \,(\text{or } \hat{\mathcal{G}})$ and vice versa. Note that the notion of orthogonality is symmetric since all the multiplication operations considered are commutative. From the definition, we have $\mathcal{K} = \mathcal{G}^\perp$ . Next, denote (commingling the notations $\mathcal{K}_S$ and $\mathcal{K}_{\subseteq S}$ )

\begin{equation*}\mathcal{K}_{(\subseteq) S}= \{ x \in \mathcal{G}_{(\subseteq)} (S) \mid \forall y \in \mathcal{G}_{(\subseteq)}(S), \, x \perp y \}= \mathcal{G}_{(\subseteq)}(S) \cap (\mathcal{G}_{(\subseteq)}(S))^\perp.\end{equation*}

Clearly, $\mathcal{K}_{(\subseteq) S}$ is an ideal in $\mathcal{G}_{(\subseteq)}(S)$ , so we can form yet another quotient algebra $\mathcal{G}_{(\subseteq)}(S) / \mathcal{K}_{(\subseteq) S}$ .

Next, we define an orthogonal projection from $\hat{\mathcal{G}}$ to the subalgebra $\hat{\mathcal{G}}_{\subseteq}(S)$ . We will show how to do it in a series of lemmas. Let $S \subseteq {\mathbb{Z}_{>0}}$ be finite. For every partial labelled graph G, let $G_S$ denote the ${\subseteq} S$ -labelled graph obtained by deleting the labels not in S from the vertices of G (unlabeling such vertices). Extending this map by linearity, we get a linear map $\pi_S \colon \mathcal{G} \to \mathcal{G}_{\subseteq}(S)$ . Note that $(\pi_S)_{| \mathcal{G}_{\subseteq}(S)} = {\text{id}}_{|\mathcal{G}_{\subseteq}(S)}$ . In particular, $\pi_S \colon \mathcal{G} \to \mathcal{G}_{\subseteq}(S)$ is surjective.

Claim 2. Let $S \subseteq {\mathbb{Z}_{>0}}$ be finite. If $x \in \mathcal{G}$ and $y \in \mathcal{G}_{\subseteq}(S)$ , then

\begin{equation*}f(x y) = f(\pi_S(x) y).\end{equation*}

Proof. For every $G_1 \in {\mathcal{PLG}}$ and $G_2 \in {\mathcal{PLG}}_{\subseteq}(S)$ , the graphs $G_1 G_2$ and $\pi_S(G_1) G_2$ are isomorphic as unlabelled graphs. Hence $f(G_1 G_2) = f(\pi_S(G_1) G_2)$ as f ignores labels. The claim follows by linearity.

Proof. Fix any $y \in \mathcal{G}_{\subseteq}(S)$ . By Claim 5.1, $f(x y) = f(\pi_S(x) y)$ so $f((x - \pi_S(x)) y) = 0$ . Thus $x - \pi_S(x) \in (\mathcal{G}_{\subseteq}(S))^\perp$ .

So for any $x \in \mathcal{G}$ , we can write $x = \pi_S(x) + (x - \pi_S(x))$ where $\pi_S(x) \in \mathcal{G}_{\subseteq}(S)$ , and $x - \pi_S(x) \in (\mathcal{G}_{\subseteq}(S))^\perp$ . This gives a decomposition $\mathcal{G} = \mathcal{G}_{\subseteq}(S) + (\mathcal{G}_{\subseteq}(S))^\perp$ . To get a direct sum decomposition, we need to pass to the quotient algebra. But to do so properly we need some more properties.

Proof. Clearly, $\mathcal{G}(S) \cap \mathcal{K} = \mathcal{G}(S) \cap \mathcal{G}^\perp \subseteq \mathcal{G}(S) \cap (\mathcal{G}(S))^\perp = \mathcal{K}_S$ , so we only need to prove the reverse inclusion. Let $x \in \mathcal{K}_S = \mathcal{G}(S) \cap (\mathcal{G}(S))^\perp$ . Take any $y \in \mathcal{G}$ . Then

\begin{equation*}f(x y) \stackrel{(1)}{=} f(x \pi_S(y)) \stackrel{(2)}{=} f(x U_S \pi_S(y)) \stackrel{(3)}{=} 0.\end{equation*}

Here step (1) uses Claim 2 as $x \in \mathcal{G}(S) \subseteq \mathcal{G}_{\subseteq}(S)$ ; (2) is true because $x = x U_S$ for $x \in \mathcal{G}(S)$ ; (3) is true as $\pi_S(y) \in \mathcal{G}_{\subseteq}(S)$ so $U_S \pi_S(y) \in \mathcal{G}(S)$ , and as $x \in (\mathcal{G}(S))^\perp$ . Then $x \in \mathcal{K}$ so $x \in \mathcal{G}(S) \cap \mathcal{K}$ , implying $\mathcal{K}_S \subseteq \mathcal{G}(S) \cap \mathcal{K}$ .

Proof. Clearly, $\mathcal{G}_{\subseteq}(S) \cap \mathcal{K}= \mathcal{G}_{\subseteq}(S) \cap \mathcal{G}^\perp \subseteq \mathcal{G}_{\subseteq}(S) \cap (\mathcal{G}_{\subseteq}(S))^\perp= \mathcal{K}_{{\subseteq} S}$ , so we only need to prove the reverse inclusion. Let $x \in \mathcal{K}_{{\subseteq} S} = \mathcal{G}_{\subseteq}(S) \cap (\mathcal{G}_{\subseteq}(S))^\perp$ . Take any $y \in \mathcal{G}$ . Then

\begin{equation*}f(x y) \stackrel{(1)}{=} f(x \pi_S(y)) \stackrel{(2)}{=} 0.\end{equation*}

Here step (1) uses Claim 2; (2) is true as $\pi_S(y) \in \mathcal{G}_{\subseteq}(S)$ and $x \in (\mathcal{G}_{\subseteq}(S))^\perp$ . Then $x \in \mathcal{K}$ so $x \in \mathcal{G}_{\subseteq}(S) \cap \mathcal{K}$ , implying $\mathcal{K}_{\subseteq S} \subseteq \mathcal{G}_{\subseteq}(S) \cap \mathcal{K}$ .

It follows from Claims 4 and 5 that $\mathcal{G}_{(\subseteq)}(S) / \mathcal{K}_{(\subseteq) S} = \mathcal{G}_{(\subseteq)}(S) / (\mathcal{G}_{(\subseteq)}(S) \cap \mathcal{K}) = \tilde{\mathcal{G}}_{(\subseteq)}(S)$ so the (canonical) isomorphism of algebras from Claim 1 takes the following form:

(12) \begin{equation}\tilde{\mathcal{G}}_{(\subseteq)}(S) \cong \hat{\mathcal{G}}_{(\subseteq)}(S),\quad x + \mathcal{K}_{(\subseteq) S} \mapsto x + \mathcal{K},\quad x \in \mathcal{G}_{(\subseteq)}(S).\end{equation}

Proof. Because $(\pi_S)_{| \mathcal{G}_{\subseteq}(S)} = {\text{id}}_{|\mathcal{G}_{\subseteq}(S)}$ and, by Lemma 5, $\mathcal{K}_{{\subseteq} S} = \mathcal{G}_{\subseteq}(S) \cap \mathcal{K}$ , we infer that $\pi_S(\mathcal{K}) \supseteq \mathcal{K}_{{\subseteq} S}$ . For the reverse inclusion, let $x \in \mathcal{K}$ . Fix any $y \in \mathcal{G}_{\subseteq}(S)$ . Then by Claim 2, $f(\pi_S(x) y) = f(x y) = 0$ , the last equality is true because $x \in \mathcal{K}$ . Hence $\pi_S(x) \in \mathcal{K}_{{\subseteq} S}$ so that $\pi_S(\mathcal{K}) \subseteq \mathcal{K}_{{\subseteq} S}$ .

For the linear map $\pi_S \colon \mathcal{G} \to \mathcal{G}_{\subseteq}(S) \subseteq \mathcal{G}$ by Lemmas 6 and 5, $\pi_S(\mathcal{K}) = \mathcal{K}_{{\subseteq} S} = \mathcal{G}_{\subseteq}(S) \cap \mathcal{K}$ , so that we have the well-defined linear map (which we denote by $\hat \pi_S$ )

(13) \begin{equation}\hat \pi_S \colon \hat{\mathcal{G}} \to \hat{\mathcal{G}}_{\subseteq}(S),\quad \hat \pi_S(x + \mathcal{K}) = \pi_S(x) + \mathcal{K},\quad x \in \mathcal{G}.\end{equation}

It is easy to see that $(\hat \pi_S)_{| \hat{\mathcal{G}}_{\subseteq}(S)} = {\text{id}}_{|\hat{\mathcal{G}}_{\subseteq}(S)}$ . In particular, $\hat \pi_S \colon \hat{\mathcal{G}} \to \hat{\mathcal{G}}_{\subseteq}(S)$ is surjective.

Proof. First, let $x \in \hat{\mathcal{G}}$ . Write $x = y + \mathcal{K}$ where $y \in \mathcal{G}$ . Then $\hat \pi_S(x) = \pi_S(y) + \mathcal{K}$ and $\pi_S(y) \in \mathcal{G}_{\subseteq}(S)$ . We have $x - \hat \pi_S(x) = y - \pi_S(y) + \mathcal{K}$ . By Claim 5.1, $y - \pi_S(y) \in (\mathcal{G}_{\subseteq}(S))^\perp$ so that $x - \hat \pi_S(x) \in (\hat{\mathcal{G}}_{\subseteq}(S))^\perp$ , since the bilinear form on $\mathcal{G}$ extends to $\hat{\mathcal{G}}$ in (11).

So we only need to show that $\hat{\mathcal{G}}_{\subseteq}(S) \cap (\hat{\mathcal{G}}_{\subseteq}(S))^\perp = 0 (= \{\mathcal{K}\})$ . Let z belong to this intersection. Write $z = t + \mathcal{K} = t^{\prime} + \mathcal{K}$ where $t \in \mathcal{G}_{\subseteq}(S)$ and $t^{\prime} \in (\mathcal{G}_{\subseteq}(S))^\perp$ . Then clearly $t - t^{\prime} \in \mathcal{K} \subseteq (\mathcal{G}_{\subseteq}(S))^\perp$ , and so $t = (t - t^{\prime}) + t^{\prime} \in (\mathcal{G}_{\subseteq}(S))^\perp$ . Thus $t \in \mathcal{G}_{\subseteq}(S) \cap (\mathcal{G}_{\subseteq}(S))^\perp = \mathcal{K}_{{\subseteq} S} \subseteq \mathcal{K}$ , the last inclusion holds by Claim 5. Therefore $z = t + \mathcal{K} = \mathcal{K}$ , implying that $\hat{\mathcal{G}}_{\subseteq}(S) \cap (\hat{\mathcal{G}}_{\subseteq}(S))^\perp = 0$ .

Thus Claim 7 allows us to rightfully call $\hat \pi_S \colon \hat{\mathcal{G}} \to \hat{\mathcal{G}}_{\subseteq}(S)$ an orthogonal projection of $\hat{\mathcal{G}}$ to $\hat{\mathcal{G}}_{\subseteq}(S)$ .

If $S, T \subseteq {\mathbb{Z}_{>0}}$ are finite subsets, then $\pi_S({\mathcal{PLG}}_{(\subseteq)}(T)) = {\mathcal{PLG}}_{(\subseteq)}(S \cap T)$ , where the projection is surjective because the restriction $(\pi_S)_{| {\mathcal{PLG}}_{(\subseteq)}(S \cap T)} = {\text{id}}_{|{\mathcal{PLG}}_{(\subseteq)}(S \cap T)}$ . So by linearity we get $\pi_S(\mathcal{G}_{(\subseteq)}(T)) = \mathcal{G}_{(\subseteq)}(S \cap T)$ . Going to the quotients, we conclude that $\hat \pi_S(\hat{\mathcal{G}}_{(\subseteq)}(T)) = \hat{\mathcal{G}}_{(\subseteq)}(S \cap T)$ .

Proof. For $\Rightarrow$ , it suffices to note that $\mathcal{G}_{(\subseteq)}(S)$ is closed under multiplication (in $\mathcal{G}$ ). To prove $\Leftarrow$ , note that $n - 2 \ge 0$ and for any $y \in \mathcal{G}_{(\subseteq)}(S)$ , we have $f(x y) = f(x y U^{n - 2}) = 0$ , where $U = U_\emptyset$ in the $\mathcal{G}_{\subseteq}(S)$ case and $U = U_S$ in the $\mathcal{G}(S)$ case, so $x \in \mathcal{K}_{(\subseteq) S}$ .

The primary goal of the various Claims above is to define the projection $\hat \pi_S \colon \mathcal{G} \to \mathcal{G}_{\subseteq}(S)$ to be used later and to prove Lemmas 5.1 and 5.2.

Proof. Let $x \in \hat{\mathcal{G}}_{(\subseteq)}(S)$ be an element satisfying the hypothesis of the lemma. Write $x = h_1 + \mathcal{K}$ where $h_1 \in \mathcal{G}_{(\subseteq)}(S)$ . By hypothesis, for every $y \in \hat{\mathcal{G}}_{(\subseteq)}(S)$ we have $f(x y) = 0$ . Let $h_2 \in \mathcal{G}_{(\subseteq)}(S)$ and put $y = h_2 + \mathcal{K} \in \hat{\mathcal{G}}_{(\subseteq)}(S)$ . Then $x y = h_1 h_2 + \mathcal{K}$ . By the definition of f on $\hat{\mathcal{G}}$ , $f(h_1 h_2) = f(x y) = 0$ . Hence $h_1 \in \mathcal{K}_{(\subseteq) S} \subseteq \mathcal{K}$ where the last inclusion is true by Claims 4 and 5. This implies that x is the zero element of $\hat{\mathcal{G}}_{(\subseteq)}(S)$ , which is $\mathcal{K}$ .

Lemma 5.2. Let $k, r \ge 0$ and $n \ge \max(2, r)$ . Suppose the connection tensor T(f, k, n) can be expressed as

\begin{equation*}T(f, k, n) = \sum_{i = 1}^r a_i \textbf{x}_i^{\otimes n},\end{equation*}

where $a_i \ne 0$ and $\textbf{x}_i\in \mathbb{F}^{{\mathcal{PLG}}[k]}$ are non-zero and pairwise linearly independent for $1 \le i \le r$ . Then for any $h \in \mathcal{G}[k]$ , we have $h \in \mathcal{K}_{[k]}$ iff $\textbf{x}_i(h) = 0, \, 1 \le i \le r$ .

Proof. The lemma is clearly true for $r=0$ . Let $r \ge 1$ . By Claim 8, $h \in \mathcal{K}_{[k]}$ iff $f(h h_1 \cdots h_{n - 1}) = 0$ for all $h_1, \ldots, h_{n - 1} \in \mathcal{G}[k]$ . In terms of T(f, k, n), this is equivalent to $(T(f, k, n))(h, \cdot, \ldots, \cdot) = 0$ which is the same as

\begin{equation*}\sum_{i = 1}^r a_i \, \textbf{x}_i(h) \, \textbf{x}_i^{\otimes (n - 1)} = 0.\end{equation*}

Now if $\textbf{x}_i(h) = 0, \, 1 \le i \le r$ , then this equality clearly holds. Conversely, if this equality holds, then by Lemma A.5, $a_i \textbf{x}_i(h) = 0$ but $a_i \ne 0$ so $\textbf{x}_i(h) = 0, \, 1 \le i \le r$ .

We will also need the following lemma that classifies all subalgebras of $\mathbb{F}^m$ for $m \ge 0$ . A proof is given in Subsection A.3. Recall that we allow zero algebras and require any subalgebra of an algebra to share the multiplicative identity.

Lemma 5.3. All subalgebras of $\mathbb{F}^m$ , where $m \ge 0$ , are of the following form: For some partition $[m] = \bigsqcup_{i = 1}^s \mathcal{I}_i$ , where $s \ge 0$ , and $\mathcal{I}_i \ne \emptyset$ for $i \in [s]$ , the subalgebra has equal values on each $I_i$ ,

\begin{equation*}\mathbb{F}^{(\mathcal{I}_1, \ldots, \mathcal{I}_s)}= \{ (c_1, \ldots, c_m) \in \mathbb{F}^m \mid \forall i \in [s], \ \forall j, j^{\prime} \in \mathcal{I}_i, \ c_j = c_{j^{\prime}} \}.\end{equation*}

5.2 Building an algebra isomorphism

In this part of the proof regarding f, for an arbitrary fixed $k \ge 0$ , we assume that there exists $n = n_k \ge 2$ such that $\text{rk}_{\textsf{S}} T(f, k, n) \le n - 1$ . We will pick an arbitrary such n and call it $n_k$ , and then write $r_k = \text{rk}_{\textsf{S}} T(f, k, n_k)$ . (Note that this is weaker than the uniform exponential boundedness in k for $\text{rk}_{\textsf{S}} T(f, k, n)$ in Theorem 3.2, nor do we require $f(K_0) = 1$ here.)

Then, for $n = n_k$ , we can write

(14) \begin{equation}T(f, k, n) = \sum_{i = 1}^{r_k} a_{k, n, i} \textbf{x}_{k, n, i}^{\otimes n}.\end{equation}

Then $a_{k, n, i} \ne 0$ and $0 \ne\textbf{x}_{k, n, i} \in \mathbb{F}^{{\mathcal{PLG}}[k]}$ are pairwise linearly independent for $1 \le i \le r_k$ .

Define the linear map

(15) \begin{equation}\Phi_{k, n} \colon \mathcal{G}[k] \to \mathbb{F}^{r_k},\quad \Phi_{k, n}(h)= ( \textbf{x}_{k, n, i}(h) )_{i = 1, \ldots, r_k}, \quad h \in \mathcal{G}[k].\end{equation}

We show that $\Phi_{k, n} \colon \mathcal{G}[k] \to \mathbb{F}^{r_k}$ is a surjective algebra homomorphism, after a normalization step (to be carried out later). Clearly, as $n \ge 2$ ,

\begin{equation*}h_1 \cdot h_2 \cdot h_3 \cdots h_n = (h_1 h_2) \cdot U_k \cdot h_3 \cdots h_n\end{equation*}

so

\begin{equation*}f(h_1 \cdot h_2 \cdot h_3 \cdots h_n) = f((h_1 h_2) \cdot U_k \cdot h_3 \cdots h_n)\end{equation*}

for all $h_1, \ldots, h_n \in \mathcal{G}[k]$ . (When $n = 2$ this is $f(h_1 h_2) = f((h_1 h_2) U_k)$ .) Therefore

\begin{equation*}(T(f, k, n))(h_1, h_2, \cdot, \ldots, \cdot) = (T(f, k, n))(h_1 h_2, U_k, \cdot, \ldots, \cdot)\end{equation*}

for all $h_1, h_2 \in \mathcal{G}[k]$ . In terms of the decomposition in (14), this is equivalent to

\begin{equation*}\sum_{i = 1}^{r_k} \, a_{k, n, i} \, \textbf{x}_{k, n, i}(h_1) \, \textbf{x}_{k, n, i}(h_2) \, \textbf{x}_{k, n, i}^{\otimes (n - 2)}= \sum_{i = 1}^{r_k} a_{k, n, i} \, \textbf{x}_{k, n, i}(h_1 h_2) \, \textbf{x}_{k, n, i}(U_k) \, \textbf{x}_{k, n, i}^{\otimes (n - 2)}.\end{equation*}

It follows that

\begin{equation*}\sum_{i = 1}^{r_k} a_{k, n, i} \, \left( \textbf{x}_{k, n, i}(h_1) \, \textbf{x}_{k, n, i}(h_2) - \textbf{x}_{k, n, i}(h_1 h_2) \, \textbf{x}_{k, n, i}(U_k) \right) \, \textbf{x}_{k, i}^{\otimes (n - 2)} = 0\end{equation*}

for any $h_1, h_2 \in \mathcal{G}[k]$ . The condition $r_k \le n - 1$ allows us to apply Lemma A.5. Since $a_{k, n, i} \ne 0$ for $1 \le i \le r_k$ , we obtain that

(16) \begin{equation}\textbf{x}_{k, n, i}(h_1) \, \textbf{x}_{k, n, i}(h_2) = \textbf{x}_{k, n, i}(h_1 h_2) \, \textbf{x}_{k, n, i}(U_k), \quad h_1, h_2 \in \mathcal{G}[k], \quad 1 \le i \le r_k.\end{equation}

Let $1 \le i \le r_k$ . Since $\textbf{x}_{k, n, i} \ne 0$ , there exists $h \in \mathcal{G}[k]$ such that $\textbf{x}_{k, n, i}(h) \ne 0$ . Substituting $h_1 = h_2 = h$ into (16), we infer that $\textbf{x}_{k, n, i}(U_k) \ne 0, \, 1 \le i \le r_k$ .

Therefore we can assume in (14) that each $\textbf{x}_{k, n, i}$ is normalized so that $\textbf{x}_{k, n, i}(U_k) = 1$ ( $1 \le i \le r_k$ ). Combined with this, condition (16) becomes for $1 \le i \le r_k$ ,

(17) \begin{equation}\begin{cases}\textbf{x}_{k, n, i}(h_1 h_2) &= \textbf{x}_{k, n, i}(h_1) \, \textbf{x}_{k, n, i}(h_2), \quad h_1, h_2 \in \mathcal{G}[k]; \\[4pt] \textbf{x}_{k, n, i}(U_k) &= 1;\end{cases}\end{equation}

so the linear functions $\textbf{x}_{k, n, i} \colon \mathcal{G}[k] \to \mathbb{F}, \, 1 \le i \le r_k$ are algebra homomorphisms. Then we have

\begin{gather*}\Phi_{k, n}(g h) = ( \textbf{x}_{k, n, 1}(g h), \ldots, \textbf{x}_{k, n, r_k}(g h) )= ( \textbf{x}_{k, n, 1}(g) \, \textbf{x}_{k, n, 1}(h), \ldots, \textbf{x}_{k, n, r_k}(g) \, \textbf{x}_{k, n, r_k}(h)) \\[3pt] = ( \textbf{x}_{k, n, 1}(g), \ldots, \textbf{x}_{k, n, r_k}(g) ) \cdot ( \textbf{x}_{k, n, 1}(h), \ldots, \textbf{x}_{k, n, r_k}(h) ) = \Phi_{k, n}(g) \Phi_{k, n}(h).\end{gather*}

So we have

\begin{gather*}\Phi_{k, n}(g h) = \Phi_{k, n}(g) \Phi_{k, n}(h), \quad g, h \in \mathcal{G}[k], \\[3pt] \Phi_{k, n}(U_k) = ( \textbf{x}_{k, n, 1}(U_k) \ldots, \textbf{x}_{k, n, r_k}(U_k) ) = \underbrace{(1, \ldots, 1)}_{r_k \text{ times}} \in \mathbb{F}^{r_k},\end{gather*}

and therefore $\Phi_{k, n} \colon \mathcal{G}[k] \to \mathbb{F}^{r_k}$ is an algebra homomorphism. We now prove its surjectivity. Clearly, ${\text{im}}(\Phi_{k, n})$ is a subalgebra of $\mathbb{F}^{r_k}$ . By Lemma 5.3, we may assume that ${\text{im}}(\Phi_{k, n})$ has the form $\mathbb{F}^{(\mathcal{I}_1, \ldots, \mathcal{I}_s)}$ for some partition $\{\mathcal{I}_1, \ldots, \mathcal{I}_s\}$ of $[r_k]$ . The pairwise linear independence of $\textbf{x}_{k, n, i}$ for $1 \le i \le r_k$ implies that for any $1 \le i_1 < i_2 \le r_k$ , we have $\textbf{x}_{k, n, i_1} \ne \textbf{x}_{k, n, i_2}$ , so there exists $h \in \mathcal{G}[k]$ such that $\textbf{x}_{k, n, i_1}(h) \ne \textbf{x}_{k, n, i_2}(h)$ . Since each $\mathcal{I}_i \ne \emptyset$ , it follows that $|\mathcal{I}_i| = 1$ for $1 \le i \le s$ . Hence ${\text{im}}(\Phi_{k, n}) = \mathbb{F}^{(\{ 1 \}, \ldots, \{ r_k \})} = \mathbb{F}^{r_k}$ . We have shown that $\Phi_{k, n} \colon \mathcal{G}[k] \to \mathbb{F}^{r_k}$ is surjective.

This results in the following lemma.

Next, by $r_k \le n - 1$ and $n \ge 2$ , clearly $n \ge \max(2, r_k)$ , so Lemma 5.2 applies. So we have

\begin{equation*}\ker \Phi_{k, n} = \{ h \in \mathcal{G}[k] \mid \textbf{x}_{k, n, i}(h) = 0,\, 1 \le i \le r_k \} = \mathcal{K}_{[k]},\end{equation*}

where the first equality is by the definition of $\Phi_{k, n}$ , and the second equality is by Lemma 5.2. Note that by Claim 4, we have $\mathcal{K}_{[k]} =\mathcal{G}[k] \cap \mathcal{K}$ . Then $\Phi_{k, n} \colon \mathcal{G}[k] \to \mathbb{F}^{r_k}$ factors through $\mathcal{G}[k] / \ker \Phi_{k, n} = \mathcal{G}[k] / \mathcal{K}_{[k]} = \tilde{\mathcal{G}}[k]$ , inducing an algebra isomorphism

\begin{equation*}\tilde \Phi_{k, n} \colon \tilde{\mathcal{G}}[k] \to \mathbb{F}^{r_k}, \quad \tilde \Phi_{k, n}(h + \mathcal{K}_{[k]})= ( \textbf{x}_{k, n, 1}(h), \ldots, \textbf{x}_{k, n, r_k}(h) ),\quad h \in \mathcal{G}[k].\end{equation*}

It follows that $\dim \tilde{\mathcal{G}}[k] = \dim \mathbb{F}^{r_k} = r_k$ . In particular, $\tilde{\mathcal{G}}[k]$ is a finite dimensional algebra. Applying Lemma A.3, we get $\dim \tilde{\mathcal{G}}[k] = \dim {\text{span}} \{ \textbf{x}_{k, n, i} \}_{i = 1}^{r_k}$ . Then $\dim {\text{span}} \{ \textbf{x}_{k, n, i} \}_{i = 1}^{r_k} = r_k$ implying that $\textbf{x}_{k, n, i}$ , $1 \le i \le r_k$ , are linearly independent. (Note that we started off only assuming they are non-zero and pairwise linearly independent.) We formalize some of the results obtained above.

Lemma 5.5. Let $k \ge 0$ . Assume there exists $n = n_k \ge 2$ such that $r_k = \text{rk}_{\textsf{S}} T(f, k, n_k) \le n_k - 1$ . Then the constructed map

\begin{equation*}\tilde \Phi_{k, n} \colon \tilde{\mathcal{G}}[k] \to \mathbb{F}^{r_k},\quad \tilde \Phi_k(h + \mathcal{K}_{[k]}) = ( \textbf{x}_{k, n, 1}(h), \ldots, \textbf{x}_{k, n, r_k}(h) ),\quad h \in \tilde{\mathcal{G}}[k]\end{equation*}

is an algebra isomorphism and $\dim \tilde{\mathcal{G}}[k] = r_k$ .

Composing $\tilde \Phi_k \colon \tilde{\mathcal{G}}[k] \to \mathbb{F}^{r_k}$ with the canonical algebra isomorphism between $\tilde{\mathcal{G}}[k]$ and $\hat{\mathcal{G}}[k]$ given in (12), we have an algebra isomorphism $\hat \Phi_k \colon \hat{\mathcal{G}}[k] \to \mathbb{F}^{r_k}$ . In particular, $\dim \hat{\mathcal{G}}[k] = \dim \tilde{\mathcal{G}}[k] = r_k$ .

Corollary 5.6. With the same assumption as in Lemma 5.5, the map

\begin{equation*}\hat \Phi_{k, n} \colon \hat{\mathcal{G}}[k] \to \mathbb{F}^{r_k},\quad \hat \Phi_{k, n}(h + \mathcal{K}) = ( \textbf{x}_{k, n, 1}(h), \ldots, \textbf{x}_{k, n, r_k}(h) ),\quad h \in \hat{\mathcal{G}}[k]\end{equation*}

is an algebra isomorphism and $\dim \hat{\mathcal{G}}[k] = r_k$ .

Note that if $S \subseteq {\mathbb{Z}_{>0}}$ is finite and $|S| = k$ , there are natural isomorphisms between $\tilde{\mathcal{G}}(S)$ and $\tilde{\mathcal{G}}[k]$ and also between $\hat{\mathcal{G}}(S)$ and $\hat{\mathcal{G}}[k]$ , both resulting from any bijective map between S and [k]. As a result, we conclude the following.

5.3 One n implies for all n

Let $n_k$ retain the same meaning as in Lemma 5.5, and let $r = r_k = \text{rk}_{\textsf{S}}(T(f, k, n_k))$ . For any $h \in \mathcal{G}[k]$ , clearly $h = h U_k^{n_k - 1}$ so $f(h) = f(h U_k^{n_k - 1})$ . As $\textbf{x}_{k, n_k, i}(U_k) = 1, \, 1 \le i \le r$ ,

\begin{equation*}\begin{split}f(h) &= f(h U_k^{n_k - 1}) = (T(f, k, n_k))(h, U_k, \ldots, U_k)= \sum_{i = 1}^r a_{k, n_k, i} (\textbf{x}_{k, n_k, i}(U_k))^{n_k - 1} \textbf{x}_{k, n_k, i}(h) \\&= \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}(h)\end{split}\end{equation*}

for any $h \in \mathcal{G}[k]$ . Hence $f_{|\mathcal{G}[k]} = \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}$ , i.e. $f_{|\mathcal{G}[k]}$ is a linear combination of r algebra homomorphisms $\textbf{x}_{k, n_k, i} \colon \mathcal{G}[k] \to \mathbb{F}$ , for $1 \le i \le r$ . In particular, applying to the product $h_1 \cdots h_m$ , for any $m \ge 0$ and any $h_1, \ldots, h_m \in \mathcal{G}[k]$ (note that this m is arbitrary, not constrained by $n_k$ ),

\begin{equation*}f(h_1 \cdots h_m) = \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}(h_1 \cdots h_m)= \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}(h_1) \cdots \textbf{x}_{k, n_k, i}(h_m).\end{equation*}

(When $m = 0$ , we view it as $f(U_k) = \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}(U_k) = \sum_{i = 1}^r a_{k, n_k, i}$ .) Hence

(18) \begin{equation}T(f, k, m) = \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}^{\otimes m}\end{equation}

for all $m \ge 0$ . (When $m = 0$ , (18) is still valid as $T(f, k, 0) = f(U_k) = \sum_{i = 1}^r a_{k, n_k, i} = \sum_{i = 1}^r a_{k, n_k, i} \textbf{x}_{k, n_k, i}^{\otimes 0}$ where the last equality is true as $\textbf{x}_{k, n_k, i}^{\otimes 0} = 1$ .)

As shown before, $\textbf{x}_{k, n_k, i}$ where $1 \le i \le r$ are linearly independent. Then by Lemma A.4 applied to (18), where the parameter m in the tensor power is at least 2, we get $\text{rk}_{\textsf{S}}(T(f, k, m)) = r$ for all $m \ge 2$ ; and the decomposition (18) is actually unique up to a permutation for $m \ge 3$ .

To summarize, this leads to the following.

We remark that this is a non-trivial statement: The existence of some $n_k$ has produced a uniform expression for the tensors T(f, k, n) all the way down to $n = 0$ .

5.4 Putting things together

From now on, we assume that for every $k \ge 0$ , there exist some $n = n_k \ge 2$ such that $\text{rk}_{\textsf{S}} T(f, k, n) \le n - 1$ . For every $k \ge 0$ , we pick an arbitrary such n, call it $n_k$ , and let $r_k = \text{rk}_{\textsf{S}} T(f, k, n_k)$ .

Having developed the theory in a more general setting, we can now follow the proof in [Reference Freedman, Lovász and Schrijver27] closely. As mentioned before, developing this theory in a more general setting is necessary because $f(K_1) = 0$ is possible which makes the normalization step from [Reference Freedman, Lovász and Schrijver27] infeasible. Now the main difference from [Reference Freedman, Lovász and Schrijver27] starting from this point is that many of our derivations will additionally contain units of the form $u_S$ for various finite $S \subseteq {\mathbb{Z}_{>0}}$ because we cannot ensure that $u_S = u_\emptyset$ . We will be interested in the idempotent elements of $\hat{\mathcal{G}}$ . For two elements $p, q \in \hat{\mathcal{G}}$ , we say that q resolves p, if $p q = q$ . We also say equivalently p is resolved by q. It is clear that the binary relation resolves is antisymmetric and transitive and, when restricted to idempotents, reflexive. Furthermore, it is easy to see that the binary relation resolves on $\hat{\mathcal{G}}$ has the following properties:

1. 1. The idempotent $0 = \mathcal{K}$ resolves everything and $1 = u_\emptyset = U_\emptyset + \mathcal{K} = K_0 + \mathcal{K}$ is resolved by everything.

2. 2. If $a b = 0$ and c resolves both a and b, then $c = 0$ .

3. 3. If a resolves b, then c resolves a iff c resolves a b.

In the algebra $\mathbb{F}^r$ ( $r \ge 0$ ), the idempotents are 0-1 tuples in $\mathbb{F}^r$ , and for idempotents $q = (q_1, \ldots, q_r)$ and $p = (p_1, \ldots, p_r)$ , q resolves p iff $q_i = 1$ implies $p_i = 1$ .

Let S be a finite subset of ${\mathbb{Z}_{>0}}$ with $|S|=k$ , and set $r = r_k$ as above. By Corollary 5.7, $\hat{\mathcal{G}}(S) \cong \mathbb{F}^r$ as algebras, so $\hat{\mathcal{G}}(S)$ has a (uniquely determined idempotent) basis $\mathcal{P}_S = \{p_1^S, \ldots, p_r^S\}$ such that $(p_i^S)^2 = p_i^S$ and $p_i^S p_j^S = 0$ for $i \ne j$ . These correspond to the canonical basis $\{ e_i \}_{1 \le i \le r}$ of $\mathbb{F}^r$ under this isomorphism. For $i \ne j$ , we have $\langle p_i^S, p_j^S \rangle = f(p_i^S p_j^S) = 0$ . Furthermore, for all $1 \le i \le r$ ,

(20) \begin{equation}f(p_i^S) = f((p_i^S)^2) = \langle p_i^S, p_i^S \rangle \ne 0,\end{equation}

otherwise $\hat{\mathcal{G}}(S)$ contains a non-zero element orthogonal to $\hat{\mathcal{G}}(S)$ with respect to the bilinear form $\langle \cdot, \cdot \rangle$ restricted to $\hat{\mathcal{G}}(S) \times \hat{\mathcal{G}}(S)$ , contradicting Lemma 5.1. We will call the elements $p_i^S \in \mathcal{P}_S$ basic idempotents.

We denote by $\mathcal{P}_{T, p}$ the set of all idempotents in $\mathcal{P}_T$ that resolve a given element $p \in \hat{\mathcal{G}}$ . If $p \in \mathcal{P}_S$ and $S \subset T$ and $|T| = |S| + 1$ , then the number of elements in $\mathcal{P}_{T, p}$ will be called the degree of $p \in \mathcal{P}_S$ , and denoted by $\deg(p)$ . Obviously, this value is independent of which $(|S| + 1)$ -element superset T of S we take.

For any $q \in \hat{\mathcal{G}}(T)$ , we have $q u_{T \setminus S} = q$ . (Here by definition, $u_{T \setminus S} = U_{T \setminus S} + \mathcal{K} \in \hat{\mathcal{G}}(T \setminus S) \subseteq \hat{\mathcal{G}}(T) \subseteq \hat{\mathcal{G}}$ .) It follows that for any $S \subseteq T$ and $p \in \hat{\mathcal{G}}$ , we have q resolves p iff q resolves $p u_{T \setminus S}$ , since $q p = q u_{T \setminus S} p = q p u_{T \setminus S}$ . It is also important to point out that an element in $\hat{\mathcal{G}}(S)$ is an idempotent in $\hat{\mathcal{G}}(S)$ iff it is an idempotent in $\hat{\mathcal{G}}$ .

Claim 9. Let x be any idempotent element of $\hat{\mathcal{G}}(S)$ . Then x is the sum of exactly those idempotents in $\mathcal{P}_S$ that resolve it,

\begin{equation*}x = \sum_{p \in \mathcal{P}_{S, x}} p.\end{equation*}

Proof. Let $k = |S|$ , and $r = r_k$ . By the isomorphism $\hat{\mathcal{G}}(S) \cong \mathbb{F}^r$ , every 0-1 tuple $x = (x_1, \ldots, x_r) \in \mathbb{F}^r$ is the sum $\sum_{i = 1}^r x_i e_i$ .

In particular

(21) \begin{equation}u_S = \sum_{p \in \mathcal{P}_S} p,\end{equation}

since $u_S \in \hat{\mathcal{G}}(S)$ corresponds to the all-1 tuple in $\mathbb{F}^r$ .

Proof. Let $k^{\prime} = |T|$ and $r^{\prime} = r_{k^{\prime}}$ . Consider the idempotents $p u_{T \setminus S}$ (which could be 0) under the isomorphism $\hat{\mathcal{G}}(T) \cong \mathbb{F}^{r^{\prime}}$ , where $p \in \mathcal{P}_S$ . We can write the 0-1 tuple corresponding to $pu_{T \setminus S}$ in $\mathbb{F}^{r^{\prime}}$ as the sum of those canonical basis 0-1 vectors. Recall that for any $q \in \mathcal{P}_T$ , q resolves p iff q resolves $p u_{T \setminus S}$ . Note that $p u_{T \setminus S}$ must have disjoint positions with entry 1 for distinct $p \in \mathcal{P}_S$ , and the sum $\sum_{p \in \mathcal{P}_S} p u_{T \setminus S} = u_S u_{T \setminus S} = u_T$ is the all-1 tuple in $\mathbb{F}^{r^{\prime}}$ . Thus each $q \in \mathcal{P}_T$ resolves exactly one $p \in \mathcal{P}_S$ .

Claim 11. Let T and U be finite sets, and let $S = T \cap U$ . If $x \in \hat{\mathcal{G}}(T)$ and $y \in \hat{\mathcal{G}}(U)$ , then

\begin{equation*}f(x y) = f(\hat \pi_S(x) y).\end{equation*}

Proof. For every T-labelled graph $G_1$ and U-labelled graph $G_2$ , the graphs $G_1 G_2$ and $\pi_S(G_1) G_2$ are isomorphic as unlabelled graphs. Hence $f(G_1 G_2) = f(\pi_S(G_1) G_2)$ as f ignores labels. Then we extend the equality from ${\mathcal{PLG}}$ by linearity to $\mathcal{G}$ and after that proceed to the quotient $\hat{\mathcal{G}} = \mathcal{G} / \mathcal{K}$ using the definition of f on $\hat{\mathcal{G}}$ .

We remarked in (20) that $f(p) \ne 0$ for any $p \in \mathcal{P}_S$ .

Claim 12. Let $S \subseteq T$ be two finite sets. If $q \in \mathcal{P}_T$ resolves $p \in \mathcal{P}_S$ , then

\begin{equation*}\hat \pi_S(q) = \frac{f(q)}{f(p)} p.\end{equation*}

Proof. Note that $q \in \hat{\mathcal{G}}(T)$ . Since $S \subseteq T$ , it follows that $\hat \pi_S(q) \in \hat{\mathcal{G}}(S)$ . Because the only element from $\hat{\mathcal{G}}(S)$ orthogonal to $\hat{\mathcal{G}}(S)$ with respect to the dot product $\langle \cdot, \cdot \rangle$ restricted to $\hat{\mathcal{G}}(S) \times \hat{\mathcal{G}}(S)$ is 0 (by Lemma 5.1), it suffices to show that both sides give the same dot product with every basis element in $\mathcal{P}_S$ . For any $p^{\prime} \in \mathcal{P}_S \setminus \{ p \}$ , we have $p^{\prime} p = 0$ so $p^{\prime} q = p^{\prime} p q = 0$ . By Claim 11, this implies that