Proof of Theorem 10. The overall plan is to show that as long as the hypothesised condition on $\boldsymbol{\lambda }$ holds, we have that if $M$ is any non-empty admissible subhypergraph of $G$ and $v$ is any vertex in $V_M$ then

\begin{equation*} \left |\frac {Z_M(\boldsymbol {\lambda })}{Z_{M-v}(\boldsymbol {\lambda })}\right | \geq 1-s \end{equation*}

where $s\lt 1$ may be chosen to be independent $M$ and $v$ . Iterating this over an ordering $v_1, v_2, \ldots, v_{|V(G)|}$ of the vertices of $G$ then yields, via a telescoping product,

(2.5) \begin{equation} |Z_M(\boldsymbol{\lambda })| = \prod _{i=1}^{|V(G)|} \left |\frac{Z_{G-\{v_1, \ldots, v_{i-1}\}}(\boldsymbol{\lambda })}{Z_{G-\{v_1, \ldots, v_i\}}(\boldsymbol{\lambda })}\right | \geq (1-s)^{|V(G)|} \gt 0. \end{equation}

To do this, we will need to control how the independence polynomial changes in going from $M$ to $M-v$ . It will turn out that in tandem with this we will also need to control how it changes in going from $M$ to $M+A$ . To achieve both of these tasks, we will carry out a parallel induction.

To state the induction hypothesis precisely, we introduce some notation. For an admissible subhypergraph $M=(V_M,E_M)$ , and a non-empty subset $A \subseteq V_M$ with $|A|=a$ , we define, for each positive integer $b$ , the quantity

\begin{equation*} n_{ab} = \#\left ({S \subseteq V_M, S \cap A = \emptyset, |S|=b \text { such that } \atop S \cup T \in E_M \text { for some non-empty } T \subseteq A}\right ) \end{equation*}

when $a=1$ or $a\gt 1$ and $b\gt 1$ . Note that if $A=\{v\}$ (so $a=1$ ) then $n_{ab}$ is simply the number of edges in $M$ of size $b+1$ that include $v$ .

For $b=1$ and $a\gt 1$ , we slightly modify this to

\begin{equation*}n_{a1}=\#\left ({S \subseteq V_M, S \cap A = \emptyset, |S|=1 \text { such that } \atop S \cup T \in E_M \text { for some non-empty } T \subseteq A}\right )+|A|,\end{equation*}

i.e., $n_{a1}=\#\left (2\text{-edges incident to }A\right )+|A|$ .

In the notation we suppress the dependence of $n_{ab}$ on $M$ and $A$ .

We are now ready to state the main result of this section precisely. Let $R\gt 0$ and $s_j \in (0,1)$ , $j=1, 2, \ldots$ be constants that satisfy the following conditions for every admissible subhypergraph $M=(V_M,E_M)$ of $G$ :

• • For every $A \subseteq V_M$ with $|A|=1$ (i.e., every $v \in V_M$ ) we have

  (2.6) \begin{equation} R \leq s_1 \prod _{i \geq 1} (1-s_i)^{n_{1i}}, \end{equation}

• • and for every $A \subseteq M$ with $|A|=j\geq 2$ such that $M+A$ is admissible we have

  (2.7) \begin{equation} R^j \leq s_j \prod _{i \geq 1} (1-s_i)^{n_{ji}}. \end{equation}

Here, finally, is the statement that we will prove. Let $\boldsymbol{\lambda }$ be such that $|w_x| \leq R$ for all $x \in V$ . Let $M=(V_M,E_M)$ be an admissible subhypergraph of $G$ . For $v \in V_M$ we have

(2.8) \begin{equation} |Z_M(\boldsymbol{\lambda })|\geq (1-s_1)|Z_{M-v}(\boldsymbol{\lambda })| \text{(vertex deletion identity)} \end{equation}

while for $A \subseteq M$ with $|A|=j\geq 2$ such that $M+A$ is admissible we have

(2.9) \begin{equation} |Z_M(\boldsymbol{\lambda })|\geq (1-s_j)|Z_{M+A}(\boldsymbol{\lambda })| \text{(edge addition identity).} \end{equation}

Before proving (2.8) and (2.9), we show that it is enough to complete the proof of Theorem 10. The key observations are that for any admissible $M$ and $v \in V_M$ we have

\begin{equation*} \sum _{b \geq 1} n_{1b} = \textrm {deg}_M(v) \leq \Delta \end{equation*}

and for any admissible $M$ and $A \subseteq M$ with $|A|=a\geq 2$ such that $M+A$ is admissible we have

\begin{equation*} \sum _{b \geq 1} n_{ab} \leq a\Delta . \end{equation*}

Indeed, consider $x \in A$ . This is in at most $\Delta -1$ edges of $M$ , since $M+A$ is admissible which means $x$ has degree at most $\Delta$ in $M+A$ . Thus there are at most $\Delta -1$ $S$ ’s disjoint from $A$ such that $S \cup T \in E_M$ with $T \subseteq A$ and $x \in T$ . Since $|A|=a$ , it follows that there are at most $a(\Delta -1)$ $S$ ’s disjoint from $A$ such that $S \cup T \in E_M$ with $T \subseteq A$ and $T \neq \emptyset$ . We now add the $a$ vertices of $|A|$ deleted to get $a\Delta$ as a bound.

It follows that if we let all $s_i$ ’s have a common value, $s$ say, then (2.6) and (2.7) are implied by:

\begin{equation*} R^j \leq s (1-s)^{j\Delta } \text { or } R \leq s^{1/j}(1-s)^\Delta \end{equation*}

for $j=1, 2, \ldots\,$ . Since $s \in (0,1)$ , all of these are implied by $R \leq s(1-s)^\Delta$ . We may now take $s=1/(\Delta +1)$ , leading to $R=\Delta ^\Delta/(\Delta +1)^{(\Delta +1)}$ . Theorem 10 now follows via the telescoping product (2.5).

Now, to prove (2.8) and (2.9), we will proceed by induction on $|V_M|$ , showing firstly that an $|M|=n+1$ instance of (2.8) can be deduced from (2.8) and (2.9) for $|M|\leq n$ , and secondly that an $|M|=n+1$ instance of (2.9) can be deduced from (2.8) for $|M|\leq n+1$ and (2.9) for $|M|\leq n$ .

As a base case we take $M=\varnothing$ , where both (2.8) and (2.9) are vacuously true.

For the first of the induction steps, assume that (2.8) and (2.9) both hold for all admissible $M'$ with $|V_{M'}|\leq n$ , and let $M=(V_M,E_M)$ be an admissible subhypergraph of $G$ with $|V_M|=n+1$ vertices. Let $v$ be any vertex of $M$ . To show that (2.8) holds, we begin by applying identity (2.3):

\begin{equation*} Z_M(\boldsymbol {\lambda })=Z_{M-v}(\boldsymbol {\lambda })+\lambda _vZ_{M/v}(\boldsymbol {\lambda }). \end{equation*}

Via the reverse triangle inequality and the assumption $|\lambda _v|\leq R$ , this gives

(2.10) \begin{equation} |Z_M(\boldsymbol{\lambda })|\geq |Z_{M-v}(\boldsymbol{\lambda })|-R|Z_{M/v}(\boldsymbol{\lambda })|. \end{equation}

To extract our desired lower bound on $|Z_M(\boldsymbol{\lambda })|$ we will use the induction hypotheses (for admissible subhypergraphs with at most $n$ vertices) to bound $|Z_{M/v}(\boldsymbol{\lambda })|$ in terms of $|Z_{M-v}(\boldsymbol{\lambda })|$ . Observe that $M/v$ may be obtained from $M-v$ by a sequence of at most $\Delta$ many vertex deletions and edge additions, and that at each step, we obtain an admissible subhypergraph of $G$ (allowing us to repeatedly apply the induction hypotheses). More explicitly, beginning with $M-v$ , we first delete all 2-neighbours of $v$ (i.e., vertices $u$ such that $\{u,v\} \in E_{M-v}$ ) to obtain $M-(\{v\} \cup C(v))$ , a total of $|C(v)| (=n_{11})$ many applications of vertex deletion. By (2.8), this gives

(2.11) \begin{equation} |Z_{M-v}|\geq (1-s_1)^{n_{11}}\cdot |Z_{M-(\{v\} \cup C(v))}|. \end{equation}

Then, to obtain $M/v$ from $M-(\{v\} \cup C(v))$ , we add a $j$ -edge $A$ for each $\{v\} \cup A \in E_M$ for all $j \geq 2$ . (Since $v$ is not a vertex in $M-(\{v\} \cup C(v))$ , each of these edge additions produces an admissible subhypergraph of $G$ ). The number of $j$ -edge additions performed at this step is at most the number of $(j+1)$ -edges in $M$ that include $v$ , which recall we have denoted $n_{1j}$ . By (2.9), this gives

(2.12) \begin{equation} |Z_{M-(\{v\}\cup C(v))}|\geq \prod _{i\geq 2}(1-s_i)^{n_{1i}}\cdot |Z_{M/v}|. \end{equation}

So combining (2.11) and (2.12), we obtain the following relationship between $|Z_{M/v}(\boldsymbol{\lambda })|$ and $|Z_{M-v}(\boldsymbol{\lambda })|$ :

(2.13) \begin{equation} |Z_{M-v}|\geq \prod _{i\geq 1}(1-s_i)^{n_{1i}}\cdot |Z_{M/v}|. \end{equation}

And combining (2.13) with (2.10), we see that

(2.14) \begin{align} |Z_M(\boldsymbol{\lambda })| &\geq |Z_{M-v}(\boldsymbol{\lambda })|\left (1-\frac{R}{\prod _{i\geq 1}(1-s_{i})^{n_{1i}}}\right ). \end{align}

Thus, our induction hypothesis (2.8) is satisfied for $M$ (and $v$ ) as long as the inequality

(2.15) \begin{equation} \left (1-\frac{R}{\prod _{i\geq 1}(1-s_{i})^{n_{1i}}}\right )\geq (1-s_1) \end{equation}

is satisfied, but this is equivalent to (2.6).

Now let us turn to the second part of the induction step, verifying (2.9). Again, let $M$ be any admissible subhypergraph of $G$ with $n+1$ vertices and consider any edge $\{v_1,\dots,v_{\ell -j},x_1,\dots,x_j\}$ in $G$ with $x_1,\dots,x_j\in V_M$ and $v_1,\dots,v_{\ell -j}\not \in V_M$ . (So here we are taking $A=\{x_1,\dots,x_j\}$ ; this will be convenient as we will be dealing with the elements of $A$ one after another.) First we dispense with a somewhat trivial case: if $\{x_1,\dots,x_j\}$ contains a hyperedge of $M$ then (2.9) is automatically satisfied, since $Z_M(\boldsymbol{\lambda }) = Z_{M+\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })$ , and so $|Z_M(\boldsymbol{\lambda })| \geq (1-s_j)\, |Z_{M+\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|$ for any $s_j\in (0,1)$ . So from here on we may assume that no $e\in E_M$ is contained in $\{x_1,\dots,x_j\}$ .

We begin by applying identity (2.4) to get

\begin{equation*} Z_M(\boldsymbol {\lambda })=Z_{M+\{x_1,\dots,x_j\}}(\boldsymbol {\lambda })+\lambda _{x_1}\dots \lambda _{x_j}Z_{M/\{x_1,\dots,x_j\}}(\boldsymbol {\lambda }). \end{equation*}

As before, using the reverse triangle inequality and noting that $|\lambda _{x_i}|\leq R$ for all $i$ , we obtain

(2.16) \begin{equation} |Z_M(\boldsymbol{\lambda })|\geq |Z_{M+\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|-R^j|Z_{M/\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|. \end{equation}

To obtain a lower bound on $|Z_M(\boldsymbol{\lambda })|$ , we will apply our induction hypotheses to bound $|Z_{M/\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|$ in terms of $|Z_{M+\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|$ . Notice that $M/\{x_1,\dots,x_j\}$ may be obtained from $M+\{x_1,\dots,x_j\}$ by the following sequence of steps: Starting from $M+\{x_1,\dots,x_j\}$ , we perform the vertex deletion operation $j+|C(\{x_1,\dots,x_j\})| (=n_{j1})$ many times. Then we add (as edges) all the $i$ -sets $\{a_1,\ldots, a_i\}$ such that $a_1, \ldots, a_i \in V_M$ and $\{x_{k_1},\dots, x_{k_t},a_1,\ldots, a_i\} \in E_M$ for some indices $k_1,\dots, k_t$ ; note that there are $n_{ji}$ such $i$ -sets for each possible $i$ .

Now by (2.8) and (2.9), we have

(2.17) \begin{equation} |Z_{M+\{x_1,\dots,x_j\}}|\geq \prod _{i\geq 1}(1-s_{i})^{n_{ji}}\cdot |Z_{M/\{x_1,\dots,x_j\}}|. \end{equation}

(Notice that the first application of (2.8) in this sequence is to delete a vertex from $M+\{x_1,\dots,x_j\}$ , which has $n+1$ vertices. This is a valid application, since we are assuming the $n+1$ case of (2.8) of the induction hypothesis.)

Combining (2.17) with (2.16) yields

\begin{align*} |Z_M(\boldsymbol{\lambda })| &\geq |Z_{M+\{x_1,\dots,x_j\}}|\left (1-\frac{R^j}{\prod _{i\geq 1}(1-s_i)^{n_{ji}}}\right ). \end{align*}

So we obtain (2.9) for $M$ as long as

\begin{equation*} \left (1-\frac {R^j}{\prod _{i\geq 1}(1-s_i)^{n_{ji}}}\right )\geq (1-s_j), \end{equation*}

which is equivalent to (2.7). This completes the induction.

Proof. As in the proof of Theorem 10, we will bound $|Z_G(\boldsymbol{\lambda })|$ by measuring the effect that removing a single vertex can have on the independence polynomial (and iterating this procedure until we reach the empty graph). In this proof we will carefully control which vertices are removed and which subhypergraphs can be produced as a result. We will first give the precise statements (3.4) and (3.5) that will be proved by induction (which bound the change in the independence polynomial after removing a vertex or adding an edge), before showing how the theorem follows, and finally establishing (3.4) and (3.5).

Let $M$ be any admissible subhypergraph of $G$ in which every connected component has at most one edge of size strictly less than $k$ . Let $R\gt 0$ and $s_1,\dots, s_{k-1}\in (0,1)$ be any constants satisfying

(3.2) \begin{equation} R \leq s_1 (1-s_j)\left (1-s_{k-1}\right )^{\Delta } \end{equation}

for all $2\leq j\leq k-2$ , and

(3.3) \begin{equation} R^j\leq s_j(1-s_1)^{j}(1-s_{k-1})^{j(\Delta -1)} \end{equation}

for all $2\leq j\leq k-1$ ; and let $\boldsymbol{\lambda }$ satisfy $|\lambda _v|\leq R$ for all $v\in V$ . We will inductively prove the following two bounds; notice that both bounds are proved only under careful structural assumptions.

• • First, let $v$ be any vertex of $M$ contained in an edge of size less than $k$ , or let $v$ be any vertex in a connected component of $M$ where all edges have size $k$ . Then we show that

  (3.4) \begin{equation} |Z_M(\boldsymbol{\lambda })|\geq (1-s_1)\,|Z_{M-v}(\boldsymbol{\lambda })|. \end{equation}

• • Second, let $2\leq j \leq k-1$ , and let $x_1,\dots, x_j$ be any vertices of $M$ that are (respectively) in components of $M$ where all edges have size $k$ , and that are in some $k$ -edge $\{v_1,\dots, v_{k-j},x_1,\dots, x_j\}$ of $G$ with $v_1,\dots, v_{k-j}\not \in V(M)$ . Then we show that

  (3.5) \begin{equation} |Z_{M}(\boldsymbol{\lambda })|\geq (1-s_j)\,|Z_{M+\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|. \end{equation}

Before proving (3.4) and (3.5), we show how an iterative application of (3.4) completes the proof of Theorem 12. Let $\boldsymbol{\lambda }$ be such that $|\lambda _v|\leq R$ for all $v\in V$ , where $R$ and $s_1,\dots, s_{k-1}$ are any constants satisfying (3.2) and (3.3) above. Taking the vertices of $G$ in any order $v_1,v_2,\dots, v_{|V|}$ , we write a telescoping product:

\begin{equation*} |Z_G(\boldsymbol {\lambda })| = \prod _{i=1}^{|V(G)|} \left |\frac {Z_{G-\{v_1, \ldots, v_{i-1}\}}(\boldsymbol {\lambda })}{Z_{G-\{v_1, \ldots, v_i\}}(\boldsymbol {\lambda })}\right |. \end{equation*}

We may then apply (3.4) with $M = Z_{G-\{v_1, \ldots, v_{i-1}\}}$ and $v=v_i$ ; notice that these are indeed valid choices respecting our structural constraints, since all edges in $Z_{G-\{v_1, \ldots, v_{i-1}\}}$ are of size $k$ . Thus, we get the bound

\begin{equation*} |Z_M(\boldsymbol {\lambda })| \geq (1-s_1)^{|V|}. \end{equation*}

To establish Theorem 12, it remains only to optimise the choice of $R$ . It would perhaps be difficult to precisely optimise the choices of $s_1,\dots, s_{k-1}$ to make the bounds (3.2) and (3.3) on $R$ as large as possible. But if we take $s_1=\cdots =s_{k-2} = \Delta ^{-1/(k-1)}$ and $s_{k-1} = 1/\Delta$ , we will be able to choose $R= \left (\frac{\Delta -1}{\Delta }\right )^{\Delta -1} \left ( 1- \Delta ^{ - \frac{1}{k-1}} \right )\Delta ^{ - \frac{1}{k-1}}$ and ensure that (3.2) and (3.3) are satisfied. Indeed, the constraints are now

\begin{equation*}R\leq \Delta ^{ - \frac {1}{k-1}}\cdot \left ( 1- \Delta ^{ - \frac {1}{k-1}} \right )\cdot \left (\frac {\Delta -1}{\Delta }\right )^\Delta \end{equation*}

and

\begin{equation*}R\leq s_j^{1/j}\cdot \frac {\Delta ^{1/(k-1)}-1}{\Delta ^{1/(k-1)}}\cdot \left (\frac {\Delta -1}{\Delta }\right )^{\Delta -1}\end{equation*}

for all $2\leq j\leq k-1$ . Note that for $2\leq j\leq k-2$ we have $s_j^{1/j}\lt \Delta ^{- \frac{1}{k-1}}$ , and thus the strongest of these constraints is for $j=k-1$ , i.e.,

\begin{align*} R&\leq \Delta ^{ - \frac{1}{k-1}}\cdot \frac{ \Delta ^{ \frac{1}{k-1}}-1}{ \Delta ^{ \frac{1}{k-1}}}\cdot \left (\frac{\Delta -1}{\Delta }\right )^{\Delta -1} \\ &= \left (\frac{\Delta -1}{\Delta }\right )^{\Delta -1} \left ( 1- \Delta ^{ - \frac{1}{k-1}} \right )\Delta ^{ - \frac{1}{k-1}}. \end{align*}

It is now easy to see that the above bound is lower than the one in (3.2), so we may chose $R$ to be equal to this value. Thus if $|\lambda _v|\leq R = \left (\frac{\Delta -1}{\Delta }\right )^{\Delta -1} \left ( 1- \Delta ^{ - \frac{1}{k-1}} \right )\Delta ^{ - \frac{1}{k-1}}$ for all $v\in V$ , then

\begin{equation*} |Z_M(\boldsymbol {\lambda })| \geq (1-s_1)^{|V|} = \left (1-\Delta ^{- \frac {1}{k-1}}\right )^{|V|}, \end{equation*}

finishing the proof of Theorem 12. Note also that the factor $\left (\frac{\Delta -1}{\Delta }\right )^{\Delta -1} \left ( 1- \Delta ^{ - \frac{1}{k-1}} \right )$ is increasing in $\Delta$ and $\left (\frac{\Delta -1}{\Delta }\right )^{\Delta -1} \left ( 1- \Delta ^{ - \frac{1}{k-1}} \right )\cdot k$ is decreasing in $k$ , so to obtain the conclusion it suffices that $|\lambda _v| \le \frac{\log 2}{2k} \Delta ^{ - \frac{1}{k-1}}$ , since $\lim _{k \to \infty } k(1-2^{-1/(k-1)}) = \log 2$ .

We now return to the vertex and edge deletion bounds (3.4) and (3.5), which we will prove by induction. The base case is $M=\varnothing$ , where both (3.4) and (3.5) are vacuously true.

For the induction step, assume that (3.4) and (3.5) hold for all $M$ with $|V(M)|\leq n$ , and consider any admissible subhypergraph $\Lambda$ of $G$ with $n+1$ vertices, where $\Lambda$ satisfies the assumption that every connected component has at most one edge of size strictly less than $k$ .

We first establish (3.4) for $\Lambda$ . To that end, let $v$ be any vertex of $\Lambda$ satisfying the hypotheses of (3.4). To bound $|Z_\Lambda (\boldsymbol{\lambda })|$ , we begin by applying identity (2.3):

\begin{equation*} Z_\Lambda (\boldsymbol {\lambda })=Z_{\Lambda -v}(\boldsymbol {\lambda })+\lambda _vZ_{\Lambda/v}(\boldsymbol {\lambda }). \end{equation*}

And using the reverse triangle inequality and noting that $|\lambda _v|\leq R$ by assumption, this gives

(3.6) \begin{equation} |Z_\Lambda (\boldsymbol{\lambda })|\geq |Z_{\Lambda -v}(\boldsymbol{\lambda })|-R|Z_{\Lambda/v}(\boldsymbol{\lambda })|. \end{equation}

To extract the desired lower bound on $|Z_\Lambda (\boldsymbol{\lambda })|$ , we will use our induction hypotheses to bound $|Z_{\Lambda/v}(\boldsymbol{\lambda })|$ in terms of $|Z_{\Lambda -v}(\boldsymbol{\lambda })|$ . Observe that $\Lambda/v$ may be obtained from $\Lambda -v$ by a sequence of at most $\Delta$ many vertex deletions and $j$ -edge additions (for $2\leq j\leq k-1$ ). But we must take care to verify that at each step, all the necessary conditions are satisfied to apply the induction hypotheses.

First, if $v$ is contained in a 2-edge $\{v,x\}$ of $\Lambda$ , then starting from $\Lambda -v$ , we delete $x$ . (Notice that by assumption, $v$ is contained in at most one such edge.) And we can indeed apply hypothesis (3.4) (with $M=\Lambda -v$ ), as $\Lambda -v$ has at most one edge of size less than $k$ in each connected component, and $x$ is in a component of $\Lambda -v$ where all edges have size $k$ (the edge $\{v,x\}$ is excluded from $\Lambda -v$ ). So applying (3.4), this gives

(3.7) \begin{equation} |Z_{\Lambda -v}|\geq (1-s_1)\cdot |Z_{\Lambda -v-x}| \end{equation}

if $v$ has a graph neighbour $x$ .

Then, regardless of whether $v$ is contained in a 2-edge, to obtain $\Lambda/v$ from $\Lambda -v-N_2(v)$ , we add a $j$ -edge $\{x_1,\dots,x_j\}$ for each $\{v,x_1,\dots,x_j\}\in E(\Lambda )$ , for all $2\leq j\leq k-1$ .

The number of $j$ -edge additions performed at this step is at most the number of $(j+1)$ -edges in $\Lambda$ adjacent to $v$ , which we will denote $n_{1j}$ . Note that most of the numbers $n_{1j}$ will be zero, as $v$ is adjacent to at most one edge of size less than $k$ .

And we may indeed apply induction hypothesis (3.5) for these edge additions: regardless of the order in which we add these edges, at each step, the corresponding $x_1,\dots,x_j$ are in different components from any edges of size less than $k$ added at previous steps, since the vertex $v$ is no longer present. And at each step, the hypergraph produced has at most one edge of size less than $k$ in each connected component. So we may repeatedly apply our second induction hypothesis (3.5) to obtain

(3.8) \begin{equation} |Z_{\Lambda -v-N_2(v)}|\geq \prod _{i\geq 2}(1-s_i)^{n_{1i}}\cdot |Z_{\Lambda/v}|. \end{equation}

Then combining (3.7) and (3.8), we obtain the following relationship between $|Z_{\Lambda/v}(\boldsymbol{\lambda })|$ and $|Z_{\Lambda -v}(\boldsymbol{\lambda })|$ :

\begin{equation*} |Z_{\Lambda -v}|\geq \prod _{i\geq 1}(1-s_i)^{n_{1i}}\cdot |Z_{\Lambda/v}| \end{equation*}

(where $n_{1i}$ is the number of graph neighbours of $v$ , which is either 0 or 1). And notice that, since $v$ is contained in at most one edge of size less than $k$ , and at most $\Delta$ edges total, this bound may be simplified substantially:

(3.9) \begin{equation} |Z_{\Lambda -v}|\geq (1-s_j)\cdot (1-s_{k-1})^\Delta \cdot |Z_{\Lambda/v}| \end{equation}

for some $1\leq j\leq k-2$ . (Note: we may very slightly improve this bound by considering whether or not $v$ is contained in an edge of size less than $k$ , but this will not substantially change our final answer. Also, we cannot control which $j$ is used; we must simply take the worst case.)

And combining this with (3.6), we see that

\begin{align*} |Z_\Lambda (\boldsymbol{\lambda })| &\geq |Z_{\Lambda -v}(\boldsymbol{\lambda })|\left (1-R\left (\frac{1}{1-s_j}\right )\left (\frac{1}{1-s_{k-1}}\right )^{\Delta }\right ). \end{align*}

for all $1\leq j\leq k-2$ . Thus the induction hypothesis (3.4) is guaranteed to be satisfied for $\Lambda$ as long as

(3.10) \begin{equation} 1-R\left (\frac{1}{1-s_j}\right )\left (\frac{1}{1-s_{k-1}}\right )^{\Delta }\ \geq \ 1-s_1 \end{equation}

for each $1\leq j\leq k-2$ ; this is equivalent to condition (3.2) above.

We now proceed with the induction step for (3.5), which deals with edge addition. Again, we let $\Lambda$ be any admissible subhypergraph of $G$ with $n+1$ vertices satisfying the assumption that every connected component has at most one edge of size strictly less than $k$ . Let $2\leq j\leq k-1$ , and consider any $x_1,\dots, x_j$ in $\Lambda$ satisfying the hypotheses of (3.5) – that is, that $x_1,\dots, x_j$ are (respectively) in components of $\Lambda$ where all edges have size $k$ , and they are in some $k$ -edge $\{v_1,\dots, v_{k-j},x_1,\dots, x_j\}$ of $G$ with $v_1,\dots, v_{k-j}\not \in V(\Lambda )$ . Notice that this implies $\Lambda + \{x_1,\dots,x_j\}$ is also an admissible subhypergraph of $G$ , and that each component has at most one edge of size less than $k$ (the setting of our induction hypotheses).

To bound $|Z_{\Lambda +\{x_1,\dots,x_j\}}|$ , we begin by applying identity (2.4):

\begin{equation*} Z_\Lambda (\boldsymbol {\lambda })=Z_{\Lambda +\{x_1,\dots,x_j\}}(\boldsymbol {\lambda })+\lambda _{x_1}\dots \lambda _{x_j}Z_{\Lambda/\{x_1,\dots,x_j\}}(\boldsymbol {\lambda }). \end{equation*}

Then as above, by using the reverse triangle inequality and noting that $|\lambda _{x_1}|,\dots, |\lambda _{x_j}|\leq R$ , we obtain

(3.11) \begin{equation} |Z_\Lambda (\boldsymbol{\lambda })|\geq |Z_{\Lambda +\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|-R^j|Z_{\Lambda/\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|. \end{equation}

To obtain a lower bound on $|Z_\Lambda (\boldsymbol{\lambda })|$ , we will apply our induction hypotheses to bound $|Z_{\Lambda/\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|$ in terms of $|Z_{\Lambda +\{x_1,\dots,x_j\}}(\boldsymbol{\lambda })|$ . Notice that $\Lambda/\{x_1,\dots,x_j\}$ may be obtained from $\Lambda +\{x_1,\dots,x_j\}$ by a sequence of at most $j\Delta$ many operations as follows: first, we delete $x_1, x_2,\dots,$ and $x_j$ . Note that, unlike in the general case, we do not need to delete $C(x_1,\dots, x_j)$ ; by our assumptions, this set is empty. Then add each $(k-1)$ -edge $\{y_1,\dots, y_{k-1}\}$ where $\{x_i,y_1,\dots, y_{k-1}\}$ is in $E(\Lambda )$ for some $x_i$ .

In total, starting from $\Lambda +\{x_1,\dots,x_j\}$ , we perform the vertex deletion operation $j$ times, and the $(k-1)$ -edge addition operation at most $j(\Delta -1)$ many times – once for each of the $\leq \Delta -1$ edges adjacent to $x_1,\dots, x_j$ in $\Lambda + \{x_1,\dots,x_j\}$ respectively (excluding the edge $\{x_1,\dots,x_j\}$ ). And again, we must take care to verify that at each step, all the necessary conditions to apply (3.4) and (3.5) are satisfied.

First, notice that our initial application of (3.4) in this sequence is to delete a vertex from $\Lambda + \{x_1,\dots,x_j\}$ , which has $n+1$ vertices. We are allowed to do so, since we just established the $n+1$ case of (3.4) above, and, without loss of generality, we begin by deleting the vertex $x_1$ , which is contained in an edge of size $j\lt k$ in $\Lambda + \{x_1,\dots,x_j\}$ . We may also delete $x_2,\dots, x_j$ next, as vertices in components where all edges have size $k$ . Furthermore, at each step, regardless of the order in which we add the $(k-1)$ -edges $\{y_1,\dots, y_{k-1}\}$ , the corresponding $y_1,\dots, y_{k-1}$ are in different components from any $(k-1)$ -edges added at previous steps, since $x_1,\dots,x_j$ are no longer present. Finally, at each step, the hypergraph produced has at most one edge of size less than $k$ in each connected component.

So by (3.4) and (3.5),

\begin{equation*} |Z_{\Lambda + \{x_1,\dots,x_j\}}|\geq (1-s_1)^{j}(1-s_{k-1})^{j(\Delta -1)}\cdot |Z_{\Lambda/ \{x_1,\dots,x_j\}}|. \end{equation*}

Now, combining this inequality with (3.11), we see that

\begin{align*} |Z_\Lambda (\boldsymbol{\lambda })| &\geq |Z_{\Lambda +\{x_1,\dots,x_j\}}|\left (1-R^ j\left (\frac{1}{1-s_1}\right )^{j}\left (\frac{1}{1-s_{k-1}}\right )^{j(\Delta -1)}\right ). \end{align*}

So we will obtain (3.5) for $\Lambda$ as long as

(3.12) \begin{equation} 1-R^ j\left (\frac{1}{1-s_1}\right )^{j}\left (\frac{1}{1-s_{k-1}}\right )^{j(\Delta -1)}\ \geq \ 1-s_j \end{equation}

for each $2\leq j\leq k-1$ ; this is equivalent to condition (3.3) above.

Therefore, if conditions (3.3) and (3.3) are satisfied, this completes the induction, giving the edge and vertex deletion bounds (3.4) and (3.5), as desired.

Proof. Since $Z_G(\lambda )$ is a polynomial of degree at most $n = |V(G)|$ , if $Z_G(\lambda )$ is zero-free in a disc of radius $B$ around $0$ and $|\lambda | \le (1-\delta )B$ then $\exp (T_r(\lambda ))$ gives an $\epsilon$ -relative approximation to $Z_G(\lambda )$ when $r \ge C \log (n/\epsilon )$ , where $C$ is a constant that depends only on $\delta$ .

Thus to prove Theorem 13 given Theorem 1 we are left with the task of computing $T_r(\lambda )$ for $r = \Theta ( \log (n/\epsilon ))$ in time polynomial in $n/\epsilon$ . Generalising the approach of Patel and Regts [Reference Patel and Regts47] to hypergraphs, Liu, Sinclair, and Srivastava [Reference Liu, Sinclair and Srivastava42] gave an algorithm to compute the first $r$ coefficients of the partition function of a $2$ -spin model on a bounded-degree hypergraph. Since the coefficients of the Taylor series for $\log Z$ are related to the coefficients of $Z$ through a triangular system of linear equations, this yields an algorithm to compute $T_r(\lambda )$ . Specialising their result to the hypergraph independence polynomial yields the following, which finishes the proof of Theorem 13.