Proof of Theorem 2 for odd k. For $p = \omega (1/n^{k - 1} )$ we bound the probability that $M \;:\!=\; M(\mathcal{H})$ has cokernel with torsion when $\mathcal{H} \sim \mathcal{H}_k(n, p)$ . If ${\mathrm{coker}}(M)$ has torsion then it has $q$ -torsion for some prime $q \leq \sqrt{k}^n$ . Moreover if $M$ has $q$ -torsion then so does ${\mathrm{coker}}(M^T)$ , so in that case there exists a vector $v$ so that $M^Tv = qw$ for $w$ an integer vector not in the image of $M^T$ and $v \notin (q\mathbb{Z})^n$ . In this case $v \pmod q$ is a nontrivial vector in $\ker _{\mathbb{Z}/q\mathbb{Z}}(M^T)$ . By Lemma 6 and a union bound over all $\sqrt{k}^n$ primes under consideration we see that with high probability, there is no vector in $\ker _{\mathbb{Z}/q\mathbb{Z}}(M^T)$ with support size larger than $n/2$ for any prime $q \leq \sqrt{k}^n$ . Indeed letting $T_q$ denote the event that there is a vector $v \in \ker _{\mathbb{Z}/q\mathbb{Z}}(M^T)$ with support size at least $n/2$ and taking union bound over all primes smaller than $\sqrt{k}^n$ we find the probability that there is $v \in \ker _{\mathbb{Z}/q\mathbb{Z}}(M^T)$ of large support for some prime $q$ is at most

\begin{eqnarray*} \mathbb{P}\left (\bigcup _{q \leq \sqrt{k}^n, q \text{ prime }} T_q\right ) &\leq & \mathbb{P}\left (\bigcup _{q \leq \sqrt{k}^n, q \text{ prime }} T_q \middle | |E(\mathcal{H})| \gt cn\right ) + \mathbb{P}(|E(\mathcal{H})| \leq cn)\\[5pt] &\leq & \sum _{q \leq \sqrt{k}^n, q \text{ prime }} \mathbb{P}(T_q | |E(H)| \gt cn) + \mathbb{P}(|E(H)| \leq cn) \end{eqnarray*}

where $c$ is some sufficiently large constant for the assumptions of Lemma 6 to hold. By a standard coupling argument between $\mathcal{H}_k(n, cn)$ and $\mathcal{H}_{k}(n, p)$ for $p = \omega (1/n^{k - 1})$ , we apply Lemma 6 to conclude that each term in the sum over at most $\sqrt{k}^n$ primes is at most $\exp\!(\!-\!Cn)$ for $C \gt \log\!(k)$ . Thus the large sum is at most $\exp\!(\!-\!n\log\!(k)/2)$ and the last term is $o(1)$ since the number of hyperedges of $\mathcal{H}$ is distributed as a binomial with mean $\omega (n)$ .

We can assume now that the vector $v$ with $M^Tv = qw$ has support size (over $\mathbb{Z}/q\mathbb{Z}$ ) at most $n/2$ . We know however that with high probability $M^T$ satisfies the assumption of Lemma 5 with $\delta = 1/2$ . In this case then taking $S$ to be the support of $v$ over $\mathbb{Z}/q\mathbb{Z}$ , we see that ${\mathrm{rank}}_{\mathbb{Q}}((M^T)_S) ={\mathrm{rank}}_{\mathbb{Z}/q\mathbb{Z}}((M^T)_S)$ . Therefore ${\mathrm{coker}}((M^T)_S)$ is $q$ -torsion-free. As $S$ is the support of $v$ over $\mathbb{Z}/q\mathbb{Z}$ , $v = v_1 + qv_2$ where ${\mathrm{supp}}(v_1) = S$ , we have that $qw = M^T(v_1 + qv_2) \in{\mathrm{coker}}((M^T))$ , so $M^T(v_1) = qw_1$ for some vector $w_1$ . As $v_1$ is supported on $S$ and ${\mathrm{coker}}((M^T)_S)$ is torsion-free there is some integer vector $u$ supported on $S$ with $M^Tu = w_1$ and so $M^T(u + v_2) = w$ contradicting the choice of $w$ as a torsion element of ${\mathrm{coker}}(M^T)$ .

Proof. Suppose that ${\mathrm{rank}}_{\mathbb{Q}}(M_S) \lt |S|$ . Let $U \subsetneq S$ be a maximal collection of linearly independent columns of $M_S$ over $\mathbb{Q}$ . Since $S$ is a torsion cocycle there exists a prime $q$ so that ${\mathrm{rank}}_{\mathbb{Q}}(M_S) \gt{\mathrm{rank}}_{\mathbb{Z}/q/\mathbb{Z}}(M_S)$ . By minimality of $S$ , ${\mathrm{rank}}_{\mathbb{Q}}(M_U) ={\mathrm{rank}}_{\mathbb{Z}/q\mathbb{Z}}(M_U)$ . On the other hand, we have

\begin{eqnarray*}{\mathrm{rank}}_{\mathbb{Q}}(M_U) &=&{\mathrm{rank}}_{\mathbb{Q}}(M_S) \\[5pt] &\gt &{\mathrm{rank}}_{\mathbb{Z}/q\mathbb{Z}}(M_S) \\[5pt] &\geq &{\mathrm{rank}}_{\mathbb{Z}/q\mathbb{Z}}(M_U). \end{eqnarray*}

Thus, we reach a contradiction and finish the proof of the first part of the claim.

For the second part, we observe that if $z$ is a row of $M_S$ so that the $i$ th entry of $z$ is $\pm 1$ and all other entries of $z$ are $0$ then clearly over any field column $i$ of $M_S$ is outside the span of all other columns of $M_S$ . Thus, deleting $i$ from $S$ drops the rank of $M_S$ over any field by 1. Thus when ${\mathrm{rank}}_{\mathbb{Q}}(M_S) \gt{\mathrm{rank}}_{\mathbb{Z}/q\mathbb{Z}}(M_S)$ as both ranks drop by one. However this contradicts minimality.

Proof. We prove this by counting the expected number of subsets $S$ satisfying both of the conditions. Let $t$ denote the size of $S$ . For each $1 \leq t \leq \delta n$ there are $\binom{n}{t}$ ways to pick $S$ . From here we have to choose at least $t$ hyperedges that intersect $S$ in at least 2 vertices to be included. The number of hyperedges that intersect $S$ in at least 2 vertices is

\begin{equation*}\sum _{i = 0}^{k -2} \binom {n - t}{i} \binom {t}{k - i} \leq \sum _{i = 0}^{k - 2} n^i t^{k - i} .\end{equation*}

And from this set, we choose at least $t$ hyperedges to be included. Next, we have to guarantee that all hyperedges that intersect $S$ in only one place are excluded. The number of hyperedges that meet $S$ in exactly one vertex is

\begin{equation*}t \binom {n - t}{k - 1}.\end{equation*}

Putting this all together the number of expected number of sets $S$ that satisfy the conditions we want to avoid is at most

\begin{eqnarray*} \sum _{t = 1}^{\delta n} \binom{n}{t} e^t \left ( \frac{\sum _{i = 0}^{k -2} n^i t^{k - i}}{t} \right )^t p^t \exp\!\left (\!-p \binom{n - t}{k - 1}\right )^t. \end{eqnarray*}

Taking $p = f/n^{k - 1}$ with $f \;:\!=\; f(n) \rightarrow \infty$ arbitrarily slowly, we have the above is at most

\begin{eqnarray*} \sum _{t = 1}^{\delta n} \frac{n^t e^{2t}}{t^t} \left ( \sum _{i = 0}^{k -2} n^i t^{k - i - 1} \right )^t \left (\frac{f}{n^{k - 1}}\right )^t \exp\!\left (-(f/n^{k - 1}) \frac{(1 - \delta )^{k - 1}n^{k - 1}}{(k - 1)^{k - 1}}\right )^t. \end{eqnarray*}

Thus for some positive constant $c = c_{k, \delta }$ that depends on $k$ and $\delta$ , we have that the above sum is at most

\begin{eqnarray*} \sum _{t = 1}^{\delta n} \left (e^2 n \sum _{i = 0}^{k - 2} n^i t^{k - i - 2} \right )^t \left (\frac{f}{n^{k - 1}}\right )^t \exp\!\left (-f c\right )^t &\leq & \sum _{t = 1}^{\delta n} \left (e^2 n^{2 - k} f\sum _{i = 0}^{k - 2} \delta ^{k - i - 2}n^{k - 2} \right )^t \exp\!\left (-f c\right )^t \\[5pt] &\leq & \sum _{t = 1}^{\delta n} \left ( e^2 (k - 1) f e^{-c f} \right )^t \end{eqnarray*}

And this is $o(1)$ as long as $f$ tends to infinity. Thus by Markov’s inequality we have the lemma.

Proof of Lemma 13 . For a vector $v \in (\mathbb{Z}/q\mathbb{Z})^n$ we wish to bound from below the number of $w \in V_{k, n}$ so that $w \cdot v \neq 0$ . With foresight into the calculations fix some positive $\varepsilon \lt \frac{(k - 1)!}{k^{k}}$ . We consider first the case that $v$ is $\varepsilon$ -balanced.

If $v$ is $\varepsilon$ -balanced then there are at least

\begin{equation*}\binom {|{\mathrm {supp}}(v)|}{k} - \binom {|{\mathrm {supp}}(v)|}{k - 1} \varepsilon |{\mathrm {supp}}(v)|\end{equation*}

vectors $w$ in $V_{k, n}$ over $\mathbb{Z}/q\mathbb{Z}$ so that $w \cdot v \neq 0$ . To see this we restrict only to those vectors $w \in V_{k, n}$ so that ${\mathrm{supp}}(w) \subseteq{\mathrm{supp}}(v)$ ; denote this set by $V_{k, n}^v$ . There are clearly (exactly) $\binom{|{\mathrm{supp}}(v)|}{k}$ vectors in $V_{k, n}^v$ . We claim that among these vectors at most $\binom{|{\mathrm{supp}}(v)|}{k - 1} \varepsilon |{\mathrm{supp}}(v)|$ will be orthogonal over $\mathbb{Z}/q\mathbb{Z}$ to $v$ . To construct a vector $w$ in $V_{k, n}^v$ that’s orthogonal to $v$ we first pick the first $k - 1$ positions in ${\mathrm{supp}}(v)$ to be nonzero in $w$ . Clearly there are at most $\binom{|{\mathrm{supp}}(v)|}{k - 1}$ ways to do this. If $w'$ denotes this vector (in $V_{k -1, n}^v$ ) then if $w' \cdot v = 0$ then there is no way to select the last position of the nonzero entry in $w$ so that ${\mathrm{supp}}(w) \subseteq{\mathrm{supp}}(v)$ and $w \cdot v = 0$ . On the other hand if $w' \cdot v \neq 0$ then in order for $w \cdot v$ to be zero, and the last entry of $w$ must fall into a position where $v$ is the unique non-zero element of $\mathbb{Z}/q\mathbb{Z}$ equal to $-w' \cdot v$ . By the $\varepsilon$ -balanced condition, there are at most $\varepsilon |{\mathrm{supp}}(v)|$ choices for this position. Thus $V_{k, n}^v$ has $\binom{|{\mathrm{supp}}(v)|}{k}$ vectors and at most $\binom{|{\mathrm{supp}}(v)|}{k - 1} \varepsilon |{\mathrm{supp}}(v)|$ of them are orthogonal to $v$ .

Suppose on the other hand, that $v$ is not $\varepsilon$ -balanced. Then there is some subset $S$ of ${\mathrm{supp}}(v)$ with $|S| \geq \varepsilon |{\mathrm{supp}}(v)|$ so that $v$ restricted to $S$ is a constant, nonzero function in $\mathbb{Z}/q\mathbb{Z}$ . Then any vector in $V_{k, n}$ with support contained in $S$ cannot be orthogonal to $v$ as the dot product of $v$ with such a vector is simply $x - x + x - x \cdots + x = x$ . Thus, there are at least

\begin{equation*}\binom {\varepsilon |{\mathrm {supp}}(v)|}{k}\end{equation*}

vectors in $V_{k, n}$ that are not orthogonal to $v$ when $v$ is not $\varepsilon$ balanced. Thus, we have the claim with

\begin{equation*}\gamma = \min\!\left \{\frac {1}{k^k} - \frac {\varepsilon }{(k - 1)!}, \frac {\varepsilon ^k}{k^k} \right \}.\end{equation*}

Proof of Lemma 6 . Consider the stochastic process to build $M^T$ one row at a time. Let $M^T_i$ denote the $i \times n$ matrix at step $i$ and let $Z_i$ be the random variable $\dim\!(\ker _{\mathbb{Z}/q\mathbb{Z}}(M^T_i))$ with the convention that $Z_0 = n$ . If $M^T_i$ has a vector of support size at least $\delta n$ in its kernel over $\mathbb{Z}/q\mathbb{Z}$ then the probability that $K_{i + 1} \lt K_i$ is at least

\begin{equation*}\frac {\gamma (\delta n)^{k}}{\binom {n}{k}} \geq k! \delta ^{k} \gamma,\end{equation*}

where $\gamma$ is as in Lemma 13. Thus the probability that $M^T_{cn}$ for $c$ a large constant has a kernel element over $\mathbb{Z}/q\mathbb{Z}$ of support size at least $\delta n$ is at most the probability that a binomial random variable with $cn$ trials and success probability $k! \delta ^k \gamma$ has at most $n$ successes. By Chernoff’s bound this is at most $\exp\!(\!-\!Cn)$ where $C$ can be made an arbitrarily large constant by setting $c$ large enough.

Proof of Theorem 2 for even k. As in the odd $k$ case it suffices to show that with high probability there are no $v$ , $w$ , and $q$ so that $M^Tv = qw$ , $w \notin \text{Im}(M^T)$ and $v \notin (q\mathbb{Z})^n$ . Fix a positive $\varepsilon \lt (k - 1)!/k^k$ and take $\delta = (1 + \varepsilon )^{-1}$ . By Lemma 10 the probability that $M^T$ has a torsion cocycle of size at most $\delta n$ is $o(1)$ . If $M^Tv = qw$ with $S \;:\!=\;{\mathrm{supp}}(v)$ of size at most $\delta n$ then we use the fact that with high probability ${\mathrm{coker}}((M^T)_S)$ is torsion-free to conclude that $w \in \text{Im}(M^T)$ .

For vectors of large support, by a similar union bound to the odd $k$ case we can use Lemmas 15 and 7 to rule out $\varepsilon$ -balanced vector $v \in \mathbb{Z}/q\mathbb{Z}$ with $|{\mathrm{supp}}(v)| \geq \delta n$ and $v \in \ker _{\mathbb{Z}/q\mathbb{Z}}(M^T)$ . The only case that’s different is if $v$ is not $\varepsilon$ -balanced but still has large support. If $v$ is not $\varepsilon$ -balanced then some $x \in \mathbb{Z}/q\mathbb{Z} \setminus \{0\}$ appears more than $\varepsilon \delta n$ times in $v$ . In this case then $x\textbf{1} - v$ satisfies $|{\mathrm{supp}}(x\textbf{1} - v)| \leq n - \varepsilon \delta n = \delta n$ , and $M^T(x \textbf{1} - v) = -qw$ , but then $S ={\mathrm{supp}}(x \textbf{1} - v)$ has size at most $\delta n$ so we appeal again to the fact that ${\mathrm{coker}}((M^T)_S)$ has no $q$ -torsion to conclude that $w \in \text{Im}(M^T)$ .

Proof of Theorem 4 . For any matrix $m \times n$ matrix $M$ the cokernel of $M$ can be read off of the elementary divisors of $M$ , so in particular the free part of ${\mathrm{coker}}(M)$ is $\mathbb{Z}^{n -{\mathrm{rank}}_{\mathbb{Q}}(M)}$ . Moreover the inequality ${\mathrm{rank}}_{\mathbb{Q}}(M) \geq{\mathrm{rank}}_{\mathbb{Z}/2\mathbb{Z}}(M)$ for $M$ an integer matrix always holds since $Mv =0$ over $\mathbb{Q}$ implies that $Mv = 0$ over $\mathbb{Z}/2\mathbb{Z}$ . For $c \lt k!$ by Theorem 17, with high probability the free part of ${\mathrm{coker}}(M(\mathcal{H}))$ has rank larger than $n - n^*$ . Thus we have the first part of Theorem 4.

For $c \gt k!$ , we know that ${\mathrm{rank}}_{\mathbb{Q}}(M(\mathcal{H})) \geq{\mathrm{rank}}_{\mathbb{Z}/2\mathbb{Z}}(M(\mathcal{H})) = n^*$ asymptotically almost surely, and trivially ${\mathrm{rank}}_{\mathbb{Q}}(M(\mathcal{H})) \leq n^*$ . Thus ${\mathrm{rank}}_{\mathbb{Q}}(M(\mathcal{H})) = n^*$ in this regime. Thus for $k$ odd, with high probability ${\mathrm{coker}}(M(\mathcal{H}))$ is finite, and for $k$ even, with high probability ${\mathrm{coker}}(M(\mathcal{H})) = \mathbb{Z} \oplus A$ for some finite abelian group $A$ . However we know by Theorem 2 that ${\mathrm{coker}}(M(\mathcal{H}))$ is torsion-free, so we conclude in the odd $k$ case we are left with the trivial group and in the even $k$ case we are left with $\mathbb{Z}$ .