Random Nilpotent Groups of Maximal Step

Phillip Harris

Abstract

Let $G$ be a random torsion-free nilpotent group generated by two random words of length $\ell$ in $U_{n}(\mathbb{Z})$ . Letting $\ell$ grow as a function of $n$ , we analyze the step of $G$ , which is bounded by the step of $U_{n}(\mathbb{Z})$ . We prove a conjecture of Delp, Dymarz, and Schafer-Cohen, that the threshold function for full step is $\ell=n^{2}$ .

A group $G$ is nilpotent if its lower central series

G=G_{0}\geq G_{1}\geq\dots\geq G_{r}=\{0\}

defined by $G_{i+1}=[G,G_{i}]$ , eventually terminates. The first index $r$ for which $G_{r}=0$ is called the step of $G$ . One may ask what a generic nilpotent group looks like, including its step. Questions about generic properties of groups can be answered with random groups, first introduced by Gromov [4]. Since Gromov’s original few relators and density models are nilpotent with probability 0, they cannot tell us about generic properties of nilpotent groups. Thus there is a need for new random group models that are nilpotent by construction.

Delp et al [1] introduced a model for random nilpotent groups, motivated by the observation that any finitely generated torsion-free nilpotent group can be embedded in the group $U_{n}(\mathbb{Z})$ of $n\times n$ upper triangular integer matrices with ones on the diagonal [3]. Note that, since any finitely generated nilpotent group contains a torsion-free subgroup of finite index, we are not losing much by restricting our attention to torsion-free groups. (Another model is considered in [2]).

We construct a random subgroup of $U_{n}(\mathbb{Z})$ as follows. Let $E_{i,j}$ be the elementary matrix with 1’s on the diagonal, a 1 at position $(i,j)$ and 0’s elsewhere. Then $S=\{E^{\pm 1}_{i,i+1}:1\leq i<n\}$ forms the standard generating set for $U_{n}(\mathbb{Z})$ . We call the entries at positions $(i,i+1)$ the superdiagonal entries. Define a random walk of length $\ell$ to be a product

V=V_{1}V_{2}\dots V_{\ell}

where each $V_{i}$ is chosen independently and uniformly from $S$ . Let $V$ and $W$ be two independent random walks of length $\ell$ . Then $G=\langle V,W\rangle$ is a random subgroup of $U_{n}(\mathbb{Z})$ . We have $\textrm{step}(G)\leq\textrm{step}(U_{n}(\mathbb{Z}))$ , and it is not hard to check that $\textrm{step}(U_{n}(\mathbb{Z}))=n-1$ . If $\textrm{step}(G)=n-1$ we say $G$ has full step.

Now let $n\to\infty$ and $\ell=\ell(n)$ grow as a function of $n$ . We say a proposition $P$ holds asymptotically almost surely (a.a.s.) if $\operatorname{\mathbb{P}}[P]\to 1$ as $n\to\infty$ . Delp et al. gave results on the step of $G$ , depending on the growth rate of $\ell$ with respect to $n$ .

Theorem 1 (Delp-Dymarz-Schafer-Cohen)

Let $n,\ell(n)\to\infty$ and $G=\langle V,W\rangle$ where $V,W$ are independent random walks of length $\ell$ . Then:

1.

If $\ell\in o(\sqrt{n})$ then a.a.s. $\textrm{step}(G)=1$ , i.e. $G$ is abelian.
2.

If $\ell\in o(n^{2})$ then a.a.s. $\textrm{step}(G)<n-1$ .
3.

If $\ell\in\omega(n^{3})$ then a.a.s. $\textrm{step}(G)=n-1$ , i.e. $G$ has full step.

In this paper we close the gap between cases 2 and 3.

Theorem 2

Let $n,\ell(n)\to\infty$ and $G=\langle V,W\rangle$ . If $\ell\in\omega(n^{2})$ then a.a.s. $G$ has full step.

To prove this requires a careful analysis of the nested commutators that generate $G_{n-1}$ . In Section 1, we give a combinatorial criterion for a nested commutator of $V$ ’s and $W$ ’s to be nontrivial. In Section 2, we show this criterion is satisfied asymptotically almost surely when $V,W$ are random walks.

1 Nested Commutators

Let $G=G_{0}\geq G_{1}\geq\dots$ be the lower central series of $G$ . We have

G_{i}=[G,G_{i-1}]=[G,[G,\dots,[G,G]\dots]]

In particular, $G_{i}$ includes all $i+1$ -fold nested commutators of elements of $G$ . We restrict our attention to commutators where each factor is $V$ or $W$ .

Let $\{0,1\}^{d}$ be the $d$ -dimensional cube, or the set of all length $d$ binary vectors. For $x\in\{0,1\}^{d},y\in\{0,1\}^{e}$ we define the norm $\left|x\right|=\sum_{1\leq i\leq d}x_{i}$ and the concatenation $xy\in\{0,1\}^{d+e}$ . For example if $x=(1,0,0)$ and $y=(0,1)$ then $xy=(1,0,0,0,1)=10^{3}1$ .

We define a family of maps $C_{d}:\{0,1\}^{d}\to G_{d}$ as follows.

$\displaystyle C_{1}(1)$	$\displaystyle=V$	(1)
$\displaystyle C_{1}(0)$	$\displaystyle=W$	(2)
$\displaystyle C_{d}(1x)$	$\displaystyle=[V,C_{d-1}(x)]$	(3)
$\displaystyle C_{d}(0x)$	$\displaystyle=[W,C_{d-1}(x)]$	(4)

Thus for example $C_{5}(10^{3}1)=C_{5}(10001)=[V,[W,[W,[W,V]]]]$ . We omit the subscript $d$ when it is obvious. To prove $G$ has full step it suffices to find an $x\in\{0,1\}^{n-1}$ such that $C(x)$ is nontrivial. We begin with Lemma 2.3 from [1], which gives a recursive formula for the entries of a nested commutator.

Lemma 1

Let $a\in\{0,1\},x\in\{0,1\}^{d-1}$ . Then $C(ax)\in G_{d}$ and we have

\displaystyle C(ax)_{i,i+d}

\displaystyle=C(a)_{i,i+1}C(x)_{i+1,i+d}-C(a)_{i+d-1,i+d}C(x)_{i,i+d-1}

(5)

and furthermore $C(ax)_{i,j}=0$ for $j<i+d$ .

In particular, for $d=n-1$ only the upper rightmost entry $C(ax)_{1,n}$ can be nonzero. From the formula it is clear that $C(ax)_{i,i+d}$ is a degree- $d$ polynomial in the superdiagonal entries of $V$ and $W$ . Let us state this more precisely and analyze the coefficients of the polynomial.

Lemma 2

Let $d\geq 1$ . There exists a function $K_{d}:\{0,1\}^{d}\times\{0,1\}^{d}\to\mathbb{Z}$ such that for $1\leq i\leq n-d$ we have

\displaystyle C(x)_{i,i+d}

\displaystyle=\sum_{\begin{subarray}{c}y\in\{0,1\}^{d}\\ \left|y\right|=\left|x\right|\end{subarray}}K_{d}(x,y)\prod_{i\leq j<i+d}V_{j,j+1}^{y_{j}}W_{j,j+1}^{1-y_{j}}

(6)

Furthermore, setting $K_{d}(x,y)=0$ for $\left|x\right|\neq\left|y\right|$ we have a recursion

\displaystyle K_{d}(ax,byc)=K_{1}(a,b)K_{d-1}(x,yc)-K_{1}(a,c)K_{d-1}(x,by)

(7)

with base cases

	$\displaystyle K_{1}(0,0)$	$\displaystyle=K_{1}(1,1)=1$
	$\displaystyle K_{1}(0,1)$	$\displaystyle=K_{1}(1,0)=0$

Note that $K_{d}(x,y)$ does not depend on $i$ . We also drop the subscript $d$ since it can be inferred from $x$ and $y$ .

Proof 1.3.

Abbreviate

\displaystyle U(i,d,y):=\prod_{i\leq j<i+d}V_{j,j+1}^{y_{j}}W_{j,j+1}^{1-y_{j}}

We first prove inductively that there exist coefficients $K_{d}:\{0,1\}^{d}\times\{0,1\}^{d}\to\mathbb{Z}$ such that

\displaystyle C(x)_{i,i+d}

\displaystyle=\sum_{y\in\{0,1\}^{d}}K_{d}(x,y)U(i,d,y)

The case $d=1$ is trivial. Assume it holds for $d-1$ . Let $a\in\{0,1\},x\in\{0,1\}^{d-1}$ , then we have

\displaystyle C(ax)_{i,i+d}

\displaystyle=C(a)_{i,i+1}C(x)_{i+1,i+d}-C(a)_{i+d-1,i+d}C(x)_{i,i+d-1}\textbf{}

Expanding $C(a)_{i,i+1}$ and $C(x)_{i+1,i+d}$ , the first term is

	$\displaystyle=\left[K_{1}(a,1)V_{i,i+1}+K_{1}(a,0)W_{i,i+1}\right]\left[\sum_{y\in\{0,1\}^{d-1}}K_{d-1}(x,y)U(i+1,d-1,y)\right]$
	$\displaystyle=\sum_{y\in\{0,1\}^{d-1}}K_{1}(a,1)K_{d-1}(x,y)U(i,d,1y)+K_{1}(a,0)K_{d-1}(x,y)U(i,d,0y)$
	$\displaystyle=\sum_{\begin{subarray}{c}b,c\in\{0,1\}\\ y^{\prime}\in\{0,1\}^{d-2}\end{subarray}}K_{1}(a,b)K_{d-1}(x,y^{\prime}c)U(i,d,by^{\prime}c)$

Similarly the second term is

\displaystyle=\sum_{\begin{subarray}{c}b,c\in\{0,1\}\\ y^{\prime}\in\{0,1\}^{d-2}\end{subarray}}K_{1}(a,c)K_{d-1}(x,by^{\prime})U(i,d,by^{\prime}c)

Combining we get

\displaystyle C(ax)_{i,i+d}

\displaystyle=\sum_{\begin{subarray}{c}b,c\in\{0,1\}\\ y\in\{0,1\}^{d-2}\end{subarray}}\left[K_{1}(a,b)K_{d-1}(x,yc)-K_{1}(a,c)K_{d-1}(x,by)\right]U(i,d,byc)

And setting $K_{d}(ax,byc)=K_{1}(a,b)K_{d-1}(x,yc)-K_{1}(a,c)K_{d-1}(x,by)$ the lemma is proved for $d$ . It is also easy to see inductively that $K_{d}(x,y)=0$ for $\left|x\right|\neq\left|y\right|$ , so we may add the condition $\left|x\right|=\left|y\right|$ under the sum to get Equation 6.

We now have a strategy for choosing $x\in\{0,1\}^{n-1}$ such that $C(x)$ is nontrivial. In the random model, it may happen that $V_{i,i+1}=0$ for some $i$ . Define the vector $v\in\{0,1\}^{n-1}$ by $v_{i}=1$ if $V_{i,i+1}\neq 0$ and $v_{i}=0$ otherwise. For now assume $0<\left|v\right|<n-1$ . If we choose $x$ such that $\left|x\right|=\left|v\right|$ , then Equation 6 simplifies to

C_{n-1}(x)_{1,n}=K_{d}(x,v)\prod_{1\leq i<n}V_{i,i+1}^{v_{i}}W_{i,i+1}^{1-v_{i}}

(8)

If we assume there is no $i$ such that $V_{i,i+1}=W_{i,i+1}=0$ , the product of matrix entries is nonzero. So we just need to choose $x$ such that $K_{d}(x,v)\neq 0$ . We can do this with some additional conditions on $v$ .

Lemma 1.4.

Let $v\in\{0,1\}^{n-1}$ with $0<\left|v\right|<n-1$ . Write $v=1^{a_{1}}01^{a_{2}}\dots 1^{a_{k}-1}01^{a_{k}}$ . Assume that $a_{i}\geq 1$ for all $i$ , i.e., there are no adjacent 0’s, and that $a_{1}\neq a_{k}$ . Then there exists $x\in\{0,1\}^{n-1}$ such that $K(x,v)\neq 0$ .

We will prove in section 2 that all assumptions used hold asymptotically almost surely.

Proof 1.5.

Using Equation 7, the following identities are easily verified by induction:

1.

If $a,b\geq 0$ , then

$K(1^{a+b}0,1^{a}01^{b})={a+b\choose a}(-1)^{b}$
2.

If $a,b\geq 1,c\geq 0$ with $c<\min(a,b)$ , then

$K(1^{c}0x,1^{a}y1^{b})=0$
3.

Let $a,b\geq 0$ . If $a<b$ then

$K(1^{a}0x,1^{a}0y1^{b})=K(x,y1^{b})$

If $b<a$ then

$K(1^{a}0x,1^{a}y01^{b})=K(x,1^{a}y)$
4.

If $a,b\geq 0$ then

$K(1^{a+b}0^{2}x,1^{a}01y101^{b})=2{a+b\choose a}(-1)^{b}K(x,1y1)$

Let $v=1^{a_{1}}01^{a_{2}}\dots 01^{a_{k}}$ . First assume $k=2\ell$ is even. Applying identity 4 repeatedly we reduce to the case $v=1^{a_{\ell}}01^{a_{\ell+1}}$ , then apply identity 1. Explicitly we have

	$\displaystyle x$	$\displaystyle=1^{a_{1}+a_{2\ell}}0^{2}1^{a_{2}+a_{2\ell-1}}0^{2}\dots 1^{a_{\ell}+a_{\ell+1}}0$		(9)
	$\displaystyle K(x,v)$	$\displaystyle=2^{\ell}(-1)^{a_{2\ell}+a_{2\ell-1}+\dots+a_{\ell+1}}{a_{1}+a_{2\ell+1}\choose a_{1}}{a_{2}+a_{2\ell}\choose a_{2}}\dots a_{\ell}+a_{\ell+1}\choose a_{\ell}$		(10)

If $k$ is odd, apply identity 3 once and proceed as before.

2 Asymptotics

In Section 1 we derived a combinatorial condition on the superdiagonal entries of $V$ and $W$ sufficient for $G$ to have full step. Define

	$\displaystyle\mathcal{V}$	$\displaystyle=\{i:1\leq i<n,V_{i,i+1}=0\}$
	$\displaystyle\mathcal{W}$	$\displaystyle=\{i:1\leq i<n,W_{i,i+1}=0\}$

Then to apply Lemma 1.4 we need that

1.

$\mathcal{V}$ and $\mathcal{W}$ are nonempty.
2.

$\mathcal{V}\cap\mathcal{W}=\emptyset$ .
3.

$\mathcal{V}$ has no adjacent elements.
4.

$\min\mathcal{V}\neq n-\max\mathcal{V}$ .

If condition (1) does not hold, then Theorem 2 follows by a modification of Lemma 5.4 in [1]. We now show that in the random model, the superdiagonal entries satisfy conditions (2)-(4) asymptotically almost surely. Recall that $V$ and $W$ are random walks

	$\displaystyle V$	$\displaystyle=V_{1}V_{2}\dots V_{\ell}$
	$\displaystyle W$	$\displaystyle=W_{1}W_{2}\dots W_{\ell}$

where each $V_{i},W_{i}$ is chosen independently and uniformly from the generating set $S=\{E^{\pm 1}_{i,i+1}:1\leq i<n\}$ . Define

\displaystyle\sigma_{j}(Z)=\begin{cases}1&\text{ if $Z=E_{j,j+1}$}\\ -1&\text{ if $Z=E_{j,j+1}^{-1}$}\\ 0&\text{ otherwise}\end{cases}

(11)

Then we have

V_{i,i+1}=\sum_{j=1}^{\ell}\sigma_{i}(V_{j})

(12)

When $\ell\gg n$ , the superdiagonal entries $V_{i,i+1}$ behave roughly like independent random walks on $\mathbb{Z}$ . We restate Corollary 3.2 from [1].

Lemma 2.6.

Suppose $\ell=\omega(n)$ . Then uniformly for $(k_{1},\dots,k_{d})\in\mathbb{Z}^{d}$ we have

\operatorname{\mathbb{P}}[k_{i}\in\mathcal{V}\text{ for all $i$}]\sim\left(\frac{n}{2\pi\ell}\right)^{d/2}

Since $\mathcal{V}$ and $\mathcal{W}$ are i.d.d, we have $\operatorname{\mathbb{P}}[i\in\mathcal{V}\cap\mathcal{W}]\ll n/\ell$ , so by the union bound we have $\operatorname{\mathbb{P}}[\mathcal{V}\cap\mathcal{W}\neq\emptyset]\ll n^{2}/\ell\to 0$ . Thus condition (2) holds a.a.s. For conditions (3) and (4) we will need a bound on the size of $\mathcal{V}$ .

Lemma 2.7.

Fix $\epsilon>0$ . Then $\operatorname{\mathbb{P}}[\left|\mathcal{V}\right|>\epsilon\sqrt{n}]\to 0$ as $n\to\infty$ .

Proof 2.8.

Define random variables

\displaystyle X_{i}

\displaystyle=\begin{cases}1&V(i,i+1)=0\\ 0&V(i,i+1)\neq 0\end{cases}

So $\left|\mathcal{V}\right|=\sum_{i}X_{i}$ . From Lemma 2.6 we have $\operatorname{\mathbb{E}}[X_{i}]\ll\sqrt{n/\ell}$ and $\operatorname{\mathbb{E}}[X_{i}X_{j}]\ll n/\ell$ for $1\leq i<j<n$ . Hence $\operatorname{\mathbb{E}}[\left|\mathcal{V}\right|]\ll\sqrt{n^{3}/\ell}$ and $\operatorname{Var}[\left|\mathcal{V}\right|]\ll n^{3}/\ell$ . By Chebyshev’s inequality

	$\displaystyle\operatorname{\mathbb{P}}[\left\|\mathcal{V}\right\|\geq\epsilon\sqrt{n}]$	$\displaystyle\leq\operatorname{\mathbb{P}}\left[\left\|\mathcal{V}\right\|-\sqrt{n^{3}/\ell}\geq\sqrt{n}(\epsilon-\sqrt{n^{2}/\ell})\right]$
		$\displaystyle\leq\frac{1}{(\epsilon-\sqrt{n^{2}/\ell})^{2}(\ell/n^{2})}\to 0$

Observe that the distribution of $\mathcal{V}$ is invariant under permutation. In other words, for a fixed set $\mathcal{S}\subset\{1,\dots,n-1\}$ and a permutation $\pi$ on $\{1,\dots,n-1\}$ we have

\displaystyle\operatorname{\mathbb{P}}[\mathcal{V}=\mathcal{S}]=\operatorname{\mathbb{P}}[\mathcal{V}=\pi\mathcal{S}]

and hence

\displaystyle\operatorname{\mathbb{P}}[\mathcal{V}=\mathcal{S}]=\frac{1}{{n-1\choose\left|\mathcal{S}\right|}}\operatorname{\mathbb{P}}[\left|V\right|=\left|\mathcal{S}\right|]

Let $A(k)$ be the number of sets $\mathcal{S}\subset\{1,\dots,n-1\}$ of size $k$ with at least one pair of adjacent elements. We have

A(k)\leq(n-2){n-3\choose k-2}

Let $B(k)$ be the number of sets $\mathcal{S}$ for which $\min\mathcal{S}=n-\max\mathcal{S}$ . Summing over the possible values of $\min\mathcal{S}$ we have

B(k)\leq\sum_{1\leq a\leq n/2}{n-1-2a\choose k-2}

One easily checks

\frac{A(k)+B(k)}{{n-1\choose k}}\leq\frac{2k^{2}}{n}

For $k\leq\epsilon\sqrt{n}$ this is $\leq 2\epsilon^{2}$ . On the other hand $\operatorname{\mathbb{P}}[\left|V\right|>\epsilon\sqrt{n}]\to 0$ , so we are done.

Acknowledgements. We thank Tullia Dymarz for suggesting this problem and for many helpful discussions.

References

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