Bound on the number of bounded primitive $k$-tuple for all unimodular lattices

Question

Let $X_d$ be the set of unimodular lattices in $\mathbb R^d$ (lattices with covolume one). For a lattice $\Lambda$ in $X_d$ and $1\le k \le d$, let $P^k(\Lambda)$ denote the set of all $k$-tuples $(v_1,\cdots,v_k) \in \mathbb R^{dk}$ such that $v_1,\cdots,v_k$ can be extended to a basis of $\Lambda$ (called primitive set or primitive tuple in some literature). Let $B(0,\delta)$ denote the ball centered at $O$ with radius $\delta$. I wonder what upper bound estimate can we give for the quantity

$$U(k,d,\delta):=\sup_{\Lambda \in X_d} |P^k(\Lambda) \cap B(0,\delta)^k|.$$

I found it is even hard to tell whether $U(k,d,\delta) <\infty$. I wish the bound could show how this quantity depends on $\delta$. I have a strong feeling that for $k<d$, the dependence of this quantity on $\delta$ is very subtle since it is always possible to shrink $k$ linearly independent vectors for a unimodular $\Lambda$ so that they become extremely small and thus fall into $B(0,\delta)$. If it is not dependent on $\delta$ then that constant for upperbound should only depend on $d$.

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Update: this number seems to be infinite. Consider the lattice $Span \{\frac{1}{n}e_1,\frac{1}{n}e_2,n^2e_3\}$. Let $k=1$ and $n >> 1$. Then $\frac{m}{n}e_1+\frac{1}{n}e_2$ will provide infinitely many combinations in that intersection (they are all extendable to a basis of this lattice).

Answer 1

After some research...

An integral lattice, $\Lambda\in X_n$ $\big($in $\mathbb{R}^n\big)$, is like a (discrete) vector space. You take a basis $\mathscr{B}:= \{\beta_1,...,\beta_n\}\subseteq\mathbb{R}^n$ and then:

$$\Lambda := Span_{\mathbb{Z}}(\mathscr{B}) = \bigg\{\sum\limits_{i=1}^n\lambda_i\beta_i\text{ }\bigg|\text{ }\text{For }\lambda_i\in \mathbb{Z}\text{ }\text{ and }\text{ }\beta_i\in \mathscr{B}\bigg\}.$$

We say an integral lattice (spanned by $\mathscr{B}$) is unimodular if:

$$det[\mathscr{B}] = det\big[\beta_1 ... \beta_n\big] = 1.$$

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Now to bring in the metric ball:

$$B(0,\delta) \equiv B_{\delta}(0):= \{p\in \mathbb{R}^n\text{ }|\text{ }d(0,p)<\delta\}$$

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In this problem, we are mainly concerned with counting the $\color{red}{\text{number of grid points inside a metric ball}}$. Then, comparing this number across all lattices in the space $X_n$.

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Using a given norm, $||\cdot||$, in $\mathbb{R}^n$, we may quantify this number as:

$$\color{red}{N(\Lambda,\delta)}:= \sum\limits_{i=1}^nN_i,$$

where

$$\color{red}{N_i}:= floor\bigg(\frac{\delta}{||\beta_i||}\bigg).$$

Elaborating, we take the radius, $\delta$, from the metric ball around the origin, and see how many lengths of a basis vector, $||\beta_i||$, fit inside this radius (i.e. divide). Then the floor just makes it the nearest integer (in $\mathbb{N}$).

Assumption: This sum doesn't account for all linear combinations, but gives a skeleton of points to compare for each basis. If $\Lambda_1$ has more skeleton points than $\Lambda_2$ inside the ball, it will also have more linear combinations of points inside the ball. [Pf: Exercise]

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Define the $\color{red}{\text{standard lattice}}$:

$$\color{red}{\Lambda_0} := Span_{\mathbb{Z}}\big(\{\partial_1,...,\partial_n\}\big)$$

as the integer span of standard unit vectors in $\mathbb{R}^n$ (trivially $det\big[\partial_1 ...\partial_n\big] = 1$, making $\Lambda_0$ unimodular).

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Lastly, take arbitrary unimodular $\Lambda\in X_n$. Then:

$$\exists T: \text{ }\mathscr{B} = T\mathscr{B}_0,$$

where $T\in SL_n(\mathbb{R})$ is an invertible, unit determinant, real matrix. [Pf: Exercise]

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So, we may compare basis vectors with the standard basis vectors:

$$\forall i, \exists j:\text{ }\beta_i = T(\beta_{0,j}).$$

This implies:

$$N(\Lambda,\delta) = \sum\limits_{i=1}^n floor\bigg(\frac{\delta}{||\beta_i||}\bigg) = \sum\limits_{i=1}^n floor\bigg(\frac{\delta}{||T(\beta_{0,j})||}\bigg).$$

Considering this term-wise, we have by Submultiplicativity of the norm:

$$\frac{\delta}{||T(\beta_{0,j})||}\leq ||T||\cdot\frac{\delta}{||\beta_{0,j}||}$$

This gives us:

$$\color{blue}{N(\Lambda,\delta) \leq ||T||\cdot N(\Lambda_0,\delta)}.$$

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Depending on the two bases involved, $T$ will vary. There are two cases:

$$\color{red}{||T||\leq 1} \implies ||T||\cdot N_0 \leq N_0$$

by the Positive Scaling property the inequality relation. Then by Transitivity it follows:

$$N\leq N_0.$$

So we have a set of lattices that are comparable with the standard lattice (they all have less points in the ball). It makes sense here to say:

$$\forall \Lambda:(||T||\leq 1)\implies \text{ }\text{ }|P^k(\Lambda)\cap B(0,\delta)^k|\leq \bigg(\begin{matrix}\widetilde{N_0}\\k\end{matrix}\bigg).$$

where $\widetilde{N_0}\geq N_0$ adds the meat to the skeleton.

In the other case, we have no conclusive inequality because they conflict. So we are unable to compute the supremum for $U(k,n,\delta)$, but we get close!

$$\sup\limits_{\Lambda\in X_n^{\leq 1}}|P^k(\Lambda)\cap B(0,\delta)^k|\leq \bigg(\begin{matrix}\widetilde{N_0}\\k\end{matrix}\bigg)$$

using the definition of least upper bound.