How to formally prove that an abelian group generated by $k$ linearly independent vectors $v_1,\cdots, v_k$ in $\mathbb R^d (k\le d)$ is discrete?

Question

There is a statement which I have taken for granted for a long time but not able to formally prove it

How to formally prove that an abelian group $A$ $\mathbb Z$-spanned by $k$ linearly independent vectors $v_1,\cdots, v_k$ in $\mathbb R^d (k\le d)$ is discrete?

I know this is equivalent to saying that there is a ball centered at $O$ that only intersect the origin of the lattice.

To prove this statement, it suffices to prove that any ball only contains finitely many points in the lattice. This statement looks intuitively so obvious to me but I could never write down a formal proof.

If unnecessary, please do not identify the lattice with $GL(d,\mathbb R)/GL(d,\mathbb R)$. The topology in the latter is more complicated to me. Please just try to use elementary linear algebra and topology to prove this.

Answer 1

The setup and desired condition are both invariant under change of coordinates (because $GL_n(\mathbb{R})$ acts by homeomorphisms) so up to a suitable change of coordinates we can assume that the $v_i$ are the first $k$ vectors of the standard basis $e_i$ of $\mathbb{R}^n$. Now it's really easy: the lattice they span is a sublattice of $\mathbb{Z}^n$ so every point is at least a distance $1$ from every other point, so we can consider balls of radius $\frac{1}{2}$.

If that argument is distasteful to you we can rephrase it as follows: complete the $v_i$ to a basis of $\mathbb{R}^n$ and consider the inner product making the $v_i$ an orthonormal basis; then as above every lattice point is a distance of at least $1$ from every other in the norm induced by this inner product, and now we can appeal to the fact that every norm on $\mathbb{R}^n$ induces the Euclidean topology.

Alternatively, we can show that every ball contains at most finitely many lattice points by completing the lattice to a full rank lattice, then showing that every ball contains at most finitely many translates of a fundamental parallelepiped of this completed lattice. This follows because balls have finite volume and the fundamental parallelepiped has positive volume.

Answer 2

If you can have an integer linear combination that comes arbitrary close to the origin you can always divide by the maximal coefficient, to get a bounded set of coefficients so that the linear combinations have a point of accumulation at zero. Therefore there is a subsequence of the coefficient vectors the limit of which gives a vanishing linear combination (which is NOT zero, since the maximal coefficient is $1$).

Answer 3

An approach using linear algebra only.

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Let $V$ be the $\Bbb{R}$-span of $A$. The familiar Gram-Schmidt process (but without the normalization step!!) gives us an orthogonal basis $u_1,u_2,\ldots,u_k$ of $V$ with the property that for all $i$ $$v_i=u_i+\sum_{j<i}c_{ij} u_j.\qquad(*)$$ That is, the change of bases matrix is triangular and "the diagonal coefficients" $c_{ii}$ are all equal to one. We don't know whether the non-diagonal coefficients are integers or not, but we don't care. Normally Gram-Schmidt is described by a recursive formula giving $u_i$ as a linear combination of $v_i$ (with coefficient $1$) and the earlier vectors $u_j, j<i$. But we can solve for $v_i$ from that equation.

The discreteness follows from the following lower bound to the length of non-zero vectors $a\in A$: $$||a||\ge\min\{||u_i||\mid i=1,2,\ldots,k\}.$$ This lower bound is strictly positive, because it amounts to selecting the smallest among $k$ positive numbers.

To see this, let $a=\sum_i m_i v_i$ be an arbitrary non-zero element of $A$. Let $\ell$ be the last index $i$ such that $m_i\neq0$. It follows from $(*)$ that $$a=m_\ell u_\ell+\sum_{j<\ell} a_ju_j.$$ Therefore the norm of $a$ is at least $$||a||\ge |m_\ell|\cdot||u_\ell||\ge ||u_\ell||.$$

Answer 4

Say we have $k$ (column) vectors in a space $F^d$, where $F$ is a field. They are linearly independent over $F$ if and only if the $d\times k$ matrix $M$ formed with these $k$ columns has rank $k$, that is, if and only if there exists a $k\times k$ minor of this matrix that is invertible.

Assume now that the vectors $v_i$ are linearly independent. Let $1\le i_1< \ldots < i_k\le d$ such that the minor $M = M_{(i_1, \ldots, i_k)}$ is invertible.

Consider $w$ a linear combination of the $v_i$

$$w = M \cdot a$$

where $a$ is a $k \times 1$ vector. We get

$$w_{(i_1, \ldots, i_k) }= M_{(i_1, \ldots, i_k)} \cdot a$$

and so

$$ a = ( M_{(i_1, \ldots, i_k)})^{-1} \cdot w_{(i_1, \ldots, i_k) }$$

So, knowing a linear combination $w$ of the $v_i$, we get the coefficients $a$ using a simple formula. When the linear combination $w$ is "small", so are the coefficients $a$. At some point, the $a_i$'s being so small, if moreover integers, will be forced to be $0$.