# Lattices generated by "small" matrices

Let $$\Lambda$$ be a lattice in $$\mathbb R^n$$ generated by the columns of a square matrix $$B$$ of order $$n$$.

If $$\Lambda$$ is full rank, then the quantity $$d(\Lambda)=|\det B|$$ has a simple geometrical meaning: it is the volume of a fundamental parallelepiped of $$\Lambda$$. Thus, for instance, the condition $$|\det B|\le 1$$ means that the volume of the fundamental parallelepiped of $$\Lambda$$ is at most $$1$$.

Suppose however that the rank of $$\Lambda$$ is $$r; therefore, $$B$$ is degenerate. What does it mean that every square submatrix of $$B$$ of order $$r$$ has determinant not exceeding $$1$$ in absolute value?

• Have you drawn a lattice in $\Bbb R^2$ generated by a rank-1 matrix? That'll tell you what degeneracy means. Your last sentence seems to be an entirely orthogonal question, but I may be misunderstanding something. Commented Aug 17, 2020 at 14:45
• @JohnHughes: I am afraid I do not understand your comment. Anyway, I have edited the question, hope it it more clear now. Commented Aug 17, 2020 at 15:05

A rank-$$r$$ matrix will span a dimension-$$r$$ subspace of $$\Bbb R^n$$; since $$r < n$$, the volume of the fundamental parallelipiped is $$0$$.
As an example, the vectors $$\pmatrix{1\\2}, \pmatrix{2\\ 4}$$ generate the lattice $$\{ \pmatrix{k\\2k} \mid k \in \Bbb Z \},$$ and the fundamental parallelipiped for that lattice consists of just the line segment from the origin to $$\pmatrix{1\\2}$$, whose two-dimensional volume (i.e., area) is zero.
• Sure, for a lattice of rank $r<n$, the matrix $B$ is singular; that is, $\det B=0$. However, $B$ contains nonsingular submatrices of order $r$. Given that the determinant of every such submatrix does not exceed $1$ in absolute value, what can we say about $\Lambda$? Commented Aug 17, 2020 at 16:43
• It means that if you pick any $r$ independent columns of $B$ to generate a sub-lattice $H$, the fundamental parallelipiped for $H$ (within the $r$-dimensional subspace $Q$ spanned by all the columns of $B$) will have $r$-dimensional volume less than $1$. Commented Aug 17, 2020 at 16:59
• You're welcome. If you can formulate what you're asking a little more precisely, that'd be great. "What does it mean that..." is pretty vague. If a calculus student says "What does it mean that $f$ is a polynomial in $x$?" the answer could be "$f$ is continuous," or "$f$ is smooth", or "the difference-quotient for defining the derivative of $f$ can always be simplified to a polynomial expression", or any of a bunch of things. When you do do so, please ask a new question rather than modifying this one. Commented Aug 17, 2020 at 17:17
• I agree that "what does it mean that" can be vague, but (1) in the present case, this is the essence of my question, and I do not see a way to avoid it, (2) this depends on the context: anybody will answer the question "what does it mean for two vectors to have their scalar product equal to $0$" the very same way, and (3) there is an indication in my question what kind of answer is expected. Anyway, I do plan to ask yet another question of this sort. Commented Aug 17, 2020 at 18:38