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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 $r<n$; 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?

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  • $\begingroup$ 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. $\endgroup$
    – John Hughes
    Commented Aug 17, 2020 at 14:45
  • $\begingroup$ @JohnHughes: I am afraid I do not understand your comment. Anyway, I have edited the question, hope it it more clear now. $\endgroup$
    – W-t-P
    Commented Aug 17, 2020 at 15:05

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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.

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  • $\begingroup$ 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 $\Lambda$? $\endgroup$
    – W-t-P
    Commented Aug 17, 2020 at 16:43
  • $\begingroup$ 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 $1$. $\endgroup$
    – John Hughes
    Commented Aug 17, 2020 at 16:59
  • $\begingroup$ Thank you for the last comment. There seems to be something fundamental that I was missing. I'll have to think on it a little more. $\endgroup$
    – W-t-P
    Commented Aug 17, 2020 at 17:13
  • $\begingroup$ 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. $\endgroup$
    – John Hughes
    Commented Aug 17, 2020 at 17:17
  • $\begingroup$ 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. $\endgroup$
    – W-t-P
    Commented Aug 17, 2020 at 18:38

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