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Let $\Lambda$ be an $n$-dimensional lattice in $\mathbb R^n$ and let $\cal B$ be the set of all bases that generate $\Lambda$. For a basis $\mathbf{B}=[\mathbf{b}_1, ... ,\mathbf{b}_n]\in {\cal B}$, define $\mathbf{B}^\dagger = [\mathbf{b}^\dagger_1, ... ,\mathbf{b}^\dagger_n]$ to be the Gram-Schmidt orthogonalization of $\mathbf B$ (i.e., $\mathbf{B} = \mathbf{B}^\dagger\mathbf{U}$, for some upper-triangular matrix $\mathbf{U}$ with unit diagonal).

I'm interested in the existence of a lattice basis $\mathbf{B}\in{\cal B}$ that has the following property: $$||\mathbf{b}^\dagger_i||\leq||\mathbf{b}^\dagger_j||,\; \mbox{for all }\; 1\leq i < j \leq n.$$

Does such a basis exist for any $\Lambda$? Is this related to any problems already studied in the literature? Any pointers/insight will be very much appreciated. Thanks!

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up vote 2 down vote accepted

$\def\b{{\bf b}}$ No, not any lattice has such a basis.

Notice that $||\b_n^\dagger||$ is the distance from $\b_n$ to the hyperplane $H_{n-1}=\langle \b_1,\dots,\b_{n-1}\rangle $; thus it is not greater than the minimal length of a lattice vector outside $H_{n-1}$. Moreover, they can be equal only in the case when this shortest vector is orthogonal to $H_{n-1}$.

Now, take a lattice spanned by the equilateral triangle with unit side length in the plane. Then $||{\bf b}_1||\geq 1$, and $||{\bf b}_2^\dagger||\leq 1$; the equality in the second case is achieved only if $H_1$ is orthogonal to some unit lattice vector; but then $||{\bf b}_1||=\sqrt3$. So in any case $||{\bf b}_1^\dagger||>||{\bf b}_2^\dagger||$.

More generally, we see that a lattice $\Lambda$ admits such a basis if it has some "layered" structure: the distance between two neighboring lattice hyperplanes parallel to $H_{n-1}$ is at least the smallest length of a lattice vector in $H_{n-1}$, and the same condition inductively holds for $\Lambda\cap H_{n-1}$. But even this condition seems to be quite weak.

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