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!



Lattice basis with Gram-Schmidt vectors of increasing length
$\def\b{{\bf b}}$
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.


