Let $\mathcal L$ be the space of lattices in $\mathbb R^d$. The **$i$-th successive minimum** of $L\in \mathcal L$, denoted $\lambda\_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$.


Let $v\_1$ denote any vector in $L$ with $\lVert v\_1\rVert=\lambda\_1(L)$. Now project $L$ along $v\_1$ onto its orthogonal complement $v\_1^{\perp}$. It is easy to see that the projection image of $L$, denoted $L\_1$, is again a lattice in the hyperplane $v\_1^{\perp}$.



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> I speculate that $\lambda\_i(L\_1) \asymp\_d \lambda\_{i+1}(L)$.
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It is easy to see $\lambda\_i(L\_1) \le \lambda\_{i+1}(L)$ but what about the other direction?



Successive minima of a lattice and projection along the the shortest nonzero vector
By taking a shortest lift from $L\_1$ to $L$ you obtain that $\lambda\_{i+1}(L)\le \lambda\_i(L\_1)+\frac{1}{2}\lambda\_1(L) \le \lambda\_i(L\_1)+\frac{1}{2}\lambda\_{i+1}(L)$. This gives $\frac{1}{2}\lambda\_{i+1}(L) \le \lambda\_i(L\_1)$, i.e. $\lambda\_{i+1}(L) \le 2\lambda\_i(L\_1)$.


