# The successive minima of a lattice

I am new to lattice theory. I hope(will be grateful) that one could explain to me this claim 7 in REGEV course(this claim appears in this file page 6 : https://cims.nyu.edu/~regev/teaching/lattices_fall_2004/ln/introduction.pdf) which states that : The successive minima of a lattice are achieved i.e., for every 1 ≤ i ≤ n, there exists a vector vi ∈ Λ with ‖v_{i}‖ = λi(Λ).

Thank you,

Claim 7: The successive minima of a lattice are achieved i.e., for every $$1 \leq i \leq n$$, there exists a vector $$v_i \in \Lambda$$ with $$\lVert v_{i}\rVert = \lambda_i(\Lambda)$$.

There is a proof of that claim

Proof: By Corollary 6, the ball of radius (say) $$2\lambda_i(\Lambda)$$ contains only finitely many lattice points. It follows from the definition of $$\lambda_i$$ that one of these vectors must have length $$\lambda_i(\Lambda)$$.

This suggests we should look at two places

Corollary 6: Let $$\Lambda$$ be a lattice. Then there exists some $$\epsilon > 0$$ such that $$\lVert x − y\rVert > \epsilon$$ for any two non-equal lattice points $$x, y \in\Lambda$$.

Definition 7: Let $$\Lambda$$ be a lattice of rank $$n$$. For $$i \in \{1, \dots , n\}$$ we define the $$i$$th successive minimum as $$\lambda_i(\Lambda) = \inf \{r \mid \dim(\mathsf{span}(\Lambda ∩ \overline{B}(0, r))) \geq i\}$$ where $$\overline{B}(0, r) = \{x \in\mathbb{R}^m \mid \lVert x\rVert \leq r\}$$ is the closed ball of radius $$r$$ around $$0$$.

How does this proof follow? Consider $$\overline{B}(0, \lambda_i(\Lambda))\cap \Lambda$$. By definition, this contains at least $$i$$ linearly independent lattice vectors (and is the smallest ball such that this occurs). This is to say that shrinking the ball leads to a set containing at most $$i-1$$ linearly independent lattice vectors, i.e. the $$i$$th linearly independent lattice vector is on the surface of this ball.

Why does the claim state

the ball of radius (say) $$2\lambda_i(Λ)$$R contains only finitely many lattice points $$\dots$$

$$\lambda_i(\Lambda)$$ is defined with an infimum. This is an "infinite" version of $$\min$$ (similarly to how $$\sup$$ is an "infinite" version of $$\max$$). In particular, it is important to know that for a sequence $$a_i$$, $$\inf a_i = a$$ does not mean there is any particular index $$i^*$$ such that $$a_{i^*} = a$$. For example, $$a_i = i^{-1}$$ (for $$i > 0$$) has an infimum of $$0$$ (as $$i\mapsto 0^+$$), but never achieves this limit.

This means that there was some risk that there is some infinite sequence of lattice points such that $$\lVert \vec v_i\rVert\mapsto \lambda_i(\Lambda)$$ in the limit, but there is no index $$i$$ such that this equality holds. The proof handles this via noting that infimum (supremum) and min (max) agree on finite sets. Since we are restricting to a finite set, we can replace the $$\inf$$ with a $$\min$$, and note that it must be achieved by some element of the finite set.

• what a great, unambiguous answer ! thanks a lot. Feb 14 at 14:40