Given a sequence of unimodular lattices $(\Lambda_n)_n$ in $\mathbb{R}^3$ $\big($i.e. a $3$-dimensional $\mathbb{Z}$-submodule of $\mathbb{R}^3$ with covolume $1$, meaning for any $\mathbb{Z}$-basis $u_n , v_n , w_n$ of $\Lambda_n$ we have $\det (u_n , v_n , w_n )=1\big)$ assume there is a sequence of vectors $z_n \in \Lambda_n \setminus \{0\}$ with $\Vert z_n \Vert_2\to 0$. I want to show that there are vectors $y_n , x_n \in \Lambda_n$ such that $x_n,y_n$ forms a basis over $\mathbb{Z}$ of some two-dimensional submodule of $\Lambda_n$ and also $\Vert x_n \times y_n \Vert \to 0$ where $x\times y $ is the usual cross product: $$\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} \times \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \begin{pmatrix} v_2 w_3 – v_3 w_2 \\ v_3 w_1 – v_1 w_3 \\ v_1 w_2 – v_2 w_1 \end{pmatrix}$$
It is clear that if we have a $\mathbb{Z}$-basis $u_n , v_n , w_n$ of $\Lambda_n$ with $\Vert u_n \Vert \to 0$ and one of the other two is bounded, then we are done. So if both of the others tend to $\infty$, then not both of them can be faster than $\frac{1}{\sqrt{\Vert u_n \Vert}}$ goes to $\infty$ (intuitively), because the whole thing has covolume $1$. But I can't really prove that, because one has to take angles into account and it starts to get messy pretty quickly…
Also I know, that the dual basis vector defined by $$\langle u_n , u_n^* \rangle = 1 , \langle v_n , u_n^* \rangle = 0 = \langle w_n , u_n^* \rangle$$(and similar for $v_n^* , w_n^*$), contained in the dual lattice, is equal to $u_n^* = \pm ~ v_n \times w_n$, if that helps.
Thanks in advance for any helpful comments!
Best Answer
I just realized it's quite trivial but I'm gonna post an answer in case someone is looking for it:
I want to show that if the shortest non-zero vector $v_n \in \Lambda\setminus \{0\}$ is bounded below by some $\delta>0$, then also $\Vert x_n \times y_n\Vert$ is bounded for $x_n,y_n$ as in the question. So let $\Lambda \subset \mathbb R^3$ be a lattice with $\min_{v \in \Lambda\setminus\{0\}} \Vert v \Vert> \delta$ for some $\delta >0$. Note, that given any plane $P \subset \mathbb R^3$, the expression $\Vert x \times y \Vert$ gives the same for any $x,y \in P \cap \Lambda$ with $\langle x ,y \rangle_{\mathbb Z} = P \cap \Lambda$.
So let $P \subset \mathbb{ R}^3$ be any plane. We define $v , w \in \Lambda$ with $$\Vert v\Vert := \min_{x \in P \cap \Lambda \setminus \{0\}} \Vert x \Vert, \quad \Vert w \Vert = \min_{x \in P \cap \Lambda \setminus \mathbb R v} \Vert x \Vert.$$Then indeed we have $\langle v,w\rangle_{\mathbb Z}= P \cap \Lambda$. By a rotation we can assume $v = \big( \Vert v \Vert , 0 , 0 \big)$. Write $w = \big( w_1 , w_2 , w_3\big)$ because for any $k \in \mathbb Z$ the system $(v , w + k v)$ is still a basis over $\mathbb Z$ for $P \cap \Lambda$ we may further assume $$w_1 \in \left[ - \frac{\Vert v \Vert}{2} , \frac{\Vert v \Vert}{2}\right).$$Let $\varphi := \measuredangle(v,w)$ be the angle in between $v$ and $w$. Then we can prove $-\frac{1}{2} \le \cos \varphi < \frac{1}{2}$. That is because $$\cos \varphi = \frac{\langle v , w \rangle}{\Vert v\Vert \cdot \Vert w \Vert}=\frac{\Vert v \Vert w_1}{\Vert v \Vert \cdot \Vert w\Vert}<\frac{\Vert v \Vert }{2 \Vert w \Vert} < \frac{1}{2}$$ and analogously for the other inequality. For the last $<$ we user the minimality of $\Vert v \Vert$. From this we get $\vert \sin \varphi \vert > \frac{1}{2}$ and now we can calculate $$\Vert v \times w \Vert = \Vert v \Vert \cdot \Vert w \Vert \cdot \vert \sin \varphi \vert > \frac{\delta^2 } {2}.$$For the other direction, where the expression $\Vert x \times y \Vert $ is bounded below for all $x,y \in \Lambda$ s.th. $x \times y \neq 0$ and we want to show that $\Vert z \Vert $ is bounded below for all $z \in \Lambda \setminus \{0\}$ we have to consider the dual lattice as mentioned in the question and can use the same proof again. Therefore the norms of the cross products are bounded below iff the norms of the vectors are. That answers the question.