Continuity of the mapping of a lattice in $\mathbb{R}^{d}$ onto the length of the shortest vector

Question

I have the following problem:

Let $X_{d}:=\{\mathbb{Z}^{d}g;g\in\operatorname{SL}_{d}(\mathbb{R})\}$ and define $\lambda:X_{d}\to(0,\infty)$ by: $$ \lambda(\Lambda):=\min\left\{r>0;\Lambda\cap \overline{B_{r}^{\mathbb{R^{d}}}(0)}\neq\emptyset\right\}\quad\forall\Lambda\in X_{d}$$ Let $g\in\operatorname{SL}_{d}(\mathbb{R})$ and $(h_{n})_{n\in\mathbb{N}}\in\operatorname{SL}_{d}(\mathbb{R})^{\mathbb{N}}$ such that $h_{n}\to1$ $(\in\operatorname{SL}_{d}(\mathbb{R}))$. Then $$\lambda(\mathbb{Z}^{d}gh_{n})\stackrel{n\to\infty}{\longrightarrow}\lambda(\mathbb{Z}^{d}g)$$

(In the entire discussion think of vectors as $1\times d$ matrices.)

$\lambda$ is well-defined thanks to Minkowski's first theorem. It is clear that $\lim\sup\lambda(\mathbb{Z}^{d}gh_{n})\leq\lambda(\mathbb{Z}^{d}g)$. So what I wanted to show was $\lim\inf\lambda(\mathbb{Z}^{d}gh_{n})>\lambda(\mathbb{Z}^{d}g)-\epsilon$ for all $\epsilon>0$.

What I tried was the following: assuming otherwise, we can without loss of generality (switching to a subsequence) assume that there is a sequence $(v_{n})_{n\in\mathbb{N}}\in(\mathbb{Z}^{d})^{\mathbb{N}}$ such that: $$\lVert v_{n}gh_{n}\rVert< \lambda(\mathbb{Z}^{d}g)-\frac{\epsilon}{2}\leq\lVert v_{n}g\rVert-\frac{\epsilon}{2}$$ Now I wanted to get the contradiction out of this. The problem was that the $v_{n}$ depended on $n\in\mathbb{N}$, so this did not directly yield the desired contradiction as they could diverge in norm to $\infty$.

I arrived at the point where I thought there was no way around it but trying to show that the $v_{n}$ stabilize. So I guessed the following:

Let $(h_{n})_{n\in\mathbb{N}}\in\operatorname{SL}_{d}(\mathbb{R})^{\mathbb{N}}$ such that $h_{n}\to 1$, $(v_{n})_{n\in\mathbb{N}}\in\mathbb{R}^{d}$ an $M>0$ such that $\lVert v_{n}h_{n}\rVert\leq M$ for all $n\in\mathbb{N}$. Then $(h_{n})_{\mathbb{N}}$ has compact closure.

I did not manage to show that. Can you help me on how to prove this or can you tell me whether the guess is wrong (which would astonish me)?

Answer 1

So I have decided to prove the following:

Let $K\subseteq\operatorname{SL}_{d}(\mathbb{R})$ compact and $A\subseteq\mathbb{R}^{d}$. If $\sup_{a\in A,k\in K}\lVert ak \rVert<\infty$, then $A$ is bounded.

It turns out that after having arranged the things in my head a bit, the proof seems rather easy. Clearly $\phi:\mathbb{R}^{d}\times \operatorname{SL}_{d}(\mathbb{R})\to\mathbb{R}^{d}$ given by $(v,g)\mapsto vg$ is continuous and $K^{-1}=\{k^{-1};k\in K\}$ is compact. As $\sup_{a\in A,k\in K}\lVert ak \rVert<\infty$, $\overline{AK}:=\overline{\{ak;a\in A, k\in K\}}\subseteq\mathbb{R}^{d}$ is compact and thus $\phi(\overline{AK}\times K^{-1})$ has compact image and thus is bounded. As $A\subseteq\phi(\overline{AK}\times K^{-1})$, $A$ is bounded.

Do you agree that this solves my problem?