Show that a vector satisfies some properties after basis change

Question

This is the exercise 12-5 from Introduction to Riemannian manifolds of John Lee.

Let $(v,w)$ be a basis of $\mathbb{R}^2$ and let $\Lambda$ be the lattice generated by them. Let $\tilde v$ be the smallest vector in $\Lambda\setminus\{0\}$ and $\tilde w$ be the smallest one of $\Lambda\setminus\mathbb{Z}\tilde v$ such that the angle between $\tilde v$ and $\tilde w$ is less than $\pi/2$. I want to show that we can send this new basis $(\tilde v, \tilde w)$ to a basis $\big((\alpha, 0), (\beta, \gamma)\big)$ such that $\alpha >0, 0 \leq \beta \leq \alpha/2, \gamma >0$ and $\beta^2 + \gamma^2 \geq \alpha^2$.

My guess is to use a rotation of minus the angle between $(1, 0)$ and $\tilde v$ that will send $\tilde v$ to $(\alpha, 0)$. For the first condition it is by definition of an isometry that $\alpha = \lVert \tilde v\rVert$. And for the last condition it is because $\lVert\tilde w\rVert \geq \lVert\tilde v\rVert$.

But now I don't see how to show that $0 \leq \beta \leq \alpha/2$.

Answer 1

I assume you mean to achieve these only through orthogonal transformations, otherwise you can always find a linear map sending a basis to a basis.

Suppose $v,w,\tilde{v},\tilde{w}\in \mathbb{R}^2$ are as given. Like you said, there is an orthogonal map $\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ which rotates the plane till $\tilde{v}$ lies on the positive $x$-axis, so that $\phi(\tilde{v})=(\alpha,0)$ for $\alpha=||\tilde{v}||>0$. Let $\phi(\tilde{w})=(\beta,\gamma)$.

• Composing $\phi$ with a reflection about the $x$-axis if necessary, we can assume $\gamma>0$.
• Since the angle between $\tilde{w}$ and $\tilde{v}$ was less than $\pi/2$, we have $\langle \tilde{v}, \tilde{w}\rangle=\alpha\beta \geq 0$, and hence $\beta\geq 0$.
• Note that $\tilde{w}-\tilde{v}=(\beta-\alpha,\gamma)$ is an element of $\Lambda\setminus \mathbb{Z}\tilde{v}$, and we have $$||\tilde{w}-\tilde{v}||=||\tilde{w}||+2\alpha\left(\frac{\alpha}{2}-\beta\right).$$ Since $\tilde{w}$ was chosen to have minimal length, we must have $\frac{\alpha}{2}-\beta\geq 0$, i.e. $\beta\leq \frac{\alpha}{2}.$
• Since $\tilde{v}$ was chosen to be of minimal positive length, we have $\beta^2+\gamma^2 = ||\tilde{w}||^2 \geq ||\tilde{v}||^2 = \alpha^2$.