Homeomorphism & inverse, between $U=\{ (x,y) \in \mathbb{R^2} :|x|+|y|\leqslant 2 \}$ and $V=\{(x,y) \in \mathbb{R^2} : \max(|x|, |y|)\leqslant 3\}$

Question

Find a homeomorphism, and its inverse, between $U$ and $V$ where:

$U= \{ (x,y) \in \mathbb{R}^2 : |x|+|y| \leqslant 2 \}$

$V= \{(x,y) \in \mathbb{R}^2 : \max (|x|, |y|) \leqslant 3 \} $

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I have sketched these regions and I believe $U$ is square with lengths $2$ rotated $45$ degrees through the origin, and $V$ is the square centred at the origin with length $3$

So $U \rightarrow V$ would involve a $45$ degree rotation and then a scaling of $1.5$

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But how do I use this information to explicitly construct an homeomorphism? I am having difficulty constructing these. Also, I want to know how to construct the inverse homeomorphism too.

Many thanks for your help

Answer 1

This is more of a hint and a correction than an answer. You need to be careful with your set $U$. They should look like the following: $U$

and $V$

I would agree with User8128 in this and say that you do want to use the combination of two invertible matrices, one a rotation and the other a scaling, which you can find many places on the internet, but it is not a scaling by 1.5. Be careful and see if you can figure out what the proper amount you need to scale by is, and the corresponding matrices for both transformations.

Answer 2

Here is another way. You have $U = \overline{B}_1(0, 2)$, $V = \overline{B}_\infty(0, 3)$, so you could take $\phi:U \to V$ given by $\phi(x) = {3 \over 2} {\|x\|_1 \over \|x\|_\infty} x$, for $x \neq 0$ and $\phi(0) = 0$.

This is not as nice as the map you suggest as it is not smooth, but the general idea works between any two balls with different norms.

Answer 3

It is not that hard to come with rotation matrix. Just as with any linear map, you need to see what it does to basis vectors.

Let us denote rotation by angle $\varphi$ with $R_\varphi$.

If we rotate $(1,0)$ by $\varphi$, by appealing to polar coordinates (after we sketch unit circle) we immediately get that $R_\varphi(1,0)=(\cos\varphi,\sin\varphi)$.

Since $(0,1)$ is orthogonal to $(1,0)$, and rotations preserve angles, it follows that $R_\varphi(0,1) = (\cos(\varphi+\frac\pi 2),\sin(\varphi+\frac\pi 2)) = (-\sin\varphi,\cos\varphi)$.

Thus, $R_\varphi = \left( \begin{array}{cc} \cos\varphi & -\sin\varphi\\ \sin\varphi & \cos\varphi \end{array} \right)$

Considering scaling factor, I agree that $1.5$ is wrong, but I won't spoil N. Owad's intention. You can't just compare $3$ and $2$, think what those lengths represent geometrically (or just look at pictures provided in the answer by N. Owad).

Finally, your homeomorphism is $f(x) = \lambda R_\varphi x$, where $\lambda$ is the appropriate scaling factor.