# Diffeomorphism and homeomorphism between open ball and polydisc in $\mathbb{R}^n$

Recently I came to know that for $$n\geq 2$$, the open ball $$B(0,1) = \{z\in \mathbb{C}^n: |z|<1\}$$ and the polydisc $$\Delta(0;1) = \{z\in \mathbb{C}^n: |z_j|<1, j=1,\ldots, n\}$$ are not biholomorphic.

This made me wonder whether $$B(0,1) = \{x\in \mathbb{R}^n: ||x||<1\}$$ and the polydisc $$\Delta(0;1) = \{x\in \mathbb{R}^n: |x_j|<1, j=1,\ldots, n\}$$ are diffeomorphic?

I know that the product topology on $$\mathbb{R}^n$$ is equivalent to the metric topology, i.e. topology generated by open polydiscs is same as the one generated by open balls. Does this imply that $$B(0,1)$$ is homeomorphic to $$\Delta(0;1)$$?

• @Dap Wouldn't it be worth to write an official answer even it is just a link? Feb 26, 2019 at 12:41

I know that the product topology on $$\mathbb{R}^n$$ is equivalent to the metric topology, i.e. topology generated by open polydiscs is same as the one generated by open balls. Does this imply that $$B(0,1)$$ is homeomorphic to $$\Delta(0;1)$$?

Not if it's for a topological reason. For example, the topology on $$\mathbb R^n$$ is also generated by the sets which are unions of two open balls of radius $$r$$ and whose centers are at distance $$5r$$. But a disjoint union $$B(0, 1) \sqcup B(0, 1)$$ is definitely not homeomorphic to $$B(0, 1)$$.

• We have a smooth map \begin{align*} B(0, 1) & \to \mathbb R^n \\ x & \mapsto \frac{x}{\sqrt{1- \|x\|^2}} \end{align*} with smooth inverse \begin{align*} \mathbb R^n & \to B(0, 1) \\ y & \mapsto \frac{y}{\sqrt{1 + \|y\|^2}} \end{align*}

• If $$g : (-1, 1) \to \mathbb R$$ is any diffeomorphism, then \begin{align*} \Delta(0, 1) & \to \mathbb R^n \\ (x_1, \ldots, x_n) & \mapsto (g(x_1), \ldots, g(x_n)) \end{align*} is a smooth map with smooth inverse \begin{align*} \mathbb R^n & \to \Delta(0, 1) \\ (y_1, \ldots, y_n) & \mapsto (g^{-1}(y_1), \ldots, g^{-1}(y_n)) \end{align*}

They are diffeomorphic since they are both open star shaped sets, see A star shaped open set in $\mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^n$

• How to prove that a polydisc is star-shaped? I expected that there will be a direct proof of diffeomorphism; similar to the proof of open ball being diffeomorphic to Euclidean space. Also, with my present knowledge level, I couldn't understand the proof given in the link. I will be grateful if you could also give a brief outline of the proof. Feb 26, 2019 at 16:49
• @rationalbeing: it's star shaped at zero because $z\in \Delta(0;1)$ implies $\lambda z\in\Delta(0,1)$ for $\lambda\in[0,1].$ The answer by "40 votes" is quite conceptual and just relies on an intuitive fact about ODEs.
– Dap
Feb 26, 2019 at 18:57
• And yes there is probably a much simpler proof for this case
– Dap
Feb 26, 2019 at 19:02

You can indeed show that two balls for two (equivalent) norms are homeomorphic :

Homeomorphism between open unit balls from equivalent distances