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Let $a > 0$, and set $B_a = \{x \in \mathbb{R}^n : |x|^2 < a \}$. Let $\phi : B_a \to \mathbb{R}^n$ be given by $\phi(x) = \frac{ax}{\sqrt{a^2 - |x|^2}}$. Prove that $\phi$ is a diffeomorphism of $B_a$ onto $\mathbb{R}^n$.

This problem is from Guillemin and Pollack's book on Differential Topology. I'm working through it as a self-study.

It's clear that $\phi(x)$ is invertible, one can directly compute the inverse and find it to be $\phi^{-1}(y) = \frac{ay}{\sqrt{a^2 + |y|^2}}$ (check out this post: What is the inverse of the function $x \mapsto \frac{ax}{\sqrt{a^2 - |x|^2}}$?). So, to see that $\phi$ is a diffeomorphism, I just need to see that it's smooth (i.e., it's component functions $\frac{ax_i}{\sqrt{a^2 - |x|^2}}$ have continuous partial derivatives of all orders).

Now, it's mostly obvious to me that each $\frac{ax_i}{\sqrt{a^2 - |x|^2}}$ is indeed smooth. Practically speaking, we should be able compute partial derivatives of this expression for as long as we like. We'd just have to use the quotient rule over and over, and we'd never have any singularities causing us trouble.

But I'm looking for a more formal method for showing that $\phi$ is smooth. Any suggestions? Maybe there is a general technique/theorem for showing that "nice" component functions like these, which contain nothing more than fractions or radicals, are smooth?

Thanks in advance for solutions and hints.

Community
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JZS
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2 Answers2

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The map $\mathbb R^n\to\mathbb R$, $x\mapsto |x|^2$ is smooth, constants are trivially smooth, adding, subtracting, multiplying, dividing is smooth as map $\mathbb R\times\mathbb R\to \mathbb R$ (or as map $\mathbb R\times (\mathbb R\setminus\{0\})$ for division), as well as scalar multiplication as map $\mathbb R\times \mathbb R^n\to\mathbb R^n$. Taking square roots is smooth on $\mathbb R_{>0}$. Finally, the composition of smooth functions is smooth. Thus your function is smooth as well: It is the composition of $ x\mapsto (x,|x|^2)$ with $(u,v)\mapsto(au,a^2-v)$, then $(u,v)\mapsto(u,\sqrt v)$, then $(u,v)\mapsto(u,\frac1v)$, and finally $(u,v)\mapsto vu$.

In essence, virtually anything written as a simple expression is smooth, just look out for $|x|$ without squaring and $\sqrt{}$ at zero (and of course the domain of definition).

Hagen von Eitzen
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Try to prove each points :

  1. $x\mapsto |x|^2$ is smooth.
  2. $\psi:]-a,a[\to\mathbb{R}_+$, $t\mapsto \dfrac{a}{\sqrt{a^2-t^2}}$ is smooth.
  3. Compositions of smooth functions are smooth.
  4. If $U$ is an open subset of $\mathbb{R}^n$, $f:U\to\mathbb{R}^m$ and $\lambda:U\to\mathbb{R}$ are smooth, then $\lambda.f$ is smooth.

Hence $\phi$ is smooth.

Philippe Malot
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