MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello, everyone.

As we know that in an $n$-dimensional Euclidean space $\mathbb{R}^n$, there exists a continuous bijective mapping from a subset $V^{n-1}\subseteq\mathbb{R}^{n-1}$ to a unit $(n-1)$-sphere $S^{n-1}=\{\mathbf{x}\in\mathbb{R}^n:\|\mathbf{x}\|=1\}$ since the $(n-1)$-sphere is indeed an $(n-1)$-dimensional manifold.

The most straightforward example is obtained from the spherical coordinate transformation by setting $r=1$, which bijectively maps a $(n-1)$-dimenional cuboid $V^{n-1}=[0,\pi]\times\ldots\times[0,\pi]\times[0,2\pi)$ to $S^{n-1}$. (Refer to

Then I wonder if there exist $n$ such mappings $\{\mathbf{p}_i:V^{n-1}\mapsto S^{n-1}\subset\mathbb{R}^n\}\_{i=1}^n$ so that the $n$ images $\mathbf{p}\_1(\theta),\ldots,\mathbf{p}\_n(\theta)$ corresponding to any same inverseimage $\theta\in V^{n-1}$ are orthogonal to each other.

Consider the case where $n=2$. In $\mathbb{R}^2$, we can map a line segment $[0,2\pi)\subseteq\mathbb{R}^1$ to a unit circle $S^1=\{\mathbf{x}\in\mathbb{R}^2:\|\mathbf{x}\|=1\}$ by either of the following two maps:

\begin{eqnarray} \mathbf{p}_1(\theta)&=&\left[\cos(\theta),\sin(\theta)\right]^\top;\\\ \mathbf{p}_2(\theta)&=&\left[-\sin(\theta),\cos(\theta)\right]^\top; \end{eqnarray} where $\mathbf{p}_1,\mathbf{p}_2:[0,2\pi)\mapsto S^1$ are bijections. What is more important, it is easy to verify that $\mathbf{p}_1^\top(\theta)\mathbf{p}_2(\theta)=0,\forall\theta\in[0,2\pi)$, and thus we obtain such a group of orthogonal mappings in $\mathbb{R}^2$.

Then, what about when $n>2$? Although we can map $V^{n-1}$ to $S^{n-1}$ using the spherical coordinate transformation as mentioned above, I cannot construct $n$ such mappings with orthogonality yet. Does there exist such a group of mappings?

Any suggestion will be welcome and thank you very much!


I asked another question yesterday about if it is possible to obtain a group of orthogonal vectors also orthogonal to a given one by orthogonal transformation. If there was such a transformation, this question would be solved as follows:

We can define $\mathbf{p}_1$ as the spherical coordinate transformation and then generate $\mathbf{p}_i(i\geq 2)$ by these orthogonal transformations. The orthogonality of $\mathbf{p}_i$'s is guaranteed by the definition of these transformations. Then, since $\mathbf{p}_i(i\geq 2)$ is obtained by an orthogonal transformation on $\mathbf{p}_1$, if $\mathbf{p}_1(\theta)$ is a bijection to $S^{n-1}$, so does $\mathbf{p}_i(i\geq 2)$.

Unfortunately, there does not exist such a orthogonal transformation in general. Then, can the question I asked today be solved in another way? Thank you very much!

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.