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Is it possible to place a sphere $S^m$ and another sphere $S^n$ in Euclidean $k$-dimensional space $R^k$ in such a way that the distance from any point of the first sphere to any point of the second sphere be constant?

For the case $m=n=0$ it is obviously possible for $k=2$: the solution is to place the two points of the second $0$-sphere as the two vertices of a rhombus with the other two vertices being the two points of the first $0$-sphere. Here we observe that $k=m+n+2$. Notice that the centres of two "spheres" coincide and the diameters are orthogonal.

For the case $m=1,n=0$, again, the solution is obvious and requires $k=3$: you just have to make the two centres coincide and make the diameter of the $0$-sphere to be orthogonal to all diameters of the $1$-sphere. Again, $k=m+n+2$, which suggests a conjecture that it holds for all $m,n$.

But how to solve this for arbitrary $m$ and $n$?

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Let $S^n$ be a unit sphere in $\mathbb{R}^{n+m+2}$ realized in the first $n+1$ coordinates and $S^m$ be similarly in the last $m+1$ coordinates. Then the distance of any $(s_1,0)$ to $(0,s_2)$ is $$ s_1^2+s_2^2=2. $$ I doubt that you can realize such spheres in smaller dimension $<m+n+2$, but don't see how to prove you can't.

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  • $\begingroup$ Thank you, yes, you are right. And I spent hours with generalized spherical coordinates on the first sphere and moving the second sphere with a composition of parallel shift + rotation and then trying to analyse the resulting distance between two points in parametric representation of both spheres (+ parameters of shift + angles of rotation)! This was horrendously complicated (except cases $m=n=0$ and $m=1,n=0$) and so I am very grateful for providing such a clear and easy to understand solution. Thank you! :) $\endgroup$ May 25, 2015 at 14:33
  • $\begingroup$ You are welcome. I have myself the experience that asking here is often helpful. $\endgroup$ May 25, 2015 at 15:01
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    $\begingroup$ (+1) Your final assertion is correct: If $S^{n}$ is a sphere centered at $p$ in $\mathbf{R}^{N}$, let $V$ be the smallest containing affine subspace ($\dim V = n + 1$), and let $W$ be the maximal orthogonal affine subspace through $p$, a.k.a. the set of points equidistant from each point of $S^{n}$. Any "complementary" sphere must be contained in $W$, and so has dimension$$m \leq \dim W - 1 = N - \dim V - 1 = N - n - 2.$$That is, $m + n + 2 \leq N$. :) $\endgroup$ May 28, 2015 at 11:33
  • $\begingroup$ @user86418 Yes, thank you. $\endgroup$ May 28, 2015 at 20:30

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