I want to specify a subspace of a high dimension Euclidean space. The subspace consists of those points whose closest proximity to a specified hyperplane is a fixed value r. Here are some examples.

Example 1. Consider a 3D space, and a line (the hyperplane) running though it. The subspace is the surface of cylinder of radius r whose axis is the line.

Example 2. Consider a 3D space, and a plane (the hyperplane) running though it. The subspace consists of the two planes parallel to the original one and a distance r from it.

I want this subspace specified in the form of a locus of points. In other words, a set of parameters uniquely specifies any point in the subspace. For instance in example 1, any point in the subspace could be specified by 3 parameters, e.g.

  • one parameter would specify a point on the line
  • r would specify a circle radius r, centred on the point and orthogonal to the line
  • theta (an angle) would specify a point on that circle. (You would have to define where theta=zero was and the direction of rotation of theta but I don't think you would need more parameters for that, just the adoption of some convention.)

I've been able to specify a b dimensional hyperplane in a b + c dimensional euclidean space (b < b+c), but am now stuck.

I appreciate any help you can give me.

What about this. One could start with the polar coordinates for a point on a b dimensional sphere. Covert from polar to Cartesian, to express these b parameters as a b-vector in b dimensional Euclidean space. Place the b dimensional sphere in a b + c dimensional Euclidean space. Place it at position zero in each of the c new dimensions. Now add c parameters which are the positions of the b-sphere in the c new dimensions. You have now created a parameterisation for one example of the desired subspace. The c dimensions are the hyperpane. You can then rotate and translate your c+b parameterisation in c+b dimensional space to make it coincide with any specific c dimensional hyperplane. You set the radius of the b-sphere to a specific value. Your final parameterisation has b+c-1 parameters.

  • $\begingroup$ For example 1 would need to specify where the zero for the angle $\theta$ is, depending on the point on the line. $\endgroup$ – coffeemath Jul 1 '18 at 14:31
  • $\begingroup$ Both of your examples can be represented implicitly by second-degree Cartesian equations. $\endgroup$ – amd Jul 1 '18 at 17:50
  • $\begingroup$ What, specifically, is your question? $\endgroup$ – amd Jul 1 '18 at 17:50
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    $\begingroup$ @coffeemath - I have edited the question to reflect your point. $\endgroup$ – Desmond Campbell Jul 1 '18 at 21:42
  • $\begingroup$ @amd - I want a parameter space such that each point in the parameter space maps to a single point in the subspace (defined above). I guess the inverse is not necessary, a point in the subspace could map to multiple points in the parameter space, e.g.for an angle parameter values 360 degrees apart might map to the same subspace point. However I'd probably restrict the parameter space so as to make the mapping one-to -one in both directions, e.g. 0 <= theta < 360 degrees. $\endgroup$ – Desmond Campbell Jul 2 '18 at 11:01

You could think about stereographic projection: If you have a linear subspace $U$ of dimension $m$ of a Euclidean vector space $V$ of dimension $n$, you could pick a vector in $U^\bot$, call it $w$, and consider (orthonormal) co-ordinates $(x_1,\ldots,x_{n-1})$ on the space $w^\bot$ orthogonal to $w$, chosen such that $(x_1,\ldots,x_m)$ give co-ordinates on $U$. Now apply the inverse stereographic projection map (you can look up the formula I guess) to the co-ordinates $(x_{m+1},\ldots, x_{n-1})$, and leave the variables $x_1,\ldots,x_m$ fixed. That gives you a map from $\mathbb{R}^{n-1}$ to your locus. Unfortunately this is not quite surjective, and misses out a translate of $U$ in the direction $w$. Still, the formulas are nicer than finding spherical co-ordinates I think.

One small point: 'hyperplane' generally refers to a subspace of dimension one less than the ambient vector space.

  • $\begingroup$ Would I end up with a point expressed in the Cartesian coordinate system? Let's call the Cartesian axes C_1, C_2, ... C_n. A point would be an n-vector. Each element of the vector would be function of (probably all) the n-1 parameters. Yes? $\endgroup$ – Desmond Campbell Jul 4 '18 at 22:30
  • $\begingroup$ Sorry for the delay - but yes, that's right, inverse stereographic projection will give a vector each element of which is a rational function (a fraction with polynomials in numerator and denominator) in the n-1 variables. The first m coordinates are just the coordinates on U, so the entry is just x_i, but the rest of the co-ordinate entries use everything from x_{m+1} to x_{n-1}. $\endgroup$ – tprince Jul 14 '18 at 12:56

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