Given a circle in a 3D space centered on the plane $x+y+z=0$, its parametric equation can be represented as:
$$
\left( \sqrt{\frac{1}{2}}\cos(t) + \sqrt{\frac{1}{6}}\sin(t), -\sqrt{\frac{1}{2}}\cos(t) + \sqrt{\frac{1}{6}}\sin(t), -\sqrt{\frac{2}{3}}\sin(t) \right)
$$
where $t \in [0, 2\pi]$. This equation is derived using two orthogonal unit vectors, $v_1$ and $v_2$, lying in the plane $x+y+z=0$.
However, I am finding difficulties in generalizing this to higher dimensions. Specifically:
How can one describe a parametric equation for an (n-2) dimensional
sphere in an n-dimensional space if that sphere is centered on the
hyperplane defined by $x_1 + x_2 + … + x_n = 0$?
Any insights or pointers would be greatly appreciated!
p.s. Visualization for better understanding of the circle talked about above:
Best Answer
The equation of the hypersphere in $\mathbb{R}^n$ that is centered at the origin is
$ r^T r = R^2 \hspace{20pt}(1)$
where $r = [x_1, x_2, \dots, x_n]^T $ and $R$ is the radius of the hypersphere.
Now, we have the hyperplane
$ \mathbf{1}^T r = 0 \hspace{20pt}(2) $
intersecting the hypersphere. Where $\mathbf{1} = [1, 1, \dots, 1]^T $
The first step is find the solution of $(2)$, and this is given by
$ r = V u \hspace{20pt}(3)$
where $V$ is $n \times (n-1)$. It $k$-column has $-1$ at the top row and $1$ at the $(k+1)$-th row. So, for example, if $n = 4$, then
$V = \begin{bmatrix} -1 && -1 && -1 \\ 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Now, substitute $(3)$ into $(1)$, you'll get,
$ u^T V^T V u = R^2 $
Note that $V^T V$ is $(n-1) \times (n-1)$.
The next step is to diagonalize $V^T V$ and express it as follows
$ V^T V = P D P^T $
where $P$ is the orthogonal matrix whose columns are the mutually orthogonal eigenvectors of $V^T V$ corresponding to the eigenvalues listed as the diagonal entries of $D$. It is easy to show that $V^T V$ has one eigenvalue equal to $n$ and the rest of the eigenvalues are $1$.
Now we can write
$ u^T P D P^T u = R^2 $
So that if $ w = D^{1/2} P^T u $ is any vector in $\mathbb{R}^{n-1}$ of length $R$, then we have
$w^T w = R^2 $
It follows that,
$ u = P D^{-1/2} w $
Finally, substituting back,
$ r = V u = V P D^{-1/2} w $
Note that $w$ is the vector parameter here, and it is just a unit vector in $\mathbb{R}^{n-1} $ multiplied by the radius of the hypersphere $R$.
Hence,
$ r = R V P D^{-1/2} e $
where $ e \in \mathbb{R}^{n-1}$ and $e^T e = 1 $.
For example, if $n = 4$, and $R = 1$, then we can take
$ e = [ \sin t \cos s , \sin t \sin s , \cos t ]^T $
And we will have
$ r = A e $
where $ A = R V P D^{-1/2} $, and is given by
$ A = \begin{bmatrix} 0 && 0 && - \dfrac{\sqrt{3}}{2} \\ - \dfrac{1}{\sqrt{2}} && - \dfrac{1}{\sqrt{6}} && \dfrac{1}{\sqrt{12}} \\ \dfrac{1}{\sqrt{2}} && - \dfrac{1}{\sqrt{6}} && \dfrac{1}{\sqrt{12}} \\ 0 && \sqrt{ \dfrac{2}{3}} && \dfrac{1}{\sqrt{12}} \end{bmatrix}$
Note that $\mathbf{1}^T A = 0 $, i.e. the sum of the entries of each of the three columns is $0$, and in addition, the three columns are of unit length, and mutually orthogonal to each other. And therefore,
$ A^T A = I_3 $
From $\mathbf{1}^T A = 0$ it follows that $ \mathbf{1}^T r = 0 $ , i.e $r$ as given above is on the given hyperplane. And in addition,
$ r^T r = e^T A^T A e = e^T e = 1 $
Hence $r$ is also on the given unit hypersphere.