The distribution for the Kronecker product of two uniform random vectors in unit sphere? Suppose that two random vectors $x$ and $y$ are uniformly distributed on unit sphere $S_{n-1}$. Is it possible to show that the Kronecker product of $x$ and $y$ is uniformly distributed on a subset of the unit sphere $S_{n-1}\otimes S_{n-1}$?
 A: Yes.  This becomes clear upon working through the definitions.
The Kronecker Product
The unit sphere $S^{n-1}$ is the set of unit vectors in the Euclidean space $E^n = \left(\mathbb{R}^n, ||\cdot ||\right)$ where $$||(x_1,x_2,\ldots,x_n)^\prime|| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}$$ is the Euclidean norm.
The "Kronecker product" is the usual tensor product. There are several ways to think about it and compute with it.  One is to define it as an $n\times n$ matrix
$$x \otimes y = x\, y^\prime = \pmatrix{x_1y_1 & x_1y_2 & \cdots & x_1y_n \\
x_2y_1 & x_2y_2 & \cdots & x_2y_n \\ 
\vdots & \vdots & \cdots & \vdots \\
x_ny_1 & x_ny_2 & \cdots & x_ny_n} \in \operatorname{Mat}\left(\mathbb{R}^n, \mathbb{R}^n\right).$$
Another equivalent way unravels the components of this matrix into a vector with $n^2$ components, allowing us to view $x\otimes y$ as an element of $\mathbb{R}^{n^2}$.  Notice that the Euclidean metric for $\mathbb{R}^{n^2}$ can be written
$$||Z||^2 = \operatorname{Tr}( Z Z^\prime ).\tag{1}$$
Both are ways of writing the sum of squares of all $n^2$ components.
The Kronecker product is compatible with the Euclidean metrics on $E^n$ and $E^{n^2}$ in the sense that
$$||x \otimes y||^2 = ||x||^2\,||y||^2.$$
This is easily demonstrated, because the left hand side is defined as the sum of the squares of all the $x_i y_j$ while the right hand side is the product of their sums of squares.  Just expand that product:
$$||x \otimes y||^2 = \sum_i\sum_j (x_iy_j)^2 = \left(\sum_i x_i^2\right)\left(\sum_j y_j^2\right) = ||x||^2\,||y||^2.$$
In particular, when both $x$ and $y$ have unit length, $x \otimes y$ has unit length.  Therefore
$$S^{n-1}\otimes S^{n-1} \subset S^{n^2-1}.$$
Uniform distributions
The Euclidean spheres inherit a measure from the usual measure on the Euclidean spaces (the Lebesgue measure, ultimately determined by the Euclidean distance).  That measure is preserved by any isometry of a sphere, because (by definition) an isometry preserves distances and the measure ultimately is determined by distances.  The group of isometries of the unit sphere in $\mathbb{R}^m$ is denoted $O(m)$, the orthogonal group.  It is a classical result, and straightforward to show, that it consists of the linear transformations represented by all $m\times m$ matrices $P$ for which $PP^\prime = P^\prime P = \mathbb{I}_m$.
The orthogonal group $O(n)$ acts transitively on $S^{n-1}$.  (Here's a proof Euclid might have made: pick any two distinct points $x$ and $y$ on the sphere.  Draw the line segment $xy$ between them.  It determines a unique hyperplane perpendicular to $xy$ passing through the midpoint of $xy$.  The reflection in that hyperplane maps $S^{n-1}$ to itself and preserves all distances, whence it is in $O(n)$.  That reflection swaps $x$ and $y$, showing there exists an orthogonal transformation sending $x$ to $y$.)
To say that "vectors are uniformly distributed on $S^{n-1}$" means that the distribution is invariant under a transitive group of isometries like $O(n)$.
Here's the punchline: the "hypertorus" $S^{n-1}\otimes S^{n-1} \subset S^{n^2-1}$ enjoys a transitive group of isometries isomorphic to a quotient of $O(n)\times O(n)$.  Indeed, given an arbitrary $x\otimes y\in S^{n-1}$ and another $x_2\otimes y_2\in S^{n-1}$, pick $P,Q\in O(n)$ for which $Px=x_2$ and $Qy=y_2$.  Let $Z=(z_{ij})$ be any $n\times n$ matrix and define
$$(P\otimes Q)(Z) = PZQ^\prime.\tag{2}$$
This is an isometry because, using the formula $(1)$,
$$\eqalign{
||(P \otimes Q) (Z)||^2 &= \operatorname{Tr}\left((PZQ^\prime)(PZQ^\prime)^\prime\right)  = \operatorname{Tr}\left(PZ Z^\prime P^\prime\right) = \operatorname{Tr}\left(PP^\prime Z Z^\prime \right) = \operatorname{Tr}\left(ZZ^\prime \right) \\
&= ||Z||^2.}$$
These steps exploited the orthogonality of $P$ and $Q$ and the fact that $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$ for any square matrices $A$ and $B$.
Because the isometries of $S^{n-1}$ induce a transitive group of isometries of $S^{n-1}\otimes S^{n-1}$ via $(2)$, the uniform (that is, group-invariant) distribution on $S^{n-1}$ is mapped to a uniform distribution on $S^{n-1}\otimes S^{n-1}$, QED.
A: As long as $x$ and $y$ are independent, then what you say is true.
It's easiest to think about angles rather than points in space.
For example, in the 2-D sphere in $\mathbb R^3$, we can generate points uniformly on the surface of that sphere by sampling $\theta_1$ and $\theta_2$ independently from $U[0,2\pi]$ and taking those as zenith and azimuth respectively.
Thus, we can think of $x$ as $(\theta_1,...,\theta_{n-1}) \sim U[0,2\pi]^{n-1}$ and $y$ as $(\phi_1,...,\phi_{n-1}) \sim U[0,2\pi]^{n-1}$. Their KP is then $(\theta_1,...,\theta_{n-1},\phi_1,...,\phi_{n-1})$ and it's clear from the construction that that follows a uniform distribution $U[0,2\pi]^{2n-2}$ as well.
