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Let $\mathcal{O}$ be the set of all orthogonal bases of the vector space $\mathbb{R}^d$. That is, every element in $\mathcal{O}$ is a single orthogonal basis of $\mathbb{R}^d$. To make this more precise, I consider two orthogonal bases $o_1$ and $o_2$ as equal if and only if they consist of exactly the same vectors (irrespective of order). If at least one vector belongs to one and not the other, then I consider them different.

Notice that I am talking about orthogonal bases, not orthonormal bases, which means that I don't restrict the vectors to be unit length. I don't even restrict all the vectors of a basis to have equal length. I only restrict the vectors to be orthogonal and to form a basis of $\mathbb{R}^d$.

Now, let's look at some probability distribution $f'(o)$ over $\mathcal{O}$. In other words, $f'(o)$ assigns a probability density to every orthogonal basis $o$.

Let's look at the following two step process:

(1) sample a basis $o$ from the distribution $f'(o)$.

(2) sample a vector $v$ out of the basis $o$ at uniform (each vector with probability $\frac{1}{d}$).

This process induces a probability distribution $f(v)$ over $\mathbb{R}^d$.

Let $g(v)$ be a probability distribution over $\mathbb{R}^d$. I say that $g(v)$ is Induced by an Orthogonal Basis Distribution (IOBD) if there exists a probability distribution $g'(o)$ over $\mathcal{O}$ such that $g(v) $ can be represented by the above two-step process using $g'(o)$.

Now suppose that I am given a probability distribution $g(v)$ over $\mathbb{R}^d$, and I want to deremine whether it is IOBD.

My question is: what are some simple criteria, that can be applied to such a $g(v)$, to determine whether it is IOBD?

(My only idea is that isotropy of $g(v)$ implies IOBD. But this is not a neccesary condition. Ideally, I would like to have a sufficient and neccesary condition that can be applied to $g(v)$).

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    $\begingroup$ It might be helpful to frame this process in terms of the Stiefel manifold $\endgroup$ – Omnomnomnom Apr 18 at 16:49

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