This is a cross-post from [this other question](https://math.stackexchange.com/q/4726158/610053) that I asked ~1 month ago in the mathematics forum, with no reaction. I am still stuck on this, looking for references or approaches to proofs. I hope I have a bit more luck in this forum! let me know if any clarifications would be needed as I am not a professional mathematician.




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### Definitions:


Consider the set of orthogonal matrices in the compact Stiefel manifold:


$\mathcal{M}\_{N, k} := \{o \in \mathbb{R}^{N \times k} : \quad o^\top o = I\_k \}$


We also define the projector $P\_i$ of a matrix $o\_i \in \mathcal{M}\_{N, k}$ as:


$P\_i = o\_i o\_i^\top$


And the *principal angles* $\theta\_{i\rightarrow j}$ between 2 matrices $o\_i, o\_j$ (see e.g. [Absil, Mahony, Sepulchre, 2003](https://www.cis.upenn.edu/%7Ecis6100/Diffgeom-Grassmann.Absil.pdf) or [Qiu, Zhang, Li, 2005, s.2](https://eeqiu.people.ust.hk/wp-content/uploads/2021/04/Unitarily-Invariant-Metrics-on-the-Grassmann-Space.pdf)) as:


$SVD[o\_i^\top o\_j] = U \quad cos(\theta\_{i\rightarrow j}) \quad V^T$


Finally, consider the following (squared and normalized) Frobenius projective metric between any 2 matrices $(o\_i, o\_j)$ in $\mathcal{M}\_{N, k}$ (see e.g. [Edelman, Arias, Smith, 1998, section 4.3](https://arxiv.org/abs/physics/9806030)):


$d\_{pF}^2(o\_i, o\_j) = \frac{1}{k} \lVert P\_i - P\_j \rVert\_F^2 = \frac{1}{k} \lVert sin^2(\theta\_{i\rightarrow j}) \rVert\_F^2 = 1 - \frac{1}{k} \lVert cos ^2(\theta\_{i\rightarrow j}) \rVert\_F^2 = 1 - \frac{1}{k} \lVert o\_i^\top o\_j \rVert\_F^2 \in [0, 1]$




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### Setup:


1. Draw a random pair $(o\_1, o\_2) \sim (\mathcal{M}\_{N, k} \times \mathcal{M}\_{N, k})$ uniformly from the Haar measure (e.g. using the [QR method](https://finmath.stanford.edu/%7Ecgates/PERSI/papers/subgroup-rand-var.pdf))
2. Compute $d\_{pF}^2(o\_1, o\_2)$
3. Repeat steps 1 and 2 for different random pairs, and average the results. We observe that $\frac{1}{k} \lVert o\_i^\top o\_j \rVert\_F^2$ converges to $\frac{k}{N}$. Or equivalently, $\mathbb{E}[d\_{pF}^2] = 1 - \frac{k}{N}$




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### Question:


I imagine this is a well known result in the literature. Where could I find it?
Alternatively, how could I prove it?


The approach I thought of so far seems rather cumbersome, and I am not sure if it would be correct:


1. Take the joint distribution $\mathcal{X}$ of principal angles for a Haar-uniform random orthogonal matrix in $\mathbb{R}^{N \times N}$ (presented e.g. in [Rummler, 2002](https://core.ac.uk/download/pdf/159153606.pdf)).
2. Truncate $\mathcal{X}$ to $k$ entries, and calculate the normalized squared cosine, i.e. $\mathcal{Y} = \frac{1}{k} cos^2(\mathcal{X}\_k)$
3. Then, $\mathbb{E}[\mathcal{Y}]$ should equal $\frac{k}{N}$ (?)


Again, this is probably not necessary since I assume this result is known and I'm just missing the reference. Thanks in advance!



Expected value of the projective metric between random orthogonal Stiefel matrices in $\mathbb{R}^{N \times k}$ equals $1 - \frac{k}{N}$
Fortunately, there's nothing deep going on here.


We'll use slightly different notation. Let $\mathbf{Q} \in \mathbb{R}^{N \times k}$ be a random matrix drawn uniformly from the Stiefel manifold of $N \times k$ orthonormal frames. In particular,


* The columns of $\mathbf{Q}$ are orthonormal.
* The marginal distribution of each column $\mathbf{q}\_i$ is the uniform distribution over the Euclidean unit sphere, hence is isotropic: $\mathbb{E} [\mathbf{q}\_i \mathbf{q}\_i^\* ] = N^{-1} \mathbf{I}\_N$, where $\mathbf{I}\_N$ is the $N \times N$ identity matrix.


The orthogonal projector onto the range of $\mathbf{Q}$ takes the form $\mathbf{P} = \mathbf{QQ}^\*$, where $\*$ is the transpose. The key observation here is that
$$
\mathbb{E}[ \mathbf{P} ] = \sum\_{i=1}^k \mathbf{E}[ \mathbf{q}\_i \mathbf{q}\_i^\* ] = \sum\_{i=1}^k \frac{1}{N} \mathbf{I}\_N = \frac{k}{N} \mathbf{I}\_N.
$$


Given two independent realizations $\mathbf{Q}\_1, \mathbf{Q}\_2$, we form the associated orthogonal projectors $\mathbf{P}\_1, \mathbf{P}\_2$. Write $\mathbb{E}\_1, \mathbb{E}\_2$ for the expectations with respect to the randomness in $\mathbf{Q}\_1, \mathbf{Q}\_2$ respectively. Using the definition of the Frobenius norm in terms of the trace, linearity, and independence, we find that
\begin{multline\*}
\mathbb{E} \Vert \mathbf{P}\_1 \mathbf{P}\_2 \Vert\_{\mathrm{F}}^2
= \mathbb{E} \operatorname{trace}[ \mathbf{P}\_2 \mathbf{P}\_1^2 \mathbf{P}\_2 ]
= \mathbb{E} \operatorname{trace}[ \mathbf{P}\_2 \mathbf{P}\_1 \mathbf{P}\_2 ] \\
= \mathbb{E}\_2 \operatorname{trace}[ \mathbf{P}\_2 \mathbb{E}\_1[ \mathbf{P}\_1] \mathbf{P}\_2 ]
= \frac{k}{N} \mathbb{E}\_2 \operatorname{trace}[ \mathbf{P}\_2^2 ]
= \frac{k^2}{N^2} \operatorname{trace}[ \mathbf{I}\_N ]
= \frac{k^2}{N}.
\end{multline\*}
This is the core part of the computation.


Finally, we compute the expected Frobenius projective distance between the pair:
\begin{multline\*}
\mathbb{E} \mathrm{dist}\_{\mathrm{F}}^2(\mathbf{Q}\_1, \mathbf{Q}\_2)
= \mathbb{E}\big[ 1 - k^{-1} \Vert \mathbf{Q}\_1^\* \mathbf{Q}\_2 \Vert\_{\mathrm{F}}^2 \big] \\
= \mathbb{E}\big[ 1 - k^{-1} \Vert \mathbf{P}\_1 \mathbf{P}\_2 \Vert\_{\mathrm{F}}^2 \big]
= 1 - k^{-1} (k^2 / N)
= 1 - k / N.
\end{multline\*}
That's it.


