**Setting:** Consider sampling two orthonormal vectors $\mathbf{u},\mathbf{v}$ in $\mathbb{R}^p$ (where $p\ge2$) from a "uniform" distribution over the $p$-dimensional sphere (alternatively, sample $\mathbf{u},\mathbf{v}\sim\mathcal{S}^{p-1}$ and then orthogonalize and normalize $\mathbf{v}$).


Let $\lceil p/2\rceil\le m\le p$.  

**Question:** Can we upper bound the following expectation?
$$ \mathbb{E}\_{\substack{\mathbf{u},\mathbf{v}\sim\mathcal{S}^{p-1}:\\\mathbf{u}\perp\mathbf{v}}}
\left(\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{I}\_{m}\\
 & \mathbf{0}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2
=
\mathbb{E}\_{\substack{\mathbf{u},\mathbf{v}\sim\mathcal{S}^{p-1}:\\\mathbf{u}\perp\mathbf{v}}}
\left(\sum\_{i=1}^{m}u\_iv\_i\right)^2
$$


**Example:** when $m$ is very close to $p$, the expectation should be very small, since the vectors are nearly orthogonal.


**Direction:** We thought about approximating this with just two independent Gaussian vectors and look for the concentration bounds when $p\to\infty$, but this seems perhaps too loose?


**Observation:** Since the vectors are orthogonal, we have $
\left(\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{I}\_{m}\\
 & \mathbf{0}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2
=
\left(\mathbf{u}^\top\mathbf{v}-\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{I}\_{m}\\
 & \mathbf{0}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2
=
\left(\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{0}\_{m}\\
 & \mathbf{I}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2$, and thus we can equivalently focus on $0\le m\le \lfloor p/2\rfloor$.


Any help and ideas would be greatly appreciated!



Expectation of the inner product of a subset of two random orthonormal vectors
Denote $\alpha=\mathbb{E} u\_1^2v\_1^2$, $\beta=\mathbb{E} u\_1v\_1u\_2v\_2$. Then by the symmetry and linearity of expectation we have $$f(m):=\mathbb{E} (u\_1v\_1+\ldots+u\_mv\_m)^2=m\alpha+m(m-1)\beta.$$
We have $f(p)=0$, thus $\beta=-\alpha/(p-1)$, and $f(m)=\alpha m(p-m)/(p-1)$.


It remains to bound $\alpha$. Choose a vector $w=(w\_1,\ldots,w\_p)\in \mathcal{S}^{p-1}$ independent of $u,v$, then $\alpha=\mathbb{E} \langle u,w\rangle^2\cdot \langle v,w\rangle^2$, as the conditional expectation clearly does not depend on $w$ (here $\langle \cdot,\cdot\rangle$ stands for the inner product). On the other hand, the conditional expectation does not depend on the pair $u,v$ of orthogonal vectors $u,v$, thus we may take $u=(1,0,\ldots,0)$, $v=(0,1,\ldots,0)$, and $\alpha=\mathbb{E} w\_1^2w\_2^2$.


Next, we have
$$1=\mathbb{E} \left(\sum\_{i=1}^p w\_i^2\right)^2=\alpha\cdot p(p-1)+p\cdot \mathbb{E} w\_1^4,$$
thus $$\alpha=\frac{1-p\cdot \mathbb{E} w\_1^4}{p(p-1)}.$$
To find $\mathbb{E} w\_1^4$, we may think that $w=(w\_1,\ldots,w\_p)=(\xi\_1,\ldots,\xi\_p)/\sqrt{\sum \xi\_i^2}$ where $\xi\_i$ are i.i.d. standard normal. Then
$$
\mathbb{E} w\_1^4=\mathbb{E} \frac{\xi\_1^4}{(\sum \xi\_i^2)^2}=:\Theta
$$
By some form of law of large numbers like Chernoff bound, the probability that $\sum \xi\_i^2<p/2$ is exponentially small in $p$, that gives exponentially small contribution to the expectation $\Theta$. If $\sum \xi\_i^2>p/2$, then $$\frac{\xi\_1^4}{(\sum \xi\_i^2)^2}<\frac{4}{p^2}\xi\_1^4,$$
and since $\mathbb{E} \xi\_1^4$ is a finite constant, we conclude that $\Theta=O(1/p^2)$. Thus $\alpha=1/p^2+O(1/p^3)$.


