TITLE: Are almost all $k$-tuples of vectors in high dimensional space almost orthogonal?
QUESTION [9 upvotes]: I would like to know, in high dimensions, if you pick $k$ vectors (not necessarily distinct) are they likely to be $\textit{almost}$ orthogonal. I make this precise below but I am also open to other interpretation of the question. 
Since we only care about orthogonality, we can assume the vectors are unit vectors. Thus I want to consider a set $\textbf{V}=\{x_1,...,x_k\}\subset S^n$, where $S^n$ is the unit sphere. We can thus view $\textbf{V}\in \underbrace{S^n\times\cdots\times S^n}_{k-times}=\mathcal{S}_{n,k}$. Let $\mathcal{O}\subset\mathcal{S}_{n,k}$ denote the subset of $k$-tuples of orthogonal vectors. Let $\mathcal{U}_\epsilon=\{x\in\mathcal{S}_{n,k} : d(x,\mathcal{O}) <\epsilon\}$ be the $\epsilon$ neighborhood of $\mathcal{O}$ in $\mathcal{S}_{n,k}$ (using, say, the $l^2$ product metric, I don't think this matters). Let $\mu$ be the obvious probability product measure on $\mathcal{S}_{n,k}$. My questions is then:
Is it true that $$\forall \epsilon > 0\textrm{ and } \forall k\textrm{ }\lim_{n\rightarrow\infty}\mu(\mathcal{U}_\epsilon) = 1$$
If not what can we say about this limit?

REPLY [3 votes]: It is true, and it can be seen as a consequence of the concentration of measure on the sphere.
Let $P$ be the uniform probability measure on $\mathbb S^{n}$. Given a subset $A\subset \mathbb S^n$, call $A_t:=\{y\in\mathbb S^n:dist(y,A)<t\}$ its $t$-neighbourhood. Then:
Concentration of measure: If $P(A)\geq\frac12$, then $P(A_t^c)\leq 2e^{-nt^2/64}$. (the specific constants are not important here)
Apply this to two complementary emispheres to obtain that for any $\nu\in\mathbb S^n$ we have $P(|x\cdot \nu|\geq t)\leq 4e^{-nt^2/64}$, i.e. the measure concentrates near any equator when the dimension increases.
In particular it follows immediately that the result you ask for is true if $k=2$: once you have chosen the first vector $x_1$, the probability that the second you choose will be near the hyperplane $x_1^\perp$ goes to $1$ as $n\to\infty$ (by the rotational invariance of the measure on the sphere it doesn't matter which specific $x_1$ you chose in the first step).
To prove it for general $k$ use induction. I'll sketch an argument. The case $k=2$ (or $k=1$ if you want) is the base case and it's proved. Suppose now the statement is true for $k$. With probability close to $1$ the first $k$ vectors are $\epsilon$-orthogonal. Move them one by one until $x_i$ is close to $e_i$, the $i^{th}$ standard unit base vector (by the rotation invariance we can compute probabilities in this new position). From the concentration of measure
$$ P(\{|x\cdot e_1|\geq t\}\cup \ldots\cup\{|x\cdot e_k|\geq t\})\leq 4ke^{-nt^2/64}$$
Therefore with probability close to $1$ the $(k+1)^{th}$ vector has small scalar products with the previous $x_i$'s. This implies that with probability close to $1$ the family is $\epsilon$-orthogonal.