TITLE: More approximately orthogonal vectors than the dimension of the space
QUESTION [10 upvotes]: It is impossible to find $n+1$ mutually orthogonal unit vectors in $\mathbb{R}^n$. 
However, a simple geometric argument shows that the central angle between any two legs of a simplex goes as $\theta = \mathrm{arccos}(-1/n)$. This approaches $90$ degrees as $n \rightarrow \infty$, so since there are $n+1$ vertices of a simplex in $n$-dimensional space, we can conclude

Given $\epsilon > 0$, there exists a $n$ such that we can find $n+1~$ approximately mutually orthogonal vectors in $\mathbb{R}^n$, up to tolerance $\epsilon$. (Unit vectors $u$ and $v$ are said to be approximately orthogonal to tolerance $\epsilon$ if their inner product satisfies $\langle u,v \rangle < \epsilon$)

My question is a natural generalization of this - if we can squeeze $n+1$ approximately mutually orthogonal vectors into $\mathbb{R}^n$ for $n$ sufficiently large, how many more vectors can we squeeze in? $n+2$? $n+m$ for any $m$? $2n$? $e^n$? 

Edit: I think one can squeeze at least $n+m$ for any $m$, via the following construction. Given $\epsilon$, one finds the $k$ such that you can have $k+1$ $\epsilon$-approximate mutually orthogonal unit vectors in $\mathbb{R}^k$. Call these vectors $v_1, v_2, ..., v_k$. Then you could squeeze $mk+m$ vectors in $\mathbb{R}^{mk}$, by using the vectors
$$\begin{bmatrix}
v_1 \\
0 \\
\vdots \\
0
\end{bmatrix},
\begin{bmatrix}
v_2 \\
0 \\
\vdots \\
0
\end{bmatrix},
\begin{bmatrix}
v_{k+1} \\
0 \\
\vdots \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
v_1 \\
\vdots \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
v_2 \\
\vdots \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
v_{k+1} \\
\vdots \\
0
\end{bmatrix}, \dots
\begin{bmatrix}
0 \\
0 \\
\vdots \\
v_1 \\
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
\vdots \\
v_2 \\
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
\vdots \\
v_{k+1} \\
\end{bmatrix}.
$$
So, setting $n = mk$, we have found an $n$ such that we can fit $n + m$ $\epsilon$-orthogonal unit vectors in $\mathbb{R}^n$.

REPLY [3 votes]: This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Lucia below.

A set of points on the unit sphere in ${\Bbb R}^n$ with $\langle x,y\rangle \le \cos \theta$ for all distinct $x$ and $y$ is called a spherical code with minimum angle $\theta$.  For $0<\theta < \pi/2$, Kabatiansky and Levenshtein gave an exponential upper bound (of the form $\exp(C(\theta)n)$) for the maximum number of points in such a spherical code.  There is also an exponential lower bound.  This is related to sphere packings.  See for example the recent paper by Cohn and Zhao, which will have more references:  http://arxiv.org/pdf/1212.5966v2.pdf