Lines in $n$ dimensional space Suppose we have $n+1$ lines in $\mathbb{R}^n$ and let $\gamma$ be the smallest angle across all pairs of lines. I am wondering what the upper bound on $\gamma$ is (probably depending on $n$). 
When $n=2$ we have $\gamma\leq \pi/3$, but even in the case $n=3$ I am having a hard time figuring it out. I think it might be $\arccos (1/n),$ but this is just a guess. Any ideas?
 A: Let $(x_i)_{i=1}^{n+1}$ be unit vectors in the directions of the lines.  Let $B$ be the $n\times(n+1)$ matrix whose $i$th column is $x_i$.  Since $B^TB$ is symmetric, its eigenvalues are real; let them be $(\lambda_i)_{i=1}^{n+1}$.  But note that since $B$ has rank at most $n$, so too $B^TB$ has rank at most $n$, and so at most $n$ of the eigenvalues are nonzero, so we can consider them to be $(\lambda_i)_{i=1}^n$.  Now,
\begin{align*}
\max_i \max_{j\ne i}\ \langle x_i,x_j\rangle^2
&\ge \frac1{n(n+1)} \sum_i \sum_{j\ne i} \langle x_i,x_j\rangle^2
&&\text{(max $\ge$ avg)} \\
&= \frac1{n(n+1)} \left(\sum_i \sum_j \langle x_i,x_j\rangle^2 - \sum_i |x_i|^4\right) \\
&= \frac1{n(n+1)} \left(\sum_i \sum_j \langle x_i,x_j\rangle^2 - (n+1)\right) \\
&= \frac1{n(n+1)} \left( \|B^TB\|_{\text{HS}}^2 - (n+1)\right)
&&\text{($\|\cdot\|_{\text{HS}}$ is Hilbert-Schmidt norm)}\\
&= \frac1{n(n+1)} \left( \sum_i \lambda_i^2 - (n+1)\right) \\
&\ge \frac1{n(n+1)} \left( \frac1n\Big(\sum_i \lambda_i\Big)^2 - (n+1)\right)
&&\text{(Cauchy-Schwarz)}\\
&= \frac1{n(n+1)} \left( \frac1n(\mathop{\text{tr}} B^TB)^2 - (n+1)\right) \\
&= \frac1{n(n+1)} \left( \frac1n\Big(\sum_i |x_i|^2\Big)^2 - (n+1)\right) \\
&= \frac1{n(n+1)} \left( \frac{(n+1)^2}{n} - (n+1)\right) \\
&= \frac1{n^2}
\end{align*}
Thus $\max_i \max_{j\ne i}\ |\langle x_i,x_j\rangle| \ge \frac1n$ and so the minimum angle between the lines is at most $\arccos\frac1n$, as you suspected.  This value is achieved with the lines from the centre of a regular simplex through its vertices (and through the centres of its facets).
This argument can be generalized for $m$ vectors, though I don't know any reason to suspect it'll retain its sharpness for larger $m$.
