Out there

Suppose that we can find $n$ lines in $\mathbb{R}^k$ which are inclined at the same angle $\alpha$ to each other. If $\alpha = 0$ then $n=1$ , and so certainly $n \le \tfrac12k(k+1)$ . Assume now that $0 < \sin\alpha < 1$ . This means that we can find $n$ unit vectors $v_1,v_2,\ldots,v_n \in \mathbb{R}^k$ such that $v_i \cdot v_j \,=\, \pm\cos\alpha$ for $1\le i < j < n$ . Consider the $k \times k$ symmetric matrix $V_i \; = \; v_i v_i^{\mathsf{T}} \qquad 1 \le i \le n$ We note that $V_iV_j \,=\, (v_i\cdot v_j)v_iv_j^{\mathsf{T}}$ for all $1 \le i,j \le n$ , so that $\mathrm{Tr}(V_iV_j) \; = \; (v_i \cdot v_j)^2 \; = \; \left\{ \begin{array}{lll} 1 & \qquad & i = j \\ \cos^2\alpha & \quad & i \neq j \end{array} \right.$

Suppose now that $c_1,c_2,\ldots,c_n \in \mathbb{R}$ and that $W \; = \; \sum_{i=1}^n c_i V_i \; = \; 0$ Then $\begin{array}{rcl} 0 \; = \; \mathrm{Tr}(W^2) & = & \displaystyle\sum_{i,j=1}^n c_ic_j \mathrm{Tr}(V_iV_j) \\ & = & \sum_{i=1}^n c_i^2 + \cos^2\alpha\sum_{i \neq j} c_ic_j \\ & = & \displaystyle\sin^2\alpha \sum_{i=1}^n c_i^2 + \cos^2\alpha\left(\sum_{i=1}^n c_i\right)^2 \end{array}$ and so we deduce that $c_1=c_2=\cdots=c_n=0$ . Thus we have shown that the collection of matrices $\{V_1,V_2,\ldots,V_n\}$ is a linearly independent set in the vector space of symmetric $k \times k$ matrices, which is $\tfrac12k(k+1)$ -dimensional. Thus we deduce that $n \le \tfrac12k(k+1)$ .

Putting $k=7$ , we can find at most $28$ lines which are equally inclined to each other. It remains to find a specific set of $28$ lines in $\mathbb{R}^7$ .

Consider the $28$ vectors in $\mathbb{R}^8$ consisting of $\tfrac{1}{\sqrt{24}}\big(\begin{array}{cccccccc}3&3&-1&-1&-1&-1&-1&-1\end{array}\big)^{\mathsf{T}}$ and all vectors obtained by permuting its components. Each of these $24$ vectors is a unit vector, and the inner product of any distinct two of them is equal to $\tfrac13$ if the two vectors have a $3$ in the same position, and is equal to $-\tfrac13$ if the two vectors do not have a $3$ in the same position. Thus each of these $24$ vectors define a line, and any two of these lines are inclined at the angle $\cos^{-1}\tfrac13$ to each other. Since all $24$ vectors are orthogonal to $(1\;1\;1\;1\;1\;1\;1\;1)^{\mathsf{T}}$ , these $28$ vectors reside in a $7$ -dimensional subspace of $\mathbb{R}^8$ which is isomorphic to $\mathbb{R}^7$ , and hence we have found $28$ lines in $\mathbb{R}^7$ which are all inclined at the same angle to each other.