Minimizing the maximum dot product among k unit vectors in an n-dimensional space
Answer: $\alpha = - 1/(k-1)$. This value is attained when vectors $v_i$ are the vertices of a regular simplex, centered at the origin. 

This follows from symmetry: given a set of vectors $u_i$ consider new vectors $u_i'= \frac{1}{\sqrt{k}}\left( u_i \oplus u_{i+1} \oplus \dots \oplus u_{i-1} \right)$ (where we add up all vectors $u_i$ in the cyclic order starting from $u_i$). It is easy to show that vectors $u'_i$ form a better (or equal) solution than vectors $u_i$, and that vectors $u_i$ are vertices of a regular simplex (not necessarily centered at $0$). Symmetry arguments (like this one) are very useful in general for finding extreme configurations of vectors. In this case, it is very easy to find the value of $\alpha$ directly.

I. Let us first prove that $\alpha \geq -\frac{1}{k-1}$. We have,
$$0 \leq \|v_1 + \dots + v_k\|^2 = \sum_i \|v_i\|^2 + 2\sum_{i &lt; j} v_i \cdot v_j \leq \sum_i 1 + 2\sum_{i&lt;j} \alpha = k + k(k-1) \alpha.$$
We get that $\alpha \geq - 1/(k-1)$.

II. Now we prove that  $\alpha \leq -\frac{1}{k-1}$. Consider a $k\times k$ matrix $A = (a_{ij})$ with $a_{ii} = 1$ and $a_{ij} = {-1/(k-1)}$ for $i\neq j$. This matrix is (non-strictly) diagonally dominant and thus is positive semidefinite. (Note also that $A$ is the normalized Laplacian matrix of $K_k$.) Therefore, there exist a set of vectors $v_i$ with $v_i \cdot v_j = a_{ij}$. We get that 
$$\alpha \leq \max_{i\neq j} a_{ij}  = -1/(k-1).$$

We proved that $\alpha = -1/(k-1)$. Note that vectors $v_i$ are vertices of a regular simplex since the distance $\|v_i - v_j\| = \sqrt{2- 2v_i \cdot v_j} = \sqrt{2k/(k-1)}$ is the same for all pairs $i,j$ (where $i\neq j$), and $\sum v_i =0$ since $\|\sum v_i\|^2 = 0$.

What we computed equals 
$$-\frac1{\vartheta(\text{empty graph on } k \text{ vertices})-1},$$
where $\vartheta$ is the  Lovász Theta Function;
see the Wikipedia article on the Lovász Theta Function for details.