Are basis vectors imaginary in special relativity? It sounds absurd, but it seems to come naturally:
I know that $g_{\mu \nu }=e_\mu \cdot e_\nu $ (source)
and that the diagonal elements of $\eta_{\mu \nu}$ are typically $\pm(1, -1, -1, -1)$. 
This seems to very clearly suggest that the basis vectors corresponding to negative metric components could be $(0, i, 0, 0)$ or similar.  
If these are not the basis vectors in SR, I would like to ask:


*

*What are the basis vectors, then? (of course they're variable, but what would a Cartesian basis look like?)


2) If $g_{\mu \nu }=e_\mu \cdot e_\nu $ does not hold for those vectors, why not?
 A: The basis vectors are exactly what you'd expect,
$$(1, 0, 0, 0), \quad (0, 1, 0, 0), \quad (0, 0, 1, 0), \quad (0, 0, 0, 1).$$
However, the inner product, i.e. the way we combine two vectors into a number, is not the same as the usual dot product. Using your notation, we're changing the definition of $\cdot$, not the definition of the $e_{\mu}$. 
You are correct that it's possible to continue working with the dot product formally if we define some of the basis vectors to have imaginary components. That's how it was done in the past, but it's a bad idea: time and lengths just aren't complex numbers. They're perfectly real, so moving to a complex vector space doesn't make physical sense. (Moreover, the dot product itself is unnatural in a complex vector space, where the Hermitian inner product fits better.)
A: There are several approaches to this question.  A long time ago people did use the hack of making some of the basis vectors complex in an attempt to make things look 'more natural'.  But this doesn't actually solve any problems worth solving: in fact it makes them worse.
If you wanted to treat the metric as corresponding to an inner product, then it should actually do so.  In particular, for a vector space over any field we need $\langle \vec{u},\vec{u}\rangle \ge 0$ with equality only when $\vec{u}=\vec{0}$.  This is just as true for a vector space over $\mathbb{C}$ as it is for one over $\mathbb{R}$.
But this simply is not true for the metric of relativity: the whole structure of the thing rests on their being null vectors: vectors which are not zero but whose 'lengths' are zero.  There is no way out of this problem other than to realise that the metric in relativity does not correspond to an inner product, because it is not positive  definite.
That being the case it is far simpler to abandon any hacky approach involving complex numbers, and to accept the metric as what it is: not actually a metric at all, but something which has all the properties of a metric except positive-definiteness.
(An earlier version of this called it a 'pseudometric', but as pointed out by John Davis in a comment, it is not even that: a pseudometric is nonnegative, but can be zero for distinct vectors.)
