Everything I read on tensors makes it clear that using the metric matrix $g_{ab}$ and its inverse $g^{ab}$ to respectively lower and raise indices of a tensor is very important. As far as I know (and I might be wrong) a tensor is defined to be something whose representation in coordinates $x^\alpha$ and $x'^\alpha$ is related either by a specific combination of partial derivatives of the new coordinates w.r.t. the old coordinates or vice-versa, the combination given by (for [1,1] tensors, for example)
$T'^a_b=\dfrac{\partial x'^c}{\partial x^a } \dfrac{\partial x^b}{\partial x'^d}T^c_d$ where the upper indices are called contravariant and the lower covariant. I understand that given a covariant vector say $x^\mu$, we define $x_\mu:=g_{\mu \nu}x^{\nu}$. My first problem is that I'm not sure why $x_\mu $ defined in this way should be covariant i.e. I'm not sure that this is well defined.
Secondly, if I write something like $V^\alpha$, am I implying that this is contravariant? Say I have a vector that is neither contravariant nor covariant, that is, just a collection of $n$ measurements in a vector $(a_1,a_2,\dots, a_n)$ and I write this as $a_i$. In a general relativity setting, am I implying that this is covariant?
I'm new to general relativity, so it would be really helpful if in addition to the two questions I had, someone could confirm that what I've written otherwise is correct (or tell me that I've got it completely wrong!). Thanks for any help.
Edit: If $g_{ab}$ complies with the tensor notation then in the new coordinates
$g'_{ab}=\dfrac{\partial x^c}{\partial x'^a } \dfrac{\partial x^d}{\partial x'^b}g_{cd}$ and so
$g'_{ab}x'^b=(\dfrac{\partial x^c}{\partial x'^a } \dfrac{\partial x^d}{\partial x'^b}g_{cd})(\dfrac{\partial x'^b}{\partial x^e}x^e$)
and since $\dfrac{\partial x^d}{\partial x'^b } \dfrac{\partial x'^b}{\partial x^e}=4\delta^d_e$ (where $\delta^d_e$=1 iff $d=e$, otherwise $\delta^d_e=0$), the coefficient of 4 because 4 goes from 0 to 3, I get to
$g'_{ab}x'^b=4\dfrac{\partial x^c}{\partial x'^a }g_{cd}\delta^d_e x^e=4\dfrac{\partial x^c}{\partial x'^a }g_{cd}x^d$
I'm quite close but there's a factor of 4 that shouldn't be there?
Hint for part I: Consider how the product $g_{\mu\nu} \; x^\nu$ transforms, since you know how the individual terms in the product transform under a change of basis. You will see that it transforms exactly like $x_\mu$.
For the second part of your question: "a vector that is neither contravariant or covariant" is not a vector (in the sense used in physics). A simple collection of n quantities is not a vector or a co-vector.
Clearly, $g_{\mu\nu}$, being covariant components transform as
$$ g^{'}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial x'^{\alpha}}\frac{\partial x^{\nu}}{\partial x'^{\beta}}g_{\mu\nu} $$
The contravariant components $x^{\nu}$ transform the other way $$ x^{'\alpha} = \frac{\partial x^{\alpha}}{\partial x'^{\nu}}x^{\nu} $$
So that $$ g^{'}_{\alpha\beta} x^{'\alpha} = \frac{\partial x^{\mu}}{\partial x'^{\alpha}}\frac{\partial x^{\nu}}{\partial x'^{\beta}}\frac{\partial x'^{\alpha}}{\partial x^{\rho}} g_{\mu\nu} x^{\rho} \\ = \delta^{\mu}_{\rho} \frac{\partial x^{\nu}}{\partial x'^{\beta}} g_{\mu\nu} x^{\rho} \\ = \left( \frac{\partial x^{\nu}}{\partial x'^{\beta}} \right) g_{\rho\nu} x^{\rho} $$
The bracketed term in the last equation shows clearly that the product transforms as covariant components do, i.e. like $x_\mu$ No extra factors of 4 anywhere.
Also, as Steven Gubkin says, this is not a good way to think about tensors at all. The reason is that we were talking about components of the tensors (in a basis) all the time, rather than the tensors themselves. This is a historical artifact of how tensors were used in physics (especially GR).
To see why you won't have a 4 or 3, think about the term $$\frac{\partial x^{\mu}}{\partial x'^{\alpha}} \frac{\partial x'^{\alpha}}{\partial x^{\rho}}$$
If you expand the sum you'll have
$$\frac{\partial x^{\mu}}{\partial x'^{1}} \frac{\partial x'^{1}}{\partial x^{\rho}} + \ldots$$ But if you remember the chain rule, you need to add all 4 (or 3) terms to replace this sum by $\delta^{\mu}_{\rho}$ $$\frac{\partial x^{\mu}}{\partial x^{\rho}}$$