vector/tensor covariance and contravariance notation As I looked over the Wikipedia article on covariance and contravariance of vectors and $\mathbf{v}=v^i\mathbf{e}_i$ is said as a contravariant vector while $\mathbf{v}=v_i\mathbf{e}^i$ is said as covariant vector (or covector).
However, in the latter part, the article says:

Then the contravariant coordinates of any vector $\mathbf{v}$ can be obtained by the dot product of $\mathbf{v}$ with the contravariant basis vectors: $q^1=\mathbf{v}\cdot \mathbf{e}^1$, $q^2=\mathbf{v}\cdot \mathbf{e}^2$, and $q^3=\mathbf{v}\cdot \mathbf{e}^3$. Likewise, the covariant components of $\mathbf{v}$ can be obtained from the dot product of $\mathbf{v}$ with covariant basis vectors, viz. $q_1=\mathbf{v}\cdot \mathbf{e}_1$, $q_2=\mathbf{v}\cdot \mathbf{e}_2$, and $q_3=\mathbf{v}\cdot \mathbf{e}_3$.

I am getting confused. It seems that the location of the indices (up or down) of contravariant or covariant vectors is different in these two different parts.
Can anyone show me what the heck is this?
Thanks.
 A: To get the components of the contravariant vector $v = v^i e_i$, where $e_i$ is the natural basis, we dot with the basis vectors $e^i$ for the dual space,
$$v\cdot e^j = v^i e_i\cdot e^j = v^i \delta_{i}^j = v^j.$$
Likewise, to find the components of a covariant vector $w = w_i e^i$ we dot with basis vectors from the natural basis, 
$$w\cdot e_j = w_i e^i\cdot e_j = w_i \delta^{i}_j = w_j.$$
Sometimes the natural basis vectors are called covariant (since their indices are downstairs) and the dual basis vectors contravariant (since their indices are upstairs). 
With this convention a contravariant vector, with contravariant components, is written in terms of the covariant basis! 
After a while, you get used to this sort of nonsense. 
Addendum: The terms contravariant and covariant refer to how an object transforms under coordinate transformation, $x\to x'$. 
In physics, where one is often dealing with coordinates, this is especially vivid. 
Does the thing transform contravariantly with $\frac{\partial {x'}^j}{\partial x^i}$ or covariantly with $\frac{\partial {x}^j}{\partial {x'}^i}$?
That is why the terminology is not so bad.
$e^i$ really does transform contravariantly.
This has to be the case so that 
$$\begin{eqnarray*}
v &=& v^i e_i \\
&=& v^i \delta_i^j e_j \\
&=& v^i 
\frac{\partial {x'}^k}{\partial x^i} 
\frac{\partial {x}^j}{\partial {x'}^k} e_j \\
&=& {v'}^i {e'}_i.
\end{eqnarray*}$$
To add another wrinkle, physicists also often say that an object that is invariant under transformation is covariant!
A: I'm a bit late to this thread, but I want to clarify one point.  The vector, ${\bf v} = v^i{\bf e}_i = v_i{\bf e}^i$, is neither contravariant nor covariant.  It is the components that are either contravariant ($v^i$) or covariant ($v_i$).
A given vector, ${\bf v}$, can be represented by the components it has in some basis or dual basis, and those components will generally change if the basis is changed, but the vector does not change.  This also extends to tensors with more indices.  So "contravariant" and "covariant" refer to how the components of the object will change when changing from one basis to another, but the object itself does not change.  
In practice, it is common to think about and talk about the components as if they are the vector, which is fine if you just want to calculate something, but the distinction can be important when talking about the abstract properties of the vector.
EDIT: The equality $v^i{\bf e}_i = v_i{\bf e}^i$ is more a correspondance than and equality, as is discussed on this post.
