# 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.

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Looks like the author switched the two up accidentally... – anon Apr 27 '12 at 7:32
@anon: I think it's okay. I address this below. – user26872 Apr 27 '12 at 8:02
FYI, the meaning of "covariant" and "contravariant" has swapped a few decades back, but not everybody uses the new meaning. – Hurkyl Apr 27 '12 at 9:03

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, $$w\cdot e_j = w_i e^i\cdot e_j = w_i \delta^{i}_j = w_j.$$
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!