# Co and contravariant: tensors or components?

I am learning Special Relativity and have a question: given a four vector $\vec{x}$ whose contravariant components are $x^\mu$, do the covariant components $x_\mu = g_{\mu\nu}x^\nu$ make reference to a different physical/geometrical object other than $\vec{x}$?

I mean, for the physical/geometrical object $\vec{x}$ we can say

$$\vec{x} \underset{\text{has components}}{\longrightarrow} x^\mu$$

Then, who is $\vec{?}$ in the following expression?, is it $\vec{x}$ to?

$$\vec{?} \underset{\text{has components}}{\longrightarrow} x_\mu$$

• "$\vec{?}$ doesn't exist, because it is a one-form. Actually, choosing $x$ to denote the 4-vector is a little unfortunate. Oct 23, 2017 at 14:19

Yes, the contravariant components make reference to a different geometrical object than the covariant components. The covariant components are components of a vector from the dual space to the vector space the contravariant components come from. The two vector spaces are isomorphic, so we can identify contravariant elements of vectors with matching elements of the covectors from their space using a simple linear map, or "metric".

When dealing with matrices, we often identify matrices that have one column as vectors, and matrices that have one row as covectors. So, the raising and lowering of indices is analogous to the matrix transpose operation, when the vector's components are real. When they aren't real, we usually use the complex conjugate transpose in order to make the length of any vector a positive semi-definite number.

In quantum mechanics, we usually treat ket-vectors as the vector space, and bra-vectors as the covector space.

So, why do we call them covariant and contravariant? It has to do with the fact that we want the scalars produced by applying the covariant vector to a contravariant one to be invariant under some symmetry transformation (e.g. rotation, Lorentz transformations). Say, we have a contravariant vector that varies as: $$x^{i} \rightarrow M^{i}_{\hphantom{i}j} x^j.$$ Then the corresponding covariant vector, $x_i \equiv g_{ij}x^j$, will vary as: $$x_i \rightarrow M_{i}^{\hphantom{i}j} x_j.$$ That is, covariant vectors change with the transpose of the symmetry matrix, relative to how the contravariant version changes.

The reason for this nomenclature is because we like to think of the vectors themselves as being invariant under the transformation, so we write: $$\vec{x} = x^i \hat{e}_i,$$ where $\hat{e}_i$ are the basis vectors of the vector space. In order for $\vec{x}$ to be invariant, the components $x^i$ have to transform in the opposite sense as the basis vectors (hence, 'contra' for against). Likewise, if we use $$\vec{y} = y_i \hat{e}^i,$$ then in order to be invariant the components $y_i$ have to vary in the same sense (using the same matrix) as the basis vectors of the original vector space.

For more, see the tensor formulation section of Wikipedia's Lorentz transformation article.

For orthogonal coordinate systems, the covariant and contravariant components are the same. The difference shows up when you have oblique coordinate systems. Here's a pretty good explanation of the difference with some illustrations:

http://www.farmingdale.edu/faculty/peter-nolan/pdf/relativity/Ch04Rel.pdf

Since we are dealing with a finite dimensional metric space, it's fine to just think of a field $\mathbf v$ and two bases $\mathbf e_\mu$ and $\mathbf e^\nu$ such that $\mathbf e_\mu \cdot \mathbf e^\nu = \delta_\mu^\nu$. Then $v^\mu$ and $v_\nu$ are just the coefficients for $\mathbf v$ in the two bases: $\mathbf v = v^\mu \mathbf e_\mu = v_\nu \mathbf e^\nu$.