What is the difference between covariant and contravariant tensors?

Question

What is the difference between covariant and contravariant tensors?

I have been seeing in a lot of problems but I´m not sure what is the difference or if is only a equivalent notation.

Answer 1

Tensors are invariant, by their very nature.

You should talk of covariant ("meaning varying like something"), or contravariant (meaning "varying in the inverse way w.r.t. something") base vectors, and components.

Now, we need to define what is "that something" that we use as a reference, to say if another object transform in its same way (and thus, we'll define it as a covariant object) or in the opposite way (and thus, we'll define as a contravariant object).

Using a set of coordinates $q^i$ to describe a space, i.e. it's possible to write a point in that space as $\mathbf{x}(q^i)$. If the parametrization $\mathbf{x}(q^i)$ is regular, it's possible to define what we call the local natural basis as $\mathbf{b}_i := \frac{\partial \mathbf{x}}{\partial q^i}$.

If we change the sets of parameters, $\mathbf{x}(\hat{q}^i)$, we get the natural basis for this new set of coordinates as

$\mathbf{\hat{b}_k} := \dfrac{\partial \mathbf{x}}{\partial \hat{q}^k} = \dfrac{\partial q^i}{\partial \hat{q}^k} \dfrac{\partial \mathbf{x}}{\partial q^i} = \hat{T}^{i}_{k} \mathbf{b}_i$,

finding the law of transformation of the natural basis due to a change of coordinates. This is often used as the reference transformation to determine if an object is covariant or contravariant.

As an example, let's write a vector field explicitly, using the natural base vectors of the sets of parameters $q^i$ and $\hat{q}^k$. We used 2 basis to write the same invariant object, so the basis and the components in these basis must be somehow related.

$\mathbf{v} = v^i \mathbf{b}_i = \hat{v}^k \mathbf{\hat{b}}_k = \underbrace{\hat{v}^k \hat{T}^{i}_{k}}_{v^i}\mathbf{b}_i$.

Thus, taking a look at the rule of transformation of the components in the natural bases, $v^i = \hat{v}^k \hat{T}^i_k$, it's possible to realize, introducing the inverse matrix $T^{n}_i$ of the matrix $\hat{T}^i_k$, the base vectors and the components follow inverse transformations,

$\mathbf{\hat{b}_k} = \hat{T}^{i}_{k} \mathbf{b}_i$,

$\hat{v}^k = T^{k}_{i} v^i$.

Now, we define all the objects that transform like the vector of the natural basis as covariant objects (with the convention used so far, with subscripts), the objects transforming with the inverse transformation as contravariant objects (with the convention used so far, with superscripts).

Further observations.

Obs #1. With tensors of higher order, it's possible to write a tensor using a mixed basis, i.e. using some vectors of the natural base $\mathbf{b}_i$, and some vectors of the reciprocal base $\mathbf{b}^k$ (that are contravariant), as an example

$\mathbb{T} = T^{ij}_{\ \ \ k} \mathbf{b}_i \otimes \mathbf{b}_j \otimes \mathbf{b}^k$.

Obs #2. The reciprocal base of a reciprocal base is defined as the set of vectors s.t. $\mathbf{b}^i \cdot \mathbf{b}_k = \delta^i_k$. It's possible to "change the nature of an index" using the components of the metric tensor $g_{ij} = \mathbf{b}_i \cdot \mathbf{b}_j$, $g^{ij} = \mathbf{b}^i \cdot \mathbf{b}^j$, as

$\mathbf{b}_i = g_{ij} \mathbf{b}^j$,
$\mathbf{b}^i = g^{ij} \mathbf{b}_j$,
$v_i = g_{ij} v^j$,
$v^i = g^{ij} v_j$,
$T^{ij}_{\ \ \ k} = g_{kl} T^{ijk}$,
and so on.