Covariant and Contravariant Coordinates

Prerequisites: Linear Algebra, Tensor Calculus. Full path (4 pages)

• Why is it interesting?
• Intuitive
• Concrete
• Abstract

Why is it interesting?

Covariant and contravariant coordinates are commonly used in special relativity.

Intuitive

Describing an arrow by numbers requires a set of reference directions (axes). When the axes are perpendicular and of unit length there is only one natural way to do it. When the axes are slanted, or stretched, there are two, and they give different numbers.

Picture a flat plane with two axes that meet at an angle of, say, 60 degrees. Draw an arrow from the origin.

• First way: ask how many steps along the first axis and how many along the second axis we need to reach the tip, walking parallel to the axes (the parallelogram rule). These are the contravariant components.
• Second way: drop a perpendicular from the tip onto each axis and read off how far along that axis it lands, or more precisely, take the dot product of the arrow with each axis vector. These are the covariant components.

For perpendicular unit axes both recipes give the same numbers, which is why the distinction never appears in elementary physics. For slanted axes they differ.

The names describe what happens when the axes change. If we make the axis vectors twice as long, we need half as many steps to reach the same tip, so the contravariant components shrink: they vary "contrary" to the axes. The dot products with the longer axis vectors double: they vary "with" the axes, covariantly.

Both sets of numbers describe the same arrow. The rule that converts one into the other is the metric, which encodes the lengths of the axis vectors and the angles between them. In special relativity the metric has a minus sign in it, so even for the standard axes the two kinds of components differ by signs.

Concrete

Let $\vec e_1, \dots, \vec e_n$ be a basis, not necessarily orthonormal, of a vector space with a scalar product. A vector is expanded as

$$\vec v = v^i \vec e_i \qquad \text{(sum over } i\text{)} ,$$

and the numbers $v^i$ (upper index) are its contravariant components. The scalar products of the basis vectors define the metric

$$g_{ij} = \vec e_i \cdot \vec e_j ,$$

and the covariant components (lower index) are the projections

$$v_i = \vec v \cdot \vec e_i = g_{ij} v^j .$$

With the inverse matrix $g^{ij}$ ($g^{ij} g_{jk} = \delta^i_k$) we go back: $v^i = g^{ij} v_j$. The scalar product of two vectors takes the form $\vec v \cdot \vec w = g_{ij} v^i w^j = v_i w^i$.

It is useful to introduce the dual basis $\vec e^{\,i} = g^{ij} \vec e_j$, which satisfies $\vec e^{\,i} \cdot \vec e_j = \delta^i_j$. Then $\vec v = v_i\, \vec e^{\,i}$ and $v^i = \vec v \cdot \vec e^{\,i}$.

Example. In the plane take $\vec e_1 = (1, 0)$ and $\vec e_2 = (\cos\alpha, \sin\alpha)$. Then

$$g_{ij} = \begin{pmatrix} 1 & \cos\alpha \\ \cos\alpha & 1 \end{pmatrix}, \qquad v_1 = v^1 + v^2 \cos\alpha, \qquad v_2 = v^1 \cos\alpha + v^2 .$$

For $\alpha = 90^\circ$ the metric is the unit matrix and $v_i = v^i$.

Change of basis. Let $\vec{e}^{\,\prime}_j = A^i{}_j\, \vec e_i$. From $\vec v = v'^j \vec{e}^{\,\prime}_j = v'^j A^i{}_j \vec e_i$ we read off $v^i = A^i{}_j v'^j$, so

$$v'^j = (A^{-1})^j{}_i\, v^i, \qquad v'_j = \vec v \cdot \vec{e}^{\,\prime}_j = A^i{}_j\, v_i .$$

The covariant components transform with the same matrix as the basis vectors, the contravariant ones with the inverse.

Coordinates. On a manifold with coordinates $x^\mu$, the components of a velocity $dx^\mu/dt$ are contravariant, and under a change of coordinates $x \to x'$

$$V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu , \qquad W'_\mu = \frac{\partial x^\nu}{\partial x'^\mu} W_\nu .$$

The gradient $\partial_\mu \phi$ of a scalar function transforms like $W_\mu$, so it is naturally covariant. The contraction $W_\mu V^\mu$ is the same in all coordinate systems.

Special relativity. With the Minkowski metric in the signature $(+,-,-,-)$, $\eta_{\mu\nu} = \mathrm{diag}(1, -1, -1, -1)$, the position four-vector $x^\mu = (ct, \vec x)$ has covariant components $x_\mu = \eta_{\mu\nu} x^\nu = (ct, -\vec x)$. For the four-momentum $p^\mu = (E/c, \vec p)$ the invariant contraction is $p_\mu x^\mu = Et - \vec p \cdot \vec x$, the combination that appears in the phase of a plane wave, and $p_\mu p^\mu = E^2/c^2 - \vec p^{\,2} = m^2 c^2$.

• The best explanation can be found in Chapter 4 in "A Student's Guide to Vectors and Tensors" by Daniel A. Fleisch
• See also http://www.project-tartarus.com/2017/05/what-are-covariant-and-contravariant-coordinates/

Abstract

Let $V$ be a finite-dimensional real vector space and $V^* = \mathrm{Hom}(V, \mathbb{R})$ its dual. A basis $e_i$ of $V$ determines a dual basis $\theta^i$ of $V^*$ by $\theta^i(e_j) = \delta^i_j$. Under a change of basis $e'_j = A^i{}_j e_i$ with $A \in GL(n, \mathbb{R})$, the dual basis changes by $\theta'^j = (A^{-1})^j{}_i\, \theta^i$. Components of vectors $v = v^i e_i$ therefore transform with $A^{-1}$ and components of covectors $\alpha = \alpha_i \theta^i$ with $A$. In the physics terminology, elements of $V$ are contravariant vectors and elements of $V^*$ are covariant vectors. A tensor of type $(r, s)$ is an element of $V^{\otimes r} \otimes (V^*)^{\otimes s}$, with $r$ upper and $s$ lower indices.

Without further structure there is no basis-independent isomorphism $V \cong V^*$. A non-degenerate symmetric bilinear form $g$ provides one, the musical isomorphisms

$$\flat : V \to V^*, \quad v \mapsto g(v, \cdot), \qquad \sharp = \flat^{-1} ,$$

which in components are lowering and raising of indices, $v_i = g_{ij} v^j$ and $\alpha^i = g^{ij} \alpha_j$. The covariant components of a vector in the Concrete layer are the components of $v^\flat$ in the dual basis. The identification depends on $g$. In general relativity the metric is a dynamical field, so raising or lowering an index brings in the field itself, and $T^{\mu\nu}$, $T^\mu{}_\nu$ and $T_{\mu\nu}$ differ, for example, in how they vary when $g$ is varied.

On a smooth manifold $M$ the same structure exists pointwise: tangent vectors live in $T_pM$, covectors (1-forms) in $T^*_pM$, and a coordinate chart provides the bases $\partial_\mu$ and $dx^\mu$. The transformation laws under coordinate changes in the Concrete layer are the special case $A^\mu{}_\nu = \partial x^\mu / \partial x'^\nu$.

A clash of names. Under a smooth map $f : M \to N$, tangent vectors are pushed forward, $f_* : T_pM \to T_{f(p)}N$, while covectors are pulled back, $f^* : T^*_{f(p)}N \to T^*_pM$. In the language of category theory, $T$ is a covariant functor and $T^*$ (on forms) a contravariant one. So the objects that physicists call contravariant vectors are covariant in the categorical sense and vice versa. The physics names refer to how components behave under a change of basis, the categorical names to the direction of maps.

Path to this page

1. Calculus
2. Linear Algebra
3. Vector Calculus
4. Tensor Calculus