Covariant vs. contravariant

1. Dec 18, 2006

cesiumfrog

Does anyone know the physical (or historical) basis for the terms covariant and contravariant?

I'm guessing a particular class of mapping always tranforms components (of ..?) in exactly two different ways, so I'm wondering what the mappings are (Change of coordinate charts? Lorentz transform?) and what it is that "covariant" things vary in common with?

2. Dec 19, 2006

dextercioby

Such terms appear in mathematics actually and (theoretical) physics just borrows them when inventing models in which such terms cannot be omitted.

Typically such terms are defined in linear algebra. The necessity of both "covariant" and "contravariant" objects is due to the existence of the canonical isomorphism (as vector space) between a vector space and its algebraic dual.

Daniel.

3. Dec 19, 2006

masudr

Yes precisely that. A change of co-ordinates will lead to a a particular change in the vectors/covectors and this will lead to the opposite kind of change in the components. See this Wikipedia section for example.

The way I see it is that covariant vectors (note not covariant components) transform with the change in co-ordinates, and contravariant vectors transform against the the change in co-ordinates. Of course, this means that the components transform in the opposite ways (which it must be of course). This interpretation of the terms also depends on what you mean by with/against. All in all, it is, of course, more important to remember just the transformation rule outright.

4. Dec 19, 2006

dextercioby

These terms are rather vague. I'd say tangent space basis vectors for the first and cotangent space basis vectors in the second case. Generally speaking such vectors are invariant objects. Under basis change, the components transform contragradiently to the basis vectors.

Daniel.

Last edited: Dec 19, 2006
5. Dec 19, 2006

Daverz

I don't know if this is this is how the original definitions were arrived at, but the book Geometrical Vectors by Gabriel Weinreich uses the example of a change of numerical scale to show the difference between the two kinds of vectors. Imagine a parallel plate capacitor with a displacement vector d between the plates and and an electric field vector E between the plates. If you change your scale from centimeters to meters, d decreases in numerical magnitude (contravariant) and E increases in numerical magnitude (covariant).

Geometrical Vectors is an interesting little book, though I do wish Weinriech had connected his terminology with more modern usage. What he calls a sheaf, for example, is a twisted 2-form (cf. Applied Differential Geometry by Burke). But his book does give you a good intuitive sense of why these mathematical objects are necessary.

Last edited: Dec 19, 2006
6. Dec 19, 2006

cesiumfrog

Thanks, that all seems to have given me a better feel for it.

7. Dec 19, 2006

Hurkyl

Staff Emeritus
I don't remember the precise motivation for the original definition of contravariant and covariant.

But I do know the modern motivation for the terms... which, alas, happens to be exactly the opposite of the original definition. (In other words, some people use "contravariant" and "covariant" in exactly the opposite way that other people do. )

The modern definition, specialized to this circumstance, deals with maps between spaces. For example, if I have a map between differentiable manifolds

$$f: \mathbb{R}^3 \rightarrow X$$

(which, for example, might be a coordinate chart on a 3-dimensional manifold, or if $X = \mathbb{R}^3$, this might be a change of coordinates), this induces maps on the tangent and cotangent bundles:

$$\begin{equation*}\begin{split} &f_* : T\mathbb{R}^3 \rightarrow TX, \\ &f^* : T^*X \rightarrow T^*\mathbb{R}^3. \end{split}\end{equation*}$$

(one might call the first of these the derivative of f)

One would then say that tangent vectors transform covariantly, because $f_*$ points in the same direction as f. And similarly, that cotangent vectors transform contravariantly, because $f^*$ points in the opposite direction.

($\mathbb{R}^3$ is not special in the above discussion; you could replace it with any differentiable manifold)