Covariant and Contravariant Tensors

1. Jun 15, 2011

benk99nenm312

Hey everyone, I am reading a Schaum's Outline on Tensor Calculus and came to something I can't seem to understand. I'm admittedly young to be reading this but so far I've understood everything except this. My question is: what is the difference between a contravariant tensor and a covariant tensor, and what do these terms mean conceptually? I've gone online and searched a variety of articles with no luck. I appreciate it, thanks in advance.

2. Jun 15, 2011

dx

First, there are covariant and contravariant vectors. A multilinear function acting on covariant vectors is a contravariant tensor. A multilinear function acting on contravariant vectors is a covariant tensor. A multilinear function which acts on both is a mixed tensor.

3. Jun 15, 2011

Rasalhague

Adding to what dx has said, a particular system/structure of tensors is defined with respect to a particular vector space. Vectors of this vector space are called contravariant vectors. Vectors of its dual space (i.e. scalar-valued linear functions of one vector each) are called covariant vectors (or covectors, dual vectors, linear functionals, etc.). As such, covariant vectors are the simplest kind of covariant tensor.

Similarly, contravariant vectors can be thought of as scalar-valued linear functions of one covariant vector each, with the following definition: If w is a covariant vector, and v a contravariant vector, then v(w) is defined as w(v). Thus contravariant vectors (often called simply "vectors") are the simplest kind of contravariant tensor.

4. Jun 15, 2011

benk99nenm312

Ok I think I'm starting to understand this, thanks guys. So what would be some typical examples of contravariant and covariant vectors?

5. Jun 15, 2011

WannabeNewton

The gradient would be a pretty good example of a covariant vector (although if you know the formalism of differential forms you could just use the more modern definition of a covariant vector as a one - form which actually makes more sense intuitively) and there are countless examples of contravariant vectors (or simply vectors) but in keeping with the previous example, one example of a contravariant vector is the directional derivative along a curve. Evaluated at some point p, this type of vector can form a basis for the tangent vector space that Rasalhague talked about. The gradient then, at that point, can form a basis for the cotangent space (this is all to some manifold).

6. Jun 15, 2011

benk99nenm312

Ok I can visualize that without trouble, thanks WannabeNewton :)

Yeah as far as I can see this book doesn't cover the modern definition of the one-form, though I do believe I've heard of that, maybe that'd clear things up a bit too..

7. Jun 15, 2011

atyy

I never know which is which, but you can think of contravariant and covariant vectors as row and column vectors. By matrix multiplication, a row vector can be thought of as a function that takes column vectors as input and produces a number as an output.

8. Jun 15, 2011

WannabeNewton

Actually it does with detail at the level of a typical Schaum's Outline book so its a good introduction to it however brief, go to the very last chapter titled Tensor Fields on Manifolds. It pretty much talks about everything I just said and then some (actually a lot more than some =D).

9. Jun 15, 2011

benk99nenm312

In this book it shows the contravariant with upper indices and the covariant with lower indices, so that helps too thanks.

10. Jun 15, 2011

benk99nenm312

Oh awesome! Lol I'm only in chapter 3 :P

11. Jun 15, 2011

Fredrik

Staff Emeritus
12. Jun 16, 2011

PhilDSP

The terminology does seem deceptive. The principle of covariance is that 2 different views or perspectives of the same phenomenon should be symmetric or inversely related to one another. Ideally the term covariant tensor would have been used for the pair of tensors which are together covariant concerning a particular phenomenon.

13. Jun 16, 2011

benk99nenm312

Thanks Fredrik, yeah so many terms to understand.. that's great to have them all lined up in a single post :)

I see, so if covariance is symmetrical or inversely symmetrical, what would contravariance be?

14. Jun 16, 2011

atyy

That's a very tangentially related use of the term. Just use the definitions in the book you are reading.

Or if we are allowed to change definitions mid-sentence, both covariant and contravariant vectors transform covariantly:P

OK, to be more serious, let's imagine we have a 2D surface covered by coordinates (x,y). Imagine that each point has a different temperature f(x,y). A vehicle, carrying a clock which reads time t moves across the surface making a curve (x(t),y(t)). The variation in temperature versus time that the vehicle experiences is df/dt which is just one dimensional calculus. At a point p, df/dt=df/dx.dx/dt+df/dy.dy.dt, a scalar which we can rewrite as a a row vector (df/dx,df/dy) multiplied by a column vector (dx/dt,dy/dt), with all derivatives evaluated at p. The column vector or contravariant vector is something like the velocity, and the row vector or covariant vector is something like the gradient of the temperature.

We don't conceive of the velocity at that point as belonging to only one curve, since many curves can have the same velocity at that point. We also conceive of the velocity of any particular curve being the same under a change of coordinates from (x,y) to (U(x,y),V(x,y)). The same temperature variation is now described by f(U,V), and the same path is now described by (U(t),V(t)). So the new column vector representing the same velocity will be (dU/dt,dV/dt)=(dU/dx.dx/dt+dU/dy.dy/dt,dV/dx.dx.dt+dV/dy.dy/dt), which is how the coordinate representation of a contravariant vector transforms. Similarly, the new row vector representing the same gradient will be (df/dU,df/dV)=(df/dx.dx/dU+df/dy.dy/dU,df/dx.dx/dV+df/dy.dy/dV), which is how the coordinate representation of a covariant vector transforms. df/dt=df/dU.dU/dt+df/dV.dV/dt remains unchanged.

Something to keep in mind for later, when the metric is introduced: at this stage, we can multiply row and column vectors, but we haven't defined what it means to "multiply" column vectors, which is a job the metric can do.

Last edited: Jun 17, 2011
15. Jun 17, 2011

benk99nenm312

This makes incredible sense! Thanks so much, that really cleared things up :)

16. Dec 31, 2011

Gadhav

1) I understand the diff between co variant and contra variant coordinate system but not sure why velocity is considered contra where as grad is considered "co"variant. What does that have to do with how we lay out coordinate system?

2) Why do we always write covariant basis for contra variant components?

17. Dec 31, 2011

Fredrik

Staff Emeritus
A coordinate system is just a function that assigns n-tuples of real numbers ("coordinates") to points in the manifold. They aren't covariant or contravariant.

It doesn't matter which coordinate system is used. Velocity is defined as a tangent vector to a curve. The gradient of $\phi$ has components $\partial_i\phi$. Those components transform covariantly (the same way as the basis vectors) because the partial derivative operators $\partial_i$ are the basis vectors for the tangent space at the relevant point. To understand this better, I recommend that you read the post I linked to above, and the three posts I linked to in that one, in particular the first one.

The convention to write cotangent vectors as $\omega_i e^i$ and tangent vectors as $v^i e_i$ is just that, a convention.

18. Jan 1, 2012

Gadhav

Thanks Fredrik.

I thought over it and it now makes sense to me that it does not matter whether they are called covariant or contravariant. I think they should probably be called system(1) and system(2).

These terms are confusing since some books insist that Contravariant component is parallel to axes and Covariant component is perpendicular. In reality vector is simply some of three independent vectors(basis). That makes the whole thing confusing as you wonder why should one of them specifically relate to gradient and other is related to velocity. I think I was missing the point that we can use either. Conversion depends on what is in the denominator when we take differential.

I also happen to come across a tensor book by Sokolnikoff which I think is the best book on tensor that I have read. I strongly suggest that everyone read that book to get the fog of these terms out of your mind. Especially first 150+ pages explain it well. It has a very logical path to go from here to SR/GR but I have yet to read that part. (Search google as "sokolnikoff tensor pdf")

BTW it also mentions on #125 why contravariant components usually have covariant basis. It may be convention but it says that the reason is primarily due to fact that change of coordinate for these basis vary as gradient.

Last edited: Jan 1, 2012
19. Jan 2, 2012

Fredrik

Staff Emeritus
I don't understand what you have in mind here. Are you talking about the two coordinate systems that are involved in the "transformation" of a tensor? You could call them something like x and y, or x and x'.

There are infinitely many coordinate systems on all the interesting manifolds, but there aren't two kinds of coordinate systems associated with the terms "covariant" and "contravariant". I'm not saying that it doesn't matter if you call a coordinate system covariant or contravariant. I'm saying that those terms don't apply to coordinate systems. It doesn't make sense to call a coordinate system covariant or contravariant.

I don't think I've seen this claim, and I don't understand it.

I don't understand what you're saying here. Algebra defines "vector" as "member of a vector space". In differential geometry, "vector" is sometimes a lazy term for "tangent vector" or "tangent vector field", meaning "member of the tangent space at a point of the manifold" and "function that takes points in the manifold to tangent vectors at that point" respectively. "Vector" can also refer to an association of an n-tuple of real numbers with each coordinate system, such that the relationship between the tuples associated with any two coordinate systems is described by the tensor transformation law.

20. Jan 2, 2012

atyy

He's probably talking about the ability to identify forms and tangent vectors in the presence of a metric (lowering and raising indices). Something like http://www.mathpages.com/rr/s5-02/5-02.htm "As can be seen, the jth contravariant component consists of the projection of P onto the jth axis parallel to the other axis, whereas the jth covariant component consists of the projection of P into the jth axis perpendicular to that axis."

That's a different approach from what you've taken where you define forms and vectors first without a metric. Then only with a metric are we allowed to identify forms and vectors. In this approach, the metric always exists, which is an assumption of classical GR.

Last edited: Jan 2, 2012
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