# Covariant vs contravariant

1. Dec 31, 2004

### StatusX

I'm starting to learn differential geometetry on my own, but I'm having a little trouble figuring out the difference between covariant and contravariant vector fields. It seems that contravariant fields are just the normal vector fields they introduced in multivariable calculus, but if so, I can't figure out what covariant fields are.

2. Dec 31, 2004

### dextercioby

I can't give u the real explanation behind it all,because i don't know how much of Diff.Geom.u know and there could be the danger of u not understanding it.My question is:are u familiar with manifolds??If so,then u'd understand it.

Daniel.

3. Dec 31, 2004

### pmb_phy

See - http://www.geocities.com/physics_world/ma/intro_tensor.htm

Pete

4. Dec 31, 2004

### mathwonk

in the expression f(x) you can vary either x or f. one of these varies covariantly and the other varies contravariantly. if you hunt around this site you wil find a large number of posts on this topic.

actually if you are in the habit of taking dot products for granted, it is hard to tell the difference between covariant and contravariant fields.

if v is an ordinary vector (contravariant?) then the expression v.( ) is a covariant vector. i.e. you have changed the role of v, from x to f, by introducing the dot product. as a vector v is something you can act on by a functional, but in the expression v.( ), you have set up v to act on another vector which goes in the parentheses.

now in abstract differential geometry, one does not always assume a dot product is given, so then you have to think of vectors and covectors as different objects. even if you do have a dot product, v and v.( ) should be understood as different objects. i.e. one of them is a vector and the other is a map from vectors to numbers.

subject to correction by pete, i will try to relate this to the indicial version on his site. a dot product is a "metric" and has representation of form gij, where ij are subscripts. this means it is covariant of order 2. a contravariant vector v (of order one) has local expression of form v^i, where here i is a superscript. then putting the two together we represent the covector v.( ) as:
summation v^i gij = aj, where we have summed over i. this object has one unsummed index j as a subscript, whereas v had one index as a superscript.

this difference in location of the index, tells us the difference between the contravariant vector v, and the covariant vector v.( ).

or as hurkyl sugests, think of vectors as columns, and think of functionals which act on vectors as rows. in this version, if v is a column vector, then v.( ) is merely the transpose of that column vector.

Last edited: Dec 31, 2004
5. Dec 31, 2004

### Hurkyl

Staff Emeritus
One rather cheesy way of looking at it is that contravariant vectors are column vectors, and covariant vectors are row vectors. (Or do I have that backwards?) Of course, this doesn't work in full generality, but IMHO it's a good start at separating the two notions... although it does force you to carefully determine when which is appropriate. (e.g. the gradient of a scalar function is a row vector)

6. Dec 31, 2004

### gazzo

covariant = row vector; because COvariant sounds like ROWvariant.
contravariant are left to columns.

hehe :) someone said that in a previous post and it's stuck. thumbs up for passive studying on physicsforums!

7. Jan 1, 2005

### marlon

Covariant also means that the properties of some "object" (like a vector) on some manifold will remain the same under certain transformations that position this object on several locations on the manifold. For example when a manifold is curved, you will always need to define some space time continuum locally on that manifold. Then you need to know that the connection between these two points on the manifold is covariant so you can be certain that for example a 0-vector in one point of this manifold will still be a 0-vector in the other point.

Or, for example, look at tensors, which are a generalization of mathematical objects like vectors, matrices,...So a tensor is basically any object with indices (like a vector or matrix or double product of two or more vectors) AND certain transformation properties. Basically, these properties are that a tensor is covariant when it is transformed. Thus, this means that for example when the elements in a tensor (like the row and colomn-elements in a matrix) are 0 in one point on the manifold, they must remain zero at some other point on the manifold.

The most famous examples are the Riemanntensor and the Riccitensor from General Relativity. The Riemanntensor can be used in order to eveluate whether a manifold is curved or not. You can use it to check that the earth is curved although in your local reference frame, everything seems flat (you can see as far as the horizon).

This tensor is calculated by using the Christoffel-symbols which are also objects with indices BUT no tensors because they are not covariant. Their structure will change as you move them along the manifold.

regards
marlon

8. Jan 1, 2005

### StatusX

Actually, I'm not sure. I mean, I could give you plenty of examples of manifolds, but I'm not exactly sure what isn't a manifold. I was told that the definition of a manifold was that it could be covered with open sets, and each of these sets could be related by a smooth, one-to-one function to some subset of En. First of all, do I have this right?

I think this excludes objects with kinks. But what about multiple objects, like two open balls separated by some distance? Or a closed ball, which can be covered with open sets relative to itself, but not relative to the ambient space? Or a cylinder or cube (as surfaces (I don't think so) or as volumes (I think so)), with kinks on the edges? Or a sheet that intersects itself? (don't think so)

Also, I heard that for two manifolds with different coordinate functions to have "the same smooth structure," there must be a smooth invertible function between the different coordinate systems (and also, ther must be one for any points where two local coordinate patches overlap on a single manifold). So for example E1 has different smooth structure with x as its function than with x3. What does this mean? Isn't E1 always just E1?

Getting back to covariant vs contravariant, I will read a little more (thanks for that site by the way Pete) and get back with more specific questions. Thanks for your help so far.

Last edited: Jan 1, 2005
9. Jan 1, 2005

### jcsd

Marlon, that's right, but it's also not quite right!

A covariant vector is specifically a vector which transforms with the basis vectors, a contravariant vector on the other hand is a vector that transforms against the basis vectors. If we talk about something like Lorentz covariancy, the meaning is sligthly different. Just to illustarte how confusing this can be: a contravariant Lorentz vector is not a covariant Lorentz vector though it is a vector that is Lorentz covariant!

10. Jan 1, 2005

### dextercioby

Okay,people,let's leave the physics aside and give the definitions:
A contravariant vector in a point "p" of a manifold (a.k.a. 'vector') is an element of the vector space tangent to the manifold in the point "p":$T_{p}$.It can be looked upon as a 'hyperplane',due to the simple case when its dimension (and implicitely the manifold's dimension) is 2.
A covariant vector in a point "p" of a manifold (a.k.a. "one form") is an element of another vector space,called cotangent space in the point "p" of the manifold:$T^{*}_{p}$.Unfortunetely it does not have a geometrical representation like the tangent space.That is we cannot visualize the cotangent space not even for the simple case when 'd=1'.
Since those 2 are vector spaces we can define bases and formulate how these 'animals' behave to a change of basis.

'Lorentz covariance' is just a confusing syntagma.One who knows Diff.Geom. will never confuse with the real meaning of "covariance".I hope... :tongue2: Anyway,i'm sure Marlon has everything clear.

Daniel.

11. Jan 1, 2005

### jcsd

And the tangent and cotangent spaces at a single point are dual vector spaces. To me that's probably the easiest way to get a basic grasp on the difference.

I think though that the catergory theorists would get pretty nasty if it was suggested that the differential geometry defintinn of covaraint and contravariant was the real deifinition.

12. Jan 1, 2005

### marlon

Correct jcsd, i admit there is confusion possible here and that's why your point is completely correct here...though when i was referring to covariance i did not make any reference to Lorentz-covariance which is ofcourse a very important issue in GTR. But it is not fundamental since it is defined purely based upon the notion of covariance...at least that is the way i have learned in college and i cannot say that no other explanation or introduction of this concept is possible...This is just the way i keep all these concepts on the right track...

regards
marlon

ps : indeed : the notion of "transforming along with basic vectors is a very good point and this was what i wanted to state in my post"

13. Jan 1, 2005

### pmb_phy

There are several ways to define these quantities. A "contravariant vector" is what you'd simply call a "vector" which comes in two flavors. There are displacement vectors, used mostly in flat spaces such as the flat spacetime of SR and then there are tangent vectors which one uses in curved spaces such as the curved spaces one often finds in GR.

A covariant vector is then defined as a multilinear operator which maps contravariant vectors into scalars. In this sense a a covariant vector is a vector in the abstract mathematical sense of the term.

Pete

14. Jan 1, 2005

### jcsd

Yes Lorentz covariance is a bit of a red herring, I used it becasue of the unfortunate terminology of covariant Lorentz vectors and thee fact taht all Lorentz vectors are Lorentz covariant where in each it is infact a different property that is being described as 'covariant'. Really I should of talked about general covaraincy as a Lorentz tensor is not necessarily a true tensor.

What confused me is that you seemed to suggets that all tensors are covariant which is true in the sense that a general tensor is general covariant, but in the context of covaraince as used in the original question is not true.

At the moment I'm trying to wrap my head round the basics of catergory theory and to muddy the water even further (though I must admit I've only got as far as the basic definition of a catergory) it appears that it catergory theory what are callled contravariant vectors would be described as covaraint objects and what are called covaraint vectors are infact contravariant objects (I'm afraid though you'll have to ask mathwonk why that is)!

15. Jan 1, 2005

### jcsd

Yes that's one way to think of them and it's the most formal defintion that I know of these objects (though it's best to call covaraint vectors linear functionals as multilinear usually implies more than one variable and the term oeprator usually implies a map from vector space to vector space)

16. Jan 2, 2005

### gvk

If you make a sketch of tangent space for manifold, you do simultaneously the sketch of cotangent space, because the tangent vector looks exactly the same as cotangent covector, e.g. for 'd=1' tangent line coinside with cotangent line, for 'd=2' tangent plane coinside with cotangent plane. Note, that this is true if you have a metric in the vector space. If you don't have a metric, strictly speaking, you can not draw tangent space.
But you can pretend that you can do it even you don't have a metric. In this case, the best way to cheat your imagination is to draw the second line, plane, hyperplane,... parallel to the tangent space (line, plane, hyperplane,...). It's not the best way, of course, but, at least, it gives a sense that those spaces are different and connected.
This 'geometrical representation' is used by many authors.
Many people emphasized that tangent and cotangent spaces are different. It is true, but, in my view, it is equally important to emphazise, that those 'animals' are tightly connected. If you change somehow the basis in tangent space, the basis of cotangent space is simultaneously changed. And these changes go in exactly oposite direction, so the scalar product of any vector from tangent space and any covector of cotangent space does not depend on the basis. And these values are called invariants. The scalar product between two vectors from different spaces (tangent and cotangent) looks very strange, if even the scalar product inside of each space is prohibited.
But it looks normal if one remember a tight connection between them.
In fact, the introduction of tangent and cotangent spaces allows to bypass the metric, which was the main tool to get the 'physical' invariants.

17. Jan 2, 2005

### mathwonk

asusual everyone "visualizes" or imagines even precise mathematical objects entirely differently. i have trouble even understanding the comments here on the ease or impossibility of visualizing tangent and cotangent vectors. here is my personal view of them.

i imagine say a two dimensional tangent space as a plane, with a distinguished point, the origin. then a non zero cotangent vector, being by definition a non zero linear function on this space with real values, is determined up to a constant multiple by the subspace of tangent vectors which are mapped to zero, hence by a line through the origin. so the projective cotangent space is merely the set of lines through the origin of the tangent space. to determine the linear function fully and hence the covector, we need to know which vectors are mapped to 1, which forms a line parallel to the previously given line through the origin. so in this representation, a non zero covector is merely a line in the plane not passing throuigh the origin.

in higher dimensions, it is a hyperplane not passing through the origin. in case one has a notion of perpendicularity, one can draw a line through the origin perpendicular to this hypwerplane, and then identify the hyperplane with the intersection point of line and hyperplane. this allows one to view the tangent space and cotangent space as the same, not otherwise.

now my grandaughter wants to, play pbskids. so goodbye

18. Jan 3, 2005

### gvk

In my understanding, 'to visualize' (in a simplest significance) means to form a mental geometrical image, which can be ploted on the paper or builded as a 3-D model.
Let's consider again how we 'visualize' the cotangent space. It is a vector space having the basis, but this basis is builded by using the basis of direct vector space. For 1-d it is $$e^1(e_1)=1$$, where $$e_1$$ is basis vector of tangent space, $$e^1$$() means a linear functional. Now we plot the straight line with origin and vector $$e_1$$ comming out from origin. It lies on the line and this line is tangent space. Now we try to plot on the same plane the cotangent space. If we do this we automaticaly introduce the metric relations between tangent and cotangent spaces (scalar product!), because we care about orientations between two lines, at least. So, we write the linear functional as the scalar product $$(e^1, e_1)=1$$. If the lines are perpendicular, the scalar product is 0 and $$e^1$$ is not basis of cotangent space. It's not what we wanted, but if the lines are coinsided, it is OK. Now we take another the basis in tangent space $$e'_1=\lambda e_1$$. The basis in cotangent spaces will be changed, $$e'^1=\frac{e^1}{ \lambda}$$, otherwise we can not satisfy the condition $$(e'^1, e'_1)=1$$.
The similar picture can be in 2-d case, where $$(e^1, e_1)=1$$ , $$(e^2, e_2)=1$$, $$(e^1, e_2)=0$$, $$(e^2, e_1)=0$$. But of course, it does not mean that pair $$e^1$$ and $$e_1$$ (or $$e^2$$ and $$e_2$$) should be parallel. Here $$e^1, e^2$$ constitute the reciprocal basis in the same plane.

Last edited: Jan 3, 2005
19. Jan 3, 2005

### StatusX

I have a question. I'm sorry if this is a little off topic, but I didn't think it deserved it's own thread. The metric is defined as:

$$g_{ij} = \hat e_i \cdot \hat e_j$$

where the $$\hat e_i$$ are the basis vectors of the local coordinate system. In terms of the ambient coordinates, this is:

$$(\hat e_i)_s = \frac{\partial y_s}{\partial x^i}$$

But this is where I get confused, because sometimes the metric is given, such as in a minkowski space where its diag[1,1,1,-c^2]. But the paper I'm reading says that:

$$g_{ij} = \hat e_i \cdot \hat e_j = \frac{\partial y_s}{\partial x^i} \frac{\partial y_s}{\partial x^j}$$

But this is implies the ordinary dot product is being used, and it can only have the normal signature (1,1,1,1). What am I missing?

Last edited: Jan 3, 2005
20. Jan 3, 2005

### jcsd

It is the dot product, in this case the dot product of the time basis vector and itself in the basis you have choosen is -c^2 and it is the metric that defines this.

Last edited: Jan 3, 2005