1. Aug 17, 2014

### davi2686

i really try understand these initial concepts before continue my work,

The necessity to use contra/covariants coordinates is ONLY to maintain invariant the module of a vector in a oblique coordinate systems? if not, what the motivations for i really need work with the two types of coordinates?

thanks, and sorry for such dumb question

2. Aug 18, 2014

### Matterwave

This is a pretty broad topic. Let me make a few comments, and then if you have further questions, we'll go from there. I don't want to write a huge wall of text that may or may not help you.

Coordinates are not contra or co variant. Coordinates are just coordinates, they are a way of smoothly assigning to a point on an n dimensional manifold a set of n numbers. That's it.

Vectors, which on a smooth manifold are the tangent vectors to that manifold, can be contra or co variant. Contra variant vectors are the tangent vectors, while co variant vectors are the one forms which are functions of vectors into real numbers. In some coordinate system, the two types of vectors can both be described at any point by, again, a set of n numbers. These n numbers change differently when we switch to a different coordinate system, and that is why we make the distinction between contra and co variance.

What I suspect you are talking about is that in a flat space, a displacement vector (point A minus point B) is a legitimate vector (in curved spaces they are not). And since a displacement vector can be used to describe a point on a manifold the co-vector (one form) associated with it (through the metric tensor) can also be used to describe the points on a manifold. This is a very confusing way to think about this. Especially because once you move to a curved space, none of this is true anymore.

Perhaps if you showed some examples of how you've seen "contra and co variant coordinates" used, it might be easier to help you further.

3. Aug 20, 2014

### Blazejr

Let me go back to finite dimensional linear algebra for a moment. For every vector space $V$ there exists dual space $V^*$, which is space of all linear functions $V\rightarrow\mathbb{R}$. If we choose basis $(e_1,...,e_n)$ in $V$ it is natural to choose basis $(\phi^1,...,\phi^n)$ in $V^*$ such that $\phi ^i (e_j)=\delta ^i _j$. That is called dual basis, and some people would note those basis vectors as $(e^1,...,e^n)$. Notice that the only difference between the notation of basis vectors from $V$ and $V^*$ is now placement of indices. It is often natural that if we are given basis in vector space and we also use dual space, we use the dual basis, and when we change basis in $V$ we also want to change basis in $V^*$ to the new dual basis. One can then show that coordinates of vectors in $V$ transform in opposite manner than those of vectors in dual space (which we would call forms). Additionally, in order to make Einstein convention work we need to write coordinates of vectors above and coordinates of forms below.

Last edited: Aug 20, 2014
4. Aug 20, 2014

### davi2686

Thanks Blazejr that is much more clearly now in my head, i will put some issues which i need to solve, in my poorly interpretation...

So co/contravariant terms refers to basis and dual basis?

Last edited: Aug 20, 2014
5. Aug 20, 2014

### davi2686

thanks a lot too Matterwave, i needed google some terms which you used, the deal is, im a undergraduate physics student and recently start work with covariant formulation of Maxwell equations with a teacher, but im not comfortable with the idea to try work with it without the comprehension of math involved... so because of my only basic math knowledge a lot of concepts become hard to me understand...

but with Blazejr post, a lot of things that you said make more sense for me, for example, if a vector A(x,y) is covariant, that means this vector is tangent to manifold, THATS IT, theres not a contravariant representation of same vector A which components (x',y'). Covariants and contra variants vectors are distinct so to speak... thats close?

6. Aug 20, 2014

### Matterwave

I'm having trouble understanding your notation. A vector field $\vec{A}(x,y,z)$ is usually not denoted with "components" $(x,y,z)$. Usually the notation means that there is a vector $\vec{A}$ at each point $(x,y,z)$ on the manifold. So I don't understand what you mean.

A contra variant vector can be thought of as a tangent vector. A co variant vector can be thought of as a linear function of contra variant vectors into real numbers. This terminology is somewhat outdated. Usually one calls contra variant vectors as simply vectors, and co variant vectors as one forms. Whether one can make an identification between the vector $\vec{A}$ and the one form $\tilde{A}$ depends on if a metric is defined on your manifold (in physics, there often is). A metric defines two things. One, it defines an isomorphism between the tangent space and the cotangent space. Two, as a consequence of the first definition, it also defines an inner product between vectors in the tangent space.

The metric is a rank 2, symmetric, non-degenerate, co variant tensor field, usually denoted $g$ (we tend to suppress the function of the points on the metric notation $g(x,y,z)$ even tho the metric is a tensor field because as you will see that might get very confusing later when we make it act on vectors). It takes as arguments two vectors, and gives one number. This is the definition of an inner product of vectors on the manifold. In Einstein summation notation: $g(\vec{A},\vec{B})=g_{ij}A^i B^j$. Because of the non-degenerate nature of the metric, it defines an isomorphism between the tangent space and cotangent space in the following manner: $g:T\rightarrow T^*|~~ g(\vec{A})=\tilde{A}=g_{ij}A^j \tilde{\omega}^{(i)}$ where $\tilde{\omega}^{(i)}$ are the basis one-forms, and the parenthesis are to remind us that the index denotes a different one form for each index $i$ and not the components of a one form.

7. Aug 28, 2014

### davi2686

thx Matterwave, let me see if i understand...

first, sorry about my notation, when i said $A(x,y)$ i really want say $\vec{A}=(x,y)$, sorry about my mistake, about all the rest, that is what i understand...

every point of a manifold have a tangent space, the elements of this tangent space we call contra-variant vectors or just vectors, and each point of a manifold also have a cotangent space, which is the dual space of the tangent space in this point, the elements of that cotangent space we call co-variant vectors or just forms. These two spaces have same dimension and the metric of a manifold works like an isomorphism.

Plz tell me what is wrong in the above text, because in present, thats what i understand about all of we said.

what is a little obscure to me is that... like you said if $\tilde{A}=g_{ij}A^j\tilde{\omega} ^{(i)}$ we obtain the covariant vector associate with $A$, but in an example, if $A$ represents the force vector of an ordinary system, so $A$ and $\tilde{A}$ represents the same force vector, right?

8. Aug 28, 2014

### WWGD

Just to make the point that differential forms are not just contravariant and multilinear (on k-ples of tangent vectors), but are also alternating.

9. Aug 28, 2014

### Matterwave

We call the elements of the co-tangent space "one-forms". "Forms" is a more general term. A k-form is a totally anti-symmetric rank (0,k) tensor. As such, higher order forms live on product spaces of the co-tangent space (i.e. $T^*_P\otimes T^*_P\otimes...\otimes T^*_P$).

Yes. If we are talking about physics, then some quantities are "naturally" a co variant or contra variant vector (or tensors). However, since we usually have a metric on our manifold in physics (like when we are working in 3-D real space), we often just use the metric to convert all the co variant vectors into contra variant ones for simplicity. For example, a gradient is "naturally" a co variant vector, but we are usually used to dealing with the contra variant version of it...and that is why we have all those weird extra terms when we express the gradient in spherical coordinates (they come from the metric!).

Because the metric in Cartesian 3-D coordinates is simply the identity matrix (i.e. it is diag(1,1,1)), we can switch between co variant and contra variant vectors without really doing anything to them (in a Cartesian coordinate system their components are the same). This is why the distinction between the two types of vectors is often not seen in many physics texts.

10. Aug 28, 2014

### Matterwave

I think so far in this thread we have not dealt with any higher order forms. We have dealt with only one forms.

11. Aug 28, 2014

### WWGD

O.K, I thought vectors includes the general case of multivectors. If not, just for the sake of having a broader context for contravariance.

12. Aug 28, 2014

### Matterwave

In standard terminology vectors (and one forms) would mean single indices (or linear maps with 1 argument only), and tensors would include vectors (and one forms, and scalars) as well as objects with several indices (higher order multi-linear functions of vectors and one forms to real numbers).

Therefore the word "tensor" would be the general term to look for. Vector should be regarded as a specific term for the specific case, otherwise things would get redundant (and confusing).

13. Sep 6, 2014

### davi2686

how can i know if a certain vector or tensor is "naturally" co or contra variant?

And the thing of tangent and co-tangent spaces when lives respectively contra and co variant vectors i think its MUCH more clearly for me, but i have some dificulty to associate that with the definitions of co/contra variant vectors which i see in some books, example, a lot of books i have tell me that

contra variant vectors transforms like $\displaystyle x'^i=\frac{\partial x'^i}{\partial x^j}x^j$
co variant vectors transforms like $\displaystyle x'_i=\frac{\partial x^j}{\partial x'^i}x_j$

my dificult resides in concatenate the ideias

14. Sep 6, 2014

### Matterwave

Well, if an object is the exterior derivative of something, then it is "naturally" a co variant vector. Like the gradient example I gave. The gradient we are used to in physics is actually the index raised version of the exterior derivative $(d\phi)^\sharp$. Similarly, for the Faraday tensor, which is the exterior derivative of the vector potential (which is naturally a one-form for this reason as well) $F=dA$.

But like I said, in physics, we almost always have a metric, so we can always just switch between the two. I'm not sure I can tell you any general statement that would tell you which objects are "naturally" of which type. Perhaps objects that are akin (analogous?) to displacement vectors are naturally contra-variant, and objects which are akin to a type of level sets (e.g. the gradient being visualized as a topological map of where the level sets of a function are) are naturally co-variant. But I'm not sure how general that statement would be!

The older book's definition usually looks at how the components of a co variant or contra variant vector or tensor transform under a change of coordinate systems. This is just a different way to look at things. It is more practical for calculational purposes, but it obscures the geometric meaning of these objects.

I am running a bit low on time right now, so I will get back to this.

15. Sep 8, 2014

### davi2686

My friend you REALLY help me a LOT, with your explanations several books that i have for some time make much more sense to me now, i really thank you.

From what i've read, i may have been wrong but i think that notation $\displaystyle x'î=\frac{\partial x'î}{\partial x_j}x_j$ is something about cartan spaces and generalisation of euclidean spaces, thats make sense or i completely wrong about this?

16. Sep 8, 2014

### Matterwave

I don't really know about "Cartan spaces" very much, but the tensor transformation law is simply the definition of what a co variant or contra variant vector on any differentiable manifold is. As I mentioned in my last post, it is a practical definition that obscures the geometric meaning of a vector.

Let's look at how a vector is defined in the "new" way (this is a contra-variant vector). Basically a vector on a manifold is the tangent vector to curves that pass through a point P. Because the vector is tangent to curves, they lie in a space called the tangent space at the point P ($T_P$). Let's look at how it is constructed:

Let's consider a curve $\Lambda:\mathbb{R}\rightarrow \mathcal{M}$ which is a map of (an interval of) the real line into the manifold which passes through a point $P$ on the manifold. In some local coordinates, this curve has the parametric equations $\{x^i(\lambda)\}$. We define an arbitrary differentiable function $f:\mathcal{M}\rightarrow \mathbb{R}$ which maps points on the manifold to some real number (assigns a real number to each point on the manifold, this should be intuitive). This function will take on the values $f(x^i(\lambda))$ along our curve. At the point $P$ we can find the derivative of a function using the chain rule:

$$\left.\frac{df}{d\lambda}\right|_P=\left.\frac{dx^i}{d\lambda} \frac{\partial f}{\partial x^i}\right|_P$$

Since this is true for all arbitrary differentiable $f$ we can writ the derivative as:

$$\left.\frac{d}{d\lambda}\right|_P=\left.\frac{dx^i}{d\lambda} \frac{\partial }{\partial x^i}\right|_P$$

We can show that these derivative operators $\left.\frac{d}{d\lambda}\right|_P$ form a vector space because they obey all of the axioms of a vector space. I will not prove rigorously here that they do, but I will motivate an intuitive explanation. Consider that these operators are in 1-1 correspondence with curves that pass through $P$ where a curve with a different parametrization is considered a different curve. Changing the length of these vectors is equivalent to changing the parametrization, changing the direction is equivalent to changing the curves, so you would intuitively expect that these differential operators form a vector space. The only non-trivial property of vector spaces that we should prove is the property of closure under linear combinations. Consider a second curve $\text{M} :\mathbb{R}\rightarrow \mathcal{M}$ which have coordinates $\{\tilde{x}^i(\mu)\}$. The tangent vector to this curve can be expressed in our coordinate system as:

$$\left.\frac{d}{d\mu}\right|_P=\left.\frac{d\tilde{x}^i}{d\mu} \frac{\partial }{\partial \tilde{x}^i}\right|_P=\left. \frac{dx^i}{d\mu} \frac{\partial }{\partial x^i}\right|_P$$

Where the last equality comes from the fact that we are evaluating at point P for which $x^i(P)=\tilde{x}^i(P)$. So we look at a linear combination of these two vectors (with the evaluation at point $P$ implicit):

$$a\frac{d}{d\lambda}+b\frac{d}{d\mu}=\left(a \frac{dx^i}{d\lambda}+b\frac{dx^i}{d\mu}\right) \frac{\partial}{\partial x^i}$$

Now there must be some curve $\Phi:\mathbb{R}\rightarrow \mathcal{M}$ with parametrization $\{\bar{x}^i(\phi)\}$ passing through the point $P$ such that:

$$\frac{dx^i}{d\phi}=a\frac{dx^i}{d\lambda}+b\frac{dx^i}{d\mu}$$

So that we know therefore:

$$\frac{d}{d\phi}=a\frac{d}{d\lambda}+b\frac{d}{d\mu}$$

Proving closure under linear combinations. We therefore make the claim (definition) that "contra variant vectors are isomorphic (really just are) to derivative operators at point $P$". We can finally now look at how the coordinates of these contra variant vectors change under a change of coordinates, and we can show that they do indeed transform as contra variant vectors according to the change of basis law. We have a new set of coordinates $x^{i'}=x^{i'}(x^i)$ (don't get this confused with the tilde and bar versions of $x^i$ earlier, those were literally the coordinates of different curves, this is the coordinates of the same curve but in a different coordinate system). Now, in new or old coordinates the vector doesn't change, we can apply the chain rule with this new coordinate system to obtain:

$$\frac{d}{d\lambda}=\frac{dx^i}{d\lambda} \frac{\partial}{\partial x^i}=\frac{dx^{i'}}{d\lambda}\frac{\partial}{\partial x^{i'}}$$

We also have that the chain rule says:

$$\frac{\partial}{\partial x^{i'}}=\frac{\partial x^j}{\partial x^{i'}} \frac{\partial}{\partial x^j}$$

In order for the above equality (involving the $\lambda$) to hold it must then be that:

$$\frac{dx^{i'}}{d\lambda}=\frac{\partial x^{i'}}{\partial x^j} \frac{dx^j}{d\lambda}$$

Which is exactly the component transformation rule given:

$$A^i\equiv\frac{dx^i}{d\lambda}$$

One can then prove that one forms, which are functions of contra variant vectors into scalars will transform "oppositely". I will not provide the proof here as this post is already getting too long lol.

Last edited: Sep 8, 2014
17. Sep 14, 2014

### davi2686

awesome, i did see that explanation in any book which i had contact, i read your post carefully in the last days and the things make sense for me now, of course still exist some issues but i feel i have to read more, now with that background you give me i think i can better enjoy the books now. i dont know how to thank you for the patience, from here that some book you might indicate for me in this subjec?

18. Sep 14, 2014

### Matterwave

The book that I read, which is at a pretty introductory level, and tailored for physicists, is Geometrical Methods of Mathematical Physics by Bernard Shutz. It's at a pretty introductory level.