# I Contracovarient and covarient basis form orthogonal basis

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1. Jul 4, 2017

### Ron19932017

Hi everyone, I am trying to self study some general relativity however I met some problem in the
contravarient and covarient basis.

In the lecture, or you can also find it on wiki page 'curvilinear coordinates',
the lecturer introduced the tangential vector ei =∂r/∂xi and the gradient vector ei =del xi
and showed them to have a relation of ei dot ej = δij .
I totally follow and agree the algebraic steps, both on the lecture and on the wiki page.

However I cannot figure it out GEOMETRICALLY ! Let consider a 2D curvilinear coordinates,
Given we have two tangential basis e1, e2 which are not parallel.
How can one geometrically (by drawing and inspection) contruct a third vector that is parallel to e1 but orthogonal to e2 ?

However from algebraically we showed the del x1 is the wanted vector !!

Lastly, I tried to explicitly see what is going on in the polar coordinate (we are familiar with polar coordinate) however the tangential vector er and eθ are orthogonal...

Appreciate you kindly help.

2. Jul 4, 2017

### DrDu

You can't. The contravariant vector isn't parallel to e1. This is not necessary for $\langle e^1| e_1\rangle=1$.
Let's introduce a vector perpendicular to e1 using Schmidt orthogonalisation : $e_\perp=e_2-\langle e_1| e_2\rangle/ \langle e_1|e_1\rangle e_1$. Then any $e=e_1/\langle e_1|e_1\rangle +\lambda e_\perp$ will fulfill $\langle e| e_1\rangle=1$.
Now from $e^1e_2=0$ we find $\langle e_1| e_2\rangle/\langle e_1| e_1 \rangle-\lambda ( \langle e_2| e_2 \rangle +\langle e_1| e_2 \rangle)=0$, we can obtain $\lambda$.

Last edited: Jul 4, 2017
3. Jul 4, 2017

### Ibix

Vectors and one-forms occupy different spaces - vectors (upper index) are in the tangent space and one-forms/co-vectors are in the cotangent space. These are both four dimensional spaces, so you have a set of four basis vectors and a set of four basis co-vectors.

I believe one way to visualise these is using arrows for vectors and a set of parallel planes for co-vectors. A larger magnitude vector has a longer arrow. A larger magnitude co-vector has planes closer together. The inner product of a vector and a co-vector is the number of planes crossed by the arrow.

I recommend Sean Carroll's lecture notes if you're looking for a free online source.

4. Jul 4, 2017

### Orodruin

Staff Emeritus
You cannot. The vector $e^1$ is generally not parallel to $e_1$. Consider the coordinates $t = x$, $s = x+y$ in two dimensions.

As long as you only deal with Euclidean space you can introduce these two bases without any problems and there is essentially only one space you need to care about. Of course, it is all due to the natural mapping between the tangent and cotangent space. I think it is actually preferable to see this construction in Euclidean space first as it takes some pressure off from the introduction of tangent and cotangent vectors in a more general manifold. (Of course, I am biased, it is the approach I take in my upcoming book.)

5. Jul 4, 2017

### Ron19932017

Thanks ! I see my problem now. I just asked a stupid question lol. You are right they are not parallel although they satisfy the eiejij.

I see my problem now, thanks. You are right. I am only considering the curvilinear coordinates in Euclidean space. I will get into cotangent space only after I can intuitively understand the very basics things..

6. Jul 5, 2017

### vanhees71

To distinguish the tangent and cotangent spaces IS the very basic thing. The cotangent space is the vector space of linear forms defined on the tangent space, one also says, it's the dual space of the tangent space.

Only if you have a scalar product (or a regular indefinite fundamental bilinear form as in relativity) you have a coordinate-independent ("canonical") way to map these spaces into each other.

7. Jul 5, 2017

### Orodruin

Staff Emeritus
Which you do in a Euclidean space. I find that it is pedagogically easier to introduce students to the concepts of covariant and contravariant bases in a setting they are familiar with. Once you have taught them that, it is easier to take the next step and say "you know these two types of bases we introduces before, they are actually the bases of two different vector spaces". What I like to do is to start by introducing the Euclidean setting in Cartesian coordinates, then go on to introduce curvilinear coordinates and define the metric as having components $\vec E_a \cdot \vec E_b$, then look at Christoffel symbols and derivatives, and only then tell the students that everything we did really was made backwards: the connection defines the derivatives, the metric defines the inner product, and the two bases are actually the bases of a vector space and its dual. I find that this approach is often useful because it gives the students something tangible to "hold on to" and some understanding of why concepts are introduced in the way they are.

Let me also just add in passing that the only requirement you need to place on the bilinear form is that it is non-degenerate. It can be symmetric as in relativity or anti-symmetric as in symplectic manifolds (or in general neither as long as it is non-degenerate).

8. Jul 5, 2017

### DrDu

I first understood this stuff when I learned about the reciprocal basis in crystallography. So your approach seems natural to me.

9. Jul 6, 2017

### Ron19932017

May I know what is your reason to push the student to use the space/ dual space concept when they already master the covarient derivative, geodesics... I want to know what is the ultimate reason we need to have the concept of dual space? Everything already seems working fine in curvilinear system.

10. Jul 6, 2017

### Orodruin

Staff Emeritus
All spaces are not Euclidean. All spaces do not have a metric (or a fundamental bilinear form). In GR you do have a metric, but there are many other cases in which these concepts come into play where it is crucial to differentiate between the tangent and cotangent space.

11. Jul 6, 2017

### DrDu

Phase space in classical mechanics?

12. Jul 6, 2017

### Orodruin

Staff Emeritus
Phase space has a fundamental bilinear form, the symplectic form. Clearly it does not define an inner product as it is anti-symmetric, but it does provide an invertible map between the tangent and cotangent spaces.

13. Jul 6, 2017

### stevendaryl

Staff Emeritus
Well, it's hard to know what's going to cause the most conceptual problems. When you're working in Euclidean space, it's hard to understand what's the point of covectors, and it's hard to understand why you need a metric to take the scalar product of two vectors (or the length of two vectors).

14. Jul 6, 2017

### Orodruin

Staff Emeritus
What I said above is based on my personal experience with students.

Note that I am talking about working in Euclidean space with general curvilinear coordinates, not Cartesian or even orthogonal coordinates. Students are used to work with orthogonal coordinates, but in general there is no such requirement and they are usually rather quick in accepting that there are two different vector bases you can use $\partial \vec x/\partial y^a$ and $\nabla y^a$, where $\vec x$ is the position vector and $y^a$ the coordinate functions. The need for the metric tensor (or something like it) becomes rather straightforward as $g_{ab} = \vec E_a \cdot \vec E_b$ and there is no reason to assume that these basis vectors are orthogonal or normalised. Everything I say (until I turn the argument around) is based on the back reference to an original Cartesian coordinate system, which makes it easier for the students to connect and imagine. The only thing that does not become obvious is that the contravariant and covariant vector components (i.e., the components relative to the two bases that are introduced) are actually components of vectors in different vector spaces. Instead, students are allowed to get a feeling for how to use these different components before that step is taken.

15. Jul 6, 2017

### Ron19932017

How can I really identify a space eg the phase space is not Euclidean? I remember in the example of Fermi-degenerate gas, or some other model we are counting the state density in the reciprocal space/momentum space, it looks like the momentum space is Euclidean because all axis are orthogonal and the algebra looks standard..

16. Jul 6, 2017

### Orodruin

Staff Emeritus
It cannot be Euclidean, there is no inner product.

Momentum space is not the same thing as phase space.