# How to define covariant basis in curved space 'intrinsicly'?

1. Oct 13, 2014

### arpon

In Euclidean space, we may define covariant basis by the partial derivative of position vector with respect to each coordinates, i.e.
$∂R/(∂z^i )=z_i$
But in curved space (such as, the two dimensional space on a sphere) how can we define covariant basis 'intrinsicly'?(as we have no position vector in curved space intrinsicly)

Last edited: Oct 13, 2014
2. Oct 14, 2014

### Matterwave

You have no position vector, but you can still define functions on your manifold. The vectors will correspond to directional derivatives of arbitrary functions. Usually we will see something like:

$$e_i=\frac{\partial}{\partial x^i}$$

Now $x^i$ is NOT the components of a vector in a curved space, but just a set of coordinates for the space.

3. Oct 14, 2014

### HallsofIvy

Staff Emeritus
One thing that often confuses beginners in "differential geometry" is that they have been used to using, in Calculus, "position vectors". "Position vectors", extending from a fixed origin to the point, can exist only in Euclidean space. (I used to worry a great deal whether "position vectors" on the surface of a sphere "curved" around the surface or went straight through the sphere!) In fact, the only vectors we ca deal with in general spaces are tangent vectors- which leads to "all vectors are derivatives". If we are given a path on the surface of a sphere, for example, a tangent vector to that path is tangent to the sphere and is a derivative. If you look at all the tangent vectors to a sphere at a give point, you have a vector space- the "tangent space" or "tangent plane" at that point. The "contravariant vectors" are simply the functions that assign a given tangent vector at each point. And when you have a vector space, you have it "dual space", the set of all functionals on that vector space. That is, the set of all functions that, to every vector assign a number. It is easy to show that the set off all such functionals on an n-dimensional vectors space is, itself, an n-dimensional vectors space. Those are the "covariant vectors"

Since vectors "are" derivatives, we can write every vector in terms of, say, $\partial/\partial\theta$ and $\partial/\partial\phi$ (or derivatives with respect to whatever parameters are used to define the surface). Functions that assign a number to derivatives are integrals so every "co-vector" or "covariant vector" can be written as, applying the functional to vector v, $\int v\cdot (f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz)$ so that we can think of the basis for the space of covariant vectors as $dx$, $dy$, and $dz$.

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