# Derivatives in topology

1. Jun 16, 2010

### RedX

Given a smooth manifold with no other structure (like a metric), one can define a derivative for a vector field called the Lie derivative. One can also define a Lie derivative for any tensor, including covectors.

Incidentally, with antisymmetric covectors (differential forms) one can define another type of derivative called 'd' that doesn't seem to have any direction (not a directional derivative).

If you add a metric tensor, then you can define another derivative for a vector field called (at least in physics) the covariant derivative, which is a partial derivative plus a connection term.

My question is which derivative is really the directional derivative of a vector field: the Lie derivative, or the covariant derivative?

Also, probably related, is the definition of a directional derivative unique?

And how can there be a derivative 'd' that has no direction? That would have to imply, roughly speaking, that 'd' is a derivative averaged around a circle at a point?

2. Jun 17, 2010

### eok20

The covariant derivative is more like a directional derivative of a vector field since this agrees with the directional derivative in Euclidean space when we take the covariant derivative corresponding to the standard metric on Euclidean space. One think to notice is that the Lie derivative is not linear over functions in either of its arguments, whereas the covariant derivative is linear over functions in the vector field you are differentiating with respect to.

It may help to think of d as the "total derivative." Indeed, when we apply it to a function f we get the covector field df that corresponds to the gradient. If we then apply that covector field to a tangent vector V, we get the directional derivative of f in the direction of V. Also, besides gradient, d also generalizes curl and divergence and so it enters into integral formulas (the generalized Stokes theorem).

3. Jun 17, 2010

### quasar987

In R³, we are familiar with the notion of taking the derivative of a vector field in the direction of some vector, not vector field. Namely, if Y:R³-->R³ is a vector field on R³, the derivative of Y at p in the direction of a vector v is the vector

$$\lim_{t\rightarrow 0}\frac{Y(p+tv)-Y(p)}{t}$$

The Lie derivative on the other hands deals with the notion of the derivative of a vector field in the direction of a vector field. This is a generalization of the above notion because in the special where X is the constant vector field X_p=v, then $L_XY|_p$ is just the derivative of Y at p in the direction of a vector v.

More generally, this is true in any vector space V, but in a general manifold M, the notion of a constant vector field is ill-defined (there is no canonical identification of the different tangent spaces)...

4. Jun 17, 2010

### LukeD

The gradient df isn't quite the same thing as a directional derivative. It's true though that in R^3, the gradient dotted with a direction gives the directional derivative (and the covariant derivative is a generalization of this idea).

The gradient of a function can be visualized in terms of the level surfaces of the function. To find the direction derivative of a function, we draw a tiny (infinitesimal) arrow in a direction and we count the level surfaces that are crossed by the arrow.
The gradient df is a function that gives the density and direction of level surfaces of f at each point. So in a sense, it can be identified with the set of level surfaces of f.

A p-form on the other hand is a bit like a set of level surfaces. You can visualize a p-form as an oriented p-dimensional grid in n-dimensional space. So a 2 form would be like having a (thick) sheet of graph paper at each point in space.

The exterior derivative of a form w is defined in terms of the gradient and the wedge product. For a simple form w = f*dx^dy^...^dz, the exterior derivative is dw = df^dx^dy^...^dz. So the derivative is the level surfaces of the function that gives a form at each point wedged together with the elements of your basis. dw is a "level surface form" that lets you approximately reconstruct your original form w.

I haven't explained what the wedge product does geometrically though.

5. Jun 18, 2010

### RedX

I thought the Lie derivative is linear in both arguments. The Lie derivative L of the tensor T in the direction of the vector field X obeys:

$$L_X (T_1+T_2)=L_X (T_1)+L_X(T_2)$$
$$L_{X_1+X_2} (T)=L_{X_1} (T)+L_{X_2}(T)$$

That has always confused me. I understand what you mean, that

$$<df,V>=V[f]=V^i \frac{\partial}{\partial x^i}f$$

But why can't the vector field V itself be called the directional derivative in the direction of V?

Calling the gradient the total derivative makes sense, since if you have the gradient you can find all the directional derivatives.

That's a good point. So in general, it's not true that $$L_XY|_p$$ is equal to $$L_{X_p}Y|_p$$ where $$X_p$$ is a constant field that has the value of the original field X at the point p. However, shouldn't it be the case that the value of X at p gives the most contribution to the Lie derivative? Imagine R^3 as your manifold, and you want the Lie derivative at the origin: why should it matter to the Lie derivative at the origin what the value of X is at far off points like infinity?

That's really interesting. The book I'm reading is very abstract and just defines forms as antisymmetric tensors, and it's all symbol manipulation.

6. Jun 19, 2010

### eok20

What I said was that the Lie derivative is not linear with respect to functions, that is

$$L_{fX} Y \ne fL_X Y, ~~ L_X fY \ne f L_XY$$

where f is a smooth function. On the other hand, it is the case that
$$\nabla_{fX} Y = f\nabla_X Y.$$
This fact basically amounts to quaser987's comment that the covariant derivative along a vector field just depends on the tangent vector at that point.

You are correct, a tangent vector is a directional derivative for functions (this is often how tangent vectors are defined for smooth manifolds). Your question was about directional derivatives of vectors. Note however, that for functions, both the Lie derivative and covariant derivative agree: $$L_X f = \nabla_X f = Xf$$.

7. Jun 19, 2010

### LukeD

Try the book Geometric Vectors by Gabriel Weinreich. It's very informal. It develops the machinery of vectors and forms in 3D space with lots of pictures. It's great.

8. Jun 19, 2010

### quasar987

In a general manifold, the notion of a constant vector field does not exist (except the notion of the vector field wich is constantly 0). For instance, if in some coordinate system,

$$X_p=\sum_ia^i\left.\frac{\partial}{\partial x_i}\right|_p$$

then how do you propose to define the "constant field that has the value of the original field X at the point p"? As

$$X_q:=\sum_ia^i\left.\frac{\partial}{\partial x_i}\right|_q$$

?? Well, if you write that field in another coordinate system, it writes

$$X_q:=\sum_j\sum_ia^i\left.\frac{\partial y^j}{\partial x_i}\right|_q\left.\frac{\partial}{\partial y_j}\right|_q$$

This is certainly not the expression of a constant vector field per our definition. In fact, given any vector field, there is a coordinate system in which it is locally constant near p: take a coordinate system in which the x^1 axis gets mapped to the integral curve of the field near p. So our definition of being constant for a vector field is meaningless

It is true that the value of the Lie derivative at p is independant of the value of either vector fields far away from p, but it seems weird and meaningless to me to say that the value of X at p "gives the most contribution". Remember that $L_XY|_p$ is just the oridinary time derivative of the pullback of Y by the flow of X:

$$L_XY|_p=\left.\frac{d}{dt}\right|_{t=0}(\phi^X_{-t})_*X_{\phi^X_{t}}$$