Differential Geometry/Ricci Calculus Question

1. Apr 15, 2006

mannyfold

I am having a problem that is glossed over in many text books but is driving me nuts.

Consider the following inner product or one-form with a vector argument:

dx_i(partial_j) = kroenicker delta ij

Here, dx_i is a one-form and partial_j (the partial wrt x_j) is a vector.

Some books say:

dx_i(partial_j) = partial x_i / partial x_j = kroenicker delta ij

The problem is that I can't see how this is true. (Well, I do know that partial x_i / partial x_j = kroenicker delta ij but I can't see the rest.) What am I missing here?

Thanks.

Last edited: Apr 15, 2006
2. Apr 16, 2006

javanse

The crux is the definition of the differential:

df(v)=vf, where v is a tangent vector.
This is of the form L(x)=ax in linear algebra:
L is a linear function, x a vector, a the covector, defined by a_i=L(e_i).

Now let df be dx and v the coordinate-tangentvector d/dx (d=partials),
then

dx^i(d/dx_j)=dx^i/dx_j=delta ij.

Last edited: Apr 16, 2006
3. Apr 16, 2006

mannyfold

AHA!

That is the problem!

I don't see how df(v) = vf. Where did you get that definition of the differential?

df = partial f / partial x^i dx^i (i = 0 .. n)

but the partials here are just the components of the 1-form.

I have the feeling that there is something very simple that I'm not seeing here.

4. Apr 16, 2006

javanse

Let's write df=(df/dx^i)dx^i (last d=partials)

What is df(v), where v is a vector?

df(v)=(df/dx^i)dx^i(v)

What is dx^i(v)?

It's simply v^i.

So you can write:

df(v)=(df/dx^i)v^i = v^i(df/dx^i).

But what is our v^i on a manifold?

It's d/dx^i!

So you can write it as df(v)=vf, where v=d/dx^i

Last edited: Apr 16, 2006
5. Apr 16, 2006

mannyfold

Hi Javanse,

I appreciate your replies, but I am going to challenge what you wrote.

We are seeking to prove df(v) = vf.

You wrote:

df(v)=(df/dx^i)dx^i(v)
What is dx^i(v)?

It's simply v^i.

But this assumes what we are trying to prove:
dx^i(v) = v^jx^i =d/dx^j (x^i) = v^i delta ij

You wrote:

df(v)=(df/dx^i)v^i = v^i(df/dx^i).

How can you commutate the differential operator v^i?

Finally, from this last statement you get df(v) = vf, but if you read it carefully you don't dispose of the df/dx^i so it appears that your result is:

df(v) = (d/dx^j) (df/dx^i) = v (df/dx^i)

6. Apr 16, 2006

mannyfold

Another perspective on this is yet another equality I've found that was not expounded upon:

dx^i (partial/partial x^j) = (partial x^i)/(partial x^j) = delta ij

I get the last equality. I can't see the first.

7. Apr 16, 2006

javanse

Okay, last try :)

Let v=v^i d/dx^i

-> v(f)=v^i df/dx^i.

This is the directional derivative.

I introduced the operator at the end.
You can also write the v^i from the beginning in front of the derivative.

But the result is the same as above: it is the directional derivative.

dxi(v)=v(xi)=vj delta ij=vi :)

Maybe someone else has another explanation.

Last edited: Apr 17, 2006
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