# Proving an Identity from Differential Geometry

1. Apr 19, 2006

### mannyfold

One often encounters the following identitiy in Tensor Analysis/Differential Geometry:

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

It's easy to see why partial x^j / partial x^i = delta ij

but how does

dx^j (partial/partial x^i) = partial x^j/partial x^i ?

I have three problems with this:

1. the dx^j involves an ordinary "d' and not a partial
2. somehow the dx^j gets moved in front of the partials
3. because when it moves we have (partial)(dx^j)/partial x^i, there are two derivatives on the top

Any explanation would be helpful. Thanks.

2. Apr 19, 2006

### javanse

Hi mannyfold.

I have read the old thread, and my second posting is wrong, i think, because here we are talking about dual basis vectors.
So dxi is the dual of d/dxj, and so there inner product ist delta ij.

When we say: dxi plugs the ith component of a vector, so dxi plugs also the ith part of the basis-vector d/dxj. This component is always =1 if i = j.
That is, with an arbitrary vector v=vi(d/dxi): dxi(v)=v(xi)=vj(dxi/dxj)=vj(delta ij)=vi

Hope that helps.

Last edited: Apr 19, 2006
3. Apr 19, 2006

### Perturbation

It looks as if that equation is is to do with covariant and contravariant basis vectors. A natural vector basis for a tangent space is the coordiante basis ${\partial_i}$ and a sensible basis dual to this coordinate basis is ${\tilde{d}x_i}$. The equation you gave $\tilde{d}x^i(\partial_j)=\delta^i_j$ is from the definition of a covariant vector, that it maps a vector into the reals via an inner product. The functional $\tilde{d}x^i(\partial_j)$ is essentially saying "what's the j-th component of the i-th dual basis vector?". Given the definition of this dual coordinate basis is $\sum_j\frac{\partial x^i}{\partial x^j}dx^j$ it is obvious that its j-th component is $\frac{\partial x^i}{\partial x^j}=\delta^i_j$. Or you could look at it through the inner product.

4. Apr 19, 2006

### mannyfold

Thanks guys. I finally got it through my head that df(v) is a function that maps vectors to the reals; hence, dx^i (partial/partial x^j) just maps all the basis vectors to 0 except the ith which it maps to 1 (orthonormality condition).

My mistake was that I was still working with the elementary definition of df being an infinitesimal change in f. Hence, the confusion.

I still think there must be a connection between the elementary definition and this definition. Anybody out there have any ideas as to how to explain or connect the two definitions of df?

PS: Perturbation, how do you get those nifty math symbols to appear on this forum?

5. Apr 20, 2006

### Nimz

Check out the LaTeX tutorial here: https://www.physicsforums.com/showthread.php?t=8997

You can also click on any of the nifty math symbols and a window will pop up where you can see exactly what code was used.