# Differential Geometry

1. Feb 13, 2009

### latentcorpse

if $\alpha, \alpha' \in \Omega^1$. Rewrite the identity,

$d(\alpha \wedge \alpha')=d \alpha \wedge \alpha' - \alpha \wedge d \alpha'$ in terms of vector calculus.

I have absolutely no idea what is going on here. So if anybody could explain to me a) what this is all about and b) how to go about doing it, that would be great.

Cheers.

2. Feb 13, 2009

### dx

The identity you're looking for is $$\nabla \cdot ( \alpha \times \alpha') =(\nabla \times \alpha) \cdot \alpha' - \alpha \cdot (\nabla \times \alpha')$$. Do you know how the exterior derivative is related to div, grad and curl? Do you know how the wedge product is related to the cross and dot products?

3. Feb 13, 2009

### latentcorpse

i don't know how the exterior derivative is related to grad div and curl

and i was under the impression that the wedge product was the cross product?

4. Feb 14, 2009

### dx

The wedge product of 1-forms is related to the cross product. The wedge product is a more general thing that is defined in all dimensions, whereas the cross product is only defined in three dimensions.

The gradient of vector calculus corresponds to the exterior derivative of functions, the curl corresponds the exterior derivative of 1-forms, and the divergence corresponds to the exterior derivative of 2-forms.

It is essential that you first learn about these ideas before attempting this problem.

5. Feb 14, 2009

### tiny-tim

Nooo … the wedge product is a 2-form, and the cross product is a pseudovector.

The dual 1-form (in 3-dimensional space) of the wedge product corresponds to the cross product.

EDIT: There's a good explanation, and a nice diagram, at http://en.wikipedia.org/wiki/Cross_product#Cross_product_as_an_exterior_product

Last edited: Feb 14, 2009
6. Feb 15, 2009

### latentcorpse

but $\alpha,\alpha' \in \Omega^1$ i.e. they are one forms. if there was an isomorphism (there is one given in the diagram that came with the question) $\Phi_1$ that maps $\Phi_1: \Omega^1 \rightarrow X$ then surely my answer should be

$\nabla \cdot (\Phi_1(\alpha) \times \Phi_1(\alpha'))= \nabla \times \Phi_1(\alpha) \cdot \Phi_1(\alpha') - \Phi_1(\alpha) \cdot \nabla \times \Phi_1(\alpha')$

note that $\Omega^1$ is the space of 1-forms and $X$ is the space of vector fields in $\maathbb{R}^3$

7. Feb 15, 2009

### tiny-tim

Hi latentcorpse!

(have an alpha: α and a phi: Φ and a del: ∇ and a dot: · )

Yes, you should use a different letter, say a, to show that it's a vector (field) …

∇·(a x a') = (∇ x a)·a' - a·(∇ x a')