# Generalisation of curl to n dimensions

1. Nov 8, 2011

### dimension10

Curl is easy to compute in 3 dimensions and if you let the third component be 0, its also easy in 2 dimensions. If you let the second and third components be 0, it is also easy in 1 dimension.

My question is,

is there a generalisation for curl to n dimensions and if there is, what is it and is it possible to use the del operator and the cross product to compute the curl.

Thanks.

2. Nov 9, 2011

### Diffeomorphic

Only in three dimensions is the geometrically defined curl of a vector field again a vector field. Since there is trouble with defining cross-product in some dimensions, it is hard to generalize. There may be some hope with the seven-dimensional cross product? http://en.wikipedia.org/wiki/Seven-dimensional_cross_product

You may also want to investigate generalizing curl on pseudo-Riemannian manifolds.

3. Nov 9, 2011

### dimension10

Ok, then if it is not a vector field, then is it a tensor field? And what if I wanted to find the curl of an 8-dimensional vector field? Then it would not be possible to use the 7 dimensional cross product either.

Also, the geometric interpretation of curl would surely exist in more than 3 dimensions, won't it? So, is it at least possible to generalise curl to n dimensions?

4. Nov 9, 2011

### homeomorphic

Actually, there is a generalization of curl to any dimension.

If you have a vector field, you can take its dual. So, the dual of 4 i + 2 j + 3 k would just be 4 dx + 2 dy + 3 dz. So, that operation allows us to turn vector fields into 1-forms. The curl of the vector field corresponds to the exterior derivative. You take the dual, then exterior derivative, then the dual of that. That gives you curl. This process works on a vector field in any dimension. All you need is a smooth manifold.

And yes, it has the "same" geometric interpretation as curl in dimension 3. Namely, the exterior derivative of a form can be thought of as the integral over the boundary of an infinitesimal parallelepiped. To spell this out, you want the exterior derivative of a p-form to be a p+1 form. A p+1 form takes in p+1 vectors and gives you a number. So you form the parallelepiped spanned by the p+1 input vectors and integrate of its boundary to find the value of the p+1 for on the p+1 input vectors.

One thing that doesn't work is the fact that the cross product gives you a vector. Instead of a vector, you get bivectors, trivectors, etc.

The analogue of the cross product would be wedge product, followed by Hodge dual. I'm not sure what the analogue of the del operator would be. At any rate, I don't think you would want to compute the exterior derivative that way. To compute exterior derivatives, you just compute exterior derivatives. Basically, you just use the Liebniz rule for exterior derivatives all over the place, plus the fact that the exterior derivative of a function is the gradient.

5. Nov 10, 2011

### dimension10

Thanks a lot, but I have a question. How does one take the exterior derivative of 4 dx + 2 dy + 3 dz in the first place?

The cross product really seems like its defined in n dimensions. The magnitude of the cross product is so I think that the cross product itself should be.

Last edited: Nov 10, 2011
6. Nov 10, 2011

### homeomorphic

Actually, there's another thing I glossed over. I should say you need a metric to do this (take duals, or, in the index notation, raise and lower indices). My discussion was in terms of Euclidean space. So, let's just forget about manifolds and stay in R^n, just to make life easier.

So, how do you take the exterior derivative? Well, I described how to do it conceptually. 4 dx + 2 dy + 3 dz is a bad example because it's constant. It's curl is zero, as you can check. And likewise, its exterior derivative will be zero.

There's a lot to be said about differential forms. 4 dx + 2 dy + 3 dz is what we would call a 1-form. It's something that you integrate over curves. Or, maybe you can integrate it over the boundary of a parallelogram.

At this point, I'm realizing it will take too long to explain exterior calculus.

It's defined if you allow it to be something other than a vector. How could the magnitude of the cross product be defined if there is no cross product? In three dimensions, the orthogonal complement of a 2-d plan (one spanned by two vectors) is one dimensional. That's what's special about 3-d. There's only one choice. But in other dimensions, the complement isn't 1-dimensional. So, if you want something like a cross-product, you need an n-2-dimensional plane element. That's the best you can do. So, it won't be a vector anymore. I mean, you have n-2 dimensions to play around with? How are you going to narrow it down? I suppose you can come up with some weird way of choosing a vector, but I don't see a way to argue for any compelling choice. Any vector in that n-2 dimensional space is just as good as any other.

There's an operation called a wedge product that does something like what I'm describing. If you have two vectors, their wedge product represents the plane-element (or bivector--think of it as the parallelogram spanned by the vectors) that they span. Then, there's another operation called the Hodge dual which takes the orthogonal complement. So, doing a wedge product, followed by Hodge dual is kind of like a cross product.

7. Nov 11, 2011

### dimension10

Thanks.

Is exterior calculus the same as exterior algebra? Or are they 2 different things? I'm asking because I know exterior algebra so I wanted to know whether it is the same as exterior calculus.

And the reason why I say that the magnitude of the cross product is defined in n dimensions is because of the identity:

$${\left(||\mathbf{a} \times \mathbf{b}|| \right)}^{2}+{\left(\mathbf{a} \cdot \mathbf{b} \right)}^{2}={\left( ||a|| \: ||b||\right)}^{2}$$

8. Nov 11, 2011

### homeomorphic

Exterior calculus uses exterior algebra. It's like complex numbers versus complex analysis. You're doing calculus with exterior algebras. And for this to work, you need a manifold and an exterior algebra structure coming from the tangent spaces at different points. Exterior algebra just refers to one vector space.

That only makes sense in R^3. I don't think that's a very enlightening way to generalize to higher dimensions. The way to do it, is, as I said, take the orthogonal complement of the plane spanned by the two vectors. And then you have a "plane element" whose magnitude is like the magnitude of the cross product. But you don't get a vector. You get a plane-element (wedge product of a bunch of vectors) with magnitude equal to the signed area of the parallelogram spanned by the two vectors, which is just like the cross product, other than the fact that it's not a vector.