# Generalizing: Curl of a Function

1. Nov 29, 2009

### pbandjay

Hello. I have tried to search all day for the answer to my question but could not find anything.

Is there a way to generalize the curl of a vector field to dimensions greater than three? It seems pretty straightforward to find the gradient and divergence of higher dimensional vector fields, but my calculus book only offers the definition: curl F := ∇×F, which appears to only work if F has three components. Or am getting something wrong?

I am also only in my first real analysis course, so I have not come across this kind of topic in that class yet.

2. Nov 29, 2009

### HallsofIvy

Not easily. The problem is that the best way to generalize the cross product itself is to use the "alternating tensor". That is the operator represented by $\epsilon_{i_1i_2\cdot\cdot\cdot i_n}$ equal to 1 if $i_1i_2\cdot\cdot\cdot i_n$ is an even permutation of $1 2 3 \cdot\cdot\cdot\ n$, -1 if it s an odd permutation, and 0 if neither of those is true.
For example, when n= 3, $\epsilon_{ijk}$ has 27 entries. $\epsilon_{123}= \epsilon_{231}= \epsilon_{312}= 1$, $\epsilon_{132}= \epsilon_{213}= \epsilon_{321}= -1$, and the other 21 entries are all 0.

With that $w_i= \sum_{j=1}^3\sum_{k=1}^3\epsilon_{ijk}u_jv_k$ gives $w_1= \epsilon_{123}u_{2}v_{3}+ \epsilon_{132}u_3v_2= u_2v_3- u_3v_2$
$w_2= \epsilon_{213}u_1v_3+ \epsilon_{231}u_3v_2= -u_2v_3+ u_3v_2$
$w_3= \epsilon_{312}u_1v_2+ \epsilon_{321}u_2v_1= u_1v_2- u_2v_3$
precisely the cross product.

But, in n dimensions, $\epsilon$ has n indices. In order to get a vector result, we would have to multiply n-1 vectors, not just two.

3. Nov 29, 2009

### hamster143

In a general space whose dimension is not necessarily three, the generalization of curl is a Hodge dual of an exterior derivative of the vector field. The vector field is a 1-form, its exterior derivative is a 2-form, and its Hodge dual is a (n-2)-form. Which is a scalar (a number) for n=2, a vector for n=3, a higher-rank tensor for n>=4.

4. Nov 30, 2009

### pbandjay

Thank you for the help! Looks like I have a few things to learn about the topic.