Let's say I want to find the cross product of two gradient fields on a manifold $M$ in any co-ordinate system (not necessarily orthonormal). So I want to define a map $C: T_{p}M$ $\times$ $T_{p}M \to N_pM$ which I guess would be a rank $3$ tensor.

Say f and g are two linear functionals on the manifold, $g_{ij}$ is the metric tensor and $\epsilon$ is the Levi-Civita symbol.

Now, $grad$ $f$ = $g^{ij}$ $\partial_{i}f$ $\partial_{i}$

So, $C(grad$ $f$, $grad$ $c$ $)$ = $C(g^{kl}$ $\partial_{k}f$ $\partial_{l},$ $g^{mn}$ $\partial_{m}f$ $\partial_{n}$ $)$

The cross product is also defined as $*(df \wedge dc)$ = $*($ $\partial_{k}f$ $dx^k$ $\wedge$ $\partial_{m}c$ $dx^m$ $)$

Where $*$ is the hodge dual and $\wedge$ is the skew symmetric tensor product(wedge operator)

Also, according to https://en.wikipedia.org/wiki/Cross_product#Index_notation_for_tensors, it seems like the cross product is:

$(a \times b)^i = C(a, b)^i = g^{ij}\epsilon_{jkl}dx^kdx^l$

Is there any way to derive this general cross product in any coordinate frame? Perhaps according to the two definitions I have above? Or is there anything that I am doing wrong?

  • $\begingroup$ Are you trying to get to Wiki's identity from your result? $\endgroup$ – IAmNoOne Dec 11 '16 at 6:10
  • $\begingroup$ Yes, I'm trying to do that but don't know how to go about it since I haven't been able to find any proof of the identity on Wiki $\endgroup$ – tushar Dec 11 '16 at 6:13
  • $\begingroup$ Read this, I think it might help ucl.ac.uk/~ucappgu/seminars/levi-civita.pdf $\endgroup$ – IAmNoOne Dec 11 '16 at 6:16
  • $\begingroup$ Basically, I want to find the Tensor that C represents in a similar fashion that the first fundamental form is defined $I = g_{ij}dx^idx^j$ $\endgroup$ – tushar Dec 11 '16 at 6:16

The usual cross product is only possible in three dimensions - its not possible in lower or higher dimensions; interestingly it makes the space into a lie algebra.

The generalisation to all (finite-dimensional) spaces is done by (differential) forms with the wedge as product.

  • 1
    $\begingroup$ ... exactly what I'm trying to find out. The cross product with differential forms is the hodge dual of wedge product of differential one forms. I simply want to see if the tensor I found on Wiki is correct or not. $\endgroup$ – tushar Dec 11 '16 at 7:49

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