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I'm learning differential geometry from a textbook, and I got stuck on a problem.

I'm supposed to calculate this for a $p$-vector $F$ in $n$ dimensions:

$(\mathrm{div}_\omega F)^{i...j} = F^{ki...j}\ _{,k}$

where

$\mathrm{div}_\omega F = (-1)^{n(p-1)} * \mathrm{d} * F$

The star is the Hodge star; the RHS operator is the codifferential.

I got as far as

$\mathrm{d}*F = \frac{1}{p!(n-p)!} \varepsilon_{i...jk...l}F^{i...j}\ _{,m} \ \mathrm{d}x^m\wedge\mathrm{d}x^k\wedge...\wedge\mathrm{d}x^l$

Clearly, $m \in {i...j}$ for all non-zero terms. But I can't get any further than that, and not for lack of trying. Any tips will be appreciated. :)

1 Answers1

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Let us denote the $p$-vector $F$ in local coordinates by $$F=F^{i_1\cdots i_p}(x)e_{i_1}\wedge\cdots\wedge e_{i_p} $$

and let us suppose to work on a flat manifold of dimension $n$: the metric tensor $g$ is just the identity matrix in $n$ dimensions. We will always use the Einstein notation for repeated indices. Using the above conventions the $n-p$-vector $\star F$ becomes

$$\star F=\frac{1}{(n-p)!}\epsilon_{i_1\cdots i_n}F^{i_1\cdots i_p}(x)e_{i_{p+1}}\wedge\cdots\wedge e_{i_n} $$ and so $$d\star F=\frac{1}{(n-p)!}\sum_{j=1}^p\epsilon_{i_1\cdots i_n}\frac{\partial F^{i_1\cdots i_p}(x)}{\partial x_{i_j}}e_{i_j}\wedge e_{i_{p+1}}\wedge\cdots\wedge e_{i_n}. $$

In the above formula the sum is up to $p$ as $ e_{i_j}\wedge e_{i_{p+1}}\wedge\cdots\wedge e_{i_n}=0$ if $j\in\{p+1,\dots, n\}$ because $e_\bullet\wedge e_\bullet =0$ by antisymmetry of the wedge product. We end up with the $n-(n-p+1)=p-1$ vector

$$\star d\star F=\frac{1}{(p-1)!(n-p)!}\sum_{j=1}^p\epsilon_{i_1\cdots i_n}\frac{\partial F^{i_1\cdots i_p}(x)}{\partial x_{i_j}}\star\left(e_{i_j}\wedge e_{i_{p+1}}\wedge\cdots\wedge e_{i_n}\right)= \\ \frac{1}{(p-1)!(n-p)!}\sum_{j=1}^p\epsilon_{i_1\cdots i_n}\epsilon_{i_j i_{p+1}\dots i_n i_1\cdots \hat{i}_j\cdots i_p }\frac{\partial F^{i_1\cdots i_p}(x)}{\partial x_{i_j}}e_{i_{1}}\wedge\cdots\wedge \hat{e}_{i_j}\wedge\cdots\wedge e_{i_p} ~~(*) $$ where $\hat{\cdot}$ denotes omission. I won't manipulate that formula further but I will provide you with an easier example.

Let us consider the $n=3$ and $p=1$ case, with $$F=F^{i_1}(x)e_{i_1}=F^1(x)e_1+F^2(x)e_2+F^3(x)e_3.$$ Then (*) becomes $(j=1=p)$

$$\star d\star F=\frac{1}{2!}\epsilon_{i_1 i_2 i_3}\epsilon_{i_1 i_2 i_3}\frac{\partial F^{i_1}(x)}{\partial x_{i_1}}= \frac{1}{2}\left[(\epsilon_{1 23}\epsilon_{123}+\epsilon_{1 32}\epsilon_{132})\frac{\partial F^{1}(x)}{\partial x_{1}} + (\epsilon_{2 13}\epsilon_{213}+\epsilon_{2 31}\epsilon_{231})\frac{\partial F^{2}(x)}{\partial x_{2}} + \right. \\ \left.(\epsilon_{3 21}\epsilon_{321}+ \epsilon_{3 12}\epsilon_{312})\frac{\partial F^{3}(x)}{\partial x_{3}} \right]=\frac{\partial F^{1}(x)}{\partial x_{1}}+ \frac{\partial F^{2}(x)}{\partial x_{2}} + \frac{\partial F^{3}(x)}{\partial x_{3}}, $$ i.e. the divergence of the vector field $F$.

Avitus
  • 14,018
  • Are you going with the convention that $$\mathrm{d}x^1\wedge\mathrm{d}x^2 = \frac{1}{2} \left(\mathrm{d}x^1 \otimes \mathrm{d}x^2 - \mathrm{d}x^2\otimes\mathrm{d}x^1\right)$$ In my original stab at the problem I used the convention that omits the fraction in the definition of the wedge product. – dom_miketa Aug 03 '14 at 09:01
  • Anyway, thanks for taking the time to help. I'm afraid I already got the simple vector case in another exercise. It's the higher $p$ case that stumped me, especially the factor of $(-1)^{n(p-1)}$. – dom_miketa Aug 03 '14 at 09:08
  • In my answer I did not explicitly relate the wedge product to the tensor one, so I used no explicit convention. In the OP the relation between the divergence $\operatorname{div}_\omega F$ and $\star d\star F$ is not clear: on the left you have a scalar quantity, while the r.h.s gives you a differential form. Could you please update / explain it? Please, feel free to upvote and flag the answer if you found it useful. – Avitus Aug 03 '14 at 09:49
  • $\mathrm{div}_\omega F$ is itself a $(p-1)$-vector, just like the RHS. The first equation in the OP gives you its alleged components, making it clear that the LHS is a tensor, too. – dom_miketa Aug 03 '14 at 11:12
  • Ok, then the notation is ok. Where are you stuck now? – Avitus Aug 03 '14 at 11:51
  • Not on the $p=1$ case, but a bit higher than that. The $(-1)^{n(p-1)}$ factor especially is a bit mysterious to me. I'm also confused what you meant by not adopting a specific definition of the wedge product; if $F(\mathrm{d}x^1...\mathrm{d}x^p)=F^{1...p}$, doesn't that necessitate a wedge product convention? – dom_miketa Aug 03 '14 at 14:29
  • Consider the algebra of p-vector (fields) as Lie with respect to the Schouten(Nijenhuis) bracket. With respect to d is it a differential graded Lie algebra? – Jim Stasheff Mar 13 '20 at 15:54