Tensor components of a Hodge dual

[guhan]

If A is a p-vector and B is a (n-p)-vector, then the hodge dual, *A, is defined by:
$$A\ \wedge\ B = (*A,B)E \ \ \forall B\in \Lambda ^{(n-p)}$$, where E=$$e_1 \wedge\ ... \ \wedge e_n$$

I am having trouble in deriving the tensor components of the dual (n-p)-vector - *A.
Specifically, I am getting stuck when I write down the components of the (n,0)-tensor on both sides and then comparing the coefficients - because, the LHS involves the antisymmetrization, $$A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] }$$.
I got stuck with the LHS even when I took B to be just a simple (n-p)-vector of basis vectors. Because, when I do that, I get the following (n,0)-tensor on LHS...
$$\frac{1}{n!} \sum_\sigma\ (-1)^\sigma\ A^{i_{\sigma (1)} ... i_{\sigma (p)}}\ \epsilon ^ {j_{\sigma (1)} ... j_{\sigma (n-p)}}\ \ e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{j_1} \otimes ... e_{j_{n-p}}$$

 

[guhan]

Just a repost of the above message with clearer typesetting (and some more):

If A is a p-vector, then the hodge dual, $$*A$$ is a (n-p)-vector and is defined by:
$$A\ \wedge\ B = (*A,B)E \ \ ,\ \forall B\in \Lambda ^{(n-p)}$$

$$\ where,\ \ E = e_1 \wedge\ ... \ \wedge e_n$$

I am having trouble in deriving the tensor components of the dual (n-p)-vector - $$*A$$ in an orthonormal basis.
Specifically, I am getting stuck when I write down the components of the (n,0)-tensor on both sides and then comparing the coefficients - because, the LHS involves the antisymmetrization, $$A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] }\ .$$

I proceed as follows, taking B to be a simple (n-p)-vector of orthonormal basis vectors...
$$i.e.\ \ Let\ B\ =\ e_{i_{p+1}}\ \wedge\ e_{i_{p+2}}\ \wedge\ ... \wedge\ e_{i_n} \ =\ \frac{1}{n!} \epsilon^{i_{p+1}, ... ,i_n}\ e_{i_{p+1}} \otimes ... \otimes e_{i_n}$$
where $$\epsilon$$ (epsilon) is the Levi-Civita symbol and $$e_{i_x}$$ (subscripted e) are the o.n. basis vectors.

$$LHS$$
$$=\ A\ \wedge\ B\$$
$$=\ A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] }\ e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n}$$
$$=\ \left(\ \frac{1}{n!} \sum_\sigma\ (-1)^\sigma\ A^{i_{\sigma (1)} ... i_{\sigma (p)}}\ \epsilon ^ {i_{\sigma (p+1)} ... i_{\sigma (n)}}\ \right)\ \ e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n}$$

$$RHS$$
$$=\ (*A,B)\ E$$
$$=\ (*A,B)\ e_1 \wedge\ ... \ \wedge e_n$$
$$=\ \left(\ (*A,B)\ \frac{1}{n!}\ \epsilon^{i_1, ... ,i_p,i_{p+1}, ... ,i_n}\ \right) e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n}$$
$$=\ \left(\ \frac{1}{n!}\ \left(\ (n-p)!\ *A^{j_1,...,j_{n-p}} \epsilon_{j_1, ... ,j_{n-p}}\ \right)\ \epsilon^{i_1, ... ,i_p,i_{p+1}, ... ,i_n}\ \right) e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n}$$

Any pointers on how to proceed further?

 

[alexgs]

It might be easier just to try to figure out how a basis of k-forms transforms. I.e. what is the dual of e_1 ^ e_2 ^ ... ^ e_k? Then you can show that that the Hodge dual is linear and write your form A in terms of the basis.