# Tensor components of a Hodge dual

## Main Question or Discussion Point

This may not be a differential geometry question (and a posting of this in Linear & Abstract Algebra forum didnt help either) but Hodge dual is used in diff geom in a slightly different form. Hence posting here:

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 proceeded 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 to get the components of $$*A^{i_1,...,i_{n-p}}$$ ?

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Hello

I am not an expert on the field, just starting.

But I think that a good approach is to try to find the Hodge dual of your basis p-vector (p-form), one by one. That is:

$$A \in \Omega^{p}$$

The basis p-forms of $$\Omega^{p}$$ are for example $$\left\{\sigma^{i}\right\}$$ with $$i=1,\dots,\mathrm{dim}(\Omega^{p})$$

Hence $$A=\sum_{i}A_{i}\sigma^{i}$$.

You have the same for the $$\star A \in \Omega^{n-p}$$ space, but the basis are $$\left\{\omega^{j}\right\}$$ with $$j=1,\dots,\mathrm{dim}(\Omega^{n-p})$$.

Hence $$\star A=\sum_{j}\star A_{j}\omega^{j}$$.

Hence if you know the duals of your basis given by a linear transformation H:

$$H:\Omega^{p}\rightarrow\Omega^{n-p}$$

that is:

$$H:\sigma^{i}\rightarrow H\sigma^{i} = \sum_{j}(\sigma^{i})_{j}\omega_{j}$$

Where your H, as a matrix is defined as:

$$H_{ji} = (\sigma^{i})_{j}$$

That is, the columns of the matrix H are the vector representations of the duals of your basis of your p-forms in the (n-p)-form basis.

As for the computation of the dual of the basis element of $$\Omega^{p}$$, I think you can do it in the following way.

Remember that in an n-dimensional space the space $$\Omega^{n}$$ of n-forms has dimension 1, hence it has only one element for the basis, say $$\left\{\nu\right\}$$ which means that every n-form is represented as a number times the basis n-form. That is:
$$\phi\in\Omega^{n} \Rightarrow\phi = c\, \nu$$.

1- start with the definition of the Hodge dual for the basis element:

$$\sigma^{i}\wedge\omega^{j} = (-1)^{s}g(\omega^{j},\star \sigma^{i})\nu$$

Where s is the signal of you metric g (I think) and $$g(\omega^{j},\star \sigma^{i})$$ is the inner product between $$\omega^{j}$$ and $$\star \sigma^{i}$$.

$$\sigma^{i}\wedge\omega^{j} = (-1)^{s}g(\omega^{j},\star \sigma^{i})\nu$$

2- express the Hodge dual of your basis element $$\sigma^{i}$$ in the basis of $$\Omega^{n-p}$$:

$$\star\sigma^{i} = \sum_{k}(\star \sigma^{i})_{k}\,\omega^{k}$$

Where basically $$(\star \sigma^{i})_{k}$$ is just the j component of the p-form $$\star \sigma^{i}$$ on the $$\left\{\omega^{k}\right\}$$ basis.

With this you can rewrite the last equation in point 1 as:

$$\sigma^{i}\wedge\omega^{j} = (-1)^{s}g(\omega^{j},\sum_{k}(\star \sigma^{i})_{k}\,\omega^{k})\nu$$

And using the linearity of the inner product you can pass to the outside the summation and the constants:

$$\sigma^{i}\wedge\omega^{j} = (-1)^{s}\sum_{k}(\star \sigma^{i})_{k}\,g(\omega^{j},\omega^{k})\nu$$

Then using the fact that the basis is orthonormal, you have that:

$$g(\omega^{j},\omega^{k})=\delta_{ik}$$

And putting this in the previous equation yields:

$$\sigma^{i}\wedge\omega^{j} = (-1)^{s}\sum_{k}(\star \sigma^{i})_{k}\,\delta_{ik}\nu= (-1)^{s}\sum_{k}(\star \sigma^{i})_{i}\nu$$

3- if for each $$\sigma^{i}$$ you compute this equation for all possible $$\omega^{j}$$ you will get a set of $$\mathrm{dim}(\Omega^{n-p})$$ equations, which is just what you need because your unknowns for each $$\sigma^{i}$$ are the $$\mathrm{dim}(\Omega^{n-p})$$ constants: $$(\star \sigma^{i})_{k}$$.

I was trying to be general, maybe I made some mistakes in the middle. Also I always prefer to see examples after a general definition , they always give a better view. Hence I will try to give you an example in $$\mathrm{R}^{2}$$.

On all this examples I assume that s is such that $$(-1)^{s}=1$$. Because it is simpler and I do not know exactly what has to happen to the metric.

Example $$\mathrm{R}^{2}$$:

In this case $$n=2$$ hence $$\nu = \mathm{d}x\mathm{d}y=-\mathm{d}y\mathm{d}x$$.

The basis functions of your 0-forms can be 1. Hence the basis of your (n-p)-forms (2-forms) is for example: $$\mathm{d}x\mathm{d}y$$:

So lets express our Hodge dual in terms of this basis:
$$\star 1 = (\star 1)_{1} \mathm{d}x\mathm{d}y$$

Remember again what as done above, $$(\star 1)_{1}$$ is just the component of $$\star 1$$ relative to the basis element 1 of the (n-p)-forms space (2-forms). In this case this looks very trivial since the dimensions are both, for 0-forms and 2-forms in 2-dimensional space, equal to 1.

Hence, using the last equation in 2 (on the general explanation):

$$\sigma^{i}\wedge\omega^{j} = \sum_{k}(\star \sigma^{i})_{i}\nu$$

That reduces to:

$$1\wedge\mathm{d}x\mathm{d}y = (\star 1)_{1} \mathm{d}x\mathm{d}y$$

The wedge product is quite simple and reduces to $$1\wedge\mathm{d}x\mathm{d}y = \mathm{d}x\mathm{d}y$$ hence:

$$\mathm{d}x\mathm{d}y = (\star 1)_{1} \mathm{d}x\mathm{d}y \Rightarrow(\star 1)_{1} = 1$$

As expected.

For 1-forms p=1,n-p=1, n=2:

our $$\left\{\sigma^{i}\right\}$$ are $$\left\{\mathm{d}x,\mathm{d}y\right\}$$ and the same for $$\left\{\omega^{j}\right\}$$ as they are $$\left\{\mathm{d}x,\mathm{d}y\right\}$$. $$\nu=\mathm{d}x\mathm{d}y=-\mathm{d}y\mathm{d}x$$ again

The Hodge dual of $$\mathm{d}x$$:

Represent the dual on the dual basis: $$\star\mathm{d}x=(\star\mathm{d}x)_{1}\mathm{d}x +(\star\mathm{d}x)_{2}\mathm{d}y$$. Write the equations:

$$\sigma^{i}\wedge\omega^{j} = \sum_{k}(\star \sigma^{i})_{i}\nu$$

That reduces to, in our case:

$$\mathm{d}x\wedge\mathm{d}x = (\star \mathm{d}x)_{1}\mathm{d}x\mathm{d}y$$

and

$$\mathm{d}x\wedge\mathm{d}y = (\star \mathm{d}x)_{2}\mathm{d}x\mathm{d}y$$

The two wedge products are: $$\mathm{d}x\wedge\mathm{d}x = 0$$ and $$\mathm{d}x\wedge\mathm{d}y = \mathm{d}x\mathm{d}y$$ hence:

$$0 = (\star \mathm{d}x)_{1}\mathm{d}x\mathm{d}y \Rightarrow (\star \mathm{d}x)_{1}=0$$

and

$$1\mathm{d}x\mathm{d}y = (\star \mathm{d}x)_{2}\mathm{d}x\mathm{d}y\Rightarrow (\star \mathm{d}x)_{2} = 1$$

Notice that I am using the short notation: $$\mathm{d}x\mathm{d}y = \mathm{d}x\wedge\mathm{d}y$$, instead of what is used in the definition of the wedge product: $$\mathm{d}x\wedge\mathm{d}y=\mathm{d}x\mathm{d}y-\mathm{d}y\mathm{d}x$$.

Hence the Hodge dual becomes:

$$\star\mathm{d}x = (\star \mathm{d}x)_{1} \mathrm{d}x + (\star \mathm{d}x)_{2} \mathrm{d}y= \mathrm{d}y$$.

Now for $$\mathrm{d}y$$

Everything goes the same way until we get the two equations:

$$\mathm{d}y\wedge\mathm{d}x = (\star \mathm{d}y)_{1}\mathm{d}x\mathm{d}y$$

and

$$\mathm{d}y\wedge\mathm{d}y = (\star \mathm{d}y)_{2}\mathm{d}x\mathm{d}y$$

The two wedge products are: $$\mathm{d}y\wedge\mathm{d}x = -\mathm{d}x\mathm{d}y$$ and $$\mathm{d}y\wedge\mathm{d}y = 0$$ hence:

$$-\mathm{d}x\mathm{d}y = (\star \mathm{d}y)_{1}\mathm{d}x\mathm{d}y \Rightarrow (\star \mathm{d}y)_{1}=-1$$

and

$$0 = (\star \mathm{d}y)_{2}\mathm{d}x\mathm{d}y\Rightarrow (\star \mathm{d}y)_{2} = 0$$

Hence the Hodge dual becomes:

$$\star\mathm{d}y = (\star \mathm{d}y)_{1} \mathrm{d}x + (\star \mathm{d}y)_{2} \mathrm{d}y= -\mathrm{d}x$$.

I hope this helps.

You have a pdf by Tevian Dray, from the Oregon State University, explaining this in a more higher level way, I think it might be good to take a look.

http://oregonstate.edu/~drayt/Courses/MTH434/2007/dual.pdf" [Broken]

Regards

-artur palha

Last edited by a moderator:
Thank you for the detailed explanation! I really appreciate it. It was very helpful.

I came across hodge duals while revising the topic of exterior algebra for tensors, and thought it might be a good exercise to derive (rather, directly write down) the tensor components of a hodge dual through generalized exterior algebra. Any pointers as to how to extend your post to write down the duals of non-simple vectors? I have either gotten myself stuck or missing something obvious, in my derivation above.

The components of the hodge dual should turn out like this:
$$*A^{j_1, j_2,..., j_{n-p}} = \frac{(-1)^s}{(n-p)!}\ \epsilon ^{i_1,...,i_p,j_1,...,j_{n-p}}\ A_{i_1,...,i_p}$$

where,
s= number of (-1) in the metric $$g_{ij}$$
$$\epsilon^{i_1,...,i_p,j_1,...,j_{n-p}}$$ = Levi-Civita symbol

""
This may not be a differential geometry question (and a posting of this in Linear & Abstract Algebra forum didnt help either) but Hodge dual is used in diff geom in a slightly different form. Hence posting here:

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 proceeded 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 to get the components of $$*A^{i_1,...,i_{n-p}}$$ ?
"

hEY GUAHN SUGUMAR R U FROM TAMBARAM.

Hi!

I was doing some more reading and I found something that might be useful to your problem.

Take a look at chapter 4 of this book:

Differential geometry for physicists

Bo-Hu Hou
and
Bo-Yuan Hou

-artur palha