Tensor components of a Hodge dual

[guhan]

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} <br /> \ =\ \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}} ?

 

[gorkiana]

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.

Lets start with 0-forms, p=0, n-p=2, n=2:

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 let's 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"

Regards

-artur palha

 

[guhan]

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:
<br /> *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}<br />

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

 

[krithika]

""

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} <br /> \ =\ \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.

 

[gorkiana]

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