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Tensor components of a Hodge dual

  1. Dec 2, 2008 #1
    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, [tex]*A[/tex] is a (n-p)-vector and is defined by:
    [tex] A\ \wedge\ B = (*A,B)E \ \ ,\ \forall B\in \Lambda ^{(n-p)} [/tex]

    [tex]\ where,\ \ E = e_1 \wedge\ ... \ \wedge e_n [/tex]

    I am having trouble in deriving the tensor components of the dual (n-p)-vector - [tex]*A[/tex] 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, [tex] A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] }\ .[/tex]

    I proceeded as follows, taking B to be a simple (n-p)-vector of orthonormal basis vectors...
    [tex] 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} [/tex]
    where [tex] \epsilon [/tex] (epsilon) is the Levi-Civita symbol and [tex] e_{i_x} [/tex] (subscripted e) are the o.n. basis vectors.

    [tex] LHS [/tex]
    [tex]=\ A\ \wedge\ B\ [/tex]
    [tex]=\ 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} [/tex]
    [tex] =\ \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} [/tex]

    [tex] RHS [/tex]
    [tex] =\ (*A,B)\ E [/tex]
    [tex] =\ (*A,B)\ e_1 \wedge\ ... \ \wedge e_n [/tex]
    [tex] =\ \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} [/tex]
    [tex] =\ \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} [/tex]

    Any pointers on how to proceed further to get the components of [tex]*A^{i_1,...,i_{n-p}}[/tex] ?
  2. jcsd
  3. Dec 4, 2008 #2

    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:

    [tex]A \in \Omega^{p}[/tex]

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

    Hence [tex]A=\sum_{i}A_{i}\sigma^{i}[/tex].

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

    Hence [tex]\star A=\sum_{j}\star A_{j}\omega^{j}[/tex].

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


    that is:

    [tex]H:\sigma^{i}\rightarrow H\sigma^{i} = \sum_{j}(\sigma^{i})_{j}\omega_{j}[/tex]

    Where your H, as a matrix is defined as:

    [tex]H_{ji} = (\sigma^{i})_{j}[/tex]

    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 [tex]\Omega^{p}[/tex], I think you can do it in the following way.

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

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

    [tex]\sigma^{i}\wedge\omega^{j} = (-1)^{s}g(\omega^{j},\star \sigma^{i})\nu[/tex]

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

    [tex]\sigma^{i}\wedge\omega^{j} = (-1)^{s}g(\omega^{j},\star \sigma^{i})\nu[/tex]

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

    [tex]\star\sigma^{i} = \sum_{k}(\star \sigma^{i})_{k}\,\omega^{k}[/tex]

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

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

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

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

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

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


    And putting this in the previous equation yields:

    [tex]\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[/tex]

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

    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 [tex]\mathrm{R}^{2}[/tex].

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

    Example [tex]\mathrm{R}^{2}[/tex]:

    In this case [tex]n=2[/tex] hence [tex]\nu = \mathm{d}x\mathm{d}y=-\mathm{d}y\mathm{d}x[/tex].

    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: [tex]\mathm{d}x\mathm{d}y[/tex]:

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

    Remember again what as done above, [tex](\star 1)_{1}[/tex] is just the component of [tex]\star 1[/tex] 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):

    [tex]\sigma^{i}\wedge\omega^{j} = \sum_{k}(\star \sigma^{i})_{i}\nu[/tex]

    That reduces to:

    [tex]1\wedge\mathm{d}x\mathm{d}y = (\star 1)_{1} \mathm{d}x\mathm{d}y[/tex]

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

    [tex] \mathm{d}x\mathm{d}y = (\star 1)_{1} \mathm{d}x\mathm{d}y \Rightarrow(\star 1)_{1} = 1[/tex]

    As expected.

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

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

    The Hodge dual of [tex]\mathm{d}x[/tex]:

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

    [tex]\sigma^{i}\wedge\omega^{j} = \sum_{k}(\star \sigma^{i})_{i}\nu[/tex]

    That reduces to, in our case:

    [tex]\mathm{d}x\wedge\mathm{d}x = (\star \mathm{d}x)_{1}\mathm{d}x\mathm{d}y[/tex]


    [tex]\mathm{d}x\wedge\mathm{d}y = (\star \mathm{d}x)_{2}\mathm{d}x\mathm{d}y[/tex]

    The two wedge products are: [tex]\mathm{d}x\wedge\mathm{d}x = 0[/tex] and [tex]\mathm{d}x\wedge\mathm{d}y = \mathm{d}x\mathm{d}y[/tex] hence:

    [tex]0 = (\star \mathm{d}x)_{1}\mathm{d}x\mathm{d}y \Rightarrow (\star \mathm{d}x)_{1}=0[/tex]


    [tex]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[/tex]

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

    Hence the Hodge dual becomes:

    [tex]\star\mathm{d}x = (\star \mathm{d}x)_{1} \mathrm{d}x + (\star \mathm{d}x)_{2} \mathrm{d}y= \mathrm{d}y[/tex].

    Now for [tex]\mathrm{d}y[/tex]

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

    [tex]\mathm{d}y\wedge\mathm{d}x = (\star \mathm{d}y)_{1}\mathm{d}x\mathm{d}y[/tex]


    [tex]\mathm{d}y\wedge\mathm{d}y = (\star \mathm{d}y)_{2}\mathm{d}x\mathm{d}y[/tex]

    The two wedge products are: [tex]\mathm{d}y\wedge\mathm{d}x = -\mathm{d}x\mathm{d}y[/tex] and [tex]\mathm{d}y\wedge\mathm{d}y = 0[/tex] hence:

    [tex]-\mathm{d}x\mathm{d}y = (\star \mathm{d}y)_{1}\mathm{d}x\mathm{d}y \Rightarrow (\star \mathm{d}y)_{1}=-1[/tex]


    [tex]0 = (\star \mathm{d}y)_{2}\mathm{d}x\mathm{d}y\Rightarrow (\star \mathm{d}y)_{2} = 0[/tex]

    Hence the Hodge dual becomes:

    [tex]\star\mathm{d}y = (\star \mathm{d}y)_{1} \mathrm{d}x + (\star \mathm{d}y)_{2} \mathrm{d}y= -\mathrm{d}x[/tex].

    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.



    -artur palha
  4. Dec 6, 2008 #3
    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}

    s= number of (-1) in the metric [tex]g_{ij}[/tex]
    [tex]\epsilon^{i_1,...,i_p,j_1,...,j_{n-p}}[/tex] = Levi-Civita symbol
  5. Apr 3, 2009 #4

  6. Oct 16, 2009 #5

    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
    Bo-Yuan Hou

    -artur palha
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