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Various problems concerning Exterior calculus

  1. Apr 8, 2006 #1
    Im currently learning some Exterior calculus which Im hoping will help me with my honours project.

    The problem I am stuck at is the following.

    [tex]\mbox{Show that } i_X\ast \omega = \ast(\omega \wedge X^{\flat})[/tex]

    where [itex]X^{\flat}[/itex] is the one-form related to the vector field [itex]X[/itex] by the metric, and [itex]\omega[/itex] is some p-form. Also, [itex]i_X[/itex] is meant to be the interior derivative. Note: this is not a homework problem, but intended as a discussion thread on the concepts concerned with problems of this type. In effect, such a thread could help me understand more about what is going on so that I may be able to show what I have written.


    But the main reason for posting here is that I would like to understand how I can use Hodge star operators, exterior derivatives, and musical isomorphisms to define an operation which is identical to all the classical vector calculus operations, and in particular curl. I believe it helps me more to discuss with other people.

    Any discussion on any of the material that I have mentioned would be greatly recieved.

    Cheers.
     
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2
    I suppose I should start with the following question:

    If I have a vector field X, then what am I doing to it by 'flatting' it, [itex]X^{\flat}[/itex]?
     
  4. Apr 9, 2006 #3
    I just read that for a semi-Riemannian manifold [itex]M[/itex], there is a canonical [itex]\mathcal{F}(M)[/itex]-linear isomorphism between tensor fields of the type [itex](r,s)[/itex] and of the type [itex](r-1,s+1)[/itex],

    [tex]T^{(r,s)} \cong T^{(r-1,s+1)}.[/tex].

    If we take some tensor field [itex]A \in T^{(r,s)}(M)[/itex] then the value of [itex]A^{\flat} \in T^{(r-1,s+1)}(M)[/itex] on arbitrary one-forms and vector fields is define by

    [tex]A^{\flat}(\theta^1,\dots,\theta^{r-1},X_1,\dots,X_{s+1}) = A(\theta^1,\dots,X^*_b,\dots,\theta^{r-1},X_1,\dots,X_{b-1},X_{b+1},\dots,X_{s+1})[/tex]

    where [itex]X^*_b[/itex] is the one-form metrically equivalent to [itex]X_b[/itex].

    So if I have a tensor, say

    [tex]R(X,Y,\omega)[/tex]

    which acts on, say, 2 vector fields and 1 one-form such as in this case. I could also say

    [tex]R(X,Y,Z^{\flat})[/tex]

    since I cannot insert a vector field into the third slot because the third slot is reserved for forms. So I can simply flat a vector field and insert it.

    I think this is also called type-changing.
     
  5. Apr 9, 2006 #4
    If you have a one-form [tex]\omega[/tex] then the components of the exterior derivative acting on it are

    [tex](d\omega)_{ab} = \partial_{[a}\omega_{b]}[/tex]

    So for example, the 12 component is [tex]\partial_1\omega_2 - \partial_2\omega_1[/tex]. This looks uncannily similar to the curl. If we then use the Hodge star operator, assuming that we have a Euclidean metric, then we send the 12 component to the 3 component etc, and so we can write the curl as [tex]*d\omega[/tex].

    Similarly we can write the divergence of a one-form as [tex]*d*\omega[/tex], and we can write the gradient of a scalar field [tex]\phi[/tex] (a function) as [tex]d\phi[/tex].

    When we're using a Euclidean metric we don't have to differentiate between forms and vectors, because the metric is [tex]g_{ab}=\delta_{ab}[/tex]. So anything that works for forms also works for vectors. This is why you'll often see all indices written as 'down' indices when you're working in Euclidean space.

    This is only a physicist's view, not a mathematician's, so I've probably missed some detail ;)
     
  6. Apr 9, 2006 #5
    What you wrote is interesting actually.

    You said that the gradient of a scalar field can be written as

    [tex]\mbox{grad}\phi = d\phi[/tex]

    but the exterior derivative d can only act on forms can they not? So we have a useful operator on forms which corresponds to the grad operator in vector calculus. So Im guessing we can also define a similar operation on vectors by simply sharping the whole thing:

    [tex]\mbox{grad}V = (d\phi)^{\sharp}[/tex]

    And you also wrote that the divergence of a vector field V is

    [tex]\mbox{div} V = \ast d \ast \omega^{\flat}[/tex]

    where [itex]\omega[/itex] is a one-form. Note: the one form in this definition is flatted because we want the divergence of a vector field. (Is this ok?) In three dimensions, the index of the one-form is 1. Then we star it and get 3-1 = 2 index. Then ext. diff. and we get (3-1)+1 = 3. Then star it again and get and index of 3-((3-1)+1) = 0. This tells us that the divergence of a one-form gives a zero-form, or a scalar.

    So with the definition of curl, we begin with a one-form. Ext. diff. it and get 1+1 =2. And then star it and get 3 - (1+1) = 1. This tells us that the curl operator on a one-form returns a one-form. Which is all good. :)
     
    Last edited: Apr 9, 2006
  7. Apr 9, 2006 #6

    George Jones

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    Yikes, I can't believe I'm doing this. I must truly be addicted to physics forums. I probably won't be able to make any more replies for at least another 48 hours.

    Let [itex]X[/itex] be a vector field. Use the metric [itex]g[/itex] to define the covector field (1-form) [itex]X^{\flat}[/itex] associated with [itex]X[/itex]: for every vector field [itex]Y[/itex]

    [tex]X^{\flat} \left( Y \right) := g \left( X , Y \right).[/tex]

    This is the abstract version of index lowering for physicists. To see this, let [itex]\left\{ e_{1}, \dots , e_{n} \right\}[/itex] be a set of basis vector fields, and let [itex]\left\{ \omega^{1}, \dots , \omega^{n} \right\}[/itex] be the associated dual basis of 1-forms. Write [itex]g_{ij} = g \left( e_{i} , e_{j} \right)[/itex].

    Write [itex]X[/itex] in terms of the basis vector fields,

    [tex]X = X^{i} e_{i},[/tex]

    and [itex]X^{\flat}[/itex] in terms of the basis 1-forms,

    [tex]X^{\flat} = X_{i} \omega^{i},[/tex].

    Then,

    [tex]X^{\flat} \left( e_{j} \right) = X_{i} \omega^{i} \left( e_{j} \right) = X_{i} \delta_{j}^{i} = X_{j}.[/tex]

    But, by definition,

    [tex]X^{\flat} \left( e_{j} \right) = g \left( X , e_{j} \right) = g \left( X^{i} e_{i} , e_{j} \right) = X^{i} g \left( e_{i} , e_{j} \right) = X^{i} g_{ij}.[/tex]

    Combining these results gives

    [tex]X_{j} = X^{i} g_{ij}.[/tex]

    This is the reason for the flat notation. Just as flatting a musical note lowers it by a semitone, flatting here lowers indices

    Regards,
    George
     
    Last edited: Apr 9, 2006
  8. Apr 10, 2006 #7
    Ok, so I must interpret the RHS of my very first equation in post #1 as the wedge product of a 1-form and a 1-form associated with X. So its a wedge product of two 1-forms?
     
  9. Apr 10, 2006 #8

    garrett

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    Hi Oxy, I'm pretty sure you're going to have to use the definition of the Hodge dual in terms of components and the permutation symbol in order to prove your equation (1). It's a bit of a mess.
     
  10. Apr 11, 2006 #9
    I'm not quite sure about that. Due to multilinearity, we can assme that omega is an elementary p-form, i.e. [itex]\omega = r{e_1}^*\wedge\dots\wedge{e_p}^*[/itex], where {e1,...,en} is an ordered, orthnormal basis of V and [itex]\{{e_1}^*,\dots,{e_p}^*\}[/itex] is the dual basis. r is some real number.

    Then there are two cases either [itex]\omega(X)=0[/itex] or not. Either case allows us to assume that X is in [itex]\{{e_1},\dots,{e_p}\}[/itex]
    or in [itex]\{{e_{p+1}},\dots,{e_n}\}[/itex].

    Note [itex]*\omega = r^{-1}{e_{p+1}}^*\wedge\dots\wedge{e_n}^*[/itex], and similarly for others...

    This should cut down the messiness considerably.
     
  11. Apr 11, 2006 #10
    This is my understanding of the Hodge star operator, it may be flawed:

    Let [itex]\Omega^p(M)[/itex] denote the space of p-forms on a manifold M. We can turn this space into a graded algebra by incorporating the wedge operation (or exterior product):

    [tex]\wedge\,:\,\Omega^p(M) \times \Omega^q(M) \rightarrow \Omega^{p+q}(M)[/tex]

    which is an operation that takes a p-form [itex]\omega[/itex]and a q-form [itex]\phi[/itex] and gives a p+q form [itex]\omega \wedge \phi[/itex]. We also have

    [tex]\omega \wedge \phi = (-1)^{pq}\phi\wedge\omega[/tex]

    The interior derivative encompasses the process of contraction and anti-symmetrizing. That is, take a p-form (which is a map on a set of p vector fields to give a function), then contract on one vector field, and antisymmetrize.

    If X is a vector field, then the interior derivative with respect to X is

    [tex]i_X \,:\,\Omega^p \rightarrow \Omega^{p-1}[/tex]

    where p-forms [itex]\omega[/itex] get mapped to [itex]i_X\omega[/itex].

    The interior derivative of a wedge product of p and q forms is

    [tex]i_X(\omega\wedge\phi) = i_X\omega \wedge\phi + (-1)^p\omega\wedge i_X \phi[/tex]

    and we also have the following property

    [tex]i_X\circ i_X = 0[/tex]


    The vector space [itex]\Omega^p(M)[/itex] is isomorphic to [itex]\Omega^{n-p}[/itex] with no natural isomorphism. By establishing a metric on the manifold we unearth a preferred isomorphism called the Hodge dual operator.

    So take a manifold M which has a metric g on it. Firstly, we have

    [tex]g = g_{ab}\theta^a\otimes \theta^b[/tex]

    which means that a covariant metric tensor gives us a contravariant metric tensor. So we also have

    [tex]g_{(1)}=g^{ab}X_a\otimes X_b[/tex]

    where [itex]g^{ab}g_{ab} = \delta^a_c[/itex], and [itex]g_{(1)}[/itex] is a metric on 1-forms. We can extend this to a metric on p-forms, [itex]g_{(p)}[/itex]:

    [tex]g_{(p)}(\alpha_1\wedge\alpha_2\wedge\dots\wedge\alpha_p,\beta_1\wedge\beta_2\wedge\dots\wedge\beta_p) = \det[g_{(1)}(\alpha_i,\beta_i)][/tex]

    that is, it is the determinant of the matrix of scalar products of the 1-form factors.

    The Hodge star is an operator which takes p-forms to (n-p)-forms, or we could say

    [tex]\omega \mapsto \ast\omega[/tex]

    Notice that [itex]\ast\omega[/itex] is a unique form such that if we act on it my some other p-form, [itex]\phi \in \Omega^p(M)[/itex] we get

    [tex]\phi\wedge\ast\omega = g_{(p)}(\phi,\omega)\Omega(M)[/tex]
     
    Last edited: Apr 11, 2006
  12. Apr 11, 2006 #11
    So if I have a p-form [itex]\omega[/itex] which can be written as

    [tex]\omega = re^*_1\wedge \dots \wedge e^*_p[/tex]

    where r is a real number, and e* are dual to the orthonormal basis e.
    Hodge starring w gives an n-p vector where n is the dimension of the manifold.

    So an arbitrary element of the manifold can be written in terms of n basis elements with metric. Then *w has n-p elements:

    [tex]\ast\omega = re^*_{p+1} \wedge \dots \wedge e^*_n[/tex]

    which is exactly what DB wrote. Now the interior derivative takes a p-form and gives a p-1 form, so

    [tex]i_X\ast\omega = re^*_{p+1}\wedge\dots\wedge e^*_{n-1}[/tex]


    So if I can show that this equals the RHS of (1) then I'd have shown it?
     
    Last edited: Apr 11, 2006
  13. Apr 11, 2006 #12
    The above is assuming that [itex]X={e_n}[/itex], i.e. [itex]\iota_X \omega=0[/itex]. There is a seperate case if [itex]\iota_X\omega\neq0[/itex]. In this case, you can assume that [itex]X={e_1}[/itex].

    Otherwise you seem to have the gist of it.

    N.B. I originally had r^{-1} as the coefficient for *w, but that's not right. Plus I forgot that w is not necessarily a 1-form, so I've changed [itex]\omega(X)[/itex] to [itex]\iota_X \omega[/itex]
     
    Last edited: Apr 11, 2006
  14. Apr 11, 2006 #13
    What I was trying to get across is that when we're working with the Euclidean metric, it doesn't matter whether we're talking about forms or vectors. Because the Euclidean metric satisfies [tex]g_{ab}=1[/tex] for [tex]a=b[/tex] and [tex]g_{ab}=0[/tex] otherwise, the i component of a flattened vector (or vector field) has the same value as the i component as the vector itself - in physics speak, it doesn't matter whether the index is up or down.

    Something else that you might like to ponder is what the operator [tex]d*d[/tex] acting on a scalar field corresponds to, in the cases where we have a Euclidean metric and a Lorentzian metric (one negative component in diagonal form, and the others positive).
     
  15. Apr 21, 2006 #14
    So I have on the RHS

    [tex]\ast(\omega\wedge X^{\flat})[/tex]

    and I have to show that this equals

    [tex]re^*_{p+1} \wedge \dots \wedge e^*_{n-1}[/tex]

    in terms of components.

    So [itex]\omega = re^*_1\wedge \dots \wedge e^*_p[/itex]

    Then [itex](\omega\wedge X^{\flat}) = re^*_1 \wedge \dots \wedge e^*_{p+1}[/itex]

    Then [itex]\ast(\omega\wedge X^{\flat}) = re^*_{p+2} \wedge \dots \wedge e^*_n[/itex]

    Can I simply subtract 1 from the indices and claim this equals

    [tex]re^*_{p+1} \wedge \dots \wedge e^*_{n-1}[/tex]?
     
  16. Apr 23, 2006 #15
    Certainly not. The problem is that you're confusing what your assumption about X is. On the RHS, you have X=e_n, but in the work above you are assuming that X=e_{p+1}.
     
  17. Apr 23, 2006 #16
    You're right. It should be

    [tex](\omega\wedge X^{\flat}) = e_1 \wedge \dots \wedge e_p \wedge e_1 \wedge \dots \wedge e_{p-1}[/tex]

    Actually I dont like this either. I missing bits. Should it be...

    [tex](\omega\wedge X^{\flat}) = \omega_{i_1\dots i_p}X^{_{i_1\dots i_p}}\mbox{d}x^{i_1}\wedge\dots\wedge\mbox{d}x^{i_p}\wedge\mbox{d}x^{i_1}\wedge\dots\wedge\mbox{d}x^{i_{p-1}][/tex]

    But how on Earth am i going to get that to be a rank (n-p-1) form? Which is the case on the LHS.
     
    Last edited: Apr 23, 2006
  18. Apr 24, 2006 #17
    no that too is incorrect. X is a vector, so X-flat will be a 1-form. I think you have forgotten the whole strategy here: first assume we can find a basis e={e1 ... en} such that X is in e and w is a simple p-form with respect to e.
    The general case will follow from multi-linearity of both sides of the equation.
     
  19. Apr 26, 2006 #18
    [tex]\omega = e_1\wedge \dots \wedge e_p[/tex]

    [tex]X^{\flat} = e_n[/tex]

    [tex](\omega\wedge X^{\flat}) = e^1\wedge\dots\wedge e^p\wedge e^n[/tex]
     
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