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Differential forms

  1. Apr 11, 2013 #1
    Hello! I think I got something wrong here, maybe someone can help me out.

    Lets consider a n-manifold. A differential n-form describing a signed volume element will then transform as:
    [tex]f(x^i) dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n = f(y^i) \;\text{det}\left( \frac{\partial x^i}{\partial y^j}\right) dy^1 \wedge dy^2 \wedge \cdots \wedge dy^n,[/tex]
    which we can compare to the unsigned volume element in Euclidean space that transforms as:
    [tex]f(x^i) dx^1 dx^2 \cdots dx^n = f(y^i) \left|\text{det}\left( \frac{\partial x^i}{\partial y^j}\right)\right|dy^1 dy^2 \cdots dy^n.[/tex]

    Clearly we can get a sign wrong when integrating and conisdering coordinate changes from systems of different orientation. What I dont see is the need for the factor of
    on Riemannian manifolds. Will this factor only fix the sign error or have I missunderstood something basic?
  2. jcsd
  3. Apr 11, 2013 #2


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    Let me try to sum up the situation so you can sort out what you understood and what you did not.

    On a n-manifold, we can meaningfully integrate n-forms because of the way they transform under change of coordinates. We do this using local charts and a partition of unity. Moreover, you need to choose an orientation on M and use only charts coherently oriented (det > 0).

    One way to meaning fully integrate functions on a manifold then is to fix once and for all a volume form (nowhere 0 n-form) dV (generally not an exact form, despite the notation), and integrate a function f:M-->R "against" dV; i.e. integrate the n-form fdV.

    The existence of a volume form is equivalent to orientability: ==>) If a volume form exist, define an orientation by saying that a basis v1,..,vn is "positively oriented" iff dV(v1,...,vn)>0.
    <==) Conversely, if an orientation has been chosen, choose a covering of M by open sets Ui on which there exists a (smooth) local basis of TM v1,...,vn. Then define a local nowhere vanishing n-form on Ui by setting Ωi(v1,..,vn)=1 (Equivalently, [itex]\Omega_i=v_1^*\wedge\ldots\wedge v_n^*[/itex] when {vi*} is the dual basis of {vi}). Then patch up the local n-forms Ωi's to a global volume form dV using a partition of unity.

    Now, on an oriented riemannian manifold (M,g), there is a natural choice of the form dVg: if v1,...,vn is a (smooth) local positively oriented orthonormal basis of TM defined on U, then define dVg over U by dVg(v1,...,vn)=1. Equivalently, [itex]dV_g=v_1^*\wedge\ldots\wedge v_n^*[/itex].) This in fact defines dVg globally. This is because if w1,..,wn is another (smooth) local positively oriented orthonormal basis of TM defined on U, then the transition matrix between the two basis is in SO(n), and so has determinant 1!

    Note that in general, you cannot take the v1,..,vn to be coordinate vector fields. If you can, then the metric is called flat. But you can write the riemannian volume form dVg in terms of a coordinate basis, and you will find (good exercice) that you get [itex]dV_g=\sqrt{|g|}dx^1\wedge\ldots\wedge dx^n[/itex] where [itex]\sqrt{|g|}[/itex] is the crazy physicist notation for [itex]\sqrt{\det(g_{ij})|}[/itex].
    Last edited: Apr 11, 2013
  4. Apr 12, 2013 #3
    Thanks! I can see the logic of what you are saying but it feels like the different explanations I've encountered about this are contradictive. If I understand you right is a matter of finding a "global convention" (i.e. our Ω) for measuring volumes.

    I will try to melt this and come back in some days. :)
    Thanks for your time quasar.
  5. Apr 12, 2013 #4
    First of all, is my transformation:
    [tex]f(x^i) dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n = f(y^i) \;\text{det}\left( \frac{\partial x^i}{\partial y^j}\right) dy^1 \wedge dy^2 \wedge \cdots \wedge dy^n,[/tex]
    correct? Carroll refers to this object as a tensor density and says that it doesnt transform as a n-form.. :/
    If this is not an n-form I guess dont know what an n-form is. After all the subspace of totally antisymmetric n-tensors is 1 dimensional so what else can t look like?
    Last edited: Apr 12, 2013
  6. Apr 12, 2013 #5


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    The transformation is correct. I don't know what Carroll means by that. The LHS in what you wrote is certainly an n-form defined on the open set where the coordinate x are defined, and if you write it in terms of the y coordinates, then you get the RHS on the region where both the x and y coordinates are defined.
  7. Apr 13, 2013 #6
    Thanks! Maybe I should consider leaving Carroll's text. It has only confused me so far.
    Does anybody know a good introduction text to curvature forms? :/
  8. Apr 13, 2013 #7


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  9. Apr 20, 2013 #8
    I think Carroll says it is not a tensor precisely because it obeys the transformation law you've written. In many conventions, tensors are defined as those objects whose components obey a certain transformation law, while those objects whose components obey the transformation law you've written - where there's a multiplication by a determinant, are just, by definition, called tensor densities or relative tensors
  10. Apr 20, 2013 #9


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    See the section starting on page 375 of Lee - Smooth Manifolds. He gives an excellent account of densities i.e. what they are, what properties they hold, and why they are used (e.g. integration). Also see appendix B of Wald's General Relativity where he talks about them using a continuous, nowhere vanishing n-form field ##\epsilon_{a_1...a_n} = \epsilon_{[a_1...a_n]}##.
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