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I Differential forms

  1. Oct 21, 2016 #1
    Why does the definition of a differential form requires a totally antisymmetric tensor?
     
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  3. Oct 21, 2016 #2

    Orodruin

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    Because that is how we define a differential form. It is a definition that turns out to be useful in describing several things, but ultimately it is just that - a definition.
     
  4. Oct 21, 2016 #3

    fresh_42

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    It doesn't require tensors.
    It requires multi-linearity (linear in each argument) and alternating. The latter is equivalent to ##d(X,X)=0##.

    Tensors are useful, when you try to make sure, that the definition of a multi-linear object actually defines something, i.e. something that exists, rather than being a funny definition, that describes the empty set.
    The reason for this is algebraic and somehow abstract, so I leave it here in the relativity section. For short: tensors are a kind of a universal multi-linear object, and special multi-linear objects like differential forms can be seen as the image under a certain projection. Even shorter: Tensors are a skeleton key.
     
  5. Oct 21, 2016 #4

    robphy

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  6. Oct 22, 2016 #5

    vanhees71

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    Well tensors are defined as multilinear forms (not necessarily antisymmetric ones as the alternating forms in Cartan calculus), i.e., a tensor of rank ##n## is a linear map from ##V^n## into the field of scalars of ##V## (field meant here in the algebraic sense).
     
  7. Oct 23, 2016 #6
    Thanks for answering. Why integration and differentiation in general is chosen to be performed with antisymmetric(0,n) tensors ? Why not on other type of tensors?
     
  8. Oct 23, 2016 #7

    Orodruin

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    Because a directed volume element spanned by n vectors is naturally described by an antisymmetric tensor of rank n.
     
  9. Oct 23, 2016 #8
    Could you give me a simple (maybe a 3-dimensional) example?
     
  10. Oct 23, 2016 #9

    vanhees71

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    Take a surface ##\vec{x}=\vec{x}(u,v)## in ##\mathbb{R}^3##. Then the surface-normal vector is determined by the parallelogram defined by the infinitesimal pieces of the coordinate lines ##v=\text{const}## and ##u=\text{const}##. The magnitude and direction is given by the cross product, i.e., you have
    $$\mathrm{d}^2 \vec{a}=\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v}.$$
    Note that the orientation, i.e., the direction of the normal vector depends in which order you choose the indepenent variables ##u## and ##v## (i.e., you have defined an oriented surface).

    In an arbitrary (pseudo-)Riemannian manifold hypersurface elements are defined in the analogous general covariant way with help of the Levi-Civita tensor,
    $$\epsilon^{\mu_1 \mu_2 \ldots \mu_d}=\sqrt{|g|} \Delta^{\mu_1 \ldots \mu_d},$$
    where ##\Delta^{\mu_1 \ldots \mu_d}## is the totally antisymmetric Levi-Civita symbol with ##\Delta^{12\ldots d}=1## and ##g=\mathrm{det} (g_{\mu \nu})## the determinant of the (pseudo-)metric components. A ##k##-dimensional (##k \leq d##) hyper-surface element is defined by some function ##q^{\mu}(u_1,\ldots,u_k)## and the corresponding integral measure by
    $$\mathrm{d}^k \Sigma^{\mu_{k+1}\ldots \mu_d}=\mathrm{d} q_1 \cdots \mathrm{d} q_d \epsilon_{\mu_1\ldots \mu_d} \frac{\partial q^{\mu_1}}{\partial u_1} \cdots \frac{\partial q^{\mu_k}}{\partial u_k}.$$
     
  11. Oct 23, 2016 #10

    robphy

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    Here's a 2-D example.... a parallelogram formed from two vectors.
    If you specify an ordered-pair with those two vectors, then you have specified an oriented parallelogram.
    The size of the parallelogram can be expressed using the determinant formed using the components of the vectors [although it could be expressed in a coordinate-free way].
     
  12. Oct 23, 2016 #11
    In your first example in R³, $$\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v}= -\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial v} \times \frac{\partial \vec{x}}{\partial u}.$$ Right? So is it a antisymmetric tensor?

    Where does the antisymmetry property comes in this situation?

    (I'm sorry for my bad english)
     
  13. Oct 23, 2016 #12

    robphy

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    If you switch the order of two vectors, it is multiplied by a minus sign.... and this shows up in the determinant.
     
  14. Oct 23, 2016 #13

    Orodruin

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    Yes, it is a contraction between the two tangent vectors ##\partial\vec x/\partial u## and ##\partial\vec x/\partial v## with the anti-symmetric tensor ##\epsilon_{ijk}##.
     
  15. Oct 23, 2016 #14
  16. Oct 24, 2016 #15

    stevendaryl

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    The simplest case to understand is integration over a curved 2-D manifold with a metric. And let me assume a Riemannian metric (so we can assume that the square of any vector is defined and positive). The case for an indefinite metric (so the square of a vector can be negative or zero) and for higher-dimensional manifolds will be left as an exercise :wink:.

    Using the usual definition of an integral, if we are integrating a scalar function [itex]F[/itex] over the 2-D space, we can approximate it by dividing space up into lots of little parallelograms, and compute

    [itex]\int F = \sum_i \mathcal{A}_i F(\mathcal{P}_i)[/itex]

    integration.jpg

    where [itex]\mathcal{A}_i[/itex] is the area of parallelogram number [itex]i[/itex], and [itex]F(\mathcal{P}_i)[/itex] is the value of [itex]F[/itex] at the point [itex]\mathcal{P}_i[/itex] somewhere inside the parallelogram. So integrating a scalar function over a 2-D manifold is reduced to the problem of computing the area of a tiny parallelogram.

    The rule for the area of a parallelogram in good old Euclidean geometry is just:

    [itex]\mathcal{A} = base \times height[/itex]

    So if [itex]A[/itex] is the displacement vector (which is approximately well-defined for small displacements, even in curved space) for one side of a small parallelogram, and [itex]B[/itex] is the displacement vector for the other side, then the area will be:
    parallelogram.png
    [itex]\mathcal{A} = |A| |B^\perp|[/itex]

    where [itex]B^\perp[/itex] is the component of [itex]B[/itex] perpendicular to [itex]A[/itex]. (And where [itex]|X|[/itex] means the square-root of [itex]X \cdot X[/itex], which is real and positive for Riemannian manifolds, which we're assuming.)

    We can compute [itex]|B^\perp|[/itex] in terms of the vector dot-product. [itex]|B^\perp| = \sqrt{B \cdot B - \frac{(A \cdot B)^2}{A \cdot A}}[/itex]. So we have:

    [itex]\mathcal{A} = |A| \sqrt{B \cdot B - \frac{(A \cdot B)^2}{A \cdot A}} = \sqrt{(A \cdot A) (B \cdot B) - (A \cdot B)^2}[/itex]

    Or squaring both sides gives:
    [itex]\mathcal{A}^2 = (A \cdot A) (B \cdot B) - (A \cdot B)^2[/itex]

    At this point, let me introduce the alternative way of writing vector dot products. The dot product [itex]X \cdot Y[/itex] can be written in terms of the metric tensor [itex]g_{\mu \nu}[/itex] as [itex]X \cdot Y = \sum_{\mu \nu} g_{\mu \nu} X^\mu Y^\nu \equiv X^\mu Y_\mu[/itex], where [itex]Y_\mu \equiv \sum_{\nu} g_{\mu \nu} Y^\nu[/itex]. So in this notation, we can write:

    [itex]\mathcal{A}^2 = A^\mu A_\mu B^\alpha B_\alpha - A^\mu B_\mu A^\alpha B_\alpha[/itex]

    Now, introduce the two-index tensor [itex]\mathcal{A}^{\mu \alpha} \equiv A^\mu B^\alpha - A^\alpha B^\mu[/itex]. We can prove the following identity:

    [itex]\mathcal{A}^{\mu \alpha} \mathcal{A}_{\mu \alpha} = (A^\mu B^\alpha - A^\alpha B^\mu)(A_\mu B_\alpha - A_\alpha B_\mu) = A^\mu A_\mu B^\alpha B_\alpha - A^\mu B_\mu A^\alpha B_\alpha - A^\alpha B_\alpha A^\mu B_\mu + A^\alpha A_\alpha B^\mu B_\mu = (A \cdot A) (B \cdot B) - (A \cdot B) (A \cdot B) - (A \cdot B) (A \cdot B) + (A \cdot A) (B \cdot B) = 2 ((A \cdot A) (B \cdot B) - (A \cdot B) (A \cdot B)) = 2 \mathcal{A}^2[/itex]
    So we have:
    [itex]\mathcal{A}^2 = \frac{1}{2}\mathcal{A}^{\mu \alpha} \mathcal{A}_{\mu \alpha}[/itex]. (The factor of 1/2 is a little worrisome, but I think it's because the implicit summing over [itex]\mu, \nu, \alpha, \beta[/itex] double-counts; the sum includes the non-summed term [itex]A^\mu A_\mu B^\alpha B_\alpha[/itex] and [itex]A^\alpha A_\alpha B^\mu B_\mu[/itex] for each [itex]\alpha [/itex] and [itex]\mu[/itex].)

    So the area tensor for the parallelogram of vectors [itex]A[/itex] and [itex]B[/itex] is antisymmetric.
     
    Last edited: Oct 24, 2016
  17. Oct 24, 2016 #16
    Thank you for your detailed explanation. So the conclusion is that [itex]\mathcal{A}_{\mu \alpha}[/itex] must be antisymmetric, since A² is (in this case) always positive, so that if we change μ and α, we have ([itex]\mathcal{A}^{\alpha \mu}[/itex])([itex]\mathcal{A}_{\alpha \mu})=[/itex] [itex](\mathcal{-A}^{\mu \alpha})[/itex][itex](\mathcal{-A}_{\mu \alpha})[/itex] in order to get the same value for A². Am I right?
     
    Last edited: Oct 24, 2016
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