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Product of a symmetric and antisymmetric tensor

  1. Aug 11, 2015 #1
    It seems there should be a list of tensor identities on the internet that answers the following, but I can't find one.

    For tensors in ##R^4##,

    ##S = S_\mu{}^\nu = S_{(\mu}{}^{\nu)}## is a symmetric tensor.
    ##A = A_{\nu\rho\sigma}= A_{[\nu\rho\sigma]}## is an antisymmetric tensor in all indices.

    The basis vectors are suppressed for simplicity.

    Parentheses ##(\ \ )## enclose symmetric indices; the tensor remains the same upon an exchange of any two indices.
    Brackets ##[\ \ ]## enclose antisymmetric indices; the tensor changes sign on an exchange of any two indices.

    Is the inner product ## SA=S_\mu{}^\nu A_{\nu\rho\sigma}## antisymmetric in all indices?

    (In a related aside, ##A_{\nu\rho\sigma}## can be expressed as a scalar times a "unit valued" antiysmmetric tensor. Call it ##\epsilon^3{}_{\mu\nu\rho}## in ##R^4##, or the 3 index Levi-Civita tensor in 4 dimensions. But I don't see any internet reference to this animal either. Are my keywords lacking?)
     
  2. jcsd
  3. Aug 11, 2015 #2

    jedishrfu

    Staff: Mentor

    While I don't have the answer to your question, I did find this article on Penrose Graphical Diagrams and Tensors that seems to use some identities:

    https://en.wikipedia.org/wiki/Penrose_graphical_notation

    Probably not too helpful in this context though. Save it for future reference. :-)

    Perhaps @HallsofIvy knows of some resource for Tensor identities.
     
  4. Aug 12, 2015 #3

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    I don't think the contraction will have any specific symmetry between the index ##\mu## and the other indices.

    Suppose you were just talking about ##n \times n## matrices. If you multiplied a symmetric matrix S with an antisymmetric matrix A, would you expect the result to have any sort of symmetry?
     
  5. Aug 12, 2015 #4
    Well...I am hoping, not expecting.

    As it turns out, ##S=\delta##, or ##{S_\mu}^\nu = \delta_\mu^\nu## obtains an antisymmetric tensor for the ##SA## product.

    In general, in the orthonormal coordinate system where S is diagonal, ##S = ({S_0}^0,\ {S_1}^1,\ {S_2}^2,\ {S_3}^3)##. The ##SA## product has 4 corresponding, independent values.

    I had thought this was sufficient to supply the exception to disprove the hypothesis, but on second glance, no.
     
    Last edited: Aug 12, 2015
  6. Aug 12, 2015 #5
    The result has multiple interesting antisymmetric properties but not, in general, is the product antisymmetric.

    In orthonormal coordinates the tensor ##\epsilon_{\mu\nu\rho}## is equal to it's symbol. Under a change of coordinates, it remains antisymmetric. Symmetric tensors likewise remain symmetric.

    We can construct a product space ##\hat{q}_\delta = \hat{\theta}_\alpha \wedge \hat{\theta}_\beta \wedge \hat{\theta}_\gamma##. Bases vectors are identified with hats. ##\hat{\theta}_\alpha## are the contravariant bases of the orthonormal 4-space. ##\hat{\theta}_\alpha## are the covariant basis of the orthonormal 4-space.

    There are for independent bases in ##q##. One choice of basis is:
    ##\hat{q}_0 = \hat{\theta}_1 \wedge \hat{\theta}_2 \wedge \hat{\theta}_3##
    ##\hat{q}_1 = \hat{\theta}_2 \wedge \hat{\theta}_3 \wedge \hat{\theta}_0##
    ##\hat{q}_2 = \hat{\theta}_3 \wedge \hat{\theta}_0 \wedge \hat{\theta}_1##
    ##\hat{q}_3 = \hat{\theta}_0 \wedge \hat{\theta}_1 \wedge \hat{\theta}_2##​

    In one choice of orthonormal coordinates, S is a diagonal matrix. Define the elements as,
    ##a = {S_0}^0##
    ##b = {S_1}^1##
    ##c = {S_2}^2##
    ##d = {S_3}^3##.​

    ##A= k\epsilon##. All 3 index antisymmetric tensors are ##k## times the corresponding 3 index Levi-Civita in ##R^4##.

    Putting it all together, with more abuse of notation (leaving out the basis vectors on the left side),

    ##{S_0}^0 A_{0\mu\nu} = ak(\hat{q}_0 - \hat{q}_1 + \hat{q}_2)##
    ##{S_1}^1 A_{1\mu\nu} = bk(\hat{q}_1 - \hat{q}_2 + \hat{q}_3)##
    ##{S_2}^2 A_{2\mu\nu} = ck(\hat{q}_2 - \hat{q}_3 + \hat{q}_0)##
    ##{S_3}^3 A_{3\mu\nu} = dk(\hat{q}_3 - \hat{q}_0 + \hat{q}_1)##.

    ##{S_\mu}^\nu A_{\nu\rho\sigma}## is a linear combination of 3 out of 4 of 3-index Levi-Civita tensors of ##R^4## in orthonormal coordinates.

    How this variegates under a general coordinate transformation, I don't know yet. I could be more literate in tensor densities.


     
    Last edited: Aug 12, 2015
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