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Symmetric Part of a Mixed (1,1) Tensor

  1. Jun 20, 2011 #1
    I have read in a couple of places that mixed tensors cannot be decomposed into a sum of symmetric and antisymmetric parts. This doesn't make any sense to me because I thought a mixed (1,1) tensor was basically equivalent to a standard linear transform from basic linear algebra. I am also having trouble with finding anything that shows how changing the order of the indexes effects a mixed tensor and how basic coordinate transforms affect mixed tensors with permuted indexes. The background is basic relativity so I usually use lorentz transforms and minkowski metric for examples. Here is my idea of the symmetric part of a mixed tensor M.

    [itex] \frac{1}{2}\left[ {M^\alpha}_\beta + {N^\alpha}_\beta \right] [/itex]

    where (i think)

    [itex] {N^\alpha}_\beta = g_{\nu\beta}{M^\nu}_\mu g^{\mu\alpha} = {M_\beta}^\alpha[/itex]

    I am having trouble seeing how to transform it to a new coordinate system because every book I have (and I have about 10) shows the coordinate transform matrices and mixed tensors with upper and lower indices directly above/below each other, so I am not sure how they apply.
    Last edited: Jun 20, 2011
  2. jcsd
  3. Jun 20, 2011 #2
    When you have a metric like the [itex]g_{\mu\nu}[/itex] here you can identify the upper and lower indices with one another by using the metric. Essentially what you always do when performing calculations in this index formalism is to "lower" or "raise" indices with the help of the metric. So in this case it make perfect sense to talk about a symmetric (1,1) tensor because you can identify it with a (0,2) tensor by using the metric and this (0,2) tensor you can use symmetrisation or anti-symmetrisation upon. So if one has a metric one usually writes indices of a symmetric (1,1) tensor directly above one another because the ordering makes no difference. Of course you cannot do this for non-symmetric tensors, e.g. assume [itex]T[/itex] non-symmetric:
    [tex] T_{\nu}^{\ \alpha}=g^{\alpha\mu}T_{\nu\mu}\neq g^{\alpha\mu}T_{\mu\nu}=T^{\alpha}_{\ \nu} [/tex]
    It is true that in the case where you do not have a metric (e.g. affine differential geometry) the "symmetrisation" of a mixed (1,1) tensor can not be properly defined. Because in this case the tensor can only be viewed as a linear map from the tangent to the cotangent space (or vice versa, depending on the ordering of indices) and not as a bilinear map from the tangent space to the reals. Without a metric you cannot canonically identify the tangent and the cotangent space.
  4. Jun 20, 2011 #3
    OK, that makes sense to me, thanks.
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