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Rank of tensor

  1. Sep 22, 2005 #1
    q1 What is rank of a tensor?
    q2 I dont know why after contraction operation (or trace of tensor) the rank of a tensor will be reduced by 2?

    q3 I can't imagiant how the fourth rank tensor, e^iklm looks like?
    q4 What does an anti-symmetric tensor e^iklm means? Is it a 4 by 4 martix or a vector?
    q5 How to find that: e^iklm e_iklm = -24?

    yuk yuk
  2. jcsd
  3. Sep 22, 2005 #2

    Physics Monkey

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    The rank of a tensor is just the total number of (free) indices that it has.

    [tex] m [/tex], scalar, rank 0

    [tex] v^\mu [/tex], vector, rank 1

    [tex] \sigma_\mu [/tex], 1 form, rank 1

    [tex] T^{\mu \nu} [/tex], tensor, rank 2 (note that while we explicitly call this thing a tensor, all of the other objects listed above are also tensors, they just have other names too)

    Alternatively, one can provide a little more information and give the number of covariant and contravariant indices.

    This is a definition of the rank of a tensor in terms of its components. We can also define the rank of a tensor more 'frame independently', if you like, as the maximum number of vectors and 1 forms that a tensor can accept as input.

    Think about what contraction does. It removes one upper index and one lower index, right? So if the total number of free indices has been decreased by 2, the rank is lower by 2.

    I like to think of fourth rank tensor as kind of a matrix of matrices. It is a 4 by 4 matrix where every entry is itself a 4 by 4 matrix. You can also think of a vector as a line of numbers, a rank 2 tensor as a square of numbers, and a rank 3 tensor as a cube of numbers. Unfortunately, this makes a rank 4 tensor a hypercube of numbers, and I happen to be unable to visualize that.
  4. Sep 22, 2005 #3


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    When there are symmetries present, the number of independent components is reduced. That is, some components are related by the symmetry, which may force some components to be zero. The e^iklm tensor is totally antisymmetric. Swapping any pair of indices introduces a minus sign: e^iklm = - e^mkli, for example. [Example from linear algebra: swapping two rows in a determinant introduces a minus sign.]

    Q5: You need to use the [metric-signature dependent] epsilon-delta (i.e. LeviCivita-Kronecker) type identities. One of the best ways to remember this is to write a 4 by 4 determinant of Kronecker deltas:
    \epsilon^{iklm}\epsilon_{iklm}=(-1)^{1}(4-4)!4! \left|
    \delta^i_i & \delta^i_k & \delta^i_l & \delta^i_m \\
    \delta^k_i & \delta^k_k & \delta^k_l & \delta^k_m \\
    \delta^l_i & \delta^l_k & \delta^l_l & \delta^l_m \\
    \delta^m_i & \delta^m_k & \delta^m_l & \delta^m_m \\
    This is a special case of a more general expression involving fewer contractions. See Wald's text, p. 433.
  5. Sep 22, 2005 #4


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    We can inter-convert vectors v^a and one-forms v_a via raising and lowering indices in the metric.

    Let us call something that is either a vector or a one-form, interchangably, a "slot".

    A rank n tensor is a map from n "slots" to a scalar.

    You probably haven't gotten this far yet, but eventually you need to be able to figure out why a map from a vector to a vector is equivalent to a map from a (vector + 1-form) to a scalar.

    As far as your fourth rank tensor goes, think of it as a 4x4x4x4 matrix. That's 256 elements, arranged in a hypercube.

    Note that in the tensor e, all indices must be different for the tensor to be nonzero.

    How many ways can you permute 4 numbers?
  6. Sep 23, 2005 #5


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    I used to play with hypercubes while writing spreadsheet programs. It was incredibly complicated. Each 'hypersquare' represented a 3d set of data generated by 6 connected 2d surfaces.
  7. Sep 23, 2005 #6


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    q1 It's how many vectors in V plus how may vectors V* (where V is the vector space that the tensor is over) that it maps to V's (and therefore V*'s) field of scalars.

    q2 A contraction creates a tensor which has fewer arguments.

    q3 You can reprensent tensors of rank (p,q) as column vectors whose compoents are rank (p-1,q) tensors or equivalently row vectors of rank (p,q-1) tensors (p=0,1,2...; q=0,1,2...). A nxn matrix can be thought of as a (1,1) tensor in this scheme.

    q4 in terms of generalized matrices antisymmetric could represnt an (anti-) symmetry that the matrix has (i..e how when you swap elements around to cretae a new matrix, the new matrix is related to the original matrix).

    q5 if the first tensor is a member of a vector space W (the indices denote a particluar basis) then the second one is a member of W*. W* is the set of linear functions that maps W onto it's associated field of scalars.
  8. Sep 24, 2005 #7
    Nice! :approve: To me this is the easiest way tyo define the rank of a tensor space. It is simply the number of vectors + number of 1-forms that are mapped into a scalar.

    Way to go pervect!

    Note - We can inter-convert vectors v^a and one-forms v_a via raising and lowering indices in the metric. - If I were trying to be a pain in the butt then I'd note that not all manifolds have a metric asscociated to them. But since I'm not trying to be then I won't. :biggrin:

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