Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trace of a tensor product

  1. Nov 9, 2008 #1
    Hey guys,

    How exactly do you take the trace of a tensor product? Do I take the trace of each tensor individually and multiply their traces?

    For example, how would I take the trace of this tensor product:

    [tex]-B^{c}_b B_{ac}[/tex]
    Last edited by a moderator: Nov 10, 2008
  2. jcsd
  3. Nov 10, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi JasonJo! :smile:

    Isn't that just -gabBbcBac = BacBac ?
  4. Nov 10, 2008 #3
    No! it's like matrices... the trace of a product is not the product of traces.
    Now if you have a product of tensors which is an 1-1 tensor the trace is the contraction of these two indices, if you have more than 2 indices you must specify over which indices the trace must be done, and then you contract...
  5. Nov 10, 2008 #4
    Yes, you can take the trace of each tensor individually and multiply their traces.
    That is, Tr(A \Otimes B) = Tr(A)Tr(B)
  6. Nov 11, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    No … it's as astros :smile: says:
    For example,
    Code (Text):
    0 1
    1 0
    has trace 0, but its square has trace 2. :smile:
  7. Nov 11, 2008 #6


    User Avatar

    If you mean the tensor product defined as how it's defined at the beginning of Ch 4 of Spivak's Calculus on Manifolds, then yes what locality said is correct.
    Let V be a Vector space over R
    A k tensor on V is a multilinear map from V^k into R.
    The tensor product [itex]\otimes[/itex] of S:V^k -> R and T:V^l -> R is the map S[itex]\otimes[/itex]T: V^(k+l) -> R defined by
    S[itex]\otimes[/itex]T(v_1,....v_k,....,v_(k+l)) = S(v_1,..v_k)*T(v_k+1,...,v_l).

    In the special case S and T are square matrices,
    tr(S[itex]\otimes[/itex]T) = tr(S)tr(T)
    Last edited: Nov 11, 2008
  8. Nov 12, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    ah, but the OP's question was clearly about tensor product in the sense of contraction (analogously to matrix product :wink:), not in the ⊗ sense of Spivak. :smile:
  9. Nov 20, 2008 #8
    That makes sense. But then what do I do this the B^ac B_ac term? What exactly is this term? Am I contracting here? When do we know something has a trace?

    And when we are taking traces of tensors, what does it mean to be tracefree? Are we referring to the antisymmetric part?

    If anyone is wondering, I am trying to derive Raychaudhuri's Equation.
  10. Nov 21, 2008 #9


    User Avatar
    Science Advisor
    Homework Helper

    BacBac is the "quadratic invariant" of the tensor.

    It's a double contraction.

    (It's a sort-of "trace of a trace", but I really don't think thinking of it like that helps at all)

    It may help to look at http://en.wikipedia.org/wiki/Raychaudhuri's_equation#Mathematical_statement … are you trying to derive ω2 and σ2? :smile:
    Tracefree simply means that the trace is zero.

    An antisymmetric tensor must be tracefree, but not vice versa.

    For example, the LHS of Einstein's field equations is Rij - (1/2)R gij, where Rij is the Ricci curvature tensor, which is symmetric, with 10 independent parameters, and R is its trace (a scalar, obviously only 1 parameter) … the tracefree part of Rij is Rij - (1/4)R gij, with 9 independent parameters, and the EFE can be written:

    Rij - (1/4)R gij = 8π(Tij - (1/4)T gij)

    and R = -8πT,

    where the first line equates the tracefree parts, with 9 independent equations, and the second line (anti-)equates the traces, with just 1.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook