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Tensor Contraction

  1. Dec 7, 2013 #1
    Hello Everyone,

    I came here with a question and hope you can shed some light.

    We know that Ricci tensor which is a contraction of Riemann tensor contains a subset of information as contained by Riemann tensor. In 3-D infact they contain the same information.

    I was wondering is it always true that Tensor obtained by contracting a primary tensor always has a subset of information ? The way I see Tensor is something which is based on the Physics/Geometry of the problem and which stay unchanged under cooridinate transformation.

    Thank You
     
  2. jcsd
  3. Dec 7, 2013 #2

    WannabeNewton

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    You seem to have affirmed the negative of your own question by yourself. For ##n\geq 3## we can decompose ##R_{abcd}## in terms of the Weyl curvature ##C_{abcd}##, the Ricci curvature ##R_{ab}##, and its trace ##R = g^{ab}R_{ab}##: ##R_{abcd} = C_{abcd}+ \frac{2}{n-2}(g_{a[d}R_{c]b}+ g_{b[c}R_{d]a}) + \frac{2}{(n-1)(n-2)}Rg_{a[c}g_{d]b}##.

    If ##C_{abcd} = 0## then ##R_{abcd}## is fully characterized by ##R_{ab}## and ##R##. In particular for ##n = 3##, ##C_{abcd} = 0## for all manifolds so you won't lose anything by contraction. Of course you won't gain any information by contraction but you also won't necessarily lose anything as your own example shows i.e. it's a sufficient condition but not a necessary one.
     
  4. Dec 7, 2013 #3
    Thank You for replying. I think I did not state the question clearly. I gave Riemann Tensor only as an example. My original question is

    If a Tensor field encodes physics/geometry of an example, will a contracted Tensor field also hold similar information but only a subset of original one ? Are there any examples other than Riemann tensor ?
     
  5. Dec 7, 2013 #4

    WannabeNewton

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    If you take an arbitrary tensor field and contract (at least) two indices then yes you will lose some information about the physics/geometry that the original tensor field described but as noted above this is a sufficient condition and not a necessary one. You can come up with an infinity of examples where contraction doesn't make you lose any information at all.

    For example, take Friedmann space-time (http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric) and consider the unit time-like vector field ##\xi^{\mu}## whose integral curves are the worldlines of observers who determine a homogenous and isotropic foliation of Friedmann space-time (the observers comoving with the Hubble flow in other words). It can be easily shown that ##\nabla_{\mu}\xi_{\nu} = \frac{1}{3}(\nabla_{\gamma} \xi^{\gamma})h_{\mu\nu}## where ##h_{\mu\nu} = g_{\mu\nu} + \xi_{\mu} \xi_{\nu}## is the spatial metric relative to the comoving observers on each spatial slice of Friedmann space-time. The comoving observers describe an irrotational, shear-free, geodesic flow. As such, all the kinematical information one can obtain from ##\nabla_{\mu}\xi_{\nu}## lies in the expansion scalar ##\nabla_{\gamma}\xi^{\gamma}##.
     
  6. Dec 7, 2013 #5
    If I may ask, what really does such contraction does ? For example say stress-energy tensor, if I contract it the trace does not seem to give any information! For Reimann Tensor I understand all these tensors are irreducible representation it. Is there any other example where some sort of contraction is an irreducible representation of original tensor.
     
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