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Has anyone seen this scalar?

  1. Mar 13, 2006 #1
    In my calculations, I come across the scalar
    [tex]R^{ji} R_{ij} - R^2[/tex]
    ([tex]R_{ij}[/tex] is the Ricci tensor, [tex]R[/tex] is the Ricci scalar)
    More specifically, I come across the integral of this scalar over a compact manifold.
    Has anyone seen it before, and does it have any nice properties?

    John Schulman
  2. jcsd
  3. Mar 13, 2006 #2


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    It seems to me...
    Up to a constant factor, that scalar is the second principal-invariant of the Ricci tensor. It is proportional to the second elementary symmetric function of its eigenvalues. How does it arise? (I've been interested in this invariant [not necessarily for Ricci] and have been searching for a geometrical interpretation [and physical interpretation] for it.)
    Last edited: Mar 13, 2006
  4. Mar 13, 2006 #3

    [tex]R^{ji} R_{ij} = R[/tex]

    it follows that

    [tex]R^{ji} R_{ij} - R^2 = R - R^2 = R(1 - R)[/tex]

    That's about all I can see about it.

  5. Mar 14, 2006 #4


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    No. [itex]R=R_{ij} g^{ij}[/itex]. The Ricci scalar has units, by the way...

    I have no idea if the original poster is looking for a purely mathematical answer or not, but if Einstein's equations hold, and you have a perfect fluid with density [itex]\rho[/itex] and pressure [itex]p[/itex], that scalar is proportional to [itex]p(\rho-p)[/itex]. Normally, [itex]\rho \gg p[/itex], so the square root of your scalar is basically a geometric average of the density and pressure.
  6. Mar 14, 2006 #5


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    In my earlier post, I was too lazy to write
    [tex]R^{ji} R_{ij} - R^2=2R^i{}_{[j}R^j{}_{i]}[/tex],
    where I've used the metric to raise and lower indices.
  7. Mar 14, 2006 #6
    Thanks for the responses, especially robphy. I'll do a little research about the eigenvalues of [tex]R^{i}_j[/tex].

    Here's how I came across this quantity:
    I'm working in a rather specific area: I'm studying the relationship between the curvature of a Kahler supermanifold with the curvature of the underlying complex manifold.
    I found that if a supermanifold with two "fermionic" dimensions satisfies [tex]R=0[/tex] then there's a scalar differential equation with some curvature tensors that the underlying complex "bosonic" manifold must satisfy. This equation has a bunch of terms in it, but if we take the integral of the expression over a compact manifold, we get
    [tex]\int (R^{ji} R_{ij} - R^2) dV=0[/tex]

    Here's the consequence of this calculation that interests me:
    If I can find a complex manifold where the above integral CAN'T equal zero regardless of metric, it is likely that I will be able to disprove a certain conjecture concerning Yau's theorem generalized to supermanifolds. Unfortunately, I don't know if such a manifold exists.
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