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Ricci form and Kahler manifolds

  1. Jun 13, 2010 #1

    OB1

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    I am confused about the different Ricci-named objects in complex and specifically Kahler geometry: We have the Ricci curvature tensor, which we get by contracting the holomorphic indices of the Riemann tensor. We have the Ricci scalar Ric, which we get by contracting the Ricci tensor. Then there is a Ricci form, and I don't see how it is at all different from the Ricci tensor (apart from it not having any indices - how can it be a 2-form without having indices?), and finally the *other* Ricci scalar R, which we get by contracting the Ricci form.
    Finally, we have the Calabi-Yau manifold, which we get by taking a "Ricci-flat" Kahler manifold. and I can't figure out for the life of me is which of these Ricci-named objects vanishes for a Ricci-flat manifold. Help in unraveling this ridiculous confusion is much appreciated!
     
  2. jcsd
  3. Jun 14, 2010 #2
    Ricci flat manifolds are manifolds for which the ricci tensor vanishes. The Ricci tensor is symmetric, but by introducing multiplication by i we can get an alternating form on a Kahler manifold.

    For Calabi-Yau, you should just take Ricci-flatness to mean the usual thing, but of course if one vanishes, the other does as well.
     
  4. Jul 8, 2010 #3
    Well, the Ricci curvature tensor R is symmetric and it is an object with 2 lower indices, but these indices are used to represent R with respect to a coordinate. As it has two lower indices it is cotensor of rank 2, so we need to feed it with 2 vectors.

    We can represent R without coordinates (thus without any indices), and then we can feed it with 2 vectors, say V and W. Then R(V,W) gives us a scalar. Now R being symmetric just means that R(V,W) = R(W,V).

    The Ricci-form, say F, is related to R and the complex structure J defined on the manifold:

    F(V,W) := R(J V,W)

    That's all. Now it is antisymmetric, thus it is a differential form. In this notation the Ricci form F indeed has no indices, but it CAN be represented with respect to a coordinate. Then it is again a symbol with 2 lower indices.

    Ricci-flatness means R = 0. Then automatically F and the Ricci scalar (say r) are zero also.

    Representing V and W in terms of a complex basis of the tangent space reveals that it is equivalent to multiplying with i. If V is a holomorphic vector, and if W is an antiholomorphic vector, then:

    F(V,W) = R(J V,W) = R(i V,W) = i R(V,W) = i R(W,V) = R(i W,V) = R(-J W,V) = -F(W,V)


    For a nice analogy, see the Kahler form: A Hermitian metric g defines a Kahler form w(V,W) = g(J V,W).


    So, to answer this question:

    Every tensor, cotensor, mixed tensor and n-form can be represented with respect to a basis of the tangent (& cotangent) space. In that case it has 'indices'. But, they can also be represented in a coordinate-free way.

    It's a pity mathematics is represented in a bit obscure manner in many texts & lectures about physics.
     
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