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Adjoint of functional derivative in superspace

  1. Feb 27, 2008 #1
    In the space of Riemannian metrics Riem(M), over a compact 3-manifold without boundary M, we have a pointwise (which means here "for each metric g") inner product, defined, for metric velocities [tex]k^1_{ab},k^2_{cd}[/tex] (which are just symmetric two-covariant tensors over M):
    [tex]\int_MG^{abcd}k^1_{ab}k^2_{cd}d^3x[/tex]
    where
    [tex]G^{abcd}=\frac{1}{2}\sqrt{g}(g^{ac}g{bd}-g^{ab}g^{cd})[/tex]
    is the Wheeler-Dewitt supermetric.
    Now, let [tex]f:Riem(M)\ra\R[/tex] be a functional of the metric which is of the form
    [tex]f[g_{ab}]=\int_Mf(g_{ab}(x))d^3x[/tex]
    i.e. it is represented by a local function [tex]f_x:T_xM\otimes_ST_xM\ra\R[/tex]
    (where the subscript S indicates symmetrized). It is known that the functional derivative [tex]\delta{f}[/tex] can be seen as a differential form in Riem(M). It can be defined at [tex]g_{ab}[/tex] as:
    [tex]\delta{f}_{g_{ab}}[k_{ab}]=\int_M\frac{d}{dt}_{t=0}f(g_{ab}(x)+tk_{ab}(x))d^3x[/tex]
    Let us suppose a one-form in Riem(M) can be defined pointwise in M by
    [tex]\lambda^{ab}_x:T_xM^*\otimes_ST_xM^*\ra\R[/tex]
    and that we have the inner product as above:
    [tex]\int_MG_{abcd}\lambda_1^{ab}\lambda_2^{cd}d^3x[/tex]
    where
    [tex]G_{abcd}=\frac{1}{det{g}}(g_{ac}g_{bd}-\frac{1}{2}g_{ab}g_{cd})[/tex]
    Now, my question is, in this instance, where the functional derivative basically acts as a exterior derivative, can we define it's adjoint with respect to the above supermetric? I.e., can we find a functional differential operator [tex]\delta^*[/tex] such that
    [tex]\int_MG_{abcd}\lambda^{ab}(x)\delta{f}^{cd}(x)d^3x=\int_M\delta^*\lambda_g^{ab}(g(x))f(g(x))d^3x[/tex]
    Thanks,
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2

    olgranpappy

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    Homework Helper

    use "[/tex]" to end your tex, not "[tex]"...
     
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