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):(adsbygoogle = window.adsbygoogle || []).push({});

[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,

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# Adjoint of functional derivative

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