1. Feb 27, 2008

HenryGomes

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 $$k^1_{ab},k^2_{cd}$$ (which are just symmetric two-covariant tensors over M):
$$\int_MG^{abcd}k^1_{ab}k^2_{cd}d^3x$$
where
$$G^{abcd}=\frac{1}{2}\sqrt{g}(g^{ac}g{bd}-g^{ab}g^{cd})$$
is the Wheeler-Dewitt supermetric.
Now, let $$f:Riem(M)\ra\R$$ be a functional of the metric which is of the form
$$f[g_{ab}]=\int_Mf(g_{ab}(x))d^3x$$
i.e. it is represented by a local function $$f_x:T_xM\otimes_ST_xM\ra\R$$
(where the subscript S indicates symmetrized). It is known that the functional derivative $$\delta{f}$$ can be seen as a differential form in Riem(M). It can be defined at $$g_{ab}$$ as:
$$\delta{f}_{g_{ab}}[k_{ab}]=\int_M\frac{d}{dt}_{t=0}f(g_{ab}(x)+tk_{ab}(x))d^3x$$
Let us suppose a one-form in Riem(M) can be defined pointwise in M by
$$\lambda^{ab}_x:T_xM^*\otimes_ST_xM^*\ra\R$$
and that we have the inner product as above:
$$\int_MG_{abcd}\lambda_1^{ab}\lambda_2^{cd}d^3x$$
where
$$G_{abcd}=\frac{1}{det{g}}(g_{ac}g_{bd}-\frac{1}{2}g_{ab}g_{cd})$$
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 $$\delta^*$$ such that
$$\int_MG_{abcd}\lambda^{ab}(x)\delta{f}^{cd}(x)d^3x=\int_M\delta^*\lambda_g^{ab}(g(x))f(g(x))d^3x$$
Thanks,