Adjoint of functional derivative in superspace

[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 [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,

 

[olgranpappy]

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

 

[Entropia]

for your question.

Yes, in this instance, we can define the adjoint of the functional derivative with respect to the given supermetric. The adjoint operator, denoted by \delta^*, can be defined as follows:

\delta^*\lambda^{ab}(g(x))=\frac{\delta}{\delta g_{ab}(x)}\left(\frac{1}{\sqrt{g}}\lambda^{ab}(x)\right)

where \frac{\delta}{\delta g_{ab}(x)} is the functional derivative with respect to the metric g_{ab}(x).

To see why this definition works, we can use integration by parts to rewrite the right-hand side of the equation as:

\int_M\delta^*\lambda_g^{ab}(g(x))f(g(x))d^3x=\int_M\frac{\delta}{\delta g_{ab}(x)}\left(\frac{1}{\sqrt{g}}\lambda^{ab}(x)\right)f(g(x))d^3x=\int_M\lambda^{ab}(x)\frac{\delta f}{\delta g_{ab}(x)}d^3x

which is equivalent to the left-hand side of the equation. Therefore, we can conclude that \delta^* is the adjoint operator of the functional derivative with respect to the given supermetric.

Furthermore, it is worth noting that this definition of the adjoint operator is consistent with the standard definition of the adjoint of a differential operator in functional analysis. In this case, the functional derivative can be seen as a differential operator acting on the space of Riemannian metrics, and its adjoint, \delta^*, can be seen as the corresponding differential operator acting on the dual space of Riemannian metrics.

In summary, the existence of the adjoint of the functional derivative in superspace allows us to define a natural inner product on the space of functionals of Riemannian metrics. This inner product is crucial in many areas of mathematical physics, such as quantum gravity and geometric analysis.