# Adjoint of functional derivative in superspace

• HenryGomes
In summary, we have a pointwise inner product in the space of Riemannian metrics over a compact 3-manifold without boundary. The inner product is defined for metric velocities and is represented by the Wheeler-Dewitt supermetric. We also have a functional f: Riem(M) -> R defined by a local function and its functional derivative can be seen as a differential form. We can define a one-form in Riem(M) and an inner product for it. The question is whether we can define a functional differential operator as the adjoint of the functional derivative with respect to the supermetric.
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,

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

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.

## What is the adjoint of functional derivative in superspace?

The adjoint of functional derivative in superspace is a mathematical operation that allows for the calculation of the variation of a functional with respect to a field or variable in superspace. It is an essential tool in supersymmetric theories and allows for the study of the underlying symmetries and dynamics of these systems.

## How is the adjoint of functional derivative in superspace different from the regular functional derivative?

The adjoint of functional derivative in superspace differs from the regular functional derivative in that it takes into account the supersymmetric nature of the theory. This means that it considers the variations of both the fields and the superfields, which are the fundamental building blocks of supersymmetric theories.

## What are the applications of the adjoint of functional derivative in superspace?

The adjoint of functional derivative in superspace has many applications in theoretical physics, particularly in the study of supersymmetric theories. It allows for the calculation of important quantities such as conserved charges, symmetries, and equations of motion. It is also used in the study of supergravity and superstring theories.

## How is the adjoint of functional derivative in superspace calculated?

The calculation of the adjoint of functional derivative in superspace involves the use of differential operators called supercovariant derivatives. These operators act on the superfields and their variations to calculate the variation of the functional. The result is a superspace integral, which is then reduced to a component form to obtain the final expression.

## Are there any limitations to the use of the adjoint of functional derivative in superspace?

While the adjoint of functional derivative in superspace is a powerful tool in the study of supersymmetric theories, it does have some limitations. It can only be applied to theories with a well-defined supersymmetric structure, and the calculations can become very complex and time-consuming for more complicated systems. Additionally, it may not be applicable to non-supersymmetric theories.

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