# Why Gateaux derivative is a distribution?

Thanks to Jan Bohr answer and comment I edited this question.

Let $$E$$ be a vector bundle , $$E^*$$ the dual bundle and $$D$$ a density bundle. Denote by $$\Gamma(E)$$ the space of section of the bundle $$E$$.

By definition the a distribution $$\omega$$ in a vector bundle $$E$$ is $$\omega \in\Gamma(E^*\otimes D)^\prime$$ where $$\Gamma(E^*\otimes D)^\prime$$ is the topological dual of $$\Gamma(E^*\otimes D)$$.

Now in this paper Algebraic Structure of Classical Field Theory: Kinematics and Linearized Dynamics for Real Scalar Fields by Romeo Brunetti, Klaus Fredenhagen, Pedro Lauridsen Ribeiro they define at page 14 the Gateaux derivative as $$F^K[\phi](\phi_1...,\phi_k)=\left=\frac{d^k}{d\lambda_1...d\lambda_k }\left(\phi + \lambda_1\phi_1+...\lambda_k\phi_k\right)$$ where $$<>$$ denotes dual pairing and $$\phi,\phi_1...,\phi_k \in\Gamma(E)$$

Now they say that $$F^K[\phi]$$ is a distribuition . But $$F^K[\phi]$$ is not in $$\Gamma(E^*\otimes D)^\prime$$ how can it be a distribution?

Thanks to Jan Bohr answer and comment I add this part . In this paper Properties of field functionals and characterization of local functionals by Christian Brouder, Nguyen Viet Dang, Camille Laurent-Gengoux, Kasia Rejzner at page 10

they define $$F^K[\phi]$$ as a d distributional section on $$M^k$$ for a general vector bundle.

Maybe they are using $$<>$$ to identify $$E$$ with $$E^*$$ but I am not sure

Consider the following example: $$M$$ is a closed manifold, $$\mu$$ is a volume form and $$F\colon C^\infty(M)\rightarrow \mathbb C$$ is defined by $$F(\varphi)=\int_M\varphi^2\mu$$. Then $$F(\varphi+\lambda\varphi_1)=F(\varphi)+2\lambda\int_M(\varphi\varphi_1)\mu+\lambda^2F(\varphi_1)$$. Taking the $$\lambda$$-derivative at $$0$$, you obtain $$dF_\varphi(\varphi_1) =\int_M\varphi_1 (2\varphi\mu).$$ For fixed $$\varphi$$ you can interpret this as the linear functional on $$C^\infty(M)$$ that takes some $$\varphi_1$$ and sends it to the integral agains the form $$2\varphi\mu$$. If we define $$\mathcal D'(M)$$ as the continuous dual of $$C^\infty(M)$$, we have all right to call $$dF_\varphi$$ a distribution. As you mentioned, sometimes $$\mathcal D'(M)$$ is defined as the dual of $$\Gamma(M,D)$$, which is convenient if one does not want to fix a particular density. However, once a volume density is chosen (e.g. $$\vert \mu \vert$$), there is an isomorphism $$C^\infty(M)\rightarrow \Gamma(M,D)$$ given by $$\psi\mapsto \psi \vert \mu \vert$$. Hence also the dual spaces are isomorphic and you can consider $$dF_\varphi$$ as a distribution also in the other sense.
The authors write $$dF_\varphi(\varphi_1)$$ as $$F^{(1)}[\varphi](\varphi_1)$$ and then also consider higher order derivatives $$F^{(k)}[\varphi](\varphi_1,\dots,\varphi_k)$$. Again for fixed $$\varphi$$ you can view this as a linear functional on the algebraic tensor product $$C^\infty(M)\otimes \dots \otimes C^\infty(M)$$ ($$k$$-times). After choosing the correct topological tensor product, this should extend to a continuous linear functional on $$C^\infty(M\times\dots\times M)$$ and hence a distribution on $$M^k$$.
• my problem is that $F^K[\phi] \in \Gamma(E^*\otimes D)^\prime$ but in your anser you are considering $F^K[\phi] \in \Gamma(E\otimes D)^\prime$ Nov 26, 2022 at 14:50
• Which vector bundle $E$ would you like to consider? In my example (and I believe also in the paper you linked), the vector bundle is trivial. Nov 26, 2022 at 15:12