# Functional derivatives as distributions

I have asked this on math stack exchange, due to its mostly mathemtical content, but aside from one upvote and minimal views it has not garnered any attention, so I am trying here as well. This isn't anything important really, but it has been stratching the back of my head while studying variational formalism in general relativity.

"Let $(M,\mathcal{S},g)$ be a smooth, $n$-dimensional manifold equipped with a Riemann metric. Let us denote the vector space of $(p,q)$-type tensor fields on $M$ as $\mathcal{T}_{q}^{p}(M)$.

Let $\Psi:\mathbb{R}\rightarrow\mathcal{T}_{q}^{p}(M),\varepsilon\mapsto\Psi(\varepsilon)$ be a smooth curve and let us use the notation where $\Psi$ denotes $\Psi(0)$.

Let $S:\mathcal{T}_{q}^{p}(M)\rightarrow\mathbb{R}$ be a functional, in such way, that $$S[\Psi]=\int_{M}\mathcal{L}(\Psi,\nabla\Psi)\sqrt{|\det(g)|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n}.$$

In this case, we say $S$ is functionally derivable at $\Psi$, if there exists a $\frac{dS[\Psi]}{d\Psi}\in\mathcal{T}_{p}^{q}(M)$ tensor field, that $$\left.\frac{dS[\Psi(\varepsilon)]}{d\varepsilon}\right|_{\varepsilon=0}=\int_M\frac{dS[\Psi]}{d\Psi}\bullet\left.\frac{d\Psi(\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}\sqrt{|\det(g)|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n},$$ where $\bullet$ denotes full contraction.

My questions are regarding technical details of this derivative. Physics books generally do not impose rigorous conditions on the space of tensor fields on which $S$ is defined.

What structures does this space need to possess for this to make sense? I assume Hausdorff-topology is a must, but does it need to be normed? If so, what norm do we use, that does not conflict with physics, or what norm makes sense in a physical context?

Wald mentions in a footnote, that in general, a tensor distribution needs exist, so that $$\left.\frac{dS[\Psi(\varepsilon)]}{d\varepsilon}\right|_{\varepsilon=0}=\left\langle\frac{dS[\Psi]}{d\Psi},\left.\frac{d\Psi(\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}\right\rangle.$$ Is there any conceivable situation within the bounds of physics, where this distribution is singular, ie. doesn't exist as an integral?"

• Crossposted from math.stackexchange.com/q/1079047/11127 – Qmechanic Dec 25 '14 at 13:44
• @Danu Why do you think so? I originally did post this on math, but my primary question is regarding whether it is possible to conceive such action within physics, that the resulting distribution is non-regular. I'm sure, that from a mathematical point of view, that is quite possible, but from a physical point of view, this is a physics question, isn't it? – Bence Racskó Dec 25 '14 at 13:50

You need not only a topology on your tensor fields but also a smooth structure. Otherwise it would not make sense to say that $\Psi: (-\epsilon, + \epsilon) \to \mathcal{T}$ is a smooth family of curves. You can try to put a Banach manifold structure on $\mathcal{T}$ by endowing it with Sobolev norms or, more naturally but also more difficult, view it as a Fréchet manifold with the usual $C^\infty$-topology. See for example here http://en.wikipedia.org/wiki/Fr%C3%A9chet_space#Examples