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Let $M$ be a manifold. There is the sheaf of distributions $\mathcal{D}'$ and the sheaf of distribution densities $\mathcal{D}(\cdot)'$ on $M$. A delta distribution density $\delta_p \in \mathcal{D}(M)'$ could be defined easily as $\delta_p(f) := f(p)$ for $f \in \mathcal{D}(M)$ and $p \in M$.

But is there a natural definition of $\delta_p$ as distribution on $M$, i.e. $\delta_p \in \mathcal{D}'(M)$?

If there is some positive smooth distribution density $\Psi \in \mathcal{D}(M)'$ given, for example by a Riemannian density, there is an isomorphism of sheaves $\mathcal{D}' \to \mathcal{D}(\cdot)'$ given by $u \mapsto \Psi u$. Via this isomorphism there would be a candidate for $\delta_p \in \mathcal{D}'(M)$. But this definition depends on the isomorphism and is therefore not natural.

The problem by defining the delta distribution $\delta_p \in \mathcal{D}'(M)$ directly as distribution is that there seems to be no natural choice of how to glue the $0$ distribution and corresponding delta distributions in the coordinate neighborhoods together. So, is it possible to define a delta distribution on a manifold in a natural way?

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    $\begingroup$ It seems to be that you've convincingly argued that the answer to your own question is no. If there were a “natural” way to define a distribution $δ_p$ for every $p$, then comparing (chartwise) it to the distribution density $δ_p$ would give a smooth density (alternatively: find the smooth density on $M$ so that the integral of the putative distribution $δ_p$ against this density gives $1$ at every $p$); and no such “natural” density exists, so neither can a “natural” distribution $δ_p$. I'm just rephrasing pretty much what you yourself explained: why does it not convince you? $\endgroup$
    – Gro-Tsen
    Commented Oct 18 at 10:59
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    $\begingroup$ This may be just a matter of terminology what you call "distribution density" is also sometimes (perhaps more often even) referred to as just a "distribution". In that sense your $\delta_p$ is already a "natural delta distribution", end of story. You can tensor distributions with the sections of any other vector bundle (including the volume form bundle). If that vector bundle has no canonical (not everywhere zero) sections, then you lose the "natural deltas" in the tensor product. $\endgroup$
    – Igor Khavkine
    Commented Oct 18 at 13:58
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    $\begingroup$ I would suggest using the language of currents, following de Rham's book Differentiable Manifolds: Forms, Currents, Harmonic Forms. $\endgroup$
    – Ben McKay
    Commented Oct 18 at 16:10

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We can give a complete classification of (candidates for) delta-distributions on a smooth manifold $M$ at point $p$.

Specifically, a delta-distribution is a smooth linear functional on the space of smooth compactly supported 1-densities $\def\Dens{{\sf Dens}}ω∈Γ(\Dens_1(M))$ on $M$.

A delta-distribution must vanish on the complement of $\{p\}⊂M$. Such distributions necessarily have the following form: take the $k$th order jet of $ω$ at point $p$ for some $k≥0$, then apply some linear functional $\def\R{{\bf R}}J^k\Dens_1(M)_p→\R$.

For a delta-distribution, as opposed to its (higher) derivatives, we must have $k=0$. Thus, candidates for delta-distributions are in a canonical bijective correspondence with linear maps $\Dens_1(M)_p→\R$, equivalently, elements of $\def\or{{\sf or}}Λ^{\dim M}T_p M⊗\or_p(M)$, i.e., unoriented volume elements on the tangent space of $M$ at $p$.

Specifying such an unoriented volume element produces a delta-distribution on $M$ supported at $p$. Conversely, given such a distribution, we get a canonical unoriented volume element on the tangent space of $M$ at $p$.

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