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On a smooth maniflod $M$ of dimension $n$, a current of degree $n-p$ is a functional on the space of compactly supported differential $p$-forms which is continuos. We denote the space of currents of degree $n-p$ by $D^{'n-p}(M)$. If we consider the functionals on $D^{'n-p}(M)$ with an approperiate comapctness and continuity assumptions, then what are these functionals? Are they just differential $p$-forms, or could be more then that?

I am asking this question because I want to understand why there is no definition for the pullback of current $T$ in general (as far as I know). Let $f: M_1 \to M_2$ be a map between manifolds. Then the pullback $f^*T$ should be defined as (formally) $$\langle f^*T, u\rangle = \langle T, f_*u \rangle.$$ Here $u$ is a differential $p$-form, in particular, $f_*u$ is well-defined as a current under some compactness assumption. Thus the problem is to make sense of $\langle T, f_*u \rangle$, i.e. can a current be a functional on the space of current? (Of course, it is enough to have $T$ be a functional on the space $\{f_*u\}$, that is why pullback of current is well-defined for submersion maps).

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  • $\begingroup$ Jenny Harrison wrote some papers on the dual of the space of currents, but I haven't looked at them for many years, so I can't remember the story: try math.berkeley.edu/~harrison $\endgroup$
    – Ben McKay
    Commented Apr 24, 2018 at 9:19

2 Answers 2

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This is a question of functional analysis: Start with a locally convex space $X$, form its dual $X^*$ consisting of continuous linear functionals on $X$. Now $X^*$ has several locally convex topologies $\tau$ with the following property:

For any $x\in X$ the linear map $L_x:X^*\to\mathbb{R}$ is $\tau$-continuous.

For such topology $\tau$ we form the bidual $X^{**}_\tau$ consisting of linear functionals on $X^*$ that are $\tau$ continuous. From the choice of $\tau$ we deduce that we have a natural map

$$X\ni x\mapsto L_x\in X^{**}_\tau. $$

The question is if there exist topologies $\tau$ on $X^*$ such that the above map is an isomorphism of locally convex spaces.

A natural topology on $X^*$ is the smallest locally convex topology such that all the linear functionals $L_x$ are continuous. Let's call this $\tau_0$. (Traditionally it is denoted by $\sigma(X^*, X)$.)

We can ask a more refined question: for which locally convex spaces the bidual $X^{**}_{\tau_0}$ coincides with $X$.

There is know a large class of such spaces namely the nuclear spaces. For details see

Gelfand & Shilov: Generalized Functions., vol 2.

For example the space $X=C^\infty(M)$, $M$ compact has this property.

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  • $\begingroup$ What is the role of nuclearity? Of course, the dual of $(X^\ast,\sigma(X^\ast,X))$ is $X$ (more precisely, $\{L_x:x\in X\}$) for every locally convex space $X$. If you ask for a topological isomorphism between $(X^\ast,\sigma^\ast)^\ast$ you have to specify the topology on that bidual. For the topology of uniform convergence on equicontinuous sets you have again topological isomorphy for all locally convex spaces. If you consider uniform convergence on bounded sets you need a barrelldness condition for $X$. For the weak$^\ast$-topology on the bidual you need that $X$ has its weak topology... $\endgroup$ Commented Apr 25, 2018 at 7:28
  • $\begingroup$ In the terminology of Gelfand and Shilov, if $X$ is a perfect locally convex space than its topology coincides with the $\sigma(X,X^*)$ topology. The nuclear spaces are perfect. $\endgroup$ Commented Apr 25, 2018 at 10:01
  • $\begingroup$ No. Whenever an $\infty$-dimensional space $X$ has a continuous norm (i.e., $p(x)=0$ implies $x=0$, seminorms need not satisfy this) then its topology is strictly finer than $\sigma(X,X^*)$ (just because the latter topology doesn't have continuous norms). The space $C^\infty(M)$ you mentioned clearly has continuous norms. $\endgroup$ Commented Apr 25, 2018 at 12:33
  • $\begingroup$ Let me rephrase: In a perfect space strong and weak convergence coincide: Gelfand & Shilov, vol.2, Section I.6.3. The theorem in Sec. I.6.2. of the same book shows that $C^\infty(M)$ is perfect if $M$ compact. $\endgroup$ Commented Apr 25, 2018 at 13:29
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    $\begingroup$ Well, in nuclear (more generally: Montel) spaces the strong and weak topologies have the same convergent sequences (or bounded nets or filters with basis in a bounded set -- just because bounded sets are strongly compact and there is no strictly coarser Hausdorff topology on a compact space). But even if two locally convex topologies have the same convergent sequences they need not have the same dual. $\endgroup$ Commented Apr 25, 2018 at 14:20
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This follows the setup of [De Rham: Differentiable manifold, Springer-Verlag 1984]. Smooth differential form with compact support form the space $\Gamma_c(\Lambda^pT^*M) = \Gamma_{C^\infty_c}(\Lambda^pT^*M)$. Its dual space is the space of distributional sections $\Gamma_{\mathcal D'}(\Lambda^pTM)$. If the manifold is orientable, smooth $(n-p)$-forms can be embedded into $\Gamma_{\mathcal D'}(\Lambda^pTM)$ by using the action $\int_M \phi\wedge\psi$. If the manifold is not orientable, passing to the orientable double cover and considering forms in the $\pm1$ eigenspaces of the pullback with the deck-transformations (these correspond to "forme pair ou impair" of De Rham) one can carry this over. Going to the completion, we can also view currents as distributional sections $\Gamma_{\mathcal D'}(\Lambda^{n-p}T^*M$.

For the pullback: you can pullback the bundle $\Lambda^{np}T^*M$, but the distributional coefficients (in a local frame) only under a diffeomorphism, or under a smooth mapping where the image of the tangent mapping contains the wave front set of each distributional coefficient.

In other words: pulling back the bundle is not the obstruction; pulling the distributional coefficients is it!

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  • $\begingroup$ Thank you! For my first question, perhaps I did not state it clearly. Let's consider a complex manifold, so the orientation is not a problem. Smooth $(n-p)$-forms are certainly lie in the set of functionals of $D^{' p}(M)$, but I am curious are they the only possibilities? $\endgroup$
    – Li Yutong
    Commented Apr 24, 2018 at 10:54

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