3
$\begingroup$

Let $\mathcal{D}(\mathbb{R})$ be the continuous dual of $C^\infty_c(\mathbb{R})$, the space of compactly-supported smooth functions. There is a nice characterization of distributions $a\in\mathcal{D}(\mathbb{R})$ that are localized at a point $r\in\mathbb{R}$; $a$ must be a linear combination of distributional derivatives of the dirac $\delta$-distribution, $\delta_r$.

Let $M$ be a compact manifold and $\Lambda\subset M$ a closed submanifold. Let $\mathcal{D}(M)$ be the continuous dual of $C^\infty_c(M)$, i.e. $\mathcal{D}$ is the collection of zero-currents (in the sense of de Rham). Is there an analogous characterization of zero-currents that are localized along $\Lambda$?

$\endgroup$

1 Answer 1

4
$\begingroup$

A special case is this:

Let $u\in\mathcal{E}'(\mathbb{R}^{n+m})$. Then $supp(u)\subseteq\mathbb{R}^n\times\{0\}$ iff $u$ is a linear combination of the form $\sum_{\substack{\alpha\in\{0\}\times\mathbb{N}^m \\ |\alpha|\leq k}} c_\alpha \partial^\alpha \iota(v_\alpha)$ for certain $v_\alpha\in\mathcal{E}'(\mathbb{R}^n)$. Here $\iota(v)$ denotes the distribution on $\mathbb{R}^{n+m}$ defined by $\langle \iota(v),\phi\rangle:=\langle v,\phi_{|\mathbb{R}^n}\rangle$.

The proof idea is to use a taylor expansion of $\phi$ orthogonal to $\mathbb{R}^n$, to show that $u$ vanishes on those functions whose orthogonal derivates up to the $k$th order vanish on $\mathbb{R}^n\times\{0\}$ and deduce the result.

(I suspect that a similar thing could be true for $u\in\mathcal{D}'$. In that case the order of $u$ is only locally bounded, to one would expect that that linear combination has to be replaced by some infinite (but locally finite) linear combination of derivatives)

By localizing and some application of change of coordinates I think this should extend to the following general result: If $u\in\mathcal{E}'(M,E)$ is a distributional section of a vector bundle $E$ on the riemannian manifold $M$ (in your case $E$ is the bundle of $n$-forms) and $S\subseteq M$ an embedded submanifold, then $supp(u)\subseteq S$ iff it is of the form $\sum_{a=0}^k \mathcal{L}_{X_{i1}} \cdots \mathcal{L}_{X_{ij_a}}\iota(v_a)$ for certain $v_a\in\mathcal{E}'(S,E_{|S})$ and certain vector fields $X_{ij} \in \Gamma(NS)$ where $NS\subseteq TM$ is the normal bundle of $S$ and $\iota:\mathcal{E}'(S,E_{|S})\to\mathcal{E}'(M,E)$ is the analogue embedding. Again, it might be the case that some version of this also holds for $\mathcal{D}'$ if one replace the finite sum by an locally finite series.

$\endgroup$
4
  • $\begingroup$ Thanks very much! I have a couple of questions about your response. (i) I'm not sure how those Lie derivatives act on sections of the general vector bundle $E$; if $E$ is not a bundle of tensors, should these be covariant derivatives relative to some connection on E? (ii) If we don't introduce a metric on $M$, do you think the global characterization can be expressed in terms of sections of the abstract normal bundle $(\iota_S^*TM)/TS$? $\endgroup$
    – Josh Burby
    Apr 17, 2014 at 15:35
  • $\begingroup$ (i) I'm not as familiar with the different notions of derivatives as I would like so that was basically guessing. I think something like $\mathcal{L}_X s := \lim_{t\to 0}\frac{1}{t}(s\circ\phi_t^X - s)$ could work where $\phi_t^X$ is the flow of $X$. (ii) I've been thinking about that as well and I realized that in the special case we don't need to restrict $\alpha$ to $0\times\mathbb{N}^m$, since any derivation in $\mathbb{R}^n\times 0$ can be absorbed into $v_\alpha$ if we'd like. Therefore we could equivalently write $u =\sum_{\alpha} \partial^\alpha \iota(v_\alpha')$. This looks like ... $\endgroup$ Apr 17, 2014 at 16:22
  • $\begingroup$ ... something that might be generalized to arbitrary smooth manifolds (one has to be careful with the dualizing process because one has no longer a canonical density against which one could integrate). Another guess would be that $u = \sum_a \mathcal{L}_{X_{i1}} \cdots \mathcal{L}_{X_{ij_i}} \iota(v_a)$ is true were $X_{ij}$ is now an arbitrary vector field $X_{ij}\in\Gamma(TM)$. But, as I said, this is just a guess. $\endgroup$ Apr 17, 2014 at 16:25
  • $\begingroup$ The hitch in this definition of $\mathcal{L}_Xs$ is that $s(\phi^X_t(m))$ and $s(m)$ live in different fibers of $E$, and so can't be subtracted. However, when $E$ is a bundle of tensors (and $\mathcal{L}_X$ is the vanilla Lie derivative), I now understand that your ansatz for $u$ is indeed the most general $u$ localized along $S$. Thank you again! $\endgroup$
    – Josh Burby
    Apr 17, 2014 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.