Here is some sort of characterization; it is not really satisfactory but may be useful for further improvements.

Let us talk in terms of nuclei on frames. For a frame $A$ let $\mathbf NA$ be the frame of its nuclei. Then the question is how to characterize those subsets of $A$ which are equal to $\varphi_*j:=\{\ a\in A\ |\ ja=1\ \}$ for some $j\in\mathbf NA$.

Each such $\varphi_*j$ is obviously a filter on $A$. Let further $\mathbf FA$ be the frame of all filters on $A$. Then it is straightforward to check that the map $\varphi_*:\mathbf NA\to\mathbf FA$ preserves all meets, so it has a left adjoint $\varphi^*:\mathbf FA\to\mathbf NA$.

To describe this $\varphi^*$ more explicitly, some shorthand terminology seems to be useful.

Let us say that a filter $\mathscr F$ on $A$ is $\neg\neg\textit{trivial}$ if all its elements are dense, that is, $\forall\ f\in\mathscr F\ \neg f=0$. And let us say that $\mathscr F$ is $\neg\neg\textit{trivial above}$ $a\in A$ if $\mathscr F\cap[a,1]$ is a $\neg\neg$trivial filter on the frame $[a,1]$. Explicitly this means$$
\forall\ f\in\mathscr F\ \ \ f\to a=a.
$$

Now in these terms we may describe, for a filter $\mathscr F$, the nucleus $\varphi^*(\mathscr F)$ by naming the set of its fixed points, which is$$
\mathrm{Fix}(\varphi^*(\mathscr F))=\{\ a\in A\ |\ \mathscr F\textrm{ is $\neg\neg$trivial above }a\ \}.
$$
This then gives the following "characterization" of filters in the image of $\varphi_*$:

A subset $\mathscr F$ of $A$ has form $\{ a\in A\ |\ ja=1\ \}$ for some nucleus $j$ if and only if it is a filter and satisfies$$
\forall\ a\notin\mathscr F\ \exists\ a\leqslant a'\notin\mathscr F\ \textrm{such that $\mathscr F$ is $\neg\neg$trivial above }a'.
$$

islocally "local" . . . – Noah S Sep 15 '12 at 21:07