TITLE: Moduli space of (all) vector bundles on $\mathbb{P}^1$
QUESTION [12 upvotes]: It is well known that, by a theorem of Grothendieck, every vector bundle (always assumed coherent in this question; and everything is over the complex numbers) splits as a direct sum of line bundles.
This is not true in families, though, not even locally on the base of the family, as it is possible to construct flat $1$-parameter families of VB that have generically the same splitting type and "jump" to a different splitting type at the special fiber. This prevents the moduli stack $\mathfrak{M}$ of all VB on $\mathbb{P}^1$ (or the corresponding functor $F$ of isomorphism classes) to be separated (in the sense of valuative criteria). Here we took the notion of isomorphism of families to mean, as usual: the families $\mathscr{E}$ and $\mathscr{E}'$ of VB on $X$ parametrized by $T$ are isomorphic if there is a line bundle $L$ on $T$ such that $\mathscr{E}\simeq\mathscr{E}'\otimes L$.
For a smooth projective curve $X$, the moduli functor $F^s$ of stable bundles is representable by a smooth separated scheme $M^s$, and the stack $\mathfrak{M}^s$ of stable bundles is a gerbe over the latter. The moduli functor $F^{ss}$ of isomorphism classes of semistable bundles has a non separated coarse moduli space with proper connected components, and collapsing $S$-equivalence classes in the sense of Seshadri fixes this: now the functor $F'^{ss}$ of such $S$-equivalence classes of semistable bundles has a coarse moduli space (each connected component of) which is a projective variety (in particular separated).
By Grothendieck's theorem, the relation of S-equivalence on $\mathbb{P}^1$ coincides with the relation of isomorphism, so we can avoid talking of $S$-equivalence in this case. Also, there are no stable VB on $\mathbb{P}^1$ (of rank $>1$).
In what follows the curve $X$ is $\mathbb{P}^1$.
Since every VB is uniquely determined by its splitting type, and there is a countable set of choices for the possible line bundles occurring as summands, each connected component of $M^{ss}$, which is a variety, has to be a reduced point $\mathrm{Spec}(\mathbb{C})$.
Recall we let $\mathfrak{M}$ be the stack in groupoids of all VB on $\mathbb{P}^1$, with isomorphisms of families as $1$-arrows, and $F$ the corresponding functor of isomorphism classes.

Q.1: How does $\mathfrak{M}$ look like?
Q.2: Does $F$ have a coarse moduli space $M$? If yes, how does it look like?

It seems possible to me that an answer to question 1 or 2 has something to do with the partial order structure on the set $\boldsymbol{\mathrm{HNP}}$ of "Harder-Narasimhan polygons" (see Shatz, The decomposition and specialization of algebraic families of vector bundles). Maybe the specialization order on (field-valued) points of $M$ (or of $F$, or of $\mathfrak{M}$) reflects somehow the structure of the lattice $\boldsymbol{\mathrm{HNP}}$?

REPLY [3 votes]: The stack $\mathfrak{M}$ was discussed in section 7 of On the motivic class of the stack of bundles by Behrend and Dhillon, where the same stratification as discussed by Alexander Braverman was described.  Moreover, the various strata are identified with classifying stacks (for the automorphism groups of the corresponding bundles).