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 $\boldsymbol{\mathrm{HNP}}$?

  • 8
    $\begingroup$ The theorem that you attribute to Grothendieck was first proved by Del Pezzo and Bertini (cf. the article of Harris on varieties of minimal degree). It was then rediscovered by Birkhoff. Grothendieck proved a vast generalization of this theorem. There is a beautiful article of Martens and Thaddeus on Grothendieck's theorem and its extensions. $\endgroup$ Apr 1, 2016 at 18:33
  • 3
    $\begingroup$ For any (split) reductive group $G$ with maximal torus $T$ and Weyl group $\mathcal{W}$, Grothendieck proves that the set of $G$-torsors over $\mathbb{P}^1$ is naturally in bijection with the quotient of the cocharacter lattice of $T$, $\text{Hom}_{\textbf{GpSch}_k}(\mathbb{G}_{m,k},T)/\mathcal{W}$, where the $\mathcal{W}$-action is the natural action on $T$. Up to choosing a Weyl chamber of the cocharacter lattice, the specialization order should be something like adding the positive coroots. $\endgroup$ Apr 1, 2016 at 18:48

2 Answers 2


Well, it is not clear to me what exactly you mean by Q.1 but here is some kind of an answer. Let's even work with $G$-bundles for any reductive group $G$ (you can take $G=GL(n)$ if you want, but that's not necessary). Consider the quotient $G((t))/G[t^{-1}]$. It has a natural scheme structure of infinite type. This scheme is isomorphic to the scheme of $G$-bundles on $\mathbb P^1$ endowed with a trivialization at the formal neighborhood of $\infty$. It is also called the thick affine Grassmannian of $G$, so we'll denote it by $Gr_G$. Now the stack of $G$-bundles on $\mathbb P^1$ is obviously the quotient $G[[t]]\backslash Gr_G$. The isomorphism classes of $G$-bundles correspond to $G[[t]]$-orbits on $Gr_G$ which are in one-to-one correspondence with dominant coweights of $G$ (this is the analog of Grothendieck's theorem for any $G$). So, for any such coweight $\lambda$ there is a stratum $\mathcal M^{\lambda}$ in $\mathcal M$ and we have $\mathcal M^{\mu}\subset \overline{\mathcal M^{\lambda}}$ iff $\lambda\leq \mu$ with respect to the standard partial order on coweights ($\lambda\leq \mu$ means that $\mu-\lambda$ is a sum of positive coroots). The geometry of $\overline{\mathcal M^{\lambda}}$ is pretty complicated (for example, it is quite singular). In fact, the closures of $G[[t]]$-orbits on $Gr_G$ are special examples of affine Schubert varieties.

  • $\begingroup$ But, in the case of $GL(n)$, doesn't Grothendieck 's theorem imply some sort of "discreteness"? For example, does every connected component of $Bun_n(\mathbb{P}^1$ have a unique closed point? What is the dimension of such components anyway (upon, say, also fixing the degree of bundles)? If this turns out to be minus the dimension of its generic stabilizer, for example, I would count this as some sort of -quotes- "discreteness" property... $\endgroup$
    – Qfwfq
    Apr 2, 2016 at 15:37
  • $\begingroup$ Also, does the functor of isomorphism classes have a coarse moduli space? (still talking of GL(n) ) $\endgroup$
    – Qfwfq
    Apr 2, 2016 at 15:38
  • $\begingroup$ No, in the case of $GL(n)$ the connected components are in one-to-one correspondence with integers. More precisely, isomorphism classes of rank $n$ bundles correspond to $n$-tuples $(k_1,\cdots, k_n)$ of integers and the connected component depends only on the sum $k_1+\cdots + k_n$. For example, if you take the connected component of the trivial bundle (i.e. all bundles when the sum of the $k_i$'s is 0) then the trivial bundle is open (and dense) there. There can't be a coarse moduli space since every component is a finite-dimensional stack of with an infinite stratification. $\endgroup$ Apr 2, 2016 at 20:58
  • $\begingroup$ Just to understand: what is the dimension of the connected component of the trivial bundle in the stack? and what is the dimension of the biggest stabilizer (I would say $n^2$)? $\endgroup$
    – Qfwfq
    Apr 3, 2016 at 9:49
  • $\begingroup$ For which reason does the presence of an infinite stratification on a finite dim stack prevent the functor of isom classes from having a (possibly badly non separated) coarse moduli space? $\endgroup$
    – Qfwfq
    Apr 3, 2016 at 9:57

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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