Does every vector bundle on a Stein space have a finite local trivialisation?


  • Stein space means either a complex analytic Stein space or a nonarchimedean Stein space in the sense of Kiehl.
  • Vector bundle means either holomorphic vector bundle or rigid analytic vector bundle (= locally free sheaf). Edit: Say, of constant rank $r$.
  • finite local trivialisation = local trivialisation where the covering consists of finitely many sets.

In the nonarchimedean case, assume moreover that the base field is spherically complete or that the Stein space is unbounded.

My thoughts so far: Usually, the space is noncompact, so the answer is probably "no". On the other hand, I could not come up with an example yet. If the Stein space is the analytification of an algebraic variety, then at least those vector bundles that are analytifications of algebraic bundles will have a finite trivialisation. Every vector bundle on a Stein space is determined by finitely many global sections. Could that help? For an example, one should probably use the non-finitely generated ideals in the algebra of global functions of the Stein space. I cannot write down any of them either.

As said, I suspect the answer to be "no". It would be great if somebody knew an example or a reference.

  • $\begingroup$ Why is every vector bundle on a Stein space global generated by finitely many elements, even if the space isn't compact? A vector bundle which is generated by finitely many global sections does indeed have a finite trivialisation. $\endgroup$
    – Ben
    Commented Sep 24, 2015 at 10:48
  • $\begingroup$ @Ben: Are you sure about that? Could you explain why, please? $\endgroup$ Commented Sep 24, 2015 at 11:15
  • $\begingroup$ In the complex analytic case, if the underlying topological $\mathbb{C}$-bundle is trivial, then there exists a global holomorphic trivialization. This is part of the Grauert-Oka principle (or is it Grauert's Oka principle?). $\endgroup$ Commented Sep 24, 2015 at 11:18
  • $\begingroup$ @Ben: For complex Stein spaces, this is Forster. In the nonarchimedean case, Gruson proved it for affine space and the general case is analogous if one uses Kiehl's Cartan theorem B. $\endgroup$ Commented Sep 24, 2015 at 11:36
  • $\begingroup$ @Ben: I thought that one maybe can deduce that every vector bundle has a finite trivialisation by that, but how? If the algebra of global functions was noetherian, it would be clear. But it isn't. $\endgroup$ Commented Sep 24, 2015 at 11:39

3 Answers 3


In the complex analytic world*, every vector bundle which is globally generated by a finite set of global sections has a finite trivialising cover:

Let $X$ be a complex space and let $E$ be a locally free (hence coherent) $\mathcal{O}_X$-module of constant rank $r$, generated by the global sections $s_1,s_2,\dots s_d$. For each subset $I=\{i_1,\dots i_r\}$ of $\{1,\dots d\}$, consider the morphism $\varphi_I\colon\mathcal{O}_X^r\to E$ given by the sections $s_{i_1},\dots s_{i_r}$.

I claim that the subsets $U_I\subset X$ consisting of the points where $\varphi_I$ is an isomorphism are open and cover $X$. Note that if the map on germs $(\varphi_I)_x\colon \mathcal{O}_{X,x}^r\to E_x$ is surjective, then it's already an isomorphism, hence $X-U_I$ is just the support of the cokernel of $\varphi_I$. This is generated by the residue classes of the finitely many sections generating $E$ and so $U_I$ is open for every $I$.

For every $x\in X$, by assumption, the $\mathcal{O}_{X,x}$-module $E_{x}$ is generated by $(s_1)_x,\dots (s_d)_x$. Since it is also free of rank $r$, a subset of $r$ germs is enough to generate $E_x$, and so $x$ is contained in some $U_I$. Thus, the $U_I$ are covering $X$. This proves everything I claimed.

  • The argument works for locally ringed spaces with coherent structure sheaf I guess.
  • 2
    $\begingroup$ I just found out that the OP has proved a theorem along these lines and called it Ben's Theorem, despite the fact that prior to today, I haven't even heard of the spaces involved in the result. I suggest to call it Helene's Theorem instead! After all, global finite generation is the much harder part of the proof, not this basic lemma. $\endgroup$
    – Ben
    Commented Nov 22, 2018 at 20:23

I assume you meant for the fiberwise rank of $E$ to be constant, say $r>0$ (or at least uniformly bounded above). The answer is "yes" when the Stein space has finite dimension (equivalently, when its analytic irreducible components, all of which must be equidimensional via consideration of their connected normalizations, have uniformly bounded dimension). There is no need to make a "global finite generation" hypothesis (which would be hard to check in practice anyway); dimension finiteness of $X$ is entirely sufficient.

In fact, for induction purposes it is better to consider more generally $E$ that is merely a coherent sheaf for which its fibers $E(x)$ at all $x \in X$ have uniformly bounded dimension, say at most some $r>0$. We will show that any such $E$ is generated by finitely many global sections. Then when $E$ is a vector bundle with fiberwise rank $\le r$ we can build a finite trivializing cover governed by where various subsets of size $\le r$ in the global generating set constitute a fiberwise frame. The main content in the argument will be the global theory of analytic irreducible components and their equidimensionality.

We may assume $X$ is non-empty. Let $d \ge 0$ be the dimension of $X$. If $d=0$ then $X$ is topologically discrete (with artinian local ring at each point) and everything is clear. Suppose $d > 0$, and let $\{X_i\}$ be the (locally finite) set of analytic irreducible components of $X$. For each $i$, choose $x_i \in X_i$ not lying in any other $X_j$ (as clearly exists by local finiteness of $\{X_i\}$ in $X$ or many other reasons). Let $\{e_j^{(i)}\}_{1 \le j \le r}$ be a spanning set of the fiber $E(x_i)$.

The set $Z$ of points $\{x_i\}$ in $X$ is discrete, and with the reduced structure we get a closed immersion $h:Z \hookrightarrow X$. Let $s_1, \dots, s_r$ be global sections of $E$ such that $s_j(x_i) = e_j^{(i)}$; this exists because of the Stein property of $X$ and the surjectivity of the map of coherent sheaves $$E \rightarrow h_{\ast}(E|_Z) = \bigoplus_i (h_i)_{\ast}(E(x_i))$$ where $h_i: \{x_i\} \hookrightarrow X$ is the natural closed immersion (surjectivity uses the discreteness of $Z$).

Now consider the natural map $\phi:O_X^r \rightarrow E$ defined by $(a_j) \mapsto \sum a_j s_j$. By design, $\phi$ is surjective between fibers at all points of $Z$, so the coherent sheaf $F = {\rm{coker}}(\phi)$ on $X$ has vanishing stalk at each $z \in Z$. Hence, the coherent ideal sheaf ${\rm{Ann}}_{O_X}(F)$ has associated closed subspace $X' \subset X$ whose intersection with each irreducible $X_i$ is a proper analytic subspace (as $x_i \not\in X'$ for all $i$). Since each irreducible component of $X'$ is contained in one of $X$ (by local finiteness considerations with analytic sets) it is therefore clear that the dimension $d'$ of $X'$ is strictly smaller than $d$. (If we are so lucky that $X'$ is empty then I suppose we declare $d' = -\infty$; whatever.)

By induction on dimension, $F$ viewed as a coherent sheaf on $X'$ (with all fibers of dimension at most $r$) is generated by a finite set of global sections. These lift to global sections of $E$ due to the Stein property of $X$, and together with the $s_j$'s above constitute a finite global generating set of $E$ due to Nakayama's Lemma.


  • $\begingroup$ @HeleneSigloch: You didn't mention the motivation for your question, but in practice how would you verify the hypothesis of finite global generation in Ben's answer without going through the argument I gave? $\endgroup$
    – grghxy
    Commented Sep 27, 2015 at 16:40
  • $\begingroup$ Use Cartan's theorem A. For complex Stein spaces there is an article by Forster. $\endgroup$ Commented Sep 28, 2015 at 7:23
  • $\begingroup$ But is that approach different from my suggested method? I don't see how to produce a finite set of global generators via the Stein hypothesis without essentially going through the above procedure. Theorem A says the coherent sheaf is generated by global sections, but one has to show a finite set of such does the job, and the usual statement of Theorem A doesn't address finiteness (and if one has a reference which does include finiteness then I guess it goes exactly by the argument I have outlined). Or is Forster's argument different? $\endgroup$
    – grghxy
    Commented Sep 30, 2015 at 7:17
  • $\begingroup$ For the exact argument I would also have to look at Forster's paper again. Sorry. $\endgroup$ Commented Sep 30, 2015 at 9:02
  • 1
    $\begingroup$ A published reference for "global finite generation" is [Kripke, Finitely generated coherent analytic sheaves, Theorem 1]. $\endgroup$ Commented Oct 14, 2018 at 9:53

I think that it is also possible to give a positive answer to the question using purely topological arguments, namely Ostrand's theorem: for every paracompact space $X$ of dimension $n$ and every open covering $\mathscr U$ of $X$, there exists $n+1$ families $\mathscr V_1,\dotsc, \mathscr V_{n+1}$ of disjoint open sets whose union is a covering that refines $\mathscr U$.

Apply this to a covering trivializing your vector bundle $E$. Then, for every $i$, $E$ is trivial on the open set that is the union of the elements of $\mathscr V_i$, and you are done.

Note that the argument does use the fact that the space is finite-dimensional, but not so much the Stein hypothesis (maybe for paracompactness).

  • $\begingroup$ If each $\mathcal{V}_i$ is Stein then this does answer the question. I'm not sure this is immediate, though. $\endgroup$
    – David Roberts
    Commented Sep 25, 2015 at 0:09
  • 2
    $\begingroup$ @DavidRoberts: I don't understand your comment. The OP did not require a finite Stein covering, only a finite covering. Anyway, every point has a basis of Stein neighborhoods, so the argument can be adapted to give this too. $\endgroup$ Commented Sep 25, 2015 at 7:04
  • $\begingroup$ I assumed the OP wanted a holomorphic trivialisation, and the opens being Stein would help with that. Just because every trivialising open can be covered by Stein charts, it's not automatic that one finds a finite cover that works. One needs to run through an argument, and the one that constructs a finite cover in the usual case would do. $\endgroup$
    – David Roberts
    Commented Sep 25, 2015 at 7:37
  • $\begingroup$ @Jérôme: Thank you. So if I want to use paracompactness, then in the nonarchimedean case I need to work with Berkovich spaces. I'm beginning to see that they do have advantages for me, too... $\endgroup$ Commented Sep 25, 2015 at 7:43
  • 3
    $\begingroup$ @DavidRoberts: Sorry, but I still fail to see the problem. Maybe it's only in the way I phrased the answer. So, the covering $\mathscr{U}$ I choose at the beginning is made of open sets (hence with a canonical analytic structure) on which the vector bundle is holomorphically trivial. Since, every $\mathscr{V}_i$ is a disjoint union of such open sets, their union $V_i$ is an open set on which the vector bundle is holomorphically trivial. $\endgroup$ Commented Sep 25, 2015 at 8:06

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.