## March 27, 2022

### Holomorphic Gerbes (Part 1)

#### Posted by John Baez

I have some guesses about holomorphic gerbes. But I don’t know much about them; what I know is a small fraction of what’s in here:

I recently blogged about the classification of holomorphic line bundles. Since gerbes are a lot like line bundles, it’s easy to guess some analogous results for holomorphic gerbes. I did that, and then looked around to see what people have already done. And it looks like I’m on the right track, though I still have lots of questions.

By ‘line bundle’ I’ll always mean a complex line bundle. These have an elegant definition, but we can describe any smooth line bundle on a smooth manifold $X$ by choosing an open cover $U_i$ of $X$ and smooth functions

$g_{i j} \colon U_i \cap U_j \to \mathbb{C}^\ast$

that are 1-cocycles, meaning

$g_{i j} g_{j k} = g_{i k}$

on all the intersections $U_i \cap U_j \cap U_k$. Here $\mathbb{C}^\ast$ is the nonzero complex numbers, made into a Lie group under multiplication.

A ‘gerbe’ is the next thing up the categorical ladder. Again these have an elegant definition, but we can describe any smooth gerbe by choosing an open cover $U_i$ of our manifold and smooth functions

$h_{i j k} \colon U_i \cap U_j \cap U_k \to \mathbb{C}^\ast$

that are 2-cocycles, meaning

$h_{i j k} h_{i k \ell} = h_{i j \ell} h_{j k \ell}$

These cocycles live in Čech cohomology. Čech cohomology lets us eliminate the dependence on the choice of open cover, and it also gives a way to define isomorphisms between line bundles, and ‘equivalences’ between gerbes. For example, two gerbes are equivalent if their cocycles differ by a coboundary.

Using some facts about Čech cohomology, we can show that line bundles on a topological space $X$ are classified by elements of its second cohomology group $H^2(X,\mathbb{Z})$. Similarly, gerbes are classified by elements of $H^3(X,\mathbb{Z})$.

Clearly there’s a lot more to be said. But I won’t go into more detail here since I’m mainly concerned about what happens when $X$ is a complex manifold. Then we can talk about holomorphic line bundles and holomorphic gerbes. To do this, we simply require that the functions $g_{i j}$ and $h_{i j k}$ are holomorphic. The notions of isomorphism and equivalence are similarly adjusted.

### Holomorphic line bundles

A smooth line bundle can often be made into a holomorphic line bundle in different ways. So for a line bundle to be holomorphic is not just an extra property, it’s an extra structure.

But we can factor this structure into its property part — can we put a holomorphic structure on this line bundle? — and its ‘pure structure’ part — if we can, what’s the set of ways we can do it? And perhaps surprisingly, if the answer to the first question is yes, the answer to the second question is always the same!

This is surprising if you’re used to structures like groups. The question can we put a group structure on this set? has the answer yes whenever the set is nonempty. But knowing just that doesn’t tell us much about how many ways we can make the set into a group!

But consider orientations of connected manifolds. If the question can we put an orientation on this connected manifold? has the answer yes, we say that manifold is orientable. And if that’s the case, there are always just two orientations!

Something like this, but a lot more interesting, happens for putting holomorphic structures on smooth line bundles. There’s a short exact sequence

$0 \to \mathrm{Jac}(X) \to \mathrm{Pic}(X) \to \mathrm{NS}(X) \to 0$

where:

• The Picard group $\mathrm{Pic}(X)$ is the set of isomorphism classes of holomorphic line bundles on $X$. It’s an abelian group, since we can tensor holomorphic line bundles and get new ones.

• The Néron–Severi group $\mathrm{NS}(X)$ is the property part of $\mathrm{Pic}(X)$. That is, it’s the group of isomorphism classes of smooth line bundles with the property that they can be given a holomorphic structure.

• The Jacobian $\mathrm{Jac}(X)$ is the pure structure part of $\mathrm{Pic}(X)$. That is, it’s the group of holomorphic structures on the trivial smooth line bundle.

The exact sequence above says a couple of things. Two elements of $\mathrm{Pic}(X)$ map to the same element of $\mathrm{NS}(X)$ iff they come from holomorphic line bundles whose underlying smooth line bundles are isomorphic. And if a smooth line bundle has any holomorphic structures at all, it has $\mathrm{Jac}(X)$ different ones. More precisely, its set of holomorphic structures is a torsor for $\mathrm{Jac}(X)$.

A torsor is a ‘group that has forgotten its identity’. More precisely, it’s a nonempty set on which a given group acts freely and transitively. Whenever we have a short exact sequence of groups

$0 \to N \stackrel{i}{\longrightarrow} G \stackrel{p}{\longrightarrow} H \to 0$

and $h \in H$ is in the image of $p$, its preimage $p^{-1}(h)$ is a torsor for $N$. Amusingly, even when $h$ is not in the image of $p$, $N$ acts freely and transitively on $p^{-1}(h)$. That’s because any group acts freely and transitively on the empty set! But the empty set is not considered to be a torsor.

I hope you see how this fits into the idea of taking a structure and factoring it into a property and ‘pure structure’. We’re wondering how to equip $h \in H$ with the structure of being $p(g)$ for some $g \in G$. But we can break this down into two questions: does $h$ have the property that it’s in the image of $p$? and then if it does, what’s the set of things that maps to it? If the answer to the first question is yes, the answer to the second question is always “an $N$-torsor”. And $N$-torsors are all isomorphic!

Another cool thing about the exact sequence

$0 \to \mathrm{Jac}(X) \to \mathrm{Pic}(X) \to \mathrm{NS}(X) \to 0$

is that you can read it off just from knowing that $\mathrm{Pic}(X)$ is a complex Lie group. Its connected component containing the identity is $\mathrm{Jac}(X)$. And its set of connected components is $\mathrm{NS}(X)$. For a tiny bit more about this, see

And another cool thing is that we can figure out the Jacobian $\mathrm{Jac}(X)$ pretty easily just knowing the topology of $X$. It’s a torus, and this torus is the real cohomology $H^1(X,\mathbb{R})$ modulo the image of the integral cohomology $H^1(X, \mathbb{Z})$. For short:

$\mathrm{Jac}(X) \cong \frac{H^1(X,\mathbb{R})}{H^1(X, \mathbb{Z})}$

This description does not immediately reveal the fact that $\mathrm{Jac}(X)$ is a complex torus; for that, see my blog article above.

The Néron–Severi group is also connected to cohomology: since $H^2(X,\mathbb{Z})$ classifies smooth line bundles on $X$, while $\mathrm{NS}(X)$ classifies smooth line bundles with the property that they come from holomorphic line bundles, we have

$\mathrm{NS}(X) \subseteq H^2(X,\mathbb{Z})$

But this is not a full description of the Néron–Severi group! To go further it seems we need to focus on some special cases, like complex tori, where Appell–Humbert theorem comes to our rescue. If someone knows more general tricks for computing the Néron–Severi group, please let me know.

### Holomorphic gerbes

I suspect that everything I just said has a very nice analogue for gerbes. This paper helps a lot:

and I’ve also found a lot of information and references here:

But I haven’t yet absorbed all the information and tracked down all the references, and it’s a lot faster to make some guesses. So let me state these guesses, and maybe some of you can confirm or correct them.

First, some definitions! Two smooth gerbes are called ‘equivalent’ if the Čech 3-cocycles defining them are cohomologous, and ditto for holomorphic gerbes.

Definition. The gerby Picard group $\mathbf{Pic}(X)$ is the group of equivalence classes of holomorphic gerbes on $X$.

It’s an abelian group since we can tensor holomorphic gerbes and get new ones. This corresponds to multiplying the Čech 3-cocycles defining them.

Definition. The gerby Néron–Severi group $\mathbf{NS}(X)$ is the group of equivalence classes of smooth gerbes with the property that they can be given a holomorphic structure.

Definition. The gerby Jacobian $\mathbf{Jac}(X)$ is the group of equivalence classes of holomorphic gerbes that are topologically trivial.

Conjecture 1. There is a short exact sequence

$0 \to \mathbf{Jac}(X) \to \mathbf{Pic}(X) \to \mathbf{NS}(X) \to 0$

coming from the natural inclusion of $\mathbf{Jac}(X)$ in $\mathbf{Pic}(X)$.

This seems easy. (I should emphasize that I’m not trying to stake out territory or say anything amazing, just state some precise guesses.)

Conjecture 2. The gerby Picard group $\mathbf{Pic}(X)$ can be made into a complex Lie group in such a way that $\mathbf{Jac}(X)$ is the identity component and $\mathbf{NS}(X)$ is the group of connected components.

Ben-Bassat has shown this when $X$ is a complex torus, but he goes much further and describes $\mathbf{Jac}(X)$ and $\mathbf{NS}(X)$ quite explicitly, mimicking the Appell–Humbert theorem, which does the same for line bundles. In general I’m hoping the higher Picard group has the structure of a Lie group, just like the Picard group does, and this should give it the desired topology.

Conjecture 3. The gerby Néron–Severi group $\mathbf{NS}(X)$ is a subgroup of $H^3(X,\mathbb{Z})$.

This seems obvious.

Conjecture 4. The gerby Jacobian is given by

$\mathbf{Jac}(X) \cong \frac{H^2(X,\mathbb{R})}{H^2(X, \mathbb{Z})}$

Ben-Bassat almost comes out and says this in the case where $X$ is a complex torus, in his Theorem 3. But it’s disguised by the fact that he uses a formula for the 2nd cohomology that works in this case.

### Holomorphic n-gerbes

As the turtle once said: while I’m sticking my neck out, let me stick it out all the way. We can define $n$-gerbes for any $n$, for example using higher Čech cohomology groups or your favorite approach to higher categories. And the patterns I’m discussing want to be continued to higher $n$.

My basic intuition is that while a line bundle has a set that’s a copy of $\mathbb{C}$ sitting over each point of our space $X$, the kind of gerbe I’m talking about has a category with one object and a copy of $\mathbb{C}$ as its morphisms, while a 2-gerbe has a 2-category with one object, one morphism and a copy of $\mathbb{C}$ as its 2-morphisms, and so on. But to actually work with these things I’ll use Čech cohomology.

At this point it’s good to introduce more systematic notation. I used $\mathrm{Pic}(X)$ for the usual Picard group and $\mathbf{Pic}(X)$ for its gerby analogue, but I can’t use bolder boldface for $2$-gerbes and so on. Besides, the whole pattern “line bundle, gerbe, 2-gerbe…” isn’t starting at the base case, and line bundles are themselves a categorification of complex-valued functions, so the numbering is all screwed up. Let’s try this:

Definition. Given a complex manifold $X$, its $n$th Picard group $\mathrm{Pic}_n(X)$ is $n$th Čech cohomology group of its sheaf of nonzero holomorphic complex functions.

This is the usual Picard group when $n = 1$, and the gerby Picard group when $n = 2$. For higher $n$ it’s supposed to be the group of equivalence classes of holomorphic $(n-1)$-gerbes.

Definition. The $n$th Néron–Severi group $\mathrm{NS}_n(X)$ of a complex manifold $X$ is the subgroup of $H^{n+1}(X)$ that’s the image of the natural map

$\mathrm{Pic}_n(X) \stackrel{p}{\longrightarrow} H^{n+1}(X,\mathbb{Z})$

This natural map comes from a long exact sequence in Čech cohomology called the exponential sequence. The $n$th Néron–Severi group is the usual Néron–Severi group when $n = 1$, and the gerby one when $n = 2$. For higher $n$ it’s supposed to be the group of equivalence classes of smooth $(n-1)$-gerbes that admit a holomorphic structure.

Definition. The $n$th Jacobian $\mathrm{Jac}_n(X)$ is the kernel of the natural map

$\mathrm{Pic}_n(X) \stackrel{p}{\longrightarrow} H^{n+1}(X,\mathbb{Z})$

This is the usual Jacobian of $X$ when $n = 1$, and the gerby one when $n = 2$. For higher $n$ it’s supposed to be the group of equivalence classes of holomorphic structures on the trivial $(n-1)$-gerbe.

Just by definition, we get a short exact sequence

$0 \to \mathrm{Jac}_n(X) \stackrel{i}{\longrightarrow} \mathrm{Pic}_n(X) \stackrel{p}{\longrightarrow} \mathrm{NS}_n(X) \to 0$

So the higher analogue of Conjecture 1 is true by definition. But this one is not quite so automatic:

Conjecture 2${}^\prime$. The $n$th Picard group $\mathrm{Pic}_n(X)$ of a complex manifold can be made into a complex Lie group in such a way that $\mathrm{Jac}_n(X)$ is the identity component and $\mathrm{NS}_n(X)$ is the group of connected components.

The $n$th Néron–Severi group $\mathrm{NS}_n(X)$ is a subgroup of $H^{n+1}(X,\mathbb{Z})$ by definition, so the higher analogue of Conjecture 3 holds. But this one is more speculative:

Conjecture 4${}^\prime$. The $n$th Jacobian of a complex manifold $X$ is given by

$\mathrm{Jac}_n(X) \cong \frac{H^{n}(X,\mathbb{R})}{H^{n}(X, \mathbb{Z})}$

I don’t know if the number $n$ is right here. But for $n = 1$ this gives a result that matches the usual Jacobian of $X$, and for $n = 2$ this matches Ben-Bassat’s calculation of the gerby Jacobian in the special case where $X$ is a complex torus.

In general, this sort of quotient is called an intermediate Jacobian, and there’s a lot of material about intermediate Jacobians and $n$-gerbes on the nLab page on intermediate Jacobians. There’s also some relation, unclear to me so far, between this conjecture and Theorem 1.5.11 in Brylinski’s Loop Spaces, Characteristic Classes and Geometric Quantization. Brylinski credits his Theorem 15.11 to Deligne but cites this reference:

• Michael Rapoport, Peter Schneider, and Norbert Schappacher, eds., Beilinson’s Conjectures on Special Values of L-functions, Academic Press, New York, 2014.

and especially the paper by Esnault and Viehweg in this volume. So I should read that.

If Conjecture 4 is true, the $n$th Jacobian is obviously a torus. But in fact Deligne came up with a way to make intermediate Jacobians into complex tori, and Griffiths came up with a different way. One of these ways should be the ‘right’ way if the $n$th Picard group is a complex Lie group and the $n$th Jacobian is the identity component of that.

So, there’s a lot left for me to think about here!

Posted at March 27, 2022 6:57 AM UTC

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### Re: Holomorphic Gerbes (Part 1)

I won’t answer your conjectures directly here for now (and I’m not sure that I immediately could anyway!), but instead I’ll point you to some things which I think might be interesting reading for you :-)

First of all, the paper that you mention by Esnault and Viehweg is “the” reference for Deligne cohomology, and if you care about these intermediate Jacobians and $n$-gerbes, then this is the sort of language I think you’ll gain a lot from. If you want a speedy introduction to this, then I’ll shamelessly (actually, a bit shamefully, because they’re not very good, and I’m not an expert) suggest some slides from a talk I gave last year.

I also think Markus Upmeier’s “The Canonical 2-Gerbe of a Holomorphic Vector Bundle” is a really good reference (it’s short but super detailed), and “exhibit[s] the precise relationship between holomorphic and smooth gerbes”.

What I’d like to know is why you’re thinking about this sort of stuff! Even though you haven’t asked me this question, let me answer it here anyway ;-) One of the goals of my thesis was to construct “the best possible” Chern classes for coherent analytic sheaves, “in a categorical way”. It turns out that “the best possible” can be argued to mean “in Deligne cohomology”, and that “in a categorical way” can be argued to mean “using some simplicial generalisation of bundle gerbes and n-gerbes”. Somehow I’ve managed to get this many years into my “career” without really learning about complex tori though, so clearly our paths are not exactly the same!

Posted by: Tim Hosgood on March 28, 2022 1:04 PM | Permalink | Reply to this

### Re: Holomorphic Gerbes (Part 1)

Thanks for all those references! I’d never guessed you were studying holomorphic gerbes until you came out and told me the other day. If you’re studying Deligne cohomology and $n$-gerbes, it sounds like you’ll inevitably need to learn what Urs Schreiber is doing with those things.

As for me, I’m studying algebraic geometry as a kind of retirement hobby, the way other people take up ballroom dancing or golf. I’m having conversations with James Dolan once a week on Zoom, and by email, and he raises lots of interesting questions that make me want to study different subjects and maybe invent a bit of new math here and there.

We started by trying to understand theta functions on abelian varieties. This made me learn the Appell–Humbert classification of line bundles on complex tori, which is nicely explained in Birkenhake and Lange’s Complex Abelian Varieties. The beautiful examples here got me pulled into studying the connection between Néron–Severi groups, Jordan algebras, and Coxeter groups. There’s a lot of stuff I want to straighten out here.

But then James asked about how all this stuff generalizes from line bundles to gerbes and $n$-gerbes, so I started reading about that — mainly in Ben-Bassat’s paper, which proves an Appell–Humbert-like classification of gerbes on complex tori. I wanted to guess how that generalizes to $n$-gerbes, but it turned out I needed to make a bunch of preliminary conjectures to even state those guesses, so that’s where this post came from!

In short, I’m just messing around, trying to learn a bunch of fun stuff. My actual ‘work’ these days is applying category theory to help create software for compositional models of epidemiology. As you may know, I’m working on that with a gang of people at the Topos Institute — including real live programmers and epidemiologists.

Posted by: John Baez on March 28, 2022 8:38 PM | Permalink | Reply to this

### Re: Holomorphic Gerbes (Part 1)

John wrote:

If someone knows more general tricks for computing the Néron–Severi group, please let me know.

‘Computing’ is not the right word for it, but I just finally realized — I think — that the Néron–Severi group $NS(X)$ is isomorphic to the intersection of the image of

$H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$

and the Dolbeault cohomology group $H^{1,1}(X) \subseteq H^2(X,\mathbb{C})$. Is this correct? If so, why haven’t I seen anyone come out and say it?

(Probably because it’s an easy consequence of famous things and because I haven’t read enough stuff. But it sure can be confusing to beginners when ‘obvious’ facts aren’t spelled out.)

On a related note, is there a standard name for the intersection of the image of

$H^{p+q}(X,\mathbb{Z}) \to H^{p+q}(X,\mathbb{C})$

and the Dolbeault cohomology group $H^{p,q}(X) \subseteq H^{p+q}(X,\mathbb{C})$?

Posted by: John Baez on March 28, 2022 9:15 PM | Permalink | Reply to this

### Neron-Severi

I just finally realized — I think — that the Néron–Severi group $NS(X)$ is isomorphic to the intersection of the image of $H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$ and the Dolbeault cohomology group $H^{1,1}(X) \subseteq H^2(X,\mathbb{C})$. Is this correct? If so, why haven’t I seen anyone come out and say it?

Well, modulo $tors(NS(X))\subset NS(X)$, it’s certainly correct.

For the canonical (pardon the pun) example of a torsion class in $NS(X)$, consider the canonical line bundle, $K_S$ of an Enriques surface, $S$.

Posted by: Jacques Distler on March 29, 2022 8:53 AM | Permalink | PGP Sig | Reply to this

### Re: Holomorphic Gerbes (Part 1)

As with most theorems in algebraic geometry, I imagine that this will be true if you stick enough adjectives in front of whatever $X$ is. Here, for example, it seems to be the case that this holds if $X$ is projective (and a weaker “up to torsion” version holds if $X$ is merely compact Kähler) — see the Lefschetz theorem on (1,1)-classes or this MSE question.

Posted by: Tim Hosgood on March 29, 2022 12:34 PM | Permalink | Reply to this

### Re: Holomorphic Gerbes (Part 1)

Great — thanks Jacques and Tim! I want to understand why the results you mentioned are true, but at least now I have some leads.

Yeah, I’ve been thinking about complex tori where the Néron–Severi group has no torsion, so we get

$NS(X) = H^2(X, \mathbb{Z}) \cap H^{1,1}(X) \qquad \qquad (1)$

I have another related question. Ben-Bassat gives a similar description of the gerby Néron–Severi group, which I called $NS_2(X)$ above, in the case where $X$ is a complex torus.

He reminds us that the exponential sequence gives us a map

$H^n(X, \mathcal{O}^\ast) \to H^{n+1}(X, \mathbb{Z})$

and then he says

… since $X$ is a Kähler manifold, the Hodge decomposition tells us that the image of $H^2(X, \mathcal{O}^\ast)$ in $H^3(X,\mathbb{Z})$ is the kernel of the map $H^3(X, \mathbb{Z}) \to H^3(X, \mathcal{O})$, which is the following intersection:

$H^3(X, \mathbb{Z}) \cap (H^{1,2}(X) \oplus H^{2,1}(X))$

So, at least for a complex torus he’s asserting that the gerby Néron-Severi group

$NS_2(X) = im [ H^2(X, \mathcal{O}^\ast) \to H^3(X,\mathbb{Z})]$

is given by

$NS_2(X) \cong H^3(X, \mathbb{Z}) \cap (H^{1,2}(X) \oplus H^{2,1}(X)) \qquad \qquad (2)$

But he doesn’t say why the image of $H^2(X, \mathcal{O}^\ast)$ in $H^3(X,\mathbb{Z})$ is this intersection: he seems to be using some well-known fact that I don’t know.

I really want to know what’s going on, and how equations (1) and (2) generalize for

$NS_n(X) = im[H^n(X, \mathcal{O}^\ast) \to H^{n+1}(X,\mathbb{Z})]$

Like for the next case: when $X$ is a complex torus do we have

$NS_3(X) \cong H^4(X, \mathbb{Z}) \cap H^{2,2}(X)$

or

$NS_3(X) \cong H^3(X, \mathbb{Z}) \cap (H^{1,3}(X) \oplus H^{2,2}(X) \oplus H^{3,1}(X))$

or what? (And hopefully for some more general class of spaces, at least mod torsion.)

I feel this is some well-understood issue that I’ve had no luck finding references about yet.

Posted by: John Baez on March 29, 2022 10:54 PM | Permalink | Reply to this

### Re: Holomorphic Gerbes (Part 1)

I’m afraid that the moment the words “Hodge theory” appear in the same sentence as “Kähler” my brain just stops working, so I’m of no help here at all (which is sad, because “exponential exact sequence” is one of my favourite things to hear!). I would be almost certain that you would find answers in Claire Voisin’s two (fundamental) books: Hodge Theory and Complex Algebraic Geometry I and Hodge Theory and Complex Algebraic Geometry II (and almost certainly in the former).

Posted by: Tim Hosgood on March 31, 2022 2:27 PM | Permalink | Reply to this

### Francis

I think you can argue as follows (when $X$ is compact and Kahler):

We have the exponential sequence

(1)$0 \to \mathbb{Z} \to \mathcal{O} \to \mathcal{O}^* \to 0$

which induces a long exact sequence in cohomology. Therefore

(2)$NS_{n}(X) = \ker[H^{n+1}(X, \mathbb{Z}) \to H^{n+1}(X, \mathcal{O})].$

The map $H^{n+1}(X, \mathbb{Z}) \to H^{n+1}(X, \mathcal{O})$ factors:

(3)$H^{n+1}(X, \mathbb{Z}) \to H^{n+1}(X, \mathbb{C}) \to H^{n+1}(X, \mathcal{O}).$

By Hodge theory we have the decomposition

(4)$H^{n+1}(X, \mathbb{C})= \bigoplus_{p + q = n+1} H^{(p,q)}(X),$

and $H^{n+1}(X, \mathcal{O}) \cong H^{(0,n+1)}(X)$. Then using the Dolbeault resolution (for example), we can see that the map

(5)$H^{n+1}(X, \mathbb{C}) \to H^{(0,n+1)}(X)$

is just given by the projection, and the kernel is therefore

(6)$\bigoplus_{p + q = n+1, p \neq 0} H^{(p,q)}(X).$

The kernel of the original map is therefore (ignoring torsion)

(7)$H^{n+1}(X, \mathbb{Z}) \cap (\bigoplus_{p + q = n+1, p \neq 0} H^{(p,q)}(X)).$

Note that the elements of $H^{n+1}(X, \mathbb{Z})$ are real, and so cannot be purely of type $(n+1,0)$, so we can just throw away this summand above.

Posted by: Francis on March 31, 2022 2:54 PM | Permalink | Reply to this

### Re: Holomorphic Gerbes (Part 1)

Thanks, Francis! This is great! I’m so happy!

And what I really like is that it doesn’t use any massive sledgehammers that I’m completely ignorant of.

Posted by: John Baez on March 31, 2022 5:24 PM | Permalink | Reply to this

### Francis

Another reference you might find useful is :

Specifically, Sections 1.2 and 1.3 discuss holomorphic and Hermitian structures on gerbes.

Posted by: Francis on April 1, 2022 12:23 PM | Permalink | Reply to this

### Francis

Another reference you might find useful is : https://arxiv.org/pdf/1007.3485.pdf

Specifically, Sections 1.2 and 1.3 discuss holomorphic and Hermitian structures on gerbes.

Posted by: Francis on April 1, 2022 12:23 PM | Permalink | Reply to this

### Re: Holomorphic Gerbes (Part 1)

I should add the comment that the topological and holomorphic classification of line bundles and gerbes (etc) are the same over Stein manifolds, because of vanishing theorems for certain sheaf cohomology groups. This is very far from the case of compact complex manifolds, but if you are thinking about various edge cases, it’s good to know.

Posted by: David Roberts on March 29, 2022 1:42 AM | Permalink | Reply to this
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