## April 10, 2022

### Holomorphic Gerbes (Part 2)

#### Posted by John Baez

Thanks to some help from Francis, I’ve converted some of my conjectures on the classification of holomorphic $n$-gerbes into theorems!

Remember the stars of the show:

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

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

$\mathrm{Pic}_n(X) \xrightarrow{c_n} H^{n+1}(X,\mathbb{Z})$

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

$\mathrm{Pic}_n(X) \xrightarrow{c_n} H^{n+1}(X,\mathbb{Z})$

We can think of these groups in terms of $(n-1)$-gerbes — or even better for the numbering scheme, complex line $n$-bundles. A line 1-bundle is just a line bundle, a line 2-bundle corresponds to a gerbe, and so on. If we take all line $n$-bundles as complex by default, here is the significance of these groups:

• $\mathrm{Pic}_n(X)$ is the group of equivalence classes of holomorphic line $n$-bundles on $X$.

• $H^{n+1}(X,\mathbb{Z})$ is the group of equivalence classes of smooth line $n$-bundles on $X$.

• $\mathrm{NS}_n(X)$ is the group of equivalence classes of smooth line $n$-bundles on $X$ that admit a holomorphic structure.

• $\mathrm{Jac}_n(X)$ is the group of equivalence classes of holomorphic structures on the trivial line $n$-bundle.

I tried to rapidly give some intuition for these ideas last time.

Smooth line $n$-bundles are classified up to equivalence by their Chern class, which is an element of $H^{n+1}(X,\mathbb{Z})$. Above I’m just identifying an equivalence class of line $n$-bundles with its Chern class, to avoid a digression into the theory of line $n$-bundles. There’s a map

$\mathrm{Pic}_n(X) \xrightarrow{c_n} H^{n+1}(X,\mathbb{Z})$

sending each holomorphic line $n$-bundle to its Chern class, and this map arises as follows. The exponential sequence is a short exact sequence of sheaves on $X$:

$0 \xrightarrow{\phantom{\mathrm{exp}(2 \pi}} \mathbb{Z} \xrightarrow{\phantom{\mathrm{exp}(2 \pi}} \mathcal{O} \xrightarrow{\mathrm{exp}(2 \pi i -)} \mathcal{O}^\ast \xrightarrow{\phantom{\mathrm{exp}(2 \pi}} 0$

and it gives a long exact sequence

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

where the connecting homomorphism is the Chern class $c_n$.

You may wonder why holomorphic line $n$-bundles are classified by $H^n(X,\mathcal{O}^\ast)$ while smooth ones are classified by something that looks so different, namely $H^{n+1}(X,\mathbb{Z})$. In fact smooth line $n$-bundles work a lot like holomorphic ones: they’re classified by elements of the $n$th Čech cohomology of the sheaf of nonzero smooth complex functions on $X$! But in the smooth case this Čech cohomology group is isomorphic to $H^{n+1}(X,\mathbb{Z})$, thanks to the smooth version of the long exact sequence above. The reason is that $H^n$ of the sheaf of smooth complex functions vanishes for $n \ge 1$, because it’s a fine sheaf.

Now let’s prove three theorems describing the $n$th Picard group, Jacobian and Néron–Severi group:

Theorem 1. For any complex manifold $X$ there is a short exact sequence

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

Theorem 2. For any complex manifold $X$ we have

$\mathrm{Jac}_n(X) \cong \frac{H^{n}(X,\mathcal{O})}{ \mathrm{im}( H^n(X,\mathbb{Z}) \to H^{n}(X,\mathcal{O}) ) }$

Theorem 3. When $X$ is a compact Kähler manifold,

$\mathrm{NS}_n(X) \cong H^{n+1}(X,\mathbb{Z})_T \oplus \big[\mathrm{im}\big(H^{n+1}(X,\mathbb{Z}) \to H^{n+1}(X,\mathbb{C})\big) \; \cap \bigoplus_{p+q = n+1, \; p,q \ge 1} H^{p,q}(X) \big]$

Here $H^{n+1}(X,\mathbb{Z})_T$ is the torsion subgroup of $H^{n+1}(X,\mathbb{Z})$, while $H^{p,q}(X)$ are the Doulbeault cohomology groups of $X$.

The first theorem is so easy that it doesn’t deserve to be called a theorem! But I’m short on theorems today, so:

Theorem 1. For any complex manifold $X$ there is a short exact sequence

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

Proof. Use the map $c_n$ to form a short exact sequence:

$0 \to \mathrm{ker}(c_n) \to H^n(X,\mathcal{O}^\ast) \xrightarrow{c_n} \mathrm{im}(c_n) \to 0$

Then note that

$\begin{array}{ccl} \mathrm{ker}(c_n) &=& \mathrm{Jac}_n(X) \\ \\ H^n(X,\mathcal{O}^\ast) &=& \mathrm{Pic}_n(X) \\ \\ \mathrm{im}(c_n) &=& \mathrm{NS}_n(X) \end{array}$

all by definition.    ⬛

The second result follows easily from the long exact version of the exponential sequence:

Theorem 2. For any complex manifold $X$ we have

$\mathrm{Jac}_n(X) \cong \frac{H^{n}(X,\mathcal{O})}{ \mathrm{im}( H^n(X,\mathbb{Z}) \to H^{n}(X,\mathcal{O}) ) }$

Proof. Let’s give the maps in the exponential long exact sequence some names:

$\cdots \to H^n(X,\mathbb{Z}) \xrightarrow{i_n} H^{n}(X,\mathcal{O}) \xrightarrow{e_n} H^n(X,\mathcal{O}^\ast) \xrightarrow{c_n} H^{n+1}(X,\mathbb{Z}) \to \cdots$

By definition $\mathrm{Jac}_n(X)$ is the kernel of $c_n$, so it’s also the image of $e_n$, so it’s also isomorphic to

$\frac{H^{n}(X,\mathcal{O}) }{\mathrm{ker}(e_n)}$

which is the same as

$\frac{H^{n}(X,\mathcal{O}) }{\mathrm{im}(i_n)}$

This is what we wanted to show.    ⬛

The third result is a bit more substantial: it uses Hodge theory to describe the Néron–Severi group for line $n$-bundles on a compact Kähler manifold. This result is well-known for $n = 1$ and shown by Ben–Bassat for $n = 2$.

Theorem 3. When $X$ is a compact Kähler manifold,

$\mathrm{NS}_n(X) \cong H^{n+1}(X,\mathbb{Z})_T \oplus \big[\mathrm{im}\big(H^{n+1}(X,\mathbb{Z}) \to H^{n+1}(X,\mathbb{C})\big) \; \cap \bigoplus_{p+q = n+1, \; p,q \ge 1} H^{p,q}(X) \big]$

Proof. Remember the exponential long exact sequence:

$\cdots \xrightarrow{e_n} H^n(X,\mathcal{O}^\ast) \xrightarrow{c_n} H^{n+1}(X,\mathbb{Z}) \xrightarrow{i_{n+1}} H^{n+1}(X,\mathcal{O}) \xrightarrow{e_{n+1}} \cdots$

By definition $\mathrm{NS}_n(X) = \mathrm{im}(c_n)$, so it’s also $\mathrm{ker}(i_{n+1})$. But

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

factors as

$H^{n+1}(X, \mathbb{Z}) \xrightarrow{f} H^{n+1}(X, \mathbb{C}) \xrightarrow{g} H^{n+1}(X, \mathcal{O})$

where $f$ comes from change of coefficients. Since $H^{n+1}(X, \mathbb{Z})$ is a finitely generated abelian group, it splits as a sum of its torsion subgroup $H^{n+1}(X, \mathbb{Z})_T$ and a finitely generated free abelian group. The kernel of $f$ above is the torsion subgroup, so

$\mathrm{ker}(i_{n+1}) \cong H^{n+1}(X, \mathbb{Z})_T \oplus \left[ \mathrm{im} (f) \cap \mathrm{ker}(g) \right]$

But by Hodge theory we have the decomposition

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

and we can identify $H^{n+1}(X, \mathcal{O})$ with $H^{(0,n+1)}(X)$. Using the Dolbeault resolution we can see that

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

is projection onto this summand, so

$\mathrm{ker} (g) = \bigoplus_{p + q = n+1, p \neq 0} H^{(p,q)}(X)$

Putting this all together we get

$\mathrm{NS}_n(X) \cong H^{n+1}(X, \mathbb{Z})_T \oplus \left[ \mathrm{im} (i_{n+1}) \cap \bigoplus_{p + q = n+1, p \neq 0} H^{(p,q)}(X) \right]$

But $\mathrm{im}(i_{n+1})$ consists of real cohomology classes, and complex conjugation on complex de Rham cohomology switches $H^{(p,q)}(X)$ and $H^{(q,p)}(X)$, and nothing in the intersection above can lie in $H^{n+1,0}(X)$, so nothing can lie in $H^{0,n+1}(X)$ either. So we can refine our description to

$\mathrm{NS}_n(X) \cong H^{n+1}(X, \mathbb{Z})_T \oplus \left[ \mathrm{im} (i_{n+1}) \cap \bigoplus_{p + q = n+1, p,q \neq 0} H^{(p,q)}(X) \right]$

which is what we were trying to prove!    ⬛

The simplicity of these results and their proofs — in particular, the way they fall straight out of famous ideas like the exponential sequence and Hodge theory — shows that holomorphic line $n$-bundles, or holomorphic $(n-1)$-gerbes, are just a geometrical way of thinking about a classic topic: the $n$th cohomology of the sheaf $\mathcal{O}^\ast$, which I have discussed using Čech cohomology.

Indeed since I’ve deliberately withheld any way of thinking about line $n$-bundles except for Čech cohomology, you can easily get the impression that they are just a fancy way of talking about Čech cohomology. I suppose there’s some truth to that. But in fact, just as holomorphic line bundles really do clarify the meaning of $H^2(X,\mathcal{O}^\ast)$ and let us do interesting things with it, holomorphic line $n$-bundles do the same for the higher cohomology groups of $\mathcal{O}^\ast$.

But exploring that was not my goal here! All I wanted to do is figure out how some classic results for line bundles generalize to line $n$-bundles, or $(n-1)$-gerbes.

Posted at April 10, 2022 7:08 AM UTC

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## 1 Comment & 1 Trackback

### Re: Holomorphic Gerbes (Part 2)

I think the nLab doesn’t have all this.

There’s a stub Néron-Severi group (but no $n$th group) and plenty of ‘Pickard’ entries (group, 2-group, 3-group, $\infty$-group, stack, scheme).

Then there’s the intermediate Jacobian for each $n$, a quotient of cohomology in odd dimension, but that’s not your $Jac_n$.

Maybe some things to add when editing returns to the nLab.

Posted by: David Corfield on April 12, 2022 11:04 AM | Permalink | Reply to this
Read the post Conversations on Mathematics
Weblog: The n-Category Café
Excerpt: Videos of math conversations between John Baez and James Dolan.
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