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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 nn-gerbes into theorems!

Remember the stars of the show:

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

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

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

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

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

We can think of these groups in terms of (n1)(n-1)-gerbes — or even better for the numbering scheme, complex line nn-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 nn-bundles as complex by default, here is the significance of these groups:

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

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

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

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

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

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

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

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

0exp(2πexp(2π𝒪exp(2πi)𝒪 *exp(2π0 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

H n(X,)H n(X,𝒪)H n(X,𝒪 *)c nH n+1(X,) \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 nc_n.

You may wonder why holomorphic line nn-bundles are classified by H n(X,𝒪 *)H^n(X,\mathcal{O}^\ast) while smooth ones are classified by something that looks so different, namely H n+1(X,)H^{n+1}(X,\mathbb{Z}). In fact smooth line nn-bundles work a lot like holomorphic ones: they’re classified by elements of the nnth Čech cohomology of the sheaf of nonzero smooth complex functions on XX! But in the smooth case this Čech cohomology group is isomorphic to H n+1(X,)H^{n+1}(X,\mathbb{Z}), thanks to the smooth version of the long exact sequence above. The reason is that H nH^n of the sheaf of smooth complex functions vanishes for n1n \ge 1, because it’s a fine sheaf.

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

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

0Jac n(X)Pic n(X)NS n(X)0 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 XX we have

Jac n(X)H n(X,𝒪)im(H n(X,)H n(X,𝒪)) \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 XX is a compact Kähler manifold,

NS n(X)H n+1(X,) T[im(H n+1(X,)H n+1(X,)) p+q=n+1,p,q1H p,q(X)] \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,) TH^{n+1}(X,\mathbb{Z})_T is the torsion subgroup of H n+1(X,)H^{n+1}(X,\mathbb{Z}), while H p,q(X)H^{p,q}(X) are the Doulbeault cohomology groups of XX.

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 XX there is a short exact sequence

0Jac n(X)Pic n(X)NS n(X)0 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 nc_n to form a short exact sequence:

0ker(c n)H n(X,𝒪 *)c nim(c n)0 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

ker(c n) = Jac n(X) H n(X,𝒪 *) = Pic n(X) im(c n) = NS n(X) \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 XX we have

Jac n(X)H n(X,𝒪)im(H n(X,)H n(X,𝒪)) \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:

H n(X,)i nH n(X,𝒪)e nH n(X,𝒪 *)c nH n+1(X,) \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 Jac n(X)\mathrm{Jac}_n(X) is the kernel of c nc_n, so it’s also the image of e ne_n, so it’s also isomorphic to

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

which is the same as

H n(X,𝒪)im(i n) \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 nn-bundles on a compact Kähler manifold. This result is well-known for n=1n = 1 and shown by Ben–Bassat for n=2n = 2.

Theorem 3. When XX is a compact Kähler manifold,

NS n(X)H n+1(X,) T[im(H n+1(X,)H n+1(X,)) p+q=n+1,p,q1H p,q(X)] \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:

e nH n(X,𝒪 *)c nH n+1(X,)i n+1H n+1(X,𝒪)e n+1 \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 NS n(X)=im(c n)\mathrm{NS}_n(X) = \mathrm{im}(c_n), so it’s also ker(i n+1)\mathrm{ker}(i_{n+1}). But

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

factors as

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

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

ker(i n+1)H n+1(X,) T[im(f)ker(g)] \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,)= p+q=n+1H (p,q)(X) H^{n+1}(X, \mathbb{C})= \bigoplus_{p + q = n+1} H^{(p,q)}(X)

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

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

is projection onto this summand, so

ker(g)= p+q=n+1,p0H (p,q)(X) \mathrm{ker} (g) = \bigoplus_{p + q = n+1, p \neq 0} H^{(p,q)}(X)

Putting this all together we get

NS n(X)H n+1(X,) T[im(i n+1) p+q=n+1,p0H (p,q)(X)] \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 im(i n+1)\mathrm{im}(i_{n+1}) consists of real cohomology classes, and complex conjugation on complex de Rham cohomology switches H (p,q)(X)H^{(p,q)}(X) and H (q,p)(X)H^{(q,p)}(X), and nothing in the intersection above can lie in H n+1,0(X)H^{n+1,0}(X), so nothing can lie in H 0,n+1(X)H^{0,n+1}(X) either. So we can refine our description to

NS n(X)H n+1(X,) T[im(i n+1) p+q=n+1,p,q0H (p,q)(X)] \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 nn-bundles, or holomorphic (n1)(n-1)-gerbes, are just a geometrical way of thinking about a classic topic: the nnth cohomology of the sheaf 𝒪 *\mathcal{O}^\ast, which I have discussed using Čech cohomology.

Indeed since I’ve deliberately withheld any way of thinking about line nn-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,𝒪 *)H^2(X,\mathcal{O}^\ast) and let us do interesting things with it, holomorphic line nn-bundles do the same for the higher cohomology groups of 𝒪 *\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 nn-bundles, or (n1)(n-1)-gerbes.

Posted at April 10, 2022 7:08 AM UTC

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3 Comments & 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 nnth group) and plenty of ‘Pickard’ entries (group, 2-group, 3-group, \infty-group, stack, scheme).

Then there’s the intermediate Jacobian for each nn, a quotient of cohomology in odd dimension, but that’s not your Jac nJac_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.
Tracked: July 17, 2022 12:26 AM

Re: Holomorphic Gerbes (Part 2)

There is one interesting thing I always wanted to do but never did with higher holomorphic gerbes on complex tori. I wanted to define a higher linear category of their “sections” (sections are allowed to have zeros). For an complex torus of dimension 1 over the complex numbers you can find line bundles so that the torus embeds into its projectivized space of sections. This proves that the complex torus is algebraic given by an elliptic curve. Suppose we now take a non-algebraic complex torus of complex dimension 2. I was hoping that sometimes it is still algebraic in some higher sense, in that it embeds into some categorical analogue of projective space. This projective space would be associated to some linear higher category of sections of a holomorphic gerbe with certain properties (e.g. ample) on the complex torus.

Posted by: Oren Ben-Bassat on May 5, 2023 8:52 AM | Permalink | Reply to this

Re: Holomorphic Gerbes (Part 2)

If anyone is interested I wrote a follow up article: https://arxiv.org/abs/1102.2312

Posted by: Oren Ben-Bassat on May 5, 2023 11:22 AM | Permalink | Reply to this

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