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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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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 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
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