## March 19, 2022

### Line Bundles on Complex Tori (Part 2)

#### Posted by John Baez

Last time I explained how the job of classifying holomorphic line bundles $L$ on a complex torus $X$ breaks into two parts:

• the ‘discrete part’: the underlying topological line bundle of $L$ is classified by an element of a finitely generated free abelian group called the Néron–Severi group $\mathrm{NS}(X)$.

• the ‘continuous part’: the holomorphic line bundles with a given underlying topological line bundle are classified by a complex torus called the Jacobian $\mathrm{Jac}(X)$.

Today I want to talk more about the discrete part: the Néron–Severi group. Studying examples of this leads to beautiful pictures like this one by Roice Nelson:

Before I can talk about these examples I want to go through some prerequisites, since I’m still learning them myself. In math you sometimes have to eat your vegetables of theory before you can have your dessert of examples!

### Complex tori

The term complex torus actually means a couple of different things, but here I mean a compact connected complex Lie group. All these are automatically abelian, and they’re all of the form

$X = V / L$

where $V$ is a complex vector space and $L$ is a lattice in $V$. (Today all vector spaces will be finite-dimensional.)

Note that we can recover $V$ and $L$ knowing just $X$ as a complex Lie group, because the vector space $V$ is the universal cover $\tilde{X}$, and the lattice $L \subseteq V$ is the kernel of the projection

$p \colon \tilde{X} \to X$

So, a complex torus is really ‘the same’ as a complex vector space with a lattice in it! In fact there’s an equivalence between the category where

• an object is a compact connected complex Lie group $X$ and a morphism is a complex Lie group homomorphism $f: X \to X'$

and the category where

• an object is a pair $(V,L)$ consisting of a complex vector space $V$ and a lattice $L \subseteq V$, and a morphism $f \colon (V,L) \to (V',L')$ is a linear map $f: V \to V'$ that maps $L$ into $L'$.

This is one reason that complex tori are worth studying! We’re really studying linear algebra enhanced with lattices: a wonderful blend of the continuous and the discrete. The same philosophy works for real tori, and if I were writing a textbook I’d start by working out the theory of those. But bringing the complex numbers into the game adds an extra level of subtlety. For example, all real tori of a given dimension are isomorphic, but this is far from true for complex tori, because there are different ways the lattice $L \subseteq V$ can get along with multiplication by $i$.

### The Néron–Severi group

Last time I talked about the set $\mathrm{Pic}(X)$ of isomorphism classes of holomorphic line bundles on our complex torus $X$. This set $\mathrm{Pic}(X)$ is an abelian group, because we can tensor line bundles. It’s called the Picard group of $X$. And it’s a topological group, because there’s a way to say what it means for two holomorphic line bundles to be close to each other.

The Picard group has a subgroup $\mathrm{Jac}(X)$, namely the connected component of the identity. It’s called the Jacobian of $X$.

The Picard group also has a quotient group

$\mathrm{NS}(X) = \frac{\mathrm{Pic}(X)}{\mathrm{Jac}(X)}$

Its elements are the connected components of $\mathrm{Pic}(X)$. It’s called the Néron–Severi group. It’s actually a finitely generated free abelian group!

So, the Picard group fits into a short exact sequence

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

together with the Jacobian and the Néron–Severi group. I will focus on the latter today!

Last time I gave one rather concrete description of the Néron–Severi group. Today I want to give two more. The first was in terms of alternating bilinear forms. The second will be in terms of hermitian forms. And the third will be in terms of maps from the complex torus $X$ to its dual.

The first two descriptions take advantage of the philosophy that a complex torus is secretly a complex vector space with a lattice in it.

### Description 1: alternating bilinear forms

The set of isomorphism classes of topological line bundles over $X$ forms a group under tensor product of line bundles, and this group is isomorphic to $H^2(X,\mathbb{Z})$: the second cohomology with integer coefficients of $X$ as a topological space. We can identify the Néron–Severi group $NS(X)$ with the subgroup of $H^2(X,\mathbb{Z})$ coming from holomorphic line bundles.

Using the fact that $X$ is a complex torus, one can show that $H^2(X,\mathbb{Z})$ is isomorphic to the group of alternating bilinear maps

$A \colon L \times L \to \mathbb{Z}$

where $L$ is the lattice associated to $X$. So, $\mathrm{NS}(X)$ is some subgroup of this: some bunch of alternating bilinear maps closed under addition and subtraction. It’s defined using the fact that $X$ is a complex torus. Because $X$ is a complex torus, its universal cover $V = \tilde{X}$ is a complex vector space, and we can extend $A$ to an alternating real-bilinear map of vector spaces, which I’ll call by the same name:

$A \colon V \times V \to \mathbb{R}$

The Néron–Severi group $NS(X)$ then turns out to consisting of alternating bilinear maps

$A \colon L \times L \to \mathbb{Z}$

whose extension obeys

$A(i v,i w) = A(v,w)$

for all $v,w \in V$. In summary:

Theorem $1$. The Néron–Severi group $NS(X)$ consists of alternating bilinear maps $A \colon L \times L \to \mathbb{Z}$ whose extension to $V$ is preserved by multiplication by $i$.

Or equivalently:

Theorem $1'$. The Néron–Severi group $NS(X)$ consists of alternating bilinear maps of vector spaces $A \colon V \times V \to \mathbb{R}$ that are integer-valued on the lattice $L \subseteq V$ and preserved by multiplication by $i$.

### Description 2: hermitian forms

The inner product on a Hilbert space is sesquilinear: it’s conjugate-linear in the first variable and complex-linear in the second. For any alternating bilinear form

$A \colon V \times V \to \mathbb{R}$

such that

$A(i v, i w) = A(v, w)$

there’s a unique sesquilinear map

$H \colon V \times V \to \mathbb{C}$

whose imaginary part is $A$:

$A(v,w) = \mathrm{Im} H(v,w)$

To show this, we just dream up a formula for $H$, namely

$H(v,w) = A(v, i w) + i A(v,w)$

and check that it really is sesquilinear and its imaginary part is $A$. We can also show that

$H(w,v) = \overline{H(v,w)}$

In this game, a sesquilinear form with this extra property is called a hermitian form. For example, the inner product on a complex Hilbert space is a hermitian form, but there are also hermitian forms that aren’t positive definite.

Conversely, any hermitian form $H$ has as its imaginary part an antisymmetric form.

Thus, there’s a correspondence between

• antisymmetric real-valued bilinear forms $A$ with $A(i v, i w) = A(v, w)$

and

• hermitian forms $H$

given by the condition that $A$ is the imaginary part of $H$. And under this correspondence $A$ is in the Néron–Severi group, meaning it takes integer values on $L$, if and only if the imaginary part of $H$ takes integer values on $L$.

In short, we’ve shown:

Theorem 2. The Néron–Severi group $NS(X)$ consists of hermitian forms $H \colon V \times V \to \mathbb{C}$ whose imaginary part takes integer values on the lattice $L \subseteq V$.

This puts a new spin on things. As I mentioned, a complex torus is secretly the same thing as a complex vector space with a lattice in it. Now we’re seeing that the Néron–Severi group of the complex torus consists of hermitian forms on the complex vector space that take integral values on the lattice!

### Description 3: maps from a torus to its dual

We can also go in the other direction and try to describe the Néron–Severi group using complex tori as much as possible, eliminating all mention of complex vector spaces and lattices. This approach is also used a lot.

The trick is to start with Theorem 2, and turn the hermitian form

$H \colon V \times V \to \mathbb{C}$

into a linear map

$\begin{array}{rcl} h \colon V &\to& V^\ast \\ v &\mapsto & H(-,v) \end{array}$

where $V^\ast$ is — beware! — the space of conjugate-linear maps from $V$ to $\mathbb{C}$. That’s because $H$ is conjugate-linear in the first slot.

Now, you may or may not recall that last time I used $V^\ast$ to mean the space of real-linear maps from $V$ to $\mathbb{R}$. Luckily there’s no deadly inconsistency in my notation here, because that is naturally isomorphic to the space of conjugate-linear maps from $V$ to $\mathbb{C}$! Indeed, one thing I had to realize while learning this subject is this:

Lemma. For any complex vector space $V$ the following three real vector spaces are naturally isomorphic:

1. the space of real-linear map from $V$ to $\mathbb{R}$

2. the space of complex-linear maps from $V$ to $\mathbb{C}$

3. the space of conjugate-linear maps from $V$ to $\mathbb{C}$.

Proof. We can get from 2) or 3) to 1) by taking the real part. We can get from 2) to 3) or vice versa by taking the complex conjugate. So it’s enough to check is that we can get from 1) to 2). Here we use the fact that any real-linear map $f\colon V \to \mathbb{R}$ extends uniquely to a complex-linear map $g \colon V \to \mathbb{C}$ given by

$g(v) = f(v) - i f(i v)$

One can check that the real part of $g$ is $f$, and that $g$ is complex-linear since $g(i v) = i g(v)$ (this is a little calculation).    ⬛

Next, suppose a hermitian form $H \colon V \times V \to \mathbb{C}$ lies in the Néron–Severi group as described in Theorem 2.

This means that $H \colon L \times L \to \mathbb{Z}$, which is equivalent to the requirement that the corresponding map

$\begin{array}{rcl} h \colon V &\to& V^\ast \\ v &\mapsto & H(-,v) \end{array}$

maps $L$ into its dual lattice:

$L^\ast = \{ f \colon V \to \mathbb{R} : f \; is\; real–linear \; and\; f(v) \in \mathbb{Z} \; for \; all \; v \in L \}$

Thanks to the lemma, we can think of $L^\ast$ as a subset of $V^\ast$. $L^\ast$ as defined above consists of real-linear functionals, and our ‘official’ definition of $V^\ast$ is that it consists of conjugate-linear functionals — but the Lemma bridges the gap.

This gives an alternative statement of Theorem 2:

Theorem $2'$. The Néron–Severi group $NS(X)$ consists of linear maps $h \colon V \to V^\ast$ that map $L$ into $L^\ast$ and have $h^\ast = h$.

The last condition, $h^\ast = h$, is a translation of this condition that a hermitian form $H \colon V \times V \to \mathbb{C}$ must satisfy: $H(w,v) = \overline{H(v,w)}$.

Next we can translate this result into the language of complex tori. To do this we define the dual torus $X^\ast$ by

$X^\ast = V^\ast/L^\ast$

Note that a linear map $h \colon V \to V^\ast$ that maps $L$ into $L^\ast$ induces a map from $V/L = X$ to $V^\ast/L^\ast = X^\ast$, which I’ll call

$\theta \colon X \to X^\ast$

This is a map of complex tori. Conversely any map of complex tori $\theta \colon X \to X^\ast$ gives rise to a map between their universal covers, which is a complex linear map

$h \colon V \to V^\ast$

and this maps $L$ to $L^\ast$.

Using this correspondence, everything about Theorem $2'$ can be translated into the language of complex tori, and we get this:

Theorem $3$. The Néron–Severi group $NS(X)$ consists of maps of complex tori $\theta \colon X \to X^\ast$ such that $\theta^\ast = \theta$.

Here we use the fact that every map of complex tori $f \colon X \to Y$ has an adjoint $f^\ast \colon Y^\ast \to X^\ast$. The condition $\theta^\ast = \theta$ is the translation into the language of complex tori of the condition $h^\ast = h$ in Theorem $2'$.

That’s it, our third and final description of the Néron–Severi group!

### The Rosati involution

I’m basically done, but I can’t resist pointing out that a trick we used in Theorem 3 has a name. Given a map of complex tori

$\theta \colon X \to X^\ast$

$\theta^\ast \colon X \to X^\ast$

In fact we have

$(\theta^\ast)^\ast = \theta$

So taking the adjoint gives an involution

$(-)^\ast \colon hom(X,X^\ast) \to hom(X,X^\ast)$

called the Rosati involution.

I’ve usually seen the Rosati involution described in a somewhat different way, which works only when $X$ is equipped with an extra structure called a ‘principal polarization’, which gives an isomorphism $X^\ast \cong X$. That more common approach has a lot of advantages, and maybe I’ll talk about it someday! But we don’t really need it to state Theorem 3, unless I’ve made some sort of mistake here.

The Rosati involution is a wonderful thing. The Rosati Involution should be a title of a novel, like The Eiger Sanction or The Bourne Identity. But it’s also the key to understanding the Néron–Severi group more deeply, and getting into examples like this:

Posted at March 19, 2022 2:08 AM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/3388

## 1 Comment & 1 Trackback

### Re: Line Bundles on Complex Tori (Part 2)

There are a number of calculations to check in my post. Just for completeness, and to while away the long and dreary evenings, I’d like to check them here.

Here’s one:

For any alternating bilinear form $A \colon V \times V \to \mathbb{R}$ such that $A(i v, i w) = A(v, w)$ there’s a unique sesquilinear map $H \colon V \times V \to \mathbb{C}$ whose imaginary part is $A$: $A(v,w) = \mathrm{Im} H(v,w)$ To show this, we just dream up a formula for $H$, namely $H(v,w) = A(v, i w) + i A(v,w)$ and check that it really is sesquilinear and its imaginary part is $A$. We can also show that $H(w,v) = \overline{H(v,w)}$ In this game, a sesquilinear form with this extra property is called a hermitian form.

It’s obvious that $Im H = A$, but let’s check that $H$ is a hermitian form. First let’s show it’s complex-linear in the second argument. Since it’s real-linear we just need to check

$H(v,i w) = i H(v, w)$

So:

$\begin{array}{ccl} H(v,i w) &=& A(v, i^2 w) + i A(v, i w) \\ &=& - A(v,w) + i A(v, i w) \\ &=& i (A(v, i w) + i A(v,w)) \\ &=& i H(v,w) \end{array}$

Given that it’s complex-linear in the second argument, $H$ will automatically be conjugate-linear in the first argument if we can show

$H(w,v) = \overline{H(v,w)}$

So let’s do that. Again we just compute:

$\begin{array}{ccl} H(w,v) &=& A(w, i v) + i A(w,v) \\ &=& - A(i v, w) - i A(v,w) \\ &=& A(v, i w) - i A(v,w) \\ &=& \overline{H(v,w)} \end{array}$

Here in the third step we finally used the fact that $A$ is preserved by multiplication by $i$: this gives $A(i v, w) = -A(v, i w)$.

This was worth checking because it helped me catch a small flaw in my logic and fix my post.

Posted by: John Baez on March 21, 2022 8:08 AM | Permalink | Reply to this
Read the post Holomorphic Gerbes (Part 1)
Weblog: The n-Category Café
Excerpt: Can we generalize a bunch of basic results about Picard groups and Néron--Severi groups from line bundles to gerbes?
Tracked: April 10, 2022 8:34 PM

Post a New Comment