## Projectivity (continued)

So what does it mean for a variety to be projective? Well, that’s easy: A variety is projective if you can embed it in projective space.

That’s easy, but that’s not particularly helpful.

What are the benefits of being projective? Why is it something that we should care about?

The way I see it, the main advantage of projectivity is that any analytic projective variety is in fact algebraic i.e. it can be described in terms of zero sets of polynomials, and not just analytic functions. This is essentially a loose paraphrase of Chow’s theorem.

So this explains why projectivity is a good thing, but it doesn’t tell us how to detect it. To help with this, let’s consider what we do get if a variety (we will only really care about tori, but for now we will be more general) is projective.

On $\mathbb{P}^N$, we have the line bundle $\mathcal{O}(1)$. Thus, given a morphism $f : X \to \mathbb{P}^N$, we can pull this line bundle back to obtain a line bundle $L := f^*\mathcal{O}(1)$. This is an ample bundle; that is, if we take a sufficiently high power $L^{\otimes n}$, then sections of this new bundle will in fact yield an embedding into projective space of some dimension. More specifically, choose a basis $s_0, \ldots, s_N$ for $H^0(X, L^{\otimes n})$. Then as this line bundle is base-point free (it comes from a map into projective space), we can consider the map

$X \to \mathbb{P}H^0(X, L^{\otimes n}) \qquad x \mapsto (s_0(x) : \cdots : s_N(x))$

then this map will be an embedding.

Given such a pair $(X, L)$ consisting of a complex manifold $X$ together with an ample line bundle $L$, then we can see that it must be projective. Such a pair is called a polarized variety*.

Now, many manifolds come with natural choices of polarizations; for example, all non-elliptic curves have either their canonical or anti-canonical bundle which are ample, and so they are just naturally polarized. Elliptic curves are as well, but you can’t use their canonical bundle, since it is trivial.

The same is of course true with complex tori; their canonical bundles are trivial, and so these do not provide us with a projective embedding. So let’s see what else a polarization gives us.

Let’s consider the first chern class of our line bundle. We have (since we are working with the complex numbers) the exponential sequence of sheaves

$0 \to \mathbb{Z} \to \mathcal{O} \to \mathcal{O}^\times \to 0$

which yields the long exact sequence some of whose low degree terms are

$\cdots \to H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^\times) \to H^2(X, \mathbb{Z}) \to \cdots$

where $H^1(X, \mathcal{O}^\times)$ is the Picard group of $X$ (denoted $Pic(X)$); that is, the group of line bundles on $X$. The map to $H^2(X, \mathbb{Z})$ is the map which takes a line bundle to its first chern class $c_1(L)$. It is this that we use to understand what makes a manifold projective.

In the case of tori, we know very well what $H^2(X, \mathbb{Z})$ (henceforth we will omit the coefficient ring if it is the integers) is. In fact, due to the Künneth theorem and the fact that topologically, a complex torus is simply a product of circles, we have the isomorphisms

$H^2(X) \cong \Lambda^2 H^1(X) \cong \Lambda^2 H_1(X)^\vee \cong \big(\Lambda^2 H_1(X)\big)^\vee$

Exercise: Check these!

That is, an element of $H^2(X)$ can be though of as an alternating bilinear form $E$ on the underlying lattice $H_1(X)$. In particular, the first chern class of a polarization on a complex torus $X = \mathbb{C}^k / \Gamma$ is an alternating form on its underlying lattice $\Gamma = H_1(X)$.

Now, it is not too hard to see (and you should check this) that there is a bijective correspondence between alternating bilinear forms $E$ on a lattice $\Gamma \subset \mathbb{C}^k$ which satisfy

$E(iv, iw) = E(v,w)$

and hermitian forms $H$ on $\mathbb{C}^k$ which satisfy $\mathfrak{Im}\, H(\Gamma, \Gamma) \subset \mathbb{Z}$; this is given by the bijection

$E(-,-) \qquad \iff \qquad E(i-,-) + iE(-,-)$

Another way to say this is that alternating bilinear forms on $\Gamma$ which are compatible with the complex structure on $\mathbb{C}^k$ are (essentially) the same as hermitian forms on $\mathbb{C}^k$ whose imaginary parts take integer values on $\Gamma$.

And the magic about this is that an element $E \in H^2(X)$ is the first chern class of an ample line bundle if and only if this latter condition is satisfied.

*Well, that’s not exactly correct. It isn’t the line bundle $L$ that is the polarization, but the class of the line bundle in the Neron-Severi group of $X$. But it’s close enough.