# Maps to projective space determined by a line bundle

The following should be pretty standard for any algebraic geometer.

Let $X$ be a compact complex variety, and let $L$ be a line bundle on $X$. We say $L$ is 'generated by global sections' if for every point $p$ in $X$, there is a global section of $L$ which doesn't vanish. If this is true, then $L$ determines a map to a projective space in the following way. The global sections of $L$ are finite dimensional, so choose a basis $(a_i)$. Then send a point $p$ in $X$ to the projective point

$$[a_1(p):a_2(p):...:a_n(p)]$$

It should be noted that $a_i(p)$ is only a point in the fiber of $L$ over $p$, and not a complex number. By choosing an isomorphism from the fiber over $p$ to $\mathbb{C}$, the $a_i(p)$ can be identified with complex numbers. The ambiguity introduced in choosing this isomorphism dissappears when taking the projective coordinates.

This constructions is remarkably ad hoc for something that ends up being foundational in algebraic geometry. It requires the variety be over $\mathbb{C}$, and requires that some inconsequential choices be made enroute.

My question is, what are nicer, more intrinsically algebraic ways to construct this map? Three ways this construction could be nicer:

• It could work over other fields, or possibly even over $\mathbb{Z}$ (though then its less clear what a line bundle should be).
• It could give a good intuitive justification of why this map is a natural and powerful thing to look at.
• It could lend itself to generalizations in different directions. For instance, a rank n vector bundle V with 'enough global sections' (in some sense) should determine a map from X to $Hom_{\mathbb{C}}(\Gamma(V),\mathbb{C}^n)//GL(n)$ (where this is a GIT quotient).

Aside. Is there even a good name for this construction? The "map to projective space determined by a line bundle" is a bit long-winded.

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I think it's usually called the "complete linear system" determined by the line bundle. It works over all fields and even Z. If you want to know what a line bundle is in this context, read Hartshorne (or any graduate-level alg geom book), which gives an intrinsically algebraic way to construct the map. –  Kevin Buzzard Nov 7 '09 at 22:59
Great question. Re: "This constructions is remarkably ad hoc", see Anton's answer, which I think makes it easiest to see how non- ad hoc these morphisms really are! –  Andrew Critch Nov 8 '09 at 0:02
I'm still confused by what you say :-) At the minute you seem to be saying "I don't believe in algebraic geometry over finite fields"... –  Kevin Buzzard Nov 8 '09 at 8:25
sigh I'm not saying that; what I was trying imply (crudely) is that the construction seemed to be very 'variety-theoretic', and not 'scheme-theoretic', in that you are defining a map on closed points and calling it a day. I know this construction works in scheme-y generality, but to even think about it over things like finite fields, there are better ways to think about the construction. I was certainly not trying to impune arithmetic algebraic geometry. –  Greg Muller Nov 8 '09 at 15:30
I don't get the point. The general definition of projective space in the functorial approach to algebraic geometry (see e.g. EGA I) just says that maps into projective space are given by invertible sheaves + global generators. The description of the points follows from it (plug in fields). If you pick a universal element, i.e. the serre twist $\mathcal{O}(1)$ with its chosen generators, then the bijection is on field-valued points exactly what is described above, but it should not be confused with a complete definition of the map into projective space. –  Martin Brandenburg Jul 18 '11 at 19:56
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This is one of the most fundamental questions possible. Hence although it is old and well answered, I venture to add something, hoping to make it seem as transparent as possible.

I would suggest the way to understand this construction is to look at it backwards. I.e. by its very definition, projective space carries a tautological line bundle, whose dual bundle has as sections the linear coordinates. These sections have no common zeroes because the hyperplanes have no common points. Hence any subvariety of projective space also has by restriction a line bundle whose sections have no common zeroes.

Moreover a point of projective space is determined by the set of hyperplanes through it, so any subvariety is determined by the restricted line bundle, since each point is recovered from the set of sections vanishing on it. Moreover the projective space it self is dual to the space of hyperplanes, hence to the space of global sections of the bundle. Hence the bundle on the subvariety determines both the ambient projective space and the embedding.

Now one sees immediately that one can imitate this to give a map, not necessarily an embedding, from any variety with a line bundle whose sections have no common zeroes, to the dual projective space of its space of sections, by sending each point to the subset of its sections vanishing at that point, as Anton said.

In a nutshell, since projective space has a line bundle whose sections have no common zeroes, and line bundles and sections pull back under maps, having such a line bundle is a necessary condition for a map to projective space. Then one asks whether it is also sufficient, and it is, as above.

It is easy also to recover the properties that determine whether the map is an embedding.

Looked at this way, there is nothing mysterious about this construction - it is in fact the defining property of projective space.

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I'm sure somebody will put up the stacky answer to this question, but let me try to give a more down to earth one.

The point is that projective space is a defined using a functor $\mathrm{Proj}$. $\mathrm{Proj}$ has the pleasant property that if $X$ is any scheme, and $L$ is a line bundle, then there is a canonical map from $X_0$ to $\mathrm{Proj}(\oplus_{n\geq 0}\Gamma(X;L^{\otimes n}))$, where $X_0$ is the subset of $X$ where at least one section of $L^{\otimes n}$ is nonvanishing for some n. This doesn't use any choices or algebraically closed base; for any section of $L^{\otimes n}$, we just take the map to the affinization of the non-vanishing set and glue all of these together.

So, if $L$ is globally generated, we have a map from $X$ to $\mathrm{Proj}(\oplus_{n\geq 0}\Gamma(X;L^{\otimes n}))$, and this is the universal polarized projective variety such that the pullback of $\mathcal{O}(1)$ is $L$. Now, we just need to find maps of this variety to $\mathbb{P}^n$ that give the right line bundle.

So we need to think about what maps between two Projs (preserving the line bundle) are, and that's quite simple, it's maps of graded rings the other way which hit all non-irrelevant ideals.

Since $\mathbb{P}^n$ is Proj of a polynomial ring, a map of a Proj to projective space picking $n$ degree 1 sections who generate an algebra that hits all non-irrelevant ideals, which is exactly the description you gave.

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I was playing with a construction similar to this, though not as clear what the objects involved were. If you take the sheafify Proj of Sym_X(L), you get X again, because you are projectifying the total space of a line bundle. There should then be a map of sheaves of graded algebras Sym_C(\Gamma(L)) to Sym_X(L), such that Proj of this map gives X -> P^n. –  Greg Muller Nov 7 '09 at 23:41

Intrinsically, the map from $X$ to $\mathbb P(\Gamma(L))$ (the space of hyperplanes in $\Gamma(L)$) is given by sending a point $x\in X$ to the hyperplane of sections that vanish at $x$ (you can see this is a hyperplane by observing that given two non-proportional sections of $L$, there is always some non-trivial linear combination of them which vanishes at $x$). If all sections of $L$ vanish at $x$, then the map is not defined at $x$.

This construction works over a field. If you want to work over $\mathbb Z$, you should use the language of Proj of a graded ring, but then it gets harder to imagine the geometry.

The right place to learn about different versions of line bundles "having enough sections" is Lazarsfeld's Positivity in Algebraic Geometry I. The right place to learn about generalizations to vector bundles is probably Positivity in Algebraic Geometry II.

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First, the stacky interpretation. A choice of a line bundle $\mathcal L$ on a scheme $X$ is equivalent to a morphism to the stack $\def\GG{\mathbb G} B\GG_m$; the $\GG_m$-torsor is the complement of the zero section in the total space $\mathbb V(\mathcal L)$. A choice of a line bundle together with $n$ sections is equivalent to a morphism $f:X\to\def\AA{\mathbb A} [\AA^n/\GG_m]$; the $\GG_m$ torsor is the complement of the zero section of $\mathcal L$, and the $n$ map to $\AA^n$ is given by the $n$ regular functions which are the pullbacks of the sections. The condition that $f$ misses the one stacky point of $[\AA^n/\GG_m]$ (i.e. the point where the $\GG_m$ doesn't act freely)—and therefore factors through the open subscheme $[(\AA^n\smallsetminus 0)/\GG_m]=\mathbb P^{n-1}$ —is precisely the condition that the sections do not all vanish at the same point.

Now the generalization. A rank $k$ vector bundle $\def\E{\mathcal E} \E$ on a scheme $X$ is equivalent to a morphism to the stack $BGL_k$; the $GL_k$-torsor is the sheaf $\def\O{\mathcal O} Isom(\O^k,\E)$. A vector bundle together with a choice of $n$ sections is equivalent to a morphism $f:X\to [(\AA^k)^n/GL_n]$; again, the $GL_k$-torsor is $Isom(\O^k,\E)$, and the $k\cdot n$ regular functions on the torsor are given by the pullback of the $n$ sections of $\E$, and the fact that the pullback of $\E$ is canonically identified with $\O^k$. Regarding $(\AA^k)^n$ as the space of $k\times n$ matrices, the stacky locus of $[(\AA^k)^n/GL_n]$ (i.e. the points with non-trivial stabilizers) is the locus where the rank of the matrix is less than $k$. So the condition that $f$ misses the stacky locus is equivalent to the condition that the $n$ given sections span the fiber of $\E$ at any point. The non-stacky locus is the open subscheme $[${$k\times n$ matrices of rank $k$}$/GL_k]$, the Grassmannian of $k$-planes in $n$-space, $Gr(k,n)$. We therefore have the following interpretation.

A vector bundle $\E$ on a scheme $X$, together with $n$ sections $s_1,\dots, s_n$ which span every fiber is equivalent to a morphism $X\to Gr(k,n)$.

If you don't want to dirty things up by choosing the sections, you can replace $(\AA^k)^n$ by $Hom(\Gamma(\E),\mathbb C^k)$ everywhere. The non-stacky locus is then the space of surjective linear maps. Then it looking at kernels, it looks like you naturally get $Gr(n-k,n)$ instead of (the isomorphic) $G(k,n)$. I'm sure this has something to do with a dualization involved in forming the total space of a locally free sheaf; it still regularly confuses me, so I'll just leave this thread loose.

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For the underinitiated: I assumed that you know that a morphism from $X$ to a quotient stack $[Y/G]$ is equivalent to the data of a $G$-torsor $P\to X$, and a $G$-equivariant morphism $P\to Y$. –  Anton Geraschenko Jul 18 '11 at 20:36

Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginnning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sectionsfor a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $H^0(L)\times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb P(H^0(L))^*$. As Anton points out, this map takes a point $x$ to the hyperplane of sections that vanish at $x$ (and it fails iff all sections vanish at $x$).

As you vary $x$, you get the map to $\mathbb P(H^0(L))^*$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

Now if take arbitrary scheme (skip if you're not familiar with schemes), they locally look like $\mathrm{Spec}\\,A$. Now it's easy to define a line bundle: it's something that maps into the scheme and the map locally looks like $$\mathrm{Spec}\\,A[t] \to \mathrm{Spec} \\,A.$$ You can define the cohomology as usual, so the space $\mathbb P(H^0(L))^*$ continues to prove a good target to map into. It's slightly more unusual to draw pictures, but you can start with tensoring everything by $\mathbb F_q$ to gain the intuition.

Return to curves over field $k$. There the situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

Finally, a line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.

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Right, this is where I was going when I was talking about a map $X\rightarrow Hom_\mathbb{C}(\Gamma(V),\mathbb{C}^n)//GL(n)$. In the case of a rank n vector bundle, which is locally modeled on $\mathbb{C}^n$, then points give maps $\Gamma(V)\rightarrow\mathbb{C}^n$ (modulo the action of $GL(n)$). –  Greg Muller Nov 7 '09 at 23:32
But this still doesn't satisfy the algebraist in me... Ideally, I'd want a construction that was naturally occuring in the module category, and used projectivity & (something else). –  Greg Muller Nov 7 '09 at 23:34
Yes, it should be so. –  Ilya Nikokoshev Nov 7 '09 at 23:35
Depends on how you define your projective variety and bundle. If you define your variety as Proj A (A has special properties) then bundle is a module over A and you can take Proj \Gamma(\oplus_n\otimes_n(M)). But now you should really look at the Hartshorne --- that's what the book is good at. –  Ilya Nikokoshev Nov 7 '09 at 23:38