Dimension counting via incidence varieties

Quick description

There are many situations in algebraic geometry in which one wishes to compute dimensions of various spaces, such as the space of all lines contained in a given surface in projective space. One approach to doing this is via so-called incidence varieties. An incidence variety is essentially the graph of a relation (for example, the collection of all ordered pairs of lines and surfaces of a given degree $d$ , where the line lies on the surface), but thought of itself as a variety. Since the incidence variety is a set of ordered pairs, it admits two natural projections, onto the first or second member of each ordered pair. Playing off the information that one obtains from considering these two projections, it is often possible to make non-trivial deductions.

Prerequisites

Basic algebraic geometry

Example 1

Let $G$ denote the space of all lines contained in $\mathbb{P}^3$ ( $3$ -dimensional projective space). (The letter $G$ is chosen to stand for Grassmannian.) What is the dimension of $G$ ? There are many ways to compute this dimension; we present here two approaches via incidence varieties.

In the first approach, we use the fact that a line is a $1$ -dimensional collection of points, together with the fact that if $P$ is any point of $\mathbb{P}^3$ , then the space of lines through $P$ is isomorphic to $\mathbb{P}^2$ , and hence is $2$ -dimensional. (To see this, note that by choosing our coordinates on projective space appropriately, we can imagine that $P$ is the origin of the affine space $\mathbb{A}^3$ sitting inside projective space $\mathbb{P}^3$ . Now just recall that one definition of $\mathbb{P}^2$ is that it is the space of lines through the origin of $3$ -dimensional affine space.)

Consider the incidence variety

$= \{(P,l) \, | \, P \in \mathbb{P}^3, \, l \in G, \, P \in l \} ;$

that is, $I$ is the incidence variety consisting of pairs of points $P$ and lines $l$ , with $P$ lying on $l$ . We have maps $I \rightarrow \mathbb{P}^3$ and $I \rightarrow G,$ given by projecting onto the first or second coordinate respectively (i.e. $(P,l) \mapsto P,$ while $(P,l) \mapsto l$ ).

We use the map $p$ to elucidate the structure of $I$ , as follows: we have already noted that for a fixed $P \in \mathbb{P}^3$ , the space of lines $l$ passing through $P$ is a $\mathbb{P}^2$ . We can rephrase this by saying that the fibre of $p$ over each point $P$ of $\mathbb{P}^3$ is a $\mathbb{P}^2$ . (Or again, we could say that the map $p$ realizes $I$ as a $\mathbb{P}^2$ -bundle over $\mathbb{P}^3$ .) In particular, the dimension of $I$ must be $5$ (i.e. $2 + 3$ ).

Now consider the map $I \rightarrow G$ . The fibre of any point (i.e. line!) $l \in G$ is precisely the set of points $P$ that lie on $l$ , i.e. the line $l$ itself, but now thought of, not as a point in $G$ , but as a subset of $\mathbb{P}^3$ . As already remarked, $l$ is $1$ -dimensional. (We could say that $q$ realizes $I$ as a $\mathbb{P}^1$ -bundle over $G$ .) Thus the map $I \rightarrow G$ has $1$ -dimensional fibres, and a $5$ -dimensional domain. Its target thus has dimension $4$ (i.e. $5 - 1$ ).

So the answer to the question is that $G$ has dimension $4$ .

Here is another way to make the same computation, using a different incidence variety. In this argument, we again use the fact a line is a $1$ -dimensional collection of points. But rather than using any information about the space of lines passing through a fixed point, we instead use the fact that a line is determined by any two distinct points on it, together with the converse, namely that any two distinct points determine a line.

This time we consider a different incidence variety (which we will again label as $I$ ), namely

$I = \{(P,Q,l) \, | \, P,Q \in \mathbb{P}^3, \, l \in G, \, P, Q \in l\};$

so $I$ consists of pairs of points $(P,Q)$ , together with a line $l$ containing both of them. We consider two projections, as above: $I \rightarrow \mathbb{P}^3 \times \mathbb{P}^3,$ defined by $p(P,Q,l) = (P,Q),$ and $I \rightarrow G,$ defined by $q(P,Q,l) = l$ .

If $(P,Q)$ is a point of $\mathbb{P}^3 \times \mathbb{P}^3$ with $P \neq Q,$ then there is a unique line $l$ containing both $P$ and $Q$ . Thus $p$ is generically 1-1, i.e. it is a birational map, and so we see that its source and target have the same dimension. Considering the target, we see that this dimension equals $6$ (i.e. $3 + 3$ ). Thus $I$ has dimension $6$ .

Now the fibre of $q$ over any line $l$ consists of pairs $(P,Q)$ with both $P$ and $Q$ lying on $l$ , i.e. this fibre is the product $l\times l,$ thought of as lying in $\mathbb{P}^3 \times \mathbb{P}^3$ . In particular, this fibre is $2$ -dimensional. Since $q$ has a $6$ -dimensional source, and $2$ -dimensional fibres, its target has dimension $4$ (i.e. $4 = 6 - 2$ ). Thus $G$ has dimension $4$ .

General discussion

Suppose that $X$ and $Y$ are two algebraic varieties, and that $R$ is a relation between $X$ and $Y$ . Thinking of $R$ as being a set of ordered pairs, we can regard $R$ as being a subset of $X \times Y$ . (Sometimes, to be explicit, we refer to this subset as the graph of the relation; but in most theoretical treatments of the notion of relation, a relation is simply identified with the collection of ordered pairs that it determines, i.e. with its graph, and that is what we are doing here.) Now suppose that the relation $R$ is described by an algebraic condition. Then we can think of $R$ (which is a priori just a subset of $X \times Y$ ) as being an algebraic subvariety of $X \times Y$ . (And indeed, we could take this as being the definition of what it means for the relation to be described by an algebraic condition.)

The subvariety $R$ comes equipped with two important structures, namely the projection maps onto the first and second factor, i.e. $R \rightarrow X$ and $R \rightarrow Y.$ One can then hope to study these maps (in particular, their fibres), to pass geometric information back and forth between $X$ and $Y$ , via the intermediary of $R$ .

An incidence variety is just a particular instance of the above general set-up, when $X$ and $Y$ are varieties that themselves parameterize other geometric objects (e.g. spaces of points, lines, curves, surfaces, etc.), and the relation $R$ is given by some geometric condition of ``incidence'', e.g. ``lies on'', ``has non-empty intersection with'', ``is tanget to'', etc. In our first example, $X$ was the space either of points in $\mathbb{P}^3$ (i.e. $\mathbb{P}^3$ -itself) or else the space of pairs of points in $\mathbb{P}^3$ (i.e. $\mathbb{P}^3\times\mathbb{P}^3$ ), and $Y$ was the space $G$ of lines in $\mathbb{P}^3$ (an example of a Grassmannian).

Example 2

Suppose that we wish to compute the dimension of the space of lines contained in a surface of some given degree $d$ in projective space $\mathbb{P}^3$ . Again, let $G$ denote the space of all lines contained in $\mathbb{P}^3$ ; this is a $4$ -dimensional space, as we saw above.

The space of surfaces of degree $d$ in $\mathbb{P}^3$ is itself a projective space, of dimension $\binom {d+3}{3} - 1$ . (There are $\binom{d+3}{3}$ monomials of degree $d$ in $4$ -variables.) We take this space to be $X$ , and will in fact denote it by $X$ . We take $Y$ to be the space $G$ of lines. We define the incidence variety $I$ as follows:

$I = \{ (\Sigma, l) \, | \, \Sigma \in X, \, l \in G, \, l \subset \Sigma\};$

thus $I$ consists of the ordered pairs of a surface $\Sigma$ of degree $d$ and a line $l$ , for which the line $l$ lies on the surface $\Sigma$ .

We have the projections $I \rightarrow X$ and $I \rightarrow G$ . Our goal is to compute the dimensions of the fibres of $p$ ; i.e. the dimension of the space of lines lying on a given degree $d$ surface $\Sigma$ . Since we know the dimension of $X$ , what we need is the dimension of $I$ .

To compute the dimension of $I$ , we turn to a consideration of the map $q$ . It has a $4$ -dimensional target. What are the dimensions of its fibres? To answer this, we have to compute how many surfaces $\Sigma$ of degree $d$ pass through a given line $l$ . We may as well choose our (homogeneous) coordinates $X,Y,Z,W$ so that $l$ is cut out by the equations $X = Y = 0.$ For $\Sigma$ to contain $l$ , we need every monomial appearing in the equation of $\Sigma$ to be divisible by at least one of $X$ or $Y$ . There are $d+1$ monomials of degree $d$ divisible by neither $X$ nor $Y$ (i.e. involving just $Z$ or $W$ ), and so there are $\binom{d+3}{3} - d - 1$ monomials divisible by at least one of $X$ or $Y$ . Thus each fibre of $q$ has dimension equal to $\binom{d+3}{3} - d - 2,$ and so $I$ has dimension equal to this fibre dimension plus $4$ (the dimension of $G$ ), i.e. the dimension of $I$ equals $\binom{d+3}{3} - d + 2$ .

We now return to the map $p$ . We now know that is maps from a variety of dimension $\binom{d+3}{3} - d + 2$ to a variety of dimension $\binom{d+3}{3} - 1$ . What can we conclude?

Well, if $d > 3$ , the dimension of the source is less than that of the target, and hence the map $p$ cannot be surjective. Thus a generic surface of degree $> 3$ contains no lines on it at all. (Of course, some $\Sigma$ will, since $X$ contains $\Sigma$ which are reducible, e.g. which are the union of $d$ planes, and such a $\Sigma$ contains plenty of lines.)

When $d = 3$ , we see that the source and target have the same dimension, and so we might expect that a generic fibre of $p$ is $0$ -dimensional, i.e. finite, and hence that a generic cubic surface contains a finite number of lines. This is indeed true: a smooth cubic surface contains exactly $27$ lines (if we are working over an algebraically closed field).

When $d = 2$ , we see that the source has dimension $1$ greater than the target, and so we might expect that the typical degree $2$ surface (or quadric, as one says) contains a one-dimensional family of lines. This is indeed true of any smooth quadric: such a quadric is a double-ruled surface. (Again, over an algebraically closed field; but one can see this for a one-sheeted hyperboloid plotted over the real numbers.)

When $d = 1$ , we see that the source has dimension $2$ greater than the target. And in this case we know that the fibres are all of dimension $2$ : a degree $1$ surface is just a plane, and a plane contains a $2$ -dimensional family of lines on it (parameterized by the dual plane); one can prove this, for example, following the method of Example 1 above.

This example brings out a general feature that will tend to arise in any careful analysis of a situation involving incidence varieties: namely, the maps $p$ and $q$ need not always have fibres of constant dimension (this is a general feature of maps between varieties), and so to draw precise conclusions, one sometimes has to investigate the structure of these maps in more detail. (This is discussed at greater length on the page How to compute the dimensions of the fibres of a map of varieties.)

This is a general feature of the method of counting constants: often to make precise conclusions, one has to consider more of the geometry of the situation than just the dimensions of the varieties involved. But even without this extra work, rough computations of the type given in this example can be very suggestive, and can be well-worth making before one proceeds with more careful geometric analyses.

Remark

The method of incidence varieties can be thought of as an algebro-geometric version of the method of double counting in combinatorics (but one is now counting dimensions, rather than actual numbers of elements).