The Enumerative Geometry of Hyperplane Arrangements
Abstract
We study enumerative questions on the moduli space of hyperplane arrangements with a given intersection lattice . Mnëv’s universality theorem suggests that these moduli spaces can be arbitrarily complicated; indeed it is even difficult to compute the dimension . Embedding in a product of projective spaces, we study the degree , which can be interpreted as the number of arrangements in that pass through points in general position. For generic arrangements can be computed combinatorially and this number also appears in the study of the Chow variety of zero dimensional cycles. We compute and using Schubert calculus in the case where is the intersection lattice of the arrangement obtained by taking multiple cones over a generic arrangement. We also calculate the characteristic numbers for families of generic arrangements in with 3 and 4 lines.
1 Introduction
Enumerative geometry has a monumental history and continues to be an inspiration for many different fields of research (for example see Katz [13] and Kontsevich and Manin [17]). Solutions to enumerative problems have given deep insight into the geometric nature of various algebraic varieties and important spaces.
Realization spaces of matroids (or moduli spaces of hyperplane arrangements) give a beautiful connection between combinatorics and algebraic geometry. In particular, Mnëv’s universality theorem presents matroids as a kind of dictionary for quasiprojective varieties (see Mnëv [18] or Vakil [27]). These realization spaces have been studied from many different viewpoints. Kapranov [12] showed the Hilbert compactification of the moduli space of generic arrangements could be viewed as a Chow quotient. Hacking, Keel, and Tevelev [10] applied the relative minimal model program to the moduli space of arrangements. Terao [25] studied the closure of this moduli space in a product of projective spaces and an associated logarithmic GaussManin connection. Speyer [24] proved that the theory class (actually a push forwardpullback) of the inclusion of the moduli space into the appropriate Grassmannian is actually the 2variable Tutte polynomial of the associated matroid. Then Fink and Speyer [6] generalized this result to nonrealizable matroids. In this note we study some of the geometry of this moduli space and find another use of the Tutte polynomial.
One of our motivating problems is Terao’s conjecture which concerns the subset of the realization consisting of free arrangements (see Orlik and Terao [19] for a general reference on hyperplane arrangements). Yuzvinsky [28] showed that this subset was Zariskiopen. It is not known if this subset is also closed. Our focus here is on computing the dimension and essentially the degree of this realization space by using both combinatorial and geometric methods.
To each hyperplane arrangement in we associate an intersection lattice , the poset whose elements are intersections ordered by reverse inclusion, . Two arrangements and are combinatorially equivalent if their intersection lattices are isomorphic, that is, if there is a bijection that preserves the lattice order, . Let be the set of arrangements that are combinatorially equivalent to the arrangement . Identifying each arrangement with its orbit under the permutation group , we obtain an embedding and we give the induced topology coming from the Zariski topology on the quotient space. It is clear that depends only on the intersection lattice of . If is the dimension of , then let be the number of arrangements in passing through points in general position in .
Question 1.
Given compute and . Ideally these answers should be given in terms of the combinatorics of the lattice .
In the next section we show that when is a generic arrangement of hyperplanes in , then and we compute . The characteristic number measures the number of arrangements combinatorially equivalent to that pass through points and are tangent to lines in general position (with ). We use intersection theory on the correspondence between hyperplane arrangements and their dual arrangements to compute the characteristic numbers for generic arrangements of three or four lines in (a good reference for the intersection theory that we use is Eisenbud and Harris [4] or Fulton [7]). This seems to be the first computation of these characteristic numbers for line arrangements though characteristic numbers were computed for smooth curves of degrees 3 and 4 by Zeuthen [29] in the 19 century. As reported in Kleiman [15], the 19 century methods lacked adequate foundations prompting Hilbert to ask for a rigorous computation of these characteristic numbers. Aluffi [1] and Kleiman and Speiser [16] verified Zeuthen’s degree 3 predictions using intersection theory. Vakil [26] used intersection theory on the moduli space of stable maps to verify the degree 4 predictions. The importance of the characteristic numbers is suggested by a theorem originally due to Zeuthen [30] (also see Fulton [7, section 10.4]) that shows that the characteristic numbers for a family of curves determine the number of such curves tangent to smooth curves of arbitrary degrees. We close Section 2 by interpreting this theorem for line arrangements. It would be interesting to use intersection theory on the moduli space of stable maps to recover our results.
In Section 3 we consider cones over generic arrangements of hyperplanes.
Definition 2.
A coned generic arrangement in is an arrangement of hyperplanes obtained from a generic hyperplane arrangement in a linear subspace by taking the cone with a dimensional linear space (equivalently, with general points). That is, there exists a linear space of dimension , disjoint from so that each hyperplane in is the linear span of both a hyperplane in and .
In Section 3 we answer the enumerative problems from Question 1 for coned generic arrangements. Here the methods of Schubert Calculus come into play and the Catalan numbers make a cameo appearance.
Our approach to computing is to first compute the number of labeled arrangements with intersection lattice isomorphic to that pass through points in general position in . Dividing by the number of ways to label the hyperplanes in gives . This allows us to work in a product of polynomial rings rather than its quotient. A recent paper by Fehér, Némethi, and Rimányi [5] also studies enumerative problems involving hyperplane arrangements; they embrace quotient varieties and work with equivariant cohomology.
Though we were not able to answer Question 1 in terms of the combinatorics of , we remain optimistic about this possibility for special families of arrangements despite several warning signs that the question may be very difficult in general. Mnëv’s Universality Theorem says that each variety appears as the closure of for some hyperplane arrangement , so the geometry of can be arbitrarily complicated. Another warning sign appears if we take a naive approach to the dimension problem in . For each line arrangement with labeled lines and labeled points of intersection among these lines, form the parameter space
Note that . If the conditions are algebraically independent, then this dimension is (sometimes this is called the virtual dimension of ). Such a formula holds for generic line arrangements and pencils as well as for many other arrangements; however, it fails for the Pappus arrangement pictured in Figure 1, since one of the conditions is implied by the others. Indeed, the dimension of depends on the syzygies among these incidence conditions and so any potential algorithms need to be sensitive enough to recognize such syzygies from the combinatorial information in , a task that appears to be quite difficult. This example (more precisely, its projective dual) was studied in detail by Fehér, Némethi, and Rimányi [5] and by Ren, RichterGebert and Sturmfels [21].
Remark 3.
We use the notation (sometimes omitting the braces) to refer to the multiset consisting of copies of , copies of , et cetera. Similarly, we use to refer to the multinomial that counts the ways of choosing groups of objects, groups of objects, , and groups of objects from labeled objects. That is, .
2 Generic Arrangements
An arrangement of hyperplanes in is said to be generic if and no point is in the intersection of more than of the hyperplanes. Carlini discovered the following fact while studying the Chow variety of zero dimensional degree cycles in [3, Proposition 3.4].
Theorem 4.
When is a generic arrangement of hyperplanes in then the dimension of is and the number of arrangements with lattice type isomorphic to that pass through points in general position in is
Proof.
The moduli space is a to 1 cover of the complement of a closed set in that parameterizes the sets of hyperplanes that contain a set of linearly dependent hyperplanes. So . We count the ordered (or labeled) generic arrangements passing through points in general position. Since the points are in general position, no more than can lie on any one hyperplane. So by the pigeonhole principle, each hyperplane contains precisely of the points and each hyperplane is completely determined by these points. There are clearly ways to distribute the points among the labeled hyperplanes, but each hyperplane arrangement can be labeled in ways so dividing the product of binomial coefficients by gives . ∎
Remark 5.
We remark that when the formula for reduces to . Aside from the nice simplicity of this result, this formulation allows us to interpret the result in terms of the multivariate Tutte polynomial of the lattice . The multivariate Tutte polynomial of (see Ardila [2] or Sokal [23]) is defined as
where if then When is a generic arrangement with hyperplanes,
In particular,
Remark 6.
The intersection ring (or Chow ring) of a variety can be defined as the ring of equivalence classes of algebraic subvarieties of modulo rational equivalence, graded by codimension [7]. Here . Also, if and intersect transversely. The intersection rings that we consider can also be interpreted in terms of cohomology; for more on this point of view, see Katz [13].
When is a product of projective spaces then
where the are the pullbacks of the classes of hyperplanes on each factor to the product.
Next we prove a Lemma that will be used in many of the following arguments.
Lemma 7.
Fix an arrangement in with and . The conditions that the arrangement contain specified points in general position are transverse and the class corresponding to the intersection of these conditions is
Proof.
The class of the condition that a point lies on the th hyperplane is just . Then the class of the condition that a point lies on one of the hyperplanes is . Now, the action of is transitive on . Since the point conditions are assumed to be generic we can use this action to move one point condition to another. Hence Kleiman’s Transversality Theorem [14] says that these conditions are transverse and the corresponding Chow class is the product of the classes. ∎
Note that Theorem 4 could also be easily proved by Lemma 7. Since we are not putting any conditions on the generic lines we have nothing but point conditions. Then the degree of the Chow class given in Lemma 4 counts the number of ordered generic hyperplane arrangements passing through points in general position. Dividing by gives the number of such unordered generic arrangements.
Continuing to focus on the case where , we will compute the characteristic numbers of generic line arrangements with or lines. We define the characteristic number to be the number of arrangements with the same lattice type as that pass through points and are tangent to lines in general position (with ). When is a generic arrangement of lines, we simplify the notation to . To get these numbers we compute the degree of a related variety which is a delicate intersection calculation. We do this in two different ways. First for the case we preform an explicit calculation using the method of undetermined coefficients. Then for the case we argue that the corresponding intersection class is a product using transversality considerations.
Remark 8.
The only way that an arrangement of 3 lines can be tangent to a given line is if the cubic defining the arrangement meets the line at a point of multiplicity strictly greater than 1. If the arrangement is generic this means that an intersection point of two of the 3 lines must lie on . This motivates the introduction of the points in the following lemma: a generic arrangement is tangent to a line if and only if there exists a point on .
Lemma 9.
The set
is a quasiprojective variety of dimension 6 in . The Chow ring is
where the are the classes of the lines and the are the classes of the points. The class of in is
Proof.
Since has codimension 6 the class is a degree 6 polynomial in the Chow ring . We write this class as the sum of all degree 6 monomials in :
Now we determine all the coefficients. First we show which coefficients are zero. To do this we first partition the set of degree6 monomials that appear in the expanded product into types. In the definitions that follow the notation denotes a degree monomial in the variables appearing in the quotient ring .
The first type of monomials we define are
The second type we define are
The third type we define are
The fourth type we define is
Note that , , , and .
For any monomial we define the complement of by the unique monomial such that . Set . Notice that has the property that . Hence to show that the monomials in have the coefficient of zero in we just check that multiplying by each of the monomials in gives zero. Also, notice that within each type there is a union of smaller sets where each of those sets can be obtained from one another by a permutation of the coordinates. Hence if we can show that the coefficient is zero for one of the subsets then we have completed the task for the entire type. In the following arguments we use duality on often.
First we argue where . Let be in the first subset so . Then in this monomial represents fixing a generic line for in the first factor of and a generic point for . Set to be the projection to the factor corresponding to the line and to be the projection to the factor corresponding to the point . With this notion we can say where we will not need the conditions from . Using the Moving Lemma (see [4, Theorem 5.4])
In the point must lie on and this cannot happen if and are generic. Hence
The other types have similar arguments. We include them for completeness, but with briefer arguments. For type 2 we look again at the first subset and put . Then equals
This means that we have fixed generic lines for lines 1 and 2. Hence the point is fixed using , but must also lie on a fixed line. We cannot find such a configuration; hence .
For type 3 we again look at the first subset . Then equals
Line must satisfy the condition coming from , which can be interpreted as saying that line lies in a fixed pencil of lines; however, this is inconsistent with the conditions requiring points and to lie in fixed (and general) positions. Hence .
For type 4 we just look at the first monomial . Then equals
In this case must be a fixed generic line and must be a fixed generic point. Since line must lie in a fixed pencil and since , the line is fixed too. Then is fixed and so it cannot satisfy the generic linear condition in the class . Hence .
In this case we have fixed a generic line for line 1 and force to be fixed point in general position. The condition corresponding to forces line 2 to lie in a fixed pencil. Then line 2 is determined (since it passes through ) and so is . However, the condition corresponding to forces to lie on a fixed general line (that is, not through ). Thus, we cannot find such a configuration and so .
Now we define another set of monomials and prove that each of these monomials have coefficient equal to 1 in . Again we define this set as a union of a few different types. Unfortunately, there are many more types in this set even though there are less total monomials. We define the types in the following table.
Type  Monomials 
Now put . In the next table we compute the intersection of where . We briefly describe, for each type, how this intersection results in a unique arrangement of three lines with the marked points . Again we use the argument that within each we only need to check one monomial since all the others within that class can be obtained from permuting coordinates.
Type  Monomial  Intersect this class with gives a unique point 
Fixing three lines fixes the three intersection points.  
Fixing lines 1 and 2. Then fixing a linear condition  
point 13 gives a unique point and then putting  
a linear condition on line 3 fixes it.  
Fixing lines 1 and 2. Then giving linear conditions on points  
13 and 23 fixes those points which fixes line 3.  
Fix line 1 and point 23. Then putting linear conditions on lines  
2 and 3 fixes them.  
Fix line 1. Then Putting a linear condition on points  
12 and 13 fixes them on line 1.  
Then putting linear conditions on lines 2 and 3 fixes them.  
Fixing line 1 and giving a linear condition on point 12  
fixes point 12. Together with putting a linear condition on  
line 2 fixes it. Then putting a linear condition on point 23  
fixes it and line 3.  
Fixing line 1, point 23, and putting a linear condition on  
line 2 fixes it. Then putting a linear condition on point 13  
fixes it on line one and hence fixes line 3.  
Fix line 1 and putting liner conditions on points  
12 and 13 fixes these points. Then putting a linear  
condition on line 2 fixes it. And then finally fixing a linear  
condition on point 23 fixes it and line 3.  
Fix line 1 and point 23. Putting linear conditions  
on points 12 and 13 fixes them on line 1. Then this  
fixes lines 2 and 3.  
Fixing point 12 with putting linear conditions  
on lines 1 and 2 fixes them. Then putting a linear condition on  
point 13 fixes it on line 1. Then putting a linear condition  
on line 3 fixes it.  
Fixing point 12 with putting linear conditions  
on lines 1 and 2 fixes them. Then putting linear conditions on  
points 13 and 23 fixes them and hence line 3.  
Fixing point 13 and putting a linear condition on  
line 1 fixes it. Then putting a linear condition on point 12 fixes  
it on line 1. Then the linear condition on line 2 fixes it.  
Then the linear condition on point 23 fixes it and line 3.  
Fixing points 13 and 23 fixes line 3. Then putting linear  
conditions on lines 1 and 2 fixes them  
Fixing points 12 and 23 fixes line 2. Then giving a linear  
condition on line 1 fixes it. Then putting a linear condition  
on point 13 fixes it on line 1 which also fixes line 3.  
The points 12, 13, and 23 are fixed. This fixes lines 1,2, and 3. 
Again since the complement we have shown that the coefficients of all these monomials in are equal to 1. There is just one more monomial to consider. Let . This set also has the property . Now we compute its coefficient. The class puts a linear condition on each line and point. For the lines we can think of this as fixing three points , , and in general position and requiring that , for . Similarly, to interpret the linear conditions on the points we fix three lines , , and in general position, and require that .
Suppose now that we choose an arbitrary point on line . Fix a embedding defined by where and each is homogeneous of degree 1. Then there exists such that . Now since the points and are fixed we have that line 1, , is fixed. Line fixed means that we can get . Next we find is determined by the points and . With we can get . Now we make the line through the points and . Since we chose randomly at the beginning of this process we have that may not be equal to . Each step in the above process to determine each line and point is a cross product computation with a vector of constants (the line or point that is fixed before the choice of ) with a vector of homogenous degree 1 in the functions . Hence there exist linear functions and such that . Now if and only if
Since this a homogeneous quadratic polynomial of two variables there must be 2 solutions up to multiplicity. Hence the coefficient of the monomial is 2.
The total number of terms in 141 (this can be calculated by a standard inclusionexclusion argument). Since , , and we have computed the coefficients of all possible terms. Finally with any computer software system one can expand the polynomial and see that the coefficients match what we have just determined. This completes the proof. ∎
Remark 10.
A generic element of can be thought of as a generic arrangement of 3 lines. The complement inside of this generic locus consists of several 5 dimensional subvarieties. These consist of arrangements where two lines are the same and arrangements consisting of a triple line, together with their marked points.
Now we determine the characteristic numbers for 3 lines.
Theorem 11.
The characteristic numbers for a generic arrangement of 3 lines are given in the following table.
0  1  2  3  4  5  6  


15  30  48  57  48  30  15 
Proof.
If we intersect with 6 generic conditions then we will obtain a finite number of points none of which will lie on the 5dimensional nongeneric subvarieties. So, when we multiply the class with that from Lemma 7 we will only be counting generic arrangements of 3 lines. From Lemma 9 we have
Then the condition that the arrangement passes through a given point has class . Similarly the condition that a given line contains one of the has class . Using the Moving Lemma, Lemma 7, and Remark 8 the degree of the class in counts the number of labeled 3generic arrangements that pass through general points and are tangent to general lines in . To remove the effect of the labeling, we divide these numbers by to obtain the characteristic numbers. ∎
Remark 12.
The symmetry of is easily explained: the dual of a 3generic arrangement through points and tangent to lines is a 3generic arrangement through the dual points and tangent to the dual lines . This symmetry does not occur in the 4 lines case as one can see below.
Now we consider the 4 line case. We compute the class as before but we use transversality arguments in place of the method of undetermined coefficients.
Lemma 13.
The set
is a quasiprojective variety of codimension 12 in . The Chow ring of the ambient space is
where the correspond to the lines and the correspond to the intersection points . In this ring the class of is
Proof.
As in Lemma 9 set to be the projection to the factor corresponding to the line and to be the projection to the factor corresponding to the point . The condition