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Background

I am interested in the projective classification of reduced curves of degree four in $\mathbb{P}^3(\mathbb{R})$ (and more generally of degree $n+1$ in $\mathbb{P}^n(\mathbb{R})$). More precisely, I am looking at the case where the curve is a union of four distinct lines. I need this classification because I want to make sure that I consider all possible cases in a problem in interpolation theory.


For instance, there are two types of configurations of three lines in $\mathbb{P}^2$. Either three lines meet in a single point, or three lines meet in three distinct points. More generally, according to this integer sequence, there are 3 configurations of four lines in $\mathbb{P}^2$, 5 configurations of five lines in $\mathbb{P}^2$, and 18 configurations of 6 lines in $\mathbb{P}^2$. These configurations are shown in this figure (except for the configurations in which all lines are concurrent).

I believe there are six configurations of three lines in $\mathbb{P}^3$: Two configurations for which the three lines lie in a plane, three configurations for which precisely two of the three lines lie in a plane, and one configuration where none of the lines intersect.

My (related) questions are now as follows:

  1. How many configurations are there of four lines in $\mathbb{P}^3$ (and more generally of $n+1$ lines in $\mathbb{P}^n$)?
  2. Is there a convenient way to enumerate these?
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Four lines through a point in $\mathbb P^2$ is the same as four points on $\mathbb P^1$. These configurations are not all projectively equivalent as the $j$-invariant (symmetrization of the cross-ratio) will distinguish them. Thus there are infinitely many projectively distinct configurations of $n+1$ lines in $\mathbb P^n$ when $n \ge 3$. –  jvp Aug 13 '10 at 12:48
    
@jvp: You beat me while I was typing! I had started counting configurations by intersection possibilities, and then realized the above, and made it an answer. –  Charles Siegel Aug 13 '10 at 12:54
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Are you sure you want to study configurations up to projectivities? Charles Siegel's answer shows that this is not a combinatorial problem. Maybe you want instead to look at configuration of lines up to homotopies that preserve crossing (along the homotopy, lines that initially cross should cross all the time and lines that does not must not cross at any time). –  Benoît Kloeckner Aug 13 '10 at 13:00
    
This is really helpful: Now I know that projective invariance is not the right notion for my problem. I need to reconsider. –  Georg M. Aug 13 '10 at 13:19
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I don't understand where does the configuration of the 3 coordinate axes (projectively completed) falls into your classification of triples of lines in $\mathbb{P}^3:$ each pair of lines lies in a plane, yet all lines do not. Also, if $l_1$ and $l_2$ are skew lines and $l_3$ intersects both of them then there are precisely two pairs of coplanar lines. –  Victor Protsak Aug 14 '10 at 5:10
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2 Answers

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Up to projectivities, there are uncountably many configurations. Let's do the naive dimension count: The Grassmannian of lines in $\mathbb{P}^3$ is four dimensional, so the parameter space for four lines is 16 dimensional. The automorphism group of $\mathbb{P}^3$, the projections, is made up of four by four matrices modulo the diagonal matrices, so is dimension 16-1=15. So we should get a whole curve worth of possible configurations.

This is analagous to how you can use cross ratio to distinguish different configurations of four points in $\mathbb{P}^1$.

EDIT: To make this more general and distinguish it from jvp's comment, if you look at $n+1$ lines in $\mathbb{P}^n$, then you have $(n+1)^2-1=n^2+2n$ automorphisms, and the space of lines is $2(n-1)$ dimensional, so in general you have $2(n-1)(n+1)-(n+1)^2+1$ dimensions worth of configurations, and this simplifies to $n^2-2n-2$.

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This is one of my favourite projective geometry examples; it is a case of classification involving moduli and if done right most of the arguments can be done with a combination of geometry and linear algebra with no explicit calculation.

I assume that we are talking about an ordered quadruple of lines in $\mathbb P^3(k)$ (I do it over any field). The trick is to not think of $\mathbb P^3$ as the set of lines in $k^4$ but rather of lines in some $4$-dimensional vector space V and then use the date to get closer to an adapted coordinatisation. Assume first that no two of the lines are skew. Then a simple geometric argument shows that either all lines lie in a plane or pass through a common point. Both of those cases are reduced to the problem of four points in $\mathbb P^2$ which I skip. We can then assume that the first two lines are skew and think of $V$ as $V_1\bigoplus V_2$, where the projectivisations of $V_1$ and $V_2$ are the two first lines. Assume then that the third line is skew with the first and second line. This means that it is the projectivisation of the graph $V_3\subset V_1\bigoplus V_2$ of an isomorphism. Hence, we may assume that $V_1=V_2$ and $V_3$ is the diagonal in $V_1\bigoplus V_1$. If we also assume that the fourth line is skew with the first two, then it is also the graph $V_4\subset V_1\bigoplus V_1$ of an automorphism $V_1\rightarrow V_1$. Hence, four lines, the first two of which are skew and the last two are skew with the first two up to projective transformations correspond to isomorphism classes of pairs $(V_1,\varphi)$ where $V_1$ is a two-dimensional vector space and $\varphi$ is an automorphism of it distinct from the identity. This is the same thing as conjugacy classes of $\mathrm{GL}_2(k)$ distinct from the identity element. The condition that the last two lines be skew is exactly that $\varphi$ does not have $1$ as an eigenvalue. Of course the characteristic polynomial distinguish between conjugacy classes so there are continuous families of configuration (i.e., it has non-trivial moduli).

The remaining case of two skew lines and two lines which are not both skew with respect to both of the first lines is easy but a little bit tedious; given a pair of non skew lines one looks at the plane spanned by them and the position of the other lines with respect to it.

Addendum: I did not mean to suggest that it is the simplest problem with proper moduli. Of course the classification of four points in $\mathbb P^1$ has the cross ratio is its moduli. (The classification is proven almost word for word in the same way as the above case.)

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Thank you, that was very helpful. –  Georg M. Aug 18 '10 at 10:38
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