Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:27:19Z http://mathoverflow.net/feeds/question/35479 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35479/up-to-projectivities-which-configurations-of-four-lines-in-mathbbp3-can-on Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish? Georg M. 2010-08-13T11:48:19Z 2010-08-13T13:49:24Z <h2>Background</h2> <p>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.</p> <hr> <p>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 <a href="http://www.research.att.com/~njas/sequences/A132346" rel="nofollow">this integer sequence</a>, 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 <a href="http://www.research.att.com/~njas/sequences/a048872.pdf" rel="nofollow">this figure</a> (except for the configurations in which all lines are concurrent).</p> <p>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.</p> <p>My (related) questions are now as follows:</p> <blockquote> <ol> <li>How many configurations are there of four lines in $\mathbb{P}^3$ (and more generally of $n+1$ lines in $\mathbb{P}^n$)?</li> <li>Is there a convenient way to enumerate these?</li> </ol> </blockquote> http://mathoverflow.net/questions/35479/up-to-projectivities-which-configurations-of-four-lines-in-mathbbp3-can-on/35490#35490 Answer by Charles Siegel for Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish? Charles Siegel 2010-08-13T12:53:09Z 2010-08-13T12:53:09Z <p>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.</p> <p>This is analagous to how you can use cross ratio to distinguish different configurations of four points in $\mathbb{P}^1$.</p> <p>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 $n^2-2n-2$.</p> http://mathoverflow.net/questions/35479/up-to-projectivities-which-configurations-of-four-lines-in-mathbbp3-can-on/35494#35494 Answer by Torsten Ekedahl for Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish? Torsten Ekedahl 2010-08-13T13:39:50Z 2010-08-13T13:49:24Z <p>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.</p> <p>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 <em>and</em> 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).</p> <p>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.</p> <p><b>Addendum</b>: 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.)</p>