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 mathbbP3(k) (I do it over any field). The trick is to not think of mathbbP3 as
the set of lines in k4 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 mathbbP2 which
I skip. We can then assume that the first two lines are skew and think of V as
V1bigoplusV2, where the projectivisations of V1 and V2 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 V3subsetV1bigoplusV2 of an isomorphism. Hence, we may assume that V1=V2 and
V3 is the diagonal in V1bigoplusV1. If we also assume that the fourth
line is skew with the first two, then it is also the graph V4subsetV1bigoplusV1 of an automorphism V1rightarrowV1. 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
(V1,varphi) where V1 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 mathrmGL2(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 mathbbP1 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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