cross-ratios of the base points and base lines of two conics

11 views
Skip to first unread message

David Madore

unread,
Jun 28, 2007, 12:43:34 PM6/28/07
to

Hi. Here's a problem of classical plane projective geometry
concerning certain invariant cross-ratios defined by two conics:

Let C and D be two plane conics (in sufficiently general position),
p1, p2, p3, p4 the four intersection points of C and D, and l1, l2,
l3, l4 the four common tangents (i.e., intersections of the dual
concics, C* and D*). Is the following statement true? After possibly
reordering, the cross-ratio of p1, p2, p3, p4 on C is equal to the
cross-ratio of l1, l2, l3, l4 on D* (which is the same as the
cross-ratio of the points of tangency of l1, l2, l3, l4 on D).

I found the following proof that the answer is "yes", but I'm not
fully convinced it's correct and, even if it, it's certainly *not* the
right way to do it:

Consider the set E of pairs (p,l) where p is on C and l is on D*
(i.e., l is a line tangent to D) such that p lies on l. Then E
(birational to) the intersection of two sufficiently general quadrics
in P^3, so E is a curve of genus 1. Now consider the two projections
f:E->C and g:E->D* (sending (p,l) to p and l respectively): both are
of degree 2. It is easy to see that the criticial values of f are p1,
p2, p3, p4, and those of g are l1, l2, l3, l4. So if we let gamma and
lambda be the respective cross-ratios, the j-invariant of E can be
computed as 256 (gamma^2-gamma+1)^3/(gamma^2(gamma-1)^2) (because the
critical points of f on E define a divisor of degree 4 which gives a
Weierstrass form of E) or exactly the same formula with lambda instead
of gamma: so both are equal up to action of S_4, QED.

(The curve E is classically introduced in the study of Poncelet's
theorem which deals, essentially, with order of the element of J(E)
given by the composition of the involutions defined by f and g. So
this doesn't come from nowhere. But it would seem very strange if
this couldn't be proved without appealing to the j-invariant of E.)

Bonus question: There are some other natural cross-ratios which we can
naturally associate to two general conics C and D, such as this:
parametrize the pencil to which C and D belong so that the three
singular members correspond to 0, 1 and infinity, then the parameters
of C and D in the pencil give natural cross-ratios associated to the
situation - or we can do the same for the duals. How can these
cross-ratios be related to those defined above?

--
David A. Madore
(david....@ens.fr,
http://www.dma.ens.fr/~madore/ )

David Madore

unread,
Jun 28, 2007, 3:25:45 PM6/28/07
to

Sorry for replying to myself. I can make an additional remark which
sheds some light on the problem:

David Madore in litteris <f60ofm$7uj$1...@dizzy.math.ohio-state.edu>
scripsit:


> Let C and D be two plane conics (in sufficiently general position),
> p1, p2, p3, p4 the four intersection points of C and D, and l1, l2,
> l3, l4 the four common tangents (i.e., intersections of the dual
> concics, C* and D*). Is the following statement true? After possibly
> reordering, the cross-ratio of p1, p2, p3, p4 on C is equal to the
> cross-ratio of l1, l2, l3, l4 on D* (which is the same as the
> cross-ratio of the points of tangency of l1, l2, l3, l4 on D).

<snip>


> Bonus question: There are some other natural cross-ratios which we can
> naturally associate to two general conics C and D, such as this:
> parametrize the pencil to which C and D belong so that the three
> singular members correspond to 0, 1 and infinity, then the parameters
> of C and D in the pencil give natural cross-ratios associated to the
> situation - or we can do the same for the duals. How can these
> cross-ratios be related to those defined above?

Here goes the "bonus" part. Let C and D be two conics. Then we can
show that the cross-ratio of the four intersection points (p1, p2, p3,
p4) on C is the same as the cross-ratio of C, X, Y, Z in the linear
pencil L of conics spanned by C and D, where X, Y, Z are the three
degenerate conics in the pencil L (and, more precisely, X is the union
of the lines p1 p2 and p3 p4, Y of p1 p3 and p2 p4 and Z of p1 p4 and
p2 p3). To see this, consider the set p1* of lines through p1: if we
map p1* to C by taking a line through p1 to its second intersection
point with C, we see that the cross-ratio of p1, p2, p3, p4 on C is
the same as the cross-ratio of the lines c1, p1 p2, p1 p3 and p1 p4
through p1, where c1 is the tangent to C through p1. Now giving a
line l through p1 determines a unique conic in L, namely the unique
conic through p1, p2, p3, p4 and having tangent l at p1: this map
takes l=c1 to C, l = p1 p2 to X and so on, so the cross-ratio of c1,
p1 p2, p1 p3, p1 p4 is the cross-ratio of C, X, Y, Z in L. QED.

One might think that this concludes the original question, but it does
not: if we call L* the set of dual conics to those of L (i.e., dual
conics which are tangent to the four dual lines p1*, p2*, p3*, p4*),
then L* is a rational curve in the space of (dual) conics but it is
not a line. On the other hand we also have the linear pencil M*
spanned by C* and D*, and M the set of dual conics: so M is the set of
conics simultaneously tangent to l1, l2, l3, l4. The argument above
shows that the cross-ratio of l1, l2, l3, l4 on D* is the cross-ratio
of D and three degenerate conics (the double lines passing through two
of the intersection points of l1, l2, l3, l4), so to conclude we would
need, for example, to find a way to relate L and M, but I can't see
that.

Ilya Zakharevich

unread,
Jun 28, 2007, 7:45:01 PM6/28/07
to

[A complimentary Cc of this posting was sent to
David Madore
<david....@ens.fr>], who wrote in article <f60ofm$7uj$1...@dizzy.math.ohio-state.edu>:

> Let C and D be two plane conics (in sufficiently general position),
> p1, p2, p3, p4 the four intersection points of C and D, and l1, l2,
> l3, l4 the four common tangents (i.e., intersections of the dual
> concics, C* and D*). Is the following statement true? After possibly
> reordering, the cross-ratio of p1, p2, p3, p4 on C is equal to the
> cross-ratio of l1, l2, l3, l4 on D* (which is the same as the
> cross-ratio of the points of tangency of l1, l2, l3, l4 on D).

a) It is enough to show this on a Zariski open subset of the moduli space;
a') It is enough to show this on a subset the PSL(3)-"orbit" of which is
Zariski open.

Actually, we use a following modification of (a):

b) On a Zariski open subset of the moduli space, the pair (C,D) is
projectively equivalent to the pair (D*, C*).

Obviously, (b) implies your statement. Moreover, one change (b) to
the "primed" variant, so it enough to consider pairs C,D of the form

A x^2 + B y^2 = 1
B x^2 + A x^2 = 1

- and for these pairs the statement is obvious.

Hope this helps,
Ilya

David Madore

unread,
Jun 29, 2007, 7:55:00 AM6/29/07
to

Ilya Zakharevich in litteris <f61h5t$1dc$1...@dizzy.math.ohio-state.edu>
scripsit:

> the pair (C,D) is
> projectively equivalent to the pair (D*, C*).

Indeed, this is a very nice way of stating the problem! And I agree
that it then becomes pretty obvious (I would tend to state it in terms
of simultaneous reduction of quadratic forms of rank 3). Thanks for
the enlightenment.

Geometrically this is still a bit unsatisfactory, because we would
want to construct an explicit projective map taking a point of C to a
tangent to D and vice versa. But I guess this involves choosing one
of four possible mappings of the intersection points of C and D to the
common tangents to C and D so it can't be canonical. Or one would
want to map a conic with the four common intersection points with C
and D to one with the four common tangents - or something.

Ilya Zakharevich

unread,
Jun 29, 2007, 8:45:00 PM6/29/07
to

[A complimentary Cc of this posting was sent to
David Madore
<david....@ens.fr>], who wrote in article <f62ruk$357$1...@dizzy.math.ohio-state.edu>:

>
> Ilya Zakharevich in litteris <f61h5t$1dc$1...@dizzy.math.ohio-state.edu>
> scripsit:
> > the pair (C,D) is
> > projectively equivalent to the pair (D*, C*).
>
> Indeed, this is a very nice way of stating the problem! And I agree
> that it then becomes pretty obvious (I would tend to state it in terms
> of simultaneous reduction of quadratic forms of rank 3). Thanks for
> the enlightenment.
>
> Geometrically this is still a bit unsatisfactory, because we would
> want to construct an explicit projective map taking a point of C to a
> tangent to D and vice versa. But I guess this involves choosing one
> of four possible mappings of the intersection points of C and D to the
> common tangents to C and D so it can't be canonical.

I'm not yet convinced. Basically, you choose an isomorphism (C,D) -->
(D*,C*) in one point of moduli space of pairs (C,D); then one needs to
calculate the corresonding monodromy map on the moduli space. If it
is non-trivial, THEN your argument works.

However, if monodromy is trivial, there should be a canonical map.
Unfortunately, I can't find a simple model (like the pair of C and D I
initially wrote) which works near the boundary of the moduli space, so
one can visualize the monodromy...

Hope this helps,
Ilya

Reply all
Reply to author
Forward
0 new messages