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

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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/ )

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.

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

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.

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

to

[A complimentary Cc of this posting was sent to

David Madore

>

> 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

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