Plucker's $\mu$

Question

In Jürgen Richter-Gebert's book "Perspectives on Projective Geometry", he talks about Plucker’s $\mu$ in Section 6.3. He says that this trick was used by Plucker quite often.

Plucker's trick involves finding the equation of a curve in a plane which passes through the intersection of two geometric objects of the same type. For example, if $f(x,y) = ax+by+c$, then $f \equiv 0$ represents a straight line. If we want the equation of a line passing through the intersection of two lines $f_1$ and $f_2$ and a point $(u,v)$, then according to Plucker, it is evidently $$f_1(u,v)f_2(x,y) - f_1(x,y)f_2(u,v).$$

One can do this for conics as well. If $f_1$ and $f_2$ represents two conic equations, then the above form represents the conic passing through the intersections of the two conics and passing through the point $(u,v)$.

I want to know some more applications of this trick. For example, if two conics do not intersect at four points, Plucker's trick yields a conic. What is the meaning of this conic?

As a special case, if two circles do not intersect, then Plucker's trick gives a circle passing through the point $(u,v)$. But I am not able to understand the significance of such a circle.

Answer 1

Two conics in general position will always intersect at four points in the complex projective plane, and the conic you get out of the trick will (by purely algebraic considerations) go through $(u,v)$ and those four (possibly complex) points.

In the degenerate case where there are points of tangency instead of merely intersection, the trick conic will -- by continuity -- share the common tangent of the two original conics at those points.

Answer 2

If two conics do not intersect at four points, Plucker's trick yields a conic. What is the meaning of this conic?

As Henning pointed out, two conics have four points in common if you also consider complex solutions and if you take multiplicities into account, i.e. contact points are double points of intersection.

If two circles do not intersect, then Plucker's trick gives a circle passing through the point $(u,v)$. But I am not able to understand the significance of such a circle.

One thing this circle would have in common with the two defining circles is a shared axis of symmetry.

Also, as one special case, the pencil of circles obtained as linear combinations of two non-intersecting circles includes the radical axis of the circles, as a degenerate circle of infinite radius. Projectively speaking, that radical axis would be described as a conic that factors into two lines, namely that radical axis and the line at infinity. The pencil of circles would also include two circles of radius zero, which are in fact each the sole real solution of a pair of otherwise complex lines.

I want to know some more applications of this trick.

I'm using Plücker's $\mu$ regularly. Here are some examples I found:

• Given an implicit 3D plane, how do I find the orthogonal projection matrix…
• Projecting a point on a plane through a matrix which is similar to the first
• Inverse points concurrence on the circumcircle with a pencil of circles
• If a hyperbola whose foci are (–2, 4) and (4, 6) touches y–axis… using dual conics
• A chain of six circles associated with a cyclic hexagon applied to Möbius geometry

Bonus section: I also often think of the following as a generalization of this trick: suppose you want to intersect a conic $A$ with a line $P\vee Q$ and you already know that $P$ is on the conic, i.e. $P^TAP=0$. Then the second point of intersection is some $R=\lambda P+\mu Q$ such that

$$0=R^TAR=(\lambda P+\mu Q)^TA(\lambda P+\mu Q)=\lambda^2\underbrace{(P^TAP)}_{=0}+2\lambda\mu(P^TAQ)+\mu^2(Q^TAQ)\\ 0=2\lambda(P^TAQ)+\mu(Q^TAQ)\\ \lambda=Q^TAQ\qquad \mu=-2P^TAQ\\ R=(Q^TAQ)P-2(P^TAQ)Q$$

Knowing that one term is zero turns the quadratic problem into a linear one, and the same trick can be applied in a slightly less symmetric fashion to again obtain a simple description of the result with no divisions or case distinctions required. See e.g. Beautiful triangle problem for this trick in action.

Answer 3

Another famous application is a simple proof of Pascal's Theorem:

Take $C_2$, a plane conic, with six points $P,Q,R,P',Q',R'$ and the lines $l_1 = PQ'$, $l_1'=PR'$, $l_2'=QP'$, $l_2=QR'$, $l_3=RP'$, $l_3'=RQ'$.

Then $f = l_1 l_2 l_3 - \mu l_1' l_2' l_3' = 0$ (where the line names stand for their linear equations) describes a family of cubics $Y_\mu$, which have the six given points in common with $C_2$. Alltogether there are nine points where one $l_i$ intersects one $l'_j$. Now take a special $\mu_0$ such that $Y'=Y_{\mu_0}$ has a seventh point in common with $C_2$. By Bezout's theorem $Y'$ contains $C_2$ and so splits into $C_2$ and a line. This line contains the three $l_i \cap l'_j$ which are not on $C_2$ and this is exactly the statement of Pascal's theorem.

I took this example from Felix Klein, "Vorlesungen über die Entwicklung der Mathematik im 19. Jahrhundert", Springer Reprint 1979, p.122,123 - an extraordinary interesting book which gives a supreme panoramic view on the mathematics of the 19th century and the mathematicians that built it.