I'm stuck on a small detail in Proposition 1.2.8 in Geiges' Introduction to Contact Topology.

Let $C$ be a conic in $\mathbb{R}P^2$ given by $q^tAq=0$, where $A$ is a nonsingular, symmetric 3x3 matrix. Assume $A$ is index 2, that is, $C$ is an ellipse.

Define the correlation $\varphi:\mathbb{R}P^2\rightarrow(\mathbb{R}P^2)^\ast$ by $\varphi(q)=q^tA$ and $\varphi^*:(\mathbb{R}P^2)^\ast\rightarrow\mathbb{R}P^2$ by $\varphi(p)=A^{-1}p^t$.

If $x$ is a point on $C$, we would like to show that $\varphi(x)$ is a line tangent to $C$ at $x$. Observe that $\varphi(x)x=x^tAx=0$. Geiges says, "[therefore] $x$ lies on its polar $\varphi(x)$." This is the point I am confused about. Here's why:

One can imagine $\mathbb{R}P^2$ as a northern hemisphere in $\mathbb{R}^3$ (with antipodal boundary points identified). Given a point $p$ on this hemisphere, consider the tangent plane to the sphere at $p$, and now parallel transport it to intersect the origin of $\mathbb{R}^3$. The points of intersection of this plane with the hemisphere should be exactly the set $p^\perp=\{q\in\mathbb{R}P^2\mid q^tp=0\}$. My reasoning for this is that the lines through the origin of the tangent plane are exactly the lines perpendicular to the line through $p$ and the origin of $\mathbb{R}^3$, so moving the tangent plane to the origin of $\mathbb{R}^3$ does not change the vectors, and we projectivize them by looking at the intersections with the hemisphere.

The problem is that, by this construction, it's obvious that $p\not\in p^\perp$, so how could $x$ lie on the polar $\varphi(x)$ if $\varphi(x)x=0$?

  • $\begingroup$ What does your visualization of the tangent plane of the sphere in $\Bbb R^3$ have to do with points in $(\Bbb RP^2)^*$? We're interested in the (projective) tangent line of the curve at $p$. So you want to visualize the tangent plane to the surface obtained by lifting $C$ to $\Bbb R^3$. $\endgroup$ – Ted Shifrin Sep 16 '14 at 2:20
  • $\begingroup$ Sure, that's what I'd like the result to be (for an index-2 $A$, the lift is just a cone, and then the tangent plane at $p$ defines the projective line). I'm trying to see this from the formulas though: specifically, why does $\varphi(x)=x^tA$ and $x\in C$ give that $\varphi(x)x=0\implies x\in\varphi(x)$? It's clear that it equals 0, just not why $x\in\varphi(x)$. Unless, as the answerer below suggested, it should be $\varphi(x)=(x^tA)^\perp$, in which case it's definitely true. $\endgroup$ – gmoss Sep 16 '14 at 21:03

If I misinterpreted something, let me know: the presentation above is a little foreign to me.

It seems to me that we're discussing the conic $C=\{q\mid q^tAq=0\}$, and that the pole-polar correlation in question is the one where $q\mapsto (q^tA)^\perp$. That is, the right hand side is a plane in $\Bbb R^3$ that determines a projective line in $\Bbb R P^2$.

If $q\in C$, then $q^tAq=(q^tA)q=0$, but that is saying that $q\in (q^tA)^\perp$, so that $q$ lies on its own polar.

  • $\begingroup$ I'd like this to be true, but the correlation is defined as $\varphi(x)=x^tA$, and not as you wrote. Perhaps a typo? That would be nice... $\endgroup$ – gmoss Sep 16 '14 at 21:13
  • $\begingroup$ @gmoss: But $\varphi(x)$ corresponds to the normal vector as an element of $(\Bbb R^3)^*$, i.e., as a linear functional. $\endgroup$ – Ted Shifrin Sep 16 '14 at 21:39
  • $\begingroup$ @gmoss but a correlation should send points to lines, not to points. That's why I was guessing the $\ast$ indicated we should pass to the orthogonal complement. I think this is what Ted suspects as well, although I don't want to put words in his mouth. $\endgroup$ – rschwieb Sep 17 '14 at 0:26
  • $\begingroup$ @TedShifrin and rschweib: OK that makes perfect sense. I neglected to pay attention to the fact that the point $q\in(\Bbb RP^2)^\ast$ corresponds to the line $p\in \Bbb RP^2$ so that $xq=0 \forall x\in p$. Thank you both! $\endgroup$ – gmoss Sep 18 '14 at 1:15

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