## Classification of Conic Sections

Apollonius of Perga (262 BC) wrote an exhaustive treatise exploring conics. He presented a classification of conic sections by angle. I’ll show you a summary of what he did, and then a conceptually more pleasing and suggestive way to think about it.

Note that Apollonius of Tynea, the Greek philosopher, is a different dude — a contemporary of Christ.

Let’s get the definitions out of the way so we’re all on the same page:

• A cone is the shape generated by considering all lines in space that pass through a fixed circle $$C$$ and a fixed point $$O$$ (the vertex) not in the plane of the circle.
• A conic section is the intersection of a plane with a cone.

Here’s the “angles” way to think about the different sorts of conic sections:

We see that the plane can either intersect the point $$O$$, or not intersect $$O$$. If the plane intersects $$O$$, then it can either intersect no other part of the cone, intersect one lobe of the cone, or intersect both lobes of the cone.

The above picture also makes it clear that there are three categories of thing. Appollonius named these hyperbola, parabola, and ellipse (angle greater than alpha, equal to alpha, or less than alpha). Then we have the “degenerate” hyperbola, parabola, and ellipse. What do I mean by degenerate? I mean the intersecting lines, the double line, and the degenerate ellipse — these guys touch the origin O.

This is the classic method of classification. I find it a bit unsatisfying. It leaves so many of the suggestive questions in the dark!

The cleanest way to classify conic sections is the following observation.

A line can intersect a circle 0 times, 1 time, or 2 times.

Picture the real plane compactified by a point at infinity, we’re in projective space, so it’s safe to think of this as a line at infinity.

Picture the ellipse — it is a circle which doesn’t intersect infinity,

Picture the parabola — it is a circle which intersects infinity once (with a double root).

Picture the hyperbola — it is a circle which intersects infinity twice.

What I’m trying to show you is that the best way to understand the conics is projectively!

This was first appreciated by Girard Desargues, in the late 1630s. Desargues was motivated by very practical considerations. He was trying to develop a complete geometry of the visual world. He wanted a comprehensive framework into which the tricks and techniques of perspective drawing, which had been around since the early fifteenth century, would fit naturally and perfectly.

Unfortunately, he was a terrible expositor. Instead of “straight line,” and “point of intersection,” he spoke of “palms,” and of “fronds,” and all sorts of other things. He used an arboreal vocabulary to speak of geometric ideas.

He wanted to address himself to craftsmen and artisans, who lacked a classical education. People who couldn’t read Euclid. But he was like other great mathematicians who have had profoundly new ideas: he was almost incapable of imagining what it was like not to know what he himself knew. It took a long time for other mathematicians really to assimilate Desargues’ ideas. The process didn’t really get far until about the middle of the 19th century — 200 years after Desargues’ death. The biggest single step was taken by Plücker and Möbius.

Rather than tell you how Desargues thought about it, I’ll give you my take. Let’s start with: conic sections are projective transformations of the circle.

We begin with a circle

\$$x^2 + y^2 – z^2 = 0\$$

Which we may rewrite, for ease of manipulation, in the following form:

\$$\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0\$$

We may then projectively transform the circle:

\$$(\begin{pmatrix} x & y & z \end{pmatrix} \cdot M^T) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \Bigg(M \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix}\Bigg) = 0\$$

Isolating the center matrix, let’s rewrite $$M^T \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} M \text{ as } \begin{pmatrix} a & b_1 & d_1 \\ b_2 & c & e_1 \\ d_2 & e_2 & f \end{pmatrix}$$.

\$$\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} a & b_1 & d_1 \\ b_2 & c & e_1 \\ d_2 & e_2 & f \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0\$$

\$$= ax^2 + (b_1 + b_2) xy + cy^2 + (d_1 + d_2)xz + (e_1 + e_2)yz + fz^2\$$

Tada! We see that any projective transformation of a circle is a quadratic form in 3 variables!

How is this related to the intersection of a conic with a plane? Well, we can view the 3-variable quadratic form

\$$ax^2+bxy + cy^2 + dxz + eyz – fz^2\$$

as the intersection of the 2-variable quadratic form \$$ax^2 + bxy^2 + cy^2 – z^2\$$ with the plane $$z = -(dx + ey + fz)$$

Alternatively, we may examine the intersection of the 3-variable quadratic form, \$$ax^2 + bxy^2 + cy^2 + dxz + cyz – fz^2\$$ with the plane $$z=0$$. We see that this intersection is $$ax^2 + bxy^2 + cy^2$$.

A conic section is a projectively transformed circle. We can classify it by counting the number of times it intersects the line at infinity.

Observe that the number of roots of the polynomial $$Q(x,y) = 0$$ is equal to the number of times the quadratic form $$Q(x,y)$$ intersects the origin.

So, we’ve reduced the problem of classifying conic sections to the problem of counting the solutions of $$ax^2 + bxy^2 + cy^2 = 0$$. Let’s do it!

Since we are in projective space, and currently in $$[x: y: 0]$$, we may rescale $$ax^2 + bxy + cy^2 = 0$$ to the coordinates $$[x: 1: 0]$$ to get a univariate polynomial:

\$$ax^2 + bx + c = 0\$$

How do we solve for the number of roots the polynomial $$ax^2 + bxy^2 + cy^2 = 0$$ has over $$\mathbb{R}$$?

We assume that the polynomial factors — then, we solve for the roots in terms of the coefficients of the polynomial.

\begin{align*}
ax^2 + bxy^2 + cy^2 &= a(x- \beta)(x – \gamma) \\
x^2 + \frac{b}{a}xy^2 + \frac{c}{a}y^2 &= (x- beta)(x – \gamma) \\
&= x^2 – (\beta + \gamma)x + \beta \gamma \\
\frac{b}{a}xy^2 + \frac{c}{a}y^2 &= -(\beta + \gamma)x + \beta \gamma
\end{align*}

\begin{align*}
– \frac{b}{a} &= \beta + \gamma \\
\frac{c}{a} &= \beta\gamma
\end{align*}

Note that we have both the sum $$a+b$$ and the product $$ab$$ of two numbers. How do we find the individual numbers?

We find the square of their difference! As can be seen pictorally:

We proceed undaunted to solve for $$\gamma$$ and $$\beta$$ in terms of the coefficients $$a, b, c$$. Apologies, it’s a bit messy.

\begin{align*}
(\gamma – \beta)^2 &= (\gamma + \beta)^2 – 4\beta \gamma \\
&= \Big(\frac{b}{a}\Big)^2 – 4\frac{c}{a} \\
\gamma – \beta &= \Big(\Big(\frac{b}{a}\Big)^2 – 4 \frac{c}{a}\Big)^{1/2} \\
\end{align*}

So,
\$$(\gamma + \beta) + (\gamma – \beta) = 2\gamma\$$

\$$-\frac{b}{a} +\Big(\Big(\frac{b}{a}\Big)^2 – 4 \frac{c}{a}\Big)^{1/2} = -\frac{b}{a} + \Big(\Big(\frac{b}{a}\Big)^2 – 4 \frac{c}{a}\Big)^{1/2}\$$

Thus, $$\gamma = \frac{-\frac{b}{a} + ((\frac{b}{a})^2 – 4 \frac{c}{a})^{1/2}}{2}$$, and $$\beta = \frac{-\frac{b}{a} – ((\frac{b}{a})^2 – 4 \frac{c}{a})^{1/2}}{2}$$.

Note that we made the the arbitrary choice for $$\beta – \gamma$$ to be the negative root, and $$\gamma – \beta$$ to be the positive root. Until then, everything is symmetric.

Usually, the square $$(\gamma – \beta)^2 = (\frac{b}{a})^2 – 4 \frac{c}{a}$$ is referred to as the discriminant. It’s either one, zero, or two, corresponding to the number of real solutions of the quadratic $$ax^2 + bxy + cy^2$$.

That this conincides with the determinent of the matrix associated to the quadratic form is no accident!

\$$\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\$$

It it is a theorem that the sign of the $$\det \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix} = ac – \frac{b^2}{4} >0$$ implies that the evaluation of $$ax^2 + bxy^2 + cy^2$$ at any point (x,y) is always positive.

In other words:

The determinent of the upper k-minors is positive $$\Rightarrow$$ the quadratic form is positive definite.

In this case, we pick the standard flag, so the upper minors are $$(a)$$ and $$\begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}$$. (We see that the degenerate conics correspond to the projective transformations which have determinant 0.)

Thanks to Laurens Gunnarsen for telling me about Desargues, Ronno Das and Nir Gadesh for helping me to understand the discriminant, and Inna Zakharevich for telling me to write this post.

## A quick comparison of Lie algebras and formal group laws

This post assumes that you are familiar with the definition of Lie group/algebra, and that you are comfortable with the Lazard ring. Note: This is less of an expository post and more of an unfinished question.

Why care about formal group laws? Well, we want to study smooth algebraic groups, but Lie algebras fail us in characteristic p (for example, $$\frac{d}{dx}(x^p) = 0$$), so, rather than a tangent bundle, we take something closer to a jet bundle.

Lie algebras are to smooth groups over $$\mathbb{R}$$ or $$\mathbb{C}$$ as formal group laws are to smooth algebraic groups over any ring $$R$$.

I want to apply this analogy! I want this deeply. I’m trying to puzzle out how to see if this analogy is deep or superficial. How deep does the rabbit hole go? Let’s look at an example.

Given the universal enveloping algebra of a Lie algebra $$\mathfrak{g}$$, we might think of this as a deformation of the Symmetric algebra:

\$$U_{\epsilon}(\mathfrak{g}) := T(\mathfrak{g})/(x \otimes y – y \otimes x – \epsilon[x, y])\$$

\$$\mathbb{C}[\mathfrak{g}^*] \simeq Symm[\mathfrak{g}] := T(\mathfrak{g})(x \otimes y – y \otimes x)\$$

Action on all of this is the adjoint action, that is, the action which takes an element $$g$$ of a Lie group $$G$$ sends $$X \mapsto gXg^{-1}$$. Orbits of this action stratify the dual Lie algebra, and there is a symplectic form that lives on each orbit.

I want to think of the adjoint action as directly analogous to the compositional conjugation action on the spectrum of the Lazard ring (over a ring $$R$$).

This action takes an invertible power series $$u$$ and applies it in the following manner to a formal group law $$F$$ over $$R$$.

\$$F(x, y) \mapsto u^{-1}(F(u(x), u(y)))\$$

This is an isomorphism on formal group laws of the same height (assuming that we are over a seperable field). Thus, the action of the invertible power series on the Lazard ring stratifies by height of the formal group law. But as far as I know, each of these geometric points carries no analogue of a symplectic form on each orbit. Do they? How would I find them? What could possibly impose a symplectic structure on the collection of of formal group laws of a given height?

What is our universal enveloping algebra? Perhaps, it is the contravariant bialgebra of the formal group law? Perhaps, this analogy is not as deep as I thought.

## The Cup Product of the Klein Bottle ($$\mathbb{Z}/2$$)

I didn’t want to use a brute force method, so I thought for a while about how to compute the cup product of the Klein bottle. This is what I came up with. I thought I’d share it.

Take coefficients in $$\mathbb{Z}/2$$. We compute the cup product by computing the intersection form (which defines the intersection product), then Poincare duality.

The Klein bottle is compact and $$\mathbb{Z}/2$$-orientable, so Poincare duality applies: \$$a^* \smile b^* = [a.b][K]\$$

where $$[K]$$ is the fundamental class of the Klein bottle, $$[a.b]$$ is the number of intersections of the cycles $$a$$ and $$b$$ (thought of as sub-manifolds embedded in the Klein bottle). Note that $$a^*$$ and $$b^*$$ denote the cocycles corresponding to the cycles $$a$$ and $$b$$ respectively.

We view the Klein bottle as the fiber bundle \$$S^1 \to K \to S^1\$$ Fix a fiber circle, $$F$$, and denote the base circle as $$B$$.

By pictoral argument: $$F$$ has zero transverse intersection with itself and intersects the base circle $$B$$ once. By wiggling the base circle by an infinitesimal amount $$\epsilon$$, we see that $$B$$ intersects transversely with itself once.

Another perspective:

We may more succinctly state the intersection product as the following matrix:

\$$\begin{pmatrix} \beta \smile \beta & \beta \smile \gamma \\ \gamma \smile \beta & \gamma \smile \gamma \end{pmatrix} \simeq \begin{pmatrix} [B.B] & [B.F] \\ [F.B] & [F.F] \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\$$

From the above matrix, we see that $$\beta^2 = \beta\gamma = 1[K]$$, and $$\gamma^2 = 0[K]$$. And, of course, $$\beta^3 = 0$$, because $$H^3(K) = 0$$. This gives us the ring structure of $$H^*(K; \mathbb{Z}/2)$$: \$$(\mathbb{Z}/2\mathbb{Z})[\beta, \gamma]/(\beta^2 + \beta\gamma, \gamma^2, \beta^3)\$$

P.S. If we wanted to be totally rigorous, we’d have to show that the cocycles $$\beta$$ and $$\gamma$$ form the basis of the $$H^1(K; \mathbb{Z}/2) \simeq \mathbb{Z}/2 \oplus \mathbb{Z}/2)$$. Until then, we don’t quite have that the intersection matrix gives us the cup product of $$H^*(K; \mathbb{Z}/2)$$.