1 Lecture 1: $${\mathbb{A}}^1$$ Enumerative Geometry

Enumerative geometry counts algebro-geometric objects, and in order to actually obtain an invariant number at the end of the day one uses an algebraically closed field $$k$$ or $${\mathbb{C}}$$. This is essentially because the conditions imposed are polynomial, and polynomials of degree $$n$$ over a closed field always have $$n$$ roots.

The goal here is to record information about the fields of definition. However, since we may no longer have invariant numbers as solutions to polynomial equations, we replace this with a notion of weights to get an “invariance of bilinear form” principle instead. Over characteristic not 2, we can use quadratic forms, which ties to Lurie’s first talk.

1.1 Example: Lines on a Smooth Cubic Surface

Joint work with Jesse Kass

A cubic surface $$X$$ consists of the $${\mathbb{C}}$$ solutions to a polynomial in three variables, i.e. \begin{align*} X = \left\{{(x,y,z) \in {\mathbb{C}}^3 {~\mathrel{\Big|}~}f(x,y,z) = 0}\right\}, \end{align*}

where $$f$$ is degree 3. In general, we want to compactify, so we view $$X \hookrightarrow{\mathbb{CP}}^3$$ as \begin{align*} {\mathbb{CP}}^3 = \left\{{\mathbf{x} = [w,x,y,z]\neq \mathbf{0} {~\mathrel{\Big|}~}\forall \lambda \in {\mathbb{C}}^\times,~\mathbf{x} = \lambda \mathbf{x}}\right\} \end{align*}

and so \begin{align*} X = \left\{{[w,x,y,z] \in {\mathbb{CP}}^3 {~\mathrel{\Big|}~}f(w,x,y,z) = 0}\right\} \end{align*} where $$f$$ is homogeneous. The surface $$X$$ is smooth if the underlying points form a manifold, or equivalently if the partials don’t simultaneously vanish.

If $$X$$ is a smooth cubic surface, then $$X$$ contains exactly 27 lines.

The Fermat cubic $$f(w,x,y,z) = w^3 + x^3 + y^3 + z^3$$. We can find one line, given by \begin{align*} L = \left\{{[s,-s,t,-t] {~\mathrel{\Big|}~}s,t \in {\mathbb{CP}}^1}\right\}, \end{align*}

and in fact this works for any $$\lambda, \omega$$ such that $$\lambda^3 = \omega^3 = -1$$, yielding \begin{align*} L' = \left\{{[s,\lambda s,t,\omega t] {~\mathrel{\Big|}~}s,t \in {\mathbb{CP}}^1}\right\}. \end{align*}

We can also permute $$s,t$$ around to get more lines, and by counting this yields 27 distinct possibilities: 3 choices for $$\lambda$$, 3 choices for $$\omega$$, and $$\frac 1 2 {4\choose 2}$$ ways to pair them with the $$s,t$$ in the original $$L$$.

There is a proof in the notes that these are the only lines, which is relatively elementary.

1.1.1 Modern Proof

We’ll use characteristic classes, which we’ll later replace by an $${\mathbb{A}}^1$$ homotopy theory variant. Let $${\operatorname{Gr}}(1,3)$$ be the Grassmannian parametrizing 1-dimensional subspaces of $${\mathbb{CP}}^3$$, where the $${\mathbb{C}}$$ points of this space parameterize 2-dimensional subspaces $$W \subseteq {\mathbb{C}}^4$$. This is a moduli space of the lines we’re looking for. Let \begin{align*}S \to {\operatorname{Gr}}(1,3)\end{align*} be the tautological bundle where the fiber is simply given by $$S_W = W$$. We can also form the bundle \begin{align*} (\operatorname{Sym}^3 S)^\vee\to {\operatorname{Gr}}(1,3) \end{align*} where the fiber over the point corresponding to $$W$$ is all of the cubic polynomials on $$W$$, i.e.  \begin{align*}(\operatorname{Sym}^3 S)^\vee_W = (\operatorname{Sym}^3 W)^\vee.\end{align*} Explicitly, we have the following two bundles to work with:

Our chosen $$f$$ determines an element of $$(\operatorname{Sym}^3 {\mathbb{C}}^4)^\vee$$, which is thus a section $$\sigma_f$$ of the second bundle above, where \begin{align*} \sigma_f(W) = {\left.{{f}} \right|_{{W}} } . \end{align*} We thus have \begin{align*} {\mathbb{P}}W \in X \iff \sigma_f(W) = 0, \end{align*} i.e. the line corresponding to $$W$$ is in our surface exactly when this section is zero. We now want to count the zeros of $$\sigma_f$$, which is exactly what the Euler class does. To be precise, the Euler class counts the zeros of a section of a properly oriented vector bundle with a given weight. Let $$V\to M$$ be a rank $$r$$ $${\mathbb{R}}{\hbox{-}}$$ vector bundle over a dimension $$r$$ real manifold where we assume that $$V$$ is oriented.

We choose $${\mathbb{R}}$$ here because $${\mathbb{C}}$$ is slightly too nice and gives us a preferred orientation (which we’ll want to track later.)

For any section $$\sigma$$ with only isolated zero, we’ll assign a weight to each zero which comes from the topological degree function \begin{align*} \deg: [S^{r-1}, S^{r-1}] \to {\mathbb{Z}}, \end{align*}

where we use the brackets to denote homotopy classes of maps.

Let $$p\in M$$ where $$\sigma(p) = 0$$, and define $$\deg_p(\sigma)$$ in the following way: Choose local coordinates near $$p$$. Since the zeros are isolated, we can choose a ball $$B_\varepsilon(p)$$ such that $$x\in B_\varepsilon(p) - \left\{{p}\right\} \implies \sigma(x) \neq 0$$. Choose a local trivialization of the total space $$V$$. This allows us to view $$\sigma: {\mathbb{R}}^r \to {\mathbb{R}}^r$$ as a real function. We can choose coordinates such that $$p = 0$$ in the domain, so $$\sigma(0) = 0$$, and moreover the image $$\sigma(B_\varepsilon(p)) = {\mathbb{R}}- \left\{{0}\right\}$$. We can then form a function \begin{align*} \mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu: {\partial}B_\varepsilon(p) = S^{r-1} &\to S^{r-1} = {\partial}\sigma(B_\varepsilon(p)) \\ x &\mapsto \frac{\sigma(x)} {{\left\lVert {\sigma(x)} \right\rVert}}, \end{align*} and so we can take $$\deg_p(\sigma) \mathrel{\vcenter{:}}=\deg \mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu$$.

There is indeterminacy here up to elements of $$\operatorname{GL}(r, {\mathbb{R}})$$ which could possibly affect the sign, however, but this can be fixed using the assumption that $$V$$ is oriented and choosing local trivializations for which the orientations are compatible. This gives us a well-defined local degree of a section at a zero.

The Euler class, which only depends on the bundle and not the section, is given by \begin{align*} e(V) = \sum_{\left\{{p{~\mathrel{\Big|}~}\sigma(p) = 0}\right\}} \deg_p(\sigma). \end{align*}

It can be shown that because $$X$$ is smooth, the zeros are all simple and so in the complex case, the degrees are all 1. We thus obtain \begin{align*} {\left\lvert {\left\{{\text{Lines on } X}\right\}} \right\rvert} = e((\operatorname{Sym}^3 S)^\vee), \end{align*}

where the RHS is independent of $$X$$ and can be computed using the splitting principle and the cohomology of $${\operatorname{Gr}}$$.

1.2 What about $${\mathbb{R}}$$?

$$X$$ can have $$3,7,15$$ or $$27$$ lines.

So it’s not constant, and thus there’s not an invariant number here, but Segre (1942) distinguished between hyperbolic and elliptic lines. Recall the characterization of elements in $${\operatorname{Aut}}L$$ for $$L={\mathbb{RP}}^1$$ (real lines) as elliptic/hyperbolic: we have $${\operatorname{Aut}}L \cong \mathrm{PGL}(2, {\mathbb{R}})$$, so pick some $$I$$ corresponding to a matrix \begin{align*} [I] = \begin{pmatrix}a & b \\ c & d\end{pmatrix},\quad z\mapsto \frac{az+b}{cz+d} \end{align*} where the second formulation above shows that there are two fixed points, since solving for $$z\mapsto z$$ yields a quadratic equation. So we have \begin{align*} \mathrm{Fix}(I) = \left\{{z \in {\mathbb{C}}{~\mathrel{\Big|}~}cz^2 + (d-z)z + b = 0}\right\}, \end{align*} and we characterize $$I$$ by the following cases:

• $$\mathrm{Fix}(I)$$ contains two real points: hyperbolic
• A complex conjugate pair: elliptic

So we’ll associate an involution to $$L$$, and port over these notions of hyperbolic/elliptic. As we’ll see later, for each point on $$L$$, there will be a unique other point that has the tangent space, and this involution will swap them. Let $$p\in L$$, and consider $$T_pX \cap X$$ . Since $$L$$ is in both of the varieties we’re intersecting here, and we can apply Bezout’s theorem, we know that its complement will some degree 2 variety $$Q$$ (since the total degree is 3).

So we can write $$T_pX \cap X = L \cup Q$$. We know that $$L \cap Q$$ will be the intersection of a degree 1 and a degree 2 curve, which will have 2 points of intersection. At one of these points, say $$p$$, $$Q$$ and $$L$$ will intersect transversally, and so the tangent vectors $$T_pQ$$ and $$T_pL$$ give a 2-dimensional frame, which yields a plane $$P \subseteq T_pX$$. Since $$X$$ is smooth, we get equality and $$P = T_pX$$. This also holds for the second point of intersection, $$p'$$, and so we take the involution $$I(p) = p'$$ and vice-versa. We then say $$L$$ is elliptic/hyperbolic exactly when $$I$$ is. A natural way to see that there should be a distinction between two types of lines is to use spin structures. Consider a physical cubic surface sitting inside $${\mathbb{R}}^3$$, and push the tangent plane alone a line. There are two things that can happen – one is a twisting by a nontrivial element of $$\pi_1 SO_3({\mathbb{R}})$$, the other is no twisting at all.

Look at the Fermat cubic surface $$x^3+y^3+z^3=-1$$

Interpretation of this image: $$X \subset {\mathbb{R}}^3$$ is a surface, which has 3 lines that are contained in a plane. We this view $$X$$ from above this plane, marking a plus/minus to denote the relative height of the surface within each bounded region. Plus denotes part of the surface that bubbles up over the plane, having positive height/$$z$$ coordinates, etc.

(DZG) This took me a while to visualize – what worked for me was thinking about “egg crate” padding: