*Note:
These are notes roughly transcribed from lectures from AWS 2019.
*

dzackgarza@gmail.com

Last updated: 2021-02-26

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.

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.

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 \({\operatorname{Gr}}\).

\(X\) 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: