# Lecture 1: $\AA^1$ Enumerative Geometry Tags: #motivic/homotopy 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 $\CC$. 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. ## Example: Lines on a Smooth Cubic Surface > Joint work with Jesse Kass :::{.definition title="Smooth Cubic Surfaces"} A **cubic surface** $X$ consists of the $\CC$ solutions to a polynomial in three variables, i.e. $$ X = \theset{(x,y,z) \in \CC^3 \suchthat f(x,y,z) = 0}, $$ where $f$ is degree 3. In general, we want to compactify, so we view $X \injects \CP^3$ as $$ \CP^3 = \theset{\vector x = [w,x,y,z]\neq \vector 0 \suchthat \forall \lambda \in \CC^\times,~\vector x = \lambda \vector x} $$ and so $$ X = \theset{[w,x,y,z] \in \CP^3 \suchthat f(w,x,y,z) = 0} $$ 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. ::: :::{.theorem title="Salmon-Cayley, 1849"} If $X$ is a smooth cubic surface, then $X$ contains exactly 27 lines. ::: :::{.example title="?"} The Fermat cubic $f(w,x,y,z) = w^3 + x^3 + y^3 + z^3$. We can find one line, given by $$ L = \theset{[s,-s,t,-t] \suchthat s,t \in \CP^1}, $$ and in fact this works for any $\lambda, \omega$ such that $\lambda^3 = \omega^3 = -1$, yielding $$ L' = \theset{[s,\lambda s,t,\omega t] \suchthat s,t \in \CP^1}. $$ 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. ::: ### Modern Proof We'll use characteristic classes, which we'll later replace by an $\Af^1$ homotopy theory variant. Let $\Gr(1,3)$ be the Grassmannian parametrizing 1-dimensional subspaces of $\CP^3$, where the $\CC$ points of this space parameterize 2-dimensional subspaces $W \subseteq \CC^4$. This is a moduli space of the lines we're looking for. Let $$S \to \Gr(1,3)$$ be the tautological bundle where the fiber is simply given by $S_W = W$. We can also form the bundle $$ (\sym^3 S)\dual \to \Gr(1,3) $$ where the fiber over the point corresponding to $W$ is all of the cubic polynomials on $W$, i.e. $$(\sym^3 S)\dual_W = (\sym^3 W)\dual.$$ Explicitly, we have the following two bundles to work with: \begin{tikzcd} W \ar[r] & S \ar[r] & \Gr(1, 3) \\ \qty{\sym^3 W}\dual \ar[r] & \qty{\sym^3 S}\dual \ar[r] & \Gr(1, 3) \end{tikzcd} Our chosen $f$ determines an element of $(\sym^3 \CC^4)\dual$, which is thus a section $\sigma_f$ of the second bundle above, where $$ \sigma_f(W) = \restrictionof{f}{W}. $$ We thus have $$ \PP W \in X \iff \sigma_f(W) = 0, $$ 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 $r$ $\RR\dash$ vector bundle over a dimension $r$ real manifold where we assume that $V$ is oriented. > We choose $\RR$ here because $\CC$ 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 $$ \deg: [S^{r-1}, S^{r-1}] \to \ZZ, $$ where we use the brackets to denote homotopy classes of maps. :::{.definition title="Degrees"} 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 $x\in B_\varepsilon(p) - \theset{p} \implies \sigma(x) \neq 0$. Choose a local trivialization of the total space $V$. This allows us to view $\sigma: \RR^r \to \RR^r$ as a real function. We can choose coordinates such that $p = 0$ in the domain, so $\sigma(0) = 0$, and moreover the image $\sigma(B_\varepsilon(p)) = \RR - \theset{0}$. We can then form a function $$ \bar\sigma: \del B_\varepsilon(p) = S^{r-1} &\to S^{r-1} = \del \sigma(B_\varepsilon(p)) \\ x &\mapsto \frac{\sigma(x)} {\norm{\sigma(x)}}, $$ and so we can take $\deg_p(\sigma) \definedas \deg \bar \sigma$. ::: :::{.remark} There is indeterminacy here up to elements of $\GL(r, \RR)$ 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. ::: :::{.definition title="Euler Class"} The Euler class, which only depends on the bundle and not the section, is given by $$ e(V) = \sum_{\theset{p\suchthat \sigma(p) = 0}} \deg_p(\sigma). $$ ::: 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 $$ \abs{\theset{\text{Lines on } X}} = e((\sym^3 S)\dual), $$ where the RHS is independent of $X$ and can be computed using the splitting principle and the cohomology of $\Gr$. ## What about $\RR$? :::{.theorem title="Schlafli, 19th century"} $X$ 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 $\Aut L$ for $L=\RP^1$ (real lines) as elliptic/hyperbolic: we have $\Aut L \cong \mathrm{PGL}(2, \RR)$, so pick some $I$ corresponding to a matrix $$ [I] = \begin{pmatrix}a & b \\ c & d\end{pmatrix},\quad z\mapsto \frac{az+b}{cz+d} $$ 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 $$ \mathrm{Fix}(I) = \theset{z \in \CC \suchthat cz^2 + (d-z)z + b = 0}, $$ 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 $T_pX \intersect X$ . 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 $T_pX \intersect X = L \union Q$. We know that $L \intersect Q$ 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 $\RR^3$, and push the tangent plane alone a line. There are two things that can happen -- one is a twisting by a nontrivial element of $\pi_1 SO_3(\RR)$, the other is no twisting at all. :::{.example title="?"} Look at the Fermat cubic surface $x^3+y^3+z^3=-1$ ![](assets/2019-03-03-22-11-15.png) Interpretation of this image: $X \subset \RR^3$ 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: ![](assets/2019-03-03-22-27-49.png) ::: After thinking about what physically happens as you push a plane around, it becomes clear that these three lines are all hyperbolic. Note that this question is the same as asking if a path in the frame bundle lifts. Although the number of lines isn't a constant, we can take a "signature" sort of formula to obtain an invariant. In this case, the number hyperbolic lines minus the number of elliptic lines *is* constant. In this case, the constant is 3. :::{.slogan} If you have a result that works over $\CC$ and $\RR$, it may be a result in $\Af^1$ homotopy theory that has realizations recovering the original results. ::: ## $\Af^1$ Homotopy Theory This will allow us to do with schemes much of what we can do in $\mathbf{Top}$. Smooth schemes behave like manifolds, where there are balls around points. The convention here will be that we're working over smooth schemes, denoted $\mathrm{Sm}_k$ where $k$ is a field. > Remark: in my notation I use $\RP^n, \CP^n$, and $\PP^n(k)$ to denote various projective spaces. I'll adopt Kirsten's convention here and just denote $\PP^n(k)$ as $\PP^n$. We'll get spheres from $S_\Af^n \definedas \PP^n/\PP^{n-1}$. One nice result due to Morel is that there is a degree map $$ [S_\Af^n, S_\Af^n] \to GW(k), $$ where the target is not the integers in this case, but rather a group of bilinear forms that are quadratic in characteristic not equal to 2. It is the Grothendieck-Witt group, whose elements are formal difference of bilinear forms. Thus the group itself is the group completion of nondegenerate symmetric isomorphism classes of bilinear forms $V^2 \to k$ where $V$ is a finite-dimensional $k\dash$vector space. The group structure arises because if we have two bilinear forms $B, B'$ on vector spaces $V, W$ respectively, then we can define a new form on $V \oplus W$ by working in components and declaring orthogonality between any of the factors. We then take formal differences of these, and inherit a ring structure from the tensor product of forms. Bilinear forms over fields can all be diagonalized, although in characteristic 2, this only holds in a stable sense. ## The [[Grothendieck-Witt Group]] Since we can diagonalize, the group $GW(k)$ has a presentation coming from the one dimensional forms. Any of these work as a generator, so we have - Generators: $\generators{a}$ where $a\in k^\times$, corresponding to the form \[ \generators{a} : k^2 &\to k\\ (x,y) &\mapsto axy .\] - Relations: if we change the basis of $k$ using a multiplication by $b\in k^\times$, we get $\generators{ab^2} = \generators{a}$. > This means that $a \in k^\times/(k^\times)^2$ - We also get $\generators{a} + \generators{b} = \generators{a+b} + \generators{ab(a+b)}$ There are many concrete computations of this known for global fields, local fields, finite fields, function fields, etc. :::{.example title="The Complex Numbers"} Computing $GW(\CC)$: The generators are in bijection with $k^\times/(k^\times)^2$, but since every element of $\CC$ is a square, so there's only one element here. We thus obtain $$ GW(\CC) &\mapsvia{\sim} \ZZ \\ \beta &\mapsto \dim V $$ which is realized by taking the rank. ::: :::{.example title="The Real Number"} Computing $GW(\RR)$: We still have the rank, but now we can also take the signature, so we have $$ GW(\RR) \mapsvia{\text{rank} \times \text{signature}} \ZZ^2, $$ although a minor parity issue crops up here that can be fixed without damaging the isomorphism type. ::: :::{.example title="Finite Fields"} Computing $GW(\FF_q)$: We can make a matrix out of how $\beta$ acts on basis elements and take the determinant of it to obtain an invariant called the *discriminant*, and so $$ GW(\FF_q) \mapsvia{\text{discriminant} \times \text{rank}} \FF_q^\times/(\FF_q^\times)^2 \times \ZZ $$ Note that the quotient is needed because we can change basis in $\FF_q$, which amounts to conjugating by a matrix $A$, and so this discriminant is only well-defined up to squares. ::: ## Euler Class There is an Euler class in this setting, $$ e(V) = \sum_{p\suchthat \sigma(p) = 0} \deg_p(\sigma). $$ Letting $X$ be a smooth cubic surface over $k$, then a line $L\subset X$ will be a closed point of the Grassmannian $\Gr(1,3)$, so we can think of it as points of the form $$ L = \theset{[a,b,c,d]s + [a',b',c',d']t \suchthat s,t \in \PP^1(k(L} $$ where the extension field $k(L) = k(a,b,c,d,a',b',c',d')$ is obtained by adjoining the coefficients to $k$. > DZG: I think these are always separable, mentioned later in the talk. We thus get $$ \PP^1(k(L)) \cong L \underset{\substack{\text{closed}\\\text{subscheme}}}\subseteq X_{k(L)} \subseteq \PP^3(k(L)). $$ Given such a line $L\subseteq X$, similar to the real setting, we obtain an involution $I \in \Aut L \cong PGL(2, k(L))$ after choosing coordinates. We also find that $\mathrm{Fix}(L)$ again falls into two cases: - $2 k(L)$ points, or - 2 conjugate points in some quadratic extension $k(L)[\sqrt D]$ where $D \in k(L)^\times / (k(L)^\times)^2$. These correspond to the oddities in the tangent plane in the real case. We then define $$ \mathrm{Type}(L) = \generators{D} \in GW(k(L)), $$ or equivalently $D=ab-cd$ when $I = \begin{pmatrix}a&b\\c&d\end{pmatrix}$, in which case $\mathrm{Type}(L) = \gs{-1} \deg I$. ## An Analogous Trace Formula :::{.theorem title="?"} Supposing $X$ is a smooth cubic surface over $k$ of characteristic not equal to 2, we then have $$ \sum_{L \in X}\mathrm{Tr}_{k(L) / k}\mathrm{Type}(L) = \text{One fixed quadratic form} = 15 \gs{1} + 12 \gs{-1} $$ where the trace/transfer maps are defined as $$ \mathrm{Tr}_{k(L) / k}: GW(k(L)) &\to GW(k) \\ (V^2 \mapsvia{\beta} k(L)) &\mapsto (V^2 \mapsvia{\beta} k) \circ \mathrm{Trace}_\text{Galois} $$ where $\mathrm{Trace}_\text{Galois}$ comes from summing the conjugates. Note that we can do this because we can view $V$ as a vector space over either $k$ or $k(L)$, so we end up with a quadratic form over $k$. ::: :::{.remark} Note: we have a well-defined map in the other direction, since the $GW$ is a stable homotopy group of spheres. ::: :::{.example title="Complex Number"} Let $k=\CC$, then apply rank to get $15+12=27$ on the RHS, while since every element is a square, the Type is just 1, so we get 27 total. ::: :::{.example title="Reals"} Let $k=\RR$, apply signature. If $L$ is defined over $C$, so the type is 1, and we're just left with the trace of $\CC/\RR$ -- but this contributes a $+1$ and $-1$, so there is no contribution. What's left are the lines of $\RR$, and since we set it up so type 1 lines are hyperbolic, we just get the trace $15-12=3$. ::: :::{.example title="Finite Fields"} Let $k=\FF_q$. We can define lines in $\FF_q^n$, and the "begin a square" partitions $(\FF_q^n)^\times$ into two disjoint subsets, we can assign types and we let squares be the hyperbolic elements. We thus get $$ \mltext{ \text{Elliptic lines $L$ } \\ \text{with $k(L) = \FF_\text{odd}$} } - \mltext{ \text{Hyperbolic lines $L$} \\ \text{with $k(L) = \FF_\text{even}$} } \equiv 0 \mod 2 $$ which follows from computing the discriminant of the given form. :::