You are currently browsing the monthly archive for January 2012.

One of the basic things we can do in algebraic geometry is take a surface, take two curves lying on that surface, and ask how many points of intersection there are. In ‘intuitive’ situations, it is obvious how to do this, but it is also not hard to find situations where common sense might break down.

For example, if we take a line tangent to a conic in a projective plane over an algebraically closed field ${k}$, does it intersect once or twice? Super-naive intuition would suggest ‘only once’, but if you think a little harder this answer seems kind of ridiculous, because for any other configuration of the line and the circle, the answer is ‘twice’, so by applying naive intuition we have created a disgusting discontinuity in our calculus of intersections.

Luckily, especially with the theory of schemes, this is easy to fix. If we have two curves ${C,D}$ intersecting at a (closed) point ${P}$ in some surface ${X}$ (everything smooth and projective just to be safe), look at the local function ring ${\mathcal{O}_{X,P}}$ and you can write equations ${f,g}$ in this ring that locally describe the curves ${C}$ and ${D}$. Thus in some sense the geometric space obtained by intersecting ${C}$ and ${D}$ at ${P}$ is best described as a scheme by ${\mathcal{O}_{X,P}/(f,g)}$ which (assuming ${C,D}$ are distinct) is a local artinian ring over ${k}$, hence of finite ${k}$-dimension, and if you look at a couple of examples (e.g. the first question on Burt Totaro’s commutative algebra examples) it’s easy to convince yourself that this dimension is the correct notion of intersection multiplicity, denoted ${(C.D)_P}$. So we have managed to justify a formal procedure for letting the answer to our original question be 2′, and thus remove our ugly discontinuity.

It turns out we can define a nice object that computes things like this for us. It is algebraically convenient to, instead of merely intersecting two irreducible curves, to intersect arbitrary ${{\mathbb Z}}$-linear combinations of irreducible curves. Such a ${{\mathbb Z}}$-linear combination is called a divisor and the group of all divisors on a surface ${X}$ is written ${Div(X)}$.

In Hartshorne V,1 it is proved that there is a unique symmetric bilinear pairing ${Div X \times Div X \rightarrow {\mathbb Z}}$ satisfying the following two properties. Firstly, that if ${C,D}$ are smooth curves meeting transversally, then ${C.D = |C \cap D|}$. The second property is a kind of continuity property which says that if two divisors are linearly equivalent then they behave identically with respect to the intersection pairing. It turns out that this pairing satisfies a more general version of property 1: that for any two curves ${C,D}$ with no common irreducible component,

$\displaystyle C.D = \sum_{P \in C \cap D} (C.D)_P.$

In particular, notice how this pairing agrees with us’ about the correct definition of intersection multiplicity.

You may well now be feeling fairly happy about what we have achieved, but there is one apparently silly but actually very important case we still have not come close to addressing. How many times does a curve intersect itself? The naive answer should be infinitely often’, or thereabouts, but again think about how ugly this would be. If you take your curve and apply a small algebraic deformation, it will then intersect in some finite number of points. In fact, if there is a different curve in the same linear equivalence class, then our pairing is guaranteed to spit out a finite number (since it’s well-defined up to linear equivalence).

How do we explain what’s going on? In particular, is there a way of getting hold of these numbers and maybe finding a geometrical interpretation for them? To really get a grasp of this, we will need some of the tools of basic algebraic geometry (Hartshorne II and III). The plan is to continue working with ${C}$ and ${D}$ distinct curves (maybe even transversally intersecting) but to massage our definition of ${C.D}$ until it doesn’t look completely mad to consider ${C.C}$.

Firstly, a divisor on a curve is just a formal ${{\mathbb Z}}$-linear combination of points on the curve, and in fact in the situation we are studying there is a natural intersection divisor

$\displaystyle V = \sum_{P \in C \cap D} (C.D)_P P.$

This can be interpreted either as a divisor on ${C}$ or on ${D}$ and in each case ${C.D}$ is the degree of this divisor.

It is one of those cool facts in algebraic geometry that there is a surjective homomorphism ${Div(C) \rightarrow Pic(C)}$, where ${Pic(C)}$ is the group of invertible sheaves on ${C}$ (with tensor product as operation), and this map has kernel precisely equal to the principal divisors, so in particular it remembers what the degree of a divisor is. Thus instead of studying ${V}$, we switch our focus to studying the corresponding invertible sheaf ${\mathcal{L}(V)}$.

In fact, this game works on surfaces and their divisors too, so in particular an irreducible curve ${D}$ in ${X}$ gives rise to an invertible sheaf ${\mathcal{L}(D)}$ on ${X}$, and such sheaves have the property that their duals ${\mathcal{L}(-D)}$ are canonically isomorphic to the ideal sheaf of ${D}$, defined by ${\mathcal{I} = Ker(\mathcal{O}_X \rightarrow i_* \mathcal{O}_D)}$ (where ${i:D \hookrightarrow X}$). A huge advantage to this is that these sheaves have nice functoriality. In particular (returning to our intersecting curves ${C,D}$), if ${j:C\rightarrow X}$ is the embedding of ${C}$ into ${X}$, then we can pull back ${\mathcal{L}(-D)}$ to an invertible sheaf on ${C}$, and it is not difficult to prove exactness of:

$\displaystyle 0 \rightarrow j^* \mathcal{L}(-D) \rightarrow \mathcal{O}_C \rightarrow \mathcal{O}_{C \cap D} \rightarrow 0.$

From what we were just saying about ideal sheaves, this sequence says precisely that ${j^* \mathcal{L}(D) \cong \mathcal{L}(V)}$. In particular, we could accurately define the intersection number of two distinct irreducible curves ${C}$ and ${D}$ to just be the degree of ${j^*\mathcal{L}(D)}$ (considered as a divisor on ${C}$), and our above argument shows that the apparent asymmetry of this formula is not a problem.

For our purposes, it also suggests the following possible answer to our question. Could it be that ${C.C = deg_C(j^*\mathcal{L}(C))}$? Miraculously, though it would be unthinkable to try pulling back the divisor ${C}$ on ${X}$ to a divisor on ${C}$, we can always pull back a line bundle and get out another line bundle, so by switching from studying divisors to studying invertible sheaves, we really do get an element of ${Pic(C)}$ that corresponds to ${C}$ intersected with itself, and indeed taking the degree of this is what the intersection pairing does.

Awesome. What does this look like? We’ll compute a pretty important example (which is the first step in the proof of the Riemann Hypothesis for curves over finite fields – something which I’ll probably say something about soon). Consider a smooth irreducible projective curve ${C}$ over an algebraically closed field ${k}$, and let ${X = C \times_{Spec k} C}$, a surface. Let ${D = \Delta(C)}$ be the embedding of ${C}$ in the diagonal of ${X}$. We shall compute the intersection number ${D.D}$, and note that it’s pretty interesting.

The main result of this post convinces us that our task is to calculate the degree of ${\Delta^*(\mathcal{L}(D))}$. But ${\mathcal{L}(-D)}$ is the ideal sheaf ${\mathcal{I}}$ of ${D}$, that is to say the kernel of ${\Delta^*: \mathcal{O}_X \rightarrow \mathcal \Delta_*{O}_D}$, so as ${\mathcal{O}_X}$-modules,

$\displaystyle \mathcal{L}(-D) \otimes_{\mathcal{O}_X} \mathcal{O}_D \cong \mathcal{I}/\mathcal{I}^2.$

But we know what this sheaf is. Recall (Hartshorne II,8) that the sheaf of differential forms ${\Omega_{C/k}}$ is defined precisely as ${\Delta^*(\mathcal{I}/\mathcal{I}^2)}$, which, since we are in dimension 1, is also the canonical sheaf. Thus our sheaf ${\Delta^*(\mathcal{L}(D))}$ is precisely the dual of the canonical sheaf ${\omega_C}$, and in particular, it has degree ${-(2g-2) = 2-2g}$. So the intersection of a curve with itself recovers the Euler characteristic.

This post has been largely improvised by me while trying to simultaneously learn some basic algebraic geometry and figure out how to prove the Weil conjectures, so is bound to contain some errors or things I could have done more clearly, which readers are encouraged to point out. The obvious reference is Hartshorne V. I may be back very soon to prove the Riemann hypothesis for curves over a finite field, which essentially boils down to computing certain intersection numbers between divisors in ${C \times C}$, and then applying the Hodge index theorem, which places constraints on how the intersection pairing may behave and lets you write down an inequality which can be massaged into the Hasse-Weil inequality, which essentially is a tight version of the prime number theorem’ for curves over finite fields that is tight enough to imply the Riemann hypothesis.

Also, this surely has connections to the algebraic topological theory surrounding the Lefschetz fixed-point theorem, in a way I haven’t properly thought about yet, so if anyone knows any interesting things about that I’d enjoy talking about it, and I vaguely recall somebody telling me that intersecting curves with themselves becomes clearer in a (Lurie style) derived algebraic geometry picture. Again, I know nothing and would be interested to know something. Thanks.

In this post we shall sketch the reduction of Fermat’s Last Theorem to Wiles’ Theorem that every semistable elliptic curve over ${{\mathbb Q}}$ is modular. I believe the ideas are due mainly to Frey, with a helpful big black box courtesy of Serre and Ribet. My main reference is the article of Stephens from the 1995 conference.

Suppose, for a fixed prime ${p \geq 5}$, that there are integers ${a,b,c}$ with ${a^p+b^p+c^p = 0}$, ${abc \not=0}$ and, assuming wlog that ${a \equiv 3 \mod 4}$ and ${b}$ is even, form the elliptic curve

$\displaystyle E_0: y^2 = x(x-a^p)(x+b^p).$

Then, by a direct computation, ${E_0}$ is semistable with minimal discriminant ${\Delta_{E_0} = 2^{-8} (abc)^{2p}}$ and conductor ${N_{E_0}=\prod_{l|abc} l}$.

Our strategy is to study the representation ${\rho_E: G_{\mathbb Q} \rightarrow GL_2(\mathbb{F}_p)}$ coming from the action of Galois on the ${p}$-torsion points of ${E}$ over ${\bar{{\mathbb Q}}}$ (note that this is the same ${p}$ as in the previous paragraph).

In general, such Galois representations can be restricted to ${G_{{\mathbb Q}_l}}$, considered as the decomposition group of the prime ${l}$ in ${G_{\mathbb Q}}$ (that this is only defined up to conjugacy, determined by a choice of embedding ${\bar{{\mathbb Q}} \rightarrow \bar{{\mathbb Q}}_l}$ is not important for us – we’ve already picked a basis for the torsion points anyway). The typical good’ local behaviour at these primes are, for ${l \not= p}$, that ${\rho_E}$ be unramified at ${l}$ (in the sense that the absolute inertia group acts trivially), and for ${l=p}$ a more general rather more technical condition which we describe by saying that ${\rho_E}$ is flat at ${p}$.

Let us (nonstandardly) call a prime ${l}$ bad (for a representation ${\rho_E}$) if ${l \not= p}$ and ${\rho_E}$ is ramified at ${l}$, or ${l=p}$ and ${\rho_E}$ fails to be flat at ${p}$. The important fact we need is that bad primes for such representations coming from elliptic curves are all flagged up by the minimal discriminant.

Proposition 1 For ${E}$ a semistable elliptic curve, ${l}$ is a bad prime for ${\rho_E}$ iff ${p}$ does not divide ${ord_l(\Delta_E)}$.

In particular, this implies for our curve ${E_0}$ that the only bad prime for ${\rho_{E_0}}$ is ${2}$. People suspected that this possibility of such an arithmetically simple’ representation coming from an elliptic curve should be pretty unlikely.

Now, as well as these representations on ${\mathbb{F}_p}$ coming from torsion points on an elliptic curve, it is possible to construct similar representations coming from certain modular forms. Indeed, from a newform ${f}$ of weight 2, whose Fourier coefficients and character are rational, it is possible to form a representation ${\rho_f: G_{\mathbb Q} \rightarrow GL_2(\mathbb{F}_p)}$. Serre made a series of conjectures about which representations could be obtained from newforms, one of which was the following result, a hard theorem proved by Ribet.

Theorem 2 (Serre’s Epsilon Conjecture) If ${f}$ is a weight two newform of conductor ${N}$ and ${\rho_f}$ is absolutely irreducible, then letting ${N'}$ be the factor of ${N}$ obtained by throwing away all the non-bad primes which divide ${N}$ at most once, it is possible to find a different weight two newform ${g}$ of conductor ${N'}$ such that ${\rho_f \cong \rho_g}$.

Finally, we need some results about modular curves. In fact, we just need the following basic fact. If ${E}$ is a modular curve, then the representation ${\rho_E}$ is equal to ${\rho_f}$ for some newform ${f}$ of conductor ${N_E}$.

Proposition 3 The curve ${E_0}$ above (constructed from a contradiction of Fermat’s Last Theorem) is not modular.

First, one shows that ${\rho_{E_0}}$ defined above is absolutely irreducible. Now, if ${E_0}$ it were modular, ${\rho_{E_0}}$ would be isomorphic to a representation ${\rho_f}$ coming from a newform of conductor ${N_{E_0}}$. But looking at ${N_{E_0}}$ and the fact that ${2}$ is the only bad prime of ${\rho_f}$, Ribet’s theorem implies that there is a newform ${g}$ of weight ${2}$ and conductor ${2}$ such that ${\rho_f \cong \rho_g}$. But the space ${S_2(\Gamma_0(2))}$ has dimension ${0}$, by the basic theory of modular forms, so no such newform exists.

Having proved proposition 3, recalling that ${E_0}$ was a semistable elliptic curve over ${{\mathbb Q}}$, Wiles’ theorem that every semistable elliptic curve over ${{\mathbb Q}}$ is modular gives a contradiction and thus proves that there are no integers ${a,b,c}$ with ${a^p+b^p+c^p=0}$, ${abc \not= 0}$.