A Second Glimpse of Spectra

Spectra come from the need for negative dimensional spheres — the need for a category where suspension has an inverse and not just an adjoint!

If you want a fantastic introduction to the stable category of spectra, and the context of various topological theorems calling for a definition of negative dimensional spheres, this might help.

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Recall that the following functor takes a pointed space $X$ to the suspension spectrum of $X$ (i.e., $\Sigma^\infty$), and that it has a left adjoint named $\Omega^\infty$.

$\text{space}_*\xrightarrow{\Sigma^\infty} \Sigma-\text{spectrum}$

$\text{space}_*\xleftarrow{\Omega^\infty} \Sigma-\text{spectrum}$

The formal definition of $S^{-n}$ is $\Omega^\infty_n S^0$ (that is, the negative dimensional sphere is $S^0$ in the nth slot), but I find it more illuminating to step back and to think of positive dimensional spheres in terms of the natural numbers that index them:

$n \in \mathbb{N}$ $S^n$

$N \times N \to N$ $T \times T \to T$
$n + m \mapsto (n + m)$ $S^n \wedge S^m \mapsto S^{n+m}$

Grothendieck complete $\mathbb{N}$:

$n + -n \mapsto 0$ $S^n \wedge S^{-n} \mapsto S^0$

which can be rewritten as $\Sigma^nS^0 \wedge \Omega^\infty_nS^0 \mapsto S^0$

However, as fun as it is to try to draw mental pictures of negative dimensional spheres, negative dimensional space not a particularly illuminating frame of reference when thinking about spectra. Let us instead take to heart the Erlangen program, which teaches us a powerful message:

Geometry is the study of properties of an object that are invariant under a chosen collection of transformations of that object.

With that in mind, let us look at the suspension functor.

What are the properties of an object that are invariant under repeated application of the suspension functor?

If you’re interested in a specific answer, here is a great overview (by Peter May) of the theory and calculations that fall out of taking this question seriously. I won’t elaborate here on the implications of this viewpoint.

Let us restrict our attention to properties invariant under repeated application of the suspension functor.

Note that $\Sigma$ raises dimension by one; let $X$ be an $n$-dimensional space, $\Sigma X$ is $(n+1)$-dimensional.

If we are only looking at stable invariants (i.e., if we’re thinking about properties invariant under repeated application of the suspension functor), then the map $S^4 \to S^3$ is indistinguishable from, say, the map $\Sigma^{16}S^4 \to \Sigma^{16} S^3$.

The notion of dimension is immaterial.

Suspension is functorial and it adds a dimension to both sides…and we can’t tell the difference.

However, we can tell the difference between the map $S^4 \to S^3$ and $S^6 \to S^3$. We can’t apply suspension to both sides of $S^4 \to S^3$ and get $S^6 \to S^3$.

We have a notion of relative dimension.

So $S^{-n}$ is obtained by “shifting” the dimension of $S^0$ (relative to $S^n$) by $n$, i.e. moving $S^0$ into the “nth slot.”

Now that we’re starting to get the hang of this point of view, let’s think about a category of spectra.

Let’s look at the classical definition of a spectrum (a collection of indexed topological spaces $\{E_n\}_{n \in \mathbb{Z}}$, equipped with a suspension map $\Sigma E_n \to E_{n+1}$).

A while ago, Akhil mentioned to me that thinking of spectra as an indexed collection of spaces together with the suspension map is like thinking of a real number as a Cauchy sequence that converges. For example, there are multiple sets of spaces which are the same spectrum (just like there are different Cauchy sequences that converge to the same real number).

So what is a “real number” definition of a spectrum?

There are all sorts of ways to define a model category of spectra (which is confusing as hell when reading the literature and not being aware of this fact), but there’s a comforting theorem that these categories of spectra are Quillen equivalent model categories so it’s not a problem.

We might expect that a category of spectra $S$ would satisfy these perfectly reasonable axioms:

  1. The category $S$ is a symmetric monoidal category (wrt $\wedge$)
  2. The functor $\Sigma^\infty$ is left adjoint to the functor $\Omega^\infty$.
  3. The unit for the smash product in $S$ is the sphere spectrum $\mathbb{S} := \Sigma^\infty S^0$
  4. There is one of the two following natural transformations:
    $\Omega^\infty D \wedge \Omega^\infty E \to \Omega^\infty (D \wedge E)$
    $\Sigma^\infty X \wedge \Sigma^\infty Y \to \Sigma^\infty (X \wedge Y)$
  5. There is a natural weak equivalence
    $$\Omega^\infty\Sigma^\infty X \xrightarrow{\delta} QX$$ and the following diagram commutes (in $\text{Top}_*$):
    Screen Shot 2015-05-15 at 2.56.30 PM

In the paper Is there a convenient category of spectra? there’s a no-go theorem. No category of spectra exists which satisfies Axioms 1-5. It is perfectly healthy to rise from your seat and pace about in frustration at this point. When you’ve calmed down, I’ll brush over the proof of his theorem.

Assume that $S$ satisfies 1-4, and $E$ is a strict ring spectrum.

$\Rightarrow$ $\Omega_1^\infty E$ is product of Eilenberg-MacLane spaces. [May, Proposition 3.6]

The unit for the smash product must be a strict ring spectrum and $\Sigma^\infty {S}^0$ is the unit for the monoid operation in $S$ — so $\Sigma^\infty {S}^0$ must be a strict ring spectrum.

People like to blur the line between saying that a diagram commutes in $C$ or commutes in $hoC$ (i.e., commutes up to homotopy). When we want to specifically say that a diagram commutes in $C$, we say that is satisfies said property “on the nose,” or that the object itself is “strict.”

$\Sigma^\infty S^0 \to E$ is the unit for the smash product in $S$.

$\Rightarrow$ $\Omega_1^\infty \Sigma^\infty S^0$ must be a product of Eilenberg-MacLane Spaces.

If we assume that $S$ also satisfies 5, then the following diagram commutes:

Screen Shot 2015-05-16 at 10.53.24 AMApply $\Omega^\infty$ to the unit map $\Sigma^\infty S^0 \to E$ in $S$.

$\Omega^\infty\Sigma^\infty S^0 \to \Omega^\infty E$ is the corresponding unit map in $Top_*$

$\Rightarrow$ The path component of the unit in $QS^0$ is a product of Eilenberg-MacLane spaces. Lewis states that this is false! I’ve not understood why this is false yet, I have to think about it a bit more.

I hate to leave you in disappointment. Worry not! In this same paper that Lewis shatters our hopes and dreams, he lists 4 axioms that are more reasonable.

Afternote: I’m not going to go over the technicalities in defining the smash product here, because I don’t understand them, I’ll just mention that they arise because naive indices are a bother and require arbitrary choices that we have to keep track of. I do want to mention that a common method of indexing arises from using the relative dimension of inner product spaces $V$ and $W$. $$EV \wedge FW \to (E \wedge F)(V \oplus W)$$ There is still choice involved — we choose an isogeny from $\mathbb{R}^\infty \oplus \mathbb{R}^\infty \to \mathbb{R}^\infty$.

Thank you to Eric Peterson for directing me to Lewis’s paper.

Bordism with singularities construction of elliptic homology

This post assumes familiarity with the Landweber exact functor theorem, elliptic genera, and bordism theories.

An ongoing desire of mine is to geometrically approach elliptic spectra.

Note that I’m not talking about geometric cocycles for tmf. I’m talking about the vague goal of understanding elliptic spectra in their own right using “geometric” techniques (properties of the object that are invariant under a chosen collection of transformations of that object).

There is a presentation of an elliptic homology theory as a bordism theory with singularities outlined by Landweber in Elliptic Cohomology and Modular Forms (based on this paper which presents $H_*(-;\mathbb{Z})$ as a bordism theory with singularities).

This seemed like the beginning of the answer to my desire, but I now think that this construction is basically a less intuitive version of $MSO_*(-) \otimes_{MSO_*} R$, thinking about “tensoring out” classes in $MSO$ as coning them off. I’ll explain what I mean by “coning them off” in a bit, first let me outline the construction:

Outline:

  1. Start with an elliptic genus $\pi_*(MSO) \xrightarrow{\phi} R$
  2. Mod out the ring spectrum $MSO$ by $ker(\phi)$ to get a spectrum “$MSO/ker(\phi)$” whose homotopy groups are $R$
    i.e., construct $F: \pi_*(MSO/ker(\phi)) \to R$ s.t. $ker(F) = 0$
  3. Check that the spectrum $MSO/ker(\phi)$ is a ring spectrum

Let’s go through it!

Before we begin, you might wonder why we’re working with $MSO$. Landweber works with $MSO_*[\frac{1}{2}]$ and not $MU$; the map from $MU$ to our elliptic spectrum $E$ factors through $MSO$ (we can forget structure on our bordisms).

Notational side, $MSO_* := \pi_*(MSO)$, and $\mathbb{S}^n$ means the the suspension spectrum with $S^n$ as it’s 0th space, i.e. $\Sigma^\infty S^n_+$.

Reason to kill the generators

Start with an elliptic genus, a concrete ring homomorphism $MSO_* \to R$ that satisfies some properties. Let’s look at $R := \mathbb{Z}[\frac{1}{2}][\delta, \epsilon]$.

What generators of $MSO_*$ are killed by this elliptic genus?

In other words, what is $ker(MSO_* \xrightarrow{\phi} \mathbb{Z}[\frac{1}{2}][\delta, \epsilon])$? Let’s say $\delta$ has degree 2, and $\epsilon$ has degree 4.

As we know (thanks Milnor), $MSO_*[\frac{1}{2}] \simeq \mathbb{Z}[x_2, x_4, x_6, x_8, …]$ ($MSO_*$ has some 2-torsion, adjoining 2 allows us to think of $x_i$ as $CP^{i}$) so we can rewrite $\phi$ as:

$$\mathbb{Z}[\frac{1}{2}][x_2, x_4, x_6, x_8, …] \xrightarrow{\phi} \mathbb{Z}[\frac{1}{2}][\delta, \epsilon]$$

$\mathbb{Z}[\frac{1}{2}][x_2, x_4, x_6, x_8, …]/(x_6, x_8, …) \simeq Z[\frac{1}{2}][x_2, x_4]$

We require that mult-by-$\alpha$ acts injectively in $E$ s.t. $C(\alpha)$ is just the cokernel of $\times \alpha$.

If mult-by-$\alpha$ acts injectively, then $0 \to \pi_k(\Sigma^E) \xrightarrow{\times\alpha} \pi_k(E) \to \pi_k(C(\alpha)) \to 0$ is an exact sequence for all $k$, and $\pi_*(C(\alpha))$ will give us the ring we’d expect, $\pi_*(E)/\alpha$.

$$MSO_*/(x_6, x_8, …) \simeq R$$

But can we do this on the level of spectra? Can we construct $MSO/(x_6, x_8, …) \simeq E$ s.t. $\pi_*(E) \simeq \mathbb{Z}[x_2, x_4]$ by coning off unwanted generators?

Well, before we get caught up in fantasy, what does “$MSO/\alpha$” even mean?

Understanding the quotient

Let’s use our intuition for what $R/\alpha$ should be when $R$ is a ring, that is, the exact sequence $0 \to (a) \to R \to R/\alpha \to 0$. I point this out to motivate that the quotient of a ring spectra by the ideal of a ring will be a cokernel (in nice cases).

Given an element $\alpha \in \pi_n(E)$, we get a map $\Sigma^n M \xrightarrow{\times \alpha} M$ where $M$ is an $E$-module.

How? We have an element $\alpha \in \pi_n(E)]$, that is:

$$\mathbb{S}^n \xrightarrow{\alpha} E$$, apply the functor $- \wedge M$

$$\mathbb{S}^n \wedge M \to E \wedge M$$

Note that a ring spectrum $M$ is a monoid object in the category of spectra (which is a monoidal category), we have $M \wedge M \to M \leftarrow \mathbb{S}$, with associativity and identity. Since $M$ is an $E$-module, we have the monoid action $E \wedge M \to M$, so we compose and get

$$S^n \wedge M \to E \wedge M \to M$$

We return our attention to the map: $\Sigma^n M \xrightarrow{\times \alpha} M$

Take the mapping cone of $\times \alpha$:

$$\Sigma^nE \xrightarrow{\times\alpha} E \to C(\alpha)$$

Taking homotopy groups:

$$… \to \pi_k(\Sigma^nE) \to \pi_k(E) \to \pi_k(C(\alpha)) \to \pi_{k+1}(\Sigma^nE) \to …$$

We require that mult-by-$\alpha$ acts injectively in $E$ s.t. $C(\alpha)$ is just the cokernel of $\times \alpha$. If mult-by-$\alpha$ acts injectively, then $0 \to \pi_k(\Sigma^E) \xrightarrow{\times\alpha} \pi_k(E) \to \pi_k(C(\alpha)) \to 0$ is an exact sequence for all $k$, and $\pi_*(C(\alpha))$ will give us the ring we’d expect, $\pi_*(E)/\alpha$).

Conceptually (and formally), we might think of the members of $ker(\phi)$ as stratifolds.

Coning off the members of the kernel (i.e. killing the appropriate generators of $MSO$)

So, back to the main story: killing the appropriate generators of $MSO_*[\frac{1}{2}] \simeq \mathbb{Z}[\frac{1}{2}][x_2, x_4, x_6, x_8, …]$ until we get $\mathbb{Z}[\frac{1}{2}][x_4, x_6]$ — but now we’re pulling this on the level of spectra. We now know that when we say “$MSO/x_6$” what we mean is:

We’re looking at $x_6 \in \pi_n(MSO)[\frac{1}{2}]$, and we take the mapping cone: $\Sigma^nMSO \xrightarrow{\times x_6} MSO \to C(x_6) =: MSO/x_6$

We can keep going in this vein, next looking at the sequence $\Sigma^n C(x_6) \xrightarrow{\times x_8} C(x_6) \to C(x_6, x_8)$, and iterate: say $x_i \in \pi_t(MSO)$, we’ll eventually come to:

$\Sigma^t C(x_6, …, x_{i-2}) \xrightarrow{\times x_i} C(x_6, …, x_{i-2}) \to C(x_6, x_8, …, x_i)$

We’ve constructed the map $MSO \to C(x_6, …, x_i)$ — requiring that the ideal $(x_6, …, x_i)$ acts injectively (s.t. the mapping cone is just the cokernel of the map).

Taking the homotopy groups of $C(x_6, …, x_i)$, we find $\pi_*(C(x_6, …, x_i)) = \mathbb{Z}[x_2, x_4]$.

We know that $C(x_6, …, x_i)$ is a spectrum, but is it still a ring spectrum? If the genus is Landweber-exact (i.e. tensoring out $x_6, …, x_i$ gives us a multiplicative homology theory), the resulting spectrum $MSO/(x_6, …, x_i)$ a ring spectrum.

This feels a bit hacky, but I’m not sure how else to prove that $C(x_6, …, x_i)$ is a ring spectrum besides some messy obstruction theory which I know nothing about.

I mentioned at the beginning that this construction is less intuitive than tensoring the underlying coefficient ring as an $MSO_*-module$, this is because we don’t have an obvious criterion for when the resultant spectrum is a ring spectrum!

Some remaining uncertainties

Let’s say we wanted to “geometrically” interpret inverting 2, even though this is a very algebraic thing, in the case of $MSO[\frac{1}{2}]$? I’m not quite sure how to do this precisely — the best I can currently do is outline a picture. Recall that $MU_n(X) := \{M^n \to X/ \sim\}$, which is the bordism classes of maps to X from an n-dimensional stably almost-complex manifold, so the $\times 2$ map is then the disjoint union of bordant classes of manifolds. Perhaps this gives us a way to invert $2$ geometrically? I’m really not sure how to make this precise.

Assuming that all of this works, this bordism with singularities $C(x_6, …, x_i) \simeq E$ is a presentation of the spectrum associated to our elliptic genus.

If we instead start with an elliptic genus $MU_* \to R$, can we get the same result? How does the grading differ or is it the same because we are factoring through $MSO$?

As a last comment, I’ll address my vague desire for the geometry of elliptic curves to come into this story. Looking, say, at the 2-torsion points of our elliptic curves — if this construction is functorial — we might expect a 2-torsion action on the bordism with singularities presentation. This might be close to what “tmf with level-2 structure” means, but I’m really not sure.

Thanks to Chris Sommer-Preis, Achim Krause, and Tomer Schlank for answering some of my questions on the topic.

Elliptic Curve Formal Group Laws: Philosophy and Derivation

Eine deutsche Übersetzung des folgenden Abschnitts befindet sich hier.

Philosophical Motivation

In the study of groups with topological structure, we commonly replace the global object (the group) with a local object (the infinitesimal group). We play the following game.

  1. Start with a space.
  2. Define a method of adding two points to get a third point of this space (which is associative, unital, commutative, and has inverses).
  3. Derive an infinitesimal group.

Examples:

  • Lie group (group internal to the category of smooth manifolds)
    $\to$ Lie algebra (group internal to the category of infinitesimal spaces).
  • elliptic curve formal group
    $\to$ elliptic curve formal group law.
  1. Start with an elliptic curve over a scheme $Spec R$.
  2. An elliptic curve is a group scheme (over $Spec R$) whose underlying object is 1-dimensional, proper, and smooth.
    In less abstract terms, it comes with an addition law and a marked point for free.
  3. Formally complete the elliptic curve along the origin
    In less abstract terms, take the Taylor series expansion of the addition law.

The phrase “formal completion of E along 0″ looks like the Taylor series expansion of an elliptic curve about the origin. In some sense, ‘completion at a point’ does give you the Taylor series of a map in the algebraic-geometric setting, but it’s more general than this. That is, a Taylor series is usually of the form $f(x+y) = f(x) + yP_1(x) + y^2P_2(X) + …$, where $P_i(x) = \frac{f^{(i)}(x)}{i!}$.

For more on this, Aaron Mazel-Gee wrote a great paper on Dieudonne modules.

We can view formal completions as key to our search for an algebraic version of a Lie Algebra. In differentiable Lie theory, the Baker-Campbell-Hausdorff theorem describes the group-law in a small neighborhood of the identity. However, this uses the exponential, which fails to make sense, even as a formal power series, in characteristic $p$.

This justifies our quest: describe the group law in a small neighborhood of the identity of the curve without using the exponential map. First, we must go on an excursion to understand completion.

You Complete Me

Recall the method of obtaining $\mathbb{Z}_p$, the p-adic completion of the integers.
\begin{align*}
& \mathbb{Z}/p \hookleftarrow \mathbb{Z}/{p^2} \hookleftarrow \mathbb{Z}/p^3 … \\
\mathbb{Z}_p & = \text{lim } \mathbb{Z}/p^n
\end{align*}

Let $R$ be a ring. We have a polynomial ring, $R[t]$. How do we make a ring of formal power series from this? Well, note that $R[t]/t = R$, so we get:
\begin{align*}
& R \hookleftarrow R[t]/t^2 \hookleftarrow R[t]/t^3 \hookleftarrow … \\
R[[t]] & = \text{lim } R[t]/t^n \end{align*}

Similarly,
\begin{align*}
& \text{Spec } A \hookrightarrow \text{Spec } A[t]/{t^2} \hookrightarrow \text{Spec } A[t]/t^3 \hookrightarrow … \\
\text{Spf } A[[t]] & = \text{colim }\text{Spec } A[t]/t^n
\end{align*}

Completion is the process of picking a ring $R$ together with a maximal ideal $m$ and forming the limit of $R/m^n$.

As an example, we consider the group scheme $E$ (for example, an elliptic curve) over $\text{Spec } A$.

$\text{Spec } A[t]/t^2$ is the infinitesimal neighborhood of the first derivative/first infinitesimal neighborhood of $\text{Spec }A$. This is like truncating the Taylor series after the information given by the first derivative, e.g. Lie algebra. Similarly, $\text{Spec } A[t]/t^3$ tells us about 2nd derivative, and we need a larger infinitesimal neighborhood for the second derivative. In the case of a formal group law, we want to write down all of the infinitesimals, so we take the colimit of $\text{Spec } A[t]/t^n$ and get a formal scheme.

Zariski locally on $S$, the formal completion $\hat{E}$ of $E$ along the zero section is of the form:

$$\hat{E} \simeq \text{Spf } (A[[t]])$$ where $\text{Spf } A[[t]] := \text{colim} \text{Spec } A[t]/t^n$

For example, Zariski-locally over $Spec(A)$, we can put an elliptic curve in Weierstraß normal form $y^2 + a_1 xy = x^3 + a_2 x^2 + a_4 x + a_6$ (I think outside of characteristic 2 and 3, we may take $a_1 = 0$).

In this form, the variable $z = x/y$ is a uniformizer at the identity ($y$ has a pole of order 3 at the identity, $x$ has one of order 2, so $x/y$ has zero of order 1). It’s reasonable to expect $Spf(A[[z]])$ to be the formal completion.

Note that we are choosing an isomorphism $G \simeq \text{Spf }R[[t]]$ here, that is, choosing where to send $t$ in $G$ (sidenote: if $G$ is assoicated to a cohomology theory, our choice of where to send $t$ corresponds to a choice of complex orientation for the cohomology theory).

It is worth noting the difference between $\varinjlim \text{Spec } A[t]/t^n$ and $\text{Spec} \varinjlim A[t]/t^n$.
$$\varinjlim \text{Spec } A[t]/t^n =: \text{Spf } A[[t]]$$
$$\text{Spec} \varinjlim A[t]/t^n =: \text{Spec } A[[t]]$$
Formal schemes live in a category of limits of schemes (affine formal schemes live in a category of rings with the $I$-adic topology for some ideal $I$). So Spec($A[[t]]$), considered as a formal scheme, would just be the trivial limit of Spec of the non-topologized ring.

Think about it: with Zariski topology, Spec($k[[t]]$) is dense in Spec($k[t]$), while Spec($k[t]/(t^n)$) is not, for any $n > 0$.

Our elliptic curve is a projective beast: reviewing the necessary coordinate changes.

Projective polynomials are defined by the following property: $P(\lambda x, \lambda y, \lambda z) = \lambda^k P(x,y,z)$

We can make any polynomial into a polynomial with this property by homogenizing (e.g. $y^2z = x^3 + axz^2 + bz^3$).

We ask for the completion of the elliptic curve at the identity wrt the group structure. That is, at the “point at infinity” if we write it in Weierstraß form. “Point at infinity” is the intuition, but “homogenize and look at the projective variety” is how its formalized. (Affine things are slightly awkward, and we can always homogenize equations by adding in another variable, so there’s a canonical way to compactify. Sadly that’s often implicit in the literature.)

  1. Input elliptic curve in homogeneous coordinates ($y^2z = …$)
  2. Coordinate transform (1) s.t. our elliptic curve contains the origin. ($f(t) = t…$)
  3. We want an equation for $S$ in terms of $T$, so we find the fixed point $T = \phi(T)$ of (2) by repeatedly substituting $t…$ for $f(t)$. This gives us a power series of in terms of $T$.

Let’s talk about step 2: we can’t talk about the point at infinity in our staring chart of the elliptic curve since it doesn’t include it , that is, $(0,0)$ doesn’t even generally lie on $y^2 = x^3+ax+b$ so we must coordinate change to an affine coordinate CONTAINING that point.

Screen Shot 2015-03-28 at 6.27.53 PM

There’s three maps from projective coordinates back to affine coordinates. All are defined on a certain open subset of $\mathbb{P}^2$ (namely the one where a certain coordinate doesn’t vanish), and the corresponding map gives a coordinate chart for that open subset).

For an elliptic curve in Weierstraß form, the $z \neq 0$ chart doesn’t contain the identity of the elliptic curve, but the $y \neq 0$ has the identity at $(0,0)$. We’re basically changing coordinates from the classical affine equation $(y^2=..)$, which lives on the $z \neq 0 thing$, to the other one living on the $y \neq 0$ thing.

Think about it as follows: the elliptic curve lives in $\mathbb{P}^2$, but we can describe its intersection with each of the coordinate charts. That’ll of course be different equations. The $z\neq 0$ and $x\neq 0$ are not suitable to talk about the neighbourhood of the identity, because they simply don’t CONTAIN the identity.

Derivation (for simplified Weierstrass):

The explicit derivation of elliptic curve formal group law from an elliptic curve is performed with a series of coordinate changes:

Screen Shot 2015-05-01 at 3.07.40 PM

By plugging the expression for $f(T)$ into itself repeatedly, $f(T)$ stabilizes to a power series with $T$ as the only variable.

It is sometimes tiresome (and sometimes meditative) to go through the coordinate changes and explicit solving for $f(T)$ in terms of $T$ on paper. sage (available free online) automates this process:

sage: E = EllipticCurve([2,3]).formal_group(); E}
Formal Group associated to the Elliptic Curve defined by y^2 = x^3 + 2*x + 3 over Rational Field
sage: F = E.differential(15); F
1 + 4*t^4 + 9*t^6 + 24*t^8 + 120*t^{10} + 295*t^{12} + 1260*t^{14} + O(t^{15})

An afternote on Cartier’s curves

I’ll wrap this up by mentioning Cartier’s curves. Cartier’s method relies very heavily on the fact that a formal group law is the formal spectrum of a power series ring. The previous method relied heavily on the fact that a formal group law $G$ is the inductive limit of the finite subgroupschemes $ker(p^n)$, where $p^n$ is the $n$th Frobenius morphism: $G \to G$, $x \mapsto x^{p^n}$. This relies on all modules being ind-finite, which Lubin mentions to be hopless in the case of Dieudonne modules.

Let $B$ be any commutative $A$-algebra which is separated and complete over the topology defined by $I^n$ where $I$ are ideals of $B$ (i.e. let $B$ be a formal $A$-scheme).

For example, take $B = A[[t]]$, the formal power series ring in one variable. The ideal $I = tA[[t]]$, the set of $n$-tuples of elements (type: power series) in $I$ may be equipped with a composition law $\textbf{F}(a, b) = a*b$ (this composition will be finite, as we are working with nilpotent elements). Cartier calls such maps $\text{Spf }A[[t]] \xrightarrow{F} \text{Spf } A[[t]]$ “curves in \textbf{F}.”

This name choice makes sense. Recall that we can define a polynomial over $\mathbb{C}$ as $\mathbb{C}[t] \to \mathbb{C}[t]$, similarly, we can define a curve over over $\text{Spec } A$ as $\text{Spf } A[[t]] \to \text{Spf } A[[t]]$.

Thank you to Akhil Mathew (for explaining the definition of formal schemes), Jesse Silliman (for answering my various questions on completion and formal schemes), and Achim Krause (for walking me through the derivation of the group law).

What does the sphere spectrum have to do with formal group laws?

This post assumes that you’re familiar with the definition of a prime ideal, a local ring, $R_{(p)}$, the sphere spectrum, $\mathbb{S}$, and the Lazard ring, $L$.

During a talk Jacob Lurie gave at Harvard this April, he labeled the moduli space of (1-d commutative) formal group laws as $\text{Spec }\mathbb{S}$.

Eric Peterson kindly explained why $\text{Spec } \mathbb{S} \simeq \text{Spec } L$ and I found his answer so lovely that I wish to share (all mistakes are due to me).

Why is Spec L iso to Spec $\mathbb{S}$?

This is part of the story of geometers working with higher algebra asking “what is an ideal of a ring spectrum?”

A ring $R$ ——————-> category $Mod_R \supseteq Perf_R$ (finitely presented)

Note that $Perf_R$ is the category of perfect complexes of $R$-modules. A perfect complex of $R$-modules is a chain complex of finitely generated projective $R$-modules $P_i$, and is thus of the form $$0 \to P_s \to … \to P_i \to 0$$

The ring spectrum $\mathbb{S}$ ——————-> category $Mod_{\mathbb{S}} \supseteq Perf_{\mathbb{S}}$

Note that $Mod_\mathbb{S} \simeq$ Spectra, and $Perf_\mathbb{S} \simeq$ Finite Spectra

A finite spectrum is a spectrum which is the de-suspension of $\Sigma^\infty F$, where $F$ is a finite CW -complex.

There’s a theorem by Balmer answering “what is an ideal in this context”, which points out this analogue:

$\text{Spec }R$ as a space $p$ as a point (an element of $\text{Spec }R$)

$Perf_R$ as a space $\mathcal{P}$ as a point (a subcategory of $Perf_R$)

satisfying that $\mathcal{P}$ is:

  1. $\otimes$-closed against R-modules $$a \in Perf_R; b \in \mathcal{P} \Rightarrow a \otimes b \in \mathcal{P}$$
  2. a thick subcategory of $Perf_R$ (i.e. it’s closed under cofiber sequences and retracts i.e. closed under extension)

A “prime ideal” of $Perf_R$ is a “proper thick tensor-ideal” $P$ ($\subsetneq Perf_R$) s.t.

$$a \otimes b \in \mathcal{P} \Rightarrow a \in \mathcal{P} \text{ or } b \in \mathcal{P}$$

So, if $K_*(-)$ is a homology theory with Künneth isomorphisms $$K_*(X \wedge Y) \simeq K_*(X) \otimes_{K_*} K_*(Y))$$

$\Rightarrow \mathcal{P} = \{X | K_*(X) = 0\}$ must be a “prime ideal”.

Sanity check:

\begin{align*}
K_*(X \wedge Y) & \simeq K_* X \otimes K_*Y \\
& \simeq 0 \otimes K_*Y \simeq 0 \\
\end{align*}

Here’s the surprising theorem that ties this prime ideal excursion into our original question (Periodicity Theorem: Hopkins and Smith):

  1. Any $C \subset Perf_{\mathbb{S}}$ arises in this way
  2. All homology theories with Künneth isomorphisms are Morava K-theories
    including $Hk$ where $k$ is a field, which is just the infinite Morava K theory $K(\infty)_{(p)}$.

The proof of this is currently beyond my grasp, so I’m afraid I can’t talk you through it.

Taking this theorem’s proof as a black box, we’ve scraped together enough context to parse the answer of why $\text{Spec }L \simeq \text{Spec }\mathbb{S}$.

Let’s look at $\text{Spec }Z$:

Screen Shot 2015-05-01 at 1.33.04 AM

let’s look at the residue classes of $Z$:

Screen Shot 2015-05-01 at 1.33.00 AM

and at $Spec HZ \simeq Spec Z$, where the ring spectra $HR$ represent $H^*(-;R)$;

Screen Shot 2015-05-01 at 1.32.56 AM

By the nilpotence theorem, the ideals of $\mathbb{S}$ are the Morava K’s (one for each height and each prime)…

2015-04-30 18.57.27

…so, $\text{Spec } \mathbb{S}$ looks like $Spec L$ (by a theorem of Lazard, 1-d formal group laws over separably closed fields of char p are classified up to iso by their height).

To be absolutely clear: for $K(n)_{(p)}$; $(p)$ corresponds to the characteristic of the field (over which the formal group law is defined), while $n$ corresponds to the height of the formal group law.

Afternote:

A comment of Lennert Meier’s on MO caught my interest. He mentioned that as the spectrum $Ell$ (associated to a supersingular elliptic curve) is Bousfield equivalent to $K(0) \vee K(1) \vee K(2)$ (with an implicit localization at a prime), we have $Ell_*(K(A,n)) = 0$ for $A$ finite abelian and $n \geq 3$.

Note that $F$ and $E$ are Bousfield equivalent if for every spectrum $X: F_*(X)$ vanish iff $E_*(X)$. This is an equivalence relation on spectra.

Any elliptic cohomology is Bousfield equivalent to a wedge of Morava K-theories. Before we discard looking at individual elliptic cohomology theories, in favor of their “universal” counterpart with nice automorphisms, let’s look at the difference between $K(0) \vee K(1)$ and complex K-theory, and try to lift these differences to those of $K(0) \vee K(1) \vee K(2)$ and supersingular $Ell$. It was pointed out to me that this is like comparing a local ring to its residue field.

To compute the Atiyah Hirzebruch spectral sequence of $E^*(X)$, we need to know both the attaching maps in the space $X$, and the attaching maps in the spectrum $E$ (which I believe are called its Postnikov tower), both are hard (in most cases).

We currently only know how to compute the AHSS of $Ell^*(X)$ when we have some map from $CP^\infty \to X$ (since we define $Ell$ using $CP^\infty$), for this map induces a map between spectral sequences.

Landweber-Stong-Ravenel Construction Flowchart

Here’s a flowchart I made while preparing for an upcoming talk. I fear that it may be hard to follow without being already familiar with the story, but there’s little harm in posting it. Maybe it’ll help someone navigate the literature.

11137755_10205356925075246_137597741_oSome context: 11147916_10205356927235300_167198395_o

Note that $MU^*(-)$ is the universal complex-orientable cohomology theory.

Summary of the relationship of $MU^*$ and $L$:

11092660_10205356929435355_1213751590_n

 

The Landweber Exact-Functor Theorem

This post assumes familiarity with formal group laws, the definition of exact sequences, the motivation of the Landweber-Ravenel-Stong construction, that the exactness axioms is one of the generalized Eilenberg-Steenrod axioms, and the fact that formal group laws over $R$ are represented by maps from the Lazard ring to $R$. I am learning these concepts, and constructive criticism is highly encouraged.

Recall the the Landweber-Ravenel-Stong Construction: $MU^*(X) \otimes_{L} R \simeq E^*(X)$, where $MU^* \simeq L$ and $R \simeq E^*(pt)$.

We know that in general, tensoring with abelian groups does not preserve exact sequences (e.g., applying $-\otimes_{\mathbb{Z}} \mathbb{Z}/2$ to $0 \to \mathbb{Z} \xrightarrow{\times p} \mathbb{Z} \to \mathbb{Z}/p \to 0$).

11136786_10205364668548828_1871863654_n

So, when does the functor $-\otimes_L R: MU^*(X) \to E^*(X)$ preserve exact sequences?

11157286_10205364655788509_750654639_o

An object M in an abelian tensor category $C$ is ‘flat’ if for all $X \in Obj(C) $, the functor $X \to X \otimes M$ preserves exact sequences.

Because arbitrary $MU_*$-modules do not occur as the $MU$-homology of spaces, the requirement of flatness over all of $MU_*$ can be relaxed.

It is Landweber-exact if it preserves exactness when applied to things in the range of the homology theory, though it doesn’t have to preserve exactness on things that aren’t in the range of the homology theory. (summarization by Alex Mennen)

Sidenote: All we require is that the map $\text{Spec } R \to M_{FG}$ is flat. Why to $M_{FG}$ and not to $M_{FGL}$? Every (complex orientable) cohomology theory corresponds to a formal group; picking a complex orientation corresponds to a choice of coordinate of our formal group law.

I say $MU_*$ instead of $MU^*$ for technical reasons, namely, colimits don’t behave nicely in cohomology. But I don’t understand the implications of that well enough to talk about it coherently.

Flatness and torsion are intimately related. It’s a theorem of Lazard that an object is flat iff it’s a filtered colimit of free modules.

Let’s say we have a map from the Lazard ring $L \xrightarrow{F} R$ representing our formal group law $F$. What is this exactness condition, precisely?

Let’s back up a bit and look at what it might mean for a formal group law to be “flat.”

We wish to “add” a point to itself via the formal group law $p$ times and look at its general form (this is called the $p$-series of our formal group law). This allows us to detect stuff like points of order $2$ in elliptic curves, that is, the points that when added to themselves give us the origin. Note that $p$ is a prime in $\mathbb{Z}$, not necessarily a prime in $R$.

In general, we can talk about adding a point $x$ to itself $n$ times using the following recursive definition:

Screen Shot 2015-04-10 at 9.50.25 PM

It doesn’t seem like it at first glance, p-series, or “multiplication by p” map, is EXTREMELY IMPORTANT — it allows us to find periodic phenomena. Let’s look at the map Screen Shot 2015-04-10 at 9.57.41 PMThe kernel of $\lambda_n$ will be the roots of unity in $C$.

Similarly, let’s look at the group $G$.

Screen Shot 2015-04-10 at 9.57.44 PM

The kernel of this map will be the points in $G$ with an order that divides $p$. If $p$ is a prime, it’s just the points of order $p$.

Examples of $p$-series: Additive formal group law $F(x, y) = x + y$

Screen Shot 2015-04-10 at 9.49.31 PM

Multiplicative formal group law $F(x, y) = x + y + cxy$, where $c = 1$; we’ll be working with $c = -1$ at some point later in the post, sorry for the inconsistency!

Screen Shot 2015-04-10 at 9.49.53 PM

The $p$-series of $F$ will always be of the form:

$$[p](x) = px + … + v_1x^{p^1} + … + v_nx^{p^n} + … $$

where $v_k$ is simply the name we give the coefficient of the expression $x^{p^k}$.

We’re interested in the tuple of coefficients relevant to the powers of $p$, that is $(p, v_1, …, v_k, …)$. Let’s mod out the $p$-series by $p$, then by $v_1$, etc. until we get to $0$. We are trying to check that $v_n$ acts injectively in $R/(v_0, …, v_{n-1})$ for all $n$.

Note that $p$ is a prime in $\mathbb{Z}$ and $v_1, …, v_n \in$ the image of $MU^*$ in $R$. The condition of “regular” wards away zero divisors because they are nasty.

For example, our tuple for the multiplicative formal group law is $(p, 1)$, since $v_1 = 1$, and the rest of the coefficients are 0, so we have:

11086663_10205364819992614_1026298151_n

Tada! It’s Landweber exact so you can bet your muffins that $MU^*(X) \otimes_L \mathbb{Z}$ is a cohomology theory, in fact, it’s iso to $K^*(X)$.

That’s a lot to take in, I know, so let’s back up a bit and examine the map $$Vect^1(X) \xrightarrow{c_1} K^2(X; \mathbb{Z})$$ where the group operators are the tensor product of line bundles and the tensor product of virtual line bundles.

Note that the dimensions multiply when you do the tensor product, so $L_1 \otimes L_2$ is still a line bundle.

Let’s say that $c_1(L) = 1-L$, then we’d expect $c_1(L_1 \otimes L_2) = 1 – L_1 \otimes L_2$. So, how do we express $c_1(L_1 \otimes L_2)$ in terms of a formal group law $F(x,y)$ where $x = c_1(L_1)$, and $y = c_1(L_2))$?

In other words, using the slightly clearer notation $F(x,y) \equiv x +_F y$…for what $F$ does $x +_F y = 1-xy$?

Screen Shot 2015-04-09 at 5.49.07 PM

Now that I’ve gotten you riled up, let’s back away from $K$-theory. I wrote this post because I was really excited about the following: What if we wish to look at the cohomology theory associated to a supersingular elliptic curve in characteristic $p$?

We can’t have torsion in the Landweber-Stong-Ravenel construction, however we can consider the elliptic curve over p-adic completion of $\mathbb{Z}$, and adjoin $v_1$, such that $(\mathbb{Z}_p[[v_1]]/p)/v_1 = \mathbb{Z}/p$.

WOO! Actually, this isn’t too surprising. Recall that the p-adics are the limit of $\mathbb{Z}/p^n$, thus by definition it comes with maps to all the guys in the limit.

An elliptic curve formal group law must be of either height 0, 1, or 2 (1 if singular, 2 if super singular). Why can’t it be of higher height?

I’d also like to mention something slightly more obvious but still awesome: Tensoring with $\mathbb{Q}$ gets rid of torsion. Recall that every formal group law over $\mathbb{Q}$ is isomorphic to the additive formal group law. The difference between cohomology theories arises due to torsion! $E \otimes \mathbb{Q}$ simply gives us singular cohomology with some coefficient ring!

(It might be tempting to thing that all even periodic cohomology theories (that is, $E^{n}(*) \otimes_{E^*(*)} E^2(*) \simeq E^{n+2}(*)$, and $E^n(*) = 0$ if n is odd) are associated to the multiplicative formal group law. However, we must recall that elliptic cohomology is even periodic.)

This is true more generally: rational spectra (i.e. all the homotopy groups are $\mathbb{Q}$ – vector spaces) are always determined by their homotopy groups. More precisely: there is a functor from rational spectra to graded $\mathbb{Q}$ – vector spaces given by taking homotopy groups, and it’s an equivalence. I’m not yet sure why this is.

Thank you to Akhil Mathew for kindly answering my questions on Landweber-exactness, Achim Krause for talking with me about rational spectra, and the lovely people of Math Overflow for answering my question on Landweber-exactness.

What is an elliptic curve?

An elliptic curve is a curve given by an equation of the form:
$y^2$ = polynomial of degree $3$ in $x$ without a multiple root

Sidenote: This is a bit misleading, for the elliptic curve is not actually the affine variety associated to $y^2=x^3+ax+b$. An “elliptic curve” refers to the projective variety associated to the corresponding homogeneous polynomial $y^2z=x^3+axz^2+bz^3$.

Elliptic curves are really quite extraordinary. Allow me to boast a few beautiful facts about elliptic curves, before we whip out the language of schemes and make some structural statements.

  1. The set of points defined over $R$ of an elliptic curve over $R$, together with a marked point $O$, forms an abelian group.
    If you haven’t encountered this fun fact, this slidedeck by Silverman discusses it pretty nicely so I’ll refer you there.
  2. The set of rational points of an elliptic curve defined over $\mathbb{Q}$ together with $O$, forms a fintely generated abelian group.
  3. In order to study rational points on an elliptic curve, it is important to use properties of the height of a rational point.
  4. An elliptic curve (over the complex numbers) is a torus.

The aim of this post is to dip the reader’s toes into some of the language people use to talk about elliptic curves. In upcoming posts, we’ll talk about Mordell’s theorem, why the notion of height is vital, the genus of a curve, and why an elliptic curve is a torus. In the meantime, I highly recommend Silverman and Tate.

When we say that $B$ is a ring over $A$, we mean that we have a fixed ring homomorphism $\phi: A \to B$.

When we say that $E$ is a curve over $R$, we really mean that we have a fixed scheme homomorphism $p: E \to \text{Spec R}$.

It might be refreshing to note that $E \xrightarrow{p} S$ is a fibration!

You might think to yourself, why isn’t the map in question $E \to R$? We must remember that $E$ is a geometric object, so it’s a bit dirty to map a geometric object to a ring, when we have a perfectly good way to look at rings as geometric objects (i.e. applying the Spec functor)!

Let’s take a look at an example fiber (let’s say $R$ is the reals) and we’ve chosen our marked pointed (in this case, the origin of our group) to be the point at infinity.

Sidenote: “point at infinity” is the intuition, but “homogenize and look at the projective variety” is how its formalized). Affine things are slightly awkward, and we can always homogenize equations by adding in another variable, so there’s a canonical way to compactify. Sadly that’s often implicit in the literature.

unnamedq

The zero section “$0$” is the section of this family of elliptic curves which consists of the marked point of each fiber (that is, it selects the “origins” of each fiber).

unnamed-2

When I say “elliptic curve with coefficients in R,” I mean a family of elliptic curves $E$, where each elliptic curve corresponds to a selection of coefficients in $R$.

Each fiber is a group, so we might think of the total space as a groupoid.

Screen Shot 2015-03-22 at 2.14.42 PM

In the vein of abstraction, we can think of an elliptic curve as a special group scheme. In the same spirit of comparing smooth variety to a smooth manifold, a group scheme can be compared to a topological group.

Now that we’ve peeked down the rabbit hole, let’s go a bit further.

In the language of algebraic geometry, an elliptic curve $E$ (over a scheme $S$) is a smooth projective proper curve of genus 1 with a marked section (i.e. “$0$” $: S \to E$).

Screen Shot 2015-03-21 at 2.13.04 PM

$S$ := $\text{Spec }R$.

It was pointed out to me that this marked section is important! This means, for example, that the automorphism group of the elliptic curve is finite.

A proper scheme can be conceptually compared to a compact closed manifold.

I’ve been told that the “proper map” can be thought of as “If I can map a punctured disk into my space, and then map a point into my space (where the puncture in the disk is), and I now have a disc in my space, then this is a proper map” — but take this with a grain of salt and a copy of Hartshorne.

But what is the definition of a proper scheme? A proper scheme is a scheme such that the map to the terminal object is proper (depending on whether you’re talking just schemes or ‘schemes over $K$’, the terminal guy will be $\text{Spec } \mathbb{Z}$ or $\text{Spec } K$).

It turns out that in algebraic geometry, a lot of properties of objects are special cases of properties of maps in this way.

A Closing Remark on Classifying Stuff

As we desire to classify line bundles
Screen Shot 2015-03-21 at 2.17.17 PM

…we desire to classify elliptic curves.

We have some collection of objects that we wish to parameterize, say, curves (i.e., 1-dimensional projective complex varieties) and so we define the corresponding representable functor (i.e. we define a “family of curves” and we want to find a space that represents it).

Turns out that to do this nicely, we need the theory of stacks, which account for the fact that some of these objects may have nontrivial automorphisms. The moduli stack of elliptic curves $\mathcal{M}_{ell}$ = upper half plane / modular group.

Screen Shot 2015-03-21 at 2.17.23 PM

I leave a question mark where the “universal elliptic curve” should be, for I have yet to really understand what it means to be a universal elliptic curve.

Thank you to Barry Mazur (for answering my question on the “0” section), Achim Krause (for patiently explaining precisely why elliptic curves are projective varieties), Alan Weinstein (for answering my conceptual questions on how to think about the map $E \xrightarrow{p} S$), Semon Rezchikov (for talking with me about properties of schemes), and David Yang (for pointing out the importance of the marked section).

Group Law on the (Punctured) Affine Line

There are likely inaccuracies in this post, as I am just beginning to learn the basics of algebraic geometry. Constructive criticism is strongly encouraged.

There once was a line…

Let’s look at the affine line over $\mathbb{C}$. This is just the complex line with no distinguished element (i.e., a plane which forgot it’s origin — it’s 2 real dimensions, or, equivalently, 1 complex dimension).

$\mathbb{A}^1 \simeq \text{Spec } \mathbb{C}[x]$

As we know, the affine line over the field $\mathbb{K}$ is isomorphic to the spectrum of a ring of single variable polynomials (with coefficients in $\mathbb{K}$). If you aren’t familiar with this isomorphism, I recommend popping over to Spectrum of a Ring. For simplicity, let’s work with the field $\mathbb{C}$, although, I’m pretty sure the rest of this post still works for any $\mathbb{K}$.

Is there a reasonable way to take in two points, and ask for a third?

This is generally a fun question to answer when you’re handed a space. So, how do we add two points of $\mathbb{A}^1$ to get a third point in $\mathbb{A}^1$?

The same way we usually add two complex numbers! But wait, we need an identity! Where does the identity come from? I think it comes from looking at the affine plane as $\text{Spec }\mathbb{C}[x]$, the maximal ideal $x$ is a distinguished point.

Screen Shot 2015-03-17 at 4.34.22 PM

It’s worth considering why $\text{Spec }\mathbb{C}[x] \times\text{Spec }\mathbb{C}[y] \simeq \text{Spec }\mathbb{C}[x,y]$. It seems that this is because $\text{Spec }$ is functorial, that is:

$\text{Spec }\mathbb{C}[x, y] = \text{Spec }(\mathbb{C}[x] \otimes {C}[y]) = \text{Spec }\mathbb{C}[x] \times \text{Spec }\mathbb{C}[y]$.

Another way to think about this is that the cartesian product $\mathbb{A}^1 \times \mathbb{A}^1= \mathbb{A}^2 = \text{Spec }\mathbb{C}[x,y]$.

$\mathbb{A}^1 – \{0\} \simeq \text{Spec } \mathbb{C}[x, x^{-1}]$

Can we analogously put a group structure on the punctured affine line? Hell yes we can, here’s one way to do so:

Screen Shot 2015-03-17 at 3.56.56 PM

Recall that $\text{Spec }\mathbb{C}[x, x^{-1}] \times \text{Spec }\mathbb{C}[y, y^{-1} =\text{Spec }\mathbb{C}[x, y, x^{-1}, y^{-1}]$.

Screen Shot 2015-03-17 at 3.54.05 PM

$\text{Spec }$ is a contravariant functor, so we get a corresponding diagram in $\text{Ring}$:

Screen Shot 2015-03-17 at 3.53.58 PMThe explicit maps that define this diagram reveal the desired group law:

Screen Shot 2015-03-17 at 4.03.30 PM

That is, $z \mapsto (x+1)(y+1) -1 = xy + x + y$, which is a form of the multiplicative formal group law that we know and love. Intuitively, we can think of it as wanting to multiply $a$ and $b$, but first needing to shift to the multiplicative identity $1$, and afterward translating back again.

$\mathbb{A}^1 – \{0\} – \{1\} \simeq \text{Spec } \mathbb{C}[x, x^{-1}]$

What I don’t yet see, dear reader, is how to put group structure on $\text{Spec } \mathbb{C}[x, x^{-1}, (1-x)^{-1}]$ (which is isomorphic to the affine line minus two points). What are your thoughts?

Thanks to Yifei Zhao for kindly helping me derive the multiplicative formal group law from the group law on the punctured affine line.

$Pic(X)$ vs. $CP^\infty$

There are likely inaccuracies in this post, as I wrote it quickly and am just beginning to learn the basics of algebraic geometry. Constructive criticism is strongly encouraged.

As we saw in a Precursor to Characteristic Classes, $CP^\infty$ is the classifying space of complex line bundles over $X$.

11029826_10205043262913888_1448706408_o

$CP^\infty$ is, in some sense, the moduli space of line bundles over a point. There’s only one isomorphism class of line bundles over a point — but then this one line bundle has automorphism group $C^\times$ (which is homotopy equivalent to $U(1)$).

Allow me to introduce you to something that looks a LOT like $CP^\infty$.

11012318_10205043263553904_1786268063_n

What is this map, $p \times C \to Pic(C) \times C$, you might ask. Choose a point on our curve $C$ and this defines a line bundle over $S$ corresponding to a choice of the class of line bundles in $Pic(C)$. In other words, we take a point on a (not sure if I require smoothness here) algebraic curve and turn it into a line bundle on that curve.

11004347_10205043264553929_659910229_nWarning: I’ve been told that there is a difference between topological line bundles and algebraic line bundles, unfortunately, I don’t know why or what it is! I mention this, for $Pic(X)$ usually corresponds to *algebraic* line bundles over $X$.
11001318_10205043268794035_1055016622_o

Why is the multiplicative formal group getting involved? Let’s briefly review what the multiplicative formal group law is (as a group scheme).
11016577_10205043272954139_1694898481_n

11008983_10205043275634206_1397363895_nThank you to Edward Frenkel for kindly explaining the difference between $CP^\infty$ and $Pic(X)$ (both classifying spaces of line bundles), and to Qiaochu Yuan for explaining why on earth $CP^\infty$ is the moduli space of line bundles over a point. Any errors are mine, not theirs.

A Precursor to Characteristic Classes

I’ll assume that you know what a line bundle is, and are comfortable with the following equivalences; if you aren’t familiar with the notation in these equivalences, John Baez might help. Note that integral cohomology := cohomology with coefficients in $\mathbb{Z}$.

$U(1) \simeq S^1 \simeq K(\mathbb{Z}, 1)$

$BU(1) \simeq CP^\infty \simeq K(\mathbb{Z}, 2)$

The aim of this post is to give you a taste of the beautiful world of characteristic classes and their intimate relationship to line bundles via the concrete example of how the second integral cohomology group of a space is actually the isomorphism classes of line bundles over that space.

That’s right! $H^2(X; Z) \simeq$ the isomorphism classes of (complex) line bundles over X. It is in fact, a group homomorphism — the group operations being tensor product of line bundles and the usual addition on cohomology. This isn’t something that I understood at first glance. I mean, hot damn, it’s unexpectedly rich.

Let’s talk about line bundles.

  • $RP^1$ consists of all lines that intersect the origin of $R^2$.
  • $CP^1$ consists of all complex lines that intersect the origin of $C^2$.

Let’s look at $C^2$: Draw a line $r$ parallel to one of the axes, each line $L$ through the origin will intersect this line at a unique point $x$. This point characterizes $L$.

image-5

Only one line, the line parallel to our line $r$ will not intersect $r$ — we can say that this line is characterized by the point at $\infty$.

image-4

$CP^1$, the collection of all complex lines through the origin of $C^2$, is then isomorphic to all of the points $x$ (including the point at infinity). In other words, $CP^1 \simeq S^2$.

Recall that the complex line has 2 real dimensions — this powerful isomorphism is simply due to the one-point compactification of $\mathbb{R}^2$ that we know and love. Note that we can also get from $CP^1$ to $S^2$ via stereographic projection.

image-7

“Canonical” line bundles

The elements of $CP^1$ are the points $x$, thus we can describe a line bundle over $CP^1$ as follows: its points are the pairs $ (a,x)$, where $a$ is a point on the line $L$ (characterized by x) $\in \mathbb{C}^2$. The base space is $C^1$ + $pt$ at infinity, and each fiber is $L$.

This line bundle is called the canonical line bundle of $CP^1$.

This story holds for all $n$. In general, each point $x$ in $\mathbb{CP}^n$ is line $L$ through the origin in $\mathbb{C}^{n+1}$. Let $\ell^n$ := the canonical line bundle of $CP^n$.

I hope we can agree that we can describe a line bundle over $X$ as follows: to each element of $X$ (a point), we associate an element of $\mathbb{CP}^\infty$ (a line).

Saying that the line bundle over $X$ we know and love is a way to associate a line to every point in $X$ seems obvious and trivial — but asking “where do lines live?” has some beautiful consequences. I want you to feel this in your bones, so I’ll spell it out a bit more explicitly.

What are line bundles over a topological space $X$?

A line bundle $f$ is:

  • a map from each point in $X$
  • to a line (i.e. an element of $\mathbb{CP}^\infty$).

I’ll repeat this again: a line bundle $f$ is a map of type $X \to \mathbb{CP}^\infty$.

*todo: add a bit about $\ell^\infty$ here*

In other words, any complex line bundle $L$ over $X$ is a pullback of $\ell^\infty$ by the map $f$.

image

Cohomology is a Representable Functor

The homotopy classes of maps from a space $X$ to the nth Eilenberg-MacLane space $B^n(G)$ of a group $G$ is isomorphic to the $n$th cohomology group of a space $X$, with coefficients in the group $G$. In other words:

$[X, B^n(Z)] \simeq H^n(X; Z)$

This is a special case of a theorem, the Brown Representability Theorem, which states that all cohomology theories are represented by spectra, and vice versa. But that’s a whole ‘nother story! Let’s see how this connects to line bundles:

  • $[X, B^n(Z)] \simeq H^n(X; Z)$
  • $[X, B^2(Z)] \simeq H^2(X; Z)$

As you’ll recall from the equivalences listed at the beginning of this post, $B^2(Z)$,the 2nd Eilenberg-MacLane space of the integers as a group, is isomorphic to $CP^\infty$, thus:

  • $[X, CP^\infty] \simeq H^2(X; Z)$

image-2

Warning: $B^2(\mathbb{Z})$ is non-standard notation, and is usually written as $K(\mathbb{Z},2)$; here’s a paper that explains how to compute Eilenberg-MacLane spaces.

Hey, you said that there would be characteristic classes! Where do those come in?

I did say that this post was a precursor to characteristic classes, but let’s look at a piece of the map. ($\to$ := correspond to)

  • complex line bundles $\to$ elements of $H^2$ over $Z$ (“Chern classes”)
  • real line bundles $\to$ elements of $H^2$ over $Z/2$ (“Stiefel-whitney classes”)
  • quarternionic line bundles $\to$ elements of $H^4$ over $Z$ (“Pontryagin classes”)

This works for BU(n) when n=1. What about other n?

This section is also a teaser. I’d like to suggest that the story generalizes from complex line bundles to complex vector bundles. I don’t quite understand the details of this generalization, but I wish to share with you what I do understand.

Note that a complex n-dimensional vector bundle + a choice of hermitian metric = a $U(n)$-principle bundle.

So, it makes sense that classifying complex n-dimensional vector bundles (which I’ll denote $E \to X$) is closely related to the story of classifying their their associated principle $U(n)$-bundles (which I’ll write as $\hat{E} \to X$).

This motivates us considering that our previous picture…
image
…might just be a special case of a more general phenomena! But how?
Well we have a complex vector bundle — so what is our associated frame bundle?

unnamed

Let the classifying map of $\hat{E} \to X$ be $f: X \to BU(n)$.

image-3

Note that $BU := \text{colim}_n BU(n)$.

If you’d like to learn more about characteristic classes, I’ve found Milnor and Stasheff to be of great help.

Thank you to Peter Teichner for patiently explaining why $\mathbb{CP}^1 \simeq S^2$ and consequently the cell decomposition of $\mathbb{CP}^n$.