## Spf $$E^*[[x]]$$: Your walk through a flower garden

Inspired by the extraordinary expository style of Dr. Kazuya Kato, I’ve started reading parts of a (translated) Japanese children’s book when I’m stuck on a tough paper or concept — revisiting the concept with such a dreamlike world in mind usually unfolds an illustrative perspective. A misty world which begs to be put into firm ground via prolonged formal and concrete afterthought.

He embraces that teaching can be poetic and tantalizing, providing not a definition but a deep and creative hint that causes an exploratory shift in perspective, allowing you to walk down the path to the conclusion yourself. I wanted to try to exposit with this philosophy: confusion is expected and encouraged as impetus for reaching understanding. With that in mind, step into your flower garden.

Planted in a line of earth ($$\text{Spec }R$$)
there are flowers, $$C$$, whose heads are smooth projective genus 1 curves
with stems that can retract into the ground,
s.t. the flower meets the earth at one point (a marked point).

Cutting the flowers off at their stems $$C \xrightarrow{p} \text{Spec }R$$
$$\Rightarrow$$ you’re left with the line of earth ($$\text{Spec }R$$)

Cutting into the petals a small ring around their stems (formal disk at marked point)

$$\Rightarrow$$ you’re left with the remaining (infinitely-layered) base of the flower sitting on top of $$\text{Spec }R$$

Feeling frustrated that you can’t see clearly, you use your hands to move each disk lying flat on the ground to lay on its side, s.t. these bases are now stacked on top of each other like CDs and form a loose layered cylinder, centered around the line of earth.

1st layer = $$\text{Spec }R[x]/x^2$$, first infinitesimal neighborhood
1st&2nd layer = $$\text{Spec }R[x]/x^3$$, second infinitesimal neighborhood;…

You stare at this line of earth, adorned with (infinitely-layered) flower bases on top forming a layered tube around the line of earth. Looking closer, you see how the layers fit together, $$\text{Spf }R[[x]]$$.

(If you’d cut out the disk and forgotten how the layers fit together, you’d find yourself with $$\text{Spec }R[[x]]$$ — an awfully boring topology.)

Glance away, toward a different line of earth $$\text{Spec }E^*$$
with (infinitely-layered) flower bases on top, $$CP^\infty_E := \text{Spf }E^*(CP^\infty)$$.

Someone already cut the flowers down to their bases, before you had a chance to see them!

Flustered, you remember that $$CP^\infty$$ is the colimit of $$CP^n$$.

You reach into your pocket for your book-keeping device, and use it to look at the connectivity rings of each $$CP^n$$.

Content, you label the layers of the flower base:

1st layer is Spec $$E^*(CP^1)$$ = Spec $$E^*[x]/x^2$$
1st&2nd layer is Spec $$E^*(CP^2)$$ = Spec $$E^*[x]/x^3$$, …

Given the flower bases ($$CP^\infty_E$$), can you tell what flowers were over $$\text{Spec }E^*$$?

That is, is there a group object with a map to $$\text{Spec }E^*$$ whose formal completion along its 0-section is $$CP^\infty_E$$?

I thought the answer was no, but I think the answer is instead tautological. There is a H-space flower which we can trim to $$CP^\infty_E$$. What is it?

The notation for $$CP^\infty_E$$ itself is extremely suggestive.

The H-space in question is $$CP^\infty$$, and we’ve $$E$$-localized it!

You have enough to precisely decipher this story, each of your loose ends can be tied.

## 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.

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 $$\text{Top}_*$$):
MISSING IMAGE

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:
MISSING IMAGE

Apply $$\Omega^\infty$$ 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.

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 $$\text{Spec }R$$.
2. An elliptic curve is a group scheme (over $$\text{Spec }R$$) whose underlying object is 1-dimensional, proper, and smooth.
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 associated 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.

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:

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 subgroup schemes $$ker(p^n)$$, 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 hopeless 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).