# Lecture 2: ? Today's lecture: computational, and tomorrow we'll revisit some of the theory and geometric interpretations. Last time we talked about a formal group, its universal deformation, and how the automorphism group of the former acts on the latter. Let's start with the example where the formal group is the multiplicative group $\GG_m$ with formal sum $$ x+y-xy = 1-(1-x)(1-y).$$ We'll work over $\ZpZ$ -- what is the automorphism group of the multiplicative formal group? A reasonable guess is that since we're just multiplying things, we should just raise those things to a power. That is, $\Aut \GG_m = \Zp^\times$, where $\lambda \in \Zp^\times$ acts on this via $x \mapsto 1- (1-x)^\lambda$ where we interpret this using the binomial theorem and expand into a power series. What is the Lubin-Tate universal deformation here? It is the ring $\Zp$ since there are $n-1$ deformation parameters and this is height 1. So we have some ring action of $\Zp^\times \actson \Zp$, but that's generated by one element and thus $\Aut\GG_m$ has to act trivially. So we could calculate the cohomology of the group $\Zp^\times$ on something, but that wasn't the whole computation we wanted. Recall that we had the graded ring $$ E_* = W[[u_1, \cdots, u_{n-1} ]][u^{\pm 1}],\quad \abs u = 2 \\ E_0 = W[[u_1, \cdots, u_{n-1} ]] $$ where we think of $u$ as an invariant differential at the identity. In our case, we really want to look at the ring $\Zp[u, u\inv]$ and we want to figure out how $\lambda \in\aut\GG_m = \Zp\units$ acts on $u$. We know $u\inv$ is an invariant differential on $\GG_m$, so $u\inv = dx + \cdots$ for some higher terms which are rigged so that when you translate by an element in $\GG_m$, we get the same differential back. If you work it out, you get $u\inv=(1-x)dx$. We don't really have to work with this, we can just check at the identity and use the fact that it is translation invariant. We can compute $1-(1-x)^\lambda = \lambda x + \cdots$ and so $u\inv\mapsto \lambda u\inv$. If you're an algebraic topologist, you may notice that this doesn't look very good because $\lambda$ is instead supposed to act on $u$ by $u\mapsto \lambda u$. So we have to be careful and decide on conventions here, particularly whether $\Aut\GG_m$ acts on the left or the right -- in topology, the convention is that this is acting on the other side, which is why there's a difference here. Now we want to compute $H^*(\Zp\units; ~\Zp[u,u\inv]$), and we can use the fact that for $p>2$ it is the case that $\Zp\units$ is topologically generated by one element. This is because we can write it as the $p-1$st roots of unity cross the elements congruent to 1 mod $p$, and by choosing a generator for each we get a generator for the product. So we pick $\lambda\in\Zp\units$ such that $\generators{\lambda} = \Zp\units\mod p$, so $\lambda^{p-1} = 1\mod p$, and such that $\lambda^{p-1} \neq 1\mod p^2$. Because of the conventions chosen earlier, we'll actually write this as $\lambda\inv$. We then find that for $n\neq0$, \[ H^1(\Zp\units;~u^n\Zp) &= \Zp / \generators{\lambda^n-1}\\ H^0(\Zp\units;~u^n\Zp) &= 0 \] If you play with these groups, you find that if $\abs{n_1 - n_2}_{(p)}$ is small (i.e. these numbers are close $p\dash$adically) then the resulting cohomology groups are close, and there is a way of $p\dash$adically interpolating these groups. In other words, we can replace the map $\lambda \mapsto \lambda^n$ appearing in the denominator on the RHS with any element of $\hom(\Zp\units, \Zp\units)$. It is useful to do this, because it's useful to understand this family of groups as a $p\dash$adically continuous family of groups. This is familiar to those who have studied Iwasawa theory -- the question raised in lecture 1 about equivariant line bundles over the Lubin-Tate space has to do with having an Iwasawa-style picture for the cohomology of these groups. We now want to try the case $n=2$. We'll let $\Gamma$ be a formal group of height 2 over $k$ an algebraically closed field, $W$ will be the Witt vectors of $k$, and we'll have a formal power series ring $W[[u_1]]$. We want to compute $$ H^*(\Aut\Gamma;~W[[u_1]]). $$ There's a problem right away -- we don't know what $\Aut\Gamma$ is, or really even what $\Gamma$ is, so we don't know what its action is. So we need to figure these out. --- **An Important Question in Algebraic Topology** It turns out that $$ H^*(\Aut\Gamma;~W) \mapsvia{\cong} H^*(\Aut\Gamma;~W[[u_1]]). $$ where $\Aut\Gamma\actson W$ trivially and the LHS is $\Lambda[x_1, x_3]$, an exterior algebra on classes of degree 1 and 3. This looks like $H^*(U(2))$, and in fact if you take the quotient of $W[[u_1]]$ by the constant functions, you obtain a module with no invariants and no cohomology. It would be interesting to have a conceptual explanation for why this is -- the explicit, complicated computation is carried out in the notes. An important question is why this is true, if it's true in higher heights, and how one can conceptually approach or explain this. --- Continuing, we need to describe what $\Gamma$ is and find out how to compute its automorphisms. We approach this via the Dieudonne module. # Dieudonne Modules Suppose $k$ is a perfect field. The motivation here is that the category of formal groups over $k$ of finite dimension is equivalent to another category that is easier to describe and work with. To introduce some terminology, let $W$ be the Witt vectors of $k$, and note that $k$ has an automorphism $$ \varphi:k\selfmap\\ x \mapsto x^p $$ called the Frobenius. This has a unique lift to $W$ which sends roots of unity to their $p$th power. So for us, a Dieudonne module will be a free $W\dash$module of finite rank equipped with two operators: $$ F: \varphi^*M \to M\\ F(am) = a^\varphi F(m) $$ for any $a\in M$, which is an additive map $M\selfmap$ that is not $W\dash$linear but is instead semi-linear. The second operator is given by $$ V \to \varphi^* M \\ V(a^\varphi m) = aV(m) $$ where these two operators satisfy $FV=VF=P$. We'll also require that $M$ is $V\dash$adically complete, so $V$ is pro-nilpotent. This is detailed in the full notes, but we won't worry about that for the moment. Now formal groups have a dimension -- we've only considered dimension 1, but we could have considered a formal group structure on the Lie variety of dimension $n$ -- and they also have a height. We have the following equivalence: $$ \pmatrix{\text{Formal groups over }k} \iff \pmatrix{\text{Suitable Dieudonne modules}} \\ \text{height} \iff \dim_W(M)\\ \text{dimension} \iff \dim_k(M/VM) $$ So we make a guess. We want a formal group of height 2, so we want our module $M$ to be free of rank 2. The dimension was 1, so $M/VM$ needs to be one-dimensional over the ground field. There is one obvious thing to write here, we take $M$ with basis $\gamma, V\gamma$, and since we have to say what $F$ is, we'll take $F\gamma = V\gamma$. This determines the rest of the structure. We now have an $M$ which corresponds to a height 2, dimension 1 formal group, and this will be our Lubin-Tate group. More generally, for height $n$ we could take the basis $\gamma, V\gamma, \cdots V^{n-1}\gamma$ with $F\gamma = V^{n-1}\gamma$. Somehow, this is all there is to this story: over a algebraically closed perfect field like $\overline F_q$, any Dieudonne module is equivalent to one of these. Constructing these equivalences becomes a problem in semi-linear algebra. We now have a description of our formal group $\Gamma$, so what about $\Aut\Gamma$? (Back in height 2.) The most general automorphism is something of the form $$ \gamma \mapsto a\gamma + bv\gamma, $$ which has to commute with the operators $F$ and $V$, so $$ V\gamma \mapsto a^{\varphi\inv}V\gamma + b^{\varphi\inv}V^2\gamma = a^{\varphi\inv}V\gamma +b^{\varphi\inv}P\gamma $$ where we use the fact that $F\gamma=V\gamma$ to deduce that $P\gamma = FV\gamma = V^2\gamma$. We actually need to check that this relation holds, so we see $$ F\gamma = a^\varphi V\gamma + Pb^\varphi\gamma $$ and equating coefficients yields $a^\varphi = a^{\varphi\inv}$, or in other words, $a,b \in W(\FF_{p^2})$. We thus finally deduce that $$ \Aut\Gamma = \left\{ \pmatrix{a & pb^{\varphi\inv} \\ b & a^{\varphi\inv} } \suchthat ~a,b \in W(\FF_{p^2}) \right\} $$ There are many way to recognize this matrix group -- it is the maximal order in a certain division algebra -- but for our purposes, the matrix representation will suffice. We now want to understand how $\Aut\Gamma$ acts on the Lubin-Tate ring. To do that, we need to introduce a new concept. # The Tapis de Cartier Tapis is the French word for "carpet", so you're supposed to think of this as a magic carpet of some sort which elevates you from a formal group in characteristic $p$ to one in characteristic zero. Here we have a formal group in characteristic $p$, and we want to lift it to the Witt vectors $W \covers k$. The Tapis will provide a way of doing that entirely in terms of the Dieudonne module $M$. We can look at $M/VM$, which has dimension 1 over $k$. We'll need to make a lot of choices for the purposes of calculation, which we'll need to undo later. We get a correspondence: $$\left\{ \array{ M &\overset{\exists~T}\dashrightarrow & W \\ \downarrow & ~ & \downarrow \\ W/VM & \underset{g}{\xrightarrow{\cong}} &k } \right\} \iff \pmatrix{\text{Lifts of }\Gamma\text{ to } W} $$ where $T$ is $W\dash$linear. So all we have to do is parameterize the space of lifts on the left, give it some coordinates, and we can write down how $\Aut \Gamma$ acts on the Lubin-Tate ring. For a quick dimension count/reality check, recall that the Lubin-Tate space had dimension 1, i.e. 1 parameter. How many parameters do we have here? $M$ had dimension 2, where if we push it through $f$ and down to $k$ it's equal to 1 mod $p$, so it still has an entire copy of $W$ to move through. The other element can go to something arbitrary, so we have 2 dimension. Thus we've already encountered something artificial about the choice we've made, so we'll want to normalize these isomorphisms classes on the LHS in a certain way. We can make more choices here; suppose $g: W/VM \to k$ maps $\gamma$ to 1. We'll get rid of the choice later, but by the Tapis, the lifts of $\Gamma$ will exactly correspond to the homomorphisms $g$. We now want to get the Lubin-Tate deformation parameter, $U_i$, which is supposed to be a function on the space of lifts. What's the obvious function? Well, given a lift, we know where $\gamma$ goes, we now just need to say where $\V\gamma$ goes. We'll proceed by making a choice here, which *a priori* might work but *a posteriori* is known to be erroneous. So let us guess a definition of $u_1$ (which will later turn out to be $w_1$), where $u_1(T) = T(V\gamma)$, which we might hope is our Lubin-Tate ring. The Tapis tells us how our matrix element acts on the Lubin-Tate ring $W[[u_1]]$, so let's work that out. We have an element in the Lubin-Tate ring which we'll write as $$ g: \gamma \mapsto a\gamma + bV\gamma $$ which is an automorphism. Then what is $g(u_1)$? This is an issue, because acting by $g$ yields a new map $M\to W$, which doesn't map $\gamma$ to 1 (it now goes to $a\gamma + bV\gamma$). We also have $$ V\gamma \mapsto a^{\varphi\inv}V\gamma + Pb^{\varphi\inv}\gamma. $$ You might think you just need to evaluate $T(V\gamma)$, but this doesn't work because $g(\gamma)\neq 1$. If you work it out, you get an action by fractional linear transformations, $$ g(u_1) = \frac{a^{\varphi\inv} u_1 + pb^{\varphi\inv}}{bu_1 + a} $$ So this looks good, now we just need to calculate the cohomology. The problem is that this formula is not correct, although it serves as a good model. A fractional linear formula is the ratio of two linear formulas, where each linear formula here is actually happening in the ring $E_{-2}$. # Crystalline "Approximation" The idea is that the Crystalline "approximation" is supposed to be a map \[ M \to &E_{-2} = uW[[u_1, \cdots, u_{n-1}]]\\ \gamma \mapsto &u\\ V\gamma \mapsto &uu_1 \\ \vdots &\\ V^{n-1}\gamma \mapsto &uu_{n-1} \] which is $\Aut\Gamma\dash$equivariant. The Tapis almost works, but requires a modification to make it work which leads to a beautiful picture of this group action. The Crystalline approximation is not correct -- we will soon rename the parameters to make it correct -- but just notice that this would give us (in closed form) a formula for the action of $\Aut\Gamma$ on $E_{-2}$. The former is always expressible as a matrix algebra, and the images appearing on the RHS of the formula are (up to completion) algebra generators for $E_{-2}$. This would mean that this Lubin-Tate ring is a suitable completion of the localization of a symmetric algebra on the Dieudonne module. A theorem of Hopkins-Hill-Ravenel for any finite subgroup of $\Aut\Gamma$ whose $p\dash$Sylow subgroup is cyclic, you can always find parameters for the RHS such that the crystalline approximation holds. The proof of this ends up being nasty, so it'd be nice to have a conceptual understanding of why this is. # Some Justifications The prior steps asked you to believe many different things, such as the Tapis, so we want to provide some justification for them. First we have to say how a formal group is actually given to you. The most convenient way to give a formal group over $k$ is to instead give a formal group over the Witt vectors $W \covers k$, but instead of giving it there, we give it in $W \tensor \QQ$. In reality, $k$ is any ring of characteristic $p$, $W$ is any torsion-free ring, which embeds into its rationals $W\tensor \QQ$: $$ \array{ W & \rightarrow & W \tensor \QQ \\ \downarrow & & \\ k } $$ Over a $\QQ\dash$algebra, any formal group is isomorphic to the additive formal group. There is a simple way to do this which is explained in the notes, which essentially amounts to integrating the invariant differential. In terms of elements, we'll do the following: $$ \array{ G & \rightarrow & G \cong_f G_a \\ \downarrow & & \\ \Gamma } $$ where $$ f(x) = \log_a(x) \definedas x + \cdots \in W[[x]] \tensor \QQ, $$ which yields the identity $f\inv(f(x) + f(y)) = x \underset{G}+ y$, the formal sum. **Example:** You can work out $\log_{\GG_m}(x) = \sum \frac {x^n} n$, which explains the weird negative sign conventions for $\GG_m$ earlier, since it's $-\log (1-x)$. How does the Tapis work? In particular, how do we go from something like the diagram for the Tapis to the log of a formal group? If we do this, we actually know how to go back from a Dieudonne module to a formal group, because given any Dieudonne module you can just choose a $T$, use the above formulas to get the log of a formal group, and this yields a functor back to Dieudonne from formal groups. Here's how you do it it. Given $T: M \to W$, show that $$ f(x) = \sum T(F^n\gamma)\frac{x^{p^n}}{p^n} $$ is the log of a formal group. **Example:** Let $G=\GG_m, M = W\theset{\gamma}$ so the Dieudonne module is free of rank 1. Then $V$ has to be topologically nilpotent and we must have $FV=P$, so this forces $F\gamma = \gamma$, and thus $$ \log = \sum \frac{x^{p^n}}{p^n}. $$ This looks bad, because this has only the $p$ powers and our prior sum had terms for all $n$. But it turns out that if you have a formal group and throw out everything but the $p$ powers, you still get a formal group. In practice, you want formal groups to have as few terms as possible, so you only look at ones like this. > In this case, this is the theory of the Artin-Hasse exponential. **Example (of height 2)**: We need to pick a $T$, so let $T(\gamma) =1$ and $T(V\gamma) = 0$; then $\log = \sum {x^{p^{2n}}\over p^{2n}}$. This is because applying $f$ yields $T\gamma$, which is annihilated by $T$, and applying $f$ again multiplies by $p$ -- so we'll only see even powers of $p$, and $2n$th power will be multiplied by $p^n$. This is a common formula, so we'll define $$ \ell(x) \definedas \sum {x^{p^{2n}}\over p^{2n}}. $$ > This is the log of the Lubin-Tate group, which you can actually get from Lubin-Tate's original construction. What is the universal one? We'll work over $W[[w_1]]$ and define $$ T(\gamma) = 1 \\ T(V\gamma) = w_1 $$ What's the log of the new formal group? What you find is that $$ f(x) = \ell(x) + {w_1 \over p} \ell(x^p), $$ and it turns out that the coefficients of $f\inv(f(x) + f(y))$ are not elements of $W[[w_1]]$, which is why the Tapis doesn't quite work. It does however have coefficients in the divided power completion $W\langle\langle w_1 \rangle\rangle$, which consists of formal sums of the form $\sum {w_1^n\over n!}$ Next time: Hasse-Winkel's Functional Equation. It's famously hard to explain.