Month: January 2015

We have a functor Spec from Ring to Schemes:

$\text{Ring} \xrightarrow{\text{Spec}} \text{Schemes}$

Schemes look like Fun(Ring, Set). So Spec sends a ring to a functor from Ring to Set.

$\text{Ring} \xrightarrow{\text{Spec}} (\text{Ring} \to \text{Set})$

Honestly, whenever I see “scheme” I replace it mentally with “algebraic curve,” but this is just because I don’t really get what a scheme is yet.

$R \xrightarrow{\text{Spec} (-)} \text{Spec(R)}$

$R \xrightarrow{\text{Spec} (-)} (S \mapsto \text{hom}(R,S))$

$\text{Spec}$ $R$ $S$ = $\text{Hom}_{\text{Ring}}(R, S)$

Let say we have:

1. a ring $\mathbb{Z}[t]$ (this is just the ring of polynomials with $t$ as a variable and coefficients in $\mathbb{Z}$)
2. an arbitrary ring $S$.

What is $\text{hom}(\mathbb{Z}[t], S)$?

Notice that we’re in Ring, so the members of $\text{hom}(\mathbb{Z}[t], S)$ must be Ring homomorphisms. Let’s say we’re looking at a ring homomorphism $\phi: \mathbb{Z}[t] \to S$.

I learned from Cris Moore that demanding examples is a useful practice, so: What is $\phi(7t^2-4t+3)$ in simpler terms?

$\phi$ is a ring homomorphism, so we know that $\phi(1) = 1_s$. This implies that $\phi(n) = n_s$.

$\phi(7t^2-4t+3) = \phi(7t^2) – \phi(4t) + \phi(3)$
$= \phi(t)(\phi(7t) – \phi(4)) + \phi(3)$
$= \phi(t) (\phi(t) 7_s – 4_s) + 3_s$

$\phi(t)$ is not determined! We can pick it to be any element of S!

$\text{hom}(\mathbb{Z}[t], S) \simeq \{\phi(t) \in S\} = S$

In other words, $\text{Spec}(\mathbb{Z}[t])$ is a forgetful functor from $S$ as an object in $\text{Ring}$ to its underlying set.

Exercise: show that $\mathbb{Z}[t]$ is an initial object in an appropriate category, and that the trivial ring is the terminal object in the appropriate category.

But what is Spec?

It’s the spectrum of a ring! Intuitively, $\text{Spec} R$ is the geometric object canonically associated to $R$.

For example:

• $\text{Spec}(\mathbb{Z})$ is “The Point”
• $\text{Spec}(\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))$ is “The Circle”
• $\text{Spec}(\mathbb{Z}[x,x^{-1}]$ is “Pairs of Invertible Elements” (e.g. a field with the origin removed)

Let’s look at $\text{Spec}(\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))(S)$ when $S$ is $\mathbb{R}$, it’s obviously the circle BUT IT WORKS FOR (almost) ANY S. The concept of there being a canonical geometry associated to every ring is very exciting!

Sidenote:

When people say “elliptic curve” they might mean “a family of elliptic curves”, this is commonly written $C \to \text{Spec} R$ where $R$ is the underlying coefficient ring.

For example, $y^2=4x^3+ax+b \mapsto a, b \in R$ is the family of elliptic curves of the form $y^2=4x^3+ax+b$ over the coefficient ring $R$.

I’m trying to figure out what happens when $S$ is not something nice like $\mathbb{R}$ or $\mathbb{C}$. What is a circle in the coefficient ring of an elliptic curve?

Thanks to Aaron Mazel-Gee for walking me through the concept of $\text{Spec}$ (all errors in this post are mine and not his).

I noticed an informal “recipe” for taking a type of object and constructing invariants (of the object). It’s been useful for removing the feeling of “what, why? where did that come from?” when learning new constructions that fit this recipe. Hopefully it will help you!

1. Take in an object
2. Look at a collection of structures defined over the object (up to isomorphism)
3. Define a binary operation closed over this collection
4. Optional: Formally append inverses
5. Output: a useful algebraic invariant (used to study the object)

Some examples of this, and comments you are welcome to skip over (you can get a good sense of what I’m saying by just looking at the pictures).

This Burnside ring is the analogue of the representation ring in the category of finite sets, as opposed to the category of finite-dimensional vector spaces (over a field $F$). The Segal theorem (proved by Gunnar Carlsson, who is a wonderful human being and patient teacher) is something that I wish to understand, which relates the Burnside ring of a finite group $G$ to the stable cohomotopy of the classifying space $BG$.Let’s fill in the recipe for K-theory:

1. Take in a space $M$ (compact Hausdorff to avoid pathologies)
2. Look at (complex/real) vector bundles over $M$ (up to iso)
3. Equip the collection with the fiberwise direct sum, and the fiberwise tensor product of bundles.
4. Formally append rank $-n$ vector bundles (a formal entity, defined as an object which, when directly summed with a rank $n$ vector bundle, reduces to a point)
5. Output: Topological (complex/real) K-theory

Let’s do an example: What’s the K-theory of a point? Well $\text{Vect}(pt) \simeq \mathbb{N}$, formally appending inverses gives us $K^0(X) \simeq \mathbb{Z}$.

It’s worth stating explicitly the relationship between algebraic and topological K-theory. The algebraic K-theory of (a ring of complex valued $C^\infty$-maps on a space) = the topological $K$-theory of (the space).

Josh Grochow kindly pointed out that the representation ring and topological K-theory are similar for a good reason! Intuitively, one can think of both as “bundles of representations.”

More specifically, assuming $G$ is finite, one can define the representation ring to be the ($G$-equivariant) K-theory of a point.

$K_G(pt) \simeq R_F(G)$

In other words, finite dimensional linear virtual $F$-representations of $G$ in $R_F(G)$ correspond to virtual equivariant bundles over a point. Note that the term “virtual” is short for “isomorphism classes of formal differences of” and that $R_F(G)$ is sometimes written as $\text{Rep}(G)$.

Along similar lines, Jacob Lurie mentions in Loop Spaces, p-Divisible Groups, and Character Theory that

$R_F(G) \to K(BG)$

(where $K(BG)$ is the topological $K$-theory of the classifying space $BG$ of $G$-principal bundles) is almost an isomorphism, and becomes an isomorphism under p-adic completion.

This fits into the picture of comparing the following two processes:

• completing a $G$-space by making the action free (a geometrical process)
• completing with respect to an ideal (an algebraic process)

Edit: I just began reading Lurie’s Higher Algebra (excerpt below), and it seems that the derived category $\mathcal{D}(R)$ of a ring $R$ fits into ‘the recipe’.

1. Take in the ring $R$
2. Look at the collection of chain complexes of modules over $R$
3. Equip the collection with chain complex homomorphisms (a.k.a chain maps)
4. Formally make quasi-isomorphisms into isomorphisms
5. Output: The derived category $\mathcal{D}(R)$

Thank you to Semon Rezchikov (for explaining to me what a virtual vector bundle was a few weeks ago, it was a lovely and intuitive description that led me to draw the $K$-theory picture above), and to Qiaochu Yuan (for pointing out and correcting the errors in the recipe).

When I define a polynomial, I am simply handing you an indexed collection of coefficients.

A polynomial with two variables, $x, y$ and coefficients $c$, is of the form:

$F(x, y) = \sum\limits_{ij} c_{ij} x^i y^j$

The coefficients of a polynomial form a ring. In other words, the coefficients $c_{ij}$ are members of a coefficient ring $R$. When we say $F$ is over $R$, we mean that $F$ is a formal power series with coefficients in $R$.

Example: The polynomial
$F(x,y) = 7 + 5xy^2 + 2x^3$ can be written out as
$F(x,y) =7x^0y^0 + 5x^1y^2 + 2x^3y^0$ such that
$c_{00} = 7$, $c_{12} = 5$, $c_{30} = 2$, and the rest of $c_{ij} = 0$.

Alright, now let’s change the coefficients; reassign $c_{00} = 4$, $c_{78} = 3$, and all other $c_{ij} = 0$.

Out pops a very different polynomial $P(x,y) = 4 + 3x^7y^8$.

In other words, By altering the coefficients $c_{ij}$ of $F(x,y)$ via a ring homomorphism $u: R \to R’$ (from the coefficent ring $c_{ij} \in R$ to a coefficient ring $u(c_{ij}) \in R’$)…

…we can get from $F(x,y)$ to any other polynomial $F’(x,y)$.

What’s a Group-y Tuple?

Intuitively, a group-y tuple is a polynomial + a constraint on our coefficients that forces the polynomial to satisfy group laws.

Concretely, a group-y tuple is an operation of the form $F(x,y) = \sum\limits_{ij}c_{ij}x^iy^j$ such that

1. $F(x, 0) = F(0, x) = x$
2. $F(F(x,y), z) – F(x, F(y,z)) = 0$

We can make sure that our polynomial satisfies the constraints of a group-y tuple by modding out our coefficient ring $c_{ij}$ by the ideal $I$ generated by the relations amoung $c_{ij}$ defined above. The ring of coefficients that results is called the Lazard ring $L = \mathbb{Z}[c_{ij}]/I$.

What does this have to do with cobordism?

Let $F^L(x,y)$ be a group-y tuple over $L$.

The premise of $F^L(x,y)$ is that any group-y tuple $F(x, y)$ can be generated from a ring homomorphism $u: L \to R$ such that $F(x,y) = u \circ F^L(x,y)$.

Quillen’s theorem: For any group-y tuple $F(x,y)$ over any commutative ring $R$ there is a unique ring homomorphism $u: MU^*(pt) \to R$ such that $F(x, y) = u \circ F^Q(x, y)$.

Quillen’s proof technique was to show that: $F^L \simeq F^Q$.

For your ventures ahead…

In this post, I have committed two semantic sins in the name of pedagogy.

1. group-y tuple = “1-dimensional abelian formal group law”
2. polynomial = “formal power series”

Conventionally, a “polynomial” is a special case of a formal power series (in which we expect that our variables evaluate to a number – useful if we care about convergence).

polynomials $\subset$ formal power series

The polynomial ring $R[x]$ is the ring of all polynomials (in two variables) over a given coefficient ring $R$.

The ring of formal power series $R[[x]]$ is the ring of all formal power series (in two variables) over a given coefficient ring $R$.

polynomial ring $\subset$ ring of formal power series
R[x] $\subset$ R[[x]]

Sources:

Groupes de Lie formels à un paramètre
Groupes analytiques en caractéristique 0
Groupes de Lie algébriques (travaux de Chavelley)
Formal Group Laws – Ravenel