Sunday, December 30, 2012

The Quantum Group $SL_q(2)$ and its coaction on the Quantum Plane

The algebra $SL(2)$ and its coproduct.

Recall our earlier discussion about a universal group structure on algebras. In particular, consider

$\displaystyle \text{Hom}_\text{Alg}(k[a,b,c,d], A) \cong A^4 \cong M_2(A)$

as a vector space. Let $M(2)$ denote the polynomial algebra $k[a,b,c,d]$. Last time we pulled back addition on $A$ to a map $\Delta k[x] \rightarrow k[x] \otimes k[x]$. This time, we're going to follow the same pattern to take matrix multiplication (a map $ m : M_4(A) \otimes M_2(A) \rightarrow M_2(A)$) back to $\Delta M(2) \rightarrow M(2) \otimes M(2)$.

In particular, we're looking for a $\Delta$ so that when we have $\alpha \in \text{Hom}_\text{Alg}(M_2(A) \otimes M_2(A), A)$, $\alpha \circ \Delta = m (\alpha)$

The above notation is slightly confusing, let me try to explain more clearly: We have that the matrix algebra $M_2(A)$ is isomorphic as a vector space to $\text{Hom}_\text{Alg}(M(2),A)$ by the map taking

$\displaystyle f \mapsto \begin{pmatrix} f(a) & f(b) \\ f(c) & f(d) \end{pmatrix} = \hat{f}$

We also have that matrix multiplication is a map $m: M_2(A) \otimes M_2(A) \rightarrow M_2(A)$, and that $M_2(A) \otimes M_2(A)$ is isomorphic as a vector space to $\text{Hom}_\text{Alg}(M(2) \otimes M(2),k)$ by the map

$\displaystyle \alpha \mapsto \begin{pmatrix} \alpha(a) & \alpha(b) \\ \alpha(c) & \alpha(d) \end{pmatrix} \otimes \begin{pmatrix} \alpha(a') & \alpha(b') \\ \alpha(c') & \alpha(d') \end{pmatrix} = \tilde{\alpha}$

I write $a'$, et al, just so it's clear that $\alpha$ varies on different elements of the tensor product basis $a \otimes a$, et al. So for a map to implement multiplication on the $M(2)$ side, it must be $\Delta: M(2) \rightarrow M(2) \otimes M(2)$ and we want $m(\tilde{\alpha}) = \alpha \circ \Delta$.

From this requirement it's obvious what $\Delta$ needs to be; $\Delta(M)$ for $M\in M(2)$ needs to ensure that the generators $a,b,c,d$, et cetera, get mapped to the items that will correspond to matrix multiplication after they are acted upon by $\alpha$. So $\Delta(a) = a \otimes a + b \otimes c$, et cetera. In matrix notation, we can write this

$\displaystyle \Delta \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \otimes \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

This looks group-like! But it's not! The matrix isn't actually an element of $M(2)$, it's just a convenient way for us to write the action of $\Delta$. The action elements are $k$-linear combinations of $a,b,c$, and $d$.

Quantizing things

I bet the reader is guessing that $\Delta$ is a coproduct, making $M(2)$ into a bialgebra. Such a reader would be correct. But before we discuss this further, let's take a quotient of $M(2)$ by the relation $ad -bc =1$. Call this new bialgebra $SL(2)$. We can make it into a Hopf algebra by introducing the antipode:

$\displaystyle S\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix}$

We're going to take a ``quantum deformation'' of this new Hopf algebra. It's actually simple to do, we're going to modify the commutativity of $a$, et al, by an element $q \in k^\ast$. In particular, let $ca = qac$, $ba =qab$, $db = qbd$, $dc = qcd$, $bc = cb$, $da -ad = (q-q^{-1})bc$, and the ``q-determinant'' relation $ad -q^{-1}bc = 1$. The coproduct remains the same.

The Quantum Plane and the coaction

The classical group we're mimicking, $SL_2(\mathbb{R})$, acts on the affine plane $\mathbb{R}^2$ by transforming it in a way that preserves orientation and area of all geometric shapes on the plane. Since $SL(2)$'s role is as the base of a set of homomorphisms to the algebra $A$, we expect any equivelant ``action'' to have the arrows reversed, we'll discuss this later. But first, we're going to need a notion of an affine plane in our polynomial algebra-geometry language:

This isn't so bad, as $\text{Hom}_\text{Alg}(k[x,y],A) \cong A^2$, hence we call $k[x,y]$ the affine plane. Quantizing it is easy, too: we define the quantum plane $\mathbb{A}^2_q$ to be the free algebra $k\langle x,y\rangle$ quotiented by the relation $yx = qxy$, id est, it's $k[x,y]$ but with a multiplication deformed by the element $q \in k$.

Now back to are ``arrow reversed'' version of an action, or a coaction. One can arrive at this definition by reversing the arrows in the commutative diagram that captures the axioms of an algebra acting on a vector space. In particular, we say that a Hopf algebra $H$ coacts on an algebra $A$ by an algebra morphism $\beta: A \rightarrow H \otimes A$ such that $(I \otimes \beta) \circ \beta = (\Delta \otimes I)\circ \beta$, and $I = (\epsilon \otimes I) \circ \beta)$, where $I$ is the identity map and $\Delta$ and $\epsilon$ are the coproduct structure maps for $H$.

We can define the coaction $\beta$ on the generators $x$ and $y$ and extend it as an algebra morphism should, so the reader can check that $\beta(x) = x\otimes a + y\otimes c$ and $\beta(y) = x\otimes b + y\otimes d$ defines a coaction. In matrix notation, we have:

$\displaystyle \beta(x,y) = (x,y) \otimes \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

I do not have a geometric interpretation for this.

Tuesday, December 18, 2012

What I learned today: Coadjoint action of group Hopf algebras

This is another post in my ``psuedo-daily'' series ``What I learned Today''

Yesterday we talked about actions of a Hopf algebra $H$ on an algebra $A$. Today, let's talk about some examples of this (I didn't cover as much ground as I hoped to today, but we're trying to make these blogs daily!). Every Hopf algebra acts on itself by

$\displaystyle h \triangleright g = \sum_h h_{(1)} g Sh_{(2)} $

This is the adjoint action. It's not hard to prove that it's 1) an action and 2) a Hopf action (that it plays well with the Hopf algebra structure of $H$ and the algebra structure of $A$). As a more concrete example, let's look at the group Hopf algebra $H=kG$ for a finite group $G$. The coproduct is given by $\Delta g = g \otimes g$ and the antipode is group inversion: $Sg = g^-1$. In this case, the adjoint action becomes:

$\displaystyle h \triangleright g = hgh^{-1} $

Which is just group conjugation.

Hopf algebras can act just as well on coalgebras - in this case we require that the action commutes with the coproduct of $A$. Now if $H'$ and $H$ are dually paired, then we also have adjoint action, or rather, a coadjoint action of $H'$ on the coalgebra $H$, given by:

$\displaystyle \phi \triangleright h = \sum_h h_{(2)} \langle \phi, (Sh_{(1)})h_{(3)} \rangle $

We know that the algebra of functions on $G$, $k(G)$ is the Hopf algebra dual of $kG$, where $\langle f, g \rangle = f(g)$, so let's see what the this action becomes for $H' = k(G)$ and $H = kG$:

$\displaystyle f \triangleright g = g \langle f, (Sg)g \rangle = f(1) \cdot g$

Additionally, we also have the notion of a coaction. Generally, if $H$ is a Hopf algebra and $A$ an algebra, we say $\beta : A \rightarrow H \otimes A$ is a coaction of $H$ on $A$ (or that $A$ is a $H$-comodule algebra if $\beta$ is a algebra morphism, $(I \otimes \beta) \circ \beta = (\Delta \otimes I) \circ \beta$ and $I = (\epsilon \otimes I) \circ \beta$, where $I$ is the identity and $\Delta$ and $\epsilon$ are the coalgebra structure maps.

The most interesting case of this is the quantum group $SL_q(2)$ coacting on the quantum plane $\mathbb{A}^2_q$. This is a topic for another post, though.

Monday, December 17, 2012

What I learned today: Weird differences between the action of a Hopf algebra and its dual

This is another post in my ``psuedo-daily'' series ``What I learned Today''

Let $H$ be a Hopf algebra. Today we're going to talk about its actions (and its coactions, time permitting). A Hopf algebra is just an algebra, which is just a ring, and ring's have actions (equiv. modules), and we're comfortable with that already. For this to make sense on a Hopf algebra, we merely require that the action agrees with all the necessary units, products, coproducts, et cetera.

More precisely, we say that $H$ acts on an algebra $A$ (or that $A$ is an $H$-module algebra) if we have a linear map $\triangleright : H \otimes A \rightarrow A$, written $h \triangleright a$ such that

$\displaystyle h \triangleright (ab) = m(\Delta(h) \triangleright (a \otimes b))$

with $\triangleright$ extended to tensor products and $m$ the product of $A$, and

$\displaystyle h\triangleright 1_A = \epsilon(h) 1_A$

Some weirdness emerges here, for instance, if $H=kG$, the group hopf algebra of a finite group $G$ (the vector space having elements of $G$ as a basis, the product being the group operation, and the coproduct defined by $\Delta(g) = g \otimes g$, antipode, units, et cetera are the obvious choice), then a hopf action collapses to a typical group action:

$\displaystyle g\triangleright(ab) = (g\triangleright a) (g\triangleright b)$

But if we forget the Hopf algebra structure for a second, we know that $kG$ is exactly the same as $k(G)$, that is, as functions defined on $G$ with values in $k$, as any ``vector'' in $kG$ defines a function and vice versa. Hence, as vector spaces $kG$ and $k(G)$ are naturally isomorphic. This is not the case with the Hopf algebra structure, however (though we have seen how the two are dually paired already), as the coproduct on $k(G)$ is $\Delta(f)(x,y) = f(xy)$, where we identify $k(G \times G)$ with $k(G) \otimes k(G)$ (this is exactly what a tensor product is for, actually.) In this case, the hopf algebra action on an algebra $A$ is the same as a $G$-grading on $A$. Let me explain this (with help from a kind person on math.stackexchange):

Say $A$ is $G$-graded and for each $a \in A$ let $|a|$ denote the element in $G$ such that $a \in A_{|a|}$. We can get a $k(G)$ action from this by $f \triangleright a = f(|a|) a$. To go the other way, start from a Hopf algebra action $\triangleright: k(G) \otimes A \rightarrow A$. Then for each $g \in G$, let $A_g$ be the set of all elements of $A$ on which every $f\in k(G)$ acts via scalar multiplication by $f(g)$. One then has to prove that these sets are nonempty, that they are subspaces, that $A$ decomposes into direct sums of them, and that they obey the axioms of a grading. I've been assured by various sources that it can be done!

Thursday, December 6, 2012

What I learned today: Dually Paired Hopf Algebras

To help keep me motivated and mathematically active, I will be blogging about "what I learned today" in my various projects. This is the second post in this "psuedo-daily" series.

I'm still trying to catch up on Quantum Groups. Any reader who is familiar with the field will be able to tell from these posts that I'm a long way away from doing research. But we're slowly getting somewhere!
Let $H = kG$ be the group Hopf algebra for a finite group $G$. In particular, the product map $m: H \otimes H \rightarrow H$ is just the group operation $g \otimes h \rightarrow gh$ for $g, h\in G$, the coproduct is $g \mapsto g\otimes g$, the antipode is group inversion $x \mapsto x^{-1}$, the group idenity is the unit and $g \mapsto 1$ is the counit. We want to talk about its dual $H^\ast = \text{Hom}(H,k)$.

Kassel's Quantum Groups text gives us several propositions to prove that the dual of a finite dimensional Hopf algebra is also a Hopf Algebra. Since $G$ is a finite group, $H$ is finite dimensional, and we'll follow Kassel in proving that $H^\ast$ in particular is a Hopf Algebra.

Now, $H^\ast$ is the space of linear functions from $kG$ to $k$. Any such linear functional will be determined by basis elements that themselves are determined by their action on $G$, hence $H^\ast$ is the space $k(G)$ of functions on $G$ with values in $k$. Clearly this space has pointwise multiplication to make it into an algebra - but we want to derive this from the dual of the coproduct map $\Delta: H \rightarrow H \otimes H$.

So let's talk about $\Delta^\ast: (H\otimes H)^\ast \rightarrow H^\ast$. This is, by definition, the map that takes each $\alpha\otimes\beta \in (H\otimes H)^\ast$ to another linear functional $\gamma = \Delta^\ast(\alpha\otimes\beta)$ on $H$ such that $\gamma(h) = \alpha\otimes\beta(\Delta(h))$. But before we can work out how this becomes a product map, we need to work out some technicalities:

Can we say that $(H\otimes H)^\ast \cong H^\ast \otimes H^\ast$ so that $\Delta^\ast$ is actually a product? Yes! this is provided by a theorem in Kassel's book which says that the map $\lambda: \text{Hom}(U, U') \otimes \text{Hom}(V,V') \rightarrow \text{Hom}(V \otimes U, U' \otimes V')$ by $(f\otimes g)(v \otimes u) = f(u)\otimes g(v)$ is an isomorphism when one of the pairs $(U, U')$, $(V, V')$, or $(U,V)$ consists of only finite dimensional spaces. We won't discuss the proof here, but I do want to point that the theorem does require finite dimensionality .

Back to the dual of comultiplication on $H$: We have that $\Delta(h) = h \otimes h$. So $\alpha \otimes \beta (h\otimes h) = \alpha(h) \beta(h) = \gamma(h)$. Hence $\Delta^\ast(\alpha \otimes \beta)(h) = \alpha(h)\beta(h)$ for $\alpha, \beta \in H^\ast$ and $h\in H$. Hence the dual of comultiplication of a group Hopf algebra is just pointwise multiplication, as we expected.

Now what about the dual of multiplication? Same definition as above, $m^\ast: H^\ast \rightarrow H^\ast \otimes H^\ast$ must be the map taking $\alpha \in H^\ast$ to an $m^\ast(\alpha)=\beta \otimes \gamma$ such that $\beta\otimes\gamma(h_1\otimes h_2) = \alpha(m(h_1,h_2)) = \alpha(h_1 h_2)$. Hence we let $m^\ast$ be the function $m^\ast(\alpha)(x\otimes y) = \alpha(xy)$

Continuing in this fashion, you'll see that the antipode is the map $S(\alpha(x)) = \alpha(x^{-1})$, the unit is the identity function, and the counit the map $\alpha \mapsto \alpha(1)$.

Hence we have a Hopf algebra $H^\ast = k(G)$ that is dual to $H = kG$. If we let $\langle, \rangle: H^\ast \otimes H$ be the evaluation map $\langle \alpha, x \rangle = \alpha(x)$, we can see write down the behavior of this map and see if we can come up with a more general situation to more varied sets of Hopf algebras. For instance, extending the map to tensor products pairwise, note that

$\displaystyle \langle \alpha\beta, h \rangle = \langle \alpha\otimes\beta, \Delta(h) \rangle$
$\displaystyle \langle \Delta (\alpha), h\otimes g \rangle = \langle \alpha, hg\rangle$
$\displaystyle \langle S(\alpha), h \rangle = \langle \alpha, S(h) \rangle$
$\displaystyle \langle 1, h \rangle = 1 $
$\displaystyle \langle \alpha, 1 \rangle = \alpha(1)$
Replacing those last two conditions with the same thing expressed in units and counits, we have
$\displaystyle \langle 1, h \rangle = \epsilon(h)$
$\displaystyle \langle \alpha, 1 \rangle = \epsilon(\alpha)$

Thus now we have a definition: Let $H_1, H_2$ be Hopf algebras. We say that they are dually paired if there is a linear map $\langle, \rangle : H_2 \otimes H_1 \rightarrow k$ that satisfies the above 5 conditions.

To make things explicit, $k(G)$ and $kG$ are dually paired by the evaluation map. Majid's book gives a more exotic example of a ``quantum group'' that is paired with itself:

Let $q \in k$ be nonzero and let $U_q(b+)$ be the $k$-algebra generated by elements $g, g^{-1}$, and $X$ with the relation $gX = qgX$. Majid's book assures me we'll see how to find this ``quantum group'' in the wild later on, but for now he gives it a Hopf algebra structure with the coproduct $\Delta X = X\otimes 1 + g\otimes X$, $\Delta g = g \otimes g$, $\epsilon X = 0$, $\epsilon g = 1$, $SX = -g^{-1} X$, and $Sg = g^{-1}$

This Hopf algebra is dually paired with itself by the map $\langle g, g \rangle = q$, $\langle X, X \rangle = 1$, and $\langle X, g \rangle = \langle g, X \rangle = 0$

Wednesday, November 21, 2012

What I learned today: universal group structure of algebras

To help keep me motivated and mathematically active, I will be blogging about "what I learned today" in my various projects. I will endeavor to post to this series daily, or as close to daily as I can. This is the first post in that series, and today we're talking about what I learned in universal algebra/hopf algebras.

So I'm taking a break from my finite fields project and am focusing on a new one, purely in quantum groups. I "started" about three months ago, but between procrastination, my job, and other projects (exempli gratia, learning class field theory) I haven't actually learned any quantum group theory.

I had made some progress in learning quantum groups in July, but due to some personal issues that came up in August, I took a long break from mathematics and largely forgot everything I had learned. So I'm trying to learn it all again. Now, Majid's Quantum Groups Primer covers the material exactly needed for my project, but I'm having trouble with many of the algebraic subtleties. In fact, I'm pretty weak in algebra (or maths in general). I'm using Kassel's Quantum Groups as a companion text, and I spent an hour or two today going over some basic universal algebra. Here's what I learned: Let $A$ be an associative algebra over a field $k$. Then:

$$\text{Hom}_\text{Alg}(k[x_1,\ldots,x_n], A) \cong A^n$$

You can see this because an algebra map from the polynomial algebra to $A$ is exactly determined by its action on the generators $x_1,\ldots,x_2$, hence a choice of an algebra map is a choice of a set map from $\{1,\ldots, n\}$ to $A$, id est, $A^n$.

This allows us to write the additive group structure of $A$ universally in terms of maps on $k[x]$ and $k[x', x'']$. That is, we will express addition on $A$ without reference to $A$ itself. The group addition on $A$ is a map $+: A^2 \to A$. Now, from above, $A^2 \cong \text{Hom}_\text{Alg}(k[x',x''],A)$. So if we let $f \in \text{Hom}_\text{Alg}(k[x',x''],A)$, we will reach our goal by expressing $+(f)$ in terms of the polynomial algebras.

$f$ is entirely determined by its action of $x'$ and $x''$, hence $f(x')$ and $f(x'')$ define $f$. Now let $\Delta: k[x] \to k[x', x'']$ by $x \mapsto x' + x''$. Then $f \circ \Delta (x) = f(x' + x'') = f(x') + f(x'') = +(f)$. Hence $\Delta$ corresponds to our group addition.

Similarly, the map $S:k[x] \to k[x]$ by $x \mapsto -x$ and $\epsilon: k[x] \to k$ by $x \mapsto 0$ correspond to the group inversion and the group identity, respectively, of $A$.

Sunday, August 19, 2012

Progress (or lack thereof) on NCG de Rham Cohomology of Finite Field Extensions

I've been working on calculations of the aforementioned entity, in particular, I want $H^0$, id est, the kernel of the exterior derivative, to be $\mathbb{F}_p$ when I pass to the limit leading to the algebraic closure.

I'm going to make that more precise, but first let me clear up some notation: since $H^0$ does not capture the isomorphism class of field extensions, and since the differential calculi of a field $K$ are in one-to-one correspondence with the monic irreducible polynomials in $K[x]$, we shall write $H^0(m(\mu), K)$ to mean the 0th cohomology of the field extension $K[\mu] (m(\mu))$, or simply $H^0(m(\mu))$ when the field is understood. Also, I shall write $\langle m(x) \rangle_K$ to mean the $K$-span of the set $\{ m(x)^n : n\in \mathbb{N}\} = \{ f(m(x)) : f\in K[x]\}$.

More precisely, we know that the algebraic closure of $\mathbb{F}_2$ is the colimit of fields $\mathbb{F}_{2^k}$, $n\in\mathbb{N}$, with field morphisms $\varphi_{kj} : \mathbb{F}_{2^k} \rightarrow \mathbb{F}_{2^j}$ whenever $k$ divides $j$. It is hoped that $H^0$ is a contravariant functor, yielding a morphism $H^0 (\varphi_{kj}) : H^0(m_j(\mu)) \rightarrow H^0(m_k(\mu))$ for monic irreducible polynomials $m_j(\mu)$ and $m_k(\mu)$ of degrees $j$ and $k$, respectively. This will give us a projective system and thus a projective limit:

$ \displaystyle H^0(\varinjlim_{k\in\mathbb{N}} m_k(\mu) ) = \varprojlim_{k\in\mathbb{N}} H^0(m_k(\mu))$

So our goal is to answer two questions:

  1. What conditions are needed on the polynomials $m_k(x)$, $k\in\mathbb{N}$ so that $H^0(\varphi_{jk})$ is well defined?
  2. When it is defined, when does the above limit equal $\mathbb{F}_2$?
(I'm starting to think the first question is vacuous, but I didn't until recently.)

1. Calculating $H^0(m_k(\mu), \mathbb{F}_2)$

So first I tried to calculate the 0th cohomology for the simplest case: $H^0(\mu^2 + \mu + 1)$. Finding the cohomology amounts to finding which polynomials satisfy $f(x +\mu) = f(x)$. After staring at the equation for awhile, it's pretty easy to see that if $f(x) \in H^0$, then $\text{deg} f = 2^k$ for some $k$.

After some effort (and help), I was able to prove that $H^0(\mu^2 + \mu + 1) = \langle x^4 - x \rangle_{\mathbb{F}_2}$. The proof went like this:

  1. Find the smallest polynomial $f(x)$ in $H^0$, in this case, $f(x)=x^4 - x$.
  2. Let $g(x) \in H^0$, prove that $\text{deg} f$ divides $ \text{deg} g$. (This requires some effort - I've only done it in specific cases by exhaustion).
  3. Hence $\text{deg} \, g = r \, \text{deg}\, f$. Then $g(x) - f(x)^r$ has degree divisible by $\text{deg}f$, so it's also in $H^0$, and it's also of smaller degree, so continuing in this fashion we have to end back at $f(x)$, and we're done.

NOTE: I don't know that there is always only one polynomial of smallest degree in $H^0$, but I've not yet found a case where this isn't true, either.

It's not hard to prove that $x^{p^k} - x$ is always in $H^0(m_k(\mu), \mathbb{F}_p)$. However, $H^0$ can contain a lot more than just $x^{2^k} - x$. But since this is a particularly nice polynomial (its splitting field is $\mathbb{F}_{p^k}$) and $\varprojlim \langle x^{p^k} - x \rangle_{\mathbb{F}_p}$ looks like it's just $\mathbb{F}_p$ (I think, I really have no idea how to calculate that limit, I've not thought about it much yet), one might desire to choose polynomials for the algebraic closure such that $H^0(m_k(\mu)) = \langle x^{p^k} - x\rangle$ .

Note that in this case ($k$ divides $j$, so $j = kq$), we have:

$ \displaystyle x^{p^j} - x = - \sum_{i=0}^{q-1} \left( x^{p^{j - k(i+1)}} -x^{p^{j - ik}} \right) = - \sum_{i=0}^{q-1} (x^{p^k} - x)^{p^{k(q-i)}} $

So that $H^0(\varphi_{kj})$ is simply the identity on $H^0(m_j(\mu))$ embedding it into $H^0(m_k(\mu))$

So, as a guess, we're going to try to find a specific set of polynomials such that $H^0(m_k(\mu)) = \langle x^{p^k} - x\rangle$. Also as a guess, our first candidates are...

2. Conway's polynomials and fields

Conway polynomials of degree $k$ for $\mathbb{F}_p$ are the least monic irreducible polynomial in $\mathbb{F}_p[x]$ under a specific lexicographical ordering. To the best of my knowledge, their primarily use is to provide a consistent standard for the arithmetic of $\mathbb{F}_{p^k}$ for portability across different computer algebra systems. Since these are the ``standard convention'' for Galois Fields, we try them first.

Sadly, they let us down. The Conway polynomial of degree 4 for $\mathbb{F}_2$ is $\mu^4 + \mu + 1$. Using a computer algebra system (sage), I found that $H^0( \mu^4 + \mu + 1) = \langle x^8 + x^4 + x^2 + x^1 \rangle_{\mathbb{F}_2}$.

In addition to those polynomials, John Conway has also described the algebraic closure of $\mathbb{F}_2$ as a subfield of $\text{On}_2$, the set of all ordinals with a field structure imposed by his ``nim-arithmetic''. The description of this field, from his ``On Numbers and Games'' does not treat $\overline{\mathbb{F}_2}$ as a direct limit over various simple extensions. Instead, he shows that the ``quadratic closure'' of $\mathbb{F}_2$ lies in $\text{On}_2$, and then extends the quadratic closure to a cubic closure, also in $\text{On}_2$, et cetera.

This means that for us to use his description, we first have calculate things like the minimal polynomial for $\omega$ and $\omega^\omega$ (where $\omega$ is the least infinite ordinal) over $\mathbb{F}_2$, these calculations are quite difficult (I haven't been able to do a single one), in fact, it's quite a bit of work just figuring out which ordinal corresponds to which $\mathbb{F}_{2^k}$.

But fortunately for us, the quadratic closure only relies on finite ordinals, namely $2^k$. With some help from the internet, we have polynomials $m_{2^k}(\mu)$ describing Conway's quadratic closure. They are given by the recursive relations $m_{2^k}(\mu) = m_{2^{k-1}} ( \mu^2 + \mu)$ with $m_2(\mu) = \mu^2 + \mu + 1$, obviously.

So $m_4(\mu)$ corresponds to the Conway polynomial (this is not the case in general), and we've already used Sage to show that this polynomial doesn't have the cohomology we're looking for.

3. So what next?

Well, for one, we could just define the polynomials $m_k(\mu)$ to be such that $H^0 = \langle x^{2^k} -x \rangle$. But is this choice unique?

Turns out no: again using Sage, I have that both $H^0(1 + \mu + \mu^2 + \mu^3 + \mu^4)$ and $H^0(1+\mu^3+\mu^4)$ are $\langle x^{16} -x \rangle$.

So let's revisit the requirement on $H^0$. I was probably being a dunce insisting on it in the first place. After all, if $H^0$ really is a functor (as it should be, though I've not tried to prove it, or even defined the categories), then the projective system is always well-defined, and we really just need to calculate the limit.

In fact, what we really want is that the chain of polynomial subalgebras $H^0(m_k(x))$ becomes coarser as $k$ tends towards larger integers (thus $m_k(x)$ tends towards larger degrees), and that we have sane maps between $H^0(m_j(x))$ and $H^0(m_k(x))$ whenever $k$ divides $j$.

If my assumption that $H^0$ is always ``spanned'' by a single polynomial is correct, and if my sketch of a proof of $H^0 = \langle f(x) \rangle$ always works, then we might be able to find said map. Say that $H^0(m_k(x)) = \langle \tilde{m}_k(x) \rangle$. We know that $x^{2^k} - x \in H^0(m_k(x))$, so that $\langle x^{2^k} - x \rangle \subset \langle \tilde{m}_k(x) \rangle$, we also know that $\langle x^{2^j} - x \rangle \subset \langle x^{2^k} -x \rangle$ as, from before:

$ \displaystyle x^{p^j} - x = - \sum_{i=0}^{q-1} (x^{p^k} - x)^{p^{k(q-i)}} $

Since $x^{2^j} -x \in H^0(m_j(x))$, we know that $f(m_j(x)) = x^{2^j} - x$ for some polynomial $f \in \mathbb{F}_2[x]$. Let $g(x) \in H^0(m_j(x))$, then

$ \displaystyle g(x) = \tilde{m}_j(x)^s + g_{s-1} \cdot \tilde{m}_j(x)^{s-1} + \ldots + g_0$

by definition. Define a map from $H^0(m_j(x)) \rightarrow H^0(m_k(x))$ by

$ \displaystyle g(x) \mapsto f(\tilde{m}_j(x))^s + g_{s-1} \cdot f(\tilde{m}_j(x))^{s-1} + \ldots + g_0$

Assuming that makes sense, and I haven't made an other stupid errors (a big if!), then I guess that leaves the following for a TODO to show that $H^0$ goes to $\mathbb{F}_2$ for any algebraic closure.:

  1. Check that $H^0$ is indeed ``spanned'' by a single polynomial, id est, that $H^0(m(x)) = \langle f(x) \rangle$ for all $m$.
  2. Define the categories for which $H^0$ is a functor, check that the above map works (or find a new one?)
  3. Prove that the subalgebras $H^0$ become coarser, as stated above. (Intuitively, it makes sense that they do, but I haven't thought about a proof yet. Also, if they do become coarser, then it seems ``obvious'' that the limit goes is $\mathbb{F}_2$, as those are the only elements left in an series of smaller and smaller subalgebras.

Monday, April 16, 2012

You should re-read the stuff you think you know.

I started my PhD in June 2011. While I've only worked on a few problems since then, a pattern is emerging: I always spend a couple months stuck on something that looks obvious and silly (both in fore- and hindsight.) Most recently I was stuck on a problem involving basic arithmetic over finite fields. I felt that it was something I should be able to do, but nonetheless was stuck on it for a few weeks.

Maybe it's that I'm doing the PhD thing wrongly, or maybe it's that I'm not spending enough time on my mathematics (I've got a job doing predictive modeling, too!). In any case, the story of how I got unstuck might interest some people:

I've thought it important to continue doing maths reading independent of my research projects. But while there are several subjects I wish to learn about (and do research in), I have not engaged in much outside study. When I approach a subject, I have two separate reactions: Either I find that the subject is too advanced, and I give up quite early - not wanting to get lost in background material. Or I think the material I have repeats too much of stuff I already know, and I give up - not wanting to waste my time on things I've seen before.

After doing this for a few months, I decided that I should read something that I think I already know - just to get in the habit of things. I started reading the field/Galois theory chapters in the abstract algebra book by Dummit and Foote. Pretty light reading as far as mathematics goes. It's stuff I should know.

Those chapters contain some straightforward proofs of basic statements on algebraic field extensions. Exempli gratia, if $\mu$ is the root of a polynomial $f$, and $\sigma$ is an element of the Galois group, then one can prove that $\sigma(\mu)$ is also a root of the polynomial by observing the action of $\sigma$ on the polynomial $f$. Not particularly new or deep material, but it turns out just the stuff I needed...

Now back to my problems with arithmetic over finite fields: I've spent the last few weeks trying to decompose an operator on polynomials into separate components. My supervisor and I both had our own ideas of writing out matrices for these operators. I spent a few late nights on it, wrote down some messy results, and went to bed hoping I'd have an epiphany on how to clean it up. I had been hoping for one for weeks! But soon after I started re-reading the basics I got it. I was just reading about applying the action of the Galois group to polynomials, and here I was staring at an operator on polynomials. So I abandoned the ``matrix approach'', naively started looking at the action of the Galois group, and a few minutes later I had the result I was looking for. It seemed so obvious in hindsight, but it often seems obvious only after you've understood it.

I should probably keep reading the basics.

Sunday, April 15, 2012

Cohomology of Finite Fields Part III

I gave a proof (with help from math.stackoverflow) in my last blog for the 0th cohomology of the field extension ${\mathbb{F}_4/\mathbb{F}_2}$. While I have sketches for descriptions of ${H^0(\mathbb{F}_{p^m} / \mathbb{F}_p)}$ for any prime ${p}$ and positive integer ${m}$, I've not yet been able to describe ${H^1}$ for even the simplest cases. I shall describe the problem in this post.

First, recall that I mention in my first post on NCG cohomologies of field extensions that a first order differential calculus gives rise to a differential graded algebra. Let's briefly describe this:

Given our [possibly noncommutative] algebra ${A = \Omega^0}$, we've already described it's first order differential calculus as an ${A}$-bimodule ${\Omega^1}$ together with a linear map ${d: \Omega^0 \rightarrow \Omega^1}$ that obeys the rules 1) ${d(ab) = a \cdot db + da \cdot b}$ (where ${\cdot}$ is the action of the algebra on the bimodule) and 2) ${\Omega^1 = \{a \cdot db : a, b \in \Omega^0\}}$. The differential graded algebra is then defined as graded algebra ${\Omega = \bigoplus_n \Omega^n}$ together with the map ${d : \Omega^n \rightarrow \Omega^{n+1}}$ and the rules: ${\Omega^0 = A}$, ${\Omega^1}$ as before, ${d^2 = 0}$, and ${\Omega}$ generated by ${\Omega^0}$ and ${\Omega^1}$. The product of the differential graded algebra is the wedge product ${\wedge}$, as in classical differential geometry.

Putting the above together, we get a complex ${A = \Omega^0 \xrightarrow{d^0} \Omega^1 \xrightarrow{d^1} \ldots}$. Here ${d^n}$ is the restriction ${d^n = d |_{\Omega^n}}$. The cohomology is then defined as ${H^n(A) = \text{ker}(d^n)/\text{Im}(d^{n-1})}$, with ${H^0(A) = \text{ker}(d^0)}$ as we saw in the last two posts.

In our case of the field extension ${\mathbb{F}_4/\mathbb{F}_2}$, we have as our algebra the polynomial ring ${A = \mathbb{F}_2[x]}$. We've already defined a differential calculus on it, the polynomial ring ${\Omega^1 = \mathbb{F}_4[x]}$, where ${\mathbb{F}_4 = \mathbb{F}_2(\mu)}$, together with the differential map ${df = \frac{f(x + \mu) - f(x)}{\mu}}$. Using this, how can we describe the differential graded algebra ${\Omega = \bigoplus_n \Omega^n}$ ?

First, recall that ${\mathbb{F}_4}$ is a two dimensional vector space over ${\mathbb{F}_2}$ with basis ${\{1, \mu\}}$. This means that the polynomial ring ${\Omega^1}$ can be ``spanned'' by ${\{1, \mu\}}$ over the polynomial ring ${\mathbb{F}_2[x]}$. In other words, every ${f \in \Omega^1}$ can be written as ${f = f_1 \cdot 1 + f_\mu \cdot \mu}$, with ${f_1, f_\mu \in \Omega^0}$. In order for us to get a graded algebra, ${\Omega^2}$ must have a basis ${\{1 \wedge \mu\}}$, hence it's a one dimensional space over ${\Omega^0}$ and thus is isomorphic to ${\Omega^0}$. ${\Omega^n = 0}$ for ${n > 2}$, as we're using a classical anticommutative wedge product.

But what is the map ${d^1: \Omega^1 \rightarrow \Omega^2}$ ? Well, from our description of ${\Omega^1}$, we can easily express the action of ${d}$ on ${\Omega^1}$ as an action on ${\Omega^0}$, which is well defined. Id est, for ${f \in \Omega^1}$,

$\displaystyle d^1 f = d^0(f_1) \wedge 1 + f_1 \wedge d^0(1) + d^0(f_\mu) \wedge \mu + f_\mu \wedge d^0(\mu)$

Now, ${d^0(a) = 0}$ for any constant polynomial, and ${d^0(g) \in \Omega^1}$ for any ${g \in \mathbb{F}_2}$. So ${d^0g = \partial_1 g \cdot 1 + \partial_\mu g \cdot \mu}$. Hence we have linear maps ${\partial_1, \partial_\mu : \mathbb{F}_2[x] \rightarrow \mathbb{F}_2[x]}$. We return to ${d^1}$ using this idea:

$\displaystyle d^1 f = (\partial_1 f_1 \cdot 1 + \partial_\mu f_1 \cdot \mu) \wedge 1 + (\partial_1 f_\mu \cdot 1+ \partial_\mu f_\mu \cdot \mu) \wedge \mu$

The wedge product is anticommutative, so this leaves us with:

$\displaystyle d^1 f = (\partial_1 f_\mu - \partial_\mu f_1) \cdot 1 \wedge \mu$

And ${d^2 f = 0}$, as the rest of the ${\Omega^n}$'s are 0. Now back to the cohomologies: ${H^1(\mathbb{F}_4/\mathbb{F}_2) = \text{ker}(d^1)/\text{Im}(d^0)}$, where ${\text{ker}(d^1) = \{f \in \Omega^1 : \partial_1 f_\mu = \partial_\mu f_1 \}}$ and ${\text{Im}(d^0) = \{f_1 \cdot 1 + f_\mu \cdot \mu \in \Omega^1 : \exists f \in \mathbb{F}_2[x] \; \text{such that} \; df = f_1 \cdot 1 + f_\mu \cdot \mu\}}$. Hence the we need to find a description of the ``partial derivatives'' ${\partial_1}$ and ${\partial_\mu}$. In other words, we need to describe ${d^0f}$ as:

$\displaystyle \frac{f(x + \mu) -f(x)}{\mu} = \partial_1 f + \partial_\mu f \cdot \mu$

By noting that ${\mu^3 = 1}$, I was able to write down a formula for ${\partial_i x^{3n}}$, and then formulas for ${\partial_i x^{3n+1}}$ and ${\partial_i x^{3n+2}}$ in terms of ${\partial_i x^{3n}}$. But the binomial expansion makes these formulas just too messy to work with. I then tried finding a basis under which the maps would have a nice, neat formula, but I had no luck there. Next I tried looking at the action of the Galois group on $d$.

Now, the only non-trivial element of the $\text{Gal}(\mathbb{F}_4/\mathbb{F}_2)$ is the Frobenius automorphism $\sigma(x) = x^2$, exempli gratia, $\sigma\mu=\mu^2=\mu+1$. The action of the Galois group extends to an action on the polynomial ring - one that leaves $\mathbb{F}_2$ fixed. Hence we have

$\displaystyle \sigma d f = \sigma(\partial_1 f) + \sigma(\partial_\mu f \cdot \mu) = \partial_1 f + \partial_\mu f \cdot (\mu + 1) = df + \partial_\mu f$

This allows us to write the ``partial derivatives'' in terms of $d$ and the Frobenius automorphism. Id est,

$\displaystyle $$\partial_\mu f = \sigma (d f) - d f$


$\displaystyle \partial_1 f = df - \sigma(d f) \mu + (df) \mu$

It remains to be seen if these formulas for the partials will actually help me calculate the remaining cohomologies.

Wednesday, April 11, 2012

Cohomology of Finite Fields Part II

Last time we described a theory of cohomology for finite field extensions, but we fell short of calculating ${H^0(\mathbb{F}_4 / \mathbb{F}_2)}$ as I had discovered a mistake in my proof shortly after I had posted the blog. Thanks to the kind folks on math.stackoverflow I have a correct proof (and result!). Let's go through it.

Recall that ${H^0(\mathbb{F}_{4} / \mathbb{F}_2) = \text{ker}(d)}$, where ${\mathbb{F}_{4} = \mathbb{F}_2(\mu)}$, ${\mu}$ a root of a monic irreducible polynomial of degree ${2}$ in ${\mathbb{F}_2}$, and

$\displaystyle d: \mathbb{F}_2[x] \rightarrow \mathbb{F}_{4}[x] \; \text{by} \; df = \frac{f(x + \mu) - f(x)}{\mu}$

Hence ${H^0(\mathbb{F}_{4}/\mathbb{F}_2)}$ is the set of all polynomials ${f \in \mathbb{F}_2[x]}$ such that ${f(x + \mu) = f(x)}$. To find it, first note that if ${f(x) \in H^0}$, then so is ${g(f(x))}$ for any ${g \in \mathbb{F}_2[x]}$, as ${g(f(x + \mu)) = g(f(x))}$. If we can show that some polynomial ${f}$ is the only polynomial in ${H^0}$ with smallest degree, then we know that every other ${g(f(x)) \in H^0}$ is spanned by ${\{f(x)^n : n\in \mathbb{N}\}}$ and further show that every other ${h \in H^0}$ has degree divisible by the degree of ${f}$, then we have that ${\{f(x)^n : n\in \mathbb{N}\}}$ is the basis for ${H^0}$.

Notice also that ${f(x) = x^{4} + x \in H^0}$. This follows from the fact that ${\mathbb{F}_{4}^{\ast}}$ is the cyclic multiplicative group of order ${3}$, hence ${\mu^{3} = 1}$. So we have

$\displaystyle f(x + \mu) = (x + \mu)^{4} + x + \mu = x^{4} + \mu^{4} + x + \mu = x^{4} + x + \mu + \mu = f(x)$

It's not hard to show (exempli gratia, by exhaustion) that ${x^4 + x}$ is the smallest degree polynomial in ${H^0}$, and that it's the only polynomial in ${H^0}$ of degree ${4}$.

Now let ${f \in H^0}$, and let ${\text{deg}f = n}$, say ${f(x) = x^n + a_1 x^{n-1} + \ldots + a_n }$. Clearly ${n}$ cannot be odd, as then ${\binom{n}{1}}$ is also odd, so in the expansion

$\displaystyle f(x + \mu) = (x+\mu)^n + \ldots + a_n = x^n + \mu x^{n-1} + \ldots + \text{lower order terms}$

has a ${\mu}$ as a coefficient of the ${x^{n-1}}$ term, so it can't possibly equal ${f(x)}$.

Now, we want that ${4 | \text{deg}f}$. Say ${\text{deg}f = n = 4k + 2}$. Let's look at the coefficient of ${x^{4k}}$ in ${f(x + \mu)}$. We have ${(x+\mu)^{4k + 2} = (x+\mu)^{4k}(x+\mu)^2 = (x^4 + \mu^4)^k(x^2 + \mu^2) = x^{4k+2} + \mu^2 x^{4k} + \text{lower order terms...}}$. Also ${(x+\mu)^{4k + 1} = (x^4 + \mu^4)^k (x + \mu) = x^{4k+1} + \mu x^{4k} + \text{lower order terms...}}$. Hence in ${f(x + \mu) - f(x)}$, ${x^{4k}}$ has a coefficient ${(\mu^2 + \mu)}$. Thus, if ${f}$ is not divisble by 4, it's not in ${H^0}$.

So say ${f \in H^0}$ has degree ${4k}$. Then ${f(x) - (x^4 + x)^k}$ is also in ${H^0}$ and divisible by 4. In this way we see that polynomials in ${t = x^4 + x}$ form ${H^0}$.

Monday, March 12, 2012

Noncommutative de Rham cohomology of finite fields

As my lack of blogs might suggest, I've not made much progress on the Ph.D. front for the past several weeks. I've been reading papers by Drinfeld, et al, trying to crack the Grothendieck-Tiechm\"{u}ller group. I've not understood much. While I'm not giving up on those ideas, I am taking my supervisor's suggestion to work more closely with him on a smaller project. I'm computing the Noncommutative de Rham cohomology of extensions of finite fields.

Finite Fields and Differential Calculi

As a specific example of this, let's look at the finite field ${\mathbb{F}_2}$ and a field extension of degree 2. Readers familiar with basic field theory should know that, in order to extend this field, we need a degree 2 irreducible polynomial in ${\mathbb{F}_2}$, say ${\mu^2 + \mu + 1}$. This polynomial will generate a prime ideal in the polynomial ring ${\mathbb{F}_2[\mu]}$, and the quotient ring ${\mathbb{F}_2[\mu] / (\mu^2 + \mu + 1)}$ will be our new field.So what does this field look like? Essentially, elements of this field are polynomials with the condition that ${\mu^2 = \mu + 1}$. The reader should see that all the elements of the new field are ${0, 1, \mu, \; \text{and}\; \mu+1}$. What we've really done here is adjoined the root of the polynomial ${\mu^2 + \mu + 1}$ to our field. Call this new field ${\mathbb{F}_4}$. It can also be thought of as a vector space over ${\mathbb{F}_2}$ with basis ${\{1, \mu \}}$.Our ``space'' is the finite line, which we're modeling with ${A = \mathbb{F}_2[x]}$. Now how do we get to noncommutative de Rham cohomology? First, we need the differential calculus of 1-forms. Recall from my early post on NCG black holes that we can define a differential calculus over an algebra ${A}$ as the pair ${(d, \Omega^1)}$, with ${\Omega^1}$ a bimodule over ${A}$ and ${d: A \rightarrow \Omega^1}$ a linear map that obeys the product rule ${d(ab) = d(a)\cdot b + a\cdot d(b)}$. Additionally, we require that the set ${\{a \cdot d(b) : a, b \in A\}}$ spans our calculus ${\Omega^1}$.Now, ${\mathbb{F}_2[x]}$ has an obvious action on the polynomial ring ${\mathbb{F}_4[x]}$, so we can think of ${\Omega^1 = \mathbb{F}_4[x]}$ as our calculus of one forms. For an ${f \in A = \mathbb{F}_2[x]}$, the derivative looks like:

$\displaystyle df = \left( f(x + \mu) - f(x) \right) \mu^{-1}$

We also have an NCG notion of an exterior algebra, which gives us spaces of n-forms extended from ${\Omega^1}$. This is how we get a de Rham cohomology. In our case we have a complex:

$\displaystyle \mathbb{F}_2[x] = \Omega^0 \xrightarrow{d^0} \Omega^1 = \mathbb{F}_4[x] \xrightarrow{d^1} \Omega^2 \xrightarrow{d^2} \ldots $

Where ${d^n = d |_{\Omega^n}}$ is the derivative restricted to the calculus of n-forms (exempli gratia, ${d^0}$ is the ${d}$ we defined above.) The n-th de Rham cohomology is then ${H^n(A) = \text{ker}(d^n) / \text{Im}(d^{n-1})}$.But what are ${\Omega^2, d^1}$, et cetera? We'll save that topic for another post. The zeroth cohomology group is just ${H^0 = \text{ker}(d^0)}$ and for now we'll just worry about calculating that.

Finding ${H^0}$

So ${H^0}$ will consist of exactly the polynomials ${f \in F_2[x]}$ such that:

$\displaystyle f(x+u) - f(x) = 0$

Obviously constants fit this. Because we're working in a field of characteristic ${p=2}$, we know that ${(x + a)^p = x^p + a^p}$ is the Frobenius automorphism. This causes me to suspect that only polynomials of the form:

$\displaystyle f_n(x) = x^{2^n} + a_{n-1} x^{2^{n-1}} + \ldots + a_0 x$

can lie in the kernel. We'll prove this later, but first let's see what exactly happens with such a polynomial:

$\displaystyle f_n(x + \mu) = (x+\mu)^{2^n} + a_{n-1} (x+\mu)^{2^{n-1}} + \ldots + a_0(x+\mu) $
= x^{2^n} + \mu^{2^n} + a_{n-1}( x^{2^{n-1}} + \mu^{2^{n-1}}) + \ldots + a_0 x + a_0 \mu $

Hence we have

$\displaystyle f_n(x + \mu) - f_n(x) = \mu^{2^n} + a_{n-1} \mu^{2^{n-1}}+ \ldots + a_0 \mu$

From earlier, we have that ${\mu^2 = \mu + 1 = \mu^{-1}}$. So ${\mu^3 = 1}$, more explicitly,

$\displaystyle \mu^r = \begin{cases} 1 & r \equiv 0 \mod 3 \\ \mu & r \equiv 1 \mod 3 \\ \mu^{-1} & r \equiv 2 \mod 3 \end{cases} $

Moreover, we have that ${2^r \equiv 1 \mod 3}$ when ${r}$ is odd, and ${2 \mod 3}$ when ${r}$ is even. Combing all this, we have that for even ${n}$:

$\displaystyle f_n(x+\mu) - f_n(x) = \mu (a_{n-1} + a_{n_3} + \ldots + a_1) + \mu^{-1} (1 + a_{n-2} + \ldots + a_0)$

Hence ${df_{2k} = 0}$ when ${\sum_{i=1}^{k} a_{2(k-i) + 1} = 0}$ and ${\sum_{i=1}^{k} a_{2(k-i)} = 1}$. Similarly, when ${n= 2k + 1}$, we have that ${df_{2k+1} = 0}$ when ${\sum_{i=1}^{k} a_{2(k-i) + 1} = 1}$ and ${\sum_{i=0}^{k} a_{2(k-i)} = 0}$.We can use these formulas to write down a basis for some things in kernel by concentrating on the shortest polynomials with the highest powers. The above formulas say that the shortest possible polynomial in the kernel will have two elements of degree 2 to the power of the same parity. Hence our basis is ${\{ x^{2^n} + x^{2^{n-2}} : n \geq 2\}}$.Now we need to prove that such things are the only thing in the kernel. To see this, let's calculate the action of...

Update 18 Mar 2012

The proof I originally posted here was wrong. I made two mistakes: first, my calculation of $f(x + \mu) - f(x)$ neglected constant terms. Second, I never bothered to check that the coefficient of $x^k$ is zero when $f$ isn't spanned by our powers-of-two basis (I only checked to see if there is a nonzero term, but since we're working in a field of characteristic 2, the sum of nonzero terms can still be zero.) I spent several days trying to correct this proof, but I can't. The statement itself is wrong. ${\{ x^{2^n} + x^{2^{n-2}} : n \geq 2\}}$ is not the basis. I have several counter examples, exempli gratia $f(x) = x^{12} + x^9 + x^6 + x^3$ and $f(x) = x^{20} + x^{17} + x^5 + x^2$ and I believe I can find polynomials in the kernel that have 8, 16, et cetera terms that aren't the sum of smaller things in the kernel. So...I still have some work to do before I can find $H^0$.

Wednesday, January 18, 2012

Don't Censor the Web.

Congress is considering two Orwellian-named laws, SOPA and PIPA, that are threatening free speech, internet security, and innovation.

This is a reminder to call your representatives in Congress and/or donate to the EFF today to help stop internet censorship.

Saturday, January 7, 2012

Constructing the Grothendieck-Teichmuller Group

So for the past six or seven months I've been trying to get a copy of the paper ``On quasitriangular Quasi-Hopf algebras and a group closely connected with ${\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})}$'' by V.G. Drinfel'd. My online search uncovered only a Russian copy (which, unfortunately, I don't read). My library searches were equally unfruitful: the British library's copy is ``unavailable'' and I was unable to find a copy at another UK institution. My supervisor was sure he had a copy, but when he discovered that it mysteriously disappeared he encouraged me to request an inter-library loan with my university. I did, and my university said they found a copy in Leeds, but the last ten pages are missing. So - if any readers have a copy of the aforementioned paper: I'd greatly appreciate your lending me it!

In the meanwhile, I've been reading ``On Associators and the Grothendieck-Teichmuller Group'' by Dror Bar-Natan. Among other things, that paper described the construction of said group via a ``Parenthesized Braid Category''. I'll discuss that construction in this post.

1. Braids and Categories

I really don't want to give a precise definition of a braid (I've not yet seen one that doesn't make my head hurt), but hopefully I won't be too wrong if I propose the following:

Def1 1 A braid on ${n}$ strands is a set of ${n}$ oriented curves with distinct starting and ending points where the starting points lie on the same line, each endpoint lies on a separate line, and no curve contains a loop.

For example, the following is a braid:

Clearly we can enumerate the starting and ending points, and treat a braid as a permutation on them. In this way, braids form a category where the braid itself is the morphism and the objects are simply sets to be permuted.

We're going to change our class of objects slightly - we're going to consider parenthesizations of sets. Id est, pretend that the sets have some underlying multiplicative structure. We want to order the sets and denote the order of multiplication. Exempli gratia, the ordering ${(1 \, 2 \, 3)}$ has two parenthesizations: ${((1 \, 2) \, 3)}$ and ${(1 \, (2 \, 3))}$. Our braid morphisms are allowed to act on any parenthesizations.

We aren't done complicated things yet: we're going to modify the morphisms themselves so that they become formal sums:

$\displaystyle \sum_{i=1}^{k} \beta_i B_i$

Where each ${B_i}$ is a braid on ${n}$ strands, all ${B_i}$'s have the same effective permutation, and ${\beta_i}$ belongs to some ${\mathbb{Q}}$-algebra, usually ${\mathbb{Q}}$ or ${\mathbb{C}}$. We're forming the ``algebriod'' over ${\mathbb{Q}}$ or ${\mathbb{C}}$ over the set of braids on ${n}$ strands, with composition defined to be a bilinear map in the same way you would do for a group algebra.

This category is our Parenthesized Braid Category ${\textbf{PaB}}$.

2. Fibred Linear Categories

Our category is fibred. To see this, consider the category ${\textbf{PaP}}$ of parenthesized objects as before, but with simple permutations as the morphisms. Then consider the functor ${\textbf{S} : \textbf{PaB} \rightarrow \textbf{PaP}}$ that maps each object to itself, but takes each formal sum of braids to that one effective permutation. Clearly the set of morphisms in ${\text{PaB}}$ from ${O_1}$ to ${O_2}$ is:

$\displaystyle \text{Hom}_\textbf{PaB}(O_1, O_2) = \textbf{S}^{-1} \left( \text{Hom}_\textbf{PaP}(O_1, O_2) \right)$

Where P is the effective permutation.

One might also note that for a fixed permutation ${P}$, the fibre of morphisms ${\textbf{S}^{-1}(P)}$ is a vector space (actually, its an algebra, because we can compose morphisms as our product). This allows us to define some very ``linear'' notions, for instance:

We can define subcategories: a category is a subcategory of ${\textbf{PaB}}$ if it shares the same collection of objects and each ${\text{Hom}(O_1, O_2)}$ is a subspace of the corresponding ${\text{Hom}_\textbf{PaB}(O_1, O_2)}$.

A subcategory is an ideal if whenever one of two composable ${B_1}$ or ${B_2}$ belongs to it then the composition belongs to it, too. Exempli gratia, let ${\textbf{I}}$ be the ideal where all morphisms ${\sum \beta_i B_i}$ have ${\sum \beta_i = 0}$. Then we can define the quotient ${\textbf{PaB}/\textbf{I}}$ in an obvious way: each set of morphisms ${\text{Hom}(O_1, O_2)}$ is the corresponding quotient in the bigger set of morphisms.

Additionally, we can define idealic powers: ${\textbf{I}^n}$ has morphisms that can be written as compositions of ${n}$ morphisms in ${\textbf{I}}$. We can continue this and take an inverse systems with an inverse limits, in fact, we can even do the ${\textbf{I}}$-${adic}$ completion:

$\displaystyle \widehat{\textbf{PaB}} = \varprojlim_n \textbf{PaB}/\textbf{I}^n$

Finally, we can do a tensor product of ${\textbf{PaB}}$ with itself (in the same way we would do tensor powers of an algebra). For ${\textbf{PaB}^{2\otimes}}$ we define the morphism as the disjoint union:

$\displaystyle \text{Hom}_{\textbf{PaB}^{2\otimes}}(O_1,O_2) = \coprod_{P \in \text{Hom}_{\textbf{PaB}}(O_1, O_2)} \textbf{S}^{-1}(P) \otimes \textbf{S}^{-1}(P)$

Now we can even do a coproduct ${\Delta: \textbf{PaB} \rightarrow \textbf{PaB}^{2\otimes}}$ by ${B \mapsto B \otimes B}$. This coproduct is a functor. (I suspect that there is a bialgebra/hopf algebra structure hidden somewhere in here but I've not thought about it enough yet. Not for this post.)

3. Some Functors

In addition to ${\textbf{S} : \textbf{PaB} \rightarrow \textbf{PaP}}$, we also have a few functors ${\textbf{PaB} \rightarrow \textbf{PaB}}$. We can describe these by their action on a single braid and one can expand them as a linear action on the formal sums we use. First, we have the extension functors ${d_0}$ and ${d_{n+1}}$ that add a single straight strand to the left or right of the braid, respectively.

Second, we have cabling functors ${d_i}$ for ${1 \leq i \leq n}$ on a braid of n strands. This functor simply doubles the ${i}$-th strand.

Finally, we have a strand-removal functor ${s_i}$, which, as you might guess, removes the ${i}$-th strand.

We can now define the Grothendieck-Teichmuller group ${\widehat{\text{GT}}}$. This is the group of all invertible functors ${a: \textbf{PaB} \rightarrow \textbf{PaB}}$ such that ${\textbf{S} \circ a = S}$, ${d_i \circ a = a \circ d_i}$, ${s_i \circ a = a \circ s_i}$, ${\Delta \circ a = a \otimes a \circ \Delta}$. We also require that $a$ leaves the braid constant:

Right! There we are. Sooner or later I actually do some maths, but for now I'm stuck understanding trying to understand definitions, so hence this post.

4. Source

This is all just a reiteration of the material in the aforementioned paper by Bar-Natan.