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 $H^0$.