The K-theory of finite fields: a synopsis

In my previous post, I proved that if $F$ is a finite field, then multiplicative group $F^\times$ is a cyclic group. This fact will play a small part in our description today of the $K$-theory of $F$. We will start by describing the classical $K$-theory of $F$ and then briefly talk about Quillen’s computation of all the higher $K$-groups of $F$.

Classical here means the first three $K$-functors: $K_0, K_1,$ and $K_2$. These are the functors that can be described algebraically and were studied before Daniel Quillen wrote down the definitions of the higher $K$-functors using topology (technically though, one does not need to use topology to define higher $K$-theory, as one can use acyclic binary multicomplexes due to Nenashev for $K_1$ and due to Dan Grayson in general. See [1]).

$K_0$

For any associative ring $R$, the group $K_0(R)$ is the free abelian group on the finitely generated projective $R$-modules, modulo the relation $[A\oplus B] = [A] + [B]$ for every finitely-generated projectve $A$ and $B$. Then $K_0(F) \cong \Z$ for any field $F$, not necessarily finite. This is because any finitely-generated $F$-module in an $F$-vector space, or free. The class $[F]$ itself generates $K_0(R)$ as a free abelian group as every finite-dimensional vector space is of the form $F^n$ for some natural number $n$.

In fact, $K_0(\Z)\cong\Z$ for the same reason. To whet the reader’s appetite, we should mention one computation that is a little more involved: for a Dedekind domain, $K_0(R) = \Z\oplus{\rm Cl}(R)$ where ${\rm Cl}(R)$ is the class group of $R$.

$K_1$

For an associative ring $R$, the group $K_1$ is defined by
$$K_1(R) = {\rm GL}(R)/[{\rm GL}(R),{\rm GL}(R)].$$ Here, ${\rm GL}(R)$ is the infinitely general linear group defined by ${\rm GL}(R) = \varinjlim {\rm GL}_n(R)$ where the inclusions are the obvious ones. The notation $[G,G]$ denotes the commutator subgroup of a group $G$: that is, the normal closure of the smallest subgroup containing all the commutators of $G$. Since we are modding out by all the commutators, we see that $K_1(R)$ is an abelian group.

In general, it is a little involved to compute $K_1$. One thing that can help us is Whitehead’s lemma, which says that
$$[{\rm GL}(R),{\rm GL}(R)] = E(R)$$ where $E(R)$ is the group generated by the elementary matrices of ${\rm GL}(R)$, these are the matrices $e_{ij}(\lambda)$ for $i\not= j$ where $e_{kk} = 1$ and $e_{ij} = \lambda$. For a field, $E(R) = {\rm SL}(R)$ and so $K_1(F) = F^\times$.

$K_2$: Definition

Of the three classical $K$-functors, $K_2$ is the most involved to define. But it is not much more complicated than the other two, and in fact like $K_1$ also involves elementary matrices.

Let $n\geq 3$. Define the Steinberg group ${\rm St}_n(R)$ as the group generated by $x_{ij}(r)$ for $1\leq i\not=j\leq n$ and $r\in R$. These generators satisfy the Steinberg relations $x_{ij}(r)x_{ij}(s) = x_{ij}(r+s)$ and
$$[x_{ij}(r),x_{k\ell}(s)] = \left\{
\begin{matrix}
1 & \text{if $j\not=k$ and $i\not=\ell$}\\
x_{i\ell}(rs) &\text{if $j=k$ and $i\not=\ell$}\\
x_{kj}(-sr)& \text{if $j\not=k$ and $i=\ell$}
\end{matrix}
\right.$$ where do these strange relations come from? In fact, the earlier defined elementary matrices $e_{ij}(\lambda)$ satisfy all these relations. Therefore, there is a group homomorphism ${\rm St}_n(R)\to E_n(R)$, which passes to a homomorphism ${\rm St}(R)\to E(R)$ where ${\rm St}(R)$ is the direct limit of all the Steinberg groups.

Typically, elementary matrices satisfy more relations than just the Steinberg relations, but not always. Therefore, we define $K_2(R) = \ker({\rm St}(R)\to E(R))$. This group measures how far off the Steinberg relations are in describing the elementary matrix group of $R$.

$K2$: Matsumoto’s Theorem

How can we compute $K_2(R)$? As usual, for a general ring there are all sorts of different techniques. However, Hideya Matsumoto derived a result that simplifies the computation for fields.

Theorem (Matsumoto). Let $F$ be a field. The group $K_2(F)$ is the abelian group generated by the symbols $(x,y)$ for $x,y\in F^\times$ with the relations

  1. $(x,y)(x’,y) = (xx’,y)$ and $(x,y)(x,y’) = (x,yy’)$ for all $x,x’,y,y’\in F$
  2. $(x,1-x) = 1$ whenever $x\not=1$

Equivalently, we can describe $K_2(F)$ as the quotient of $F\otimes_\Z F$ by the subgroup generated by $x\otimes(1-x)$. We need to be careful and not to get messed up with notation here because typically in working with the symbols in Matsumoto’s theorem and Steinberg symbols in general, we write the group multiplicatively and with the tensor product we write the multiplication additively. It makes more sense to write it multiplicatively because the group operation in $F^\times$ is so written.

Before we actually compute anything with Matsumoto’s theorem, let us prove a basic property of $K_2$ as presented in Matsumoto’s theorem.

Theorem. For any $x\in F$,
$$(x,-x) = 1.$$
Proof. First, we note that for any $x\in F^\times$, we have $(1,x)(1,x) = (1,x)$ and so $(1,x)$ is trivial, and similarly $(x,1)$ is trivial. Therefore, $(x,y)^{-1} = (x,y^{-1})$. We compute:
$$(x,-x) = (x,(1-x)/(1-x^{-1})) = (x,1-x^{-1})^{-1}).$$ On the other hand,
$$(x,1-x^{-1}) = (x,1-x^{-1})(x^{-1},1-x^{-1}) = (1,1-x^{-1}) = 1.$$

This property allows us to show that the Matsumoto symbols are skew-symmetric. Indeed, we observe that $(x,y) = (x,y)(x,-x) = (x,-xy)$. Therefore, $(x,y)(y,x) = (x,-xy)(y,-xy) = (xy,-xy) = 1$. In particular, $(x,x)^2 = 1$.

Now let us return to Matsumoto’s theorem. First, let us do a special case: $\F_3$, the finite field with three elements with $\F_3^\times = \{1,2\} \cong \Z/2$. However, $(1,1),(1,2),(2,1)$ are all trivial. And, $(2,2)(2,1) = (1,1)$, so $(2,2)$ is also trivial. Therefore, $K_2(\F_3) = 0$. The same is true in general.

Theorem. If $F = \F_q$ is a finite field then $K_2(F) = 0$.
Proof. Let $x$ be a generator for $F^\times$, which exists because the multiplicative group of a finite field is cyclic. We just need to show that $(x,x)$ is trivial in Matsumoto’s presentation. If $q$ is even (i.e. $F$ has characteristic two) then $(x,x) = (x,-x)$, which we know is trivial. Therefore, we assume that $q$ is odd.

By bilinearity, $(x,x)^{mn} = (x^m,x^n)$ for any natural numbers $m$ and $n$. But since $(x,x)^2 = 1$, we get
$$(x,x) = (x^m,x^n)$$ whenever $m$ and $n$ are odd. Therefore, if we can find odd numbers $m$ and $n$ such that $x^m = u$ and $x^n = 1-u$, we will have finished the proof. Since odd powers are the same as nonsquares, it suffices to find a $u\in F^\times$ such that $u$ is not a square and $1-u$ is not a square. Now, the map $x\mapsto 1-x$ is a bijection $F^\times-\{0,1\}\to F^\times-\{0,1\}$. Moreover, we observe that in the set $F^\times-\{0,1\}$ there are $(q-1)/2$ nonsquares and $(q-1)/2-1$ squares. Therefore, some nonsquare $u$ is taken to a nonsquare under $x\mapsto 1-x$, and this is the required element.

The Higher $K$-theory of Finite Fields

Dan Quillen provided the correct definition of higher $K$-theory in terms of the plus construction and also the Q-construction. It would take us too far afield in this short post to explain all these concepts, so we will just state the result here. Recall that we have computed $K_1(\F_q)\cong \Z/(q-1)$ and $K_2(\F_q) = 0$. A pattern is repeated in the higher $K$-group:

Theorem (Quillen). Let $\F_q$ be the finite field with $q$ elements. For $n\geq 1$,
$$K_n(\F_q)\cong \left\{
\begin{matrix}
\Z/(q^i-1) & \text{if $n=2i – 1$}\\
0 & \text{if $n$ is even}
\end{matrix}
\right.$$

[1] Grayson, Daniel R. Algebraic K-theory via binary complexes. J. Amer. Math. Soc. 25 (2012), no. 4, 1149—1167.

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