$K_2$ over finite fields and polynomials over finite fields

Question

I am interested in presentations of the group $SL_n(\mathbb{F}_q)$ (and eventually $SL_n(\mathbb{F}_q[t])$).

The standard "Chevalley" presentation of $SL_n(R)$ for a ring $R$ has generators $\{x_{i,j}(r) : i \neq j \in [n], r \in R\}$ and three types of relations:

1. Linearity relations, $x_{i,j}(t) x_{i,j}(u) = x_{i,j}(t+u)$.
2. Commutator relations, $[x_{i,j}(t),x_{j,k}(u)] = x_{i,k}(tu)$ (for $k \neq i$) and $[x_{i,j}(t),x_{k,\ell}(u)] = 1$ (for $i,j,k,\ell$ all distinct).
3. "Diagonal" relations, $h_{i,j}(t) h_{i,j}(u) = h_{i,j}(tu)$ where $h_{i,j}(t) := g_{i,j}(t) g_{i,j}(-1)$ and $g_{i,j} := x_{i,j}(t) x_{j,i}(-t^{-1}) x_{i,j}(t)$ (only defined when $t,u$ are invertible).

There is a related "Steinberg" group which is defined only by the relation sets (1) and (2). According to the following paper:

S. Splitthoff, “Finite presentability of Steinberg groups and related Chevalley groups,” in Contemporary Mathematics, vol. 55.2, S. J. Bloch, R. K. Dennis, E. M. Friedlander, and M. R. Stein, Eds., Providence, Rhode Island: American Mathematical Society, 1986, pp. 635–687. doi: 10.1090/conm/055.2/1862658.

for the cases $R=\mathbb{F}_q[t]$ and $R = \mathbb{F}_q$, the Steinberg and Chevalley groups coincide. (Actually, the former was the result of this paper, the latter was apparently already known.)

Splitthoff phrases these results as the triviality of the "$K_2$ group" for these rings ($K_2(R)$ being defined as the kernel of the surjective map of the Steinberg group onto the Chevalley group over $R$). However, I know absolutely nothing about $K$-theory, and my Googling is not looking promising. Is there any 'elementary' way to see that the diagonal relations are derivable from the linearity and commutator relations over these rings, or must one go learn some more theory?

Answer 1

For the concrete instances that you are considering, the finite field ${\mathbb F}_q$ and the ring of polynomials ${\mathbb F}_q[x]$, the following may help: (i) You have the elementary matrices $e_{ij}(\lambda)\in\text{GL}_n(R)$ for $\lambda$ in a ring $R$ and $i\neq j$, ranging from $1$ to $n$ and these satisfy the linearity condition $e_{ij}(\lambda)e_{ij}(\mu)=e_{ij}(\lambda+\mu)$ from where it follows that $e_{ij}(\lambda)^{-1}=e_{ij}(-\lambda)$. (ii) You also have commutator relations $[e_{ij}(\lambda),e_{k\ell}(\mu)]=\begin {cases} 1 &\text{if $j\neq k$ and $i\neq \ell$}\\ e_{i\ell}(\lambda\mu)&\text{if $j= k$ and $i\neq \ell$}\\ e_{kj}(-\mu\lambda)&\text{if $j\neq k$ and $i= \ell$}. \end{cases}$

Then, for $n\geq 3$ you define the Steinberg group $\text{St}(n,R)$ with generators abstract elements $x_{ij}(\lambda)$ subject to the linearity relations and the first two commutator relations. There is the obvious homomorphism $\varphi:\text{St}(n,R)\rightarrow \text{GL}_n(R)$ sending $x_{ij}(\lambda)$ to $e_{ij}(\lambda)$ whose image is the subgroup generated by all elementary matrices. Next pass to the direct limit over $n\rightarrow\infty$ to a homomorphism $\varphi:\text{St}(R)\rightarrow \text{GL}(R)$, whose kernel is the Milnor $K$-group $K_2(R)$. Its first property is that $K_2(R)$ is the center of the Steinberg group $\text{St}(R)$.

Next, for invertible elements of the ring you have defined the elements $g_{ij}(\lambda)$ and $h_{ij}(\lambda)$. In Chapter 9 of Milnor's classical "Introduction to Algebraic K-Theory" (PUP, 1971) Milnor introduces the Steinberg symbols $\{u,v\}=[h_{ij}(u),h_{ik}(v)]$ and shows that they are skew-symmetric and bimultiplicative. A calculation shows that if both $u$ and $1-u$ are units, then $\{u,1-u\}=1$ and for any unit $\{u,-u\}=1$. A little more work shows that for any field $K_2(F)$ is generated by the symbols $\{u,v\}$. For a finite field ${\mathbb F}_q$ its group of units ${\mathbb F}_q^{\times}$ is cyclic of order $q-1$ and half of the units are squares and the other half non-squares. A quick counting implies that there is a non-square $u$ such that $1-u$ is also a non-square. Now, if $v$ is a generator of the cyclic group ${\mathbb F}_q^{\times}$, write $u=v^i$ and $1-u=v^j$ (both non-squares and thus $i$ and $j$ are both odd). Then, $\{v,v\}^{ij}=\{u,1-u\}=1$ and since $\{v,v\}^2=1$ by skew-symmetry it follows that $\{v,v\}=1$. Hence, the group $K_2({\mathbb F}_q)$ is trivial. I hope this gets you started and then you find the computation for the ring ${\mathbb F}_q[x]$, which is not as straightforward as for the field.

Answer 2

Weibel explains, in Example 5.2.2 of chapter 3 of "The K-book," how to use the Euclidean algorithm and some very concrete linear algebra to show that $K_2(\mathbb{Z})$ is cyclic of order $2$. Weibel also remarks, in Example 5.2.3, that the same technique shows that $K_2(F[t])$ agrees with $K_2(F)$ for every field $F$. This seems like what you're looking for.

Weibel's book is available freely for download from his webpage, here: https://sites.math.rutgers.edu/~weibel/Kbook.html