Steinberg relations and symbol computations

Given a field F, Quillen defined K-groups $K_n(F)$ to be the higher homotopy groups of a space $K(F)$. Though apparently the space itself can be easily and explicitly constructed, the K-groups are mysterious gadgets. However the homology groups of the connected component of K(F) containing its base point coincide with the homology groups of the general linear group.

Anyway, back to Quillen K-groups. For $n\leq 2$, these groups coincide with the Milnor K groups $K^M_n(F)$ which are more algebraic in nature and much easier to define. The n-th Milnor K group of a field F is defined as follows :

Let $F^*$ denote the abelian group of non-zero elements of the field F. Then

$K^M_n(F) = (F^*\otimes_{\mathbb{Z}} F^*\otimes_{\mathbb{Z}}\otimes_{\mathbb{Z}} \ldots F^*)/J$

where $J$ is the subgroup generated by tensors $a_1\otimes a_2\otimes \ldots \otimes a_n$ where for some $i\neq j$, $a_i+a_j=0$. The class of the tensor $a_1\otimes a_2\otimes \ldots \otimes a_n$ in $K^M_n(F)$ is denoted $\{a_1,a_2,\ldots,a_n\}$.

Thus for $a\in F^*$, we have $\{\ldots, a, \ldots, 1-a, \ldots \} = 0$. These are the the so-called Steinberg relations.

Thus, we have

• $K^M_0(F) = \mathbb{Z}$
• $K^M_1(F) = F^*$
• $K^M_2(F) = (F^*\otimes_{\mathbb{Z}} F^*)/(a\otimes 1-a)$

Let us do some symbolic computations in $K^M_2(F)$ to get a feel for this object (and also because it is fun!)

The Rules

1. Since $\otimes$ is bilinear, $\{a, b\} + \{a, c\} = \{a, bc\}$.
2. Similarly $\{a, b\} + \{c, b\} = \{ac, b\}$.

So  $\{a,1\} + \{a,1\} = \{a,1\}$ which implies $\{a,1\} =0$. Similarly $\{1,b\}=0$. Using bilinearity repeatedly, $n\{a,b\} = \{a^n, b\} = \{a,b^n\}\forall n\in \mathbb{Z}$.

If $n<0$, maybe this needs some further explanation. For instance if $n=-1$ say, then $0= \{1,b\} = \{a,b\} +\{a^{-1},b\}$..

The Game

I. Show $\{x,-x\}=0$.

This is because  $\{x,-x\} = \{x,-x\} - \{1-x,x\} = \{\frac{x}{1-x},x\}= -\{\frac{x+1}{x}, x\} = - \{1+\frac{1}{x}, x\} = \{1+\frac{1}{x}, \frac{1}{x}\} = 0$.

II. Show $-\{a,b\}=\{b,a\}$.

This is because  $\{a,b\}+\{b,a\} = \{a,b\} + \{a,-a\}+\{b,-b\} + \{b,a\}=\{a,-ab\}+\{b, -ab\}=\{ab,-ab\}=0$.

III. Show $\{a,-1\}=\{a,a\}$.

This is because  $\{a,a\}=\{a,a\}-\{a,-a\} = \{a,a\} + \{a,\frac{-1}{a}\} = \{a,-1\}$.