Grothendieck Group – 解析タネル

To give a functor from a category of commutative monoid to the category of abelian group, namely,

$Gr: CMon\to Ab$

let us construct the abelian group for given $M\in CMon$ , define $Gr(M):=(M\times M)/\sim$ with equivalent relation of

$(x,y)\sim (x',y')\Leftrightarrow \exists k\in M\ s.t.\ x+y'+k=x'+y+k$

and universal property characterized by the following commutative diagram (for given and constructed monoid homomorphism $i_M$ and for any monoid homomorphism $f$ , there uniquely exists abelian homomorphism $Gr(M)\to N$ that commutes the diagram).

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By this construction, there’s the identity element $[0,0]\in Gr(M)$ and for each element $[x,y]\in Gr(M)$ we have unique inverse $[y,x]$ and exposes other functorial properties.

To confirm all those properties are valid on this construction, we begin with checking the compatibility of additive operation in the Cartesian product of commutative monoid $M\times M$ and in that of $Gr(M)$ , namely,

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Here q is the quotient map. To verify this, we just show the equation $[m_1,m_2]+[n_1,n_2]=[m_1+n_1,m_2+n_2]$ for any $m_j,n_j\in M (j=1,2)$ is valid.

Let $(u_1,u_2),\ (v_1,v_2), (w_1,w_2)$ represent $[m_1,m_2],[n_1,n_2],[m_1+n_1,m_2+n_2]$ respectively. Then the equation above reduces to the equation

$\exists k\in M,\ s.t.\ u_1+v_1+w_2+k=u_2+v_2+w_1+k \qued \cdots \qued (*)$

Let $l_1,l_2,l_3$ be corresponding elements of M to $w_1, u_1, v_1$ in such a way that:

$w_1+m_2+n_2+l_1 = w_2+m_1+n_1+l_1$ , $u_1+m_2+l_2=u_2+m_1+l_2$ and $v_1+n_2+l_3=v_2+n_1+l_3$ . Set $l=l_1+l_2+l_3$ and $k=m_1+n_1+l$ , then we have the equation (*) ■

With this compatibility, we can immediately verify $[m_1,m_2]+[m_2,m_1]=[0,0]$ .

Now define $i_M: M\to Gr(M)$ by $i_M(x)=[x,0]$ . We will now verify the following proposition.

Prop. $i=i_M$ is injective if and only if $M$ has cancellative property, i.e.

$\forall a\in M,\forall (x,y)\in M\times M,(a+x,a+y)\in\Delta\Leftrightarrow (x,y)\in\Delta$

Proof.
Suppose M is cancellative, then assuming $[a,0]=[a',0]$ is equivalent condition to $a=a'$ . Conversely when $i_M$ is injective, then if there exists a non-cancellative element $l\in M$ for $a,a'$ , i.e. $a+l=a'+l$ but $a\neq a'$ . $i_M$ being injective can be restated by $\exists k\in A,a+k=a'+k\Leftrightarrow a=a'$ , obviously contradict ■

Next let’s show another important property that the Grothendieck construction is idempotent on an abelian group, to be precise, restate in a form of proposition as following.

Prop. $A\in Ab\Leftrightarrow A\overset{i_A}{\to} Gr(A)$ is bijective
Proof.
$(\Leftarrow)$ $\forall [x,y]\in Gr(A),\exists a\in A,[x,y]=[a,0]$ . Then $\exists k\in A$ s.t. $x+k=y+a+k$ . k can be canceled and we have $x=y+a$ . Put $x=0$ , then a is the inverse of y.
$(\Rightarrow)$ Let $(u,v)$ represents $[x,y]\in Gr(A)$ . We have $x+v+k=y+u+k (\exists k\in A)$ . By canceling k, we have $x=y+u-v$ , which indicates $[x,y]=[u-v,0]$ ■

Another construction:

We cannot avoid another construction of the Grothendieck group by means of completion of group, because not only of the historical importance, but we can also need this for ring construction for the basis of K-theory.

For given $M\in CMon$ , denote the free abelian group $F(M)$ generated by the elements of M as the free basis, in which the element is denoted by $[m]$ for $m\in M$ . Let $R(M)$ be the subgroup of $F(M)$ generated by an element of the form $[m+n]-[m]-[n]$ . We define the quotient group $Gr'(M)=F(M)/R(M)$ .
Our claim is $Gr(M)\cong Gr'(M)$ .

This isomorphism (and the universal property from the former construction) is verified by proving following proposition.

Prop. The properties of free generator constructed Grothendieck group $Gr'(M)$ .
(a) $Gr'(M)$ is composed of a form of element $[m]-[n]$ .
(b) For $m,n\in M$ , $[m]=[n]\in Gr'(M)\Leftrightarrow m+p=n+p\ (\exists p\in M)$

Proof.
(a) By expressing the element $[m]\in Gr'(M)$ as the finite sum $[m]=\sum [m_j]$ , we can immediately get the equation

$[m]=[\sum m_j]$

And then adding up the differences of generators compose the given form.

(b) When there exists $p\in M$ s.t. $m+p=n+p$ , this is obvious by the cancellation of $[p]$ . Conversely if $[m]-[n]=0$ , then

$\begin{array}{lcl} [m]-[n] &=& \sum ([a_i+b_i]-[a_i]-[b_i]) - \sum ([c_i+d_i]-[c_i]-[d_i]) \\ \therefore && [m] + \sum([a_i]+[b_i]) + \sum[c_i+d_i]= [n] + \sum [a_i+b_i] + \sum ([c_i]+[d_i]) \end{array}$

Observing the number of generators of free abelian group of the form $\sum ([a_i]+[b_i])$ and $\sum [a_i+b_i]$ , we conclude that we can remove brackets and obtain the results ■

Now consider the following commutative diagram:

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Here $f:M\to Gr'(M)$ is defined by $f(m)=[m]$ , and $h:Gr(M)\to Gr'(M)$ is defined by $h[m,n]=[m]-[n]$ .
With the previous proposition, we can see $h$ is surjective by (a) and injective by (b), then we have $Gr(M)\cong Gr'(M)$ .

Finally, we’ll view $Gr:CMon\to Ab$ as a functor. Let A,B be elements of $CMon$ . For $id:A\to A$ , $Gr(id)$ is induced identity in $Gr(A)$ because of the uniqueness property:

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Also the composite property is verified with the uniqueness property by replacing the map and range and extending the diagram above.