Introduction to K Theory 1: Grothendieck Group and Ring for modest category with binary operation

Grothendieck group of semigroupUniversal property of Grothendieck groupExistenceUniquenessSome basic propertiesGrothendieck Group of modest category with binary operationGrothendieck ring of semiringUniversal property of Grothendieck RingGrothendieck Ring of modest category with binary operation

 

Grothendieck group of semigroup

Let $(M,\cdot )$$(M,\cdot)$ be a semigroup, we define the Grothendieck group of $M$$M$ to be $K_0(M):=F(M)/⟨D⟩$$K_0(M):=F(M)/\langle D\rangle$, where $⟨D⟩$$\langle D\rangle$ is the subgroup of the free $\mathbb{Z}$$\mathbb{Z}$ module $F(M)$$F(M)$, generated by $x\cdot y-x-y$$x\cdot y-x-y$.

Universal property of Grothendieck group

Proof.

Let $(M,\cdot )$$(M,\cdot)$ be a semigroup, $K_0(M)=F(M)/⟨D⟩$$K_0(M)=F(M)/\langle D\rangle$ its Grothendieck group, and
$i:M\to K_0(M)$$i\colon M\to K_0(M)$ the canonical map sending $m\mapsto [m]$$m\mapsto [m]$.

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Existence

Let $A$$A$ be an abelian group and $f:M\to A$$f\colon M\to A$ a semigroup homomorphism. We define a map

$$\phi :F(M)\to A$$

on the free abelian group $F(M)$$F(M)$ by

$$\phi (\sum _j{n}_j{m}_j)=\sum _j{n}_{j}f(m_j),$$

and extend linearly. For each generator

$$m\cdot n-m-n$$

of the subgroup $⟨D⟩$$\langle D\rangle$, we have

$$\phi (m\cdot n-m-n)=f(m\cdot n)-f(m)-f(n)=f(m)+f(n)-f(m)-f(n)=0,$$

since $f$$f$ is a semigroup homomorphism into an abelian group. Hence

$$⟨D⟩\subseteq \ker \phi ,$$

and $\phi$$\varphi$ descends to a well‑defined group homomorphism

$$\tilde{f}:K_0(M)=F(M)/⟨D⟩\to A$$

satisfying

$$\tilde{f}([m])=f(m).$$

By construction, $\tilde{f}\circ i=f$$\widetilde{f}\circ i = f$.

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Uniqueness

If $\psi :K_0(M)\to A$$\psi\colon K_0(M)\to A$ is any group homomorphism with $\psi \circ i=f$$\psi\circ i = f$, then for each generator $m\in M\subset F(M)$$m\in M\subset F(M)$ we have

$$\psi ([m])=f(m).$$

Since $K_0(M)$$K_0(M)$ is generated by the classes $[m]$$[m]$ and the relations in $⟨D⟩$$\langle D\rangle$ are sent to zero in $A$$A$, $\psi$$\psi$ must coincide with $\tilde{f}$$\widetilde{f}$ on all of $K_0(M)$$K_0(M)$. Thus $\tilde{f}$$\widetilde{f}$ is the unique such homomorphism.

$\boxed{{\displaystyle }}$$\boxed{}$

This universal property tells us that $K_0$$K_0$ is the left adjoint of the forgetful functor from $\mathrm{Ab}$$\mathrm{Ab}$ to category of semigroup.

Some basic properties

Proposition. Each element of the Grothendieck group $K_0(M)$$K_0(M)$ has the form $\hat{x}-\hat{y}$$\hat{x}-\hat{y}$ with $x,y\in M$$x,y\in M$.

Proof. Each element of $K_0(M)$$K_0(M)$ is the linear combination of cosets:

$$m=a_1{\hat{x}}_1+...+a_n{\hat{x}}_n$$

Consider the subset of the index set $I_{+}=\{1\le i\le n:a_i>0\}$$I_+=\{1\le i\le n:a_i> 0\}$ and $I_{-}=\{1\le i\le n:a_i<0\}$$I_{-}=\{1\le i\le n:a_i< 0\}$.

Then

$$m=\sum _{i\in I_{+}}{a}_i{\hat{x}}_i-\sum _{j\in I_{-}}(-a_j){\hat{x}}_j=i\left(\prod _{i\in I_{+}}{x}_i^{a_i}\right)-i\left(\prod _{j\in I_{-}}{x}_j^{-a_j}\right)◻$$

Definition.

Let $\mathcal{C}$$\mathcal{C}$ be a category, the isomorphism class of an object $X$$X$ is denoted as $[X]$$[X]$. A binary operation $\ast$$*$ on $\mathcal{C}$$\mathcal{C}$ is, for each $X,Y\in \mathrm{Ob}(\mathcal{C})$$X,Y\in\mathrm{Ob}(\mathcal{C})$, we have a unique $X\ast Y\in \mathrm{Ob}(\mathcal{C})$$X*Y\in\mathrm{Ob}(\mathcal{C})$. And it is well defined for the isomorphism class, i.e. if $X\cong X^{\prime },Y\cong Y^{\prime }$$X\cong X',Y\cong Y'$, then $X\ast Y\cong X^{\prime }\ast Y^{\prime }$$X* Y\cong X' *Y'$. The operation $\ast$$*$ is associative if $(X\ast Y)\ast Z\cong X\ast (Y\ast Z)$$(X*Y)*Z\cong X*(Y*Z)$. It is commutative if $X\ast Y\cong Y\ast X$$X*Y\cong Y*X$.

It has identity if there exists an $E$$E$ such that $E\ast X\cong X\ast E$$E*X\cong X*E$. It is unique up to isomorphism since if you have $E^{\prime }$$E'$ as well, then $E\cong E\ast E^{\prime }\cong E^{\prime }$$E\cong E*E'\cong E'$.

Example. Consider tensor product on a tensor category, for example, category of set with product.

Grothendieck Group of modest category with binary operation

Definition. Let $\mathcal{C}$$\mathcal{C}$ be a category such that there exists a set $S\subset \mathrm{Ob}(\mathcal{C})$$S\subset\mathrm{Ob}(\mathcal{C})$, for all $X\in \mathrm{Ob}(\mathcal{C})$$X \in \mathrm{Ob}(\mathcal{C})$ , $S\cap [X]\ne \mathrm{\varnothing }$$S\cap[X]\neq\emptyset$, then we say $\mathcal{C}$$\mathcal{C}$ is a modest category. Then all the isomorphism classes form a set, denoted as $[\mathcal{C}]$$[\mathcal{C}]$, with an associative binary operation $\ast$$*$, forming a monoid.

We define $[X]\cdot [Y]:=[X\ast Y]$$[X]\cdot [Y]:=[X*Y]$.

We define the $K_0([\mathcal{C}],\ast )$$K_0([\mathcal{C}],*)$ to be the Grothendieck group of the category with binary operation.

Example.

https://marco-yuze-zheng.blogspot.com/2025/04/compact-oriented-surfaces-as-symmetric.html

https://marco-yuze-zheng.blogspot.com/2025/01/grothendieck-group-of-category-of.html

Let us define category $\mathcal{M}$$\mathcal{M}$ as follows:

The object class are all the modest categories with binary operation, and the morphism class are functors preserving the binary operation. The group homomorphism is defined by

$$F[X\ast Y]:=[F(X\ast Y)]=[F(X)\ast F(Y)]=[F(X)]\cdot [F(Y)]=F[X]\cdot F[Y]$$

Then $K_0$$K_0$ is a functor from $\mathcal{M}$$\mathcal{M}$ to $\mathrm{Ab}$$\mathrm{Ab}$.

Let us consider the category of finite sets with $\coprod$$\coprod$, denoted as $+$$+$.

Then easy to see that $X\cong Y⟺|X|=|Y|$$X\cong Y\iff |X|=|Y|$. The cardinality induces a group isomorphism of $K_0(\mathrm{FinSet},+)\cong \mathbb{Z}$$K_0(\mathrm{FinSet,+})\cong\mathbb{Z}$.

The free vector space functor $F$$F$ preserves colimit, hence we have $[F(X+Y)]=[F(X)\oplus F(Y)]=[F(X)]+[F(Y)]$$[F(X+Y)]=[F(X)\oplus F(Y)]=[F(X)]+[F(Y)]$.

Easy to see it is an isomorphism.

Grothendieck ring of semiring

Definition. Let $(M,+,\cdot )$$(M,+,\cdot)$ be a monoid with a product $\cdot$$\cdot$, both $(M,\cdot )$$(M,\cdot)$ and $(M,+)$$(M,+)$ are commutative monoids and we have $a\cdot (x+y)=a\cdot x+a\cdot y$$a\cdot(x+y)=a\cdot x+a\cdot y$.

Example. Consider $\mathbb{N}$$\mathbb{N}$ and any distributive lattice and any commutative ring.

Proposition. Define $K_0(M,+,\cdot ):=F(M)/⟨D⟩$$K_0(M,+,\cdot):=F(M)/\langle D\rangle$ again, then $K_0(M,+,\cdot )$$K_0(M,+,\cdot)$ is a ring, the multiplication is defined to be

$$[a]\cdot [b]:=[a\cdot b]$$

It is well defined since for $\mathrm{\forall }x,y\in ⟨D⟩,a=a^{\prime }+x,b=b^{\prime }+y,a\cdot b=a^{\prime }{b}^{\prime }+a^{\prime }y+x{b}^{\prime }+xy$$\forall x,y\in \langle D\rangle ,a=a'+x,b=b'+y,a\cdot b=a'b'+a'y+xb'+xy$ and $a^{\prime }y+x{b}^{\prime }+xy\in ⟨D⟩$$a'y+xb'+xy\in\langle D\rangle$.

Example. The Grothendieck ring of $\mathbb{N}$$\mathbb{N}$ is $\mathbb{Z}$$\mathbb{Z}$.

Universal property of Grothendieck Ring

Grothendieck Ring is the left adjoint of the forgetful functor from $\mathrm{CRing}$$\mathrm{CRing}$ to category of semiring.

Grothendieck Ring of modest category with binary operation

Definition.

Let $\mathcal{C}$$\mathcal{C}$ be a modest category with two associative binary operators such that $([\mathcal{C}],+,\cdot )$$([\mathcal{C}],+,\cdot)$ is a semiring, then the Grothendieck ring of $(\mathcal{C},+,\cdot )$$(\mathcal{C},+,\cdot)$ is $K_0([\mathcal{C}],+,\cdot )$$K_0([\mathcal{C}],+,\cdot)$.

Example. Consider the category of finite set with coproduct and product, then the Grothendieck ring is $\mathbb{Z}$$\mathbb{Z}$. Similarly, category of finite dimensional vector space with tensor product and direct sum.

In general, we could consider a topos with product and coproduct, since topos has product-exponential adjoint.

Now let us consider a quite interesting example, Burnside Ring of a Group.

Let $G$$G$ be a finite group and consider the category of $G-\mathrm{Set}$$G-\mathrm{Set}$ with coproduct and product. The $0$$0$ is $\mathrm{\varnothing }$$\emptyset$ and $1$$1$ is $G/G$$G/G$. We call the Grothendieck ring of this category Burnside ring of $G$$G$.

Notice that every $G$$G$-set is a disjoint union of orbits and each orbit is isomorphic to $G/H$$G/H$. $G/H\cong G/H^{\prime }$$G/H\cong G/H'$ as $G$$G$-Set iff $H$$H$ and $H^{\prime }$$H'$ are conjugate subgroups. Hence $[X]$$[X]$ are the classes of conjugate subgroups of $G$$G$. Hence the Grothendieck ring as an abelian group is the free $\mathbb{Z}$$\mathbb{Z}$ module of the number of conjugate subgroups.