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\begin{document}

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\section*{Grothendieck group of a commutative monoid}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{algebra}{}\paragraph*{{Algebra}}\label{algebra}

[[!include higher algebra - contents]]

\hypertarget{group_theory}{}\paragraph*{{Group Theory}}\label{group_theory}

[[!include group theory - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The \emph{Grothendieck group construction} is an explicit presentation of the [[group completion]] of a [[commutative monoid]] to an [[abelian group]]. For a [[cancellative monoid]] it reduces to the age-old construction that turns the additive monoid of [[natural numbers]] into the aditive group of [[integers]], or the multiplicative monoid of non-zero [[integers]] into the multiplicative group of non-zero [[rational numbers]]. But the construction also applies to non-[[cancellative monoids]]. The archetypical application of the construction is to monoids of [[topological vector bundles]] over some [[topological space]] $X$ under [[direct sum of vector bundles]], in which case it yields the \emph{[[topological K-theory]] group} $K(X)$ of $X$.

Motivated by the example of [[topological K-theory]] there is a vaguelly related construction of [[algebraic K-theory]] groups from [[Quillen exact categories]]. Applied to the category of [[topological vector bundles]] this coincides with the Grothendieck group of the monoid of vector bundles, and hence is also called \emph{Grotheniek group construction}. For more on this [[category theory|category theoretic]] operation see at \emph{[[Grothendieck group]]}.

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\begin{defn}
\label{GrothendieckGroupViaQuotientOfCartesianProduct}\hypertarget{GrothendieckGroupViaQuotientOfCartesianProduct}{}
\textbf{([[Grothendieck group of a commutative monoid]])}

Let $(A,+)$ be a [[commutative monoid]] (i.e. a commutative [[semi-group]]).

On the [[Cartesian product]] of underlying sets $A \times A$ (the set of ordered [[pairs]] of elements in $A$), consider the [[equivalence relation]]

\begin{displaymath}
\big(
    (a_+, a_-) \sim_1 (b_+, b_-)
  \big) 
    \;\Leftrightarrow\; 
  \left(
    \underset{k \in A}{\exists}
    \left( 
      a_+ + b_- + k = b_+ + a_- + k
    \right)
  \right)
\end{displaymath}
or equivalently the equivalence relation

\begin{displaymath}
\big(
    (a_+, a_-) \sim_2 (b_+, b_-)
  \big) 
    \;\Leftrightarrow\; 
  \left(
    \underset{k_1, k_2 \in A}{\exists}
    \left(
      (a_+ + k_1, a_- + k_1)
      =
      (b_+ + k_2, b_- + k_2)
    \right)
  \right)
  \,.
\end{displaymath}
Write

\begin{displaymath}
G(A) \coloneqq (A \times A)/\sim
\end{displaymath}
for the set of [[equivalence classes]] under this equivalence relation. This inherits a binary operation

\begin{displaymath}
+ \;\colon\; G(A) \times G(A) \longrightarrow G(A)
\end{displaymath}
by applying the addition in $A$ on representatives:

\begin{displaymath}
[a_+,a_-] + [b_+,b_-] \coloneqq [ a_+ + b_+ , a_- + b_- ]
  \,.
\end{displaymath}
This defines the structure of an [[abelian group]]

\begin{displaymath}
(G(A),+)
\end{displaymath}
and this is the \emph{Grothendieck group} of $A$.

This comes with a canonical [[homomorphism]] of [[monoids]] ([[semigroups]] with [[unitality|unit]]):

\begin{displaymath}
\itexarray{
    A &\overset{\phantom{A} \eta_A \phantom{A} }{\longrightarrow}& G(A)
    \\
    a &\overset{\phantom{AAA}}{\mapsto}& [a,0]
  }
  \,.
\end{displaymath}
\end{defn}
\begin{prop}
\label{UniversalProperty}\hypertarget{UniversalProperty}{}
\textbf{([[universal property]] of Grothendieck group)}

The Grothendieck group in def. \ref{GrothendieckGroupViaQuotientOfCartesianProduct} is well defined, and the homomorphism $A \to G(A)$ satisfies the [[universal property]] of the [[group completion]] of $A$:

Given an [[abelian group]] $B$ and a [[homomorphism]] of commutative [[semi-groups]] ([[commutative monoids]]) $f \colon A \longrightarrow B$ then there is a unique homomorphism of [[abelian groups]] $\tilde A \;\colon\; G(A) \longrightarrow B$ such that $f = \tilde f\circ \eta_A$:

\begin{displaymath}
\itexarray{
    A
    \\
    {}^{\mathllap{\eta_A}}\downarrow & \searrow^{\mathrlap{f}}
    \\
    G(A) &\overset{\exists ! \tilde f}{\longrightarrow}& B    
  }
\end{displaymath}
\end{prop}
\begin{proof}
First to see that the two equivalence relations in def. \ref{GrothendieckGroupViaQuotientOfCartesianProduct} are indeed the same:

If $a_+ + b_- + k = b_+ + a_- + k$ then take $k_1 \coloneqq b_- + k$ and $k_2 \coloneqq a_- + k$ to find that

\begin{displaymath}
\begin{aligned}
    (a_+ + k_1 , a_- + k_1)
    & = 
    ( a_+ + b_- + k, a_- + b_- + k)
    \\
    & =
    (b_+ + a_- + k, a_- + b_- + k)
    \\
    & =
    (b_+ + k_2, b_- + k_2)
  \end{aligned}
  \,.
\end{displaymath}
Conversely, if $(a_+ + k_1 , a_- + k_1) = (b_+ + k_2, b_- + k_2)$ then take $k \coloneqq k_1 + k_2$ to find that

\begin{displaymath}
\begin{aligned}
    a_+ + b_- + k 
      & = 
    a_+ + k_1 + b_- + k_2
    \\
      & =
    b_+ + k_2 + a_- + k_1
    \\
      & = 
    b_+ + a_- + k
  \end{aligned}
  \,.
\end{displaymath}
Now to see that $(G(A),+)$ is indeed an abelian group:

\begin{enumerate}%
\item the second equivalence relation also makes it immediate that the [[neutral element]] is the class

\begin{displaymath}
[0,0] =  [a,a]
\end{displaymath}
for all $a \in A$.


\item with this the second equivalence relation makes it immediate that the [[inverse element]] to any $[a_+, a_-]$ is

\begin{displaymath}
-[a_+, a_-] = [a_-, a_+]
  \,,
\end{displaymath}


\end{enumerate}
That this group is abelian is immediate from the fact that $A$ is assumed to be abelian.

Regarding the universal property: let $B$ be any abelian group and let

\begin{displaymath}
\tilde f \colon G(A) \longrightarrow B
\end{displaymath}
be a homomorphism of abelian groups. Observe from the above that then

\begin{displaymath}
\begin{aligned}
    \tilde f([a_+,a_-])
    & = 
    \tilde f( [a_+,0]  - [a_-, 0] )
    \\
    & = 
    \tilde f([a_+,0]) - \tilde f([a_-,0])
    \\
    & = 
    \tilde f(\eta_A(a_+)) - \tilde f(\eta_A(a_-))
    \\
    & = 
    f(a_+) - f(a_-)
  \end{aligned}
\end{displaymath}
by the linearity of $f$ and the definition of $\eta_A \colon A \to G(A)$.

Conversely, given $f \colon A \to B$ then this equation uniquely defines $\tilde f$ with $f = \tilde f \circ \eta_A$.

\end{proof}
\begin{remark}
\label{GrothendieckGroupForCancellativeMonoids}\hypertarget{GrothendieckGroupForCancellativeMonoids}{}
\textbf{(Grothendieck group for [[cancellative monoids]])}

If $(A,+)$ is a [[cancellative monoid]], in that

\begin{displaymath}
\underset{a,b,z \in A}{\forall}
  \left(
    \left(
      a \cdot z = b \cdot z
   \right)
     \Rightarrow
   \left(
     a = b
   \right)
  \right)
\end{displaymath}
then, as is immediate from the first of the two equivalence relations in def. \ref{GrothendieckGroupViaQuotientOfCartesianProduct}, the definition of the Grothendieck group $G(A)$ simplifies to

\begin{displaymath}
G(A) = (A \times A)/ \sim
\end{displaymath}
with

\begin{displaymath}
\big(
    (a_+,a_-)
      \sim
    (b_+,b_-)
  \big)
     \;\Leftrightarrow\;
  \big(
    a_+ + b_- = b_+ + a_-
  \big)
  \,.
\end{displaymath}
\end{remark}
\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{example}
\label{GrothendieckGroupOfNaturalNumbersUnderAdditionIsTheIntegers}\hypertarget{GrothendieckGroupOfNaturalNumbersUnderAdditionIsTheIntegers}{}
\textbf{(Grothendieck group of the [[natural numbers]] is the [[integers]])}

Let $(\mathbb{N}, +)$ be the [[commutative monoid]] of [[natural numbers]] under [[addition]]. By def. \ref{GrothendieckGroupViaQuotientOfCartesianProduct} its Grothendieck group consists of pairs $(n_+, n_-) \in \mathbb{N} \times \mathbb{N}$ subject to some equivalence relation, and since $(\mathbb{N}, +)$ is [[cancellative monoid|cancellative]], remark \ref{GrothendieckGroupForCancellativeMonoids} says that this equivalence relation is simply

\begin{displaymath}
\big(
    (a_+,a_-)
      \sim
    (b_+,b_-)
  \big)
     \;\Leftrightarrow\;
  \big(
    a_+ + b_- = b_+ + a_-
  \big)
  \,.
\end{displaymath}
Let

\begin{displaymath}
\itexarray{
    (G(\mathbb{N}),+) &\longrightarrow& (\mathbb{Z},+)
    \\
    (n_+, n_-) &\mapsto&  n_+ - n_-
  }
\end{displaymath}
be the evident [[homomorphism]] of abelian groups to the additive group of [[integers]].

This is manifestly [[surjective function|surjective]]. For it to be injective we need that

\begin{displaymath}
(a_+, a_-) \sim (b_+,b_-)
\end{displaymath}
precisely if

\begin{displaymath}
a_+ - a_- = b_+ - b_- \;\; \in \mathbb{Z}
  \,.
\end{displaymath}
The last condition holds precisely if

\begin{displaymath}
a_+ + b_- = b_+ + a_- \;\; \in \mathbb{Z}
\end{displaymath}
which is precisely the above equivalence relation. Therefore the above homomorphism is a [[bijection]] and hence the Grothendieck group of the natural numbers is the [[integers]]:

\begin{displaymath}
(G(\mathbb{N}, +) \simeq \mathbb{N}
  \,.
\end{displaymath}
\end{example}
\begin{example}
\label{}\hypertarget{}{}
Consider the commutative monoid $(\mathbb{Z}^\times, \cdot)$ of non-zero [[integers]] under [[multiplication]].

Consider the homomorphism

\begin{displaymath}
\itexarray{
     (G(\mathbb{Z}), \cdots)
       &\longrightarrow&
     (\mathbb{Q}^\times, \cdot)
     \\
     ( n_+ ,n_- )
        &\mapsto&
     n_+/n_-
  }
\end{displaymath}
to the non-zero [[rational numbers]] under multiplication.

It is immediate that this is surjective. For it to be injective we need that

\begin{displaymath}
(a_+, a_-)
    \sim
  (b_+, b_-)
\end{displaymath}
precisely if

\begin{displaymath}
a_+/ a_- = b_+ / b_- \;\; \in \mathbb{Q}
\end{displaymath}
which is the case precisely if

\begin{displaymath}
a_+ \cdot b_- = b_+ \cdot a_-
  \,.
\end{displaymath}
Since $(\mathbb{Z}^\times, \cdot)$ is a [[cancellative monoid]], this is indeed the equivalence relation on $G(\mathbb{Z}^\times)$, according to remark \ref{GrothendieckGroupForCancellativeMonoids}.

\end{example}
\begin{example}
\label{}\hypertarget{}{}
\textbf{([[topological K-theory]])}

Let $X$ be a [[topological space]] and let $(Vect(X)_{/\sim}, \oplus)$ be the monoid of [[isomorphism classes]] of [[topological vector bundles]] on $X$ with addition induced from the [[direct sum of vector bundles]]. (This is in general not a [[cancellative monoid]]). Then the Grothendieck group

\begin{displaymath}
K(X) \coloneqq (G(Vect(X)_{/\sim}), +)
\end{displaymath}
is called the \emph{[[topological K-theory]]} group of $X$.

\end{example}
\hypertarget{references}{}\subsection*{{References}}\label{references}

See also

\begin{itemize}%
\item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Grothendieck_group}{Grothendieck group}}

\end{itemize}
[[!redirects Grothendieck groups of commutative monoids]]

[[!redirects Grothendieck group construction on commutative monoids]]



\end{document}