nLab Grothendieck group of a commutative monoid

Idea

The 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 additive 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 topological K-theory group $K(X)$ of $X$.

Motivated by the example of topological K-theory there is a vaguely 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 Grotheniek group construction. For more on this category theoretic operation see at Grothendieck group.

Definition

Definition

(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

$$\big( (a_+, a_-) \sim_1 (b_+, b_-) \big) \;\Leftrightarrow\; \left( \underset{k \in A}{\exists} \left( a_+ + b_- + k = b_+ + a_- + k \right) \right)$$

or equivalently the equivalence relation

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

Write

$$G(A) \coloneqq (A \times A)/\sim$$

for the set of equivalence classes under this equivalence relation. This inherits a binary operation

$$+ \;\colon\; G(A) \times G(A) \longrightarrow G(A)$$

by applying the addition in $A$ on representatives:

$$[a_+,a_-] + [b_+,b_-] \coloneqq [ a_+ + b_+ , a_- + b_- ] \,.$$

This defines the structure of an abelian group

$$(G(A),+)$$

and this is the Grothendieck group of $A$.

This comes with a canonical homomorphism of monoids (semigroups with unit):

$$\array{ A &\overset{\phantom{A} \eta_A \phantom{A} }{\longrightarrow}& G(A) \\ a &\overset{\phantom{AAA}}{\mapsto}& [a,0] } \,.$$

Proposition

(universal property of Grothendieck group)

The Grothendieck group in def. 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 $f = \tilde f\circ \eta_A$:

$$\array{ A \\ {}^{\mathllap{\eta_A}}\downarrow & \searrow^{\mathrlap{f}} \\ G(A) &\overset{\exists ! \tilde f}{\longrightarrow}& B }$$

Proof

First to see that the two equivalence relations in def. 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

$$\begin{aligned} (a_+ + k_1 , a_- + k_1) & = ( a_+ + b_- + k, a_- + b_- + k) \\ & = (b_+ + a_- + k, a_- + b_- + k) \\ & = (b_+ + k_2, b_- + k_2) \end{aligned} \,.$$

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

$$\begin{aligned} a_+ + b_- + k & = a_+ + k_1 + b_- + k_2 \\ & = b_+ + k_2 + a_- + k_1 \\ & = b_+ + a_- + k \end{aligned} \,.$$

Now to see that $(G(A),+)$ is indeed an abelian group:

1. the second equivalence relation also makes it immediate that the neutral element is the class

   $$[0,0] = [a,a]$$

   for all $a \in A$.

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

   $$-[a_+, a_-] = [a_-, a_+] \,,$$

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

$$\tilde f \colon G(A) \longrightarrow B$$

be a homomorphism of abelian groups. Observe from the above that then

$$\begin{aligned} \tilde f([a_+,a_-]) & = \tilde f( [a_+,0] - [a_-, 0] ) \\ & = \tilde f([a_+,0]) - \tilde f([a_-,0]) \\ & = \tilde f(\eta_A(a_+)) - \tilde f(\eta_A(a_-)) \\ & = f(a_+) - f(a_-) \end{aligned}$$

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 $f = \tilde f \circ \eta_A$.

Remark

(Grothendieck group for cancellative monoids)

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

$$\underset{a,b,z \in A}{\forall} \left( \left( a + z = b + z \right) \Rightarrow \left( a = b \right) \right)$$

then, as is immediate from the first of the two equivalence relations in def. , the definition of the Grothendieck group $G(A)$ simplifies to

$$G(A) = (A \times A)/ \sim$$

with

$$\big( (a_+,a_-) \sim (b_+,b_-) \big) \;\Leftrightarrow\; \big( a_+ + b_- = b_+ + a_- \big) \,.$$

Examples

Example

(Grothendieck group of the natural numbers is the integers)

Let $(\mathbb{N}, +)$ be the commutative monoid of natural numbers under addition. By def. 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, remark says that this equivalence relation is simply

$$\big( (a_+,a_-) \sim (b_+,b_-) \big) \;\Leftrightarrow\; \big( a_+ + b_- = b_+ + a_- \big) \,.$$

Let

$$\array{ (G(\mathbb{N}),+) &\longrightarrow& (\mathbb{Z},+) \\ (n_+, n_-) &\mapsto& n_+ - n_- }$$

be the evident homomorphism of abelian groups to the additive group of integers.

This is manifestly surjective. For it to be injective we need that

$$(a_+, a_-) \sim (b_+,b_-)$$

precisely if

$$a_+ - a_- = b_+ - b_- \;\; \in \mathbb{Z} \,.$$

The last condition holds precisely if

$$a_+ + b_- = b_+ + a_- \;\; \in \mathbb{Z}$$

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:

$$(G(\mathbb{N}), +) \simeq \mathbb{Z} \,.$$

Example

Consider the commutative monoid $(\mathbb{Z}^\times, \cdot)$ of non-zero integers under multiplication.

Consider the homomorphism

$$\array{ (G(\mathbb{Z}), \cdots) &\longrightarrow& (\mathbb{Q}^\times, \cdot) \\ ( n_+ ,n_- ) &\mapsto& n_+/n_- }$$

to the non-zero rational numbers under multiplication.

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

$$(a_+, a_-) \sim (b_+, b_-)$$

precisely if

$$a_+/ a_- = b_+ / b_- \;\; \in \mathbb{Q}$$

which is the case precisely if

$$a_+ \cdot b_- = b_+ \cdot a_- \,.$$

Since $(\mathbb{Z}^\times, \cdot)$ is a cancellative monoid, this is indeed the equivalence relation on $G(\mathbb{Z}^\times)$, according to remark .

Example

(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

$$K(X) \coloneqq (G(Vect(X)_{/\sim}), +)$$

is called the topological K-theory group of $X$.

References

See also

• Wikipedia, Grothendieck group