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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.
(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
or equivalently the equivalence relation
Write
for the set of equivalence classes under this equivalence relation. This inherits a binary operation
by applying the addition in $A$ on representatives:
This defines the structure of an abelian group
and this is the Grothendieck group of $A$.
This comes with a canonical homomorphism of monoids (semigroups with unit):
(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 $f = \tilde f\circ \eta_A$:
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
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
Now to see that $(G(A),+)$ is indeed an abelian group:
the second equivalence relation also makes it immediate that the neutral element is the class
for all $a \in A$.
with this the second equivalence relation makes it immediate that the inverse element to any $[a_+, a_-]$ is
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
be a homomorphism of abelian groups. Observe from the above that then
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 $f = \tilde f \circ \eta_A$.
(Grothendieck group for cancellative monoids)
If $(A,+)$ is a cancellative monoid, in that
then, as is immediate from the first of the two equivalence relations in def. , the definition of the Grothendieck group $G(A)$ simplifies to
with
(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
Let
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
precisely if
The last condition holds precisely if
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:
Consider the commutative monoid $(\mathbb{Z}^\times, \cdot)$ of non-zero integers under multiplication.
Consider the homomorphism
to the non-zero rational numbers under multiplication.
It is immediate that this is surjective. For it to be injective we need that
precisely if
which is the case precisely if
Since $(\mathbb{Z}^\times, \cdot)$ is a cancellative monoid, this is indeed the equivalence relation on $G(\mathbb{Z}^\times)$, according to remark .
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
is called the topological K-theory group of $X$.
See also
Last revised on November 22, 2023 at 21:47:42. See the history of this page for a list of all contributions to it.