3
$\begingroup$

The group completion (aka Grothendieck group) of an abelian monoid $M$ is an abelian group $G(M)$ with a homomorphism $\iota:M \to G(M)$ of monoids satisfying the following universal property: for every homomorphism $f:M \to H$ whose target is an abelian group there exists a unique homomorphism $\phi:G(M)\to H$ of groups such that $\phi\circ\iota=f$.

By the universal property of $G$, every homomorphism $f:M \to N$ of monoids uniquely gives rise to a group homomorphism $G(f):G(M) \to G(N)$.

Of course, this concept is used to define $K$-theory: if $\mathcal V_X$ is the abelian monoid of the isomorphism classes of real vector bundle over a topological space $X$, then $KO(X):=G(\mathcal V_X)$.

In general, the homomorphism $\iota:M \to G(M)$ is not injective. (e.g. $M=\mathcal V_{\mathbb{CP}(1)}$.)

Questions:

  1. Is there any criterion which implies (or is equivalent to) the injectivity of the map $\iota: M\to G(M)$? How about the cases when $M=\mathcal V_X$?
  2. Is there any criterion for a homomorphism $f:M \to N$ of monoid which implies the injectivity of the group homomorphism $G(f):G(M) \to G(N)$? How about the cases when $f:=F^*:\mathcal V_X \to \mathcal V_Y$? (Here $F:Y \to X$ is a continuous map.)
$\endgroup$
2
$\begingroup$

Injectivity of the map $M \to G(M)$ is equivalent to cancellability in the monoid $M$: that $x + y = x + z$ implies $y = z$. Indeed, $G(M)$ may be constructed as the set of equivalence classes of pairs $(u, v)$ ("think $u - v$") where $(u, v) \sim (w, x)$ iff $\exists_t \; t + u + x = t + w + v$ in $M$; the existential condition is needed to establish transitivity. The universal map $i: M \to G(M)$ takes $u \in M$ to the class of $(u, 0)$. Given cancellabity in $M$ and $(u, 0) \sim (v, 0)$, we have $t + u = t + v$ for some $t$, whence $u = v$ by cancelling $t$; this shows injectivity. Conversely, if $M \to G(M)$ is injective and $t + u = t + v$ in $M$ for some $t$, then

$$i(t) + i(u) = i(t + u) = i(t + v) = i(t) + i(v)$$

in $G(M)$, whence $i(u) = i(v)$ since the group $G(M)$ is cancellable, whence $u = v$ in $M$ since $i$ is injective. This shows $M$ is cancellable.

I don't know of a very easy necessary and sufficient condition on $f: M \to N$ for $G(f)$ to be injective. Given that $N$ be cancellable, it is necessary and sufficient that $f$ be injective. (Injectivity of $f$ alone is not sufficient: consider the inclusion map of multiplicative monoids $(0, 1] \to [0, 1]$, where $G([0, 1])$ is the trivial group since $0$ is absorbing in $[0, 1]$. Similarly, the homomorphism $[0, 1] \to \{1\}$ shows that injectivity of $f$ isn't necessary either.) A slightly more sophisticated sufficient condition is that $f$ be injective and that $\forall_{t \in N} \exists_{s \in N, m \in M} \; s + t = f(m)$. But it's hard to say how useful that would be.

I'd have to think a little on the topological $K$-theory sub-question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.