Cherry Kearton, Bayer-Fluckiger and others have results that say the monoid of isotopy classes of smooth oriented embeddings of $S^n$ in $S^{n+2}$ is not a free commutative monoid provided $n \geq 3$. The monoid structure I'm referring to is the connect sum of knots.


Bayer-Fluckiger has a result in particular that says you can satisfy these equations $$a+b=a+c, \ \ \ \ b \neq c$$ 
where $a,b,c$ are isotopy classes of knots and $+$ is connect sum.


When $n=1$ it's an old result of Horst Schubert's that the monoid of knots is free commutative on countably-infinite many generators. 


What I'm wondering is, does anyone have an idea of how difficult it might be to compute the structure of the group completion of the monoid of knots, say, for $n \geq 3$? That's not really my question for the forum, though. 


It's this: Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious? Meaning, one where rather minimal amounts of information are required to compute the group completion? Presumably there are examples where it's painfully difficult to say anything about the group completion? For example, can it be hard to say if there's torsion in the group completion?



Computing the structure of the group completion of an abelian monoid, how hard can it be?

> 
> Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious?
> 
> 
> 


This happens all the time in K-theory $K^0(X)$, both algebraic and topological. Perhaps it is even the reason that K-theory is a useful tool.


For a striking algebraic example, take $X = \mathbb{A}^n\_k$ where $k$ is a field. Then $K^0(X)$ is the group completion of the commutative monoid $M$ of isomorphism classes of finitely generated projective modules over $R = k[x\_1, \ldots, x\_n]$. In 1955 Serre asked whether every such module was free, i.e., whether $M = \mathbb{N}$. This question became known as [Serre's conjecture](http://en.wikipedia.org/wiki/Quillen-Suslin_theorem). Serre proved in 1957 that every finitely generated projective $R$-module is *stably* free, i.e., $K^0(X) = \mathbb{Z}$. However, it was not until 1976 that Quillen and Suslin independently proved Serre's original conjecture. So between 1957 and 1976, $M$ was an example of a commutative monoid whose group completion was known but which itself was not known. This is only a historical example, because $M = \mathbb{N}$ turns out to be very simple; however, it illustrates the difficulty of the question in general.


A topological example where the commutative monoid is not so simple is given by $KO^0(S^n)$. Let us take $n$ congruent to 3, 5, 6, or 7 modulo 8, so that $KO^0(S^n) = \mathbb{Z}$ by Bott periodicity (the generator being given by the trivial one-dimensional real vector bundle). Let $T$ be the tangent bundle to $S^n$. In $KO^0(S^n)$, of course, the class of $T$ is equal to its dimension $n$. But if we let $M$ be the commutative monoid of isomorphism classes of finite-dimensional real vector bundles on $S^n$ (so that $KO^0(S^n)$ is the group completion of $M$) then the class of $T$ is not equal to the class of the trivial $n$-dimensional vector bundle unless $S^n$ is parallelizable, which only happens when $n$ is equal to (0 or 1 or) 3 or 7. So for all other values of $n$, $M$ is not simply $\mathbb{N}$; there are extra vector bundles which get killed by the group completion process. Understanding these monoids $M$ for all $n$ amounts to understanding the homotopy groups of all the groups $O(m)$, which I expect is not much easier than understanding unstable homotopy groups of spheres.


Finally, Pete's example of the monoid of cardinalities of at most countable sets and its absorbing element also makes an appearance in K-theory; here it is called the [Eilenberg swindle](http://en.wikipedia.org/wiki/Eilenberg_swindle) and it explains why we restrict ourselves to *finitely-generated* projective modules.


