Computing the structure of the group completion of an abelian monoid, how hard can it be? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:15:01Z http://mathoverflow.net/feeds/question/13942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca Computing the structure of the group completion of an abelian monoid, how hard can it be? Ryan Budney 2010-02-03T07:24:35Z 2010-02-04T13:44:27Z <p>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.</p> <p>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.</p> <p>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. </p> <p>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. </p> <p>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?</p> http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca/13961#13961 Answer by Pete L. Clark for Computing the structure of the group completion of an abelian monoid, how hard can it be? Pete L. Clark 2010-02-03T11:07:06Z 2010-02-03T11:18:52Z <p>(Warning: the examples that I give here are all quite trivial compared to the motivating example.)</p> <p>In general the group completion of a commutative monoid can have a much simpler structure than the monoid itself. An extreme example is the case of a monoid $M$ with an <a href="http://en.wikipedia.org/wiki/Absorbing%5Felement" rel="nofollow">absorbing element</a>, i.e., an element $z$ with $z*x = x*z = z$ for all $x \in M$. Then the group completion will just be the trivial group.</p> <p>There are natural examples of monoids with absorbing elements. For instance, on p.5 </p> <p><a href="http://math.uga.edu/~pete/settheorypart2.pdf" rel="nofollow">http://math.uga.edu/~pete/settheorypart2.pdf</a></p> <p>I give the example of the commutative monoid of cardinalities of at most countable sets. This is the usual natural numbers together with an additional (absorbing) point at infinity. It has the natural structure of a semiring, so it is somewhat disappointing that its ring completion is trivial.</p> <p>More generally, if you take a submonoid $M$ (in particular, a subset!) of the cardinal numbers under addition such that $M$ contains infinite cardinals, then you need not have an absorbing element but nevertheless the group completion will be trivial. </p> <p>This essentially amounts to the example of a totally ordered set $(X,\leq)$ with a least element $0$ made into a commutative monoid via $x+y = \max(x,y)$. </p> <p><b>Addendum</b>: As Yemon Choi has pointed out below, a yet weaker condition for a commutative monoid to have trivial group completion is that $x+x = x$ for all $x \in M$. There are very rich classes of monoids satisfying this property!</p> http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca/13979#13979 Answer by Reid Barton for Computing the structure of the group completion of an abelian monoid, how hard can it be? Reid Barton 2010-02-03T15:01:49Z 2010-02-03T15:01:49Z <blockquote> <p>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?</p> </blockquote> <p>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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Quillen-Suslin_theorem" rel="nofollow">Serre's conjecture</a>. Serre proved in 1957 that every finitely generated projective $R$-module is <em>stably</em> 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.</p> <p>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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Eilenberg_swindle" rel="nofollow">Eilenberg swindle</a> and it explains why we restrict ourselves to <em>finitely-generated</em> projective modules.</p> http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca/13981#13981 Answer by David Speyer for Computing the structure of the group completion of an abelian monoid, how hard can it be? David Speyer 2010-02-03T15:16:23Z 2010-02-03T15:16:23Z <p>This example is similar to Reid's: If $G$ is a finite group, then $K^0(G-\mathrm{rep})$ is just the class functions on $G$. But the question of whether a specific class function is the character of a representation, or only of a virtual representation, can be very hard. In a sense, Mark Haiman got tenure at Berkeley for proving that <a href="http://arxiv.org/abs/math.AG/0010246" rel="nofollow">certain class functions</a> on $S_n$ were characters.</p> http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca/13988#13988 Answer by Tom Leinster for Computing the structure of the group completion of an abelian monoid, how hard can it be? Tom Leinster 2010-02-03T15:54:39Z 2010-02-03T16:02:27Z <p>Here's a strange result that can help in computing the group completion of a commutative monoid.</p> <p>Let $M$ be a commutative monoid. Call an element $h \in M$ <strong>high</strong> if for all $x \in M$, there exists $y \in M$ such that $h = x + y$. Write $H(M)$ for the set of high elements of $M$.</p> <p>Examples: </p> <ul> <li>If $M$ is a group then $H(M) = M$ (and conversely).</li> <li>Any join-semilattice (i.e. a poset in which every finite subset has a least upper bound) can be viewed as a commutative monoid $M$, with the least upper bound of two elements as $+$ and the least element as $0$. Then $H(M)$ has at most one element, which is the greatest element if such exists.</li> <li>If $M = \mathbb{N}$, with the usual addition, then $H(M) = \emptyset$.</li> </ul> <p><strong>Proposition</strong> If $H(M) \neq \emptyset$ then $H(M)$ is a group, under the same binary operation $+$ as $M$, <em>but not necessarily the same zero</em>.</p> <p>For a rather trivial example of why the zero might not be the same, consider a nontrivial join-semilattice with a greatest element. For a proof and nontrivial examples, see <a href="http://arxiv.org/abs/math.CT/0212377" rel="nofollow">this paper</a> by Marcelo Fiore and me. (The proof's in section 3.)</p> <p>Now: </p> <p><strong>Theorem</strong> $H(M)$ is, if not empty, the group completion of $M$.</p> <p>How does this work? Write $z$ for the zero element of $H(M)$. Then there is a monoid homomorphism $\pi = z + (\ ): M \to H(M)$. It's not too hard to show that every homomorphism from $M$ to a group factors uniquely through $\pi$. Indeed, given a map $\phi: M \to A$, with $A$ a group, the corresponding map $\bar{\phi}: H(M) \to A$ is simply the restriction of $\phi$.</p> <p>The theorem only helps when there's at least one high element, though. There are nontrivial situations when there are no high elements, as the example above of $\mathbb{N}$ illustrates.</p> http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca/14071#14071 Answer by Oscar Randal-Williams for Computing the structure of the group completion of an abelian monoid, how hard can it be? Oscar Randal-Williams 2010-02-03T22:36:28Z 2010-02-03T22:36:28Z <p>In a paper (<a href="http://arxiv.org/abs/0905.2855" rel="nofollow">here</a>) with Soren Galatius, we compute the topological group completions of certain topological monoids made up of moduli spaces of surfaces with various structures. Taking $\pi_0$ of these statements leads to examples of such group completions of discrete monoids.</p> <p>As a particular example, take the discrete monoid <code>$\mathcal{M} := \coprod_{g \geq 0} [\Sigma_{g, 1}, \partial; Y, *] / \Gamma(\Sigma_{g, 1})$</code>. Here the square brackets denotes the set of homotopy classes of maps from the genus $g$ surface with one boundary component $\Sigma_{g,1}$ to a path connected space $Y$ (sending the boundary of the surface to a basepoint $* \in Y$), and we quotient this set by the action of the mapping class group of the surface (rel boundary). The monoid structure is by pair-of-pants gluing of surfaces. This is a fairly complicated monoid, but its group completion turns out to be $$MTSO(2)_0(Y),$$ the degree zero part of a certain homology theory applied to the space $Y$. (It is the homology theory associated to the spectrum occurring in the Madsen--Weiss theorem.) In particular it only sees $Y$ "stably", so if you do something drastic like plus-construct $Y$, the group completion does not change (but the monoid certainly does!).</p> <p>Thus for example if you take $Y$ to be the Poincare sphere (which plus-constructs to $S^3$), one finds that the group completion is simply $\mathbb{Z}$ (which is always in there: there is a surjection $\mathcal{M} \to \mathbb{N}$ that sends a surface to its genus). I don't know if one can see this directly from the monoid.</p> http://mathoverflow.net/questions/13942/computing-the-structure-of-the-group-completion-of-an-abelian-monoid-how-hard-ca/14142#14142 Answer by Allen Hatcher for Computing the structure of the group completion of an abelian monoid, how hard can it be? Allen Hatcher 2010-02-04T13:44:27Z 2010-02-04T13:44:27Z <p>An ultra-classical example: the failure of unique factorization in algebraic number fields. Here one looks at the multiplicative monoid of nonzero algebraic integers in a finite extension field of $\mathbb Q$. Factoring out units gives a quotient monoid $M$, and this is free (abelian) on the irreducible elements exactly when the unique factorization property holds. The monoid $M$ embeds in the ideal group, the free abelian group generated by the prime ideals, and the group completion of $M$ is the finite-index subgroup of the ideal group generated by the principal ideals. This subgroup is also free, but without a canonical basis. One can think of this subgroup as a somewhat skewed lattice in the ideal group, and $M$ is the intersection of this lattice with the positive "orthant" of the ideal group. </p> <p>I'm not an expert on this stuff, so please correct any inaccuracies in what I said above. I gather that the structure of $M$ as a monoid can be rather complicated, even though it's a submonoid of a free monoid. Perhaps this complexity is why the monoid structure seems rarely to be discussed explicitly. Does anyone know any references describing the monoid structure?</p>