6
$\begingroup$

Consider the category $\mathsf{FinAb}$ of finite abelian groups. The structure theorem tells us that we can write down a skeleton for this category (a set of representatives for the isomorphism classes), which consists of the groups $\mathbb{Z}/n_1 \oplus \cdots \oplus \mathbb{Z}/n_s$ with positive integers satisfying $n_1 | \cdots | n_s$. But this does not tell us anything about the morphisms between finite abelian groups.

Question. Can we improve the structure theorem in such a way that we find an explicit and easy to describe (this also includes the morphisms), countable category $\mathcal{C}$ together with an equivalence of categories $\mathcal{C} \cong \mathsf{FinAb}$?

Of course, "the full subcategory consisting of all $\mathbb{Z}/n_1 \oplus \cdots \oplus \mathbb{Z}/n_s$" is no answer.

Variants:

a) There is an equivalence of categories $\mathsf{FinAb} \cong \mathsf{FinAb}^{op}$, given by $\hom(-,\mathbb{Q}/\mathbb{Z})$. We could require that $\mathcal{C}$ also comes equipped with an explicit anti-equivalence and that $\mathcal{C} \cong \mathsf{FinAb}$ is compatible.

b) Actually $\mathsf{FinAb}$ is a symmetric monoidal (abelian) category with the usual tensor product $\otimes_{\mathbb{Z}}$ (without unit). It would be great if we can set up an equivalence of symmetric monoidal categories $\mathcal{C} \cong \mathsf{FinAb}$; in particular $\mathcal{C}$ should be symmetric monoidal.

c) Probably everything may be reduced to $\mathsf{FinAb}_p$, the category of finite abelian $p$-groups, where $p$ is a prime number.

d) I am also happy with the category of finitely generated abelian groups. This is the initial finitely cocomplete symmetric monoidal abelian category.

$\endgroup$
4
  • $\begingroup$ Isn't a homomorphism going to be specified by a matrix? I guess the hard part is knowing when two homomorphisms are the same... $\endgroup$
    – Zhen Lin
    Commented Apr 22, 2012 at 8:39
  • $\begingroup$ At least for endomorphisms of $\mathbb{Z}/n_1 \oplus \dotsc \oplus \mathbb{Z}/n_s$, a matrix representation is known. See msri.org/people/members/chillar/files/autabeliangrps.pdf $\endgroup$ Commented Apr 22, 2012 at 9:12
  • $\begingroup$ On further reflection I think it's actually easy to tell when two matrices represent the same homomorphism; the hard part is knowing whether a matrix represents a homomorphism in the first place. But isn't the only obstruction to the construction of homomorphisms the orders of the generators? $\endgroup$
    – Zhen Lin
    Commented Apr 22, 2012 at 9:22
  • $\begingroup$ @Zhen: Sure this is only application of universal properties. But this is not quite what I'm looking for. $\endgroup$ Commented Apr 22, 2012 at 11:23

1 Answer 1

7
$\begingroup$

As you notice in a comment, $FinAb$ is the direct sum of the $FinAb_p$, so we need only deal with the latter.

It should not be difficult to describe the full subcategory $S_p$ of $FinAb_p$ spanned by the set $\{\mathbb Z/p^r\mathbb Z:r\geq0\}$, which is a skeleton, by generators and relations as a category enriched in abelian groups. Once you do that, $FinAb_P$ is (equivalent to) what I would write $S^\oplus$, the category whose objects are finite sequences of objects of $S$ and whose morphisms are matrices.

The tensor product "comes" from the functor on $S$ which we can write $\min$ for hopefully obvious reasons. In the generating set for $S$ there will be one map $\alpha_k:\mathbb Z/p^r\mathbb Z\to \mathbb Z/p^{r+1}\mathbb Z$ and one map $\beta_r:\mathbb Z/p^r\mathbb Z\to\mathbb Z/p^{r-1}\mathbb Z$ (which can be taken to be the maps such that $1\mapsto p$ and $1\to 1$, respectively): the Pontryagin duality is induced by a functor on $S$ which is the identity on objects and on endomorphism rings, and which swaps the $\alpha$s and the $\beta$s.

Later. I think $S_p$ can be presented as the quotient of the $\mathbb Z$-span of the category generated by the graph $$ 0\underset{\alpha_0}{\overset{\beta_1}\leftrightarrows} 1\underset{\alpha_1}{\overset{\beta_2}\leftrightarrows} 2\underset{\alpha_2}{\overset{\beta_3}\leftrightarrows} 3\underset{\alpha_3}{\overset{\beta_4}\leftrightarrows} \cdots$$ modulo the relations $$p^rI_r=0, \qquad \alpha_{r-1}\beta_r=pI_r=\beta_{r+1}\alpha_r,$$ with $I_r$ the identity of the object $r$. I did not really check, but the tensor product seems to be the unique $\mathbb Z$-linear functor which acts like $\min$ on objects.

$\endgroup$
3
  • $\begingroup$ I like this decomposition, thanks! Is there some simplical interpretation of $S_p$? $\endgroup$ Commented Apr 22, 2012 at 13:39
  • $\begingroup$ The edit is very interesting, thanks! I will check the details later. $\endgroup$ Commented Apr 24, 2012 at 8:22
  • $\begingroup$ Viewed as a categorical poset, the tensor-esque product would be the greatest lower bound, min, the meet and the coproduct would be the lowest upper bound, max, the join and the relations would be the units and counits of the alpha vs. beta adjunction. $\endgroup$ Commented Aug 13, 2016 at 22:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .