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For a finite abelian $p$-group $G$ we have that $$ G \simeq \mathbf{Z}/(p)^{\lambda_1} \oplus \dotsb \oplus \mathbf{Z}/(p)^{\lambda_r} $$ for some positive integers $\lambda_1 \geq \dotsb \geq \lambda_r$. Note that $G$ is uniquely determined by $p$ and this partition $\lambda = (\lambda_1, \dotsc, \lambda_r)$, so let's call $\lambda$ the type of $G$. For types $\lambda$, $\mu$, and $\nu$, define the Hall number $g_{\mu,\nu}^\lambda(p)$ to be the number of normal subgroups $N \mathrel{\triangleleft} G$ of type $\nu$ such that $G/N$ has type $\mu$. These Hall numbers serve as the structure constants of an associative algebra called the Hall algebra.

It turns out that this algebra is commutative, i.e. $g_{\mu,\nu}^\lambda(p) = g_{\nu,\mu}^\lambda(p)$. The proof of this that I'm looking at, following the more general theory in MacDonald's Symmetric Functions and Hall Polynomials, goes like this: You realize that we're looking at the category of finite-length modules over $\mathbf{Z}_p$, the $p$-adic integers. The Prüfer $p$-group $\mathbf{Z}(p^\infty)$ is the injective hull of $\boldsymbol{k} = \mathbf{Z}/(p)$ in this category, and the functor $\mathrm{Hom}({-},\mathbf{Z}(p^\infty))$, via Matlis duality, gives you a bijection of the short exact sequences in question, so $g_{\mu,\nu}^\lambda(p) = g_{\nu,\mu}^\lambda(p)$.

Proving this can also be approached by developing the theory of characters of finite abelian groups, section 3 in particular. But this is really the same approach in a different language: $\mathbf{Z}(p^\infty)$ plays the role of $S_1$ in this context. But in either approach, we're introducing some heavy stuff just to prove a fact about $p$-groups and partitions. Is there a elementary way to prove that $g_{\mu,\nu}^\lambda(p) = g_{\nu,\mu}^\lambda(p)$ in the case of finite abelian $p$-groups?

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  • $\begingroup$ You can replace $\mathbf{Z}\left(p^\infty\right)$ in the duality argument by its "finite approximation" $\mathbf{Z}/p^N\mathbf{Z}$, where $p^N$ is an upper bound on the sizes of your $p$-groups. This makes everything more elementary (it is certainly a lot easier to prove that each group $\mathbf{Z}/p^k\mathbf{Z}$ is isomorphic to its "$N$-dual" group $\operatorname{Hom}\left(\mathbf{Z}/p^k\mathbf{Z},\mathbf{Z}/p^N\mathbf{Z}\right)$ when $k \leq N$). $\endgroup$ – darij grinberg Feb 22 at 3:57
  • $\begingroup$ Then we don't have to bring up the Prüfer group at all. :) But then we have to do some manual labor to show the map $\mathrm{Hom}(G, \mathbf{Z}/p^N\mathbf{Z}) \to \mathrm{Hom}(N, \mathbf{Z}/p^N\mathbf{Z})$ is a surjective. And we've still gotta define $\mathbf{Z}_p$ because these are all $\mathbf{Z}_p$ modules. ... or do you? $\endgroup$ – Mike Pierce Feb 22 at 4:30
  • $\begingroup$ You don't need to define $\mathbf{Z}_p$; you can read "finite abelian $p$-group" for "$\mathbf{Z}_p$-module" (since all of your $\mathbf{Z}_p$-modules are finite). $\endgroup$ – darij grinberg Feb 22 at 4:52
  • $\begingroup$ Do you mean $G \cong \mathbf{Z}/(p)^{\lambda_1} \oplus \dotsb \oplus \mathbf{Z}/(p^r)^{\lambda_r}$? $\endgroup$ – Lord Shark the Unknown Feb 22 at 5:41
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    $\begingroup$ Yes, that's equivalent to what he is writing. (We always have $\left(n\right)^k = \left(n^k\right)$ as ideals.) $\endgroup$ – darij grinberg Feb 22 at 5:50
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The commutativity of the Hall algebra is saying that you can 'turn short exact sequences around', so is essentially equivalent to having a duality. You don't necessarily need the Prüfer group, though. You can take the 'usual' injective $\mathbb Q/\mathbb Z$ for abelian groups. Then $\mathrm{Hom}_{\mathbb Z}(\mathbb Z/p^n\mathbb Z,\mathbb Q/\mathbb Z)\cong\mathbb Z/p^n\mathbb Z$ is clear, sending a homomorphism $f$ to the image of the cyclic generator $f(1)$.

In the 'algebraic' setting, rather than the 'number theoretic' setting, the same result holds. Here you are taking finite dimensional $k[t]$-modules on which $t$ acts nilpotently; equivalently finite dimensional $k[[t]]$ modules. In this case one can instead use the usual vector space duality $D=\mathrm{Hom}_k(-,k)$. When the field $k$ is finite, the corresponding Hall algebra is symmetric.

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  • $\begingroup$ Thank you! And I was just beginning to wonder which other categories give you commutative Hall algebras too. $\endgroup$ – Mike Pierce Mar 18 at 16:07

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