# How should I formalize that there’s no canonical isomorphism between a finite Abelian group and its Pontryagin dual?

The title says it all, but let me repeat.

We all learn that the Pontryagin dual of a finite Abelian group is abstractly isomorphic as groups, but there’s no canonical isomorphism.

I think I understand it, but I don’t know how to formalize this statement, maybe using category theory.

Could someone enlighten me?

If many thinks that this is more suitable to Math.SE, please move this there.

• Someone will probably tell you to move this to Math.SE. But in any case, it amounts to figuring out what "canonical" means. Your question is basically the same as the question: "why is there no canonical isomorphism between" a vector space and its dual", and that might be easier to think about. Oct 1, 2017 at 15:16
• The key point is that every map $f\colon V\to W$ has a dual map $f^*\colon W^*\to V^*$. Having "canonical" isomorphisms $c_V\colon V\to V^*$ should mean that the $c_V$'s are "compatible" with the $f$s and $f^*$s, in the sense that there is a certain diagram that commutes. Oct 1, 2017 at 15:18
• I think the question is fine here and does not need to be moved to MSE, as per the "MO can be for questions you yourself don't know, but think might be standard knowledge for the guy in the office down the hall, who's a specialist in something else" ethos Oct 1, 2017 at 15:20
• @ArunDebray The duality functor is contravariant, so it doesn't make sense to talk about a natural isomorphism with the identity. You can still ask for a relation such as $c_V=f^*.c_W.f$ however, which is what I was getting at. Oct 1, 2017 at 15:20
• @R.vanDobbendeBruyn canonical is not the same as natural. I would say that canonical means "natural for isomorphisms" (eg the centre of a ring is canonically identified with the endomorphism ring of the identity functor on $R$-modules). If you restrict attention to isomorphism then you can convert freely between covariant and contravariant functors. Oct 1, 2017 at 19:15

Suppose we have a family of isomorphisms $\alpha_A\colon A\to A^*$ for all finite abelian groups $A^*$. If $\phi\colon A\to B$ is an isomorphism, we have a dual isomorphism $\phi^*\colon B^*\to A^*$ and thus an isomorphism $(\phi^*)^{-1}\colon A^*\to B^*$. This makes $A^*$ a covariant functor of $A$ on the category of finite abelian groups and isomorphisms, and it is only reasonable to call $\alpha$ canonical if it is natural with respect to this structure. In other words, we should have $(\phi^*)^{-1}\circ\alpha_A=\alpha_B\circ\phi$ for all $\phi$, or $\alpha_A=\phi^*\circ\alpha_B\circ\phi$. In particular, this must hold when $B=A$ and $\phi=n.1_A$ for some $n$ that is coprime to the order of $A$. This means that $(n^2-1).\alpha_A=0$, but $\alpha_A$ is assumed to be an isomorphism, so the exponent of $A$ must divide $n^2-1$. This fails when $A=\mathbb{Z}/5$ and $n=2$, for example, so there is no natural map $\alpha$ as described.
The same argument also shows that there is no natural isomorphism $V\to V^*$ for finite-dimensional vector spaces $V$ over $\mathbb{Z}/p$ provided that $p>3$. The same conclusion holds for $p=2$ or $p=3$ but one needs to use some different choices of $\phi$ to prove it.
On the other hand, if we restrict even further to elementary abelian groups of order $4$, then you can check that there is a natural choice of $\alpha_A$. It sends each nonzero element $a\in A$ to the unique map $\theta\colon A\to\mathbb{Z}/2$ such that $\theta\neq 0$ but $\theta(a)=0$.
• Could you please explain the line "This means that $(n^2 - 1).\alpha_A = 0$"? I don't understand exactly what that is true. Oct 3, 2017 at 18:42