26
$\begingroup$

What is a simple example, without getting into the mess of triangulated categories, of an additive category that is not abelian?

$\endgroup$
16
$\begingroup$

There've been lots of mildly complicated examples given, but what about the category of even-dimensional vector spaces over a field?

$\endgroup$
10
  • $\begingroup$ The subject is old now. But why this does not work? $\endgroup$ – Sov Mar 5 '18 at 23:48
  • 2
    $\begingroup$ This is not an abelian category because a linear map with odd rank (between two even-dimensional vector spaces) does not have a kernel in this category. $\endgroup$ – Jeremy Rickard Mar 6 '18 at 9:33
  • $\begingroup$ Why can't you have a linear map of odd nullspace between even-dimensional vector spaces? $\endgroup$ – linkhyrule5 Dec 15 '20 at 3:20
  • $\begingroup$ @linkhyrule5 You can, that’s the point. But such a map doesn’t have a kernel in the category of even-dimensional vector spaces, which means that category can’t be abelian. $\endgroup$ – Jeremy Rickard Dec 15 '20 at 7:56
  • 1
    $\begingroup$ @linkhyrule5 $\begin{pmatrix}0&0\\0&0\end{pmatrix}$ doesn't factor uniquely through $k$. And you'll have the same problem for any $k$ that is not injective. $\endgroup$ – Jeremy Rickard Dec 16 '20 at 11:50
11
$\begingroup$

The category of finitely generated modules over a non-Noetherian ring.

The category of filtered modules over a ring is an example given in Gelfand-Manin.

$\endgroup$
6
  • 2
    $\begingroup$ The category of modules over any ring is abelian. You mean the category of finitely generated modules over a non-Noetherian ring. Also, the category of filtered modules (over any ring) will do; you don't need to pass to chain complexes. $\endgroup$ – Matt E Aug 10 '10 at 2:06
  • $\begingroup$ @Matt E. What's the problem with filtered modules? There aren't kernels? Images? $\endgroup$ – Agustí Roig Aug 10 '10 at 6:21
  • $\begingroup$ @Matt E: Thanks. I will make the changes. $\endgroup$ – user977 Aug 10 '10 at 10:54
  • 5
    $\begingroup$ The problem is that images and coimages are not isomorphic. Let's just talk about filtered vector spaces. Let V and W be one dimensional vector spaces. Filter them so that V_i=(0) for $i \leq 0$ and V_i = k for i>0, while W_i=(0) for i<0 and W_i=k for i \geq 0. The isomorphism V \to W is a map of filtered vector spaces. I leave it to you to compute that the cokernel of the kernel is 0 in degree 0, while the kernel of the cokernel is one dimensional. $\endgroup$ – David E Speyer Aug 10 '10 at 16:40
  • 4
    $\begingroup$ Dear Agusti, David has answered your question, so let me just add: filtered modules is a basic example of an additive category admitting kernels, cokernels, images, and coimages, but which is not abelian, because images and coimages don't coincide in general. Another such example is the category of topological vector spaces (or Banach spaces, if you like), over $\mathbb R$ or $\mathbb C$. (In general, filtrations behave a lot like topologies as a structure, and give the same kind of categorical difficulties.) $\endgroup$ – Matt E Aug 10 '10 at 20:18
10
$\begingroup$

In infinite dimensions, all hell breaks loose. For example, neither the category of Banach spaces nor the category of Hilbert spaces, although additive, are abelian.

$\endgroup$
8
$\begingroup$

There are at least two kinds of (interesting) examples.

I. When we can get an abelian category, but have to add more (co)kernels: 1) the category of free modules over a ring; 2) the category of projective modules over a ring; 3) the category of vector bundles on a topological space (if fact, 3 is a particular case of 2). (From 1 or 2 one gets abelian category of all modules over the ring, from 3 — abelian category of sheafs of vector spaces on X.)

II. When we already have (co)kernels ("category is pre-abelian") but not all mono-/epimorphisms are normal. As explained in another answer, an example is the category of filtered modules.

$\endgroup$
2
  • $\begingroup$ The problem with free modules over a ring is that submodules of free modules need not be free? If that's the case, the category of all free modules over a PID is abelian... please correct me if I'm wrong. $\endgroup$ – Bruno Stonek Jun 12 '11 at 14:17
  • $\begingroup$ @BrunoStonek The problem with free modules is the co-kernels, not the kernels. The category of free modules over a PID only has all co-kernels if the PID is actually a field. $\endgroup$ – Thomas Andrews Mar 28 '12 at 15:12
6
$\begingroup$

Another nice example is the category of finite-dimensional vector bundles over a fixed base space (with bundle maps over the identity as morphisms).

If the base space is not too simple (the interval suffices), then this category is not (pre-)Abelian because there are, in general, neither kernels nor cokernels. Intuitively speaking, the obvious candidate for the kernel of a bundle map need not form a vector bundle because its dimension need not be locally constant.

$\endgroup$
3
  • $\begingroup$ what's the problem with this one? $\endgroup$ – Soarer Aug 10 '10 at 16:38
  • $\begingroup$ Sorry if I overlooked something, but why isn't this abelian? $\endgroup$ – Soarer Aug 10 '10 at 17:26
  • $\begingroup$ @Soarer: I have added clarification to my answer. I deleted my previous comment because it contained a false statement. $\endgroup$ – Rasmus Aug 10 '10 at 18:03
5
$\begingroup$

There's a slight modification of an example that almost works but that was deleted: take the category of Hausdorff topological abelian groups. The coimage of a morphism in this category is its image in the ordinary sense, but the image is the closure of the coimage (exercise), so the two don't need to agree in general.

As Matt E mentions in a comment, adding topologies, like adding filtrations, is an easy way to cause this sort of thing to happen. Similarly we could take Hausdorff topological vector spaces, etc.

$\endgroup$
1
  • 6
    $\begingroup$ You can argue that the category is not abelian by providing a morphism which is monic and epic, but not an isomorphism: the identity from $\mathbb{R}$ with the discrete topology and $\mathbb{R}$ with the usual metric topology is such a mono-epi not iso. Any nondiscrete topological abelian group works, too. $\endgroup$ – egreg Jun 18 '14 at 17:54
2
$\begingroup$

A quite explicit example coming from quantum algebra.

Consider the ring $K_h:=\mathbb K[[h]]$ of formal power series with coefficients in the field $\mathbb K$ (take for example $K=\mathbb R)$. Let $\mathcal C_f$ be the category of topologically free $K_h$ modules, i.e. all those $K_h$-modules that are isomorphic to modules of the form $M[[h]]$, denoting by $M$ any $K$-vector space. Morphisms $\varphi : M[[h]]\rightarrow N[[h]]$ are formal power series $\sum_{i\geq 0} \varphi_i h^i$ with $\varphi_i: M\rightarrow N$ morphism of $K$-vector spaces. If $\varphi : M[[h]]\rightarrow N[[h]]$ with $\varphi=\sum_{i\geq 0} \varphi_i h^i$ and $\psi : N[[h]]\rightarrow Q[[h]]$ with $\psi=\sum_{i\geq 0} \psi_i h^i$ are morphisms in $\mathcal C_f$, then their composition is the morphism $\psi\circ\varphi$ with power series expansion

$\sum_{i+j\geq 0}(\psi_j\circ\varphi_i) h^{i+j}$.

$\mathcal C_f$ is additive with biproduct $M[[h]]\oplus_h N[[h]]:=(M\oplus N)[[h]]$; it is not abelian because the cokernel of the inclusion (which is a morphism in $\mathcal C_f$)

$i: M\rightarrow M[[h]]$, $m\mapsto i(m)=(m,0,0,\dots)$

i.e. $coker(i)=hM[[h]]$, is not an object in $\mathcal C_f$. In fact, there exists no $\mathbb K$-vector space $N$ and isomorphism $\rho: hM[[h]] \rightarrow N[[h]]$ in $\mathcal C_f$; if such a morphism $\rho$ existed, then it would not be an isomorphism as any object in $N[[h]]$ of the form $(n,0,0,...)$ does not belong to the image of $\rho$ for $n\neq 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy