# (Short) Exact sequences with no commutative diagram between them

This question was asked by a student (in a slightly different form), and I was unable to answer it properly. I think it's quite interesting.

The problem is to produce an example of the following situation: find a short exact sequence $$0 \to X_1 \to X_2 \to X_3 \to 0$$ (in some category of your choice), and a second exact sequence $$0 \to Y_1 \to Y_2 \to Y_3 \to 0$$ such that there are isomorphisms $X_n \cong Y_n$ for all $n$, BUT in such a way that there is no commutative diagram whatsoever between the two sequences, with the vertical maps being isomorphisms.

It is impossible to find such an example in the category of vector spaces, or of finitely-generated abelian groups. I don't know about the general case, though.

I would be grateful for any example, but would be disappointed if the chosen category were constructed specifically to answer the problem.

EDIT / COMMENT: in the first version of this question I was asking for sequences which could potentially be infinite. Some great examples came in the comments straightaway. I'm very thankful for them, but I recall only now that the student's original question was about short exact sequences, so I've edited accordingly. (I'm sorry for the confusion, the student asked me this question several months ago, and I was posting only now, for some reason... and got it wrong. I'm very happy to know about the examples involving infinite sequences, though.)

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I'm not sure what you mean about f.g. abelian groups, but if you consider a bi-infinite sequence with $X_n=Z/4Z$ for all integers $n\in Z$, you can choose either all morphisms to be multiplication by 2, or all morphisms to be alternatively 0 or identity, in order to get non-isomorphic exact sequences. Maybe your remark about f.g. abelian groups concerns finite sequences? –  YCor Feb 18 '14 at 15:47
And in the category of vector spaces, you can let $X_n=Y_n=V$ for all $n$ (where $V$ is a fixed vector space)and alternate $0$ and identity in two different ways to get exact sequences $X_\bullet$ and $Y_\bullet$ with no possible isomorphism between them. –  Dag Oskar Madsen Feb 18 '14 at 16:06
To both of you: oops, I remember now that the question was definitely about finite sequences. Let me edit the question (and think a little first). I have to say that the student came to me with this question several months ago, and I was only now putting on MO... –  Pierre Feb 18 '14 at 16:24
All examples satisfying the conditions in the edited version will also provide an answer to the following question: mathoverflow.net/questions/119475/… –  Dag Oskar Madsen Feb 18 '14 at 19:59

$\def\ZZ{\mathbb{Z}}$ This can happen in finitely generated abelian groups. Let $p$ be prime, set $G = \ZZ/p^2 \oplus \ZZ/p$ and set $H = \ZZ/p^3 \oplus \ZZ/p^2 \oplus \ZZ/p$. Then there are two non-isomorphic short exact sequences $0 \to G \to H \to G \to 0$. The first one is the sum of the extensions: $$\begin{matrix} 0 & \to & \ZZ/p & \to & \ZZ/p^3 & \to & \ZZ/p^2 &\to& 0 \\ 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 & \to & 0 &\to& 0 \\ 0 & \to & 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 \\ \end{matrix}$$ and the second is the sum of $$\begin{matrix} 0 & \to & \ZZ/p^2 & \to & \ZZ/p^3 & \to & \ZZ/p &\to& 0 \\ 0 & \to & 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 &\to& 0 \\ 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 & \to & 0 \\ \end{matrix}$$

Let's see that these are not isomorphic. Write $\alpha$ for the map $G \to H$. In the first extension, $\alpha(p G) \cap p^2 H = (0)$; in the second extension, $\alpha(pG) = p^2 H$.

In any reasonable category, isomorphism classes of extensions $0 \to X \to ?? \to Z \to 0$ are classified by the orbits of $\mathrm{Aut}(X) \times \mathrm{Aut}(Z)$ on $\mathrm{Ext}^1(Z,X)$. You are asking for cases where there is more than one orbit which makes the center term isomorphic as an abstract element of the category. There is no reason this shouldn't happen, so I would expect it to happen basically any time there are nontrivial extension groups available.

To add some more high level context, the analogous example works in $k[t]$-modules. (That is to say, extending $k[t]/t^2 \oplus k$ by itself to get $k[t]/t^3 \oplus k[t]/t^2 \oplus k$.) In general, for any partition $\lambda$, let $M(\lambda)$ be the $k[t]$ module $\bigoplus_i k[t]/t^{\lambda_i}$. Then $\mathrm{Ext}^1(M(\mu), M(\lambda))$ is stratified into locally closed pieces according to the isomorphism type of the extension, and the number of components where the extension is isomorphic to $M(\nu)$ is the Littlewood-Richardson number $c_{\lambda \mu}^{\nu}$. This example is $c_{(21) (21)}^{(321)}=2$. Since $Aut(M(\lambda))$ is always a connected group (it is a unipotent extension of $\prod GL(\lambda_i)$), it can't mix the components.

You can start to get context for this from the first two chapters of Schiffmann's Lectures on Hall Alegbras. Unfortunately, Schiffmann always takes $k$ to be a finite field so that he can count points, which makes it harder to talk about positive dimensional subvarieties of $\mathrm{Ext}^1$.

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PS: this seems to works with every UFD $A$ and any nonzero non-invertible element $p\in A$. –  YCor Feb 18 '14 at 22:14
the remark in your edit with the analogy with $k[t]$-modules is a particular case of my previous comment ($k[t]$ being a UFD, $t=p$). –  YCor Feb 19 '14 at 15:16

Here's an example with short exact sequences of (infinitely generated) abelian groups.

Fix a prime $p$ and let $C=\mathbf{Z}[1/p]/\mathbf{Z}$ be the $p$-Prüfer quasi-cyclic group; let $C^{(n)}$ be an isomorphic copy of $C$, and let $C^{(n)}_k\simeq\mathbf{Z}/p^k\mathbf{Z}$ be the $p^k$-torsion in $C^{(n)}$.

Consider $B=\bigoplus_{n\in\mathbf{N}}C^{(n)}$. Let $\mathbf{N}^*$ denote the positive integers. Define subgroups $A=\bigoplus_{n\in\mathbf{N}^*}C^{(n)}_{n}$ and $A_2=\bigoplus_{n\in\mathbf{N}}C^{(2n)}_{n}$. Then clearly, $A$ is isomorphic to $A_2$, and $B/A$ and $B/A_2$ are isomorphic (to $B$). On the other hand, the exact sequences $A\hookrightarrow B\twoheadrightarrow B/A$ and $A_2\hookrightarrow B\twoheadrightarrow B/A_2$ are not isomorphic, because $A$ contains the $p$-torsion of $B$ while $A_2$ does not.

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lovely answer! thanks! –  Pierre Feb 18 '14 at 17:02
I think $C_2$ should be $A_2$, I'm not allowed to edit just one symbol. Nice answer! –  Simone Virili Feb 19 '14 at 16:28

Here is an answer involving finitely generated groups in the quotient, and free groups in the kernel and the total group.

Suppose that $G$ is an infinite, finitely generated group. Let $F_n = \langle s_1,...,s_n \rangle$ be the rank $n$ free group, and let $f : F_n \to G$ be a surjective, noninjective homomorphism; this is a fancy way of saying that $f(s_1),...,f(s_n)$ is a generating set of $G$ but not a free basis. The kernel of $f$, being an infinite index, nontrivial, normal subgroup of $F_n$, must be of infinite rank. So, you get a short exact sequence $$1 \to F_\infty \to F_n \xrightarrow{f} G \to 1$$ Now we just have to dream up some $G$ and two homomorphisms $f,g : F_n \to G$ for which the two short exact sequences are inequivalent. This problem is a translation of a well-studied (but nonetheless still mysterious) problem in group theory: the two short exact sequences are equivalent if and only if the generating sets $f(s_1),...,f(s_n)$ and $g(s_1),...,g(s_n)$ are Nielsen equivalent''.

This paper (by Lars Louder with most cases attributed to Zieschang) shows that if $G$ is the fundamental group of a closed surface of genus $g \ge 2$ then any generating set of cardinality $2g$ is Nielsen equivalent to the "standard" one given by the presentation $$\pi_1(G) = \langle a_1,b_1,...,a_g,b_g \, | \,\, [a_1,b_1] [a_2,b_2] ... [a_g,b_g] = 1\rangle$$ So we won't get an answer from surface groups.

On the other hand the paper of Lustig and Moriah (in Topology (1991)) gives examples where $G$ is a Fuchsian group with torsion (fundamental group of a closed 2-D orbifold) and two generating sets of the same cardinality, both of which look "standard", but which are not Nielsen equivalent.

This paper (by Kapovich and Weidmann) has an introduction with a survey of examples.

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Louder's theorem places no restriction on the cardinality of the generating set or orientability of of the surface... –  nolte Mar 10 '14 at 17:28

Others have given examples. It might be worth noting that (in the category of finitely generated abelian groups, or more generally in the category of finitely generated modules over a noetherian ring) there can be no example with $X_2$ isomorphic to $X_1\oplus X_3$ .

Indeed suppose we have such a sequence $$0\rightarrow X_1\rightarrow X_2\rightarrow X_3\rightarrow 0$$

We want to show that any such sequence splits, so that all such sequences are isomorphic.

Apply $Hom(,X_1)$ to get

$$0\rightarrow Hom(X_3,X_1)\rightarrow Hom(X_2,X_1)\rightarrow Hom(X_1,X_1)$$

The original sequence splits if and only if this sequence is surjective on the right. For this, it suffices to check surjectivity after localizing and then completing at an arbitrary prime $P$ and for this it suffices to check exactness after modding out an arbitrary prime power $P^n$. Now everything is of finite length so to get surjectivity, all we need is for the lengths on the left and right to add up to the length in the middle, which is automatic when $X_2\approx X_1\oplus X_3$.

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A related result of Joseph Ayoub is that the same holds for finite (not necessarily abelian) groups: user.math.uzh.ch/ayoub/PDF-Files/DIRECT.PDF ( J. Group Theory 9 (2006), 307-316) –  YCor Feb 18 '14 at 20:43

It is possible to reuse my answer to this question. It gives an example in the category of $\mathbb C[x,y]$-modules.

Let $R$ be the ring $R=\mathbb C[x,y]$, and let $B$ be the $5$-dimensional $R$-module with shape like a 'W'. That is, basis elements are $a_1,a_2,a_3,b_1,b_2$ and the module structure is given by $$y \cdot a_1=b_1,$$ $$x \cdot a_2=b_1,$$ $$y \cdot a_2=b_2,$$ $$x \cdot a_3=b_2,$$ and all other products of generators and basis elements are zero.

Let $A=\mathbb C$ be the trivial $R$-module and consider the parallel morphisms $f,f' \colon A \rightarrow B$ defined by $f(z)=zb_1$ and $f'(z)=zb_2.$ Now ${\mathrm{coker}} \; f \simeq {\mathrm{coker}} \; f'$ as $R$-modules, but $f$ and $f'$ are non-isomorphic in $\mathrm{Mor}(\mathrm{Mod} \; R)$.

That $f$ and $f'$ are non-isomorphic in $\mathrm{Mor}(\mathrm{Mod} \; R)$ is another way of saying there is no commutative diagram $$\begin{array}{ccc} A & \xrightarrow{f} & B \\ \downarrow u & & \downarrow v \\ A & \xrightarrow{f'} & B \end{array}$$ with $u$ and $v$ isomorphisms. Here $f$ and $f'$ are monomorphisms, so we obtain an example satisfying the conditions given in the question.

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i like this one very much, too! I'll ask my student which answer she prefers, and one of you will get points. In the meanwhile, other answers are of course welcome. –  Pierre Feb 18 '14 at 17:02