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I'd like to have a big-list of "great" short exact sequences that capture some vital phenomena. I'm learning module theory, so I'd like to get a good stock of examples to think about. An elementary example I have in mind is the SES:

$$ 0 \rightarrow I \cap J \rightarrow I \oplus J \rightarrow I + J \rightarrow 0 $$

from which one can recover the rank-nullity theorem for vector spaces and the Chinese remainder theorem. I'm wondering what other 'bang-for-buck' short exact sequences exist which satisfy one of the criteria:

  • They portray some deep relationship between the objects in the sequence that is non-obvious, or
  • They describe an interesting relationship that is obvious, but is of important consequence.
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    $\begingroup$ Should every mathematician know what a short exact sequence is? $\endgroup$
    – Ville Salo
    Jun 21, 2020 at 16:56
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    $\begingroup$ @VilleSalo I was under the impression every mathematician does know what a short exact sequence is :) I'm a computer science student though, so I wouldn't know :D $\endgroup$ Jun 21, 2020 at 17:15
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    $\begingroup$ @SiddharthBhat I definitely do not think every mathematician knows what a short exact sequence is. While this topic probably appears in most first-year PhD courses, if a mathematician goes into, say PDEs, there is no real reason they would remember this topic. Just like in computer science, if someone goes into systems they might forget the Ford-Fulkerson algorithm, and a theoretical computer scientist might forget the soldering they learned in computer architecture. $\endgroup$ Jun 21, 2020 at 17:26
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    $\begingroup$ I once met a mathematician who had never heard of a homeomorphism. Maybe they were not a true Scotsman though. $\endgroup$
    – Ville Salo
    Jun 21, 2020 at 18:39
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    $\begingroup$ @VilleSalo : I once heard the following (probably apocryphal) story about an oral exam. Professor: "Are $M_1$ and $M_2$ homeomorphic?" Student: "$M_1$ is, but $M_2$ isn't." $\endgroup$ Jun 21, 2020 at 19:23

38 Answers 38

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There is one obvious sequence that underlies all vector analysis and a lot that builds up on it, no matter if its applied analysis, PDE, physics or the original foundations of algebraic topology. Yet it is rarely written out, as the people in the applied fields prefer to split it into its constituent statements and the people in pure mathematics are inclined to immediately write down some generalization instead. What I am talking about is of course the relationship between the classic differential operators on 3D vector fields:

$$0 \to \mathbb R\to C^\infty(\mathbb{R}^3;\mathbb{R}) \stackrel{\operatorname{grad}}{\to} C^\infty(\mathbb{R}^3;\mathbb{R}^3) \stackrel{\operatorname{curl}}{\to} C^\infty(\mathbb{R}^3;\mathbb{R}^3) \stackrel{\operatorname{div}}{\to} C^\infty(\mathbb{R}^3;\mathbb{R}) \to 0 $$

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    $\begingroup$ @SiddharthBhat "[T]he people in pure mathematics are inclined to immediately write down some generalization instead." $\endgroup$
    – user347489
    Jun 22, 2020 at 11:25
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    $\begingroup$ There is also the single variable version that every freshman is implicitly exposed to: $0 \to {\mathbb R} \to C^\infty({\mathbb R}; {\mathbb R}) \stackrel{\frac{d}{dx}}{\to} C^\infty({\mathbb R}; {\mathbb R}) \to 0$. That is to say, the fundamental theorem of calculus, complete with "+C". Incidentally the ${\mathbb R}$ is also missing on the left of your long exact sequence. $\endgroup$
    – Terry Tao
    Jun 22, 2020 at 22:05
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    $\begingroup$ ...well, strictly speaking the full fundamental theorem of calculus also comes with an important splitting of this short exact sequence (the definite integral). Anyway, I submit this sequence as a literal answer to the question title of a "short exact sequence every mathematician should know". $\endgroup$
    – Terry Tao
    Jun 22, 2020 at 22:30
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    $\begingroup$ @mlk This sequence may be short and exact, but it is not a short exact sequence. $\endgroup$ Jun 23, 2020 at 16:22
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    $\begingroup$ @SiddharthBhat To clarify my comment, a short exact sequence is an exact sequence that is of the form $0 \to A \to B \to C \to 0$. Sequences with slightly more (or fewer) terms do not quality, even if they are still short in a colloquial sense. $\endgroup$ Jun 28, 2020 at 4:19
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This might be very basic, but the short exact sequence $$ 0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0 $$ is both an injective resolution of $\mathbb{Z}$, and a flat resolution of $\mathbb{Q}/\mathbb{Z}$, making it a very useful exact sequence in many homological computations.

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The exponential sheaf sequence: $$0\to 2\pi i\,\mathbb Z \to \mathcal O_M {\buildrel\exp\over\to}\mathcal O_M^*\to 0$$ where $\mathcal O_M$ is the sheaf of holomorphic functions on the complex manifold $M$, $\mathcal O_M^*$ is the sheaf of non-vanishing functions and $$\exp : \mathcal O_M \to \mathcal O_M^*$$ is induced by the usual exponential function.

Restricting to sections over any open set $U$ we get a long exact sequence including the map $$\cdots \to H^0(\mathcal O_U^*)\to H^1(2\pi i\,\mathbb Z|_U) \to \cdots.$$ The cohomology group $H^0(\mathcal O_U^*)$ is the set of non-vanishing holomorphic functions on $U$. Roughly speaking, the cohomology group $H^1(2\pi i\,\mathbb Z|_U)$ gives us ($2π i$ times) an integer for each closed loop in U. This map essentially tells us the winding number, around zero, of each non-vanishing holomorphic function, as we go around such a loop. I say all mathematicians should know this because the winding number is so fundamental. But this short exact sequence is also the starting point for a long path through all kinds of interesting places like the Riemann-Roch theorem.

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    $\begingroup$ Also, the image of a line bundle $L\in H^1(M,\mathcal{O}^\times)$ in the next cohomology group corresponds to the first Chern class $c_1(L)\in H^2(M,\mathbb{Z})$. $\endgroup$
    – Qfwfq
    Jun 22, 2020 at 2:23
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    $\begingroup$ I've been a mathematician for 30 years and one of my most cited papers introduced a new exact sequence. Yet I have no idea what is a sheaf. So I fail to see how this is an exact sequence every mathematician should know. $\endgroup$ Jun 22, 2020 at 17:14
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    $\begingroup$ @TerryLoring I'm envious. I remember learning this stuff for the first time with great fondness (30 years ago almost exactly). I can never have that feeling again, but you can! $\endgroup$
    – Dan Piponi
    Jun 22, 2020 at 17:20
  • $\begingroup$ @DanPiponi What is a good place to learn this? This seems like a beautiful way to look at winding. $\endgroup$ Jun 26, 2020 at 23:00
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    $\begingroup$ I learnt it from Gunning's "Lectures on Riemann Surfaces" (1966). Confusingly the sequel to this book is also called "Lectures on Riemann Surfaces (Jacobi Varieties)". You want the one that gives this exact sequence (written as $0\to\mathbb Z\to\mathcal O\to\mathcal O^*\to 0$) at the bottom of page 26. $\endgroup$
    – Dan Piponi
    Jun 27, 2020 at 16:28
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I find it hard to believe that three days have gone by and no one has explicitly mentioned $$ 0 \to \Bbb Z \to \Bbb R \to \Bbb S^1 \to 0 $$

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    $\begingroup$ We can rewrite the $\mathbb{S}^1$ term of this sequence as the Pontryagin dual $\text{Hom}(\mathbb{Z},\mathbb{T})$ of $\mathbb{Z}$. Then the resulting short exact sequence is key to the Poisson summation formula. Tate gave a similar short exact sequence in his thesis: $$0\to K\to \mathbb{A}_K\to \text{Hom}(K,\mathbb{T})\to 0,$$ where $K$ is a global field and $\mathbb{A}_K$ is its adele ring. Note that these sequences are self-dual in the sense that applying the Pontryagin duality functor yields ``the same" sequences. $\endgroup$ May 10, 2021 at 22:08
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I think a short exact sequence that every teacher should know is

$$ 0 \to \mathbb R^d \to \mathrm{Isom}(\mathbb R^d) \to \mathrm{O}(\mathbb R^d) \to 0, $$

maybe for $d=2$ or $d=3$. Better still, forget about the origin and see $E=\mathbb R^d$ as an affine space.

It is a great visual helper, in the sense that many elementary properties take a concrete sense. It makes it clear that there is a copy of the group on the left hand side (the translations) in the isometry group, and that there is some other component described by the group on the right hand side, although how to perform the decomposition is not obivous (which is of course expected for working mathematicians, but a nice way to introduce it to others). Once we describe how the right hand side sits in the isometry group, it is very visual also that is not canonical, and that in fact there is a choice to be made to define such a copy, introducing what a section is and why it is interesting/important.

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    $\begingroup$ More generally: semidirect products for groups. $\endgroup$ Jun 21, 2020 at 22:21
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    $\begingroup$ (These are split, by the way.) $\endgroup$ Jun 21, 2020 at 23:30
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The short exact sequence

$$ 0 \to \mathrm{rad}({\mathfrak g}) \to {\mathfrak g} \to {\mathfrak g}/\mathrm{rad}({\mathfrak g}) \to 0$$

separates a Lie algebra ${\mathfrak g}$ into its solvable radical $\mathrm{rad}({\mathfrak g})$ and its semisimple projection ${\mathfrak g}/\mathrm{rad}({\mathfrak g})$, and is absolutely fundamental in the classification theory of Lie algebras, particularly in the characteristic zero, finite dimensional setting in which Levi's theorem is available to split the above sequence. Given the ubiquity of Lie algebras and Lie groups in mathematics, as well as the wider philosophy of separating algebraic objects into their "solvable" and "simple" components, I would submit that this sequence should be known to any mathematician.

In a somewhat similar spirit, the short exact sequence $$ 0 \to [G,G] \to G \to G/[G,G] \to 0$$ that separates a group $G$ into its commutator subgroup $[G,G]$ and its abelianisation $G/[G,G]$ is generally the first step towards understanding solvable groups $G$ (because if $G$ is solvable then $[G,G]$ is also solvable with the derived length decremented by one), while the analogous short exact sequence $$ 0 \to Z(G) \to G \to G/Z(G) \to 0$$ separating a group $G$ into its centre $Z(G)$ and the quotient $G/Z(G)$ is similarly often the first step in understanding nilpotent groups (because if $G$ is nilpotent then $G/Z(G)$ is nilpotent with the nilpotency degree decremented by one). There are of course analogues of these sequences for Lie algebras also.

In the category of topological (or algebraic) groups, one also has the short exact sequence $$ 0 \to G^0 \to G \to G/G^0 \to 0,$$ where $G^0$ is the identity component and $G/G^0$ is the totally disconnected projection. In principle, this separates the study of such groups into the connected and totally disconnected cases.

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    $\begingroup$ One might add to this list$$0\to H_2(G)\to[F,F]/[R,F]\to G\to0$$for perfect $G=F/R$ with free $F$, and similar extensions for Lie algebras, etc. $\endgroup$ Jun 22, 2020 at 14:57
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"Every mathematician should know" is too much to ask, but I do think the following is a great short exact sequence that captures a vital phenomenon: $$0 \to K(H) \to B(H) \to Q(H) \to 0.$$

$K(H)$ is the compact operators on a Hilbert space $H$, $B(H)$ is the bounded operators, and $Q(H)$ is the Calkin algebra. The "vital phenomenon" is that being invertible modulo the compacts, i.e., being Fredholm, is the same as being invertible in the Calkin algebra.

It won't help you learn module theory but IMHO it deserves to be on a big list ...

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    $\begingroup$ I voted up, even if this is not much more than a definition (of Calkin algebra). $\endgroup$ Jun 23, 2020 at 12:23
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    $\begingroup$ @DenisSerre it's a fair comment. I guess the substantive fact is that the Calkin algebra is a C*-algebra --- easy enough for us now that we have the GNS construction, but it wasn't easy for Calkin ... $\endgroup$
    – Nik Weaver
    Jun 23, 2020 at 13:38
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    $\begingroup$ @J.vanDobbendeBruyn I guess the modern approach is to define C*-algebras abstractly, and you're right, we don't need GNS to see that the Calkin algebra is an abstract C*-algebra. I should have said "concrete C*-algebra". $\endgroup$
    – Nik Weaver
    Jun 23, 2020 at 18:42
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    $\begingroup$ @NikWeaver ah yes, the historical distinction between $B^*$-algebras and $C^*$-algebras... Indeed, looking at Calkin's original paper, it seems that he does not attempt to prove the $B^*$-identity, and immediately aims for a faithful representation. (In fact, according to Wikipedia, the $B^*$-identity was only introduced a few years later, in 1946.) $\endgroup$ Jun 23, 2020 at 22:33
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    $\begingroup$ @J.vanDobbendeBruyn yes, if I remember right he uses something like an ultraproduct construction to do this. Anyway your point is well-taken. $\endgroup$
    – Nik Weaver
    Jun 24, 2020 at 0:20
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I strongly doubt there is any short exact sequence that every mathematician should know, but I certainly wish that those of them who know that for a (co)chain complex $(C,d)$ $$ 0\to\operatorname{Im}(d)\to\operatorname{Ker}(d)\to H(C,d)\to0 $$ is short exact, would also know that $$ 0\to H(C,d)\to\operatorname{Coker}(d)\to\operatorname{Im}(d)\to0 $$ is short exact too.

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    $\begingroup$ It's sad this is so far down the list, because I think it's the only short exact sequence I actually know! $\endgroup$
    – cduston
    Jun 22, 2020 at 13:45
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    $\begingroup$ The $\text{Im}(d)$ complex in the second sequence needs to be shifted by one degree, no? $\endgroup$
    – Feryll
    Jun 24, 2020 at 23:15
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    $\begingroup$ @Feryll Well, $\operatorname{Im}(d)$ can be naturally graded in two ways, shifted wrt each other: the second one comes from the isomorphism $\operatorname{Im}(d)\cong C/\operatorname{Ker}(d)$ $\endgroup$ Jun 25, 2020 at 4:18
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An example of a short exact sequence satisfying your first desiderata, but one which you probably won't fully understand till you are further along in homological algebra, is the Universal Coefficient Theorem. The homology version says: if $R$ is a PID, $A$ is an $R$-module, and $C$ is a flat chain complex over $R$, then there is a natural short exact sequence

$$ 0 \rightarrow H_n(C) \otimes_R A \rightarrow H_n(C\otimes_R A) \rightarrow Tor_1^R(H_{n-1}(C),A) \rightarrow 0 $$

Moreover, this sequence splits, and the splitting is natural in $A$ but not in $C$.

A related result is the Künneth Theorem: if $R$ is a PID and $X,Y$ topological spaces then there is a natural short exact sequence

$$ 0 \rightarrow \bigoplus_{i+j=k} H_i(X;R) \otimes_R H_j(Y;R) \rightarrow H_k(X\times Y; R) \rightarrow \bigoplus_{i+j=k-1} Tor_1^R(H_{i}(X;R),H_j(Y;R)) \rightarrow 0 $$

Furthermore, this sequence splits, but not canonically.

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    $\begingroup$ Of course, there is also a dual version, with Ext and product instead of Tor and direct sum. And a universal coefficient theorem in cohomology. Look them up in Hilton and Stammbach's book (Theorems 3.1, 3.3). $\endgroup$ Jun 21, 2020 at 17:46
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Within the category of Banach spaces and bounded linear maps,

$$0\to c_0 \to \ell_\infty \to \ell_\infty / c_0 \to 0$$

is a paradigm example of a short exact sequence that does not split, contrary to any short exact sequence

$$0\to c_0 \to X \to Z \to 0,$$

where $X$ (or $Z$) is separable. Here $c_0$ is the space of sequences convergent to 0 and $\ell_\infty$ is the space of bounded sequences, both endowed with the supremum norm.

The relevant results are Sobczyk's and Phillips-Sobczyk's theorems. See also the paper Sobczyk's Theorems from A to B by Félix Cabello Sánchez, Jesus M. F. Castillo, and David Yost.

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    $\begingroup$ The "commutative analog" of my example! Might be worth mentioning that $l_\infty/c_0$ is isomorphic to $C(\beta\mathbb{N}\setminus \mathbb{N})$. $\endgroup$
    – Nik Weaver
    Jun 21, 2020 at 17:52
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    $\begingroup$ Evaluating at a single element of $\beta {\mathbb N}/{\mathbb N}$ one obtains the short exact sequence $0 \to o({\mathbb R}) \to O({\mathbb R}) \to {\mathbb R} \to 0$, where $O({\mathbb R})$ is the ring of bounded nonstandard reals, $o({\mathbb R})$ is the ideal of infinitesimal nonstandard reals, and ${\mathbb R}$ is the standard reals. Also can't resist mentioning the variant $0 \to o({\mathbb Q}) \to O({\mathbb Q}) \to {\mathbb R} \to 0$, which one way to construct the real numbers in nonstandard analysis. More generally one has the nonstandard hull construction in metric or Banach spaces. $\endgroup$
    – Terry Tao
    Jun 21, 2020 at 20:04
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How about the short exact sequence that expresses that every group can be expressed in terms of generators and relators? For any group $G$, there is a short exact sequence (in fact many) of the form $$0\to R\to F\to G\to 0,$$ with $F$ and $R$ being free groups. This expresses $G$ as a free group of generators modulo the relations encoded in $R$.

Of course there are analogous statements in other categories, such as those of modules.

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    $\begingroup$ "Analogous" should be taken with a grain of salt, since $R$ might not be free in other categories (including many module categories). $\endgroup$ Jun 23, 2020 at 14:37
  • $\begingroup$ Good point. I was thinking about particularly nice rings. In general projective resolutions might be much longer. $\endgroup$ Jun 23, 2020 at 21:30
  • $\begingroup$ I think this was explicitly part of my first answer to this question (and, indeed, is very worth highlighting to a newcomer like the OP) mathoverflow.net/a/363723/11540 $\endgroup$ Jun 26, 2020 at 20:00
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For any abelian group $A$, there is a short-exact sequence $$0 \to T(A) \to A \to A/T(A) \to 0,$$ where $T(A)$ is the torsion subgroup of $A$, and $A/T(A)$ is torsion-free.

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The kernel-cokernel exact sequence: in an abelian category, given $A \xrightarrow{f} B \xrightarrow{g} C$, the following sequence is exact

$$ 0 \to \ker f \to \ker gf \to \ker g \to \text{coker } f \to \text{coker }gf \to \text{coker} g \to 0$$

The maps are the obvious ones. The map $\ker g \to \text{coker } f$ is the one which factors through $B$.

I don't know if this fits, because it's not short and maybe it is too trivial, but I really think that every mathematician should know. For example, at a very low level, this tells the following basic facts

  1. $gf$ is injective iff $f$ is injective and $\ker g \hookrightarrow{} \text{coker }f$
  2. $gf$ is surjective iff $g$ is surjective and $\ker g \twoheadrightarrow \text{coker }f$
  3. $gf$ is an isomorphism iff $f$ is injective, $g$ surjective and $\ker g \xrightarrow{\cong} \text{coker } f$
  4. If $f$ and $g$ are injective/surjective, so is $gf$.

I think that more cools applications are covered in the following paper by Xiong, which I found just now.

A nice picture of this sequence from Nakaoka's website is the following

enter image description here

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  • $\begingroup$ Do you remember on exactly which website you found this image? I don't find it at sci.kagoshima-u.ac.jp/nakaoka . $\endgroup$
    – LSpice
    Dec 23, 2023 at 21:50
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Given a finitely generated module $M$ over a commutative Noetherian ring $R$, there is a short exact sequence $$0\to M_1 \to R^n \to M\to 0$$ where you map $1$ in each $R$ to a generator of $M$ and $M_1$ (also finitely generated) is called a module of syzygy of $M$. Understanding this sequence (and it's repetitions) is a fundamental problem in commutative algebra and algebraic geometry and has generated countless beautiful results as well as been widely used a versatile tool on it's own. For instance, Hilbert's Syzygy Theorem says that if $R$ is a polynomial ring over a field, and you repeat this with $M_1$, etc, eventually you will get a free module.

Not sure there is one thing that "everyone should know", but for instance Stanley's proof of the Upper Bound Conjecture in combinatorics used a refinement of the Hilbert Theorem above.

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    $\begingroup$ Since the OP is just getting started, I'll add that this story is told in section 1.8 of Zimmermann's book on representation theory, where he also explains that (by definition) the $n$-th syzygy $\Omega_M^n$ fits in a short exact sequence $0\to \Omega_M^n \to P_{\Omega_M^{n-1}} \to \Omega_M^{n-1}\to 0$. This is an important reason why the stable module category in representation theory is a triangulated category. Here $P_{\Omega_M^{n-1}}$ is as in my answer, a projective module that maps onto $\Omega_M^{n-1}$ $\endgroup$ Jun 21, 2020 at 18:11
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    $\begingroup$ There is also a story of cosyzygies, i.e., cokernels of injective envelopes, which also fits into the framework of short exact sequences. $\endgroup$ Jun 21, 2020 at 18:14
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I suppose many algebraic topologists would agree that the short exact sequence

$$0\longrightarrow \mathbb Z/p \longrightarrow \mathbb Z/p^2 \longrightarrow \mathbb Z/p\longrightarrow 0$$ giving rise to the Bockstein operator in (co)homology and the exact sequence $$ 0\longrightarrow C \stackrel{\cdot p}\longrightarrow C \longrightarrow C/pC\longrightarrow 0$$ giving rise to the Bockstein spectral sequence of the form $$H(C/pC) \Longrightarrow H(C)/pH(C)$$ may fit the bill. They are quite simple and lead to remarkably interesting mathematics.

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    $\begingroup$ In addition to the Bockstein, this is the key example to see that $\mathrm{Ext}(\mathbb Z/p, \mathbb Z/p)\ne 0$, which is nice to know. I also like $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/n\to 0$, as a good first example of a non-split short exact sequence but also just a general insight into how abelian groups behave differently than vector spaces. $\endgroup$ Jun 22, 2020 at 4:20
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    $\begingroup$ In fact every first grader has to know this short exact sequence: it is how you do addition by “carrying”. $\endgroup$ Aug 27, 2020 at 7:39
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    $\begingroup$ @VivekShende Sounds a bit dramatic but OK. :) $\endgroup$
    – Pedro
    Aug 27, 2020 at 9:58
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    $\begingroup$ (I think there was a nice thread here about how carrying gives a cocycle, too...) $\endgroup$
    – Pedro
    Aug 27, 2020 at 9:58
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The Tate extension. Let $k$ be a field, and let $V$ be the space $k((t))$ be the space of Laurent series with coefficients in $k$, considered as a topological vector space. If we write $\operatorname{GL}(V)$ for the group of (topological) automorphisms of $V$, then there is a canonical central extension

$$0\rightarrow k^{\times}\rightarrow\operatorname{GL}^{\flat}(V)\rightarrow\operatorname{GL}(V)\rightarrow0.$$

For a detailed construction, see http://www.its.caltech.edu/~justcamp/notes/sila.pdf. A brief summary: let $L$ be the space $k[[t]]\subset V$. Then for any element $g\in\operatorname{GL}(V)$, $gL$ and $L$ are commensurable (their intersection is of finite codimension in both) and so one can associate to them a one-dimensional vector space, the relative determinant. $\operatorname{GL}^{\flat}(V)$ can be defined as pairs of an element $g$ and a trivialization of the relative determinant.

Here the deep fact is the existence of the object $\operatorname{GL}^{\flat}(V)$. This leads to a host of other central extensions that are central (haha....) to the study of e.g. affine Lie algebras and friends (and thus to conformal field theory.) In a different direction, this short exact sequence is equivalent to the existence of the local residue symbol; in fact, I think Tate's name got attached to this because of his use of the local residue symbol to give a new proof of Riemann-Roch.

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I saw $0\to \mathbb Z_p\to \mathbb Z_{p^2}\to \mathbb Z_p\to 0$ as an answer, with $p$ prime, but I will add with $p$ not prime and the particular choice $p=10$,

$$0\to \mathbb Z_{10}\to \mathbb Z_{100}\to \mathbb Z_{10}\to 0 $$ for the following reasons:

  1. We all know that the group in the middle is given by $\mathbb Z_{10}\times \mathbb Z_{10}$ twisted by a 2-cocycle.

  2. We know (if we don't, we should!) that the 2-cocyle is given by $f(n,m)=0$ if $n+m<10$ and $f(n,m)=1$ if $n+m\geq 10$, where $n,m\in\{0,\dots,9\}$.

  3. Every child knows (or should know) (in particular, every mathematician -even a non algebraist one-) how to sum 2-digit numbers.

Since the question is what exact sequence we "should" know, I think 3. and 2. are two good reasons, even for the two different meanings of "should" know.

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    $\begingroup$ What is the 2-cocycle? Is that the homology group that's generated by $Ker(g)/Im(f)$? $\endgroup$ Mar 21, 2021 at 7:50
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    $\begingroup$ If the Babylonians knew about exact sequences, they would certainly know $0 \rightarrow\mathbb{Z}_6 \rightarrow \mathbb{Z}_{60} \rightarrow \mathbb{Z}_{10} \rightarrow 0$. So why not also $0 \rightarrow \mathbb{Z}_p \rightarrow\mathbb{Z}_{pq} \rightarrow \mathbb{Z}_{q} \rightarrow 0$ for any natural numbers $p, q$? $\endgroup$ Jun 19, 2021 at 22:14
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I guess the quintessential example, satisfying your second desiderata, is

$$ 0 \rightarrow A \stackrel{f}{\rightarrow} B \rightarrow B/f(A) \rightarrow 0. $$

For example, if $f = \mu_n: \mathbb{Z} \to \mathbb{Z}$ is multiplication by $n$, this means the following is exact

$$ 0 \rightarrow \mathbb{Z} \stackrel{\mu_n}{\rightarrow} \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \rightarrow 0. $$

Another example of the same general result is that, if $C$ is finitely presented, then it fits in a short exact sequence

$$ 0 \rightarrow N \rightarrow P \rightarrow C \rightarrow 0 $$

where $N$ and $P$ are finitely generated, and $P$ is projective. Think of $P$ as the generators, and $N$ as the relations you quotient out by to get $C\cong P/N$.

Since you asked for a big list, I'll try to restrict myself to one example per answer.

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    $\begingroup$ Even though the $\mathbb{Z}/n\mathbb{Z}$ example is very basic, I often find it extremely helpful to think about (e.g. lets you understand the difference between a SES and a direct sum). $\endgroup$ Jun 21, 2020 at 17:46
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    $\begingroup$ @SamHopkins Is it different from a direct sum because it does not split? $\endgroup$ Jun 21, 2020 at 19:00
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    $\begingroup$ @SiddharthBhat: yes, exactly, that's the same thing. $\endgroup$ Jun 21, 2020 at 19:36
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Despite it being frequently used implicitly in papers (a classical example being Milnor's '56 paper about exotic spheres), I have never seen the following spelled out anywhere, so this might be a good place:

Let $\xi^n : E \to B$ be a real, smooth vector bundle over a manifold $B$. There is a short exact sequence of vector bundles over $E$,

$$0 \to \xi^*(\xi) \stackrel{i}{\to} \tau(E) \stackrel{d\xi}{\to} \xi^*(\tau(B)) \to 0,$$

where $i$ sends $(e_1, e_2) \in E \times_B E$ to the tangent vector starting at $e_1$ and pointing in the direction specified by $e_2$. Here, $\tau(M)$ denotes the tangent bundle of $M$. (As always, short exact sequences of vector bundles split.)

It follows from an easy dimension count. Alternatively, one can write out the gluing of the charts, giving something like

$$0 \to \text{colim } \mathbb R^n_\text{d} \times \mathbb R^n_\text{o} \times U_i \to \text{colim } \mathbb R^n_\text{d} \times \mathbb R^n_\text{o} \times TU_i \to \text{colim } \mathbb R^n_\text{o} \times TU_i \to 0,$$

where subscript $d$ and $o$ are just labels to distinguish between the coordinate which tracks the direction and that which tracks the origin, and $\{U_i\}$ is an atlas for $\xi$. This also gives a more rigorous description of $i$.

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    $\begingroup$ This must be closely related to another very important short exact sequence - the Atiyah class extension$$0\to\Omega^1\otimes E\to J^1(E)\to E\to0$$ $\endgroup$ Jun 22, 2020 at 14:23
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    $\begingroup$ I find it particularly illuminating when talking about connections. A right splitting ($H:\xi^*(\tau(B))\to\tau(E)$ such that $d\xi\circ H=\mathrm{id}$) is a choice of horizontal lift, while a left splitting ($K:\tau(E)\to\xi^*(\xi)$ such that $K\circ i=\mathrm{id}$), sometimes known as the connector, is the projection on the vertical part (hence $K(v)=0$ if $v$ is horizontal). The introduction of connections is a direct (human) result of the fact that this sequence does not canonically split. $\endgroup$
    – Pierre PC
    Jun 23, 2020 at 11:59
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This one is just too much fun to leave out. Write the braid group on $n$ strands as $B_n$. By following the strands of a braid $\sigma\in B_n$ we construct a permutation of $n$ items, which we write as $\eta(\sigma)$. This $\eta$ is an epimorphism whose kernel is the pure braid group $P_n$. The pure braids are the braids whose strands end up where they started:

$$ P_n \to B_n \overset{\eta} \to S_n $$

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Short exact sequences form a bridge of sorts between homological algebra and representation theory. For example, Maschke's theorem is the statement that, if $G$ is a finite group and $k$ is a field whose characteristic does not divide the order of $G$, then the $k$-representations of $G$ are completely reducible. This is equivalent to the statement that every short exact sequence of $k[G]$-modules

$$ 0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0 $$

splits. This is, in turn, equivalent to the statement that $k[G]$ is a semisimple ring (there are many equivalent forms of what this means; my favorite is that every module is both injective and projective). You can then hit it with the Artin–Wedderburn theorem and write $k[G]$ as a product of matrix algebras. Strictly speaking, you don't need the language of short exact sequences, but many find it clarifying.

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Another fundamental (half) short exact sequence is the Jacobi--Zariski sequence. For algebras over operads, for example, it takes the following form: for a triple $C\to B\to A$ of maps of $P$-algebras, there is a half exact sequence of functors $$0\longrightarrow \mathrm{Der}_B(A,-) \longrightarrow \mathrm{Der}_C(A,-) \longrightarrow\mathrm{Der}_C(B,-) $$ coming from the half exact sequences in Kahler differentials $$A\otimes_B\Omega_{B\mid C}^1\longrightarrow \Omega_{A\mid C}^1 \longrightarrow \Omega_{A\mid B}^1 \longrightarrow 0 $$ that is exact if $B\to A$ is a cofibration.

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Decided to turn into an answer my comment to another answer here.

The Atiyah class $\alpha_E\in\operatorname{Ext}^1(E,\Omega^1\otimes E)$ of a holomorphic vector bundle $E$ is the class of the short exact sequence $$ 0\to\Omega^1\otimes E\to J^1(E)\to E\to0, $$ where $\Omega^1$ is the cotangent bundle (corresponding to the sheaf of holomorphic 1-forms) and $J^1(E)$ is the sheaf of first order jets of sections of $E$. A good reference is "Rozansky-Witten invariants via Atiyah classes" by Kapranov (Compositio Math. 115 (1999) 71-113). Kapranov notes that there is a dual way to represent this class, using another remarkable short exact sequence $$ 0\to E\to{\mathcal D}^{\leqslant1}\otimes_{\mathcal O}E\to T\otimes E\to0. $$ Here $T$ is the tangent bundle and ${\mathcal D}^{\leqslant1}$ is the sheaf of differential operators of order $\leqslant1$. (More precisely, this gives the class corresponding to $-\alpha_E$ in view of the canonical isomorphism $\operatorname{Hom}(-,\Omega^1\otimes-)\cong\operatorname{Hom}(T\otimes-,-)$.)

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For a free product $A*B$ of groups $A$ and $B$, there is the exact sequence

$1 \to [A,B] \to A*B \to A \times B \to 1$

where $[A,B]$ is the subgroup generated by all elements $[a,b]=aba^{-1}b^{-1}$ and $A \times B$ is the direct product group. The first map is the inclusion and the second one is the intuitive one. This sequence is important for combinatorial and geometric group theory.

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A starting point in anabelian geometry (a "thème central de la géométrie algébrique anabélienne", as Grothendieck writes in his Esquisse d'un Programme) can be considered to be the following:

Let $k$ be a field with separable closure $\overline{k}$ and $X$ a quasi-compact, quasi-separated $k$-scheme. If $\overline{x}$ is a geometric point of $X$ and the base change $X_{\overline{k}}$ is connected, then there is a short exact sequence of profinite topological groups:

$$ 1 \to \pi^{ét}_1(X_{\overline{k}}, \overline{x}) \to \pi^{ét}_1(X, \overline{x}) \to \pi^{ét}_1(Speck, \overline{x}) \simeq Gal(\overline{k}/k) \to 1$$

In fact, for $X=\mathbb{P}^1_{\mathbb{Q}}\backslash{\{0,1,\infty\}}$, J. S Milne on p. 30 of his LEC course notes, calls $\pi^{ét}_1(X, \overline{x})$ "arguably, the most interesting object in mathematics" due to the deep motivic ideas and profound connections that surround it, in relation with the already mysterious absolute Galois group of the rationals.

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Let $M$ be a smooth manifold and $x:M\rightarrow \mathbb{R}$ a smooth function with $0$ as regular value, such that $X=\{x=0\}\subset M$ is a smooth submanifold. Then $$ 0\rightarrow x C^\infty(M)\hookrightarrow C^\infty(M) \xrightarrow{f\mapsto f\vert_X} C^\infty(X)\rightarrow 0 $$ is a short exact sequence and a right split corresponds to an extension map.

Where does this show up:

  • For $M=\mathbb{R}$ this is the statement that the remainder in a Taylor series can be written as $R(x)=x^kr(x)$ for a smooth function $r(x)$.
  • For $M=\mathbb{R}^d\times \bar {\mathbb{R}}^d$ (where $\bar{\mathbb{R}}^d$ is the radial compactification) and $x$ a boundary defining function of $\partial M=\mathbb{R}^d\times S^{d-1} = S^*\mathbb{R}^d$ (co-sphere bundle), this yields $$ 0\rightarrow \Psi_{\mathrm{cl}}^{m-1}(\mathbb{R}^d)\hookrightarrow \Psi_{\mathrm{cl}}^{m}(\mathbb{R}^d) \xrightarrow{\sigma_m} C^\infty(S^*\mathbb{R}^d)\rightarrow 0, $$ the shorth exact symbol sequence of pseudo-differential operators. Here $\Psi^m(\mathbb{R}^d)=\mathrm{Op}(x^{-m}C^\infty(M))$ with $\mathrm{Op}$ denoting the standard quantisation of symbols $a:\mathbb{R}^d_z\times \mathbb{R}^d_\xi\rightarrow \mathbb{C}$. One can take $x=\langle \xi \rangle^{-1}$ as bdf. of fibre-infinity. A right split is then a quantisation map. The symbol sequence (together with the multiplicativity of $\sigma_m$) allows to construct parametrices of elliptic operators and is thus the starting point of elliptic regularity theory.
  • For $M=\bar {\mathbb{R}}^d\times \bar {\mathbb{R}}^d$, which is a manifold with corners, the constructions from the previous point yields Melrose's scattering (classical) scattering pseudo-differential operators.
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    $\begingroup$ Can you please expand a bit on how the $M=\mathbb{R}$ case corresponds to remainders in Taylor series? $\endgroup$ Mar 24, 2021 at 21:53
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    $\begingroup$ On $M=\mathbb{R}$ we take $x(t)=t$, such that $X=\{0\}$. Now, if a smooth function $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies $f(0)=0$, then by exactness of the SES we can write $f(t)=t g(t)$ for another smooth function $g:\mathbb{R}\rightarrow \mathbb{R}$. Inductively, if $f(t)=O(t^k)$, then $f(t) = t^k g(t)$ for a smooth function $g:\mathbb{R}\rightarrow\mathbb{R}$. This applies to the remainder in the Taylor series, where $g$ might a priori only be bounded. $\endgroup$
    – Jan Bohr
    Mar 26, 2021 at 17:30
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Take a bundle $E \rightarrow M$ in $Diff$ and then apply the tangent functor. We get $TE\rightarrow TM$. The kernel of this is the vertical bundle, $VE$ and this all assembles into the short exact sequence:

$0 \rightarrow VE \rightarrow TE \rightarrow TM \rightarrow 0$

A splitting of this gives an Ehresmann connection, aka a horizontal bundle $HE$ such that $TE = VE \oplus HE$

The concept then descends to connections over vector bundles and principal bundles.

All this is in Michor, Kolar & Slovaks Natural Operations in Differential Geometry. It's possible to introduce curvature in this generality and which describes the local integrability of the horizontal bundle. In fact, they introduce it in greater generality over just a manifold, rather than a bundle and there they have cocurvature as well as curvature where the former describes the integrability of the vertical bundle. In the example above, the cocurvature vanishes as the vertical bundle is always integrable.

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The following is both elementary and simple, and carries a useful content, so it may qualify at least as something most mathematicians may not regret to know. In the context of formal Laurent series,

$$0 \rightarrow \mathbb{C} \rightarrow \mathbb{C}((z)) \xrightarrow{D} \mathbb{C}((z)) \;\xrightarrow{ \mathrm{Res} }\; \mathbb{C} \rightarrow 0,$$

describes the main features of the operation of formal residue, taking a formal Laurent series $\sum_k c_kz^k$ to its coefficient $c_{-1}$. Among other results these allow to give an easy proof of the Lagrange inversion formula (check e.g. Genealogy of the Lagrange inversion theorem and its answers for other remarks).

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This probably isn’t a short exact sequence that everyone needs to know, but it’s one of some significance for mathematical physics and differential geometry.

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Any smooth principal $G$-bundle $\pi : P \to B$ gives rise to a canonical short exact sequence $$ 0 \to \mathfrak{g} \times P \to TP \to \pi^\ast TB \to 0 $$ of $G$-equivariant vector bundles on the total space $P$, where $\mathfrak{g}$ carries the adjoint representation of $G$. Here, the map $\mathfrak{g} \times P \to TP$ is induced by the map that sends $X \in \mathfrak{g}$ to the corresponding fundamental vector field $X_P$ on $P$, while the map $TP \to \pi^\ast TB$ is given by $\pi_\ast$.

As Atiyah first observed, a principal connection can be identified with a splitting of this short exact sequence. Moreover, the gauge action of global gauge transformations on principal connections is compatible with this identification: if $f : P \to P$ is a gauge transformation and $\rho : \pi^\ast TB \to TP$ is a right splitting, then $f$ acts on $\rho$ to yield the right splitting $f_\ast \circ \rho$.

From the (somewhat idiosyncratic) perspective of noncommutative geometry, this short exact sequence can be viewed as relating the $G$-equivariant differential calculus on the total space $P$ to differential calculus on the base $B$ and the differential calculus along the orbits (modelled on that of the structure group $G$); a principal connection, then, tells you how to decompose the total differential calculus into a “direct sum” of basic and orbitwise calculi. Indeed, in noncommutative geometry itself, this short exact sequence—more precisely, a noncommutative analogue of its dual—actually becomes part of the definition of (algebraic) quantum principal bundle and principal connection.

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    $\begingroup$ Can you suggest some references(non-physics) which define connection as splitting of the Atiyah sequence... One that I know is the Appendix A of the book "Lie groupoids and Lie algebroids in differential geometry" by K. Mackenzie.. $\endgroup$ Jun 25, 2020 at 3:02
  • $\begingroup$ I wish I did! It's all been scattered articles and the odd MO post (including some of yours, in fact)—thanks for the reference to Mackenzie! $\endgroup$ Jun 25, 2020 at 3:06
  • $\begingroup$ Very good answer, very close to this one as well. $\endgroup$
    – Pierre PC
    Jun 26, 2020 at 21:16
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    $\begingroup$ On obvious reference is Atiyah's 1957 paper Complex Analytic Connections in Fiber Bundles, where Atiyah writes "Definition: A connection in the principal bundle P is a splitting of the exact sequence A(P)." $\endgroup$
    – Ben McKay
    Jul 5, 2020 at 12:47
  • $\begingroup$ That is exactly where Atiyah first observed all this. $\endgroup$ Jul 5, 2020 at 13:58
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A mathematician specialising in $C^*$-algebra theory should probably know the following short exact sequences. These play a central role in the $K$-theory of $C^*$-algebras.

  • Let $A$ be a $C^*$-algebra with unitisation $A'= A \oplus \mathbb{C}$. There is an obvious split short exact sequence $$0 \to A \to A' \to \mathbb{C}\to 0$$ Hence, $K_i(A') \cong K_i(A)\oplus K_i(\mathbb{C})$.

  • Let $\mathbb{A}$ be the Toeplitz algebra and $\mathbb{T}$ be the unit circle and $H^2$ be the Hardy space. Then we have a short exact sequence $$0 \to B_0(H^2) \to \mathbb{A}\to C(\mathbb{T})\to 0$$ This can be used to give a proof of the Bott-periodicity theorem.

  • As mentioned in another answer, given a Hilbert space $H$, there is the short exact sequence $$0 \to B_0(H)\to B(H)\to B(H)/B_0(H)\to 0$$ which allows us to determine the $K$-theory of the Calkin algebra (via the six-terms exact sequence).

  • If $\mathbb{S}=\{f\in C(\mathbb{T}): f(1)=0\}$, there is also a trivial split short exact sequence $$0 \to \mathbb{S}\to C(\mathbb{T})\to \mathbb{C}\to 0$$ which allows us to determine the $K$-theory of the $n$-dimensional torus $\mathbb{T}^n$, i.e. we can determine the $K$-theory of $C(\mathbb{T}^n)$.

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  • $\begingroup$ For Bott-periodicity reasons $0 \to C_0(\mathbb{R}^2) \to C(S^2) \to \mathbb{C} \to 0$ is also important. The generator of $K_0(C_0(\mathbb{R}^2))$ comes from the Bott projection associated to to Hopf line bundle on $S^2$. Tensoring by $C_0(\mathbb{R}^2)$ performs the "double suspension". $\endgroup$
    – Zorngo
    Jun 1, 2021 at 4:28
  • $\begingroup$ @Zorngo I've been once told by the late Tamaz Kandelaki that there are similar short exact sequences involving $C(S^7)$ and $C(S^8)$ that give real Bott periodicity. Cannot remember now, I think he referred to some paper by Cuntz? $\endgroup$ Jun 1, 2021 at 8:58

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