136 votes

Short exact sequences every mathematician should know

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 ...
76 votes

Short exact sequences every mathematician should know

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 $\...
75 votes

Short exact sequences every mathematician should know

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 ...
67 votes

Short exact sequences every mathematician should know

I find it hard to believe that three days have gone by and no one has explicitly mentioned $$ 0 \to \Bbb Z \to \Bbb R \to \Bbb S^1 \to 0 $$
61 votes

Short exact sequences every mathematician should know

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, ...
47 votes

Short exact sequences every mathematician should know

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 $...
46 votes

Understanding a quip from Gian-Carlo Rota

I think Abdelmalek Abdesselam and William Stagner are completely correct in their interpretation of the words "Behind" and "one immutable source" as describing one theory, the theory of symmetric ...
46 votes

Short exact sequences every mathematician should know

"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: $$0 \to K(H) \to B(H) \to Q(H) \to 0.$$ $...
45 votes

Short exact sequences every mathematician should know

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}(...
41 votes

Short exact sequences every mathematician should know

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 ...
40 votes
Accepted

Morava K-theories for dummies?

This is a result in algebraic topology, where we study the structure of topological spaces $X$. One early way to do this is to calculate a thing called $H_*(X)$, the ordinary homology of $X$. Later ...
Neil Strickland's user avatar
39 votes

Vladimir Voevodsky's works

Perhaps one of the biggest ideas that VV was pursuing is the Initiality Conjecture for Martin-Löf type theory (with universes). A rough idea is that from the rules for a type theory, one can define a ...
35 votes

Short exact sequences every mathematician should know

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 ...
29 votes

Vladimir Voevodsky's works

Aside from his work on the foundations of mathematics, which others have already elaborated on, earlier in his career Voevodsky also proved the Milnor conjecture in algebraic geometry. The Milnor ...
29 votes

Short exact sequences every mathematician should know

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 $$...
28 votes

Short exact sequences every mathematician should know

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.
27 votes
Accepted

Torsion in the Atiyah–Hirzebruch spectral sequence of a classifying space

Of course, in any spectral sequence $E_{r+1}$ is a subquotient of $E_r$ (the kernel of $d_r$ divided by the image of $d_r$). And in general new torsion can appear in the sense of torsion elements in $...
Tom Goodwillie's user avatar
27 votes

Short exact sequences every mathematician should know

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 \...
26 votes

Short exact sequences every mathematician should know

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 $...
24 votes
Accepted

H-space structures on non-sphere suspensions?

If $Y$ is a connected CW-complex of finite type which is both an H-space and a co-H-space, then $Y$ has the homotopy type of $S^1$, $S^3$, $S^7$ or a point. This is a result of Robert West: Robert W. ...
Mark Grant's user avatar
24 votes

Understanding a quip from Gian-Carlo Rota

I think you are misinterpreting the quote. In the last sentence, the word "source" does not mean "source of these theories (K-theory, categories, group representations", but "source of the theory of ...
23 votes

Short exact sequences every mathematician should know

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 ...
23 votes
Accepted

Does Waldhausen K-theory detect homotopy type?

The answer to the question Does the homotopy type of $𝐴(𝑋)$ determine the homotopy type of $𝑋$? is No in general. As you say, $A(X)$ is determined by the homotopy type of $\Sigma^\infty \Omega X_+...
Gregory Arone's user avatar
21 votes

The $K$-theory homology of the Eilenberg-MacLane spectrum

We have $KU_*(H\mathbb Z) = \pi_*(KU\wedge H\mathbb Z)$; this ring is concentrated in even dimensions and carries an isomorphism between the additive and multiplicative formal group law, hence is ...
Bertram Arnold's user avatar
21 votes
Accepted

Why does K-theory need schemes to be Noetherian?

You don't need the Noetherianness hypothesis to talk about K-theory. But the definition you propose in your question is not suited for the most general case. From a notion of K-theory we want at least ...
Denis Nardin's user avatar
  • 16.2k
21 votes

Short exact sequences every mathematician should know

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{...
21 votes

Magic behind idempotent-complete categories a.k.a. why (sometimes) be Karoubian is sexier than be Abelian

I don't know if this is going to answer your question but here's some relevant background. Splitting idempotents has a very special property from a categorical point of view: it is an absolute colimit ...
Qiaochu Yuan's user avatar
21 votes
Accepted

Problems concerning subspaces of $M_{n}(\mathbb{Q}) $

Let's call this maximal dimension function $\rho_{\mathbb{Q}}:\mathbb{N}\to\mathbb{N}$, i.e., $\rho_{\mathbb{Q}}(n)$ is the largest possible dimension of a subspace $N\subset M_n(\mathbb{Q})$ such ...
Robert Bryant's user avatar
20 votes
Accepted

Entering to the K-theory realm

I think that doing algebraic K-theory properly certainly requires a good background on stable homotopy theory, that is to say the homotopy theory of spectra. Unfortunately there are not many textbooks ...
Denis Nardin's user avatar
  • 16.2k
20 votes

The $K$-theory homology of the Eilenberg-MacLane spectrum

Since for any two spectra $E,F$ we have $$ E_n(F)=[\mathbb{S}^n,E\wedge F]\cong [\mathbb{S}^n,F\wedge E] = F_n(E), $$ you may as well ask about $H_n(KU;\mathbb{Z})$, the integral homology of the ...
Mark Grant's user avatar

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