62
votes
Accepted
What is homology anyway?
Let's take coefficients in a field $k$, for simplicity.
On 2): the singular cohomology of a topological space $X$ is the dual of its singular homology, almost by definition. But if $X$ is a space for ...
- 17.6k
41
votes
Accepted
Why study Higher Sheaf Cohomology?
I think these kinds of lifting-based statements are the wrong place to start. For me these arguments about lifting and such are most convincing as answers to questions like "What are cohomology ...
- 137k
27
votes
What is homology anyway?
For a long time (and still today), I very much shared the confusion of the OP. I think Jacob Lurie gives a very clear take on the standard perspective, but Mike Shulman does have a very valid ...
- 19.9k
27
votes
What is prismatic cohomology?
I just take a quick opportunity to share what a prism is, and why it is called like that (as I learned from Lars Hesselholt). All the theory is developed relatively to a fixed prime $p \in \mathbb{N}$....
- 1,066
27
votes
Accepted
A natural construction of real numbers?
So here is my attempt to reconstruct the construction...
Suppose $f\colon\mathbb Z\to\mathbb Z$ satisfies $|f(m+n)-f(m)-f(n)|\le M$ as $m,n$ run over $\mathbb Z$. Then setting $m=n=2^k$, we see
$|f(2^{...
- 22.5k
26
votes
What is homology anyway?
I generally think about the relationship differently than Jacob, probably because I'm coming from an algebraic topology background rather than an algebraic geometry one. I would say that if $\mathcal{...
- 65.1k
25
votes
Accepted
Image of a map on cohomology rings
No. Consider the Hopf map $\eta:S^3\to S^2$. If there were such a space $Z$, it would have $\widetilde H^*(Z)=0$, so at the very least $Z$ would be stably trivial, forcing $\eta$ to be stably trivial; ...
- 12.5k
24
votes
Accepted
Has anyone seen a nice map of multiplicative cohomology theories?
I'm not sure I understand what "the" map is here, but I'll attempt to answer
the questions that were asked in the body of the question. Sorry if I'm just
saying things that you already know.
$\...
- 5,550
23
votes
Accepted
Is cohomology always related to topology?
While it is always possible to introduce topology, it is not always the obvious or most useful thing to do. So at least in this sense, there are notions of cohomology that do not immediately connect ...
- 9,549
22
votes
A manifold is a homotopy type and _what_ extra structure?
Igor is giving a good reference to the topic. For completeness, my education, and satisfaction of other reader's laziness, I'm going to give a rough outline here.
A Poincaré complex is, very ...
- 5,545
22
votes
Hodge theory (after Deligne)
A brief answer.
First of all, the results are miraculous. Deligne's Hodge II and Hodge III give just a few example applications of the kind of results you can prove using mixed Hodge theory; these ...
- 39.3k
21
votes
Accepted
Church-Farb on the cohomology of pure braid groups and character polynomials, intuition behind proof of result?
This turns out to be a completely general phenomenon for configuration spaces on any open manifold, though we did not know this at the time; it came in the later paper "FI-modules and stability for ...
- 8,156
21
votes
Accepted
A manifold is a homotopy type and _what_ extra structure?
You are talking about the (much studied) Poincare duality spaces. For a survey, see the very nice one by John Klein: (seems to be unpublished, but dates to April 2010).
- 95.6k
20
votes
Accepted
Cohomology of $ko,tmf,MSpin,MString$ with coefficients $\mathbb{Z}/p$ for odd primes $p$
[Some folks started emailing me about this so I supposed I should click on the link and post what I knew...]
The homology of tmf at all primes as a comodule over the dual steenrod algebra is Theorem ...
- 266
19
votes
Accepted
A spectral sequence for computing cohomology of a space from that of its strata
Let $X = T_n \supset T_{n-1} \supset \cdots \supset T_{-1} = \varnothing$ be a topological space filtered by closed subspaces, where for simplicity I assumed the filtration bounded. Then there is a ...
- 39.3k
18
votes
Why study Higher Sheaf Cohomology?
I think you're absolutely right that the function $(i\in \mathbb N)\mapsto $interestingness($H^i$) is a rapidly decreasing function. I heard that Gel$'$fand compared it to the successive derivatives ...
- 27.7k
18
votes
Accepted
Wu formula for manifolds with boundary
A relative Wu formula for manifolds with boundary is discussed in Section 7 of
Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.
In particular, ...
- 35k
17
votes
Motivations to study the cohomology of the moduli space of curves
As the title to Mumford's famous paper "Toward an enumerative geometry..." suggests, knowing the cohomology / cycle theory of the moduli space of curves allows one to answer enumerative geometry ...
- 5,910
17
votes
Accepted
Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex?
If $G$ is compact, the inclusion $H(M^G) \to H(M)^G$ is an isomorphism. The inverse map is defined as follows: Take a class $\omega$ in $H(M)^G$ and lift it to a closed form $\alpha \in \Omega(M)$. ...
- 151k
17
votes
Accepted
Why Use Hypercohomology When Defining the de Rham Cohomology of a Smooth Scheme over $k$?
This is pretty much explained in the comments, but let me put it into an answer. One wants algebraic de Rham cohomology to be isomorphic to the usual de Rham cohomology (using $C^\infty$ forms) when $...
- 34.2k
17
votes
Accepted
Unifying "cohomology groups classify extensions" theorems
$\newcommand{\cA}{\mathcal{A}}\newcommand{\Ext}{\mathrm{Ext}}\newcommand{\Hom}{\mathrm{Hom}}$Let $\cA$ be an abelian category; then, $\Ext_\cA^i(A,B)$ is literally $\Hom_{D(\cA)}(A, B[i])$, where $B[i]...
- 5,550
17
votes
The (current) obstructions for a cohomological interpretation of the Riemann zeta function
One can't give a complete answer to this question without first understanding how etale cohomology does give a cohomological interpretation of the zeta function in the function field case.
Let $X$ be ...
- 137k
16
votes
Accepted
What is the $\mathbb{Z}_2$ cohomology of an oriented grassmannian?
I was surprised to learn that the ring structure of $H^*({\rm Gr}^+(k,n);\mathbb{Z}_2)$ seems to be unknown, in general. The ring structure in the case $k=2$ is given in
Korbaš, Július; Rusin, Tomáš, ...
- 35k
16
votes
"a sign that one should be computing K-theory"
Usually you are computing $H^*(X)$ or $K^*(X)$ for a reason; for example if $H^*(X)\not\simeq H^*(Y)$ then you know that $X$ and $Y$ are not homotopy equivalent, but if $H^*(X)\simeq H^*(Y)$ and this ...
- 55k
16
votes
Accepted
An intuitive explanation for group cohomology via cochains?
What I'm going to say is pretty much the same that JK34 has written in their answer, but in a more elementary approach that is hopefully adding some insight.
Suppose that you want to look at the "...
- 389
16
votes
Is there an $\mathbb{R}$-valued cohomology theory for varieties over $\mathbb{F}_p$?
Let me advertise a conjectural positive answer to a slightly different question. This actually works not only over $\mathbb F_p$ but over its algebraic closure $\overline{\mathbb F}_p$.
Consider the ...
- 19.9k
16
votes
Accepted
A non-Abelian de Rham complex?
What is being described in the main post is simply the de Rham (crossed) complex valued in a Lie group (not necessarily commutative).
See, for example, Section 6.2 in Anders Kock's Synthetic Geometry ...
- 36.3k
16
votes
Accepted
Are all degree-1 cohomology operations Bocksteins?
Yes. For $i\ge1$ you can build $K(G,i)$ from the Moore space $M(G,i)$ by adding cells of dimension $\ge i+2$, so $H_i(K(G,i); Z) = G$ and $H_{i+1}(K(G,i); Z) = 0$. Hence $Ext(G, H) \cong H^{i+1}(K(G,...
- 8,958
16
votes
Accepted
Comparing singular cohomology with algebraic de Rham cohomology
This is the subject of periods: recall that the de Rham isomorphism between $H^k_{\text{dR}}(X/K) \otimes_K \mathbf C = H^k_{\text{dR}}(X_{\mathbf C}/\mathbf C)$ and $H^k_{\text{sing}}(X(\mathbf C),\...
- 27.3k
16
votes
Accepted
Is there a ring stacky approach to $\ell$-adic or rigid cohomology?
This is an interesting question.
First, I think the [PS] reference does not give the "correct" Betti stack. In my notes on 6 functors, I define a different stack $X_B$ such that $D_{\mathrm{...
- 19.9k
Only top scored, non community-wiki answers of a minimum length are eligible