I am a PhD student in algebraic / arithmetic geometry and I never took a formal course in algebraic topology, even though I have some basic knowledge.

In algebraic geometry we deal exclusively with sheaf cohomology since we care about non-constant sheaves. But I feel, maybe in my naivety, that a lot of the important results (for usual topological spaces) are only true for singular and simplicial cohomology when they coincide with sheaf cohomology (Alexander duality and "$H^i=0$ for $i>$ the covering dimension" come to mind).

With that in mind, I wonder if it is worth for someone with a similar background to study the details of a first course in algebraic topology. (Perhaps on the level of Hatcher's book.) Do I lose something by just thinking in terms of sheaf cohomology?

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    $\begingroup$ Even as an algebraic geometer you will often care about constant sheaves, and then you will be happy about the fact there exist comparison theorems between say etale and singular cohomology. In many cases this gives you means if computing cohomology of spaces from AG using tools from AT. $\endgroup$ – Wojowu Jul 12 at 12:02
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    $\begingroup$ One very specific example I have in mind is computation of (co)homology using CW complex decomposition. It is very easy to write down such a structure for the complex projective $n$-space. I believe you can do something similar for Grassmannians. So it's not really pieces of general theory that we miss but the ability to do some specific calculations. $\endgroup$ – Wojowu Jul 12 at 13:57
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    $\begingroup$ I'm surprised no one has mentioned this yet but if you're an arithmetic geometer, it seems like the main advantage to being familiar with AT is because of etale cohomology. Many of the theorems/results one encounters in singular cohomology e.g. excision, the existence of cycle class maps, Lefschetz hyperplane theorems, Poincare duality, etc have their counterparts in etale cohomology, but the proofs are infinitely harder. To me that by itself is already justification to learn AT: You get to see how these things are proven in an easier context. $\endgroup$ – David Benjamin Lim Jul 12 at 14:41
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    $\begingroup$ @Asvin I believe there is no known algebraic proof of that fact, actually! $\endgroup$ – Raju Jul 12 at 20:19
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    $\begingroup$ I remember talking to a grad student in Bonn years ago. I was explaining something on the board, and it involved me writing down the cohomology of $S^1$. They exclaimed: "I haven't seen that computation before, but it looks like the étale cohomology of the multiplicative group"! $\endgroup$ – Geordie Williamson Jul 12 at 21:58

Sheaf cohomology is a powerful tool, but it isn't a replacement for all of basic algebraic topology. For example, fundamental groups and homology some topics that would get lost. And these topics are certainly relevant to algebraic geometry. Also, as pointed out in the comments, you would lose valuable intuition if you just stuck to the sheaf cohomology viewpoint.

Let me expand my original answer a bit. Let me focus on the simplest example, where $X$ is a smooth complex projective curve of genus $g$. One learns in topology that $X$ is obtained by identifying the sides of a $2g$-gon in the standard way. One can use this to extract 2 things about $X$.

  1. One gets the homology $H_1(X,\mathbb{Z})=\mathbb{Z}^{2g}$, with its intersection pairing equal to the standard symplectic form. For an algebraic geometer, this corresponds to the lattice of the Jacobian of $X$ together with its Riemann form. In particular, this is a principal polarization.

  2. Also one gets the familiar presentation of the fundamental group $$\pi_1(X)= \langle a_1\ldots a_{2g}\mid [a_1,a_2]\ldots[a_{2g-1}, a_{2g}]\rangle$$ But why should an algebraic geometer care about this? Answer: because it tells us what etale covers of $X$ look like. The etale fundamental group of $X$ is the profinite completion of the above group. This is also true for the prime to $p$ part if $X$ lives in positive characteristic by lifting. Note that Grothendieck uses this reduction to the topological case in SGA1.

  • $\begingroup$ @DonuArapura, of course I am wrong here, but the intuition I have is that (while more intuitive) homology is just a weaker invariant than cohomology. And the "topological fundamental group" doesn't work well in AG anyway. So if you could explain this a little further, that would be great :) $\endgroup$ – Gabriel Jul 12 at 12:39
  • $\begingroup$ Thank you for spelling out some specific examples. $\endgroup$ – Wojowu Jul 12 at 13:59
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    $\begingroup$ @Gabriel Your intuition is exactly the opposite of the truth: homology is a finer invariant of cohomology (assuming you endow both of them with all their structure - it's not fair to give the product in cohomology but not the coproduct in homology!) $\endgroup$ – Denis Nardin Jul 12 at 14:35
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    $\begingroup$ DonuArapura, those are two examples that hit particularly close to the heart. Thank you! Also, @DenisNardin, I absolutely didn't imagine that. Very cool! $\endgroup$ – Gabriel Jul 12 at 14:42
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    $\begingroup$ @Z.M You are right, under finiteness hypothesis they are the same data $\endgroup$ – Denis Nardin Jul 13 at 6:32

I am an algebraic / arithmetic geometer. I never took a formal course in algebraic topology, but I picked up bits and pieces in various places. I know I looked at Hatcher's book in grad school, but I don't think very much, and I certainly didn't study it systematically.

In my opinion, it would certainly be helpful to know enough about singular cohomology that, if you see a result you need about cohomology written by topologists in the language of singular cohomology, you are not totally lost, and can see if it is relevant for a problem you care about.

For me this involves knowing most of the definitions and some simple arguments, but not knowing the hard proofs. But my approach to a lot of fields involves learning the definitions and intuitions and skipping the hardest proofs, and this might not work as well for everyone.

The same thing, by the way, is true for de Rham cohomology - understanding it to the point that you have some intuition for why it works, can compute some examples, and will likely get what it means if someone expresses a mathematical statement in its language, will almost certainly come in handy, but you don't need to know everything about it.

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    $\begingroup$ Nice point about de Rham cohomology - though the OP doesn't talk about it explicitly, I think a same argument could go for it. And yet de Rham cohomology is fairly important to know something about, for instance because of the algebraic de Rham cohomology or Hodge theory. $\endgroup$ – Wojowu Jul 12 at 14:01
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    $\begingroup$ Perhaps we could say that the de Rham cohomology is really $H^i(X,\Omega^\bullet_X)$ but we're lucky when studying smooth manifolds because then the existence of partitions of unity implies that the $\Omega_X^i$ are soft, hence acyclic, hence we can calculate the cohomology as $H^i(\Gamma(X,\Omega_X^\bullet))$, which is its usual definition. (And which doesn't work for algebraic de Rham cohomology.) That being said, I understand how knowing the simpler definitions can be of great aid intuition-wise and for talking with other people / reading articles of other fields. $\endgroup$ – Gabriel Jul 12 at 14:40
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    $\begingroup$ @Gabriel The question is entirely whether you are sufficiently comfortable with these slightly-roundabout routes to be able to do the same fundamental computations that people can do the direct way. If you can use de Rham cohomology to calculate the cohomology of a torus, and use a series of excisions together with the Bruhat decomposition to calculate the cohomology of a Grassmanian, you're in good shape in my opinion. $\endgroup$ – Will Sawin Jul 12 at 18:08
  • $\begingroup$ I guess that it is not easy to recover the torsion data from the de Rham cohomology (various versions of étale comparison)? $\endgroup$ – Z. M Jul 12 at 20:18
  • $\begingroup$ @Z.M Correct, I don't know any way to recover torsion data from de Rham cohomology. If you're interested in torsion, you can use de Rham for intuition sometimes, but you want to use another method. $\endgroup$ – Will Sawin Jul 12 at 20:21

The very advantage of algebraic topology over algebraic geometry is the existence of a segment. From a segment it's easy to construct triangles, tetrahedra, and simplices in general. There are at least three reasons why you could enjoy simplicial stuff.

Firstly, if you end up doing complex geometry, comparison theorems allow you to compute more easily the cohomology of the space. I have seen you asked: what you can do with simplicial cohomology that you can't do with sheaf cohomology? Well, if you are lucky enough, you can find a (finite) CW decomposition of the (compactified) space you are studying. If you are not comfortable with these words, just think about triangulating a surface. From algebraic topology tools you then have an explicit finite dimensional complex that compute the cohomology, that in degree n is spanned by the free group of n-dimensional cells. A very easy consequence in the "triangulating a surface" example is that you can compute the genus by just knowing how many faces, segments and points you used. You will never have something like that in algebraic geometry, because you don't have cells! Another example: the cohomology of the (real or complex) projective space is really easy from this viewpoint, but the computations with Cech cohomology in algebraic geometry takes a few pages (the one I have seen on Hartshorne, at least).

Secondly, if you ever want to do some infinity category stuff, that is entering the algebraic geometry side of Lurie, you definitely want to learn infinity categories. While a category is intuitively made of points and directed segments, an infinity category contains all the superior cells that encode "higher deformations". In other words, an infinity category is a particular kind of space. This is the part I enjoy the most, because I believe my way of thinking about geometry has deeply changed since I have learnt this stuff. Personal taste, anyway.

At last, there is another instance of simplicial being the higher notion of an ordinary concept. The keyword is "derived" stuff. Take for example the Verdier duality: the compact support transport functor has not generally a right adjoint in the category of sheaves. That's why you have to move to the derived category of chain complexes of sheaves. We then see a chain complex as a "deformation" of the original abelian object. Maybe you know from the Cold Kan correspondence that in many cases the (positively graded) chain complexes of abelian objects correspond to simplicial such objects. Instead of having just one d, you have different $d_i$'s, which do not commute between them. I feel that such simplicial identities are the non commutative generalizations of $d^2 =0$.

This suggests that when working with a non abelian category, the right "derived" generalization of objects in a category $C$ are simplicial objects with values in $C$. That's why one talks about simplicial rings, simplicial schemes, simplicial sheaves... There is a whole field called derived algebraic geometry that makes substantial use of such generalizations. If I am not wrong, the first time I have seen the definition of a stack was in a similar context, even if I don't remember the details since now I have a more hands-on definition. You can look up Vezzosi and Calaque. Such techniques allow to address some very delicate problems (which I rarely understood, unluckily).

  • $\begingroup$ Dear @AndreaMarino, what exactly do you mean by "the existence of a segment" in algebraic geometry? $\endgroup$ – Gabriel Jul 12 at 14:30
  • $\begingroup$ Sorry, I used a bad phrasing. I mean that in algebraic topology there is an object that models $[0, 1]$, a segment between two points. There is no such thing in algebraic topology: you can only model lines. Do you agree with this statement? This is the principal failure of not having cell decompositions in AG. $\endgroup$ – Andrea Marino Jul 12 at 19:14
  • $\begingroup$ In analytic geometry (either complex or $p$-adic), one has unit polydisks. $\endgroup$ – Z. M Jul 12 at 20:11
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    $\begingroup$ Isn't that because in such fields you can use inequalities, which does not make sense in general fields? I am not an expert though $\endgroup$ – Andrea Marino Jul 12 at 21:18
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    $\begingroup$ @AndreaMarino I get it now! Thank you :) $\endgroup$ – Gabriel Jul 13 at 7:02

The most basic examples in algebraic geometry are algebraic curves. Of these, the simplest are the smooth irreducible complex algebraic curves, which correspond to compact Riemann surfaces. Understanding extremely basic results like the Riemann-Hurwitz formula requires one to understand the algebraic topology of surfaces, including classification by genus and the formula for genus in terms of a CW decomposition.

If you are going to have any chance of understanding sheaf cohomology, you should understand the proof of Riemann-Roch for complex algebraic curves. In preparation, besides knowing this rudimentary algebraic topology you should also know some complex analysis (of one variable at least), and complex de Rham cohomology. However, if you really want to understand it, the original home of sheaf cohomology is several complex variables, and especially Cartan's theorems A and B. How can you understand what a quasi-coherent sheaf is if you do not know what it is defined by analogy with?

I would suggest reading Rick Miranda's "Algebraic Curves and Riemann Surfaces", and then "Principles of Algebraic Geometry" by Griffiths and Harris. This is where the geometric intuition comes from for other "less geometric" parts of Algebraic Geometry. Maybe you can develop your own intuitions without understanding this material, but for me it would be impossible.


I'm turning my comment above into an answer, as I realized there was more that I wanted.

IMO, if you're an arithmetic geometer the main motivation for learning singular cohomology is that it is intimately related to etale cohomology, in the sense that both of these are the simplest examples of a Weil cohomology theory. I won't recall all the axioms but state the first two (at least as defined in the wikipedia article here):

Let $k$ be a field, $X/k$ a smooth projective variety of dimension $d$, and a $K$ a coefficient field of characteristic zero (for singular you often take $K = \mathbf{C}$ while for etale you take $K = \mathbf{Q}_\ell$)

  1. $H^i(X)$ is finite dimensional.
  2. $H^i(X)$ is zero for $i < 0$ and $i > 2d$.

In the topological setting, each of these is somewhat straightforward to prove because $X(\mathbf{C})$ has the structure of a real $2d$-dimensional CW-complex: One deduces (1) and (2) from the corresponding simplicial cochain complex that computes the cohomology of $X(\mathbf{C})$.

On the other hand in the etale setting these are not trivial to prove at all. For instance, the vanishing beyond degree $2d$ is a result of Artin for varieties over a separably closed field. To prove this result you need to fiber $X$ in curves to reduce to the case of a curve, and even then (may I say) the result is not trivial: Grothendieck was confused for a long time on how to compute the etale cohomology of curves!

As you keep going down the list of axioms for a Weil cohomology theory, you realize immediately that the corresponding statements in singular cohomology are much easier to prove. This alone should be enough to motivate you to care about singular cohomology.

Why is it important to realize that etale cohomology is a Weil cohomology theory? Well if you're an arithmetic geometer (with high probability) you will care about the Weil conjectures. The first two statements - rationality of zeta and the functional equation - pop out immediately as a consequence of the axioms of a cohomology theory. For instance, rationality is a consequence of the finite dimensionality of cohomology and the Lefschetz trace formula, while rationality from Poincare duality.


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