# What do higher cohomologies mean concretely (in various cohomology theories)?

Superficially I think I understand the definitions of several cohomologies: (1) de Rham cohomology on smooth manifolds (I understand this can be probably extended to algebraic settings, but I haven't read anything about it) (2) Cech cohomology on Riemann surfaces, or schemes (3) Group cohomology in number theory and I have some rough understanding of interpreting cohomology functors as derived functors.

So my question is: what do higher cohomology groups mean concretely?

Some specifics:

For (1): closed forms modulo exact forms, but is there anything more concrete? It does solve some differential equations, but...is there more to it?

For (2): Serre duality implies the 1st cohomology group is dual to linearly independent meromorphic functions satisfying certain conditions wrt a divisor. How about higher cohomologies? My primary source is Forster's book, so the Serre duality treated there might not be the most general possible.

For (3): $H^0(G,A)$ is $G$-invariant elements of $A$, 1st cohomology is the $A$-torsors, 2nd cohomology is extensions of $A$ by $G$. How about higher cohomologies? My primary source is Artin's "Algebraic numbers and algebraic functions", and "Cohomology of number fields". The latter book (p.20, 2nd Edition) states very roughly that (my interpretation, sincere apologies to the authors if I misunderstand anything), higher cohomologies may not have concrete interpretations, but they play significant roles in understanding lower cohomologies and proving results about them.

PS: My background is (in case it's needed), very very rudimentary knowledge in analysis, algebra, algebraic geometry and number theory, but have not seriously learned any algebraic topology (though have seen the proof of Brouwer fixed point theorem via singular homology). I might have a tendency for the analytical and algebraic understanding of things (e.g. my primary impression of cohomology is that it's the obstruction of exactness, the need to extend exact sequences).

A side question: is it advisable to actually seriously learn algebraic topology to get a better idea of cohomology theories?

Thank you very much.

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There are conceptual pictures available, but they're sort of complicated: ncatlab.org/nlab/show/cohomology If you think in terms of extensions, higher group cohomology measures "higher extensions," the simplest example being en.wikipedia.org/wiki/… –  Qiaochu Yuan Sep 13 '11 at 3:29
@Qiaochu : I don't think that someone struggling to understand the basics would be enlightened by a discussion of topi and higher categories. In fact, most experts get by without ever studying these things... –  Adam Smith Sep 13 '11 at 3:59
@Adam: maybe. I personally find it useful to know that a conceptual explanation exists, even if I can't currently understand it. –  Qiaochu Yuan Sep 13 '11 at 4:07
@Qiaochu : That's pretty meta. In any case, I want to remark (largely for the OP) that the people at the nlab have an amazing ability to take things which I use every day and understand at a fairly deep level and write about them in such a way that I have no idea what is going on. It's probably best not to spend too much time there unless you have a really good reason to. –  Adam Smith Sep 13 '11 at 4:22
Thank you both for your kind help, I will keep your advice in mind. –  Poldavian Sep 13 '11 at 4:35

## 4 Answers

Your last question has an easy answer : it is very desirable to learn algebraic topology! It's the most concrete setting for a cohomology theory, and it would be hard to appreciate much (for instance) sheaf cohomology without at least some understanding of the algebraic topology background.

With that said, the following theorem of Thom gives what I consider the most concrete description of higher singular homology groups, at least over $\mathbb{Q}$. This is probably the best thing to try to understand first.

THEOREM : Let $X$ be a smooth manifold. Then for any $v \in H_k(X;\mathbb{Q})$, there exists a nonzero rational number $q$, a smooth compact, oriented $k$-manifold $M^k$, and a continuous map $\phi : M^k \rightarrow X$ such that $v = q \cdot \phi_{\ast}([M^k])$. Here $[M^k]$ is the orientation class of the manifold.

In other words, you should think of $k$-dimensional homology classes as being some kind of "weighted singular $k$-dimensional submanifold" of your space. If $X$ is a manifold of dimension at least $2k+1$, in fact, you can assume that the maps $\phi$ described about are embeddings, so the homology classes are actually weighted submanifolds.

Of course, you asked about cohomology, not homology. However, for compact $n$-manifolds we have Poincare duality which gives a natural identification between $H^{n-k}(M;\mathbb{Q})$ and $H_{k}(M;\mathbb{Q})$. You can thus transfer your intuition for homology over to cohomology.

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Thank you very much! I guess I will seriously learn algebraic topology, hopefully reaching the theorem you mentioned one day. By the way, do you happen to have a reference (e.g. introductory algebraic topology books) for this theorem of Thom's? –  Poldavian Sep 13 '11 at 3:07
Alas, it's actually a fairly hard theorem! For instance, it is false over $\mathbb{Z}$, so something subtle must be going on. It originally appeared in Thom's classic paper "Quelques propriétés globales des variétés différentiables", which is available in translation in the book "Topological Library". This is a little hard to read. There are definitely other proofs available, but I don't know a good one to recommend off the top of my head. The keyword if you want to search is the "Steenrod realization problem" (the problem of representing homology classes by manifolds). –  Adam Smith Sep 13 '11 at 3:15
As far as good books to read to learn basic algebraic topology, the most popular one today seems to be Hatcher's book, which is available for free on his webpage. There are many other books (for instance, I'm rather fond of Peter May's "Concise course in algebraic topology"), but none of them hold your hand as much as Hatcher. –  Adam Smith Sep 13 '11 at 3:17
Many thanks again! –  Poldavian Sep 13 '11 at 3:21
Very nice answer, Adam. In the case $dim(M)\geq 2k+1$, I have two difficulties with your statement that a k- homology class may be seen as a submanifold: 1) $\phi$ might not be an embedding. Actually you didn't even say that $\phi$ is differentiable. 2) $q$ might not be equal to 1. –  Georges Elencwajg Sep 13 '11 at 6:00

Learning algebraic topology, at least to the level of singular homology and cohomology, and Poincare duality, is essential if you plan to use homological techniques in your work with any sort of serious understanding.

It is not just that the simplicial techniques involved in singular homology and cohomology are important in themselves (although they certainly are!), but they provide a model for much else (e.g. the definition of Cech cochains). Also, group cohomology is an instance of singular cohomology (of a classifying space for the group), and so this provides one interpretation of the higher cohomology groups, and a method for proving things about them (and also a potential reason to care about them).

As another example, de Rham cohomology is perhaps best understood in terms of the de Rham theorem, which relates it to the singular cohomology of the manifold (and shows that there is "more to it" --- namely, the de Rham theorem is one of the first examples of the phenomenon of topological obstructions to solving certain differential equations, something which is a pretty general theme in mathematics). It can also be related to certain coherent cohomology spaces (in the context of smooth projective algebraic varities, e.g. Riemann surfaces), and then one also obtains a relationship between Poincare duality and Serre duality, which sheds light on both.

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+1 for "namely, the de Rham theorem is the first example of the phenomenon of topological obstructions to solving certain differential equations" –  Grumpy Parsnip Sep 13 '11 at 17:58
The homotopical obstructions to constructing a primitive of a function of a complex variable predates that, no? –  Mariano Suárez-Alvarez Sep 13 '11 at 18:35
@Mariano: Dear Mariano, Yes, I think you're right. An edit to less categorical language will ensue! Best wishes, –  Matt E Sep 13 '11 at 18:59

Take some algebra $\Lambda$, and consider its category of finitely generated left-modules $_\Lambda \operatorname{mod}$. It is folklore that for modules $M,N$, the group $\operatorname{Ext}^1 (M,N)$ controls extensions of $N$ by $M$, i.e. modules obtained by gluing $M$ and $N$ in some way. So suppose we know all the simple modules $S_i$ and all the $\operatorname{Ext}^1$-groups between them. You might guess that this was enough to recover the whole module category up to equivalence, since any module can be built by gluing together simples. You'd be wrong though, and there are easy counterexamples: take $\Lambda_3 = k[x]/x^3$ and $\Lambda_4 =k[x]/x^4$. These algebras have only one simple module (the trivial module $k$). Their module categories are different, but they have the same ext groups: in fact, the ext rings $$\operatorname{Ext}^* (k,k) = \bigoplus \operatorname{Ext}^n (k,k) \cong k[x,y]/x^2$$ for both algebras. So the ring structure on the graded ring formed from higher ext groups is not enough to get back the module categories. However, the ext ring carries what's called an $A_\infty$-structure: a sequence of higher-order operations generalizing the Massey products and satisfying certain axioms. It is a theorem that $_\Lambda\operatorname{mod}$ can be recovered from the ext ring $\operatorname{Ext}^*(\bigoplus S_i,\bigoplus S_i)$ together with its $A_\infty$-structure: see http://arxiv.org/abs/math/9910179 and Keller's other notes on this topic.

To summarise: the higher ext groups control the structure of the module category, but only when equipped with some extra operations.

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Here are some examples of higher cohomology groups from algebra:

1. A very concrete example of a higher cohomlogy group that "means something" is the Brauer group of a field. If $K$ is a field and $K^s$ its separable closure (think algebraic closure if $K$ is perfect, e.g. if it has characteristic 0), then the Galois group $\text{Gal}(K^s/K)$ acts on the multiplicative group $(K^s)^\times$, and the Brauer group is the second group homology group $$\text{Br}(K)= H^2(\text{Gal}(K^s/K),(K^s)^\times).$$ Now, what does it mean? The Brauer group classifies central simple algebras over $K$ up to Morita equivalence. The group structure corresponds to tensor product of algebras. This is a very real and concrete interpretation. For example $\text{Br}(\mathbb{R})\cong C_2$, the cyclic group of order 2, because any central simple algebra over $\mathbb{R}$ is Morita equivalent to $\mathbb{R}$ or to $\mathbb{H}$ - the quaternion algebra (you can think of Morita equivalence as saying "well there are also matrix algebras over these guys, but we don't treat them as anything new"). Also, $\mathbb{H}\otimes \mathbb{H}$ is Morita equivalent to $\mathbb{R}$, which reflects the fact that $\mathbb{H}$ represents an element of order 2 in the Brauer group.
The Brauer group is extremely important in class field theory, a branch of algebraic number theory (there are some nice introductions, e.g. on Milne's website).

2. Another concrete example of a derived functor with concrete meaning from the world of algebra is the Ext functor. Given $R$-modules $A$ and $B$, the group $\text{Ext}^n_R(A,B)$ classifies $n$-term extensions $$0\rightarrow B\rightarrow X_n\rightarrow\cdots\rightarrow X_1\rightarrow A\rightarrow0$$ up to suitable equivalence.

3. Finally, you might also want to read about the Schur multiplier of a finite group. But to understand what it really "means", you need to know a little bit of representation theory of finite groups.

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Thank you Alex B., for the concrete examples. I guess I am in the process to learn class field theory, which might be a good chance to understand the Brauer group you mentioned. I will need to think more about 2, and need to read things to understand 3. –  Poldavian Sep 13 '11 at 4:34