TITLE: What do higher cohomologies mean concretely (in various cohomology theories)?
QUESTION [15 upvotes]: 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.

REPLY [5 votes]: 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.