Interesting contexts in which the first few cohomology groups ($H^1$ and $H^2$) show up?

Question

I am learning about $H^n(G,A)$ and as part of it I'm cataloguing interesting examples of these groups showing up "in the wild", mostly for the case $n=1,2$. I'm looking for more isolated examples, rather than entire theories built around cohomology (e.g. class field theory). What I have so far:

1. $H^1(G,L)$ and $H^1(G,L^\times)$ being trivial for $L/K$ Galois (and its connection with Hilbert $90$ and Kummer theory).
2. $H^2(G,A)$ and group extensions, $H^1(G,A)$ and split extensions.
3. $H^2(G,L^\times)$ and Brauer groups.

Answer 1

$H^1(G,A)$ with $G$ a Galois group and possibly noncommutative $A$ shows up in the theory of Galois descent. There are many "incarnations" of this, but here the general idea. Let $X$ be some kind of "object defined over a field $K$" (it might be a projective variety, an algebraic group, a finite-dimensional vector space, a finite-dimensional algebra etc.). Let $L/K$ be a Galois extension. Then say that a form of $X$ with respect to the extension $L/K$ is another object $Y$ defined over $K$ of the same kind such that the base changes $X_L \cong Y_L$ are isomorphic over $L$. In a lot of situations, the isomorphism classes of forms are classified by the Galois cohomology group $H^1(\mathrm{Gal}(L/K),\operatorname{Aut}(X_L)$. A proof for a general statement can be found in Serre's book on Galois cohomology.

Some examples: let's work in the category of vector spaces and consider the vector space $X=K^n$. In this case, we can use that vector spaces are isomorphic iff they have the same dimension to see that $X_L\cong Y_L\implies X \cong Y$. Via the above statement, this implies that we get a generalized Hilbert 90 which says that $H^1(\mathrm{Gal}(L/K),\operatorname{GL}_n(L))$ vanishes.

Here's a slight variation: one can consider a group $H$, unrelated to the Galois group and fix a representation of $H$ on a finite-diimensional $K$-vector space $V$. Then we can also consider the set of $K$-representations $W$ such that $V_L \cong W_L$. Again these are classified $H^1(\mathrm{Gal}(L/K),\operatorname{Aut}_{L[H]}(V_L))$ and again this vanishes, but this is a more nontrivial result, it follows from the Noether-Deuring theorem.

Now let's work in the category of varieties and consider $X=\Bbb P_K^n$, the projective space. Then we get that $\mathrm{Aut}(X_L)\cong \mathrm{PGL}_{n+1}(L)$. Forms of $X$ are called Severi-Brauer varieties and they are classified by $H^1(\mathrm{Gal}(L/K),\mathrm{PGL}_{n+1}(L))$. They are related to the Brauer group: we get from the exact sequence $1 \to L^\times \to \operatorname{GL}_{n+1}(L) \to \mathrm{PGL}_{n+1}(L) \to 1$. A morphism $H^1(\mathrm{Gal}(L/K),\mathrm{PGL}_{n+1}(L)) \to H^2(\mathrm{Gal}(L/K),L^\times)$ which is injective by Hilbert 90. One can show that every element of the Brauer group lies in the image of this for some $n$.

One can also take other projective varieties (though there's not necessarily a relation to the Brauer group as above).

Here's a more exotic example: consider the standard octonion algebra $\mathbb O_K$ over $K$ (same structure constants as over $\Bbb R$). Then forms of this are general octonion algebras over $K$. The automorphism group $\mathrm{Aut}(\mathbb O_L)$ is the set of $L$-rational points of the exceptional algebraic group $G_2$. We get that $H^1(\mathrm{Gal}(K^{sep}/K),G_2(K^{sep}))$ classifies generalized octonion algebras.

One can also take other finite-dimensional algebras over $K$.

Answer 2

To expand on my comment, let $X$ be an extension (not necessarily split) of the abelian group $A$ by $G$. That is, $X$ has a normal subgroup $A$ with $X/A \cong G$.

Let $Z$ be the subgroup of ${\rm Aut}(X)$ consisting of those automorphisms that fix the subgroup $A$ and induce the identity map on both $A$ and $X/A$. Then $\alpha \in Z$ maps $x \in X$ to $x\delta(x)$, where $\delta(x) \in A$ and $\delta(x)$ is constant on the coset $xA$, so $\delta$ induces a map $G \to A$.

It is routine to check that this map is a derivation, so it lies in $Z^1(G,A)$, and conversely any such derivation arises from some $\alpha \in Z$. So $Z \cong Z^1(G,A)$.

Furthermore, $\alpha \in {\rm Inn}(X)$ if and only if the corresponding derivation lies in $B^1(G,A)$, so $H^1(X,A)$ is isomorphic to the subgroup of ${\rm Out}(X)$ induced by elements of $A$.