I am coming at the 2nd cohomology group in Group Cohomology from the perspective of the Group Extension Problem (or rather the group central extension problem, which perhaps more closely related to the problem of lifting irreducible projective representations --- "From Schur's lemma, it follows that the irreducible representations of central extensions of $G$, and the irreducible projective representations of $G$, are essentially the same objects" --- than it is to the full group extension problem).
Anyways, in this handout https://math.stanford.edu/~conrad/210BPage/handouts/gpext.pdf (I'll only be working in the context of group central extensions, i.e. as Conrad says in the paragraph preceding $\S2$ taking $M$ to be an abelian group with trivial $G$-action, instead of a more general $G$-module), Brian Conrad starts with a group $G$, and an abelian group $M$, and describes an arbitrary group $E$ s.t. $M\leq Z(E) \trianglelefteq E$ and $G \simeq E/M$ (with quotient projection map $\pi: E \twoheadrightarrow E/M \simeq G$) as follows (I'm just summarizing what Conrad wrote for my own sake):
- We can describe the underlying set of $E$ by choosing a "section" $s:G\to E$ so that $s(g)$ is a unique representative of the $M$-coset $\pi^{-1}(g) = M\cdot s(g)$ of $E$ ----- (I think another terminology for $s$ is "set-theoretic right splitting of the s.e.s. $1 \to M\hookrightarrow E\twoheadrightarrow G \to 1 )$), ----- so as a set $E = \{m\cdot s(g): m\in M, g\in G\}$ (the "$\cdot$" denotes multiplication in $E$), which is isomorphic to the set $M \times G=\{(m,g): m\in M, g\in G\}$
- We can then understand the multiplication of such elements: $(m\cdot s(g)) \cdot (m' \cdot s(g')) = (m\cdot m')\cdot (s(g)\cdot s(g'))$ ----- (using that $M \leq Z(E)$; in the general case that $M$ is an arbitrary $G$-module, i.e. abelian group with arbitrary $G$-action, that will specify in advance "they way that $s(g)$ acts on $M$ by conjugation within $E$", allowing us to replace $(m\cdot s(g)) \cdot (m' \cdot s(g')$ with $m\cdot (s(g) \cdot m' \cdot s(g)^{-1} )\cdot s(g)\cdot s(g') = (m\cdot (g.m'))\cdot (s(g)\cdot s(g'))$) ----- by observing that $\pi$ being a group hom. means that $\pi(s(g)\cdot s(g'))= \pi(s(g))\pi(s(g'))=gg' \implies s(g)s(g') \in \pi^{-1}(gg') = M \cdot s(gg') \implies$ there is a function $c: G \times G \to M$ s.t. $s(g)s(g') = c(g,g')s(gg')$ so that $$(m\cdot s(g)) \cdot (m' \cdot s(g')) = (m\cdot m')\cdot (s(g)\cdot s(g')) = (m\cdot m' \cdot c(g,g'))\cdot s(gg'),$$ or through the language of the set-theoretic isomorphism $E \simeq M \times G$: $$(m,g) (m,g') := (m\cdot m' \cdot c(g,g'), gg').$$
- The associativity of this product is the 2-cocycle condition, saying that $c\in Z^2(G,M)$.
- And finally, choosing a different section $s':G \to E$ (i.e. "replacing $s$ with $s' = f\cdot s$ for a function $f : G \to M$") replaces $c$ with $c' = c + \delta f$ where $\delta$ is precisely the 2-coboundary map, and so $c-c' \in B^2(G,M)$.
- Therefore, $[c] \in H^2(G,M) := Z^2(G,M)/B^2(G,M)$ is "independent of [the choice of section] $s$ and so depends only on the isomorphism class of the given extension structure $E$ of $G$ by $M$".
- And conversely, given $c$ satisfying the 2-cocycle equation and the above multiplication law on the set $M \times G$, we do get an extension $E$ of $G$ by $M$ (with $M \leq Z(E)$; or more generally $E$ will "induce the given G-module structure on M via conjugation on the extension structure"), with any 2-cocycle $c'$ in the same cohomology class giving rise to an "extension structure ... isomorphic to the one constructed from $c$".
So this is a very hands-on, "concrete" construction and is a great first taste of cohomology and all this cocycle modded out by coboundaries business. One can then search up the general cocycle/coboundary definition of group homology (for instance on Wikipedia https://en.wikipedia.org/wiki/Group_cohomology#Cochain_complexes):
- Define $C^n(G,M)$ ("the $n$-cochains") to be the group (adding functions pointwise) of all set-functions $G^n \to M$.
- Define the "coboundary homomorphisms" $d^{n+1}: C^n(G,M) \to C^{n+1}(G,M)$ by the following formula (which looks like it dropped straight out of the sky) \begin{equation} [d^{n+1} f](g_0,\cdots,g_n) = g_0 f(g_1,\cdots,g_n) + \sum_{j = 1}^n (-1)^j f(g_0,\cdots, g_{j-1} g_{j}, g_{j+1}, \cdots, g_n) + (-1)^{n+1} f(g_0,\cdots, g_{n-1}) \end{equation}
- Define $Z^n(G,M):= \ker (d^{n+1})$ and $B^n(G,M) = \operatorname{im}(d^n)$, and then $H^n(G,M) = Z^n(G,M)/B^n(G,M)$.
Of course, I found this definition of the coboundary map very unsatisfactory/arbitrary, so I searched "group cohomology boundary definition intuition", and found this MO thread https://mathoverflow.net/questions/346765/an-intuitive-explanation-for-group-cohomology-via-cochains, which tells us that
"the group cohomology of $G$ is exactly the singular cohomology of $BG$"
(though I don't know how the modules $M$ come into play with this MO answer...like yes we do get $[df](g,h) = f(\partial(g,h))$ but what's the meaning behind this, behind its admittedly pleasing algebraic minimalism/simplicity???), where $BG$ is a topological space (cell complex) that we build out of $G$ (encoding somehow the relations between the elements of the group by homotopies):
- Elements of $G$ are loops starting and ending at one point --- perhaps we can imagine the point $*$ being the object whose symmetries are encoded by $G$, and an action $g\in G$ acting on this object is represented/encoded/witnessed by a path/edge from the object to itself --- though for the sake of visualization, I think it's best to draw/picture edges between different points instead of loops on the same point, and then keep in mind that those different points in the picture all actually mean the same point.
- And the group operation encoded by homotopies: the fact that the element $gh \in G$ equals the composition/multiplication of $g$ and $h$ means that the edge $gh$ should be homotopic, i.e. imagine it continuously deforming, to the edges $g$ and $h$ "together". Picture a triangle like $\triangle$, with the left edge traversed in the direction $\nearrow$ labelled $g$, and the right edge traversed in the direction $\searrow$ labelled $h$, and the bottom edge traversed in the direction $\to$ labelled $gh$; and the 2-simplex/interior of the triangle serving as "witness" to the homotopy between the two paths (from the bottom left vertex to the bottom right vertex).
- Equalities involving from composing/operating on 3 elements (e.g. the equation of associativity) are then homotopies between these "2-dimensional faces"/2-simplices. I have the following picture in mind: given $g,h,i\in G$, I can form the following tetrahedra
where the path $g \cdot h \cdot i$ (the 3 black paths) is homotopic to the path $ghi$ (dotted red) via the green faces --- more explicitly, the chained/composed path $h \cdot i$ is homotopic to $hi$ by the dark green 2-simplex, and $g \cdot hi$ is homotopic to $ghi$ by the light green 2-simplex. In other words the green surface "witnesses" that $g\cdot (h\cdot i)$ equals $ghi$. Of course, the other 2 faces of the tetrahedra "witness" that $(g \cdot h) \cdot i$ equals $ghi$:
Both these green and yellow surfaces are homotopies between the paths $g \cdot h \cdot i$ and $ghi$ (imagine deforming the chain of 3 black arrows through the green/yellow surface to reach the dotted red line). By inserting parentheses to $g \cdot h \cdot i$ we end up with either the green or yellow surface. But the parentheses are artificial addition --- is it possible to think of the 2 surfaces as portions/fragments of a "greater/higher homotopy" between $g \cdot h\cdot i$ and $ghi$? Well, the whole tetrahedra "witnesses" the equality between $g \cdot h \cdot i$ and $ghi$; in particular the interior of the tetrahedra is a homotopy between the green and yellow surfaces. I guess we can call this the tetrahedra/3-simplex associated with $g\cdot h\cdot i$. - With $n$ elements $g_1,\ldots, g_n\in G$, there is now an $n$-simplex "witnessing" the equality between $g_1\cdot \ldots \cdot g_n$ and $g_1\ldots g_n$ (with interior of the $n$-simplex being a homotopy between the $(n-1)$-simplices associated with $(g_1g_1)\cdot \ldots \cdot g_n$ and $g_1 \cdot \ldots \cdot (g_{n-1}g_n)$ --- all other ways of inserting parentheses, e.g. $g_1 \cdot (g_2g_3)\cdot \ldots \cdot g_n$ are already witnessed by the $(n-1)$-simplex witnessing $g_1 \cdot \ldots \cdot g_{n-1}=g_1\ldots g_{n-1}$.
Finally, there is the most abstract definition of group cohomology: the "$G$-fixed/invariant point" functor $-^G : G{\textbf{-mod}} \to \textbf{AbGrp}$ is left-exact, but not necessarily right-exact, meaning it takes a s.e.s. $0 \to A \to B \to C \to 0$ to a truncated/incomplete exact sequence $0 \to A^G \to B^G \to C^G$. We can complete exact sequence using right derived functors (in some sense the universal/canonical solution to the question of how to extend incomplete exact sequences): $$0 \to A^G \to B^G \to C^G \to H^1(G,A) \to H^1(G,B) \to H^1(G,C) \to H^2(G, A) \to \ldots$$ where these $H^i(G,M)$ are exactly the same cohomology groups as mentioned before. In other words,
group cohomology of $G$ measures how far the "$G$-fixed/invariant point functor" is from being (right-)exact.
Because $(-)^G$ can be written in terms of Hom (see Lemma 4.3 in this article) by $M^G \simeq \operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z, M)$, the derived functors above can be written in terms of Ext: $H^i(G,M) \simeq \operatorname{Ext}_{\mathbb Z[G]}^i(\mathbb Z, M)$. Finally, instead of computing Ext with injective resolutions (of $M$ in this case) --- as typically right derived functors are computed using injective resolutions --- by another algebraic miracle, one can compute it with a projective resolution of $\mathbb Z$ (see Wiki, or Where can I found an explanation of group cohomology from the point of view of invariants?).
One place where it's pretty obvious/natural that we'd care about this fixed point functor is in Galois theory/cohomology. My friend summarized this very well:
"The point is that Galois theory tells us we really really really really care about galois fixed points. For instance, the fundamental theorem of Galois theory tells us that if we take $A = L$ then $A^G = K$. That is, Galois fixed points are what let us decrease our field. Another good example is $A = L^*$, whence $A^G = K^*$
This is the general principle of "Galois descent" which roughly speaking (and rough is the best I understand it) says that to do something over $K$, you first do it over the algebraic closure and then take Galois fixed points
For instance, algebraic geometry is much easier over an ACF so a common way to define elliptic curves over some number field $K$ is to first define them over $\overline{\mathbb Q}$ then define a Galois action so you can descend it to be defined over $K$"
Summary: I have given 3 descriptions of group cohomology (or I guess in the case of the group central extension problem, only the description of the 2nd cohomology group), that are all quite natural in their respective contexts: (1) a very hands-on deconstruction of the group operation in a central extension of $G$; (2) singular cohomology of a space that is quite naturally built out of a group $G$, especially natural if one likes looking at things through a homotopical lens/perspective; and (3) a measurement of how the $G$-fixed points functor fails to preserve surjective maps, which is a major concern for in particular Galois theorists/number theorists.
Ok, time for my question:
Question: it seems utterly absurd and magical that from these 3 seemingly very different points of view, we get the exact same thing? It is as if I visited 3 different continents and saw the same person in all 3 destinations. I also get the impression that I'm reading a book and have skipped to three random chapters and noticed that the same character appears in 3 different contexts. I'm wondering if there is some linear route I can take/story I can tell, going through this world of group cohomology, that makes this feel like one unified narrative, instead of 3 separate ones.
P.S. in sort-of the spirit of the hands-on way of teasing out the 2-cocycle and 2-coboundary formulas from the group central extension problem, this MO answer does something similar for 1-cocycles in the context of modular forms: https://mathoverflow.net/a/33649/112504. Indeed that MO question asked for "Intuition for Group Cohomology", and all the answerers provided their own vision of what group cohomology was to them. It reminded me very much of the blind men and elephant parable. So my question can be rephrased as: what is the elephant?