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Group cohomology is one thing I don't seem to get my head around.

I understand $H^0$ as being the "fixed points" but when it comes to anything higher I have no idea what the notions are meant to be capturing...if anything.

For example I can see that 1-cocycles are homomorphisms with a specific property but that property doesn't seem to mean anything to me...and I think it should!

Is there an intuition or example that explains the meaning or is it just something that "stuck" and seems to work in lots of places?

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    $\begingroup$ You might have a look here $\endgroup$ Mar 8, 2012 at 18:31
  • $\begingroup$ Thanks, I shall take a look at that. $\endgroup$
    – fretty
    Mar 8, 2012 at 18:36
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    $\begingroup$ All right, @Mariano, I have pasted my answer from MO but I suggest you don't close the question, so that users here can also give their answers. I'm sure there are many examples that can help build intuition and I bet you know quite a few ones yourself ! $\endgroup$ Mar 8, 2012 at 19:15
  • $\begingroup$ But it is usually a good idea to keep this things in one place: people can add examples there, too :) I won't close this, though. $\endgroup$ Mar 8, 2012 at 19:30

3 Answers 3

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[At Mariano's suggestion, I'll copy my answer from MathOverflow.]

Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology.

Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176).

If moreover $G $ is cyclic with generator $s$, this implies that an element of $K$ has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$.

Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff $x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ .

So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$ $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$

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Group cohomology can be defined very naturally in a purely topological way. The definition of $1$-cocycles is not random, or due to historical accident.

More specifically, given a group, $G$, the Eilenberg-Maclane space $X=K(G,1)$ is defined which has $\pi_1(X)=G$, and $\pi_{\geq 2}(X)=0$. This is well-defined up to homotopy-type if you assume it is a CW-complex. Since the cohomology functor is invariant under homotopy equivalence, the groups $H^i(X)$ are well-defined abelian groups associated with the original group $G$, and this is what we call group cohomology. Now when you take coefficients in a $\mathbb Z[G]$ module $M$, this is actually equivalent to taking cohomology with ``twisted coefficients" of $K(G,1)$.

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  • $\begingroup$ Ok, so basically group cohomology is defined to in some sense match up with cohomology of an underlying space? $\endgroup$
    – fretty
    Mar 9, 2012 at 11:36
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    $\begingroup$ @fretty: Yes. It is a composition of two functors. One assigns a group its Eilenberg-Maclane space, and the other takes the cohomology of that space. $\endgroup$ Mar 9, 2012 at 11:53
  • $\begingroup$ ah right, that justifies it enough to me then! Thanks. $\endgroup$
    – fretty
    Mar 9, 2012 at 12:59
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To add to George's answer above, 1st Galois cohomology group has a natural interpretation as the set of classes of principal homogeneous spaces for a group.

Let $G$ be an algebraic group defined over $k$. Let $K/k$ be a Galois extension. The $K$-points of $G$ form a $\mathrm{Gal}(K/k)$-group (or a $\mathrm{Gal}(K/k)$-module, if $G$ is commutative). A principal homogeneous $K$-space is a $K$-variety $X$ with a free transitive action of $G$. There is a bijective correspondence between $k$-isomorphism classes of principal homogeneous $K$-spaces and cocycles of $H^1(\mathrm{Gal}(K/k), G(K))$.

Let $X$ be a p.h.s. Pick a point $x \in X$ and act on it by $\sigma \in \mathrm{Gal}(K/k)$. Since $G$ acts on $X$ freely and transitively there is a unique $g_\sigma$ such that $g_\sigma \cdot x = \sigma(x)$. One checks that $\sigma \mapsto g_\sigma$ define a 1-cocycle.

Conversely, given a 1-cocyle $\{g_\sigma\}$, consider a disjoint union $\sqcup G \times \{\sigma\}_{\sigma \in \mathrm{Gal}(K/k)}$ and define an action of $\mathrm{Gal}(K/k)$ on it: $\sigma(x,\tau)=(g_\sigma \cdot x, \sigma\tau)$. There is a natural $G$ action on the disjoint union, and the factor by the action $\mathrm{Gal}(K/k)$ inherits the action of $G$ (a highbrow term for what is happening is "Galois descent"), turning it into a p.h.s.

Moreover, if $G$ is commutative, the group law on $H^1(\mathrm{Gal}(K/k), G(K))$ has a natural geometric interpretation in terms of principal homogeneous spaces. This group is also known as Weil-Chatelet group. I recommend the original Weil's article on the subject: A. Weil, "On algebraic groups and homogeneous spaces". Amer. J. Math. , 77 (1955) pp. 493–512

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