### Crossed Homomorphisms

#### Posted by John Baez

While reading Gille and Szamuely’s *Central Simple Algebras and Galois Cohomology* I’m finding myself frustrated by my poor understanding of $H^1$ in group cohomology.

Roughly speaking, $H^2(G,A)$ classifies group extensions of a group $G$ by an abelian group $A$ on which it acts. $H^3(G,A)$ classifies 2-group extensions of $G$ by an abelian group $A$ on which it acts. And so on —this continues on up forever. I love this story: I call it the layer-cake philosophy of cohomology. But I never figured out how $H^1$ or $H^0$ fit into this story!

If you blindly follow the pattern, $H^1(G,A)$ should classify ways of extending a group $G$ by a group $A$ on which it acts to get a 0-group. But what does that mean? Is there any way to make it make sense? There must be.

(I won’t even *try* to think about $H^0$ this way. Not today anyway.)

What everyone says is this. Suppose we are given a group $G$ and a *not-necessarily abelian* group $A$ on which $G$ acts. Let’s write $g \!\triangleright\! a$ for the result of acting on $a \in A$ by $g \in G$. Then $H^1(G,A)$ consists of ‘cohomology classes’ of ‘crossed homomorphisms’ $\phi \colon G \to A$. Here a **crossed homomorphism** $\phi \colon G \to A$ is a function such that

$\phi(g h) = \phi(g) \;\; g \!\triangleright\! \phi(h)$

for all $g, h \in G$, and two crossed homomorphisms $\phi, \psi \colon G \to A$ count as **cohomologous** if for some $a \in A$ we have

$a \; \; \phi(g) = \psi(g) \; \; g \!\triangleright\! a$

for all $g \in G$.

All this reminds me a bit of semidirect products and crossed modules. Basically, a crossed homomorphism obeys the homomorphism rule $\phi(g h) = \phi(g) \phi(h)$ up to a fudge factor involving the action of $G$ on $A$. Two crossed homomorphisms are equivalent if they’re conjugate up to a fudge factor involving the action of $G$ on $A$:

$\psi(g) = a \; \phi(g) \; (g \!\triangleright\! a)^{-1}$

But for some reason I’ve never studied crossed homomorphisms, so I don’t see how they’re connected to topology… or anything else.

Well, that’s not *completely* true. Gille and Szamuely introduce them with an example. Stripping all specific details from this example, here’s what I seem to get. It helps a bit.

Suppose we have a group $G$ acting as endo-transformations, *not necessarily natural*, of the identity functor on some category $X$.

So, for any $g \in G$ and any object $x \in X$ we get an isomorphism

$g_x \colon x \to x$

and these obey

$(g h)_x = g_x h_x$

but these isomorphisms are not necessarily natural. So, for any morphism

$f: x \to y$

in our category we get a *not-necessarily-commutative* square:

$f g_x \ne g_y f$

Now let’s fix an invertible morphism $f \colon x \to y$; we’ll get a crossed homomorphism from this. Let $\phi(g)$ be the automorphism of $x$ that we get from going all the way around our not-necessarily-commutative square:

$\phi(g) = f^{-1} g_y f g_x^{-1}$

(This would be so much better with pictures!)

So, we get a map

$\phi \colon G \to \mathrm{Aut}(x)$

Now, it turns out that

$\phi(g h) = \phi(g)\, g \, \phi(h)\, g^{-1} \qquad \qquad (\star)$

Showing this requires a little calculation; I won’t do it here.

But note, $G$ acts on the group $\mathrm{Aut}(x)$ via conjugation! We can write this action as follows:

$g \!\triangleright\! a = g_x \, a \, g_x^{-1}$

With this notation, $(\star)$ becomes a special case of the equation in the definition of “crossed homomorphism”:

$\phi(g h) = \phi(g) \; \; g \!\triangleright\! \phi(h)$

In short:

**Theorem.** Suppose we are given a group $G$ acting as not-necessarily-natural transformations of the identity functor on a category $X$, and an isomorphism $f \colon x \to y$ in $X$. Then we get an action of $G$ on the group $\mathrm{Aut}(x)$ and a crossed homomorphism $\phi \colon G \to \mathrm{Aut}(x)$. If $G$ acts as *natural* transformations, this crossed homomorphism is trivial.

The hypotheses of this theorem seem a bit awkward to me, but I swear I’ve lifted them straight out of Gille and Szamuely, who do an example.

So, I’m wondering if there’s some better way to think about what’s happening here, or about crossed homomorphisms in general. For a topological understanding we might as well let $X$ be a groupoid. But what does it mean to have a group acting as not-necessarily-natural transformations of the identity functor? Well, in terms of topology, I guess it’s something like a choice of loop for each point.

If you want to see Gille and Szamuely’s example, go to *Central Simple Algebras and Galois Cohomology* and look at Section 2.3, “Galois descent”.

They actually have several examples. They always start by fixing a field $k$ and a finite Galois extension $K$ of $k$. Then here’s one of their categories $X$: it has vector spaces over $k$ as objects, but $K$-linear transformations $f \colon K \otimes V \to K \otimes W$ as morphisms from $V$ to $W$. Their group $G$ is the Galois group $\mathrm{Gal}(K|k)$. Any $g \in G$ gives, for any vector space $V$ over $k$, a linear transformation $g_V \colon K \otimes V \to K \otimes V$. But this is not natural because it does not commute, in the obvious sense, with all $K$-linear transformations.

By the way, the application they’re leading up to — which is what got me into this mess in the first place — is this: there’s a one-to-one correspondence between:

- isomorphism classes of central simple algebras over $k$ that become isomorphic to $n \times n$ matrix algebras when tensored with $K$,

and

- elements of $H^1(G, \mathrm{PGL}_n(K))$, where $G$ acts on the projective general linear group $\mathrm{PGL}_n(K)$ in the obvious way.

In this example we have the same group $G = \mathrm{Gal}(K|k)$, but a different category $X$. Now I guess this has central simple algebras over $k$ that become isomorphic to $n \times n$ matrix algebras as objects, and $K$-algebra morphisms $f \colon K \otimes A \to K \otimes B$ as morphisms from an algebra $A$ to an algebra $B$. If we fix an object $A$ in this category, we get $\mathrm{Aut}(A) = \mathrm{PGL}_n(K)$.

I was just trying to understand this… not just follow the argument step-by-step, but actually understand it. I think I understand it a bit better after writing this blog article. But I want to understand crossed homomorphisms and $H^1$ in group cohomology from many points of view now… a glaring hole in my education.

One of the things measured by $H^1(G; A)$ is the set of conjugacy classes of splittings of the split extension $A \to A \rtimes G \to G$. If you stretch enough, then splittings of the split $(n+1)$-group extension $A[n] \to A[n] \rtimes G \to G$ should be the same as $n$-group extensions $A[n-1] \to E \to G$. (Me, I haven’t done yoga in weeks.) So a splitting of the split 1-group extension should be a “0-group extensions”.

Whether this is helpful to your application is, of course, for you to decide.