The n-Category Café

First, I’ll completely eliminate the context that led me to get interested in crossed homomorphisms in the first place, since — as I suspected — something very general is at work here. So I won’t say anything out loud about ‘Galois descent’ or ‘Galois cohomology’ — at least, not at first.

Second, I will take a beautiful question in higher category theory and brutally truncate it and strictify it and skeletalize it, just to reach a conceptual explanation of ‘crossed homomorphisms’ as rapidly as possible. For a more civilized approach, see Qiaochu Yuan’s posts.

Crossed homomorphisms

The idea is that crossed homomorphisms arise naturally when we categorify the concept of ‘fixed point’.

As a warmup, suppose we have a group $G$ acting on a set $A$. Thus, we have a homomorphism

$$\alpha \colon G \to \mathrm{Aut}(A)$$

We might be interested in the set of fixed points of $\alpha$: that is, the set of $a \in A$ such that

$$\alpha(g) a = a$$

for all $g \in G$. This set is sometimes called $H^0(G,A)$, where you’re implicitly supposed to know that $A$ here is a set equipped with an action of $G$.

Why such a weird name for the set of fixed points? Well, this is a degenerate special case of group cohomology. And one way to understand the group cohomology $H^1(G,A)$ is to understand it as a categorification of this degenerate special case. One can go further and understand $H^n(G,A)$ for $n > 1$ by repeated categorification, but I won’t do that here.

How can we categorify the concept of fixed point?

There are several things we could do. The most important is to replace the set $A$ by a category. But this opens the door for other tricks. For example we could replace $G$ by a 2-group, which is the categorical analogue of a group. There’s also another, independent generalization we could try: we could ‘weaken’ the concept of group action, replacing the usual equations $\alpha(g h) = \alpha(g) \alpha(h)$ and $\alpha(1) = 1$ by isomorphisms satisfying some equations.

All these directions of generalization are interesting. We really should do all of them. But I want to go in a very classical direction. So, I’ll take a very low-budget approach! I’ll replace $A$ by a category, but I’ll keep $G$ as a group, and keep the action of $G$ on $A$ strict.

So, assume we have a group $G$, a category $A$, and a group homomorphism

$$\alpha \colon G \to Aut(A)$$

where $Aut(A)$ is just the group of invertible functors from $A$ to itself.

This is pretty simple, but now we’ll do something a bit more interesting: we’ll look at homotopy fixed points of the action of $G$ on $A$.

What are these?

For starters, a homotopy fixed point is an object $a \in A$ equipped with isomorphisms

$$\phi(g) \colon \alpha(g) a \stackrel{\sim}{\to} a$$

for each $g \in G$. But that’s not good enough: we want these isomorphisms to obey some ‘coherence laws’. Given $g,h \in G$ we have two isomorphisms from $\alpha(g h)a$ to $a$, the obvious one:

$$\alpha(g h) a \stackrel{\phi(g h)}{\longrightarrow} a$$

and the composite

$$\alpha(g h) a \stackrel{\alpha(g)(\phi(h))}{\longrightarrow} \alpha(g) a \stackrel{\phi(g)}{\longrightarrow} a$$

so for a homotopy fixed point we demand that these are equal. We also have two isomorphisms from $\alpha(1) a$ to $a$, namely

$$\alpha(1) a \stackrel{\phi(1)}{\longrightarrow} a$$

and the identity

$$\alpha(1) a = a \stackrel{1}{\longrightarrow} a$$

So, for a homotopy fixed point we also demand that these are equal.

Besides defining homotopy fixed points, we can define morphisms between homotopy fixed points, so we get a category with homotopy fixed points as its objects. Qiaochu Yuan explains this. But following the bargain-basement approach I have in mind here, let me simplify the situation further!

I’ll assume $A$ is skeletal, so isomorphic objects are equal. Then our homotopy fixed point $a$ actually has $\alpha(g) a$ equal to $a$ for all $g$. This may seem crazy, like we’re throwing out the whole idea of homotopy fixed point and going back to a plain old fixed point. But it’s not! It doesn’t imply that the isomorphism $\phi(g) \colon \alpha(g) \to a$ is the identity: it’s just some automorphism of $a$. So a homotopy fixed point is still more interesting than a plain old fixed point.

Given that $A$ is skeletal, if we’re looking at a single homotopy fixed point, or even a bunch of them with isomorphic underlying objects, we can take our category $A$ and throw out all the objects except one, say $a$, since they play no role in the story. This makes $A$ into a one-object category.

And since we’re only working with isomorphisms, we can also throw out all the noninvertible morphisms in $A$. This makes $A$ into a one-object groupoid. Such a category is called the ‘delooping’ of a group… but let’s just switch to thinking of $A$ as a group.

What are we left with now? We’re left with an action of a group $G$ as automorphisms of a group $A$:

$$\alpha \colon G \to \mathrm{Aut}(A)$$

A homotopy fixed point of this now consists of a function

$$\phi \colon G \to A$$

obeying two equations, the ‘coherence laws’ I wrote down earlier. Translated into the language we’re using now, the first, more interesting coherence law says that

$$\phi(g h) = \phi(g) \; (\alpha(g) (\phi(h)))$$

The second says that

$$\phi(1) = 1$$

And now we’ve explained crossed homomorphisms! Given a group $G$ acting on a group $A$ via $\alpha$, a crossed homomorphism is defined to be a function $\phi \colon G \to A$ obeying the equation

$$\phi(g h) = \phi(g) \; (\alpha(g) (\phi(h)))$$

If it also obeys $\phi(1) = 1$ we say it’s normalized.

So, we’ve proved:

Theorem. Suppose $G$ is a group acting strictly on some skeletal category, let $a$ be an object of that category, and let $A = \mathrm{Aut}(a)$ be its automorphism group. Then ways of making $a$ into a homotopy fixed point of the action of $G$ correspond bijectively to normalized crossed homomorphisms

$$\phi \colon G \to A$$

This is great, but we can go further. I mentioned that there’s actually a category of homotopy fixed points. So, we can talk about isomorphisms between homotopy fixed points… and figure out what look like in this simplified context, where we have a group acting strictly on a skeletal category, and we focus on one object $a$, so that homotopy fixed points can be described using crossed homomorphisms.

I think the answer is this — I haven’t checked all the details. Suppose we have two normalized crossed homomorphisms:

$$\phi, \psi \colon G \to A$$

Then a morphism from $\phi$ to $\psi$ is an element $a \in A$ such that

$$a \; \phi(g) = \psi(g) \; (\alpha(g)(a))$$

Composition of morphisms is just multiplication in $A$.

The set of isomorphism classes of normalized crossed homomorphisms is called $H^1(G,A)$. But now we have a conceptual interpretation of this:

Theorem. Suppose $G$ is a group acting strictly on some skeletal category, let $a$ be an object of that category, and let $A = \mathrm{Aut}(a)$ be its automorphism group. Then the set of isomorphism classes of homotopy fixed points of the action of $G$ having $a$ as their underlying object is in bijection with $H^1(G,A)$.

Galois descent

Now I can’t resist sketching an application, just to give a little taste of ‘Galois descent’.

Suppose we have a Galois extension $K$ of a field $k$, and let $Gal(K|k)$ be the Galois group of $K$ over $k$. Remember, this consists of automorphisms of the field $K$ that fix $k \subseteq K$.

You can usually take an algebraic gadget defined over $k$ — like a vector space over $k$, or an algebra over $k$, or a variety over $k$ — and get a gadget defined over $K$, by ‘extension of scalars’. Galois descent is all about reversing this. If you have a gadget defined over $K$, what are the different gadgets over $k$ that it can come from? There could be more than one.

To answer this question, the trick is to notice that $Gal(K|k)$ acts on the category of gadgets over $K$. And here’s the magic: we can think of gadgets over $k$ as weak fixed points of the action of $Gal(K|k)$ on gadgets over $K$!

Let’s do an easy example, where our gadgets are just vector spaces.

The Galois group $Gal(K|k)$ acts strictly on the category $Vect_K$ of vector spaces over $K$. How?

Here’s how it acts on objects. Given $V \in Vect_K$ and $g \in Gal(K|k)$, we get a new vector space $\alpha(g) (V) \in Vect_K$ that has the same underlying set and the same underlying addition, but where we multiply by scalars $c \in K$ in a new way: instead of multiplying by $c$, we multiply by $g(c)$.

Here’s how it acts on morphisms. If $f \colon V \to W$ is linear, the very same function is a linear map from $\alpha(g) V$ to $\alpha(g) W$, and we call this $\alpha(g) (f)$.

Now, it’s easy to turn a vector space $L$ over $k$ into a vector space over $K$: just tensor it with $K$, getting $K \otimes_k L$. This is called ‘extension of scalars’. This gives a functor from $Vect_k$ to $Vect_K$.

But here’s the interesting point: in this situation you can go further and make $K \otimes_k L$ into a homotopy fixed point of the action of $Gal(K|k)$. Even better, the category of homotopy fixed points of this action is equivalent to $Vect_k$.

As a consequence we get this, which is a simple case of ‘Galois descent’:

Ways of making $V \in Vect_K$ into a homotopy fixed point of $Gal(K|k)$ are ‘the same’ as ways of writing $V$ as $K \otimes_k L$ for some $L \in \Vect_k$.

‘The same’ means we have an equivalence of categories, and thus a bijection between isomorphism classes.

We can classify these isomorphism classes using common sense, or using all the fancy machinery we’ve developed.

Using common sense, there’s clearly just one. That is: all vector spaces $L \in Vect_k$ that become isomorphic to $V \in Vect_K$ when we tensor them with $K$ were already isomorphic. They had to have the same dimension, after all.

So that was easy. But now let’s use our fancy machinery!

First, I believe (but haven’t carefully checked so I’m still worried) that we can replace $Vect_K$ with a skeletal category while keeping the action of $Gal(K|k)$ strict. Given this, we can apply the general theorem from the previous section and see the following:

For any vector space $V \in \mathrm{Vect}_K$, the set of isomorphism classes of ways of making $V$ into a homotopy fixed point for the action of $Gal(K|k)$ is in bijection with

$$H^1(Gal(K|k), \mathrm{Aut}(V))$$

Using common sense, we already know this set has just one element. So what’s the payoff for bringing all the fancy machinery?

Here’s the payoff: we’ve computed the first cohomology of $Gal(K|k)$ with coefficients in $\mathrm{Aut}(V)$ without doing any real work — just thinking. And we’ve seen it’s a set with one element.

The most famous example is when $V = K$. Then $\mathrm{Aut}(V)$, the group of invertible linear transformations of $V$, is just $K^\times$, the multiplicative group of $K$. So we’ve shown

$$H^1(Gal(K|k), K^\times) = \{1\}$$

And this is called Hilbert’s Theorem 90. It’s a famous result from his Zahlbericht — his report on number theory, which came out in 1897. Needless to say, he didn’t phrase it in terms of cohomology. I read that Andreas Speiser did it in 1919, but from a brief glance I couldn’t recognize cohomology in his paper. Noether wrote about it in 1933, and her paper looks a lot easier to understand.

For other more interesting cases of Galois descent, $H^1$ becomes nontrivial. To compute it, homological algebra takes over where common sense leaves off. Read Gille and Szamuely’s Central Simple Algebras and Galois Cohomology for nice examples, starting in Section 2.3 — the section on Galois descent.