# how do automorphisms act on the right in grothendieck's galois theory

So, I'm reading through some notes on the etale fundamental group (mostly Murre, but also some other notes I have), and I find it confusing how in a galois category $\mathcal{C}$ with fundamental functor $F$, the automorphism group $\text{Aut}(X)$ of an object $X\in\mathcal{C}$ can act on $F(X)$ on the right''. I know every left-action is equivalent to a right action by acting by the inverse, but it doesn't seem right that you should have to do that. Furthermore, given an object $X$ and an element $\xi\in F(X)$, Murre describes an injection (bijection if $X$ is galois) $\text{Aut}(X)\hookrightarrow F(X)$ by sending $u\in\text{Aut}(X)$ to $F(u)(\xi)$. This implies to me that we're not (don't need to?) do this inverse-action business.

For example, suppose $X$ is a galois object, and suppose $S$ is the terminal object for $\mathcal{C}$. Then, there is a natural (though not canonical) identification of $\text{Hom}(X,X) = \text{Aut}(X)$ with $F(X)$, and so it's natural for $\text{Aut}(X)$ to act on itself (which is identified with $F(X)$) on the right (by right-multiplication).

If $X$ is not a galois object, then we can still pick a galois object $Y$ over $X$, and again we may identify $\text{Hom}(Y,X) = F(X)$. In this case, $\text{Aut}(Y)$ naturally acts on the right on $\text{Hom}(Y,X)$ and hence on $F(X)$. However, this is not an action of $\text{Aut}(X)$. In classical galois theory, we could just restrict automorphisms of $Y$ to automorphisms of $X$, and hence get a right action of $\text{Aut}(X)$ on $F(X)$, but this restriction map doesn't seem to be available in the abstract language of galois categories.

Can someone explain how I should think of Aut$(X)$ acting on the right of $F(X)$?

thanks

• will

Yes.

First notice that the "restriction map" you describe in Galois theory is actually kind of subtle. You first take the subgroup of automorphisms of the larger field that send the smaller field to itself. You then have to mod out by the subgroup that fix the smaller field. So if $L$ is the Galois closure of $K$, there is not always a map $Aut(K) \to Aut(L)$, nor is there always a map $Aut(L) \to Aut(K)$.

Group-theoretically, if $Aut(L)=G$, and $K$ corresponds in the Galois correspondence to $H$, then the automorphisms that send $K$ to itself are $N(H)$, and the automorphisms that fix $K$ are $H$, so $Aut(K)= N(H)/H$.

So we have the group-theoretic fact that $N(H)/H$ acts on $G/H$ on the right. This is easy to see!

Another picture of the right action on a Galois category is that we can view $G$ as a subset of $S_n$, $n=|G|/|H|$ the number of elements. Then the right action is exactly the centralizer of $G$ in $S_n$.

So another group theory fact is that $N(H)/H$ is the whole centralizer of $G$ in its action on $G/H$. This isn't too hard to see, using exactly Murre's injection.

Since the abstract language of Galois categories is set up precisely to be exactly the same as the category of $G$-sets for $G$ a profinite group, you can apply exactly this picture to a Galois theory case.

When you say $Aut(Y)$ acts on the right on $Hom(Y,X)$, don't you mean it acts on the left? The right action is exactly the centralizer of the left action, which turns out to be naturally isomorphic to a subquotient of that group. This is the same as the Galois theory picture.

Taking the inverse is only really necessary if you want to write your right action like a left action, so you have to switch the order of multiplication in the group.

• I do actually mean that Aut$(Y)$ acts on the right on Hom$(Y,X)$. Namely, given some $f\in\text{Hom}(Y,X)$ and $\sigma\in\text{Aut}(Y)$, we have it act by $f\circ\sigma$. – Will Chen Nov 13 '13 at 6:52