# Action on étale fundamental group is conjugation

Let $$X$$ be a quasi-compact, quasi-separated connected scheme and let $$\bar{x}$$ be a geometric point. Denote by $$\pi_1(X,\bar{x})$$ the étale fundamental group, defined as the automorphism of the fiber functor $$\bar{x}: \mathrm{FÉt}_X \to \mathrm{Set},$$ where $$\mathrm{FÉt}_X$$ is the category of schemes finite étale over $$X$$.

Suppose now that I have an element $$\sigma \in \pi_1(X,\bar{x})$$ and suppose that $$\sigma$$ lies in the center. Then for any $$Y/X$$ finite étale, $$\sigma$$ induces an automorphism of $$Y$$ over $$X$$, which we call $$\sigma_Y$$.

Since $$\sigma_Y$$ is an automorphism of $$Y$$, we get a map $$(\sigma_Y)_\ast: \pi_1(Y,\bar{y}) \to \pi_1(Y,\sigma_y(\bar{y})).$$ Composing with the choice of a path $$\gamma:\bar{y} \to \sigma_y(\bar{y})$$, we get an automorphism $$(\sigma'_Y)_\ast = \gamma_\ast \circ (\sigma_Y)_\ast : \pi_1(Y,\bar{y}) \to \pi_1(Y,\bar{y}).$$

I have seen it claimed in Curves and their fundamental groups pg. 133, proof of Lemma, by Faltings that the automorphism $$(\sigma'_Y)_\ast$$ is inner. I don't see why this is true, could someone help explain the argument?

• @R.vanDobbendeBruyn Maybe I am misinterpreting numdam.org/item/SB_1997-1998__40__131_0.pdf pg. 133, proof of Lemma? There Faltings writes that the induced automorphism $\sigma$ acts by an inner automorphism on $\pi_1(Y,y)$. Am I misinterpreting Faltings? I would want to understand his argument by working via Galois categories.
– DCM
Aug 18, 2022 at 21:05
• @R.vanDobbendeBruyn Any two natural isomorphisms of the fibre functors differ by an automorphism, and automorphisms give elements of the group, not automorphisms of the group. So if one is inner, they all are. Aug 19, 2022 at 1:27
• @Will Sawin: Thanks for helping clear up some questions. The question remains, however. I think that the map $(\sigma_Y)_\ast$ is given by some version of conjugation, but the exact relation is a bit unclear.
– DCM
Aug 19, 2022 at 8:49
• @WillSawin I posted what I believe is an answer. If you have the time, please check and see if it looks alright.
– DCM
Aug 19, 2022 at 10:58

By assumption, $$\sigma$$ lies in the center. In particular, if $$H = \pi_1(Y,\bar{y})$$ is the (open) subgroup of $$G = \pi_1(X,\bar{x})$$ corresponding to $$H$$ and $$K \subset H$$ is an open subgroup, the diagram $$\require{AMScd} \begin{CD} G/K @>{R_{\sigma}}>> G/K\\ @VVV @VVV \\ G/H @>{R_{\sigma}}>> G/H \end{CD}$$ of finite $$G$$-sets commutes, where $$R_\sigma$$ is right-multiplication with $$\sigma$$, and where the vertical maps are just the projection. Since the category of $$G$$-sets is equivalent to the category of schemes finite étale over $$X$$, we should be able to deduce that for $$Z$$ finite étale over $$Y$$, we have a commutative diagram $$\require{AMScd} \begin{CD} Z @>{\sigma_Z}>> Z\\ @VVV @VVV \\ Y @>{\sigma_y}>> Y \end{CD}$$ where $$\sigma_y$$ and $$\sigma_z$$ are the maps corresponding to $$R_\sigma$$. Note that this diagram is a pull-back as well.
For any $$Z/Y$$ étale, we have a commutative diagram of cospans $$\require{AMScd} \begin{CD} \bar{y} @>{j}>> Y @<<{p_Z}< Z \\ @VV\mathrm{id}V @V\sigma_YVV @VV \sigma_Z V \\ \bar{y} @>{\sigma_Y \circ j}>> Y @<<{p_Z}< Z \end{CD}$$ and by functoriality of pull-back, we get an induced map $$Z_{\bar{y}} \to Z_{\sigma(\bar{y})}$$, which defines a natural transformation $$F_{\bar{y}} \to F_{\sigma({\bar{y}})}$$.
$$\DeclareMathOperator\Aut{Aut}$$The induced map $$\sigma_y: \pi_1(Y,\bar{y}) = \Aut(F_\bar{y}) \to \Aut(F_{\sigma(\bar{y})}) = \pi_1(Y,\sigma(\bar{y}))$$ is now defined, up to an inner homomorphism of $$\pi_1(Y,\sigma(\bar{y})$$, as follows. We take a natural automorphism $$(\tau_Z)_{Z \in \mathrm{FÉt}_Y}$$ and map it to the natural automorphism $$(\tau'_Z)_{Z \in \mathrm{FÉt}_Y}$$ of $$F_{\sigma(\bar{y})}$$, where $$\tau'_Z = \tau_{\sigma(Z)}$$. Here $$\sigma(Z)$$ is the finite étale scheme over $$Y$$ defined as $$Z \to Y \xrightarrow{\sigma_y} Y$$, where the first map is the "original" projection of $$Z$$ to $$Y$$. One notes that $$\tau'_{\sigma(Z)} = \sigma_Z \tau_Z \sigma_Z^{-1}$$, where $$\sigma_Z$$ is the map $$Z_{\bar{y}} \to Z_{\sigma(\bar{y})}$$ induced by $$\sigma_Z: Z \to Z$$ on fibers.
Now, on the other hand, we wish to compute the map $$\pi_1(Y,\sigma(\bar{y})) \to \pi_1(Y,\bar{y})$$ induced by an étale path; we will see that there is a particularly convenient path. Indeed, the above computation on cospans gives us an isomorphism of fiber functors $$\sigma:F_{\bar{y}} \to F_{\sigma(\bar{y})}$$ and the map on automorphism groups is just given by conjugation by this isomorphism. This means that the map $$\pi_1(Y,\sigma(\bar{y})) \to \pi_1(Y,\bar{y})$$ is, up to conjugation, given as the inverse of the map $$\pi_1(Y,\bar{y}) \to \pi_1(Y,\sigma(\bar{y}))$$.
This shows that the composite $$\pi_1(Y,\bar{y}) \to \pi_1(Y,\sigma(\bar{y})) \to \pi_1(Y,\bar{y})$$ is given by conjugation.