Etale fundamental group of the circle What is the étale fundamental group of the circle $X({\bf R})$, where
$$
X(k) = \{(x,y) \in k^2 \mid x^2+y^2 = 1\}?
$$
I know that there is a sequence
$$
1 \rightarrow \pi_1^{et}(X({\bf C})) \rightarrow \pi_1^{et}(X({\bf R})) \rightarrow Gal({\bf C}/{\bf R}) \rightarrow 1
$$
with $Gal({\bf C}/{\bf R})= \{z\mapsto z, z\mapsto \bar{z}\}$. The first group is the profinite completion of $\bf Z$ since $X({\bf C})$ is the projective complex line with two points removed, but I don't know if the sequence splits or not.
More generally, what can be said about that sequence for general smooth varieties over ${\bf R}$?
 A: As Donu explained, the sequence splits by choosing an $\mathbf R$-point of $X$. So the only question remaining is what the $\operatorname{Gal}(\mathbf C/\mathbf R)$-action on $\pi_1^{\text{ét}}(X_{\mathbf C})$ is. I claim that the action is trivial, because the two points at infinity $V(x^2+y^2)$ are not defined over $\mathbf R$. (By contrast, for $\mathbf G_{m,\mathbf R}$ we get the nontrivial action where a generator $\sigma \in \operatorname{Gal}(\mathbf C/\mathbf R)$ acts by $-1$, using the same argument as below.)
Indeed, consider the tower $Y_n \to X_{\mathbf C} \to X$ where $Y_n \to X_{\mathbf C}$ is the unique cover of degree $n$. The composite $Y_n \to X$ is Galois as $\pi_1^{\text{ét}}(X_{\mathbf C}) \trianglelefteq \pi_1^{\text{ét}}(X)$ is normal and $n\mathbf Z \subseteq \mathbf Z$ is characteristic (in other words, any conjugate $Y'_n$ would contain $X_{\mathbf C}$ as $X_{\mathbf C} \to X$ is normal, hence $Y'_n \cong Y_n$ as $X_{\mathbf C}$-covers since $Y_n$ is the unique degree $n$ cover).
We need to compute $\operatorname{Gal}(Y_n/X)$. This depends on some choices of isomorphisms: we choose $X_{\mathbf C} \stackrel\sim\to \mathbf G_{m,\mathbf C}$ via $(x,y) \mapsto x+yi$, and we write $t = x+yi$. Under this identification, the generator $\sigma$ of $\operatorname{Gal}(X_{\mathbf C}/X) = \operatorname{Gal}(\mathbf C/\mathbf R)$ acts on $\mathbf G_{m,\mathbf C}$ via $x+yi \mapsto x-yi$, i.e. the $\mathbf C$-semilinear map $\sum_j c_jt^j \mapsto \sum_j \bar c_jt^{-j}$.
Then $Y_n$ can be identified with $\operatorname{Spec} \mathbf C[t^{\pm1/n}]$. The Galois group of $Y_n$ over $Y_1$ is $\mu_n(\mathbf C)$, where $\zeta \in \mu_n(\mathbf C)$ acts as the $\mathbf C$-linear map $t \mapsto \zeta t$. The conjugation action of $\sigma \in \operatorname{Gal}(X_{\mathbf C}/X)$ on $\operatorname{Gal}(Y_n/X_{\mathbf C})$ is therefore trivial, as
$$(\sigma \circ \zeta \circ \sigma^{-1})(t) = (\sigma \circ \zeta)(t^{-1}) = \sigma (\zeta^{-1}t^{-1}) = \overline{\zeta^{-1}}t = \zeta t.$$
Thus, we see that the conjugation action of $\operatorname{Gal}(\mathbf C/\mathbf R)$ on $\pi_1^{\text{ét}}(X_{\mathbf C})$ is trivial. $\square$
