Copies of topological fundamental groups inside etale fundamental groups given by different embeddings of your field into $\mathbb{C}$

Question

Let $X$ be a smooth curve over a number field $K$ (not necessarily proper). Fix an algebraic closure $\overline{K}$ of $K$.

Let $i,i' : \overline{K}\hookrightarrow\mathbb{C}$ be two abstract embeddings (ie, as $\mathbb{Q}$-algebras). Let $X_\mathbb{C},X_{\mathbb{C}}'$ be the base changes of $X_{\overline{K}}$ to $\mathbb{C}$ via $i$ and $i'$. Then, $X_\mathbb{C}(\mathbb{C})$ has the structure of a Riemann surface. Let $x\in X_\mathbb{C}(\mathbb{C})$ come from a $\overline{K}$-rational point. We may consider its topological fundamental group $\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)$. Since every finite cover of the Riemann surface $X_\mathbb{C}(\mathbb{C})$ is algebraic, for every loop in $\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)$, its monodromy action on the fibers at $x$ of its finite covers determines an automorphism of the fiber functor at $x$, and hence we obtain a homomorphism $$\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)\rightarrow \pi_1^{et}(X_\mathbb{C},x)$$ which is known to be the embedding of the first group into its profinite completion. The map $X_\mathbb{C}\rightarrow X_{\overline{K}}$ given by base change induces an isomorphism on etale fundamental groups, and composing these maps we get $$\pi_1^{top}(X_\mathbb{C}(\mathbb{C}))\longrightarrow \pi_1^{et}(X_\mathbb{C})\stackrel{\sim}{\longrightarrow}\pi_1^{et}(X_\overline{K})$$ where I've omitted the base points because I only care about these maps up to conjugacy (say, inside $\pi_1^{et}(X_{\overline{K}})$). Similarly, with $X_\mathbb{C}'$, we get a map

$$\pi_1^{top}(X'_\mathbb{C}(\mathbb{C}))\longrightarrow \pi_1^{et}(X'_\mathbb{C})\stackrel{\sim}{\longrightarrow}\pi_1^{et}(X_\overline{K})$$

Both of these maps give embeddings of the topological fundamental groups inside $\pi_1^{et}(X_{\overline{K}})$, canonical up to conjugation.

My question is:

When are the images the same (up to conjugation)?

Are there examples when the images are not the same?

I'm particularly interested in the case when $X_\mathbb{C}(\mathbb{C})$ is hyperbolic.

References would also be appreciated.

Answer 1

For affine hyperbolic curves, when $i$ and $i'$ agree on $K$, this happens only when $i$ and $i'$ are equal or complex conjugate of each other.

Let $f: \pi_1^{top}(X) \to \pi_1^{et}(X_{\mathbb C})$ be the natural dense inclusion, and $e_i,e_{i'} \pi_1^{et}(X_{\mathbb C}) \to \pi_1^{et}(X_{\overline{K}})$ be the natural isomorphisms defined by $i$ and $i'$. Let $\sigma$ be the element of the Galois group of $\overline{K}$ over $K$ that sends $i$ to $i'$, then $e_{i'}$ is $e_i$ composed with the action of $\sigma$ by outer automorphism of $\pi_1^{et}(X_{\overline{K}})$.

If the image of $e_i \circ f$ is conjugate to $e_{i'} \circ f$, then there must be some automorphism $\alpha$ of $\pi_1^{top}(X)$ such that $e_i \circ f \circ \alpha$ is conjugate to $e_{i'} \circ f = \sigma \circ e_i \circ f$ as a homomorphism. Now because $f$ is dense, we can extend $\alpha$ to an outer automorphism of $\pi_1^{et}(X_{\mathbb C})$ and thus to an outer automorphism of $\pi_1^{et}(X_{\overline{K}})$, and then the condition that these two maps are conjugate becomes the condition that these two outer automorphisms are equal in the outer automorphism group.

It is a result of Matsumoto and Tamagawa that this can only happen for the identity and comple conjugation (Mapping-Class-Group Action versus Galois Action on Profinite Fundamental Groups, Remark 2.1).

Furthermore, it looks to me that following their argument, the only case that the affineness assumption is used can be replaced with Theorem C(i) of Hoshi and Mochizuki

Answer 2

This is only a partial answer to your question, but I think I have an example where the images are not conjugate.

Assume that $K$ has a non-real embedding $\sigma : K \hookrightarrow \mathbb{C}$, and let $\iota$ denote complex conjugation, so that $\iota \circ \sigma$ is another non-real embedding. Let $X$ be the projective line over $K$ with the points $z_1, z_2, z_3, z_4 \in K$ deleted, where $\sigma(z_i) = \iota \circ \sigma(z_i) \in \mathbb{R}$. Assume that the $z_i$'s are ordered so that their images under $\sigma$ go from least to greatest. Now let $\gamma \in \pi_1^{\mathrm{top}}(X_{\mathbb{C}}(\mathbb{C}))$ be represented by a loop wrapping around the missing points $z_1$ and $z_3$ and passing below $z_2$. Then I believe it's fairly straightforward to show that

(i) its image under the automorphism $\iota_* : \pi_1^{\mathrm{top}}(X_{\mathbb{C}}(\mathbb{C})) \to \pi_1^{\mathrm{top}}(X_{\mathbb{C}}(\mathbb{C}))$ induced by $\iota$ is represented literally by the image of that loop reflected across the real axis (the reflection will wrap around $z_1$ and $z_3$ and pass above $z_2$); and

(ii) the image of $\gamma$ under the map $\sigma^* : \pi_1^{\mathrm{top}}(X_{\mathbb{C}}(\mathbb{C})) \to \pi_1^{\mathrm{\acute{e}t}}(X_{\bar{K}})$ that you described (induced by the embedding $\sigma$) is equal to the image of $\iota_*(\gamma)$ under the analogous map $(\iota \circ \sigma)^*$.

But the group elements $\gamma$ and $\iota_*(\gamma)$ are not equivalent up to conjugation; in fact, the elements they represent in $H_1(X_{\mathbb{C}}(\mathbb{C}), \mathbb{Z})$ differ by a sign. Or we can lift the loops up to the double cover which gives the complex elliptic curve ramified over $z_1, ... , z_4$ and see that they represent two distinct elements (no longer differing by a sign) of the homology group of that genus-1 torus.

If we want an example for a hyperbolic Riemann surface, we can do the same trick but with points $z_1, ... , z_{2g + 2}$ for some integer $g \geq 2$, so that we are comparing two loops representing distinct elements of the homology group of the genus-$g$ complex hyperelliptic curve ramified over the $z_i$'s.

EDIT: Hmm just realized that I didn't really answer any of your question, since you were asking about images of an entire group rather than of individual elements. I hope this "answer" is of interest anyway...