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.

  • 4
    $\begingroup$ Tangentially related: Serre has an example of a smooth projective surface where the topological fundamental groups for two embeddings $K \to \mathbb C$ are not abstractly isomorphic; in particular they do not give the same subgroup up to conjugation. See Serre, Exemples de variétés projectives conjuguées non homéomorphes. Of course for curves the topological fundamental groups are abstractly isomorphic, so this doesn't answer your question. Serre's embeddings come from the Galois theory of $K \to \bar K$. $\endgroup$ Commented Oct 19, 2017 at 5:55
  • 3
    $\begingroup$ Is the case $X=\mathbb{C}\smallsetminus\{0\}$ clear? one gets two embeddings of $\mathbb{Z}$ into $\widehat{\mathbb{Z}}$: are they always equal? $\endgroup$
    – YCor
    Commented Oct 19, 2017 at 8:03

2 Answers 2


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

  • $\begingroup$ Your answer prior to the edit: You seem to be saying that nontrivial elements of the abs Galois group can never act as conjugation on the image of the topological fundamental group. However, I don't think this implies that the images of the topological fundamental group cannot be conjugate? (e.g. there are lots of subgroups of the etale fun. group which are left invariant by the Galois action but on which the Galois action is not conjugation) $\endgroup$
    – Will Chen
    Commented Oct 19, 2017 at 16:19
  • $\begingroup$ Also I don't follow your reasoning in your edit. In your second sentence, what outer automorphism are you referring to? Are you still assuming that $i,i'$ agree on $K$? $\endgroup$
    – Will Chen
    Commented Oct 19, 2017 at 16:24
  • $\begingroup$ @oxeimon Yes. The outer automorphism coming from the Galois group element must agree with an outer automorphism of the topological fundamental group. There are only countably many outer automorphisms of the topological fundamental group (by finite generation) so that is quite restrictive. $\endgroup$
    – Will Sawin
    Commented Oct 19, 2017 at 16:56
  • $\begingroup$ @oxeimon Your first comment is exactly what I realized, and tried to correct, in my edit. The phenomenon you mention, for an arbitrary subgroup is measured by the outer automorphsim group of that subgroup. $\endgroup$
    – Will Sawin
    Commented Oct 19, 2017 at 16:58
  • $\begingroup$ So I'm still a little confused. Are you claiming to give examples where the images are not conjugate? I agree that your cardinality argument shows that there must exist uncountably many images, but this doesn't imply that they can't all be conjugate. After all the etale fundamental group is uncountable and center-free (in the hyperbolic case), so one would expect a typical non-closed subgroup to have uncountably many conjugates, right? $\endgroup$
    – Will Chen
    Commented Oct 20, 2017 at 0:23

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...


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