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Let $X$ be an irreducible smooth projective algebraic curve over $\mathbb{C}$. Then, the Abel-Jacobi map gives embedding $X \hookrightarrow Jac(C)$ of $C$ into it's Jacobian. This map induces an isomorphism on $H_1$.

Question: Can it be made obvious, using purely topological reasoning, that a connected, oriented compact 2-dimensional topological manifold of genus $g$ embeds into a torus $\phi: C \rightarrow (S^1)^{2g}$, so that $\phi$ induces an isomorphism on $H_1$?

I don't know how to see this without using the Abel-Jacobi theorem.

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Presumably you want the embedding to induce an isomorphism on $H_1$; otherwise you can just use the fact that any oriented surface embeds in $\mathbb{R}^3$. – Eric Wofsey Feb 21 '13 at 4:36
Ah, yes. Thanks Eric. – LMN Feb 21 '13 at 4:46
I like this question in part because it forces people to think in a concrete topological way about something many people learn only in the context of algebraic or complex geometry. – David Corwin Feb 21 '13 at 6:51
Thanks Davidc897, and thanks everyone for your great answers! – LMN Feb 21 '13 at 21:24

Yes. We will come up with $g$ maps from the Riemann surface to the two-dimensional torus. We can then take the product of these maps. As long as one of them is nonconstant at each point, it will be an embedding.

View the Riemann surface in the standard way as a handlebody that looks like a bunch of tori glued together. (Presumably the classification of Riemann surfaces, which we use to put it in this form, counts as purely topological reasoning!) There are $g$ tori. Pick one, and contract all but that one to a point. If that is one of the tori at the ends, then all the rest of the surface gets contracted to a single point. If it's in the middle, the left half gets contracted to a point on the left side of the torus and the right side gets contracted to a point on the right side of the torus.

This set of $g$ maps gives us the associated kind of map. Clearly, this gives an isomorphism on homology. To make sure this is an embedding, we have to make sure that no point on the torus gets contracted in every single map. We will do this by defining the maps a little generously, such that the stuff that is contracted for the leftmost torus ends a little bit to the right of where the stuff contracted for the second-leftmost torus begins.

Hopefully this explanation makes sense! My picture-drawing skills are not up to the task of realizing this.

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This is really nice! It's essentially a version of the Abel-Jacobi map, where instead of integrating holomorphic differentials, one integrates differentials supported on (or near) each of the tori you've glued together. – Daniel Litt Feb 21 '13 at 4:58
You might be interested in the discussion in the paper "Parameterized Abel-Jacobi maps and abelian cycles in the Torelli group" by Church and Farb (see, which builds on this type of construction. – Andy Putman Feb 21 '13 at 5:25
Thanks! Of course, by differentiating any such map, one can write it as the integral of some set of differentials. – Will Sawin Feb 21 '13 at 5:40
Very nice answer! Another way maybe to "see" it: picture a wedge of g tori, embedded in the "obvious" way in their product, and then smooth it by replacing the common point by a common sphere, obtaining an embedded sphere with g handles. – roy smith Feb 21 '13 at 5:49

I'll do things a little more abstractly than Will.

Of course, things are trivial in genus $1$. To simplify things a bit, I'll restrict myself to a surface $\Sigma_g$ of genus $g \geq 3$ (the case $g=2$ can be handled with a bit more care). Both the surface $\Sigma_g$ and the torus $(S^1)^{2g}$ are Eilenberg-MacLane spaces. Recall that (modulo basepoint issues) homotopy classes of maps between Eilenberg-MacLane spaces are in bijection with homomorphisms between their fundamental groups. Fixing an identification of $\pi_1((S^1)^{2g})$ with $H_1(\Sigma_g;\mathbb{Z})$, there thus exists a unique homotopy class of continuous maps $\phi : \Sigma_g \rightarrow (S^1)^{2g}$ inducing the surjection $\pi_1(\Sigma_g) \rightarrow H_1(\Sigma_g;\mathbb{Z})$; since the target is abelian, there is no need to worry about basepoints. Since the dimension of $(S^1)^{2g}$ is greater than $5=2 \cdot 2 + 1$, the usual Whitney argument shows that we can homotope $\phi$ a little bit to assure that $\phi$ is an embedding.

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