Suppose that $\gamma_1,...,\gamma_n$ are a set of disjoint simple closed curves on a closed orientable surface $\Sigma$ that all bound disks in some handlebody $H$ with $\partial H = \Sigma$. Let $\gamma$ is a simple closed curve on $\Sigma$ such that the homology class $[\gamma]$ is in the subgroup of $H_1(\Sigma; \mathbb{Z})$ generated by $[\gamma_1],...,[\gamma_n]$.

Does $\gamma$ also bound a disk in $H$?


First, it doesn't change anything to suppose each $[\gamma_i]$ is nonzero and $[\gamma_i]\neq [\gamma_j]$ when $i\neq j$. Second, it doesn't hurt to suppose $n$ is the genus of $\Sigma$ by adding in the rest of the curves.

With this setup, the kernel of $H_1(\Sigma)\to H_1(H)$ is generated by $[\gamma_1],\dots,[\gamma_n]$. Consider the universal abelian cover $H'\to H$, which is the one corresponding to the homomorphism $\pi_1(H)\to H_1(H)$, and it can be visualized in $\mathbb{R}^{n}$ with a ball at each lattice point $\mathbb{Z}^{n}$ and $1$-handles connecting the balls in axial directions. Each $\gamma_i$ lifts to $H'$ on a handle that goes in the $i$ direction, and the disk that $\gamma_i$ bounds lifts to a disk in that handle.

If $\gamma\subset\Sigma$ is a simple closed curve by your hypotheses, then $[\gamma]$ is in the kernel, or equivalently that $[\gamma]$ is zero in $H_1(H)$, which in terms of covering spaces is that $\gamma$ lifts to a closed loop in $H'$. Furthermore, $\gamma$ bounds a disk in $H$ if and only if it does in $H'$.

So, if we find a homotopically nontrivial simple closed curve in $H'$ that downstairs is still an embedded curve, we have a $\gamma$ that does not bound a disk. Here's a candidate for a curve in a $H$ a genus-$2$ handlebody, as seen from the cover $H'$:

A homotopically nontrivial simple closed curve in the universal abelian cover of a genus-2 handlebody

The image of this curve on $\Sigma\subset H$ itself is

A homotopically non-trivial simple closed curve on the surface

Notice that the curve is actually zero in $H_1(\Sigma)$ since it passes over the top of each handle once in both directions.

I personally appreciate seeing these things on a standard handlebody, so here is a $\gamma$ with $[\gamma]\in\ker(H_1(\Sigma)\to H_1(H))$ and yet $\gamma$ does not bound a disk in $H$.

Again, but in a standard surface

I was going to give the next picture before the previous one as a sort of joke, but after deforming the above one into this this form (a thickened punctured torus) it becomes manifestly obvious the curve (1) does not bound a disk and (2) is nullhomologous in $\Sigma$:

The curve in a genus-2 handlebody represented as a thickened punctured torus


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.