Talk:Simple closed curves in surfaces (Ex)

Suppose that $S=\Sigma_g\setminus(\cup_{i=1}\nu(\alpha_i))$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}S=\Sigma_g\setminus(\cup_{i=1}\nu(\alpha_i))$ is connected. We will construct a dual basis $\beta_1,...,\beta_g$$\beta_1,...,\beta_g$ as usual and then use the intersection pairing $\lambda$$\lambda$ on $H_1(\Sigma_g;\mathbb{Z}/2)$$H_1(\Sigma_g;\mathbb{Z}/2)$ to see that the $[\alpha_i]$$[\alpha_i]$ are linearly independent.

Deleting an annulus from a surface preserves Euler characteristic, so $\chi(S)=2-2g$$\chi(S)=2-2g$, and $S$$S$ has $2g$$2g$ boundary components (in pairs corresponding to each $\alpha_i$$\alpha_i$). Thus, $S\cong(S^2-2g$$S\cong(S^2-2g$ disks$)$$)$. Let $\beta’_1$$\beta’_1$ be an arc in $S$$S$ between the two boundary components of $\overline{\nu(\alpha_1))}$$\overline{\nu(\alpha_1))}$. Extend $\beta’_1$$\beta’_1$ to a loop in $\Sigma$$\Sigma$ by adding a cocore of the annulus $\overline{\nu(\alpha_1)}$$\overline{\nu(\alpha_1)}$. Now $S’=S-\nu(\beta’_1)$$S’=S-\nu(\beta’_1)$ is another planar surface, so we repeat the argument in turn to find loops $\beta_2,...,\beta_g$$\beta_2,...,\beta_g$ in $\Sigma$$\Sigma$ so that for $1\le i\neq j\le g$$1\le i\neq j\le g$, we have $\lambda(\beta_i, \beta_j)=0, \lambda(\alpha_i,\beta_j)=0,$$\lambda(\beta_i, \beta_j)=0, \lambda(\alpha_i,\beta_j)=0,$ and $\lambda(\alpha_i,\beta_i)=1$$\lambda(\alpha_i,\beta_i)=1$.

Now suppose for some $n_1,...,n_g\in\{0,1\}$$n_1,...,n_g\in\{0,1\}$, $w=\sum_{i=1}^g n_i[\alpha_i]$$w=\sum_{i=1}^g n_i[\alpha_i]$ is the identity in $H_1(\Sigma_g;\mathbb{Z}/2)$$H_1(\Sigma_g;\mathbb{Z}/2)$. Then $0=\lambda(w,\beta_i)=n_i$$0=\lambda(w,\beta_i)=n_i$ for each $i$$i$. Thus, $[\alpha_1],...,[\alpha_g]$$[\alpha_1],...,[\alpha_g]$ are linearly independent.

The converse holds as well. That is, suppose instead that $S$$S$ is disconnected. Let $F$$F$ be one component of $S$$S$, and let $X=\{i\mid\overline{\nu(\alpha_i)}$$X=\{i\mid\overline{\nu(\alpha_i)}$ meets $F$$F$ in exactly one boundary component$\}$$\}$. Since $S$$S$ has multiple components, $X$$X$ is nonempty. Let $w=\cup_{i\in X}\alpha_i$$w=\cup_{i\in X}\alpha_i$. Then $w$$w$ bounds in $S$$S$ (up to reorienting some $\alpha_i$$\alpha_i$), so $[w]=\sum_{i\in X}[\alpha_i]=[0]\in H_1(\Sigma_g;\mathbb{Z}/2)$$[w]=\sum_{i\in X}[\alpha_i]=[0]\in H_1(\Sigma_g;\mathbb{Z}/2)$. Thus, $[\alpha_1],...,[\alpha_g]$$[\alpha_1],...,[\alpha_g]$ are linearly dependent.

Alternate method: We can control the number of components by the zeroth homology. First, thicken up the circles given by the curves, so that they become 2D bands. Then use Mayer Vietoris, where the first component is the desired complement $\Sigma_g\setminus(\cup_{i=1}\nu(\alpha_i))$$\Sigma_g\setminus(\cup_{i=1}\nu(\alpha_i))$, and the second component is the union of the thickened circles. The purpose of thickening is to provide a workable intersection. Since the group is $\mathbb Z_2$$\mathbb Z_2$, we can treat the homology as a vector space, which lets us count dimensions. If the curves are linearly dependent, the boundary map from $H_1$$H_1$ to $H_0$$H_0$ has image whose dimension is smaller than the number of curves ($g$$g$), which then forces the dimension of the zero-th homology of the cut manifold to be at least 2.

Note that this alternate method works without modification, if there is a different number of curves (say, $g-1$$g-1$).