Surfaces and Manifolds

The most common spaces appearing in this theory:

• $${\mathbb{S}}^2$$,
• $${\mathbb{T}}^2 \coloneqq S^1\times S^1$$,
• $${\mathbf{RP}}^2$$
• $${\mathbb{K}}$$ the Klein bottle
• $${\mathbb{M}}$$ the Möbius Strip
• $$\Sigma_n \coloneqq#_{i=1}^n {\mathbb{T}}^2$$.

The first 4 can be obtained from the following pasting diagrams:

Pasting Diagrams for Surfaces

Classification of Surfaces

The set of surfaces under connect sum forms a monoid with the presentation \begin{align*} \left\langle{ {\mathbb{S}}^2, {\mathbf{RP}}^2, {\mathbb{T}}{~\mathrel{\Big\vert}~}{\mathbb{S}}^2 = 0, 3{\mathbf{RP}}^2 = {\mathbf{RP}}^2 + {\mathbb{T}}^2}\right\rangle = \left\{{ \Sigma_{g, n} {~\mathrel{\Big\vert}~}g, n \in {\mathbf{Z}}^{\geq 0} }\right\} .\end{align*} where $$\Sigma_{g, n}$$ is a surface of genus $$g$$ with $$n$$ discs removed to form boundary components.

Surfaces are classified up to homeomorphism by orientability and $$\chi$$, or equivalently “genus”

• In orientable case, actual genus, $$g$$ equals the number of copies of $${\mathbb{T}}^2$$.
• In nonorientable case, $$k$$ equals the number of copies of $${\mathbf{RP}}^2$$.

In each case, there is a formula \begin{align*} \chi(X) = \begin{cases} 2-2g - b & \text{orientable} \\ 2 - k & \text{non-orientable}. \end{cases} \end{align*}

Every surface can be obtained as the identification space of a polygon labeled with sides $$\alpha_i, \beta_i, \rho_i$$.

\todo[inline]{Examples, general procedure?}

Orientable?$$-4$$$$-3$$$$-2$$$$-1$$$$0$$$$1$$$$2$$
Yes$$\Sigma_3$$$$\emptyset$$$$\Sigma_2$$$$\emptyset$$$${\mathbb{T}}^2, S^1\times I$$$${\mathbb{D}}^2$$$${\mathbb{S}}^2$$
No????$${\mathbb{K}}, {\mathbb{M}}$$$${\mathbf{RP}}^2$$$$\emptyset$$

\begin{align*} X = U\cup V \implies \chi(X) = \chi(U) + \chi(V) - \chi (U\cap V) .\end{align*}

Todo

\todo{Proof.}


\begin{align*} \chi(A # B) = \chi(A) + \chi(B) - 2 .\end{align*}

Set $$U= A, B=V$$, then by definition of the connect sum, $$A\cap B = {\mathbb{S}}^2$$ where $$\chi({\mathbb{S}}^2) = 2$$

\begin{align*} {\mathbf{RP}}^2 = {\mathbb{M}}{\textstyle\coprod}_{\operatorname{id}_{{{\partial}}{\mathbb{M}}}} {\mathbb{M}} .\end{align*}

\begin{align*} {\mathbb{K}}\cong {\mathbf{RP}}^2 # {\mathbf{RP}}^2 .\end{align*}

Todo

\todo{Proof.}


\begin{align*} {\mathbf{RP}}^2 # {\mathbb{K}}\cong {\mathbf{RP}}^2 # {\mathbb{T}}^2 .\end{align*}

Todo

\todo{Proof.}


Manifolds

To show something is not a manifold, try looking at local homology. Can use point-set style techniques like removing points, i.e. $$H_1(X, X-{\operatorname{pt}})$$; this should essentially always yield $${\mathbf{Z}}$$ by excision arguments.

If $$M^n$$ is a closed and connected $$n{\hbox{-}}$$manifold, then $$H^{\geq n} X = 0$$.

If $$M^n$$ is a closed connected manifold, then $$H_n = {\mathbf{Z}}$$ and $$\operatorname{Tor}(H_{n-1}) = 0$$. More generally, \begin{align*} \begin{cases} {\mathbf{Z}}& M^n \text{ is orientable } \\ 0 & \text{else}. \end{cases} \end{align*}

For $$M^n$$ a closed orientable manifold without boundary and $${ \mathbf{F} }$$ a field, \begin{align*} H_k(M^n; { \mathbf{F} }) \cong H^{n-k}(M^n; { \mathbf{F} }) \iff M^n \text{ is closed and orientable} .\end{align*}

If $$M^n$$ is a closed orientable manifold with boundary then \begin{align*} H_k(M^n; {\mathbf{Z}}) \cong H^{n-k}(M^n, {\partial}M^n; {\mathbf{Z}}) .\end{align*}

If $$M^n$$ is closed and $$n$$ is odd, then $$\chi(M^n) = 0$$.

Todo. Uses Poincaré duality?

\todo[inline]{Proof!}


For $$M^n$$ closed and orientable, the intersection pairing is nondegenerate modulo torsion.

For any manifold $$X$$ there exists a covering space $$p: \tilde X_o\to X$$, the orientation cover, where any map $$Y\to X$$ factors through $$\tilde X_o$$. If $$X$$ is nonorientable, then $$p$$ is a double cover.

Todo

\todo[inline]{Statement of Lefschetz duality.}


3-Manifolds, and Knot Complements

Every $${\mathbf{C}}{\hbox{-}}$$manifold is canonically orientable.

Let $$M^3$$ be a 3-manifold, then its homology is given by the following (by cases):

• Orientable: $$H_* = ({\mathbf{Z}}, {\mathbf{Z}}^r, {\mathbf{Z}}^r, {\mathbf{Z}})$$

• Nonorientable: $$H_* = ({\mathbf{Z}}, {\mathbf{Z}}^r, {\mathbf{Z}}^{r-1} \oplus {\mathbf{Z}}_2, {\mathbf{Z}})$$

For $$K$$ a knot, $$S^3\setminus K$$ is a $$K(\pi, 1)$$, and $${\mathbf{R}}^3 \setminus K \simeq S^2 \vee \qty{S^3 \setminus K}$$. Moreover, if $$K$$ is nullhomologous and $$X$$ is any 3-manifold, \begin{align*} H_1\qty{X\setminus\nu(K)} \cong H_1 X \times{\mathbf{Z}} \end{align*} where $$\nu(K)$$ is a tubular neighborhood of $$K$$.

Todo

\todo[inline]{todo}


For $$K$$ a knot, \begin{align*} H_*(S^3 \setminus K) = [{\mathbf{Z}}, {\mathbf{Z}}, 0, 0, \cdots] .\end{align*}

Apply Mayer-Vietoris, taking $$S^3 = n(K) \cup (S^3-K)$$, where $$n(K) \simeq S^1$$ and $$S^3-K \cap n(K) \simeq T^2$$. Use the fact that $$S^3-K$$ is a connected, open 3-manifold, so $$H^3(S^3-K) =0$$.