Surfaces and Manifolds

remark:

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

theorem (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 $\begin{align*} \chi(X) = \begin{cases} 2-2g - b & \text{orientable} \\ 2 - k & \text{non-orientable}. \end{cases} \end{align*}$

proposition (Polygon Models for Surfaces):

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

figures/image_2021-04-08-19-40-14.png

figures/image_2021-04-08-19-40-31.png

figures/image_2021-04-08-19-40-41.png

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

fact:

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	?	?	?	?	${\mathbb{K}}, {\mathbb{M}}$	${\mathbf{RP}}^2$	$\emptyset$

proposition (Inclusion-Exclusion):

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

proof:

Todo

\todo{Proof.}

corollary (Euler for Connect Sums):

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

proof:

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

proposition (Decomposing

$\RP^2$ ):

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

proposition (Decomposing a Klein Bottle):

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

proof:

Todo

\todo{Proof.}

proposition (Rewriting a Klein Bottle):

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

proof:

Todo

\todo{Proof.}

Manifolds

remark:

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.

proposition (Dimension vanishing for homology of manifolds):

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

proposition (Top homology for manifolds):

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

proposition (Poincaré Duality for manifolds):

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

proposition (Relative Poincaré Duality for manifolds):

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

proposition (Known Euler characteristics):

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

proof (?):

Todo. Uses Poincaré duality?

\todo[inline]{Proof!}

proposition (Nondegenerate intersection pairings):

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

proposition (Orientation covers):

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.

theorem (Lefschetz Duality):

Todo

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

3-Manifolds, and Knot Complements

fact:

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

proposition (Homology of 3-manifolds):

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

proposition (Homotopy type of knot complements):

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 $K$ .

proof (?):

Todo

\todo[inline]{todo}

proposition (Homology of knot complements in

$S^3$ ):

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

proof:

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 $H^3(S^3-K) =0$ .