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?}
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*}



\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*}



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




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?


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[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 \(K\).



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