Classification of surfaces

Section 2.20 Classification of surfaces

Definition 2.20.1. Equivalent labelling schemes.

Two labelling schemes $w_1,w_2,\dots, w_n$ and $w_1', w_2',\dots, w_m'$ are equivalent if their corresponding quotient spaces are homeomorphic. In this case we write

\begin{equation*} w_1,w_2,\dots, w_n\sim w_1', w_2',\dots, w_m'\text{.} \end{equation*}

Theorem 2.20.2. Elementary operations.

Each of the following procedures, called elementary operations, when applied to a labelling scheme $w_1, w_2,\dots, w_m\text{,}$ produces an equivalent labelling scheme.

Cut/Paste.
Cut any $w_k\text{,}$ replacing it with $y_0c^{-1}, cy_1\text{,}$ where $w_k=y_0y_1\text{,}$ each $y_i$ has length at least two, and $c$ does not appear as a label in the original labelling scheme.

Past any two elements $w_j=y_0c^{-1}, w_k=cy_1$, together, replacing them with $w=y_0y_1$, provided each $y_i$ has length at least two, and $c$ does not occur anywhere else in the labelling scheme.
Relabel base/flip exponent.
Fix a label $c$ occurring in the scheme. Replace all occurrences $c^\epsilon$ with $d^\epsilon\text{,}$ where $d$ is a label not occurring in the original labelling scheme.

Fix a label $c$ occurring in the scheme. Replace all occurrences of $c^\epsilon$ with $c^{-\epsilon}\text{.}$
Cycle.
Replace any $w_k=a_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_r^{\epsilon_r}$ with $w_k'=a_{\sigma(1)}^{\epsilon_{\sigma(1)}}a_{\sigma(2)}^{\epsilon_{\sigma(2)}}\cdots a_{\sigma(r)}^{\epsilon_{\sigma(r)}}\text{,}$ where $\sigma$ is any cyclic permutation of $\{1,2,\dots, n\}\text{.}$
Flip.
Replace any $w_k=a_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_r^{\epsilon_r}$ with its formal inverse $w_k^{-1}=a_r^{-\epsilon_r}a_{r-1}^{-\epsilon_{r-1}}\cdots a_1^{-\epsilon_1}\text{.}$
Fusion/Fission.
Assuming labels $a,b$ always occur in the labelling scheme either as $ab$ or $b^{-1}a^{-1}\text{,}$ replace each instance of $ab$ with $c$ and each instance of $b^{-1}a^{-1}$ with $c^{-1}\text{,}$ provided $c$ does not occur in the original labelling scheme, and that the resulting polygons all have length at least $3\text{.}$

Replace all instances of $c$ with $ab$ and all instances of $c^{-1}$ with $b^{-1}a^{-1}\text{,}$ assuming $a$ and $b$ do not occur in the original labelling scheme.
Cancel/uncancel.
Replace any $w_k=y_0cc^{-1}y_1$ with $w_k'=y_0y_1\text{,}$ provided $c$ does not occur anywhere else in the labelling scheme and $w_k$ has length at least 5.

Replace any $w_k=y_0y_1$ with $w_k'=y_0cc^{-1}y_1\text{,}$ provided $c$ does not occur in the original labelling scheme.

Theorem 2.20.3. Abelianization of quotients.

Let $N$ be a normal subgroup of the group $G\text{,}$ and let $q\colon G\rightarrow G/N$ and $q_{\ab}\colon G\rightarrow G/[G,G]=G^{\ab}$ be the quotient homomorphisms. We have

\begin{equation*} (G/N)^{\ab}\cong G^{\ab}/q_{\ab}(N). \end{equation*}

Proof.

Let $q\colon G\rightarrow G/N$ and $q_{\ab}\colon G\rightarrow G^{\ab}=G/[G,G]$ be the quotient maps. Given $gN\in G/N\text{,}$ let $\overline{gN}$ denote its image in $(G/N)^{\ab}$ under the quotient map $G/N\rightarrow (G/N)^{\ab}\text{.}$ Similarly, given $g[G,G]\in G^{\ab}\text{,}$ let $\overline{g[G,G]}$ denote its image in $G^{\ab}/q_{\ab}(N)\text{.}$ (There will be no ambiguity in our discussion as to which map the bar notation refers to.) Using mapping properties of the quotient $G/N$ and the abelianization $G^{\ab}\text{,}$ we get the following commutative diagram

$Quotients and abelianizations diagram$

where $r$ satisfies $r(gN)=\overline{g[G,G]}$ and $s$ satisfies $s(g[G,G])=\overline{gN}^{\ab}\text{.}$ Now the mapping properties of the abelianization $(G/N)^{\ab}$ and the quotient $G^{\ab}/q_{\ab}(N)$ give rise to the commutative diagram

$Another quotients and abelianizations diagram$

where $\phi$ satisfies

\begin{equation*} \phi(\overline{gN})=r(gN)=\overline{g[G,G]} \end{equation*}

and $\psi$ satisfies

\begin{equation*} \psi(\overline{g[N,N]})=s(g[G,G])=\overline{gN}\text{.} \end{equation*}

It is then evident from the for formulas for $\phi$ and $\psi$ that $\psi\circ \phi=\id$ and $\phi\circ\psi=\id\text{.}$ We conclude that $\phi$ (and $\psi$) is an isomorphism, as desired.

Theorem 2.20.4. Non-homeomorphic surfaces.

Let $T_n$ denote the $n$-fold torus, and let $P_m$ denote the $m$-fold projective plane ($P_1=\PP^2$). We have

\begin{align*} (\pi_1(S^2,p))^{\ab} \amp \cong \{e\} \text{ for all } p\in S^1\\ (\pi_1(T_n,p))^{\ab} \amp \cong \Z^{2n}=\bigoplus_{k=1}^{2n}\Z \text{ for all } p\in T_n\\ (\pi_1(P_m,p))^{\ab} \amp \cong \Z/2\Z\oplus \Z^{m-1} \text{ for all } p\in P_m \end{align*}

It follows that the fundamental groups of the surfaces $S^2, T_1,T_2,\dots, P_1, P_2,\dots $ are pairwise non-isomorphic, and hence that the surfaces are pairwise non-homeomorphic.

Proof.

First observe that the last sentence of the theorem follows from the fact that isomorphic groups have isomorphic abelianizations, and homeomorphic spaces have isomorphic fundamental groups. We now prove the statement about the abelianizations.

To see that the listed abelianizations are pairwise non-isomorphic groups, observe first that all the groups associated to spaces $P_m$ have torsion elements, whereas those associated to the spaces $T_n$ are torsion free. Next for any $n$ the abelian group associated to $T_n$ has rank $n\text{.}$ Thus for any $n\ne n'$ the corresponding abelianizations have distinct ranks, and hence are not isomorphic. Similarly, the rank of the abelian group associated to $P_m$ is $m-1\text{,}$ and thus these groups are non-isomorphic for any pair $m\ne m'\text{.}$

We now turn to the computation of these abelianizations. Recall that we have

\begin{align*} \pi_1(S^2,p) \amp \cong \{e\} \text{ for all } p\in S^1\\ \pi_1(T_n,p) \amp \cong \angvec{a_1,b_1,a_2,b_2\dots, a_n,b_n\vert [a_1,b_1][a_2,b_2]\cdots[a_n,b_n]} \text{ for all } p\in T_n\\ \pi_1(P_m,p) \amp \cong \angvec{a_1,a_2,\dots, a_m\vert a_1^2a_2^2\cdots a_m^2} \text{ for all } p\in P_m\text{.} \end{align*}

Clearly we thus have $(\pi_1(S^2, p))^{\ab}=\{e\}\text{.}$ We will use Theorem 2.20.3 for the remaining statements. To relate this result more directly to our given finite presentations, we will make use of the following claim (which you are invited to prove): if $q\colon G\rightarrow G'$ is a surjective homomorphism and $Y\subseteq G\text{,}$ then $q(\lns(Y))=\lns(q(Y))\text{,}$ where $\lns(Y)$ denotes the least normal subgroup containing the set set $Y\text{.}$ (A hint for proving this fact: use the correspondence theorem to relate normal subgroups of $G'$ with normal subgroups of $G\text{.}$)

Now consider $T_n\text{.}$ Let $X=\{a_1,b_1,a_2,b_2,\dots, a_n,b_n\}\text{,}$ let $F(X)$ be the free group on $X\text{,}$ let $Y=\{[a_1,b_1][a_2,b_2]\cdots [a_n,b_n]\}\text{,}$ and let $q\colon F(X)/\lns(Y)\rightarrow (F(X)/\lns(Y))^{\ab}$ be the quotient map. We have

\begin{align*} (\pi_1(T_n,p))^{\ab} \amp \cong (F(\{a_1,b_1,\dots, a_n,b_n\})/\lns(Y))^{\ab}\\ \amp \cong (F(X))^{\ab}/q(\lns(Y)) \amp \knowl{./knowl/xref/th_abelianization_of_quotient.html}{\text{2.20.3}}\\ \amp \cong (F(X))^{\ab}/\lns(q(Y)) \amp \text{(see claim)}\\ \amp \cong F(X)^{\ab}\text{,} \end{align*}

where the last line follows since $q([w,w'])=e\in F(X)^{\ab}$ for any elements $w,w'\in F(X)\text{.}$ Finally, by Theorem 2.14.11 $F(X)^{\ab}$ is a free abelian group of rank $2n\text{:}$ and in fact, the image of the system of free generators $\{a_1,b_1,\dots, a_n,b_n\}$ is a basis of $F(X)^{\ab}$ as a free abelian group. We conclude that $(\pi_1(T_n,p))^{\ab}\cong \Z^{2n}\text{,}$ as claimed.

Finally, consider $P_m\text{.}$ Let $X=\{a_1,a_2,\dots, a_m\}\text{.}$ An argument very similar to the preceding one implies

\begin{equation*} (\pi_1(P_m,p))^{\ab}\cong F(X)^{\ab}/\lns(\{a_1^2a_2^2\cdots a_m^2\})\cong F(X)^{\ab}/\angvec{\overline{a_1}^2\overline{a_2}^2\cdots \overline{a_m}^2}\text{,} \end{equation*}

where $\overline{a_k}$ denotes the image of $a_k$ in the abelianization $F(X)^{\ab}\text{.}$ (Note that all subgroups in $F(X)^{\ab}$ are normal since the group is abelian. This is why we can dispence with the $\lns$ operator.) Let $\bolde_1, \bolde_2,\dots, \bolde_{m}$ be the standard basis of $\Z^m\text{.}$ (In other words, $\bolde_k$ is the $m$-tuple containing a one in the $k$-th entry, and zeros elsewhere.) Since $\{\overline{a}_k\}_{k=1}^m$ forms a basis of $F(X)^{\ab}$ as a free abelian group, there is an isomorphism

\begin{equation*} F(X)^{\ab}/(\angvec{\overline{a_1}^2\overline{a_2}^2\cdots \overline{a_m}^2})\isomto\Z^m/\angvec{2(\bolde_1+\bolde_2+\cdots +\bolde_m)} \end{equation*}

obtained by sending $\overline{a_k}$ to $\bolde_k\text{.}$ Next, it is easy to see that

\begin{align*} \bolde_1' \amp = \bolde_1+\bolde_2+\cdots +\bolde_m\\ \bolde_2' \amp = \bolde_2\\ \amp \vdots \\ \bolde_m'\amp =\bolde_m \end{align*}

is another basis of $\bigoplus_{k=1}^m\Z\text{.}$ It follows that the isomorphism of $\Z^m$sending $\bolde_k'$ to $\bolde_k$ for all $1\leq k\leq m$ gives rise to an isomorphism

\begin{equation*} \Z^m/\angvec{2(\bolde_1+\bolde_2+\cdots +\bolde_m)}\isomto \Z^m/\angvec{2\bolde_1}\cong \Z/2\Z\oplus \Z^{m-1}\text{,} \end{equation*}

as desired.

Theorem 2.20.5. Classification theorem.

Let $T_n$ denote the $n$-fold torus, and let $P_m$ denote the $m$-fold projective plane ($P_1=\PP^2$). If $X$ is a compact 2-manifold (i.e., a surface), then $X$ is homemorphic to exactly one of the following surfaces: $S^2\text{,}$ $T_n$ for some $n\geq 1\text{,}$ $P_m$ for some $m\geq 1\text{.}$

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