MATH 422 Lecture Note #9 (2018 Spring)
Surfaces and abelianization

Jae Choon Cha
POSTECH $\def\id{\operatorname{id}} \def\rel{\text{ rel }} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\R{\mathbb{R}} \def\C{\mathbb{C}} \def\sm{\smallsetminus}$

Orientable surfaces

Let $g\ge 1$. Take the 2-disk $D^2$, and regard it as a regular 4g-gon with edges labeled by symbols $a_i$, $b_i$ as follows.


Define the space $\Sigma_g$ to be the quotient space of $D^2$ obtained by identifying edged with the same label. This is called the orientable surface of genus $g$. For $g=0$, let $\Sigma_0=S^2$.


Let $q\colon D^2\to \Sigma_g$ be the quotient map. It is straightforward to verify that the $4g$ vertices on $\partial D^2$ are all identified, and so $q(\partial D^2)$ is a graph with one vertex and 2g edges $a_1,b_1,\ldots,a_g,b_g$.

Let’s compute the fundamental group of $\Sigma_g$ by using the Seifert-van Kampen theorem. Let $U=q(\operatorname{int}D^2)$ and $V = q(D^2\sm \{0\})$. $U$ and $V$ are open sets in $\Sigma_g$, $U\cup V=\Sigma_g$, and $U\cap V = q(\operatorname{int}D^2\sm \{0\})$ is path connected. Thus, fixing a basepoint $*\in U\cap V$, the Seifert-van Kampen theorem gives us the following pushout diagram:


Note that $U=q(\operatorname{int}D^2)$ is homeomorphic to $\operatorname{int}D^2$, since the restriction of $q$ on $\operatorname{int}D^2$ is 1-1. It follows that $\pi_1(U,*)$ is trivial.

The homotopy equivalence $D^2\sm\{0\}\simeq \partial D^2$ decends to a homotopy equivalence $V=q(D^2\sm\{0\}) \simeq q(D^2)$. One can verify the details as follows: Let $i\colon \partial D^2 \to D^2\sm \{0\}$ be the inclusion, and let $f\colon D^2\sm \{0\} \to \partial D^2$ be the radial projection $f(x)=x/|x|$. Then $f$ is a homotopy inverse of $i$, since $fi=\id_{\partial D^2}$, and $if\simeq \id_{D^2\sm \{0\}}$ via the homotopy $H\colon D^2\sm\{0\} \times I \to D^2\sm\{0\}$ defined by $H(x,t) = (1-t)x/|x| + tx$. Now, think of the induced maps on the quotient spaces. Namely, $\bar f \colon q(D^2\sm \{0\}) \to q(\partial D^2)$ and $\bar i \colon q(\partial D^2) \to q(D^2\sm \{0\})$ given by $\bar f(q(x))=q(f(x))$ and $\bar i(q(x)) = q(i(x))$ are well-defined maps, and furthermore, $\bar H\colon q(D^2\sm\{0\}) \times I \to q(D^2\sm\{0\})$ given by $\overline H(q(x),t) = q(H(x,t))$ is well-defined. This is essentially because $\bar f(x)=x$ and $\bar H(x,t)=x$ for every $x\partial D^2$ and because $q$ identifies points on $\partial D^2$ only. From the above, it follows that $\bar f \bar i=\id_{q(\partial D^2)}$ and $\bar H \colon \bar i \bar f \simeq \id_{q(D^2\sm\{0\})}$. This shows that $\bar f$ is a homotopy inverse of $\bar i$.

By the above paragraph, we have $\pi_1(V,*)$ is isomorphic to the fundamental group of the graph $q(\partial D^2))$, which is the free group on $2g$ generators $a_1,b_1,\ldots, a_g,b_g$. Since $q$ is injective on $\operatorname{int}D^2\sm\{0\}$, $U\cap V = q(\operatorname{int}D^2\sm\{0\})$ is homeomorphic to $\operatorname{int}D^2\sm\{0\}$, and thus homotopy equivalent to $S^1$. Therefore $\pi_1(U\cap V)=\langle x\mid \cdot\rangle$, an infinite cyclic group. Here the generator $x$ is represented by the class of the loop $\alpha(s)=\frac12 e^{2\pi i s}\in \operatorname{int}D^2\sm\{0\}$. From this it follows that the image of $x$ under $\pi_1(U\cap V) \to \pi_1(V)$ is equal to


where the commutator $[a,b]$ is defined to be $aba^{-1}b^{-1}$. So, computing the pushout, we have proven the following result:

Theorem. $\pi_1(\Sigma_g) \cong \langle a_1,b_1,\ldots,a_g, b_g \mid [a_1,b_1]\cdots[a_g,b_g] \rangle$.

In what follows we will develop some methods which are useful in distinguishing $\pi_1(\Sigma_g)$ and $\pi_1(\Sigma_{g'})$ for $g\ne g'$.


Our first approach is to pass to something simpler than the fundamental group, by taking a quotient. Some information will be lost, but on the other hand, the resulting quotient is often good enough to extract the information we wanted. We begin by recalling a definition from elementary abstract algebra.

Definition. Suppose $G$ is a group, and $A$, $B$ are subsets of $G$. Define $[A,B]$ to be the subgroup of $G$ generated by the set $\{[a,b] \mid a\in A,\, b\in B\}$. The subgroup $G'=[G,G]$ is called the commutator subgroup of $G$.

Here, recall from the above that $[a,b] = aba^{-1}b^{-1}$.

It is easy to verify that $[A,B]$ is a normal subgroup of $G$ if $A$ and $B$ are normal subgroups of $G$, using the identity $g[a,b]g^{-1} = [gag^{-1},gbg^{-1}]$. In particular, $G'$ is a normal subgroup of $G$.


  1. $G/G'$ is an abelian group.

  2. Let $p\colon G\to G/G'$ be the quotient homomorphism $p(g)=gG'$. If $f\colon G\to A$ is a homomorphism to an abelian group $A$, then there is a unique homomorphism $\phi\colon G/G' \to A$ such that $f = \phi p$.


    Moreover, if $q\colon G\to H$ is another group homomorphism, with $H$ abelian, which has this property of $p\colon G\to G/G'$, then $H\cong G/G'$.

Proof. (1) holds since $(aG')(bG') = (bG')(aG')$ is equivalent to $ab(ba)^{-1} \in G'$, which is true since $ab(ba)^{-1}=[a,b]$. To show (2), observe that if $f$ is as given above, then $f([a,b]) = f(a)f(b)f(a)^{-1}f(b)^{-1}$ is trivial in $A$ since $A$ is abelian. This shows that $G'\subset \operatorname{Ker}f$, and from this it follows that $f$ induces a desired homomorphism $\phi\colon G/G' \to A$, $\phi(gG')=f(g)$. The homomorphism $\phi$ is uniquely determined by the property $f=\phi p$ since $p$ is surjective. The last statement is proven by a standard argument using the universal property. $\quad\square$

Definition. The group $G/G'$ is called the abelianization of $G$.

Sometimes we will denote $G/G'$ by $G_{ab}$.

For a group presentation, it is very easy to compute the abelianization.

Theorem. For $G=\langle S\mid R\rangle$, $G/G' = \langle S\mid R'\rangle$ where $R' = R\cup \{[x_i,x_j] \mid x_i,x_j\in S\}$.

Proof. We will verify the universal property of the abelianization.
Let, temporarily, $H=\langle S\mid R'\rangle$ and let $q\colon G \to H$ be the homomorphism $q(x_i)=x_i$ for $x_i\in S$. (This is well-defined by the property of a presentation.) Since for every $x_i, x_j\in S$, $[x_i,x_j]\in R'$ and thus $x_i x_j = x_j x_i$ in $H$. From this it follows that $x_i^{\pm1} x_j^{\pm1} = x_j^{\pm1} x_i^{\pm1}$. Therefore, repeatedly using this, $wv=vw$ in $H$ for any two words $w,v$. This proves that $H$ is abelian.

Suppose $f\colon G \to A$ is a homomorphism and $A$ is abelian. Define $\phi\colon H \to A$ by $x_i \mapsto f(x_i)$ for each generator $x_i\in S$. Observe that the relators $R$ of $G$ are sent to the trivial element by this association, since $f$ is a homomorphism. Moreover, the additional relators $[x_i,x_j]$ of $H$ is sent to $[f(x_i),f(x_j)]\in A$, which is trivial since $A$ is abelian. Therefore, our $\phi$ is a well-defined homomorphism, and satisfies $f=\phi q$ obviously. The homomorphism $\phi$ is determined uniquely since $q$ is surjective. $\quad\square$

For free groups, in particular, the abelinization is something familiar.

Theorem. For the free group $F=\langle x_1,\ldots,x_n \mid \cdot\rangle$, the abelianization is given by $F/F' \cong \Z^n = \underbrace{\Z\times\cdots\times\Z}_n$.

Proof. Recall that

\[F/F' = \langle x_1,\ldots,x_n \mid [x_i,x_j]\,(i,j=1,\ldots,n) \rangle.\]

Define $\phi\colon F/F' \to \Z^n$ by $\phi(x_i)=e_i$, where $e_i = (0,\ldots,1,\ldots,0)$ is the $i$th standard basis for $\Z^n$. Define $\psi\colon \Z^n \to F/F'$ by $\psi(a_1,\ldots,a_n) = x_1^{a_1} \cdots x_n^{a_n}$. Then, it is routine to verify that $\phi$ and $\psi$ are well-defined homomorphisms and one is the inverse of another. We leave the details as an exercise for the readers. $\quad\square$

The abelian group $\Z^n$ is often called the free abelian group of rank $n$.

Example. Recall from the above that

\[\pi_1(\Sigma_g) \cong \langle a_1,b_1,\ldots,a_g, b_g \mid [a_1,b_1]\cdots[a_g,b_g] \rangle.\]

Observe that the relator is the product of commutators $[a_i,b_i]$. Therefore

\[\begin{aligned} \pi_1(\Sigma_g) _ {ab} &\cong \langle a_1,b_1,\ldots,a_g, b_g \mid [a_1,b_1]\cdots[a_g,b_g],\, [a_1,b_1], \ldots, [a_g,b_g] \rangle \\ &\cong \langle a_1,b_1,\ldots,a_g, b_g \mid [a_1,b_1], \ldots, [a_g,b_g] \rangle \\ & \cong F_{ab} \cong \Z^{2g} \end{aligned}\]

where $F=F\langle a_1,b_1,\ldots,a_g,b_g\rangle$ is the free group on the $2g$ generators $a_1,b_1,\ldots,a_g,b_g$.

So, once we show that $\Z^n\not\cong \Z^m$ for $n\ne m$, we would obtain that $\pi_1(\Sigma_g) \not\cong \pi_1(\Sigma_{g'})$ for $g\ne g'$. In particular, we would prove that $\Sigma_g$ is not homotopy equivalent to $\Sigma_{g'}$, and so $\Sigma_g$ and $\Sigma_{g'}$ are not homomorphic if $g\ne g'$. Perhaps you may accept it as a well-known "obvious" fact that $\Z^n\not\cong \Z^m$ for $n\ne m$. In what follows, we will discuss how to show this, since it will also be useful for our purpose later, not only for the surfaces $\Sigma_g$.

Localization of abelian groups

Recall that we produced an abelian group $G/G'$, namely the abelianization, from a given group $G$. As a secondary step, we will produce a vector space over $\Q$, from an abelian group. The construction is analogous to that of the quotient field of an integral domain, which you may have learned from the abstract algebra course.

Indeed, an abelian group $A$ is already pretty similar to a vector space over $\Q$. We have addition, and multiplication by an integer scalar: namely, for $a,b\in A$, the group operation gives $a+b$, and furthermore, for $n\in \Z$, we have

\[n\cdot a = \begin{cases} \underbrace{a+\cdots +a}_n & \text{if } n>0, \\ \underbrace{-a-\cdots -a}_{-n} & \text{if } n<0, \\ 0 &\text{if } n=0. \end{cases}\]

So, the missing feature is the scalar multiplication by $\frac1n$, or division by $n$ for $n \in \Z-\{0\}$. (Here $n$ is nonzero!) The following definition is designed for this purpose.

Definition-Theorem. Suppose $A$ is an abelian group. Consider the catesian product $A\times S$, where $S=\Z-\{0\}$.

  1. Define a relation $\sim$ on $A\times S$ by $(a,n) \sim (b,m)$ if $s(ma-nb)=0$ in $A$ for some $s\in S$. Then $\sim$ is an equivalence relation.

    Let $A\otimes \Q$ be the set of equivalence classes, and denote by $\frac{a}{n} \in A\otimes \Q$ the equivalence class of $(a,n)\in A\times S$.

  2. Define operations

    \[(A\otimes \Q) \times (A\otimes \Q) \xrightarrow{+} A\otimes\Q\]

    \[\Q \times (A\otimes \Q) \xrightarrow{\cdot} A\otimes \Q\]

    by $\frac{a}{n} + \frac{b}{m} = \frac{ma+nb}{mn}$ and $\frac{k}{\ell}\cdot \frac{a}{n} = \frac{ka}{\ell n}$, for $\frac{a}{n}, \frac{b}{m}\in A\otimes\Q$ and $\frac{k}{\ell}\in \Q$ ($k\in \Z$, $\ell\in S$).

    Then $(A\otimes\Q, {}+{}, {}\cdot{})$ is a vector space over $\Q$, with $\frac{0}{1}\in A\otimes\Q$ as the zero vector.

We call $A\otimes\Q$ the vector space of fractions. The process of constructing $A\otimes\Q$, or $A\otimes \Q$ itself, is often called the localization.

The proofs of the above facts are routine. So, we leave it as an exercise.

The following theorem describes the universal property of the vector space of fractions. Recall that we can view a vector space over $\Q$ as an abelian group, by forgetting the scalar multiplication. Define $\iota\colon A\to A\otimes\Q$ by $\iota(a)=\frac a1$ for $a\in A$.


  1. $\iota$ is a group homomorphism.

  2. If $f\colon A\to V$ is a group homomorphism to a vector space $V$ over $\Q$, then there is a unique linear transformation $\phi\colon A\otimes\Q \to V$ such that $f=\phi \iota$.

  3. The property 2 determines $A\otimes\Q$ uniquely, up to isomorphisms of vector spaces over $\Q$.

Once again, the proof is straightforward. For (1), we have $\iota(a+b)=\frac{a+b}{1} = \frac a1 + \frac b1 = \iota(a)+\iota(b)$ by definition. For (2), just define $\phi\colon A\otimes\Q \to V$ by $\phi(\frac an)=\frac1n f(a)$. Note that the last expression is defined since $V$ is a vector space over $\Q$ and $\frac 1n \in \Q$. The verification that $\phi$ is a well-defined linear transformation and that $\phi$ is uniquely determined is routine.


  1. Suppose $A=\Z^n$. Then $A\otimes \Q \cong \Q^n$, under the correspondence $\frac{e_i}{n} \mapsto \frac 1n \cdot e_i$, where $e_i$ designates the $i$th standard basis element.

  2. Suppose $A=\Z_d$ with $d\ge 1$, the cyclic group of order $d$. Then $A\otimes \Q=0$, the trivial vector space. Indeed, for any $a\in A$ and $n\in \Z\sm \{0\}$, we have $\frac an = \frac{d\cdot a}{dn} = \frac{0}{n} = 0$ in $A\otimes\Q$. This shows that every element in $A\otimes \Q$ is zero.

Using the above, now we are ready to prove the following statement:

Theorem. For two surfaces $\Sigma_g$ and $\Sigma_{g'}$, the following are equivalent.

  1. $g=g'$.
  2. $\Sigma_g$ and $\Sigma_{g'}$ are homeomorphic.
  3. $\Sigma_g$ and $\Sigma_{g'}$ are homotopy equivalent.
  4. $\pi_1(\Sigma_g)$ and $\pi_1(\Sigma_{g'})$ are isomorphic.

Proof. Obviously we have (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4). By the above, $\pi_1(\Sigma_g)_ {ab} \otimes\Q \cong \Q^{2g}$. So, if (4) holds, then

\[2g=\dim \pi_1(\Sigma_g)_{ab} \otimes\Q = \dim \pi_1(\Sigma_{g'})_{ab} \otimes\Q = 2g'.\]

It follows that (1) holds. $\quad\square$

Computation for a group presentation

In general, suppose $G=\langle S\mid R\rangle$. Let $F$ be the free group generated by $S$. Then, we have the quotient homomorphism $p\colon F\to F_{ab}=F/F'$, and the localization $\iota\colon F_{ab} \to F_{ab}\otimes \Q$.

Recall from the above that $F_{ab}$ is isomorphic to $\Z^n$ if $S$ is a finite set with $n$ elements. To generalize this, we use the following notation.

First recall the definition of the direct product of a (possibily infinite) family of abelian groups $\{A_\lambda\}_{\lambda\in\Lambda}$:

\[\prod_{\lambda\in \Lambda} A_\lambda = \{(a_\lambda)_ {\lambda\in\Lambda} \mid a_\lambda\in A_\lambda\}\]

This is the usual cartesian product of the $G_\lambda$, with group structure given by $(a_\lambda)_ {\lambda\in\Lambda} + (b_\lambda)_ {\lambda\in\Lambda} = (a_\lambda + b_\lambda)_{\lambda\in\Lambda}$. Consider the following subset:

\[\bigoplus_{\lambda\in \Lambda} A_\lambda = \Big\{(a_\lambda)_ {\lambda\in\Lambda} \in \prod_{\lambda\in \Lambda} A_\lambda \,\Big|\, a_\lambda=0 \text{ for all but finitely many }\lambda\in\Lambda \Big\}\]

This is closed under the group operation, and thus is a subgroup. We call $\bigoplus_{\lambda\in \Lambda} A_\lambda$ the direct sum of the abelian groups $A_\lambda$.

Now, returning to our case, recall that our $F$ is the free group generated by an arbitrary set $S$. Write $S=\{x_i\}_ {i\in \Lambda}$, where $\Lambda$ is an index set. For each $x_i\in S$, let $\langle x_i\rangle=\langle x_i \mid \cdot\rangle$ be the infinite cyclic group generated by the symbol $x_i$, and let $\Q\langle x_i\rangle$ be the vector space over $\Q$ which has dimension 1 and basis $\{x_i\}$. That is, $\Q\langle x_i\rangle = \{cx_i\mid c\in \Q\}$. Here we make abuse of notations: $x_i$ represents an element in either $F$, $\langle x_i\rangle$, or $\Q\langle x_i\rangle$.

Also, there are homomorphisms

\[\begin{gathered} p\colon F \to \bigoplus_{i\in \Lambda} \langle x_i \rangle \\ \iota\colon \bigoplus_{i\in \Lambda} \langle x_i \rangle \to \bigoplus_{i\in \Lambda} \Q\langle x_i \rangle \end{gathered}\]

defined by $p(x_j)=(x_i^{n_i})_ {i\in \Lambda}$ where $n_i=\delta_{ij}$ is given by the Kronecker delta, and $i((x_i^{n_i})_ {i\in\Lambda}) = (n_i x_i)_{i\in\Lambda}$.


  1. $\bigoplus_{i\in \Lambda} \langle x_i \rangle$ is the abelianization $F_{ab}$ of $F$.

  2. $\bigoplus_{i\in \Lambda} \Q\langle x_i \rangle$ is the vector space of fractions $F_{ab}\otimes\Q$ of $F_{ab}$.

Proof. We once again check the universal property. Suppose $f\colon F\to A$ be a homomorphism to an abelian group $A$. Define $\phi\colon \bigoplus_{i\in \Lambda} \langle x_i \rangle \to A$ by $\phi((x_i^{n_i})_ {i\in \Lambda}) = \sum_{i\in\Lambda} f(x_i^{n_i})$. Here the key point is that since $(a_i)_ {i\in \Lambda}$ is in the direct sum, not the direct product, the summation has only finitely many nonzero summands, and therefore the sum is well defined. From this definition it follows that $\phi p(x_i) = f(x_i)$, and consequently $\phi p = f$ since the elements $x_i$ generate $F$. Moreover, $\phi$ is uniquely determined since $i$ is surjective. This shows (1).

To show (2), suppose $f\colon F_{ab} \to V$ is a homomorphism to a vector space $V$. Define $\phi\colon \bigoplus_{i\in \Lambda} \Q\langle x_i \rangle \to V$ by $\phi(x_i)=f(x_i)$ and extending it linearly, namely $\phi((n_ix_i)_ {i\in \Lambda}) = \sum_{i\in\Lambda} n_i f(x_i)$. This is a linear transformation such that $\phi \iota = f$, and such $\phi$ is uniquely determined by the value $\phi(x_i)=f(x_i)$. $\quad\square$

Remark. In the above argument, we indeed showed that for any collection of elements $\{a_i\}_ {i\in\Lambda}$ in an abelian group $A$, there is a homomorphism $\bigoplus_ {i\in \Lambda} \langle x_i \rangle \to A$ which takes $x_i$ to $a_i$. Compare this with the universal property of the free group $F=F\langle S\rangle$. Because of this property, the abelian group $\bigoplus_{i\in \Lambda} \langle x_i \rangle$ is called the free abelian group generated by $S$.

Recall that our group $G$ is given by the presentation $G=\langle S \mid R\rangle$. Let $H$ be the subgroup of $\bigoplus_{i\in \Lambda} \langle x_i \rangle$ generated by $\{p(r_j) \mid r_j\in R \}$. Let $W$ be the subspace of $\bigoplus_{i\in \Lambda} \Q\langle x_i \rangle$ generated by $\{\iota p(r_j) \mid r_j\in R \}$. Namely $W$ is consists of all linear combinations of the vectors $\iota p(r_j)$. Then there is the following commutative diagram:


Here the vertical arrows $\alpha$, $\beta$, $\gamma$ are quotient homomorphisms. The homomorphism $q$ exists since relators $r_j\in R$ are killed under $\beta p$. The homomorphism $\jmath$ exists since the subgroup $H$ is killed under $\gamma\imath$.


  1. $\big(\bigoplus_{i\in \Lambda} \langle x_i \rangle\big) / H$ is the abelianization $G_{ab}$ of $G$.

  2. $\big(\bigoplus_{i\in \Lambda} \Q\langle x_i \rangle\big) / W$ is the vector space fractions $G_{ab}\otimes\Q$ of $G_{ab}$.

Proof. Let $f\colon G\to A$ be a homomorphism to an abelian group $A$. By the universal property of the abelianization $F_{ab} = \bigoplus_{i\in \Lambda}\langle x_i\rangle$, there is a homomorphism $g\colon \bigoplus_{i\in \Lambda}\langle x_i\rangle \to A$ such that $f\alpha= gp$. Since $g(H)$ is trivial, there is a homomorphism $\phi\colon \bigoplus_{i\in \Lambda}\langle x_i\rangle \to A$ such that $g=\phi\beta$. So $f\alpha = gp=\phi\beta p = \phi q\alpha$. Since $\alpha$ is surjective, it follows that $f = \phi q$. In addition, this property determines $\phi$ uniquely, since $q$ is surjective. This shows that $\big(\bigoplus_{i\in \Lambda}\langle x_i\rangle \big)/H$ satisfies universal property of the abelianization of $G$. The desired universal property of $\big(\bigoplus_{i\in \Lambda} \Q\langle x_i\rangle \big)/W$ is verified in the same way. $\quad\square$