Presentations for Closed Orientable Surfaces

From a Polygon to a Fundamental Group

A closed orientable surface can be described by gluing the edges of a polygon in pairs. The boundary word records which edges are glued and in which direction. Once the edges are identified, the polygon becomes a two-dimensional cell attached to a graph. The fundamental group is then obtained by taking the free group of that graph and imposing the relation represented by the polygon boundary.

We use the standard polygon model of a closed orientable surface of genus $g$, where $g$ is a nonnegative integer. The classification theorem for compact surfaces says that every connected closed orientable surface of genus $g$ is homeomorphic to this model; the classification theorem itself is a standard result whose proof is beyond this tutorial. For $g\geq 1$, the model is a $4g$-gon with oriented boundary word $a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}$. Each label occurs twice, once with each orientation. We use the convention $[a,b]=aba^{-1}b^{-1}$.

Definition: The commutator of group elements $a,b$ is $[a,b]=aba^{-1}b^{-1}$. The normal closure of a word $w$ in a group $G$, written $\langle\!\langle w\rangle\!\rangle$, is the smallest normal subgroup of $G$ containing $w$. A presentation $\langle a_1,\ldots,a_n\mid w\rangle$ means the quotient of the free group on the generators $a_1,\ldots,a_n$ by the normal closure of the relator $w$.

The edge identifications in this polygon identify all its vertices to one vertex. Each pair of equally labelled edges becomes one loop edge based at that vertex, so the edges form a wedge of $2g$ circles. The polygon interior becomes one open two-cell. Its attaching map follows the boundary in order, traversing the corresponding loop positively for a label and negatively for an inverse label.

The Attaching Relation

We will use two standard results established in the earlier material on the Seifert–van Kampen theorem. First, the fundamental group of a wedge of $n$ circles, based at the common vertex, is the free group on the $n$ oriented loops. Second, if a disk is attached to a path-connected space $X$ along a map $f:S^1\to X$, the resulting fundamental group is $\pi_1(X)$ modulo the normal closure of the element represented by the loop $f$. This is the attaching-a-disk theorem: attaching the disk makes its boundary loop null-homotopic, and the theorem says that this adds precisely that relation and all its group consequences.

Theorem (Closed orientable surface presentation): Let $\Sigma_g$ be a connected closed orientable surface of genus $g$, with a chosen basepoint at the vertex of its standard polygon model. For $g\geq 1$, $$ \pi_1(\Sigma_g)\cong \left\langle a_1,b_1,\ldots,a_g,b_g\ \middle|\ [a_1,b_1][a_2,b_2]\cdots[a_g,b_g]=1\right\rangle. $$ For $g=0$, the group is trivial; equivalently, the presentation has no generators and no nontrivial relators.

Proof. Assume first that $g\geq 1$. The one-dimensional part of the polygon model is a wedge of $2g$ circles. By the wedge-of-circles result, its fundamental group is the free group on the loops $a_1,b_1,\ldots,a_g,b_g$. The polygon is one disk attached along the boundary word $a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}$. The attaching-a-disk theorem therefore says to quotient that free group by the normal closure of this word. By the commutator definition, the word is exactly $[a_1,b_1]\cdots[a_g,b_g]$, giving the displayed presentation.

For $g=0$, the surface is the sphere, whose polygon model has one vertex and one two-cell, with no one-cells. Its one-dimensional part is a point, whose fundamental group is the trivial group. Attaching a disk to a point introduces no generators, so the resulting group remains trivial. This is the presentation with no generators and no nontrivial relators. $\square$

The word in the presentation is one relation, not $g$ separate relations: the attaching map goes once around the entire polygon boundary. Each handle contributes a consecutive commutator, and the single face makes their product equal to the identity.

Reading and Checking the Polygon Word

Worked Example: Genus Zero

The genus-zero surface is the sphere. In its one-vertex, one-face cell structure there are no edge loops. Thus the free group on the one-dimensional part has no generators and consists only of the identity. Attaching the face cannot add a generator, so $\pi_1(S^2)$ is trivial.

Check: The general word has no commutator factors when $g=0$. The product over no factors is the identity, so it imposes no additional relation on the trivial group. This agrees with the cell description.

Worked Example: Genus One

For $g=1$, the polygon is a square with boundary word $aba^{-1}b^{-1}$. Its two pairs of edges give the two loop generators $a$ and $b$. The attaching disk gives the presentation $\langle a,b\mid aba^{-1}b^{-1}=1\rangle$.

We can rewrite its relation as a commuting relation, keeping track of the group operations. Starting with $aba^{-1}b^{-1}=1$, multiply on the right by $b$: $aba^{-1}b^{-1}b=1b$, hence $aba^{-1}=b$. Now multiply this equation on the right by $a$: $aba^{-1}a=ba$, hence $ab=ba$. Conversely, if $ab=ba$, then $aba^{-1}b^{-1}=baa^{-1}b^{-1}=bb^{-1}=1$. So the relator is equivalent to $ab=ba$.

Check: The boundary word contains $a$, $b$, $a^{-1}$, and $b^{-1}$, in that order, exactly as the commutator convention requires. The resulting relation makes the two generators commute.

Worked Example: Genus Two

For $g=2$, the octagon boundary reads $a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}$. The first four traversals form $[a_1,b_1]$, and the next four form $[a_2,b_2]$. The single face therefore gives $$ \pi_1(\Sigma_2)\cong \left\langle a_1,b_1,a_2,b_2\ \middle|\ a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}=1 \right\rangle. $$

This is one relator involving all four generators. It does not say that either commutator is separately the identity: the condition is that their product is the identity. The attaching-a-disk theorem imposes the normal closure of the full boundary word, which is exactly what the displayed presentation records.

Check: There are $4g=8$ polygon sides, and the written word has eight edge traversals. Each generator $a_1,b_1,a_2,b_2$ occurs once with exponent $+1$ and once with exponent $-1$, as required by the paired edge identifications.

Worked Example: Genus Three

For $g=3$, the twelve oriented sides read in order as $a_1,b_1,a_1^{-1},b_1^{-1},a_2,b_2,a_2^{-1},b_2^{-1},a_3,b_3,a_3^{-1},b_3^{-1}$. Grouping each consecutive block of four gives the full relator $$ [a_1,b_1][a_2,b_2][a_3,b_3] =a_1b_1a_1^{-1}b_1^{-1} a_2b_2a_2^{-1}b_2^{-1} a_3b_3a_3^{-1}b_3^{-1}. $$

There are six generators, one for each pair of sides, and one relation, represented by the whole boundary. Thus $\pi_1(\Sigma_3)\cong\langle a_1,b_1,a_2,b_2,a_3,b_3\mid [a_1,b_1][a_2,b_2][a_3,b_3]=1\rangle$.

Check: The displayed expanded word has twelve factors, matching the twelve sides. Each of the six generator labels appears twice, with opposite exponents, and the three four-letter blocks are precisely the three commutators.

Why the Product and the Convention Matter

The order and signs in the boundary word encode the way the paired sides are glued. With our convention, a positively traversed edge contributes its generator, while traversing the same edge in the opposite direction contributes its inverse. The resulting product is the attaching loop itself. Changing the starting point on the boundary cyclically rearranges the word; reversing the boundary direction gives the inverse word. Neither change alters the normal subgroup imposed by the attaching disk: cyclic permutations of a word are conjugate in the free group, and a word and its inverse have the same normal closure.

Proof of the last assertion. If $w=uv$, then the cyclic permutation $vu=u^{-1}wu$ is a conjugate of $w$, so it belongs to the normal closure of $w$. Applying the same argument to the reverse cyclic permutation shows that $w$ belongs to the normal closure of $vu$, so their normal closures agree. Also $w^{-1}$ belongs to the normal closure of $w$, and $w=(w^{-1})^{-1}$ belongs to the normal closure of $w^{-1}$. Thus the two normal closures agree. $\square$

Common pitfall: an invalid algebraic simplification. From $aba^{-1}b^{-1}=1$, right-multiplying by $b$ correctly gives $aba^{-1}=b$. But left-multiplying this equation by $a^{-1}$ gives $ba^{-1}=a^{-1}b$, not $ab=ba$. To obtain the commuting relation, multiply $aba^{-1}=b$ on the right by $a$, giving $ab=ba$. In a noncommutative group, the side on which an equation is multiplied matters.

The presentation is useful because it translates the surface’s global gluing into a finite algebraic description. In genus one the single commutator relation says the two generators commute. In higher genus the relation couples the handle generators through the product of their commutators; replacing it by separate commuting relations would describe a different group.

Check Your Understanding

Use the polygon boundary and the attaching-a-disk theorem to answer each question.

Write the full boundary word and fundamental group presentation for genus four.
How many generators and how many face relations occur in the standard polygon presentation for genus $g\geq 1$?
For genus one, show directly that $aba^{-1}b^{-1}=1$ follows from $ab=ba$.
Explain why the genus-two attaching relation is a relation on the product $[a_1,b_1][a_2,b_2]$, rather than two separate commutator relations.
In the genus-three polygon word, how many times does each generator occur, and with which exponents?