I am asked to classify the compact surface obtained by pasting the edges of a polygonal region with the labeling scheme $abcdabdc$ and compute it's first homology group. I classify the space as the connected sum of 3 real projective planes using cutting and pasting. (A quick algebraic check: call our scheme $w$ and let $e = abd$. Then $w = abcb^{-1}a^{-1}e^2c$. Let $f = b^{-1}a^{-1}e^2c$, then $w = abe^{-2}abf^2$. Let $g = e^{-2}ab$, then $w = e^2g^2f^2$ so our space can be classified as the connected sum of 3 projective planes).

The first homology group of the $n$-fold connected sum of real projective planes is known to be $\mathbb{Z}^{n-1} \oplus \mathbb{Z}_2$ (see here: Homology of connected sum of real projective spaces) so I deduce $H_1(X) \cong \mathbb{Z}^2 \oplus \mathbb{Z}_2$.

Now consider $H_1(X)$ as the abelianization of the fundamental group $\pi_1(X) \cong \langle a, b, c, d | abcdabdc = 1\rangle$. Then $H_1(X) \cong \text{Ab}(\pi_1(X)) \cong \langle a, b, c, d |2(a+b+c+d)=0 \rangle$. This representation is justified here: Presentation of the abelianization of $G$. If we consider the generators of this presentation to be $\{a, b, c, a+b+c+d\}$ it seems clear that $H_1(X) \cong \mathbb{Z}^3 \oplus \mathbb{Z}_2$. (If you're not convinced that this manipulation on the generators is correct it is also shown here Presentation of abelian group that we can use the Smith Normal Form which gives the same result).

Clearly $\mathbb{Z}^2 \oplus \mathbb{Z}_2 \ne \mathbb{Z}^3 \oplus \mathbb{Z}_2$ so what went wrong?

  • $\begingroup$ Doesn't $acdadc$ give the same surface? $\endgroup$ – Angina Seng Aug 12 '18 at 19:29
  • $\begingroup$ @LordSharktheUnknown Good point, yes it does if you substitute $a = ab$ in which case the abelianization method above gives the correct result. Is it true then that you always have to make those types of substitutions before abelianizing in that way? $\endgroup$ – wyoumans Aug 12 '18 at 19:36
  • 1
    $\begingroup$ In your presentation, $a$ and $b$ do not represent loops in the surface. $\endgroup$ – Angina Seng Aug 12 '18 at 19:39
  • $\begingroup$ @LordSharktheUnknown Of course, if I had labeled the vertices on the polygon it would have been clear. The vertex where $a$ meets $b$ is distinct from the rest so it's not a loop. Thank you. If you make that an answer I'd be happy to accept. $\endgroup$ – wyoumans Aug 12 '18 at 19:45

If you identify your edges of an octagon in the pattern $abcdabdc$ then its eight vertices are identified into two points $P$ and $Q$ say, where $Q$ is the image of the vertices between the $a$ and $b$ edges and $P$ is the images of the other six vertices. Therefore the edges $a$ and $b$ are not mapped into loops in the surface, but $c$ and $d$ are mapped to loops based at $P$. If we replace the $ab$ pairs of edges with edges labelled $e$, say, then $e$ becomes a loop in the surface, and the fundamental group is $\langle c,d,e\mid ecdedc=1\rangle$, Abelianising to $\Bbb Z_2 \oplus\Bbb Z^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.