0
$\begingroup$

The torus $T$ has a CW decomposition given by one $1$-cell $p$, two $2$ cells $a,b$, and one $2$-cell $D$ being attached to the $1$-skeleton by the word $aba^{-1}b^{-1}$. I'm struggling to compute the cellular boundary homomorphism $d_2\colon\mathbb{Z}\{D\}\to\mathbb{Z}\{a,b\}$. I read elsewhere that $d_2$ is the abelianization of the word, so $d_2=a+b-a-b=0$. Could someone explain why this is true? I know the cellular boundary formula.

$\endgroup$
1
  • $\begingroup$ I think you mean a zero-cell $p$ and then two one-cells $a$ and $b$ (and finally one two-cell $D$). $\endgroup$
    – quarague
    Jun 9, 2023 at 8:20

1 Answer 1

2
$\begingroup$

Let's use the cellular boundary formula to compute $d_2(D)$. Let $\varphi : S^1 \to T^1$ (where $T^1$ is the $1$-skeleton of $T$) be the attaching map of the $2$-cell $D$. Note that $T^1$ is just a wedge sum of the two circles corresponding to $a, b$, and $\varphi$ is just the loop given by $aba^{-1}b^{-1}$. To find the coefficient of $a$ in $d_2(D)$, we want to compute the degree of the map $$ S^1 \xrightarrow{\varphi} T^1 \xrightarrow{q} S^1, $$ where $q$ is the quotient map that collapses the circle corresponding to $b$ to a single point. Note that $q\varphi$ is just the loop given by $aca^{-1}c^{-1}$, where $c$ is a constant path. In particular, $q\varphi$ is nullhomotopic, so its degree is $0$. By the cellular boundary formula, it follows that the coefficient of $a$ in $d_2(D)$ is $0$. Similarly, the coefficient of $b$ in $d_2(D)$ is also $0$, which proves $d_2 = 0$.


Edit: Here is an explanation of how $d_2(D)$ can be interpreted as the "abelianization" of the word $aba^{-1}b^{-1}$. Here's how to see it intuitively. You can view the CW structure of $T$ as a square whose opposite edges are identified:

enter image description here

You can see that the "boundary" of the $2$-cell is the word $aba^{-1}b^{-1}$. Since the cellular boundary map $d_2$ maps into the free abelian group $\mathbb{Z}\{a, b\}$, the boundary $d_2(D)$ is the abelianization $a + b - a - b$ of this word. That is, $d_2(D)$ is just the boundary of $D$ in the obvious sense, and everything is abelian since that's how we've defined things. Though this is just intuition, this is really how you should be thinking about the cellular boundary map.

A way to rigorously justify this (aside from what is above) is to define the cellular chain groups $C_n(T)$ as the relative homology groups $H_n(T^n, T^{n - 1})$ (again, $T^n$ is the $n$-skeleton), and each boundary map $d_n : H_n(T^n, T^{n - 1}) \to H_{n - 1}(T^{n - 1}, T^{n - 2})$ as the boundary map from the long exact sequence of the triple $(T^n, T^{n - 1}, T^{n - 2})$. This is how Hatcher defines cellular homology. In our case, we have $C_2(T) = H_2(T^2, T^1) \cong \mathbb{Z}$, which is generated by a $2$-chain whose image is $D$. If you work through the proof of the snake lemma, you'll find that the boundary map $d_2 : H_2(T^2, T^1) \to H_1(T^1, T^0)$ just takes the boundary of this $2$-chain, which is the $1$-chain $a + b - a - b$ (or at least homologous to it).

$\endgroup$
3
  • $\begingroup$ Thanks, this does help a lot. Though, I still am curious how this is related to "abelianization" of the word. $\endgroup$ Jun 9, 2023 at 6:44
  • $\begingroup$ @Rough_Manifolds I edited my answer with an explanation. $\endgroup$
    – Frank
    Jun 9, 2023 at 7:12
  • $\begingroup$ This is wonderful, thanks a ton! $\endgroup$ Jun 10, 2023 at 4:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .