3
$\begingroup$

Question:

In the final result of the calculation, how do we get $$\pi_1(T)\cong (\mathbb{Z}*\mathbb{Z})/N \cong N \ $$ where $ \ N = \langle a, b : aba^{-1}b^{-1}=e \rangle $?

In particular, why is $ (\mathbb{Z}*\mathbb{Z})/N \cong N \ $?


Using the square representation of the torus $T$ (with identified opposite edges), one can use the cover of $T$ consisting of an open square, $U$ say, and a closed square with a closed disc removed from it, $V$ say.

This gives the free group

$$\pi_1(U)*\pi_1(V)= \{e\}*\pi_1(V)\cong\{e\}*(\mathbb{Z}*\mathbb{Z}) \cong \mathbb{Z}*\mathbb{Z}$$

where the associated homomorphism

$$ \ \phi : \{e\}* (\mathbb{Z}*\mathbb{Z}) \longrightarrow \pi_1(T)$$

has as kernel all those words $w \in \mathbb{Z}*\mathbb{Z} $ generated by the word $aba^{-1}b^{-1} \in \mathbb{Z}*\mathbb{Z}$, and this kernel is a normal subgroup denoted by $N = \langle a, b : aba^{-1}b^{-1}=e \rangle$. (I'm not certain about this.)

So, if $e$ is the identity in $ \pi_1(T)$, then $\ e=\phi(w)=aba^{-1}b^{-1} $ (since $\phi$ is just an embedding?).

Now, if my understanding is correct, then van Kampen says

$$\pi_1(T)\cong (\mathbb{Z}*\mathbb{Z})/N$$

where $(\mathbb{Z}*\mathbb{Z})/N$ is a quotient group?

However, from several different sources I get that in fact

$$\pi_1(T)\cong N = \langle a, b : aba^{-1}b^{-1}=e \rangle . $$

Thanks for any help making sense of this.

(Also, when we say "$N$ is generated by the word $aba^{-1}b^{-1}$", we include the conjugation operation - what does conjugation mean exactly in this case?)

$\endgroup$
  • $\begingroup$ Could you clarify what you're confused about? At the moment, I don't know what you would like to have explained. $\endgroup$ – Noah Caplinger Nov 24 '19 at 21:08
  • $\begingroup$ @NoahCaplinger Yes, of course. In the final result of the calculation, how do we get $$\pi_1(T)\cong (\mathbb{Z}*\mathbb{Z})/N \cong N \ $$ where $ \ N = \langle a, b : aba^{-1}b^{-1}=e \rangle $? $\endgroup$ – Stephen Nov 24 '19 at 21:22
  • $\begingroup$ @NoahCaplinger In particular, why is $ (\mathbb{Z}*\mathbb{Z})/N \cong N \ $? $\endgroup$ – Stephen Nov 24 '19 at 21:24
5
$\begingroup$

Now, if my understand is correct, then Van Kampen says $$\pi_1(T) \cong (\mathbb{Z} \ast \mathbb{Z})/N$$

Your understanding is correct, however, the kernel of $\phi$ is not $\langle a,b : aba^{-1}b^{-1}\rangle$. Rather, the kernel is all products of conjugates of $aba^{-1}b^{-1}$, or the "normal closure" of the subgroup generated by $\{aba^{-1}b^{-1}\}$. This is what

when we say "$N$ is generated by the word $aba^{-1}b^{-1}$", we include the conjugation operation

means. Since this may be the point of confusion, I'll expand on it more:

You can think of taking the normal closure of a subgroup as a process: start with the subgroup, and add the necessary elements so that it becomes normal. If $aba^{-1}b^{-1}$ is in the subgroup, and $g$ is some other element of that group, the normal closure must also include $gaba^{-1}b^{-1}g^{-1}$ (ie $aba^{-1}b^{-1}$ conjugated by $g$) if we want it to be normal. The normal closure is basically what you get when you keep on doing this.

Hopefully that clears up what is meant by "including the conjugation operation."

With this view, we can see that $N$ is just all products of conjugates of $aba^{-1}b^{-1}$. (It may take a moment to digest that statement)

Then, when we take the quotient, we can see that $(\mathbb{Z} \ast \mathbb{Z})/N$ is the free group of rank 2 with the added relation $aba^{-1}b^{-1}$. That is,

$$\pi_1(T) \cong (\mathbb{Z} \ast \mathbb{Z})/N \cong \langle a,b : aba^{-1}b^{-1}\rangle \cong \mathbb{Z}^2$$

Edit: in response to your comments: it is not true that $(\mathbb{Z} \ast \mathbb{Z})/N \cong N$.

Second edit: As for actually computing the quotient...

First, think about the coset $aba^{-1}b^{-1}N = aNbNa^{-1}Nb^{-1}N$. This must be equal to $N$, as $aba^{-1}b^{-1} \in N$. Then, $aNbNa^{-1}Nb^{-1}N = N$, so $aNbN = bNaN$, ie the quotient is abelian. So, we add the relation $aNbN = bNaN$. Now, are there any other relations? no:

If $wN = N$, then $w \in N$, so $w$ is a product of conjugates of $aba^{-1}b^{-1}$. Then, $wN$ is a product of conjugates of $aNbNa^{-1}Nb^{-1}N$, which is just the identity (in light of the previous paragraph). Then $aNbN = bNaN$ is the only relation, and we get the presentation $$ \langle a,b : aba^{-1}b^{-1}\rangle$$

$\endgroup$
  • $\begingroup$ Thanks a lot. I still need to work out how to get the quotient from the normal closure , as that is not clear to me. Do we easily obtain it by calculating the cosets of $N$? $\endgroup$ – Stephen Nov 24 '19 at 21:51
  • 1
    $\begingroup$ I've added another edit explicitly calculating the quotient. Taking quotients is really important in algebraic topology, and I'd recommend drilling it hard if you're at all iffy on it. $\endgroup$ – Noah Caplinger Nov 24 '19 at 22:28
  • $\begingroup$ Thanks again. Indeed, quotients seem important here. The strange thing is my lecturer simply wrote down every single quotient without a single hint as to how to find it. From your working, I understand how to find this quotient. What bothers me though is, it almost felt like I had to know in advance that it was abelian, i.e. I had to know what the quotient was before trying to find it! I suppose with practice it will become easier to identify them. Thanks again for the help. $\endgroup$ – Stephen Nov 24 '19 at 22:39
  • $\begingroup$ Actually, in this case, it's not that hard come to think of it. Indeed, from $aba^{-1}b^{-1}=e \implies ab=ba$, one can easily guess that the quotient must be abelian. $\endgroup$ – Stephen Nov 24 '19 at 22:42
  • 1
    $\begingroup$ It does come much easier with practice. In cases similar to these, it’s even easier—notice that the relations of the quotient are exactly the generators of $N$. See if you can make/prove a conjecture about what happens more generally. $\endgroup$ – Noah Caplinger Nov 25 '19 at 2:39
1
$\begingroup$

In this particular case we can actually use universal properties to establish the isomorphisms we need, namely the universal properties of the pushout and of the abelianization. It's more direct to just compute the normal closure of $\langle aba^{-1}b^{-1}\rangle$ in $\mathbb{Z}\ast\mathbb{Z}$, but if you're familiar the some category theory concepts this might be interesting to you.

First I want to make my notation precise. Let $x_0$ be the basepoint of $T$, let $D$ be a small open disk containing $x_0$, and let $p$ be any other point in $D$. Then $T$ decomposes as a union of open sets $(T\setminus p) \cup D$, and the intersection $S=(T\setminus p) \cap D = D\setminus p$ is path-connected and contains the basepoint $x_0$. Notice that $T\setminus p \simeq S^1\vee S^1$, $S\simeq S^1$, $D\simeq \ast$, and the homomorphism

$$\varphi\colon\mathbb{Z}\cong\pi_1(S) \to \pi_1(T\setminus p)\cong \mathbb{Z}\ast\mathbb{Z}$$

is given by $1\mapsto aba^{-1}b^{-1}$, where $a$ and $b$ are appropriate free generators (this is seen by expressing $T$ as a quotient space of a square in the usual way).

Pushout: The Seifert-van Kampen theorem states that $\pi_1(T)$ is isomorphic to $P:=\pi_1(D)\ast_{\pi_1(S)}\pi_1(T\setminus p)$ which is defined as the pushout

$$\require{AMScd} \begin{CD} \pi_1(S) @>{\varphi}>> \pi_1(T\setminus p)\\ @V0VV @V\rho VV \\ \pi_1(D) @>0>> P \end{CD}$$

Since $\pi_1(D) = 0$, the usual universal property is simplified: given a group $G$ and a homomorphism $f\colon \pi_1(T\setminus p) \to G$ such that $f\circ \varphi = 0$, there exists a unique homomorphism $f'\colon P \to G$ such that $f = f'\circ \rho$. (As Noah Caplinger points out in his answer, we can express $P$ as $(\mathbb{Z}\ast\mathbb{Z})/N$ where $N$ is the normal closure of $im(\varphi)$.)

Abelianization: Now consider the abelianization map $$\alpha\colon \pi_1(T\setminus p)\cong \mathbb{Z}\ast\mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$$

whose kernel is the commutator subgroup $C = \{g_1g_2g_1^{-1}g_2^{-1}\ |\ g_1, g_2 \in \mathbb{Z}\ast\mathbb{Z} \} \supset \langle aba^{-1}b^{-1}\rangle$. The universal property for this map is as follows: given any homomorphism $g\colon\mathbb{Z}\ast\mathbb{Z} \to A$ where $C\subset ker(g)$, there is a unique homomorphism $g'\colon\mathbb{Z}\oplus\mathbb{Z} \to A$ such that $g = g'\circ \alpha$.


(For the rest of the argument you should draw yourself lots of diagrams so it's clear where all these functions are going, it's hard doing good commutative diagrams on stackexchange.)

Now we can use these properties to obtain natural homomorphisms $h_1\colon P \to \mathbb{Z}\oplus\mathbb{Z}$ and $h_2\colon \mathbb{Z}\oplus\mathbb{Z} \to P$, which will turn out to be isomorphisms. First, $\alpha\circ \varphi = 0$ since $im(\varphi) \subset C$, so by the universal property for $P$ there is a unique $h_1$ such that $\alpha = h_1\circ \rho$. Next, since $\rho(aba^{-1}b^{-1}) = 0$ it follows that $C\subset ker(\rho)$, so by the universal property for $\mathbb{Z}\oplus \mathbb{Z}$ there is a unique $h_2$ with the property that $\rho = h_2 \circ \alpha$. Combine these two equations to get $\alpha = h_1\circ h_2 \circ \alpha$ and $\rho = h_2 \circ h_1 \circ \rho$.

Finally we use the universal properties again and use the "uniqueness" part to show that $h_1\circ h_2 = id_{\mathbb{Z}\oplus \mathbb{Z}}$ and $h_2\circ h_1 = id_P$, and therefore they are mutually inverse isomorphisms. Notice that $h_2\circ \alpha$ is a homomorphism from $\pi_1(T\setminus p)$ to $P$ whose kernel contains $im(\varphi)$, so again there is a unique homomorphism $i\colon P\to P$ such that $h_2\circ \alpha = i\circ \rho$. Recalling from above that $\alpha = h_1\circ\rho$, our equation becomes $h_2\circ h_1 \circ \rho = i \circ \rho$. But we already know two values of $i$ which satisfy this equation: obviously $h_2\circ h_1$, and also $id_P$ as we saw in the previous paragraph. Therefore by the "uniqueness" part of the universal property for $P$ it holds that $h_2\circ h_1 = id_P$. The argument for $h_1\circ h_2$ is entirely analogous, using the universal property for the abelianization (make sure you work out the details yourself). Therefore $$ \pi_1(T) \stackrel{S-vK}{\cong} P \stackrel{h_1}{\cong} \mathbb{Z}\oplus \mathbb{Z}$$

$\endgroup$

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.