MAS61015 Algebraic Topology 10 Homology 12 Abelian groups

11. Homology of the punctured plane

Later we will prove that for all $n\geq 2$ we have

H_{k}({\mathbb{R}}^{n}\setminus\{0\})=H_{k}(S^{n-1})=\begin{cases}{\mathbb{Z}}% &\text{ if }k=0\text{ or }k=n-1\\ 0&\text{ otherwise. }\end{cases}

For this we will need the Mayer-Vietoris sequence, which is a very important and useful tool, but it will take some work to set that up. In this section, we outline a different approach which is more direct and elementary but which works only for $n=2$ . We will identify ${\mathbb{R}}^{2}$ with ${\mathbb{C}}$ and write ${\mathbb{C}^{\times}}={\mathbb{C}}\setminus\{0\}$ , so our main task will be to prove that $H_{1}({\mathbb{C}^{\times}})={\mathbb{Z}}$ .

Video (Definition 11.2 to Theorem 11.11)

Definition 11.1.

Let $z\in{\mathbb{C}^{\times}}$ be a nonzero complex number. This can be expressed as $z=r\,e^{i\theta}$ for a unique pair or real numbers $r,\theta$ with $r>0$ and $-\pi<\theta\leq\pi$ . We put $\operatorname{plog}(z)=\log(r)+i\theta$ , and call this the principal logarithm of $z$ .

Note that $\operatorname{plog}\colon{\mathbb{C}^{\times}}\to{\mathbb{C}}$ and $\exp(\operatorname{plog}(z))=z$ for all $z$ , but $\operatorname{plog}$ is not continuous (because for small $\epsilon>0$ we have $\operatorname{plog}(-1+i\epsilon)\approx i\pi$ but $\operatorname{plog}(-1-i\epsilon)\approx-i\pi$ ). This cannot be fixed by adjusting the definitions: there is no continuous map $f\colon{\mathbb{C}^{\times}}\to{\mathbb{C}}$ with $\exp(f(z))=z$ for all $z$ . To work around this we make the following definition:

Definition 11.2.

For any $z\in{\mathbb{C}^{\times}}$ we put

\operatorname{LOG}(z)=\{\widetilde{z}\in{\mathbb{C}}\;|\;\exp(\widetilde{z})=z% \}=\operatorname{plog}(z)+2\pi i{\mathbb{Z}}.

Any element of $\operatorname{LOG}(z)$ will be called a logarithm of $z$ . More generally, suppose we have a topological space $T$ and a continuous map $u\colon T\to{\mathbb{C}^{\times}}$ . By a continuous logarithm of $u$ we mean a continuous map $\widetilde{u}\colon T\to{\mathbb{C}}$ with $\exp\circ\widetilde{u}=u$ , or equivalently $\widetilde{u}(t)\in\operatorname{LOG}(u(t))$ for all $t$ .

Note that the cosets $i\pi+2\pi i{\mathbb{Z}}$ and $-i\pi+2\pi i{\mathbb{Z}}$ are the same, and that $\operatorname{LOG}(-1+i\epsilon)$ is close to this coset for all small $\epsilon$ , independent of whether $\epsilon$ is positive or negative. Thus, $\operatorname{LOG}(z)$ depends continuously on $z$ even though $\operatorname{plog}(z)$ does not.

Given a continuous map $u\colon T\to{\mathbb{C}^{\times}}$ , we could attempt to define a continuous logarithm of $u$ by $\widetilde{u}=\operatorname{plog}\circ u$ . This works provided that the image $u(T)$ does not touch the negative real axis where $\operatorname{plog}$ is discontinuous. If $u(T)$ does touch the negative real axis then it may be possible to find a continuous logarithm by a different method, but in some cases, no continuous logarithm exists.

Lemma 11.3.

Let $u\colon[0,1]\to{\mathbb{C}^{\times}}$ be continuous, and suppose that $x\in\operatorname{LOG}(u(0))$ . Then there is a unique continuous logarithm $\widetilde{u}\colon[0,1]\to{\mathbb{C}}$ with $\widetilde{u}(0)=x$ .

We will prove this properly later when we come to discuss covering maps.

Sketch proof.

If we choose $N$ sufficently large, then when $|s-t|\leq 1/N$ the points $u(s)/u(t)$ will be close to $1$ in ${\mathbb{C}^{\times}}$ and so will be far from the negative real axis where $\operatorname{plog}$ is discontinuous. We can thus define

\widetilde{u}(t)=x+\sum_{k=1}^{N}\operatorname{plog}\left(u\left(\frac{kt}{N}% \right)/u\left(\frac{(k-1)t}{N}\right)\right).

This is a continuous function of $t$ . When $t=0$ we see that all the terms in the sum are $\operatorname{plog}(u(0)/u(0))=\operatorname{plog}(1)=0$ , so $\widetilde{u}(0)=x$ . In general we have

\exp(\widetilde{u}(t))=\exp(x).\prod_{i=1}^{N}\frac{u(kt/N)}{u((k-1)t/N)}=u(0)% .\prod_{i=1}^{N}\frac{u(kt/N)}{u((k-1)t/N)},

and most of the terms in the product cancel out leaving only $\exp(\widetilde{u}(t))=u(t)$ . ∎

Corollary 11.4.

Let $K\subseteq{\mathbb{R}}^{N}$ be convex, and suppose that $k\in K$ . Let $u\colon K\to{\mathbb{C}^{\times}}$ be continuous, and suppose that $x\in\operatorname{LOG}(u(k))$ . Then there is a unique continuous logarithm $\widetilde{u}\colon K\to{\mathbb{C}}$ with $\widetilde{u}(k)=x$ . (In particular, this applies when $K=\Delta_{d}$ for some $d$ .)

Sketch proof.

For $m\in K$ we can define $v_{m}\colon[0,1]\to{\mathbb{C}^{\times}}$ by $v_{m}(t)=u(tm+(1-t)k)$ . By the lemma, there is a unique continuous logarithm $\widetilde{v}_{m}\colon[0,1]\to{\mathbb{C}}$ with $\widetilde{v}_{m}(0)=x$ . We define $\widetilde{u}(m)=\widetilde{v}_{m}(1)$ , so $\exp(\widetilde{u}(m))=\exp(\widetilde{v}_{m}(1))=v_{m}(1)=u(m)$ . We have $v_{k}(t)=u(k)$ for all $t$ so $\widetilde{v}_{k}$ must be the constant path at $x$ so $\widetilde{u}(k)=\widetilde{v}_{k}(1)=x$ . With some work one can check that $\widetilde{u}$ is continuous. ∎

Definition 11.5.

Given any path $u\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ we define

\omega(u)=(\widetilde{u}(e_{1})-\widetilde{u}(e_{0}))/(2\pi i)\in{\mathbb{C}},

where $\widetilde{u}$ is any continuous logarithm of $u$ . (This is well-defined, because any two continuous logarithms differ by a constant of the form $2n\pi i$ , and the constant cancels out when we calculate $\widetilde{u}(e_{1})-\widetilde{u}(e_{0})$ .) We then extend this linearly to get a homomorphism $\omega\colon C_{1}({\mathbb{C}^{\times}})\to{\mathbb{C}}$ , given by

\omega(n_{1}u_{1}+\dotsb+n_{r}u_{r})=n_{1}\omega(u_{1})+\dotsb+n_{r}\omega(u_{% r}).

Example 11.6.

The standard loop $u_{n}\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ of winding number $n$ is given by $u_{n}(1-t,t)=\exp(2\pi int)$ . The obvious continuous logarithm is $\widetilde{u}_{n}(t)=2\pi int$ , and using this we get $\omega(u_{n})=n$ .

Lemma 11.7.

For $u\colon\Delta_{2}\to{\mathbb{C}^{\times}}$ we have $\omega(\partial(u))=0$ . Thus, we have $\omega(B_{1}({\mathbb{C}^{\times}}))=0$ .

Proof.

For $i=0,1,2$ we put $v_{i}=u\circ\delta_{i}\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ , so $\partial(u)=v_{0}-v_{1}+v_{2}$ . By Corollary 11.4, we can choose a continuous logarithm $\widetilde{u}\colon\Delta_{2}\to{\mathbb{C}}$ for $u$ . We then note that the map $\widetilde{v}_{i}=\widetilde{u}\circ\delta_{i}\colon\Delta_{1}\to{\mathbb{C}}$ is a continuous logarithm for for $v_{i}$ , so $\omega(v_{i})=(\widetilde{v}_{i}(e_{1})-\widetilde{v}_{i}(e_{0}))/(2\pi i)$ . This gives

2\pi i\omega(\partial(u))=(\widetilde{v}_{0}(e_{1})-\widetilde{v}_{0}(e_{0}))-% (\widetilde{v}_{1}(e_{1})-\widetilde{v}_{1}(e_{0}))+(\widetilde{v}_{2}(e_{1})-% \widetilde{v}_{2}(e_{0})).

However, we have

	$\displaystyle\delta_{0}(e_{1})$	$\displaystyle=e_{2}$	$\displaystyle\delta_{1}(e_{1})$	$\displaystyle=e_{2}$	$\displaystyle\delta_{2}(e_{1})$	$\displaystyle=e_{1}$
	$\displaystyle\delta_{0}(e_{0})$	$\displaystyle=e_{1}$	$\displaystyle\delta_{1}(e_{0})$	$\displaystyle=e_{0}$	$\displaystyle\delta_{2}(e_{0})$	$\displaystyle=e_{0},$

so the above expression becomes

2\pi i\omega(\partial(u))=(\widetilde{u}(e_{2})-\widetilde{u}(e_{1}))-(% \widetilde{u}(e_{2})-\widetilde{u}(e_{0}))+(\widetilde{u}(e_{1})-\widetilde{u}% (e_{0}))=0,

so $\omega(\partial(u))=0$ as claimed. More generally, if $u\in C_{2}({\mathbb{C}^{\times}})$ then $u=n_{1}u_{1}+\dotsb+n_{r}u_{r}$ for some integers $n_{i}$ and maps $u_{i}\colon\Delta_{2}\to{\mathbb{C}^{\times}}$ , and this gives

\omega(\partial(u))=\sum_{i}n_{i}\omega(\partial(u_{i}))=\sum_{i}n_{i}.0=0

as before. Thus, if $w\in B_{1}({\mathbb{C}^{\times}})$ then $w=\partial(u)$ for some $u\in C_{2}({\mathbb{C}^{\times}})$ giving $\omega(w)=\omega(\partial(u))=0$ as claimed. ∎

Definition 11.8.

We now define homomorphisms $\beta\colon{\mathbb{C}}\to{\mathbb{C}^{\times}}$ and $\gamma\colon C_{0}({\mathbb{C}^{\times}})\to{\mathbb{C}^{\times}}$ by $\beta(z)=\exp(2\pi iz)$ and

\gamma(n_{1}z_{1}+\dotsb+n_{r}z_{r})=z_{1}^{n_{1}}z_{2}^{n_{2}}\dotsb z_{r}^{n% _{r}}.

(Here we regard ${\mathbb{C}}$ and $C_{0}({\mathbb{C}^{\times}})$ as groups under addition and ${\mathbb{C}^{\times}}$ as a group under multiplication, so to say that $\beta$ and $\gamma$ are homomorphisms means that $\beta(w+z)=\beta(w)\beta(z)$ and $\gamma(u+v)=\gamma(u)\gamma(v)$ ; it is easy to see that both of these identities are valid.)

Lemma 11.9.

The following square commutes (or in other words, $\beta(\omega(u))=\gamma(\partial(u))$ for all $u\in C_{1}({\mathbb{C}^{\times}})$ ).

Proof.

In general $u$ will be a ${\mathbb{Z}}$ -linear combination of paths in ${\mathbb{C}^{\times}}$ , but all the maps are homomorphisms, so it will be enough to consider the case where $u\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ is just a single path. We then have $\partial(u)=u(e_{1})-u(e_{0})\in C_{0}({\mathbb{C}^{\times}})$ and so $\gamma(\partial(u))=u(e_{1})/u(e_{0})\in{\mathbb{C}^{\times}}$ . Now choose a continuous logarithm $\widetilde{u}\colon\Delta_{1}\to{\mathbb{C}}$ for $u$ . By definition we have $\omega(u)=(\widetilde{u}(e_{1})-\widetilde{u}(e_{0}))/(2\pi i)$ , so

\beta(\omega(u))=\exp(\widetilde{u}(e_{1})-\widetilde{u}(e_{0}))=\exp(% \widetilde{u}(e_{1}))/\exp(\widetilde{u}(e_{0}))=u(e_{1})/u(e_{0})=\gamma(% \partial(u)).

∎

Corollary 11.10.

We have $\omega(Z_{1}({\mathbb{C}^{\times}}))={\mathbb{Z}}$ and $\omega(B_{1}({\mathbb{C}^{\times}}))=0$ , so $\omega$ induces a homomorphism $\overline{\omega}\colon H_{1}({\mathbb{C}^{\times}})\to{\mathbb{Z}}$ given by $\overline{\omega}(z+B_{1}({\mathbb{C}^{\times}}))=\omega(z)$ .

Proof.

Suppose that $z\in Z_{1}({\mathbb{C}^{\times}})$ , so $\partial(z)=0$ , so $\gamma(\partial(z))=\gamma(0)=1$ . By the Lemma we then have $\beta(\omega(z))=1$ , or in other words $\exp(2\pi i\omega(z))=1$ , so $\omega(z)\in{\mathbb{Z}}$ . As in Example 11.6 we also have standard loops $u_{n}\in Z_{1}({\mathbb{C}^{\times}})$ with $\omega(u_{n})=n$ , so the image $\omega(Z_{1}({\mathbb{C}^{\times}}))$ is the whole group ${\mathbb{Z}}$ . We saw in Lemma 11.7 that $\omega(B_{1}({\mathbb{C}^{\times}}))=0$ , and it follows that the rule $\overline{\omega}(z+B_{1}({\mathbb{C}^{\times}}))=\omega(z)$ gives a well-defined homomorphism from the quotient group $H_{1}({\mathbb{C}^{\times}})=Z_{1}({\mathbb{C}^{\times}})/B_{1}({\mathbb{C}^{% \times}})$ to ${\mathbb{Z}}$ . ∎

Theorem 11.11.

The homomorphism $\overline{\omega}\colon H_{1}({\mathbb{C}^{\times}})\to{\mathbb{Z}}$ is an isomorphism.

Proof.

We have already remarked that $\overline{\omega}(u_{n}+B_{1}({\mathbb{C}^{\times}}))=\omega(u_{n})=n$ for all $n\in{\mathbb{Z}}$ ; this shows that $\overline{\omega}$ is surjective. Now suppose we have $h\in H_{1}({\mathbb{C}^{\times}})$ with $\overline{\omega(h)}=0$ . By Proposition 10.29, we can find a loop $u\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ based at $1\in{\mathbb{C}^{\times}}$ with $h=[u]$ . By Lemma 11.3, there is a unique continuous logarithm $\widetilde{u}\colon\Delta_{1}\to{\mathbb{C}}$ with $\widetilde{u}(e_{0})=0$ . We then have $\omega(u)=(\widetilde{u}(e_{1})-\widetilde{u}(e_{0}))/(2\pi i)=\widetilde{u}(e% _{1})/(2\pi i)$ . However, we also know that $\omega(u)=\overline{\omega}([u])=\overline{\omega}(h)=0$ , so we must have $\widetilde{u}(e_{1})=0$ as well. We now define $\widetilde{v}\colon\Delta_{2}\to{\mathbb{C}}$ by

\widetilde{v}(t_{0},t_{1},t_{2})=\begin{cases}(1-t_{0})\widetilde{u}(t_{1}/(1-% t_{0}),t_{2}/(1-t_{0}))&\text{ if }t_{0}<1\\ 0&\text{ if }t_{0}=1.\end{cases}

We leave it to the reader to check that $\widetilde{v}$ is continuous even at $e_{0}$ . (A full proof of a more general fact will be given later.) We then define $v=\exp\circ\widetilde{v}\colon\Delta_{2}\to{\mathbb{C}^{\times}}$ . It is easy to see that $\widetilde{v}$ is a filling in for $\widetilde{u}$ , and thus that $v$ is a filling in for $u$ , so $[u]=0$ in $H_{1}(X)$ by Lemma 10.31, or in other words $h=0$ . This proves that $\overline{\omega}$ is also injective, and so is an isomorphism as claimed. ∎