In this post we find the Picard group of the circle using cohomology, and more generally discuss a cohomological description of $\mathrm{Pic}(M)$ .

Motivation

Let me first and foremost make a confession. People of the world are cut into two categories: those who know topology, and those that don’t. I sit squarely in this latter group (not by choice–I wish I knew so much more). Consequently, a lot of ‘obvious facts’ for people more knowledgable about topology often confound me. For a long time, a perfect example of such an ‘obvious statement’ is that there are, up to equivalence, only two (smooth) line bundles on the circle. In this post, I’d like to work through how one could prove such a statement, in a way intuitive to someone with a background in algebraic geometry (read ‘sheaf cohomology’).

Intuitive preamble

The fact that the circle only has two line bundles intuitively follows from the following basic topological reasoning. Let $E\to S^1$ be such a line bundle. The circle can be covered with two open sets $U$ and $V$ , each isomorphic to $\mathbb{R}$ . Since $\mathbb{R}$ is contractible, we know then that $E\mid_U\to U$ must be (smoothly) trivial. Similarly for $E\mid_V\to V$ . Thus, the only question is how we glue the two trivial copies $E\mid_U\to U$ and $E\mid_V\to V$ together. Well, we essentially have only one option–for each of the components of $U\cap V$ , we can choose two perform some number of half twists of one of $E\mid_U$ or $E\mid_V$ there, and then glue them like that. It’s clear that at on each connected component of $U\cap V$ , performing two half-twists is the same thing as performing $0$ half twists. Thus, the number of half twists on each component only matter modulo $2$ .

Using similar intuition we may deduce that if $M$ is, say, a smooth $n$ -manifold, then we could cover $M$ with some copies of $\mathbb{R}^n$ , say $\{U_\alpha\}$ . Then, for any bundle $E\to M$ , we’d have trivializations $E_{U_\alpha}\to U_\alpha$ . We now need to choose how to glue these together on overlaps. So, on connected components of overlaps $U_\alpha\cap U_\beta$ , we intuitively only have two options: half twist one copy, or don’t. Thus, for each connected component of an overlap we have two choices, which I will conveniently denote by $\mathbb{Z}/2\mathbb{Z}$ . Thus, the number of raw choices on $U_\alpha\cap U_\beta$ is $(\mathbb{Z}/2\mathbb{Z})^{\pi_{\alpha,\beta}}$ , where $\pi_{\alpha,\beta}$ is the number of connected components of $U_\alpha\cap U_\beta$ (and similarly define $\pi_{\alpha,\beta,\gamma}$ for triple intersections).

But, if we’re to consistently glue these bundles together, we must make sure that they agree on triple overlaps. To say what this means, let us first define a map

$\displaystyle \varphi:\prod_{\stackrel{\alpha,\beta}{U_\alpha\cap U_\beta\ne\varnothing}}\left(\mathbb{Z}/2\mathbb{Z}\right)^{\pi_{\alpha,\beta}}\to\prod_{\stackrel{\alpha,\beta,\gamma}{U_\alpha\cap U_\beta\cap U_\gamma\ne\varnothing}} \left(\mathbb{Z}/2\mathbb{Z}\right)^{\pi_{\alpha,\beta,\gamma}}$

Let’s fix a connected component $C_j$ of $U_\alpha\cap U_\beta\cap U_\gamma$ , corresponding to the $j^{\text{th}}$ component of $(\mathbb{Z}/2\mathbb{Z})^{\pi_{\alpha,\beta,\gamma}}$ . Now, there are unique components $C_k\subseteq U_\alpha\cap U_\gamma$ and $C_\ell\subseteq U_\beta\cap U_\gamma$ such that $C_j\subseteq C_k$ and $C_j\subseteq C_\ell$ . The component map of $\varphi$ in the $j^{\text{th}}$ component of $(\mathbb{Z}/2\mathbb{Z})^{\pi_{\alpha,\beta,\gamma}}$ takes a tuple $(t_{\alpha,\beta})$ , where $t_{\alpha,\beta}\in (\mathbb{Z}/2\mathbb{Z})^{\pi_{\alpha,\beta}}$ , and sends it to $t_{\alpha,\gamma}^k-t_{\gamma,\beta}^\ell$ , where these are the $\ell^{\text{th}}$ and $k^{\text{th}}$ coordinates of the portion of the tuple $t_{\alpha,\gamma}$ and $t_{\gamma,\beta}$ respectively. It’s clear then that the admissible gluings are those then in the kernel of $\varphi$ . Indeed, the above just says that we’ve consistently chosen half twists on connected components–in the sense that they agree on overlaps.

That said, while this kernel is the set of admissible gluings, it double counts some. For example, think about the gluing on the circle obtained as follows. The opens $U$ and $V$ have two conneted components $C_1$ and $C_2$ in their intersection $U\cap V$ . It’s clear then that the tuple $(1,1)\in (\mathbb{Z}/2\mathbb{Z})^2$ is obviously in the kernel of the map $\varphi$ (in this case $\varphi$ is zero–there are no triple intersections!). But, this is the same as the twist $(0,0)$ , since the twisting of the bundles at each component undoes the other. More generally, a little more thought (along the same lines of this example) shows that precisely the elements of the kernel of $\varphi$ which define the same bundles are those in the image of the map

$\displaystyle \psi:\prod_\alpha \mathbb{Z}/2\mathbb{Z}\to \prod_{\alpha,\beta}(\mathbb{Z}/2\mathbb{Z})^{\pi_{\alpha,\beta}}$

defined by $\psi((s_\alpha))=(t_{\alpha,\beta})$ where $t_{\alpha,\beta}\in(\mathbb{Z}/2\mathbb{Z})^{\pi_{\alpha,\beta}}$ has the same entry in each coordinate: the number $s_\alpha-s_\beta$ .

Thus, the above seems to suggest that the line bundles on $M$ should be in correspondence with the elements of the group $\ker\varphi/\text{im }\psi$ . But, anyone who has dealt with sheaf cohomology at length will immediately recognize the above for what it is: it is the first Čech cohomology group of the constant sheaf $\underline{\mathbb{Z}/2\mathbb{Z}}$ with respect to the open cover $\{U_\alpha\}$ !

But, using a little bit of sheaf-theoretic know-how, we can rephrase this group in even nicer terms. Namely, for any locally contractible topological space $X$ , and any constant sheaf of abelian groups $\underline{A}$ , it’s common knowledge that $H^i(X,\underline{A})=H^i_\text{sing}(X,A)$ (this is in, say, Spanier). Thus, one application of this shows that the cover $\{U_\alpha\}$ (consisting of contractible spaces!) is a Leray cover for $\underline{A}$ . Thus, by Leray’s theorem

$\check{H}^1\left(\left\{U_\alpha\right\},\underline{\mathbb{Z}/2\mathbb{Z}}\right)=H^1(M,\underline{\mathbb{Z}/2\mathbb{Z}})$

and a second application of this theorem implies that this is $H^1(M,\underline{\mathbb{Z}/2\mathbb{Z}})=H^1_\text{sing}(M,\mathbb{Z}/2\mathbb{Z})$ . Thus, the above would lead us to believe that the line bundles on $M$ should be classified by $H^1_\text{sing}(M,\mathbb{Z}/2\mathbb{Z})$ .

That said, while the above is clear intuitively, it’s a bit hard to see how put all of this on firm footing. But, below, we put the well-oiled machinery that is sheaf cohomology to good use, to show that not only is it easy to prove this theorem rigorously, but it follows largely from abstract nonsense. That said, it’s good to have the above intuitive understanding of the result, and not just the slick, heavy machinery proof that follows.

The proof

So, the goal of this section is to prove formally what we intuited above:

Theorem(Main): Let $M$ be a smooth manifold. Then, $\mathrm{Pic}(M)\cong H^1_\text{sing}(M,\mathbb{Z}/2\mathbb{Z})$ .

Here $\mathrm{Pic}(M)$ denotes the group of smooth line bundles, up to isomorphism, with tensor product as the group operation. The first key to this proof is the following general fact:

Lemma: Let $M$ be a smooth manifold, then $\mathrm{Pic}(M)\cong H^1(M,\mathcal{O}_M^\times)$ .

Here $\mathcal{O}_M^\times$ denotes the sheaf of units of the sheaf $\mathcal{O}_M$ of smooth functions.

Proof(Sketch): Unlike the main theorem, this an almost purely formal consequence which goes under the header of ‘ $H^1$ classifies torsors’. Namely, for any ringed space $(X,\mathcal{O}_X)$ and any $\mathcal{O}_X$ -module $\mathcal{F}$ , the cohomology $H^1(X,\text{Aut}_{\mathcal{O}_X}(\mathcal{F}))$ classifies $\mathcal{O}_X$ -modules $\mathcal{G}$ which are locally on $X$ isomorphic to $\mathcal{F}$ . Moreover, when $\text{Aut}_{\mathcal{O}_X}(\mathcal{F})$ is abelian, the set of $\mathcal{O}_X$ -modules $\mathcal{G}$ locally isomorphic to $\mathcal{F}$ forms a group in a natural way, and the correspondence above is actually an isomorphism of abelian groups.

Taking the ringed space to be $(M,\mathcal{O}_M)$ and the sheaf $\mathcal{F}=\mathcal{O}_M$ we see that $\text{Aut}_{\mathcal{O}_M}(\mathcal{O}_M)=\mathcal{O}_M^\times$ , and so $H^1(M,\mathcal{O}_M^\times)$ classifies $\mathcal{O}_M$ -modules on $M$ locally isomorphic to $\mathcal{O}_M$ . These locally free $\mathcal{O}_M$ -modules of rank $1$ are well-known to be in correspondence with line bundles on $M$ (the corresponding taking a line bundle $E$ to its sheaf of sections). $\blacksquare$

So, let us now consider the smooth exponential sequence:

$0\to\mathcal{O}_M\xrightarrow{\exp}\mathcal{O}_M^\times\to\underline{\mathbb{Z}/2\mathbb{Z}}\to 0$

defined in the obvious way on opens: the first map being the exponential map $f\mapsto \exp(f)$ , and the second map being $f\mapsto \text{sgn}(f)$ , thinking of $\underline{\mathbb{Z}/2\mathbb{Z}}$ as being the sheaf of locally constant functions to $\{\pm 1\}\subseteq\mathbb{R}$ .

Let us verify that this sequence is, indeed, exact. Injectivity is clear–if two (real valued!) functions have the same exponential, then they are equal by the existence of a global log function. For surjectivity, note that we, in fact, have a section $\mathcal{O}_M^\times\to\underline{\mathbb{Z}/2\mathbb{Z}}$ given by sending $\pm 1$ to their associated constant functions.

Now, from the short exact sequence, we obtain the long exact sequence

$\begin{matrix} 0 & \to & \mathcal{O}_M(M) & \to & \mathcal{O}_M(M)^\times & \to & \underline{\mathbb{Z}/2\mathbb{Z}}(M)\\ & & & & & &\\ & \to & H^1(M,\mathcal{O}_M) & \to & H^1(M,\mathcal{O}_M^\times) & \to & H^1(M,\underline{\mathbb{Z}/2\mathbb{Z}})\\ & & & & & & \\ & \to & H^2(M,\mathcal{O}_M) & \to & H^2(M,\mathcal{O}_M^\times) & \to & H^2(M,\underline{\mathbb{Z}/2\mathbb{Z}})\\ & & & & & & \\ & \to & H^3(M,\mathcal{O}_M)& \to & \cdots & & \end{matrix}$

Now, we from our above discussion, we can immediately replace $H^i(M,\underline{\mathbb{Z}/2\mathbb{Z}})$ with $H^i_\text{sing}(M,\mathbb{Z}/2\mathbb{Z})$ , and $H^1(M,\mathcal{O}_M^\times)$ with $\mathrm{Pic}(M)$ . Thus, we see that we have a map $\mathrm{Pic}(M)\to H^1(M,\mathbb{Z}/2\mathbb{Z})$ , and the impediments to this map being an isomorphism are $H^1(M,\mathcal{O}_M)$ and $H^2(M,\mathcal{O}_M)$ . So, these are the objects we must get a handle on. Luckily for us, they are extremely easy to understand.

Let us recall that a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$ is called soft if for all closed subsets $C\subseteq X$ , the natural map $\mathcal{F}(X)\to\mathcal{F}(C)$ is surjective. Just because this is possibly odd terminology, let us really quickly say what $\mathcal{F}(C)$ means (what does it mean to evaluate a sheaf on a closed subset?). Well, there are two possible approaches. Perhaps the easiest is just that $\mathcal{F}(C)=(i^{-1})\mathcal{F}(C)$ , where $i:C\to X$ is the inclusion. So, $\mathcal{F}(C)=\varinjlim \mathcal{F}(U)$ , where $U$ ranges over all opens containing $C$ . The other, is that $\mathcal{F}(C)=\Gamma(C,\mathcal{E})$ , where $\mathcal{E}$ is the sheaf of sections of the étalé space $\mathrm{Et}(\mathcal{F})\to X$ . The most important property of soft sheaves, is that they are acyclic. This is something that is proved in all texts on sheaf theory.

Now, the important thing to note is that if $M$ is a smooth manifold, then any $\mathcal{O}_M$ -module is soft. Indeed, suppose that $\mathcal{F}$ is such an $\mathcal{O}_M$ -module, and let $C\subseteq X$ be closed. Then, for any section $s\in\mathcal{F}(C)$ , we obtain a section $\phi s\in\mathcal{F}(X)$ , where $\phi\mid_C =1$ , and $\phi=0$ outside of some open set. Thus, in particular, we see that $(\phi s)\mid_C=s$ , and thus $\mathcal{F}$ is soft. In particular, $\mathcal{O}_M$ is soft, and thus $H^i(M,\mathcal{O}_M)=0$ for all $i>0$ !

From this, we may conclude that the map $w:\mathrm{Pic}(M)\to H^1_\text{sing}(M,\mathbb{Z}/2\mathbb{Z})$ we constructed above is an isomorphism thus proving our main theorem. The map $w$ is denoted as such since we call $w(\mathscr{L})$ , for a line bundle $\mathscr{L}$ , the Whitney-Stiefel class of $\mathscr{L}$ . The above then can be rephrased as saying that the Whitney-Stiefel class of a line bundle $\mathscr{L}$ on a manifold $M$ is a complete invariant.

Back to the circle

Now that we have all of the machinery developed above, talking about the circle itself becomes an easy task. Namely, we finally find that

$\mathrm{Pic}(S^1)\cong H^1_\text{sing}(S^1,\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$

and thus, as we claimed at the beginning, there are, up to isomorphism, precisely two smooth line bundles on $S^1$ .

Now that we have this number, it’s easy to just list out the line bundles. We obviously have the trivial one, $S^1\times\mathbb{R}$ . The non-trivial one can then be seen to be the Mobius strip. Indeed, it’s well-known that the Mobius strip is a line bundle over $S^1$ which can’t be isomorphic to $S^1\times\mathbb{R}$ as a line bundle, since they’re not even homeomorphic (one’s orientable, the other isn’t).

In fact, now that we know that our motivating computations at the beginning of the post were valid, we can read the isomorphism $\mathrm{Pic}(S^1)=\mathbb{Z}/2\mathbb{Z}$ straight from there. Namely, we said that $\mathrm{Pic}(M)$ should be $\ker\varphi/\text{im }\psi$ where

$\displaystyle \varphi:\left(\mathbb{Z}/2\mathbb{Z}\right)^2\to 0$

is the zero map, and

$\psi:(\mathbb{Z}/2\mathbb{Z})^2\to(\mathbb{Z}/2\mathbb{Z})^2$

is the map $(\alpha,\beta)\mapsto (\alpha-\beta,\alpha-\beta)$ . Thus, we clearly obtain

$\mathrm{Pic}(S^1)=\ker\varphi/\text{im }\psi=\left(\mathbb{Z}/2\mathbb{Z}\right)^2/\left\{(\gamma,\gamma):\gamma\in\mathbb{Z}/2\mathbb{Z}\right\}\cong\mathbb{Z}/2\mathbb{Z}$

But, as pointed out above, this was just a calculation of $H^1_\text{sing}(S^1,\mathbb{Z}/2\mathbb{Z})$ , and so really nothing new has been gained from this computation (except, perhaps, a reminder of what singular cohomology is computing!).

A consequence

While the above theorem is very nice, since it makes the computation of the Picard group of a smooth manifold tenable, it also has some unexpected consequences:

Corollary: Let $M$ be a topological manifold. Then, $\mathrm{Pic}(M)$ (smooth Picard group!) is independent of smooth manifold structure on $M$ .

This is somewhat surprising since, after all, the definition of the smooth Picard group involves the notion of smooth line bundles. It also tells us that the Picard group is not really a good ‘smooth invariant’ of a manifold $M$ . For example, suppose you were Milnor and you wanted to check that two smooth structures on $S^7$ were non-equal. Well, the first thing you might check is that perhaps some of the obvious statistics of the two structures differ. One of these obvious statistics is the Picard group. But, the above says that since the underlying topological manifold is just $S^7$ , neither have any non-trivial smooth line bundles.

Topology done wrong

I begin by saying, as I said earlier, that I am no topologist. The above approach is, if I had to guess, perhaps not the approach that a topologist would take to proving the main theorem. My guess is that topologists, unlike me, do not, as a default, think of singular cohomology as a form of sheaf cohomology. Thus, I doubt a sheaf cohomological proof is the first idea that would jump into a topologists head. If anyone reading this is a topologist, please feel free to weigh in–I would be curious to hear your opinion!

Now, with that being said, I would like to hazard a guess at what the more canonical proof of the following variant of the main theorem:

Theorem(Main, variant): Let $X$ be a topological space, then $\mathrm{Pic}_\text{cont}(M)\cong H^1_\text{sing}(X,\mathbb{Z}/2\mathbb{Z})$ .

where $\mathrm{Pic}_\text{cont}(X)$ is the group of continuous line bundles on $X$ , up to isomorphism.

So, it’s well-known that all topological (real) line bundles $\mathscr{L}$ on $X$ come from a map $X\to\mathbb{RP}^\infty$ . More specifically, there exists a map $\phi_\mathscr{L}:X\to\mathbb{RP}^\infty$ with the property that

$\mathscr{L}=\phi_\mathscr{L}^\ast\left(\mathcal{O}(-1)\right)$

Moreover, two such maps $\phi,\psi:X\to\mathbb{RP}^\infty$ pull back $\mathcal{O}(-1)$ to isomorphic line bundles, if and only if they are homotopic.

Thus, the last paragraph allows us to say that we have a canonical bijection

$\left[X,\mathbb{RP}^\infty\right]\xrightarrow{\approx}\mathrm{Pic}_\text{cont}(X):\phi\mapsto \phi^\ast\left(\mathcal{O}(-1)\right)$

But, a simple computation shows that $\mathbb{RP}^\infty=K(\mathbb{Z}/2\mathbb{Z},1)$ . Thus, since Eilenberg-Maclane spaces represent cohomology, we get

$[X,\mathbb{RP}^\infty]=[X,K(\mathbb{Z}/2\mathbb{Z},1)]=H_\text{sing}^1(X,\mathbb{Z}/2\mathbb{Z})$

Putting this together with our previous statements implies the variant of the main theorem.

Putting the variant of the main theorem, and the main theorem itself together, we get a consequence that is subtly more potent than the consequence discussed in the previous section. Namely, not only does $\mathrm{Pic}(M)$ , for a topological manifold $M$ , not depend on the smooth structure, it’s actually isomorphic to $\mathrm{Pic}_\text{cont}(M)$ .

2 comments

zarathustra says:

January 17, 2015 at 4:28 am

Hey great blog, thanks !

Tiny typo in the proof of the first lemma : it should say” classifies modules G locally isomorphic to F” not G 🙂

1. alexyoucis says:
  
  January 26, 2015 at 2:25 am
  
  Thank you very much for the kind words! 🙂
  
  I also fixed your typo–thanks for pointing it out!

Hard Arithmetic

Line Bundles on the Circle

Motivation

Intuitive preamble

The proof

Back to the circle

A consequence

Topology done wrong

2 comments

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Motivation

Intuitive preamble

The proof

Back to the circle

A consequence

Topology done wrong

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