Flat non-trivial $U(1)$-bundle? Is it possible?

Question

maybe this is a very stupid question and I'm missing something very trivial.

It's well known that $U(1)$-bundles are classified by the Euler class or the first Chern class. More precisely, the isomorphism $$c: \check{H}^1 (X, \mathscr{C}^{\infty} (-, U(1))) \xrightarrow{\sim} H^2 (X, \mathbb{Z})$$, induced by the exact sequence $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{R} \xrightarrow{exp} U(1) \rightarrow 0$, the sheaf isomorphism $\mathscr{C}^{\infty} (-, \mathbb{Z}) \cong \mathbb{Z}_X$ and the isomorphism from \v{C}ech to singular cohomology (or de Rham cohomology with integral periods), is the first Chern class or the Euler class.

It's well known too that the first Chern class is given by $$[\frac{1}{2\pi i}F_{\nabla}]$$. This equality can be accomplished by using that the first Chern class is equal to the Euler class, the global angular form $\psi$ restricts to the Maurer-Cartan form on each fiber and every connection is of the form $\psi + \pi^{*} \alpha$ for some $\alpha \in \Omega^1 (X, \mathfrak{u}(1))$, where $\pi: P \twoheadrightarrow X$ is the $U(1)$-bundle

1) From these two remarks, one can conclude that if $F_{\nabla} = 0$ on some circle bundle $P$, then $c_1 (P) = 0$ and, therefore, $P$ is trivial.

However it's well known too that in the theory of Cheeger-Simons differential characters $$H^1 (X, U(1))$$ classifies flat $U(1)$-bundles with connection by assigning the holonomy $$\text{Hol}_{\nabla} (z) = \langle c, z \rangle$$ for every $z \in Z^1 (X, \mathbb{Z})$ to each $[c] \in H^1 (X, U(1))$ and using the canonical pairing coming from the fact that $U (1)$ is divisible and, hence, $H^1 (X, U(1)) = \text{Hom} (H_1 (X, \mathbb{Z}), U(1))$.

2) Therefore it may exist non-trivial bundles with flat connection whenever the first cohomology does not vanish.

Having this in mind, why there's a non-trivial flat $U(1)$-bundle?

In other words, what's wrong in my conclusions in 1 and 2?

Thanks in advance.

score 11 · Answer 1 · answered Feb 23 '16 at 17:20

The Euler class detects topological triviality, not triviality in the finer sense of whether or not a flat connection is a product.

The prototypical example of a flat, non-trivial bundle starts with the product bundle $[0, 1] \times U(1) \to [0, 1]$ equipped with the product connection. If $\theta$ is an arbitrary real number, glue the fibre over $0$ to the fibre over $1$ by rotating through $\theta$; that is, identify $(0, e^{i\phi}) \sim (1, e^{i(\phi + \theta)})$.

The result is a flat, topologically-trivial $U(1)$ bundle $P \to S^{1}$ over the circle. If $\theta \not\in 2\pi \mathbf{Z}$, however, then $P$ admits no covariantly-constant section. When an arbitrary point of $S^{1}$ is removed, the restriction of $P$ to the resulting open interval $I$ is a product bundle $I \times U(1)$ whose covariantly-constant sections are constant maps $I \to U(1)$, and no such section extends continously to a global section of $P$.

In case it's of interest, there similarly exist topologically-trivial holomorphic line bundles over an elliptic curve $E$ that admit no non-trivial global section. For example, let $p$ and $q$ be distinct points of $E$, and let $L$ be the line bundle associated to the divisor $[p] - [q]$. A non-trivial holomorphic section of $L$ would define a meromorphic function $f$ on $E$, holomorphic off $q$ and with a simple zero at $p$ and a simple pole at $q$. Integrating the meromorphic $1$-form $$ \eta = z \frac{f'(z)\, dz}{f(z)} $$ around the boundary of a fundamental domain for $E$ and using the fact that the total residue is zero would give $p - q = 0$, which is not the case.

I still cannot understand why the residue is zero… It would be zero for $\frac{f' (z)}{f(z)} dz$, though. I've got the part $\int \eta = 2\pi i (p - q)$, however why it's zero too? — user40276, Feb 24 '16 at 07:52
Another question, why the divisor over $E$ is topologically trivial? — user40276, Feb 24 '16 at 09:09
Thanks for your reply. 1. This would be true if $\eta$ was closed, however I could see no reason for this (why it's closed?). 2. Ok, I've got it, you use that the integral is the sum of indexes plus Poincaré duality plus the fact that $H^2 (E, \mathbb{Z})$ is torsion-free (actually free of dimension 1 by the orientability). — user40276, Feb 24 '16 at 17:11

score 9 · Answer 2 · 2016-02-23T17:26:49.073

9

This is an answer to your question 1); question 2) is covered in Andrew's answer.

You're mistaking two different characteristic classes. There's the integral first Chern class $c_1(E) \in H^2(M,\Bbb Z)$ and the real first Chern class $c_{1,dR}(E) \in H^2_{dR}(M)$. This last one is the one you calculate with Chern-Weil theory. They're usually not denoted differently. The relation is that, under the canonical map (changing coefficients and then following the de Rham isomorphism), $c_1(E) \mapsto c_{1,dR}(E)$.

Having a flat connection on a $U(1)$-bundle is equivalent to $c_{1,dR}(E)$ vanishing. You've proved one direction. In the other direction, suppose it vanishes, and that $A$ is a connection on your bundle. I want to find a 1-form $a \in \Omega^1(E \otimes E*) \cong \Omega^1(M)$ such that $da = - F_A$, because of the equation $F_{A+a} = F_A + da$ for connections on a $U(1)$-bundle. Now note that the right hand side is closed by the Bianchi identity; and by your assumption that $c_{1,dR}(E)$ is zero, we see that it must actually be exact. So I can find such an $a$, and therefore my bundle is flat iff $c_{1,dR}(E)$ is torsion.

What's the difference? Torsion classes! Since $M$ is compact we can decompose its second cohomology as $H^2(M;\Bbb Z) \cong \Bbb Z^a \oplus T$, where $T$ is a finite group. That finite group is the group of flat $U(1)$-bundles up to bundle isomorphism, not necessarily preserving the flat connection, because their real first Chern classes vanish.

edited Feb 23 '16 at 17:26

answered Feb 23 '16 at 17:21

Thanks for your answer. Actually my question is just one question. Sorry if this was not clear. I just wanted to know why 1 does not contradicts 2 and vice-versa. – user40276 Feb 23 '16 at 18:00
1

@user40276: But (1) is wrong, and the other answer does not clarify this. That's the point of this answer. – Feb 23 '16 at 18:01
Ah! Ok! I've got it. I forgot the $d\alpha$ part. The integral cohomology class will actually be represented by $[\frac{1}{2\pi i}d\psi]$. So I can conclude that the integral Chern class is zero iff $[\frac{1}{2\pi i}d\psi] = 0$, i.e., the circle bundle is topologically trivial. – user40276 Feb 23 '16 at 18:10
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What is $\psi$? If what you're writing down lives in de Rham cohomology, you cannot possibly detect topological triviality, because torsion dies in de Rham cohomology. – Feb 23 '16 at 18:13
$\psi$ is the global angular form. Under the isomorphism to Cech cohomology $\frac{1}{2\pi i}d \psi $ goes to $\frac{1}{2\pi i}(\varphi_{jk} - \varphi_{ik} + \varphi_{ij}) \in H^2 (X, \mathbb{Z})$ which is exactly the connecting morphism $c$ of the exact sequence I've written above, where $\varphi_{ij} = \theta_j - \theta_i$ and $\theta_i$ is a local coordinate. So if this is zero, the bundle is really topologically trivial. – user40276 Feb 23 '16 at 18:19
@user40276: Gotcha, I understand and agree. Thanks for the clarification. – Feb 23 '16 at 18:20

score 5 · Answer 3 · answered Feb 23 '16 at 17:54

Topological line bundles are classified by $H^1(X, U(1)) \cong H^2(X, \mathbb{Z})$ where $U(1)$ has the usual topology. Flat line bundles are classified by $H^1(X, U(1)) \cong \text{Hom}(H_1(X), U(1))$ where $U(1)$ has the discrete topology; this is often denoted by $U(1)_d$. There's a natural map

$$H^1(X, U(1)_d) \to H^1(X, U(1)) \cong H^2(X, \mathbb{Z})$$

sending a flat line bundle to its first Chern class, and Chern-Weil theory implies that if $X$ is a smooth manifold then this class has image zero in de Rham cohomology. But as Mike Miller says this only implies that it's torsion, not zero.

Flat non-trivial $U(1)$-bundle? Is it possible?

3 Answers3