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Hey,

For personal exercising purposes I try to give a proof, that a U(1)-principal-bundle has curvature $\alpha$ iff the cohomology class of $\alpha$ is integral:

By the Cech-deRham-isomorphism a $[\alpha] \in H^2_{DR}(M, \mathbb{Z})$ iff $[\alpha] \in H^2_{Cech}(M, \mathbb{Z})$. Then we can use the isomorphism $H^2_{Cech}(M, \mathbb{Z}) \tilde= H^1_{Cech}(M, U(1))$ to construct an U(1)-principal bundle P up to equivalence by defining its transition functions. To define a connection 1-Form $\varphi$ use the local trivialization $h_j: U_j \times U(1) \mapsto \pi^{-1}(U_j)$ and set \begin{equation} h_j^* \varphi = \pi^* \beta_j + d\vartheta, \end{equation} where $d \beta_j = \alpha$ on $U_j$ and $\vartheta$ is the coordinate in U(1). This defines a connection with curvature $d \varphi = \alpha$.

I think this should it be. Any remarks/corrections are welcome!

But I read, that the various choices of such an principal bundle with given curvature are parametrized by $H^1(M,U(1))$, which in view of my proof makes absolutely no sense! Where lies the error?

Thank you for your help!

P.s. - a second question: How does the curvature of the principal bundle defines the curvature of an associated (line)bundle? In the special case of a U(1)-bundle with curvature $\alpha$ and the representation $x \mapsto exp(-i k x)$, is the curvature of the assoziated line bundle given by $k \cdot \alpha$??

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I suspect it's a problem of notation. Your $H_{Cech}^1(M, U(1))$ would appear to be the first cohomology of the sheaf of $C^\infty$ sections of $U(1)$, while the second $H^1(M,U(1))$ ought to be locally constant sections by the claimed interpretation. –  Donu Arapura Sep 6 '11 at 21:22
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2 Answers

up vote 3 down vote accepted

EDIT: I'm not very happy the the exposition below. But I hope you get the idea.

Your proof is correct. Given a principal bundle with connection having the given curvature, and a second connection on a possibly different with the same curvature, the difference between the two is a bundle with a flat connection (note that the category of $U(1)$-bundles with connection is a symmetric 2-group). Thus one gets an ambiguity in the choice of bundle and connection. This information is reflected in the short exact sequence

$$ 0 \to H^1(M,U(1)) \to \check{H}^1(M) \to \Omega^2_\mathbb{Z}(M) \to 0 $$

where the left cohomology group has discrete coefficients and classifies bundles with flat connections, and the group on the right is that of integral differential 2-forms on $M$. The central group is the ordinary differential cohomology of $M$. One way it may be thought of is as a moduli space (which has a group structure) of $U(1)$-bundles with connection. The right map is the map that sends a bundle with connection to its curvature form.

There is another short exact sequence dealing with the characteristic class of the bundle, see Proposition 1 at this nLab page. It describes the kernel of the (surjective) homomorphism $\check{H}^1(M) \to H^2(M,\mathbb{Z})$, sending a bundle with connection to its first Chern class (which one can think of as being a Cech class).

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How can you add two connections? –  Deane Yang Sep 7 '11 at 2:09
    
Good point. I'll edit. –  David Roberts Sep 7 '11 at 3:06
    
I think you were closer to the mark when you said that the difference between two connections with the same curvature is a 1-form. The difference between two connections with the same curvature is not a flat connection. The space of connections is an affine space (no distinguished origin), and the difference between any two $U(1)$-connections is a real-valued $1$-form. Two connections have the same curvature if and only if their difference is a closed $1$-form. Two $U(1)$- connections are gauge equivalent if and only if their difference is an exact $1$-form. –  Deane Yang Sep 7 '11 at 4:40
    
Hi Deane - that is only true if you consider connections on the same bundle. If you permit different bundles, the difference of two bundles-with-connection with the same curvature is a third bundle, equipped with a flat connection. This is what the exact sequence I give says. The statement about the difference of two connections on the same bundle that give rise to the same curvature is more connected with the other exact sequence I mention. –  David Roberts Sep 7 '11 at 5:22
    
David, sorry about that. I withdraw all of my comments. I didn't notice that you were allowing for different bundles. I don't know what a "symmetric 2-group" is, but the space of complex line bundles with $U(1)$-connections does indeed have an abelian group structure by taking the tensor product of two bundles and the naturally induced connection defined using the product rule. The group inverse is given by the dual bundle with the naturally induced connection, and the identity element is the trivial bundle with the trivial connection. So you can add two elements this way. –  Deane Yang Sep 7 '11 at 15:11
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I'm late to the party with this answer, but the question is on the front page, so whatever.

As David essentially said, given any $\mathrm{U}(1)$-bundle/connection pair over $M$ with connection $\alpha$, one can obtain any other bundle/connection pair by tensoring with a flat bundle over $M$. Flat bundles are characterized by $H^1(M, \mathrm{U}(1))$, which is isomorphic to the character group of $\pi_1(M)$ (for $M$ path-connected at least). One way to see this, which is less powerful than David's characterization, but a bit easier to grasp, is that a flat connection on a (right) $G$-bundle $P$ is equivalent to an involutive, hence integrable, distribution on $P$. We can lift any curve starting at $m\in M$ to a horizontal curve through any $p\in\pi^{-1}(m)$, where $\pi:P\rightarrow M$ is the bundle projection. The maximal integral manifold of the distribution through $p$ is actually a covering space of $M$ via $\pi$, and so it follows that lifts through $p$ of homotopic curves in $M$ are homotopic in $P$, and have the same endpoint. Call this endpoint $E_p([\gamma])$, where $[\gamma]$ is the homotopic class of the curve $\gamma$. If $\gamma$ is a closed curve, then the endpoint of its lift lies in the same fiber as $p$, and we can write $$ E_p([\gamma]) = p\cdot\chi_p([\gamma]^{-1}) $$ where $\chi_p:\pi_1(M)\rightarrow G$. It is not difficult to show that $\chi_p$ is a homomorphism. $P$ can then be reconstructed as the quotient of the trivial bundle $\widetilde{M}\times G$ with the obvious flat connection under the equivalence relation $$ ([\delta],g) \sim (\[\gamma]*[\delta],\chi_p([\gamma])g) $$ by mapping $\left[[\delta],g\right]_\sim\mapsto E_p([\delta])\cdot g$. Here $\widetilde{M}$ is the universal covering space of $M$, thought of as classes of homotopic curves starting at $m$. Hence all such flat bundles are characterized by the set of homomorphisms $\chi_p:\pi_1(M)\rightarrow G$, which in the case $G=\mathrm{U}(1)$ is just the character group of $\pi_1(M)$. (It's very easy to screw up the conventions here; hopefully I haven't.)

To answer your second question, it depends what you mean by "curvature in the line bundle". Normally one would think of the curvature $\alpha$ as a $\mathfrak{g}$-valued two-form on $P$. Maybe this is what you're getting at: given a representation $\lambda:G\rightarrow \mathrm{GL}(V)$ on the vector space $V$, we form the vector bundle $P\times_\lambda V$ associated to $P$. Similarly we can form the associated bundle $P\times_{\mathrm{Ad}}~\mathfrak{g}$, where $\mathrm{Ad}:G\rightarrow \mathrm{GL}(\mathfrak{g})$ is the adjoint representation. The infinitesimal $\mathfrak{g}$-action $\lambda'(\xi)v = \frac{d}{dt}\lambda(\exp(\xi t))v\big\vert_{t=0}$ on $V$ induces a well-defined action of $P\times_{\mathrm{Ad}}~\mathfrak{g}$ on $P\times_\lambda V$.

Using the curvature $\alpha$ we define a $P\times_{\mathrm{Ad}}~\mathfrak{g}$-valued 2-form on $M$, given by $$ \beta_m(X_m,Y_m) = [p,\alpha_p(A_p,B_p)]_{\mathrm{Ad}} $$ where $A_p, B_p$ are (arbitrary) lifts of $X_m, Y_m$ to $p\in \pi^{-1}(m)$. The fact that $\alpha$ vanishes on vertical vectors, coupled with the $G$-equivariance of $\alpha$, ensures that $\beta$ is well-defined.

It's possible (and a worthwhile exercise) to prove that $$ ([\nabla_X,\nabla_Y]-\nabla_{[X,Y]})s = \beta(X,Y)\cdot s $$ where $s$ is a section of the associated vector bundle $P\times_\lambda V$. (The left hand side appears to depend on the vector fields $X,Y$, but in fact at a particular point it just depends on their values at that point.)

If one expresses $s$ in terms of its corresponding $G$-equivariant function $\widetilde{s}:P\rightarrow V$, i.e. $$ s(m) = [p,\widetilde{s}(p)]_\lambda \qquad\textrm{any }p\in\pi^{-1}(m) $$ then this unpacks to $$ \left[p, ([X^h,Y^h]-[X,Y]^h)_p \widetilde{s}(p)\right] _ \lambda = \left[p, \lambda'(\alpha_p(X^h_p,Y^h_p)) \widetilde{s}(p)\right]_\lambda $$ ($X^h$ being the horizontal lift of $X$ etc.)

As with the curvature $\alpha$, in the case when $G$ is abelian you don't need to go to all this trouble, and can just define $\beta$ as a $\mathfrak{g}$-valued 2-form on $M$. Then as you guessed, these 2-forms are related by $\beta=\lambda'\circ\alpha$.

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