# complex vector bundles and curvature

Let us suppose that $X$, with a 2-form $\omega$. Suppose $J$ is an element of $su(2)$ such that $J^2=-e$ for $e$ the identity. Is there a necessary and sufficient condition on $\omega$ which will give the existence of $E\rightarrow X$ a $SU(2)$ bundle with a connection whose curvature is $J\omega$?

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Are you saying that $X$ is a compact Kaehler manifold with Kaehler form $\omega$? The curvature should be a 2-form valued in $E$, so I don't understand the question. –  Ben McKay Feb 24 '12 at 8:50
Please rewrite the question with everything defined precisely. –  Deane Yang Feb 24 '12 at 10:34
Sorry, let me define these things. $X$ is a smooth manifold. $\omega$ is an arbitrary 2-form on $X$. $J\omega$ is then an element of $\Omega^{2}(X)\otimes su(2)$. $E$ is a $SU(2)$ bundle over $X$. No Kähler structure is assumed. I hope that helps. –  Blake Feb 25 '12 at 1:21
Blake, thanks. I guess you meant what you said. –  Deane Yang Feb 25 '12 at 20:34
Actually, I think that this problem would be more naturally stated in somewhat different terms: What can you say about connections $\nabla$ on an $SU(2)$-bundle $P\to X$ that have the property that the curvature operator $K^\nabla:\Lambda^2(TX)\to {\frak ad}(P)$ takes values in a subbundle $\textbf{J}\subset {\frak ad}(P)$ of rank $1$? The way it's stated now, the problem seems to be assuming that the rank $3$ bundle ${\frak ad}(P)$ is trivial, but I don't see why this is a natural (or desirable) assumption. The formulation I am suggesting is gauge-invariant and doesn't make this assumption. –  Robert Bryant Mar 1 '12 at 16:06

An obvious necessary local condition is that $d\omega=0$. On the open set $U\subset X$ on which $\omega^2\not=0$, one has the further condition that the only ${\frak su}(2)$-valued connection that could have curvature $J\omega$ is one that is locally of the form $J\alpha$, where $\omega = d\alpha$. This may mean (I haven't checked the mod 2 conditions) that any such $E$ would have to restrict to $U$ to be of the form $E = L \oplus \overline L$, where $L$ is a complex line bundle with curvature $\omega$.
Here is the reasoning behind the above statements: Recall that ${\frak su}(2)$ can be regarded as the space of imaginary quaternions, with standard basis ${\bf i},{\bf j},{\bf k}$. Our curvature form can then be written as $\Omega = \omega {\bf i}$, and a potential connection form would be $\alpha = \alpha_1 \textbf{i} + \alpha_2\textbf{j} + \alpha_3 \textbf{k}$ for some (at the moment, only locally defined) $1$-forms $\alpha_i$. Now, the curvature equation $d\Omega = d\alpha + \alpha \wedge \alpha$ implies the Bianchi identity $d\Omega = \Omega \wedge \alpha - \alpha \wedge\Omega$, which unravels to the three equations $$d\omega = 0\quad\text{and}\quad \omega\wedge\alpha_2 = \omega \wedge\alpha_3 = 0.$$ Thus, $\omega$ must be closed. Moreover, if $\omega^2\not=0$, the second and third equations imply that $\alpha_2 = \alpha_3 =0$, so that one must have $\omega = d\alpha_1$, and the connection must keep the splitting of $E$ into the sum of two (conjugate) line bundles invariant under parallel translation.
There is unlikely to be a simple sufficient condition without making some constant rank assumptions on $\omega$, there are just too many possibilities.
I'll add some remarks about the case when $\omega^2=0$ but $\omega\not=0$ when I have time.
I don't think $\omega^2\neq 0$ is enough for this. Let $\omega = \sum dp_i \wedge dq_i$ on $\mathbb(R)^n$; then $\omega ^2\neq 0$ but we can take the connection on a trivial bundle given by the potential $\alpha = Kdp_i + Idq_i$ where $I,J,K$ are the matrices corr. to the $i,j,k$ quaternions. Then $d\alpha + \alpha \wedge \alpha$ = 2J\omega$so we just need to normalize it to get$J\omega$, but$d\alpha = 0$. – Blake Feb 28 '12 at 8:11 That should read$d\alpha + \alpha \wedge \alpha = 2J\omega$so we just need to normalize$\alpha$, but$d\alpha = 0$. – Blake Feb 28 '12 at 8:12 @Blake: Check your algebra. Your computation of the curvature of$\alpha$is not correct. Instead of getting$\omega$, which you have defined to be$\sum dp_i \wedge dq_i$, you get$\sigma = (\sum dp_i) \wedge (\sum dq_i)$, and$\sigma^2 = 0$. – Robert Bryant Feb 28 '12 at 13:36 The case$\omega^2=0$holds always when$M$is a surface, and I want to restrict to the case of compact oriented surfaces of genus$g\geq2.$Lets say you have a flat$SL(2,\mathbb C)$-connection$\nabla$on a topological trivial bundle$V\to M$whose underlying holomorphic structure $$\bar{\partial}=\frac{1}{2}(\nabla+i*\nabla)$$ is stable ($\star$is the Riemann surface$\star,$i.e. it is given on$1$forms by$\star dz=idz,\star d\bar z=-id\bar z).$Stability implies that there are no non-trivial holomorphic endomorphisms, i.e. $$H^0(M;End(V,\bar\partial)=\mathbb C Id.$$ This case occurs for example when$\nabla$is an irreducible flat$SU(2,\mathbb C)$connection by a Weitzenboeck argument. By Serre duality and with$End_0(V)=(End_0(V))^*$via trace $$H^1(M,KEnd_0(V))=H^0(M,End_0(V))=\{0\}$$ which means that every$2-$form with values in the trace free endomorphisms is in the image of$\bar\partial.$This implies there exists$\alpha\in\Gamma(M, KEnd_0(V))$with $$F^{\nabla+\alpha}=F^\nabla+d^\nabla\alpha=0+\bar\partial\alpha=J\omega$$ as$J\omega$is trace free. - Yes, but I meant to ask for the case where the holomorphic structure is inherent to the manifold, and is not defined by the connection. – Blake Feb 28 '12 at 8:14 @Blake: I don't understand this comment. In your comments to the question, you assert that you are interested in general$M$, which would not have any holomorphic structure. Did you mean to have this connection interact with some holomorphic structure on$M\$? Can you explain how you'd describe this interaction? –  Robert Bryant Feb 28 '12 at 15:48