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Let $(M,g)$ be a riemannian manifold and let $P\to M$ be a principal $G$-bundle with connection $A$. Let $\alpha \in \Omega^1(M;\mathrm{ad}P)$ be a one-form on $M$ with values in the adjoint bundle $\mathrm{ad} P$. Then consider the equation $$ F_{A + \alpha} = \tfrac12 [\alpha,\alpha] $$ in $\Omega^2(M;\mathrm{ad}P)$, where $F_{A+ \alpha}$ is the curvature of the connection $A + \alpha$.

This equation is reminiscent of the Hitchin equations on a Riemann surface, but of course is different.

(In the context in which I have seen this equation, it is supplemented by the condition $\nabla^A \alpha \in \Omega^2(M;\mathrm{ad}P)$, which is equivalent to the vanishing of the symmetrisation of $\nabla^A \alpha$.)


Has anyone come across such an equation before? And if so, could you point me to a reference?

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Since $F_{A+\alpha}=F_A + d_A\alpha + \frac{1}{2}[\alpha,\alpha]$, you mean $F_A + d_A\alpha=-[\alpha,\alpha]$, right? And did you mean supplemented by $d_A\alpha=0$? i.e. you dont mean just the second order obstruction to deforming a flat connection? –  Paul Jul 12 '11 at 20:26
@Paul: Perhaps I should have written the original version of the equation, which is $F_A + \nabla_A \alpha = 0$, whereas $d_A \alpha$ is the skew-symmetrisation of $\nabla^A\alpha$. In other words, the equation breaks up into two: $F_A + d_A \alpha = 0$ and the vanishing of the symmetrisation of $\nabla^A \alpha$. Does this make more sense? –  José Figueroa-O'Farrill Jul 12 '11 at 22:51
$\nabla_A$ above should be $\nabla^A$. (I continue to wish I could easily edit comments.) –  José Figueroa-O'Farrill Jul 12 '11 at 22:51
I'm still confused by the sign: $F_A+d_A\alpha=0$ is the same as $F_A+d_A\alpha+\frac{1}{2}[\alpha,\alpha]=\frac{1}{2}[\alpha,\alpha]$ which in turn is $F_{A+\alpha}=\frac{1}{2}[\alpha,\alpha]$. In any case, I've never seen this. (As you probably know) if $A$ is flat, then $d_A\alpha=0$ is the first obstruction to $A+t\alpha$ being a path of flat connections and $[\alpha,\alpha]=2d_A\alpha$ is the second obstruction. Maybe the other equation is hidden form of some gauge-fixing condition? (are you on a 2-manifold? can you relate d_A^*\alpha=0$ to $Sym(\nabla^A\alpha=0)$?) –  Paul Jul 13 '11 at 3:31
Ah... my apologies. The sign is wrong! I must have copied it wrong. I'll edit. –  José Figueroa-O'Farrill Jul 13 '11 at 14:15

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