Are there known examples of compact Riemannian manifolds with Q-curvature negative?
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$\begingroup$ It might help if you gave a definition of $Q$-curvature or a reference that defines it. $\endgroup$– Robert BryantCommented Mar 1, 2015 at 19:26
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Yes. Consult this reference, for example, for the definition of $Q$-curvature.
In dimension $2$, the $Q$-curvature is essentially just the Gauss curvature, so any compact Riemann surface with genus $g>2$ carries a conformal metric with negative Gauss curvature and hence negative $Q$-curvature.
Meanwhile, in dimension $4$, $$ Q = \left(-\tfrac16\Delta R - \tfrac12 R^{ab}R_{ab} + \tfrac16 R^2\right)\mathrm{d}vol $$ where $R$ is the scalar curvature and $R_{ab}$ represents the Ricci curvature. Obviously, any compact manifold with vanishing scalar curvature will have non-positive $Q$, and one can easily construct many examples that have $Q$ strictly negative. For example, the product of the unit $2$-sphere and a compact Riemann surface with Gauss curvature identically equal to $-1$ will have vanishing scalar curvature, but won't be Einstein, so it will have $Q$ be strictly negative.
answered Mar 1, 2015 at 19:40