Negative Paneitz constant on $n$-sphere - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:16:38Z http://mathoverflow.net/feeds/question/68243 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68243/negative-paneitz-constant-on-n-sphere Negative Paneitz constant on $n$-sphere Viktor Bundle 2011-06-19T22:55:03Z 2011-06-20T14:35:07Z <p>Let $\Pi$ be the Riemannian functional defined on the space of Riemannian metrics on $S^n$, $n>4$, as follows: $$\Pi(g) = \int_M \frac{(n-4)(n^3-4n^2+16n-16)}{16(n-1)^2(n-2)^2} R_g^2 - \frac{2(n-4)}{(n-2)^2} |Ric_g|^2 dv_g.$$ Here $R_g$ is the scalar curvature, and $|Ric_g|$ is the norm of the Ricci tensor. This functional is sometimes referred to as the Paneitz functional. It is of great interest to conformal differential geometers, because of its connection to $Q$-curvature. The coefficients, which are functions of the dimension, are positive if $n>4$ (here I am not including the negative sign as part of the coefficient). The reason the coefficients are strange in appearence is that they are carefully constructed so that the $Q$-curvature has nice conformal properties.</p> <p>Since there will be a scalar flat metric on $S^n$ it is known that $\Pi$ obtains a non-positive value on the space of Riemannian metrics on $S^n$, $n>4$. Can you show that it takes on negative values on the space of Riemannian metrics on $S^n$ as well? Note, this would trivially follow from the existence of a scalar flat metric on $S^n$, $n>4$, that was not Ricci flat. Note as well that all Einstein metrics will return a non-negative value.</p>