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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.

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.

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For those of you interested in conformal differential geometry, the above integral is the total $Q$-curvature of the manifold. Hence, we are asking if there exists a metric on $S^n$ that has a negative average value of $Q$-curvature. – Viktor Bundle Jun 20 '11 at 0:46
That's a pretty amazing normalization factor! :) – Mariano Suárez-Alvarez Jun 20 '11 at 1:51
Mariano: The coefficients seem baffling because they are at specific value that provide nice conformal properties for $Q$-curvature. Interestingly enough, there is one other combination of constants that has this property. – Viktor Bundle Jun 20 '11 at 2:17
Anton: that is correct. I'll edit the post to make that clear. – Viktor Bundle Jun 20 '11 at 14:34

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