suppose $\Omega \subset \mathbb{R}^3$ is a Riemannian manifold with $g= dx\_1^2 + dx\_2^2 + c(x\_1,x\_2,x\_3)dx\_3^2$. Is it true that $det(CY)=0$?


Thanks



Vanishing of determinant of Cotton York tensor
The answer is 'no', the expression $\det(CY)$ does not vanish identically for metrics of the specified form. 


This follows by a direct computation, which is not all that difficult to do by hand, but is made easier by a symbolic calculator, such as Maple.


If you are having difficulty doing the calculation, I suggest you implement it in your favorite symbolic manipulation program, and you'll see the result.


Unfortunately, there does not appear to be any particular identity satisfied by the Cotton York tensor for metrics of this particular form. 


What *is* true is that the vector field $\partial/\partial x\_3$ is a null vector for the Cotton York tensor of the above metric, but that's not saying very much, and there are no other pointwise algebraic identities of this kind.


