Bochner's formula and harmonic 1-forms on 2-manifold

Question

For a harmonic 1-form $\omega$ (we'll denote its dual vector by $X$), we have from Bochner's formula: $\frac{1}{2}\triangle ||X||^2 = Ric(X , X) + g(X, \nabla div X) + ||\nabla X||^2 = Ric(X , X) + ||\nabla X||^2$. Now on a 2-manifold, its Hodge dual $\star \omega$ is also a harmonic 1-form (we'll denote its dual vector by $\star X$), so we also have $\frac{1}{2}\triangle ||\star X||^2 = Ric(\star X , \star X) + ||\nabla \star X||^2$. But since $||X||^2 = ||\star X||^2, ||\nabla X||^2 = ||\nabla \star X||^2$, we must have $Ric(X , X) = Ric(\star X , \star X)$. As $X$ and $\star X$ are orthogonal and have the same magnitude, this would imply that the Ricci tensor is proportional to the metric tensor, which certainly yields a contradiction as we are considering the general case. What's the problem in my reasoning?

Answer 1

You’re forgetting that this is a surface. The Ricci curvature is just the Gaussian curvature.