First-time poster here. I don't yet know how to typeset. I am working through "Tensors, Differential Forms, and Variational Principles" by Lovelock & Rund. At the top of p156, the assertion is made that G dot (curl G) = 0 is a necessary and sufficient condition for G being the gradient of a scalar (G is a vector field).
Am I correct that this assertion is flawed? The necessary part is true, for if G is the gradient of a scalar, then its curl is zero and hence G dot (curl G) is indeed zero. However, it's not hard to find an example of a non-zero vector field which is perpendicular to its curl. For instance: if G = (-y,x,0), then curl G = (0,0,2) so that G dot (curl G) = 0. However, G is clearly not the gradient of a scalar, for if it were, its curl would be zero (which it's not).
Thank you.
Yup, you're right. If $\mathbf{G}$ is the gradient of a smooth function then $\mathbf{G}\cdot (\nabla \times \mathbf{G})=0$, but $\mathbf{G}\cdot(\nabla\times \mathbf{G})=0$ does not imply $\mathbf{G}$ is the gradient of a scalar, as your counterexample shows.
Edit:
You misquoted the book. They say "is a necessary and sufficient condition that $\mathbf{G}$ be proportional to the gradient of a scalar function", and here of course, proportionality means proportional by a function, meaning $\mathbf{G}=f \nabla \phi$ for some functions $f,\phi$. It is easily verified in this case that $\mathbf{G}\cdot (\nabla \times \mathbf{G})=0$; for the converse I'm guessing it requires Frobenius (something which I'm not entirely comfortable with so I can't elaborate too much on that). In your example, let $f(x,y)=-y^2$ and $\phi(x,y)=\frac{x}{y}$. Then, $\mathbf{G}=(-y,x,0)$ is equal to $f\nabla \phi$.