# What is Rotational on a Riemannian Manifold?

I have learned divergence, gradient and rotational in vector analysis of $\mathbb R^3$. However, when I read Riemannian Geometry, there are only definitions about divergence and gradient. So I have an idea to generalize the conception of Rotational.

In Do Carmo's book Differential Forms and Applications, rotational is defined as below: $$X\rightarrow\omega\rightarrow d\omega\rightarrow*(d\omega)$$ Let $\omega$ denote the differential one-form obtained from $X$ by the canonical isomorphism induced by the inner product $<,>$ and $d$ be exterior differential, $*$ be Hodge star operation.

My questions are:

(1) Can this definition be generalized for a Riemannian Manifold?

I think yes. But does it still mean rotational?

(2) Is there another way to define the rotational?

Although there is a definition above, I still feel it is complex or strange with reference to physics, compared with that in $\mathbb R^3$.

If I understand correctly, with Do Carmo's definition $\operatorname{curl}(X) = *(d(X^\flat))$ is a $(n-2)$-form (where $n$ is the dimension of the manifold)1.

One might want to define it instead as $$\operatorname{curl}(X) = (d (X^\flat))^\sharp$$ so that $X$ is a bivector field (or a $2$-covariant antisymmetric tensor). In a local chart $(x^1, \dots, x^n)$, this is2 $$\operatorname{curl}(X) = \sum_{1 \leqslant i,j \leqslant n} \left( {\partial X^j \over \partial x^i} - {\partial X^i \over \partial x^j} \right)\, {\partial \over \partial x^i} \wedge {\partial \over \partial x^j}$$

In dimension $3$, one may identify bivectors and vectors via the cross-product, for instance as in ${\partial \over \partial x} \wedge {\partial \over \partial y} = {\partial \over \partial z}$. This is how you get back your usual curl.

In the end I don't think it really matters how you define it, as long as you know what you're talking about (the exterior derivative of the Riemannian dual of the vector field, interpreted as some tensor field).

"Does it still mean rotational?" I'm not sure the question is that relevant. The important thing is that it has properties you would expect (such as $\operatorname{curl}(\operatorname{grad} f) = 0$). Actually, it's probably conceptually accurate to say that the curl is a bivector rather than a vector.

1 Note that I'm using the musical isomorphisms notation.

2 I hope! This should be checked.

• So two definition are same? – gaoxinge Feb 3 '14 at 0:59
• No, $*(d(X^\flat))$ is a $(n-2)$-form whereas $(d (X^\flat))^\sharp$ is a bivector field. They are strongly related but they are different objects. – Seub Feb 3 '14 at 5:37

In O'Neill's Semi-Riemannian Geometry With Applications to Relativity there's another (very similar) definition: if $(M,g)$ is a pseudo-Riemannian manifold and $X \in \mathfrak{X}(M)$ is a vector field, then ${\rm curl}(X)$ is the $2$-form on $M$ whose coordinates are $$\frac{\partial X^j}{\partial x^i} - \frac{\partial X^i}{\partial x^j},$$in any chart. It turns out that ${\rm curl}(X) = {\rm d}(X_\flat)$, where $\flat$ denotes the musical isomorphism $TM \to T^*M$, as always, and ${\rm d}$ denotes exterior derivative. The condition ${\rm d}^2 = 0$ reads in the this context as ${\rm curl}\circ {\rm grad} = 0$ and ${\rm div}\circ {\rm curl} = 0$, where the divergence is taken with respect to any of the arguments of the tensor ${\rm curl}$.