If $\operatorname{div} X = 0$ what can be said about $X^\flat$?

170 Views Asked by At

If vector field $X$ is divergent free $$\operatorname{div} X = 0$$ what are the properties of a corresponding covector field $X^\flat$ (via musical isomorphism with a metric $g$)?

Are there some special relations involving $d \, X^\flat$, maybe with a volume form $\mu$, corresponding to $g$?

1

There are 1 best solutions below

0
On BEST ANSWER

As I learned it seems all that can be said about $X^\flat$ is that it is co-closed. Though it is merely a restatement of $\operatorname{div} X = * d * X^\flat = 0$ ($*$ is a Hodge star), there is more than just terminology because of substantial related theory.

Namely, operator $\delta : \Omega^{k}(M) \to \Omega^{k-1}(M)$ on $n$-manifold $M$

$$\delta = (-1)^{n(k+1)+1}*d*$$

has its own name --- codifferential and it is adjoint to $d$:

$$\int\langle d \alpha, \beta \rangle \operatorname{vol} = \int\langle \alpha, \delta \beta \rangle \operatorname{vol},$$

$\langle \,, \rangle $ is a scalar product and $\operatorname{vol}$ is the volume form. A form $\alpha$ is said to be co-closed if $\delta \alpha = 0$. For further information one is too search for "Hodge decomposition" and "Hodge theory".

So if $X$ is divergence-free, then $X^\flat$ is co-closed. As such, due to Poincare lemma $X^\flat$ locally can be represented as codifferential of some two-form $X^\flat = \delta \eta$. This is equivalent to $X = \operatorname{curl} A$ in vector calculus.