Helicity integral in differential forms Let $V^{3}(t)$ be a compact region moving with the fluid.
Assume that at $t=0$ the vorticity $2$-form $\omega^{2}$ vanishes when restricted to the boundary $\partial V^{3}(0)$; that is, $i^{*}\omega^{2}=0$, where $i$ is the inclusion of $\partial V$ in $\mathbb{R}^{3}$.
(This does $\textit{not}$ say that $\omega^{2}$ itself vanishes, rather only that $\omega({\vec{u}},{\vec{w}})=0$ for $\vec{u}$,$\vec{w}$ tangent to $\partial V^{3}(0)$.)
Then the $\textbf{helicity}$ integral $\displaystyle{\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx\wedge dy\wedge dz}$ can be constant in time.

How you actually prove that the helicity integral is constant in time?

Here's my attempt:
$$\frac{d}{dt}\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx\wedge dy\wedge dz$$
$$= \frac{d}{dt}\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx^{1}\wedge dx^{2}\wedge dx^{3}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} {\vec{v}}\cdot{\vec{\omega}}\ \text{vol}^{3}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu^{1}\wedge \omega^{2}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu^{1}\wedge {\rm d}\nu^{1}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu\wedge {\rm d}\nu$$
$$= \frac{1}{o(x)\sqrt{g}(x)}\frac{1}{2} \frac{d}{dt}\int_{W(t)} (\nu\wedge {\rm d}\nu)+(\nu\wedge {\rm d}\nu)$$
$$= \frac{1}{o(x)\sqrt{g}(x)}\frac{1}{2} \frac{d}{dt}\int_{W(t)} (\nu\wedge {\rm d}\nu)-({\rm d}\nu\wedge \nu)$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{W(t)} {\rm d}(\nu\wedge \nu)$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\oint_{\partial W(t)} \nu\wedge \nu$$
$$= 0.$$
I think this is wrong because I haven't used the assumption that at $t=0$ the vorticity $2$-form $\omega^{2}$ vanishes when restricted to the boundary $\partial V^{3}(0)$; that is, $i^{*}\omega^{2}=0$, where $i$ is the inclusion of $\partial V$ in $\mathbb{R}^{3}$.
 A: I assume that you are trying to solve problem 4.3(5)(v) in Theodore Frankels' Geometry of Physics (adding this for metadata).

First of all:
$\nu \wedge d \nu = d \nu \wedge \nu$, not $-d\nu \wedge \nu$, because $d \nu$ is a 2-form.

Here is my attempt at solution (unfinished). Some facts first:


*

*$d \omega^2 = d \; d \nu = 0$

*$\frac{d}{dt} \omega^2 = 0$, i.e. $\omega^2$ is invariant under the flow. (By Helmholtz's theorem)

*$\frac{d}{dt} \nu = d(\frac{1}{2}|v|^2 + \phi - \int \frac{dp}{\rho}) \equiv d \psi$,
which is just Euler equation ($\psi$ is introduced for later convenience).


The unfinished solution:
By (2):
$$
\frac{d}{dt} \int_{V(t)} \nu \wedge \omega^2 =
\int_{V(t)} \frac{d}{dt} \nu \wedge \omega^2 +
  \nu \wedge \frac{d}{dt} \omega^2 =
\int_{V(t)} \frac{d}{dt} \nu \wedge \omega^2 = \cdots
$$
By (3) and (1):
$$
\cdots = \int_{V(t)} d \psi \wedge \omega^2 =
\int_{V(t)} d(\psi \wedge \omega^2) - \psi \wedge d \omega^2 =
\int_{V(t)} d(\psi \wedge \omega^2) = \cdots
$$
Now using Stokes' Theorem:
$$
\cdots = \int_{\partial{V}(t)} \psi \wedge \omega^2
$$
Given that we have to pullback $\omega^2$ to the boundary, at $t = 0$ this
integral is $0$. I have yet to prove that it stays $0$ at all times, this is
why this solution is incomplete.
