IF a 2-Form is Closed, is its Contraction along any Vector Field also Exact?

Question

If a 2-form $\omega \in \Omega^n(M)$ is closed, is its contraction $\theta = \iota_X\omega$ also closed for any smooth vector field $X$ on $M$?

Answer 1

As the comments say, the answer is no in general. Here is a simple example, done explicitly (mostly for my own practice, but I hope it's useful to see calculations in coordinates).

Consider $M=\mathbb{R}^3-\left\{(0,0,0)\right\}$, and take $\omega$ to be the pullback of the volume form $\sin\phi\,d\phi\wedge d\theta$ of the unit sphere $S^2$ under the map

$$\alpha:M\to S^{2}, \alpha(p)=\frac{p}{\|p\|}$$

Then $\omega$ is closed because the pullback commutes with exterior differentiation, and of course the volume form on the sphere is closed. Explicitly, with $r=\|p\|$ for $p\in M$,

$$\omega=\frac{1}{r^3}\left(z\,dx\wedge dy+y\,dx\wedge dz+x\,dy\wedge dz\right)$$

Now take $X$ to be the field $X:M\to M$ defined by $$X(p)=\frac{p}{\|p\|}=\frac{1}{r}\left(x\,\partial_x+y\,\partial_y+z\,\partial_z\right)$$

Then, with $X_1=x_1\partial_X+y_1\partial y+z_1\partial_z$,

$$\theta(X_1)=\iota_X\omega(X_1)=\omega(X, X_1)=\frac{1}{r^4}\big(z(xy_1-yx_1)+y(xz_1-zx_1)+x(yz_1-zy_1)\big)$$

so

$$\theta=\frac{1}{r^4}\left(-2yz\,dx+2xy\,dz\right)$$

whence

$$d\theta=-2d\left(\frac{yz}{r^4}\right)\wedge dx+2d\left(\frac{xy}{r^4}\right)\wedge dz$$

This can't vanish; the first term will have a nonzero multiple of $dx\wedge dy$, for example, but there is no multiple of $dx\wedge dy$ in the second term to cancel it.

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Intuitively, recall that Cartan's formula says

$$L_X\omega=d(\iota_X\omega)+\iota_X(d\omega)$$

The second term on the right vanishes since $\omega$ is closed. But for $X$ and $\omega$ in our example, the Lie derivative doesn't vanish: as we flow along $X$ the form $\omega$ is changing.