Does the Lie derivative of a harmonic form with respect to a killing vector field vanish

Question

On the wikipedia page for Killing vector fields under 'properties' it is stated that if $X$ is a Killing vector field and $\omega$ is a harmonic differential form then $\mathcal{L}_{X}\omega=0$. https://en.wikipedia.org/wiki/Killing_vector_field#Properties

It seems like this should be true but I can't find it stated anywhere else except for the online previews of a couple of books which I don't have access to. I have tried to show this using the formula for the Lie derivative of a form $\mathcal{L}_{X}\omega=d(i_{X}\omega)+i_{X}(d\omega)$ along with the co-closure condition on $\omega$ and Killing's equation but I'm not really sure where to go.

So, my question is: is this true and if so, what direction should I go in to try and show it?

Answer 1

Here some facts those lead you for a solution:

For any p-form $\omega$ follows:

1. $d(L_X\omega)=L_X(d\omega)$.
2. $L_X(*\omega)=(\text{div} X)*\omega+*[(L_Xg)\cdot\omega]+*(L_X\omega)$. (See here)
3. $\text{div}X=\frac{1}{2}\text{tr}(L_Xg)$. Use definition of divergence and Lie derivative. (If you prefer, here a solution)

Now, if $X$ is Killing, then itens 2 and 3 gives

4. $L_X(*\omega)=*(L_X\omega)$

But, by definition, when $(M,g)$ is compact and oriented, $\delta:=(-1)^{n(p-1)+1}*d*$, thus, from itens 1 and 4 we obtain

5. $\delta(L_X\omega)=L_X(\delta\omega)$

Since $\Delta=d\delta+\delta d$, we have from itens 1 and 5 for a Killing field

6. $\Delta(L_X\omega)=L_X(\Delta\omega)$

Now, for $\omega$ harmonic form, from item 6:

7. $\Delta(L_X\omega)=0$. Thus, $L_X\omega$ is a harmonic form as well.

Last step: Show that item 7 implies that $L_X\omega$ is equal zero.

Answer 2

For completing the proof consider the following lemma:

Lemma: If $\psi$ is exact (i.e. $\psi=d\eta$) and coclosed (i.e. $\delta\psi=0)$ then $\psi =0$.

This is easy to prove: Consider the induced inner product $\langle\!\langle., .\rangle\!\rangle$ that $d$ and $\delta$ are adjoint (i.e. $\langle\!\langle dA, B\rangle\!\rangle= \langle\!\langle A, \delta B\rangle\!\rangle$) w.r.t. that inner product. Then compute $\|\psi\|^2=\langle\!\langle\psi, \psi\rangle\!\rangle$.

(Hint for rest of proof: ${\cal L}_X\omega=d(i_X(\omega))+0=d(\text{something})\implies$ ${\cal L}_X\omega$ it is exact.).

Answer 3

Others have given a proof when the manifold is compact. The result is false in general if the manifold is non-compact. Consider for instance $$f = x^2 - y^2$$ in $\mathbb{R}^2$. Then $f$ is not preserved by any non-zero Killing field. Nor is the harmonic 1-form $\omega = df$ (which is also exact and non-zero).