This is a question that arises from my research problem. Suppose $(M,g)$ is a compact Riemannian manifold with boundary and $g$ is smooth up to the boundary (if you like, take $M$ to be diffeomorphic to a closed ball). Let $h$ be a symmetric traceless 2-tensor on $M$ (in my case this is actually the traceless Ricci $\stackrel\circ {Ric}$).
Now, I have shown that $\int_M \langle h, \mathcal L_Xg\rangle dV=0$ for any vector field $X$ on $M$. I wonder does it follow that $h=0$? (Well, may be not…)
Naturally, one would try to see if there is $X$ such that $\mathcal L_X g=h$. While this seems plausible, I have no idea when it will hold.
In case this is not always possible, instead one ask: Let $\epsilon>0$ and $p\ge 1$, does there exists functions $\lambda, \mu$ on $ M$ with $\lambda\ge 1$, and a vector field $X$ such that
$\|\mathcal L_X g -(\lambda h+ \mu g)\|_{L^p(M)}<\epsilon$?
Well, if they don't hold in general, can we impose conditions on $(M,g)$ so that one of them hold?
Best Answer
If $h\in S^2(TM^*)$ is a symmetric $(0,2)$-tensor such that $\int_M \langle h,\mathcal L_X g\rangle =0$ for all vector fields $X$ on $M$, then $h$ need not be zero. In fact, the above is precisely equivalent to $\delta h=0$, where $$\delta=\nabla^*\big|_{S^2(TM^*)} \colon S^2(TM^*)\to \Omega^1(M)$$ is the divergence operator, i.e., the formal $L^2$-adjoint of the covariant derivative $\nabla \colon \Omega^1(M)$ $\to S^2(TM^*)$.
The above claim follows from an infinitesimal version of Ebin's slice theorem for the pull-back action of the diffeomorphism group $\mathcal D:=Diff(M)$ on $S^2(TM^*)$, see this paper of Berger and Ebin, JDG '69. The statement I am referring to is the $L^2$-orthogonal decomposition: $$S^2(TM^*)=\ker \delta\oplus Im \ \delta^*.$$ Given any element $g\in S^2(TM^*)$, e.g., a Riemannian metric, the space $\ker\delta$ is the tangent space to the slice at $g$ to the $\mathcal D$-action and $Im\ \delta^*=T_g \mathcal D(g)$ is the tangent space to the orbit of $g\in S^2(TM^*)$. The latter is precisely formed by tensors of the form $\mathcal L_X g$, where $X$ is some field on $M$. In fact, consider $\phi_g\colon\mathcal D\to S^2(TM^*)$, $\phi_g(\eta)=\eta^*(g)$, and let $\eta_t\in\mathcal D$ be a curve of diffeomorphisms, with $\eta_0=id$ and $\dot\eta_0=X$. Then $$\mathrm d\phi_g(id)X=\frac{\mathrm d}{\mathrm dt}\phi_g(\eta_t)\big|_{t=0}=\frac{\mathrm d}{\mathrm dt}\eta_t^*(g)\big|_{t=0}=\mathcal L_X g=2\delta^*(X^b),$$ where $X^b=g(X,\cdot)$ is the $1$-form dual to the field $X$. The last equality follows from $\delta^*(\omega)=\tfrac12\mathcal L_{X_\omega}g$, where $X_\omega$ is the vector field dual to the $1$-form $\omega$. Note the above line also proves that $Im \ \delta^*=T_g\mathcal D(g)$.
Thus, going back to your original question, the above result of Ebin and Berger tells you that your symmetric traceless tensor $h$ is geometric, in the sense that it is tangent to the slice of the $\mathcal D$-action on $S^2(TM^*)$, or, equivalently, $L^2$-orthogonal to the tangent space to the $\mathcal D$-orbit. In some sense, this means that it ``descends'' to an object in the quotient space of tensors modulo diffeomorphisms (where, e.g., the moduli space of Riemannian metrics should live).
The above observations also answer your question regarding what symmetric $(0,2)$-tensors are of the form $\mathcal L_Xg$; namely, they are precisely the ones in $Im \ \delta^*$. Hope this helps...