# Prescribing the Lie derivative of the metric?

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?

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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...