Is every variation field a Lie derivative of the metric? $\newcommand{\Vol}{\text{Vol}}$
Let $(M,g)$ be a smooth Riemannian manifold, and let $V \in \Gamma(T^*M \otimes T^*M)$ be symmetric.
Is it true that $V=L_Xg$ for some vector field $X \in \Gamma(TM)$?
I think the answer is negative, but I have no idea how to show this.
(This is a question about whether every section is "exact" in a suitable sense).
Edit:
The global question:
The answer is in general no. To understand the question better, let's separate into cases:


*

*$M$ is compact. Let $X \in \Gamma(TM)$, and denote by $\phi_t$ its flow. Define $g_t=\phi_t^*g$. Since $$\Vol(M,\phi_t^*g)=\Vol(M,g)=\text{const},$$


we get by differentiating that
$$0=\left.\frac{d}{dt}\Vol(M,\phi_t^*g) \right |_{t=0}= \frac{1}{2}\int_M \langle g,\left.\frac{\partial g}{\partial t}\right |_{t=0}\rangle \Vol_g=\frac{1}{2}\int_M \langle g ,L_x g \rangle \Vol_g.$$
So, a necessary condition for $V \in \{ L_xg \, | \, X \in \Gamma(TM) \}$
is $ \int_M \langle g ,V \rangle \Vol_g=0$.
Is this condition sufficient?


*What happens when $M$ is non-compact? Are there any necessary conditions?


The local question:
In dimension $1$ every variation field is realized in this way. In higher dimensions I guess that is not so; my heuristic is that there are different numbers of degrees of freedom: If $\dim M=d$, then pointwise $\dim(T^*M \otimes T^*M)=d^2$ while $\dim(TM)=d$.
 A: Locally I don't know.
Globally though I think it's false.
Here's why :
Consider $M=S^1 = [0,1]/\sim$.
Take local coordinates $x$ on $[0,1[$.
Consider $g= \mathrm dx \otimes \mathrm dx$ and $V=\mathrm dx\otimes \mathrm dx$.
Now, $X\in \mathfrak{X}(M)$ can be seen as $X=f\partial_x$ where $f(0)=f(1)$ (because $0\sim 1$ in $S^1$).
Now,
$$
\mathcal L_X (g) = \mathcal{L}_X(\mathrm dx \otimes \mathrm dx) \\
= (\mathcal{L}_X \mathrm dx)\otimes \mathrm dx + \mathrm dx \otimes (\mathcal{L}_X \mathrm dx) \\
= 2\partial_x f \mathrm dx
$$
Now we want $\mathcal L_X (g) = V$. That is $2\partial_x f = 1$. So we get $f = x/2 + C$. This function doesn't live on $S^1$.
Answer : No you cannot find $X$ in the global case.
Now, in the local case : in dimension 1 yes (that's easy to compute and solve). In dimension higher I don't know.
A: This answer addresses the local question. Your guess is right; a general variation of the metric cannot be realized as the Lie derivative with respect to a vector field. This makes sense as all the metrics $\phi_t^*g$ are the same in a way, where $\phi_t$ is the flow of a vector field $X$. "In a way" means here that they differ from one another by a diffeomorphism. 
As an example, take $M=\mathbb{R}^n$, and let $g$ be the standard metric. Let $X=X^i\frac{\partial}{\partial x^i}$ be a vector field. Using Cartan's formula, for every $1\leq i\leq n$ we have $$\mathcal{L}_Xdx^i=di_Xdx^i=dX^i=\frac{\partial X^i}{\partial x^j}dx^j.$$Next, we have $$\mathcal{L}_Xdx^i\otimes dx^i=\frac{\partial X^i}{\partial x^j}\left(dx^i\otimes dx^j+dx^j\otimes dx^i\right),$$which ultimately yields $$\mathcal{L}_Xg=\sum_{i,j}\frac{\partial X^i}{\partial x^j}\left(dx^i\otimes dx^j+dx^j\otimes dx^i\right).$$ Now, the only restriction we have on the matrix $$A_{ij}:=\frac{\partial X^i}{\partial x^j}$$ is that its rows are "closed", in the sense that for every $i,j,k$, we have $$\frac{\partial A_{ij}}{\partial x^k}=\frac{\partial A_{ik}}{\partial x^j}.$$ This means that, thinking of variations of the metric as maps from $\mathbb{R}^n$ to the space of symmetric matrices, the variations which can be obtained this way are $$\left\{\left.A+A^T\right|A\mathrm{\;has\;closed\;rows}\right\},$$where "closed rows" should be understood as above. As a simple check verifies, the above set is strictly smaller than the set of all variations, even when $n=2$.
