How can even there be a non-zero BMS vector field with zero asymptotic data? Following the BMS approach, one spacetime $(M,g)$ is asymptotically flat when:


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*We can find a Bondi gauge set of coordinates $(u,r,x^A)$ characterized by $$g_{rr}=g_{rA}=0,\quad \partial_r\det\left(\dfrac{g_{AB}}{r^2}\right)=0\tag{1}.$$

*The range of the $r$ coordinate is $r_0\leq r < +\infty$ and the $x^A$ coordinates parameterize a two-sphere $S^2$

*The metric has asymptotic behavior (where $\gamma_{AB}$ is the $S^2$ round metric) $$g_{uu}=-1+O(r^{-1}),\quad g_{ur}=-1+O(r^{-2})\quad g_{uA}=O(1),\\\quad g_{AB}=r^2\gamma_{AB}+O(1)\tag{2}.$$
In that scenario a BMS vector field is a vector field in $(M,g)$ which preserves (1) and (2) when we vary the metric as $\delta g = L_X g$. The space of all such vectors is then the $\mathfrak{bms}_4$ algebra.
It is possible to show that such a vector field is identified by a pair $(f,Y)$ where $f\in L^2(S^2)$ and $Y$ is a Conformal Killing Vector on $S^2$ such that its leading behavior is: \begin{align}    X &= \left(\frac{u}{2}D_A Y^A + f\right)\partial_u + \left(-\frac{r}{2}D_A Y^A -\frac{u}{2} D_A Y^A +\frac{1}{2}D_A D^A f + O(r^{-1})\right)\partial_r\\ & + \left(Y^A -\frac{D^A f + \frac{u}{2}D^A (D_B Y^B)}{r}+O(r^{-2})\right)\partial_A.\tag{3}\end{align}
Moreover preservation of (1) still demands two conditions. Preservation of $g_{rA}=0$ demands: $$\partial_r X^A = -g_{ur}g^{AB}\partial_B X^u \tag{4}$$
and preservation of the determinant condition demands $g^{AB}L_X g_{AB} =0$ which becomes:
$$X^r g^{AB}\partial_r g_{AB}=-\bigg(X^u g^{AB}\partial_u g_{AB}+X^C g^{AB}\partial_C g_{AB}+2 g^{AB}\partial_A X^u g_{uB}+2g^{AB}\partial_A X^C g_{CB}\bigg)\tag{5}$$
which in effect fully determines $X^r$.
Now in "Advanced Lectures on General Relativity" the author says that "Trivial boundary diffeomorphisms $f=Y^A=0$ form an ideal this algebra". 
But why isn't the set $f = Y^A=0$ comprised of just the zero vector? I mean if $f = Y^A = 0$ then $X^u =0$. If $X^u = 0$ then (4) implies that $X^A = Y^A$ and therefore $X^A =0$. Finally using $X^u,X^A = 0$ in (5) implies that, since $g^{AB}\partial_r g_{AB}\neq 0$, we have $X^r = 0$.
What am I missing here? What is my misunderstanding? How can there be BMS vector fields, preserving (1) and (2), with $f = Y^A =0$ which are not identically zero?
 A: Here is the problem:

It is possible to show that such a vector field is identified by a pair $(f,Y)$ where …

vector field $X$ is not uniquely identified by this pair $(f,Y)$, only its action on the boundary data, while the vector field itself is defined not only near the boundary but everywhere “inside” the manifold. In other words, this pair $(f,Y)$ says nothing about behavior of $X$ at finite values of radial coordinate, only about its asymptotic behavior. 
For example we can choose $X$ to be arbitrary for $r<r_1$ (for some $r_1<\infty$) and identically zero for $r_2 < r <\infty$ (with $r_2>r_1$), while for $r_1<r<r_2$ we can choose $X$ interpolating between two behaviors to satisfy the necessary smoothness conditions. 
It is easy to see that for such example vector field the pair $(f,Y)$ would be zero. Thus this $X$ is a nontrivial example of the generator of a trivial boundary diffeomorphism. As an aside, such construction is related to so called Einstein's hole argument.
Also note, that fields $X$  corresponding to trivial boundary diffeomorphisms do not have to be identically zero in some vicinity of the boundary, they just have to approach zero fast enough to not alter the boundary data. 
