Transverse component of distorsion tensor in GR On pages 164-165 of Eric Gourgoulhon's lecture notes on Numerical Relativity, the author introduces the decomposition (9.49) for the distorsion tensor related to a foliation $(\Sigma_t)_{t\in \mathbb{R}}$ with induced metric $\gamma_{ij}$. This tensor is defined as
$$ Q_{ij}= \dfrac{\partial \gamma_{ij}}{\partial t} - \dfrac13 \gamma^{kl}\dfrac{\partial \gamma_{kl}}{\partial t} \gamma_{ij}.$$
Then, the author introduces the decomposition (9.49), i.e.
$$ Q^{ij}= (LX)^{ij} + Q^{ij}_{TT},$$
with $(LX)^{ij}= D^iX^j + D^j X^i - \dfrac23 D_k X^k \gamma_{ij} $, and $X$ is a vector field. By definition, the $LX$ (longitudinal) part of this decomposition is traceless. As $Q_{ij}$ is also traceless by definition, this makes $Q^{ij}_{TT}$ traceless as well. However, it is stated at the top of page 165 that $Q^{ij}_{TT}$ is also transverse, i.e. $D^iQ_{ij}^{TT}=0$.
How can I see that is statement is true? This decomposition first appeared in this paper, in which the author translates the transverse condition to a constraint on the vector field X. It does not automatically conclude that the transverse-traceless part is indeed transverse.
 A: The paper you are referring to proves that any symmetric tensor field can be decomposed into transverse-traceless, longitudinal and trace parts, eq. 2 and 7.
$$ \psi_{ab} = \psi_{ab}^{\rm TT} + \psi_{ab}^{\rm Tr} + \psi_{ab}^{L}$$
In the above,
$$ \psi_{ab}^{\rm Tr} = \frac{1}{3}\Psi g_{ab} = \frac{1}{3}\psi_{cd}g^{cd}g_{ab}\\ 
\psi_{ab}^{L} = \nabla_{a}W_{b} + \nabla_{b}W_{a} - \frac{2}{3}\nabla_{c}W^{c}g_{ab}
$$
and the transverse-traceless part is defined as:
$$ \psi_{ab}^{\rm TT}  = \psi_{ab} - \psi_{ab}^{\rm Tr} - \psi_{ab}^{L}$$
The transversality requirement on the TT part
$$ \nabla^{b}\psi_{ab}^{\rm TT} = 0  $$
is equivalent to the vector field in the longitudinal part satisfying a certain equation that involves the trace-less part of $\Psi_{ab}$.
$$ \nabla^{b}\Psi^{L}_{ab}= \nabla^{b}(\nabla_{a}W_{b} + \nabla_{b}W_{a} - \frac{2}{3}\nabla_{c}W^{c}g_{ab}) = \nabla^{b}(\Psi_{ab} -\frac{1}{3}\Psi g_{ab}) $$
If there exists a vector field $W^{a}$ such that the equation is satisfied, it means that the decomposition as outlined is possible.
To answer the last part of your question - demanding that $ \nabla^{b}\psi_{ab}^{\rm TT} = 0  $ leads to an equation for $W^{a}$. If such a $W^{a}$ exists (which is proven under certain assumptions) and we find such a $W^{a}$, we know the $\Psi^{L}_{ab}$, and thus, the TT part from  $ \psi_{ab}^{\rm TT}  = \psi_{ab} - \psi_{ab}^{\rm Tr} - \psi_{ab}^{L}$. The conclusion is that the original tensor field can indeed be decomposed into these parts.
By the transversality demand (satisfied once you find a suitable $W^{a}$) the transverse-traceless part is indeed transverse.
