Setting of the problem
Let $Q \subset \mathbb{R}^d$ a regular bounded domain. We denote
$$L^2_{sol}(Q)=\{v \in (L^2(Q))^2 \ | \ \int_Q v \cdot \nabla \varphi \ \mathrm{d}x=0 \quad \forall \varphi \in (H^1_0(Q))^3 \}$$ and $$L^2_{pot}(Q)=\{v \in (L^2(Q))^d \ | \ \int_Q v_j \partial_i \varphi - v_i \partial_j \varphi \ \mathrm{d}x=0, \quad \forall \varphi \in H^1_0(Q), \quad 1 \leq i,j \leq d \}$$ i.e the space of $L^2(Q)$ vector with $div_{H^{-1}(Q)}=0$ and $rot_{H^{-1}(Q)}=0$ respectively. We also denote $\nu^2_{pot}(Q)$ the functions of $L^2_{pot}(Q)$ with zero average on $Q$. It is well known that the following orthogonal representation holds :
$$(L^2(Q))^d= \nu^2_{pot}(Q) \oplus L^2_{sol}(Q)$$
My question
I was wondering if there exists a similar decomposition for the space
$$A= \{v \in (L^2(Q))^2 \ | \ \int_Q v \cdot \nabla \varphi \ \mathrm{d}x=0 \quad \forall \varphi \in (H^1(Q))^3 \}$$ which is almost the same as $L^2_{sol}(Q)$ except the test function are now in $(H^1(Q))^3$ rather than $(H^1_0(Q))^3$ . I'm also wondering why would one consider $A$ rather than $L^2_{sol}(Q)$ as a functional space.
Any suggestion is welcomed !