The tangent space of the moduli space of connection? I'm reading one of Floer's paper. (An Instanton-Invariant for 3-Manifold). Let $M$ be a $3$-manifold. A principal $SU_2$-bundle P over $M$ must be trivial. Fixed a trivialization $P \cong M \times SU_2$  and we can regard the space of connection as $su_2$-valued $1$-forms on $M$. Denote $L_1^4(\Omega^1(M)\otimes su_2)$ by $\mathcal{A}(M)$. The gauge group $L_1^4(M,SU_2)$ (denoted by $\mathcal{G}(M)$) atcs on $\mathcal{A}(M)$ by 
$$ g(a) = g a g^{-1} + (dg) g^{-1}.$$
Denote $\mathcal{A}(M)/\mathcal{G}(M)$ by $\mathcal{B}(M)$. For $a \in \mathcal{A}(M)$ such that $\operatorname{Fix}(a)$ is the center $\mathbb{Z}_2$. There is an identification for the tangent space
$$ T_{[a]} \mathcal{B}(M) = \{ \alpha \in L(\Omega(M) \otimes su_2) | d_a^\ast \alpha = 0 \}$$
where $d_a$ extend $d$ on $M$ by the connection $a$ and $d_a^\ast$ is its $L^2$-adjoint. 
My question is that how we can get the identification for the tangent space? Or is there any more detailed text about this type of objects? (Moduli spaces, Banach manifolds, etc.) 
Thank you.
 A: Whenever you have a quotient manifold $M/G$, $G$ some Lie group, if the tangent space to the orbit at $a$ is denoted $V_a$, you can identify the tangent space $T_{[a]}(M/G) \cong T_aM /V_a$. If $M$ further has a Riemannian metric, you can identify $V_a^\perp \cong T_a M /V_a$.
This case is no different. $\mathcal A$ is the space of connections on $E$, affine over the space $\Omega^1(\text{End}(E)) \cong \Omega^1(\mathfrak{su}(2))$ (because $E$ is trivial), and hence its tangent space at some point is $\Omega^1(\mathfrak{su}(2))$. The tangent space to the Lie group $\mathcal G$ at the identity is $\Omega^0(\mathfrak{su}(2))$. The Lie group action of $\mathcal G$ on $\mathcal A$ has derivative at $A$ equal to $\alpha \mapsto d_A \alpha$. So the image of this (the tangent space to the orbit) is equal to the image of $d_A$. So we can identify $T_{[A]} \mathcal B$ with $\Omega^1(\mathfrak{su}(2))/d_A\Omega^0(\mathfrak{su}(2))$. Now (as above, again) $\mathcal A$ has a Riemannian metric - $\langle \nu, \xi \rangle = \int \text{Tr}(\nu \wedge \ast \xi)$. (This is just to say it's the $L^2$ inner product with respect to the standard inner product on $\mathfrak{su}(2)$ - maybe up to a constant I'm forgetting). With respect to this inner product, the orthogonal complement of the image of $d_A$ is (more-or-less definitionally) the kernel of $d_A^*$. So we get our canonical identification $T_{[A]} \mathcal B = \text{ker } d_A^* \subset \Omega^1(\mathfrak{su}(2))$, as desired.
