Computing the Cartan1-form for $Sp(2)$ Context:
I must find the Cartan 1-form for $Sp(2)$ before I start dealing with the natural connection of the Hopf fibration $S^3 \hookrightarrow S^7 \overset{\mathcal P}\to S^4$. To do so, the idea is to look at the restriction of the Cartan 1-form of $G = GL(2,\mathbb H)$ to $Sp(2)$. Now if $g = \begin{pmatrix} a & b \\c & d\end{pmatrix}\in GL(2,\mathbb H)$ is any given matrix, this 1-form is written as
$$\Theta_G(g) = g^{-1} dq$$
Question: (Here I will adress to the question itself by showing what I have done so far). 
Using the Schur complement for $g$ we may write $g^{-1}$, assuming $a \neq 0$, as follows
$$g^{-1} = \begin{pmatrix}a^{-1} + a^{-1}b a_s^{-1}ca^{-1} & -a^{-1}b a_s^{-1} \\ -a_s^{-1}ca^{-1} & a_s^{-1}\end{pmatrix}$$
where $a_s^{-1} = d - ca^{-1}b$. Now, if $g \in Sp(2)$ then $g^{-1} = \bar g^T$, that is, 
$$\begin{pmatrix}a^{-1} + a^{-1}b a_s^{-1}ca^{-1} & -a^{-1}b a_s^{-1} \\ -a_s^{-1}ca^{-1} & a_s^{-1}\end{pmatrix} = \begin{pmatrix} \bar a & \bar c \\ \bar b & \bar d\end{pmatrix}$$
which in turn gives 
$$\begin{cases}a^{-1} + a^{-1}b a_s^{-1}ca^{-1} &= \bar a \\-a^{-1}b a_s^{-1}  &= \bar c\\-a_s^{-1}ca^{-1} &= \bar b \\ a_s^{-1} &= \bar d \end{cases}\tag{*}$$
Since $g$ is invertible $a_s$ is nonzero and we may write $a_s^{-1} = \frac{\bar a_s}{|a_s|^2}$. Well, by $(*)$
$$\bar d = a_s^{-1} \implies d = \overline{a_s^{-1}} = \frac{a_s}{|a_s|^2}$$
Then, $d = \frac{d - ca^{-1}b}{|d-ca^{-1}b|^2}$, and this is true only if $ca^{-1}b = 0$ and $|d|^2 = 1$, which gives that either $c=0$ or $b=0$ and $|d|^2 = 1$, and again by $(*)$, implies that $c = b = 0$ and $|d|^2=1$. Now 
$$g^{-1}dq = \begin{pmatrix}\bar a & 0 \\ 0 & d\end{pmatrix}\begin{pmatrix}dq^{11} & dq^{12}\\ dq^{21} & dq^{22}\end{pmatrix} = \begin{pmatrix} \bar a\  dq^{11} & \bar a\  dq^{12}\\ d\  dq^{21} & d \ dq^{22}\end{pmatrix} $$ 
We have that any element of $T_g (Sp(2))$ can be written as $\eta'(0)$, where $\eta$ is a smooth curve in $Sp(2)$ with $\eta(0)=g $. In terms of the coordinates $q^{ij}$ we have
$$(\iota \circ \eta) (t) = \begin{pmatrix}q^{11}((\iota \circ \eta)(t)) & dq^{12}((\iota \circ \eta)(t))\\ dq^{21}((\iota \circ \eta)(t)) & dq^{22}((\iota \circ \eta)(t))\end{pmatrix} = \begin{pmatrix}a (t) & 0 \\ 0 & d(t)\end{pmatrix}$$
But by $g\bar g^t = id$ we have $a (t) \bar a(t) = 1$, differentiating at $t= 0$ we obtain $a'(0) \bar a(0) + a(0)\bar a'(0) = 0$, so 
$$a'(0) \bar a + a \bar a '(0) = 0 $$
From which we may conclude that $\mathrm{Re} (\bar a'(0)) = 0$. We remember that $a'(0) = dq^{11} (\iota_{\ast g} (\eta'(0)))$. So 
$$\mathrm{Re} (\bar a\  dq^{11} (\iota_{\ast g} (\eta'(0)))) = \mathrm{Re} (\bar a a'(0)) = 0$$
In other words the restriction $\Theta_H(g) = \iota^* \Theta_G(g)$ is the restriction of $\mathrm{Im} (\bar q^{11}dq^{11})$ to $Sp(2)$. 
Problems:
1) This is not the general answer, because it was supposed to be the restriction of $\mathrm {Im}(\bar q^{11} dq^{11} + \bar q^{21} dq^{21})$;
2) In the case $GL(2,\mathbb C)$ I had no problem showing this, due to the much friendlier aspect of the matrix $g^{-1}$, and everything seemed to go according to plans. This case though I couldn't work it through. 
3) This is not some exercise, I'm actually working on this. This is something I need to prove to get a result. 
Could someone, please, give me some insight, maybe point me in the right direction? 
what should I do?
 A: Here is the answer. 
For each $g = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \in G = GL(2,\mathbb H)$, $\boldsymbol \Theta_G (g)$ is given by $g^{-1}dq$. Now if $g \in Sp(2)$ then $g^{-1} = \bar g^T$ and $\color{red}{\bar g^T g = id}$ (this is what I was missing). In particular, $\bar \alpha\alpha + \bar \gamma \gamma = 1$. Thus we have
$$\begin{aligned}g^{-1} dq & = \begin{pmatrix}\bar \alpha & \bar \gamma \\ \bar \beta & \bar \delta \end{pmatrix} \begin{pmatrix}dq^{11} & dq^{12} \\ dq^{21}& dq^{22}\end{pmatrix}\\&= \begin{pmatrix}\bar \alpha \  dq^{11} + \bar \gamma \  dq^{21} & \bar \alpha \ dq^{12} + \bar \gamma \ dq^{22} \\ \bar \beta \ dq^{11} + \bar \delta \  dq^{21} & \bar \beta\ dq^{12} + \bar \delta\ dq^{22}\end{pmatrix}\end{aligned}$$
Any element of  $T_g(Sp(2))$ can be written as $\eta'(0)$, where $\eta$ is a smooth curve in  $Sp(2)$ such that $\eta (0) = g$. Now, $\iota_{\ast g}(\eta'(0)) = (\iota \circ \eta)'(0)$
where 
$$(\iota \circ \eta)'(t) = \begin{pmatrix}q^{11}((\iota \circ \eta)(t)) & q^{12}((\iota \circ \eta)(t)) \\ q^{21}((\iota \circ \eta)(t)) & q^{22}((\iota \circ \eta)(t))\end{pmatrix} = \begin{pmatrix}\alpha(t) & \beta(t) \\  \gamma (t) &  \delta (t)\end{pmatrix}$$
with $\bar\alpha (t) \alpha (t) + \bar \gamma (t)\gamma(t) = 1$ for all $t$. Differentiating this last equation at $t=0$ we obtain 
$$\begin{aligned}\bar \alpha' (0)\alpha (0) & + \bar\alpha (0) \alpha'(0)+ \bar \gamma' (0)\gamma (0) + \bar \gamma (0)\gamma'(0) = 0\\&\implies \bar \alpha' (0) \alpha  + \bar \alpha \alpha'(0)+ \bar \gamma' (0)\gamma + \bar \gamma \gamma'(0) = 0\\&\implies \overline{\bar \alpha \alpha'(0)} + \bar \alpha \alpha'(0) + \overline{\bar \gamma \gamma'(0)} + \bar \gamma \gamma'(0) = 0 \\&\implies \mathrm{Re} (\bar \alpha \alpha'(0) + \bar \gamma \gamma'(0)) = 0\end{aligned}$$ 
Moreover, $\alpha'(0) = dq^{11} (\iota_{\ast g}(\eta'(0)))$ and $\gamma'(0) = dq^{21} (\iota_{\ast g}(\eta'(0)))$, thus, 
$$\mathrm{Re}(\bar \alpha \  dq^{11}(\iota_{\ast g}(\eta'(0))) + \bar \gamma\  dq^{21} (\iota_{\ast g}(\eta'(0)))) = \mathrm {Re} (\bar \alpha \alpha'(0) + \bar \gamma \gamma'(0)) = 0 $$ 
We conclude that $\bar \alpha \  dq^{11} + \bar \gamma \  dq^{21}$ is pure imaginary in $\iota_{\ast g}(\eta'(0))$. We have shown, therefore, that the 11-entry of the Cartan 1-form of $Sp(2)$ is the restriction of $\mathrm{Im}(\bar q^{11}dq^{11} + \bar q^{21}dq^{21})$ to $Sp(2)$.
