I would like to solve the following:
An element $ g \in SL(4,\mathbb{C})$ acts on the 6-dimensional complex vector space $( \mathbb{C}^6 )$ via: $A → gAg^t$ for $A \in \mathfrak{so}(4,\mathbb{C})$. Show that this induces a Lie group homomophism φ : $SL(4,\mathbb{C}) → O(6,\mathbb{C})$.
We identify the complex vector spaces $\mathbb{C}^6$ and $\mathfrak{so}(4,\mathbb{C})$ = $\left\{ A \in \mathfrak{gl}(4,\mathbb{C}) \ | \ A^t = -A \right\}$ by:
$$ \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ y_1\\ y_2\\ y_3\\ \end{array} \right] \leftrightarrow \left[ \begin{array}{cccc} 0& x_3 - iy_3 & -x_2 + iy_2 & x_1 + iy_1\\ -x_3 + iy_3 & 0 & x_1 - iy_1 & x_2 + iy_2\\ x_2 - iy_2 & -x_1 + iy_1 & 0 & x_3 + iy_3 \\ -x_1 - iy_1 & -x_2 - iy_2 & -x_3 - iy_3 &0\\ \end{array} \right] $$
Then the standard bilinear form in $\mathbb{C}^6$ (where $||v||^2 = v^2_1 +···+ v^2_6$) is equal to the Pfaffian: $ Pf(A) = A_{12}A_{34} + A_{31}A_{24} + A_{23}A_{14}$ for $ A \in \mathfrak{so}(4,\mathbb{C}).$
We may use (without proof) the following properties of the Pfaffian for $ B \in \mathfrak{so}(4,\mathbb{C})$:
$$ • Pf(B^t) = (−1)^nPf(B).\\ • Pf(λB) = λ^nPf(B).\\ • (Pf(B))^2 = detB. \\ • Pf(CBC^t) = detC ·Pf(B)$$ for any $C ∈\mathbb{C}^{2n×2n}.$
The second part of this question is showing that the kernel of the above homomorphism is $\left\{ \pm I_n \right\}$
I would be very happy to receive some guidance.