 Solution

Write $x=(x_1,x_2,x_3{)}^T$, $v=(x_4,x_5,x_6{)}^T$ and $z=x+iv$. Here products of differentials in a metric tensor are symmetric products, unlike the wedge products used later. ThenIndeed $d{z}_{k}d{\bar{z}}_k=d{x}_k^2+d{x}_{k+3}^2$. A transformation $z^{\prime }=Uz$ preserves this metric whenever $U^{†}U=I$, becauseThus the entire unitary group $U(3)$ preserves it, including its special unitary group $SU(3)$.

To obtain a real matrix embedding, write $U=A+iB$ with $A,B$ real. ThenThe unitarity equations imply $A^{T}A+B^{T}B=I$ and $A^{T}B=B^{T}A$, hence $\rho (U{)}^T\rho (U)=I$. Multiplication of the block matrices gives $\rho (UV)=\rho (U)\rho (V)$, and the blocks recover $U$, proving injectivity. FinallyFor this last identity, over the complex numbers the change of basis from $(x,v)$ to $(z,\bar{z})$ conjugates $\rho (U)$ to $\mathrm{diag}(U,\bar{U})$. This proves that the image lies in the special orthogonal group.