I am wondering if it is possible to parametrize the space of all unitary 3 by 3 matrices with complex entries.

$\displaystyle U=\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&j\end{ array}\right)$ with U being any unitary 3 by 3 matrix.

We know that the column vectors have to be orthonormal such that we have the unitary property $\displaystyle U^*U=1$.

So let $\displaystyle a=a_1+ia_2, ~ d=d_1+id_2, ~ g=g_1+ig_2$ then we can write

$\displaystyle a^2_1+a^2_2+d^2_1+d^2_d+g^2_1+g^2_2=1$

this is just a sphere in $\displaystyle \matbb{R}$ and we can easily parametrize it as follows

$\displaystyle \begin{array}{ccc}

g_2&=&\cos\phi_1\\

g_1&=&\sin\phi_1\cos\phi_2\\

d_2&=&\sin\phi_1\sin\phi_2\cos\phi3\\

d_1&=&\sin\phi_1\sin\phi_2\sin\phi_3\cos\phi_4\\

a_2&=&\sin\phi_1\sin\phi_2\sin\phi_3\sin\phi_4\cos \phi_5\\

a_1&=&\sin\phi_1\sin\phi_2\sin\phi_3\sin\phi_4\sin \phi_5

\end{array}$

using these we can find the corresponding parametrization of the first column. My question is how would we parametrize the remainder of the matrix?

If $\displaystyle v_1=\left(\begin{array}{c}a\\d\\g\end{array}\right )$ and $\displaystyle v_2,v_3$ are respectively the second and third columns of U, then we know that

$\displaystyle v_i\cdot v_j=0$ if $\displaystyle i\neq j$ and 1 otherwise

we would have to consider the space orthogonal to $\displaystyle v_1$, this space must be some two dimensional plane. The vectors $\displaystyle v_2,v_3$ must then lien in this plane.

I can solve this problem for the case of two by two matrices with complex entries and wanted to see if it was at all possible for higher dimensions, however the parametrization is a little tricky.