Let $M$ be a $3\times 3$ matrix:
$$ M=\pmatrix{A_x & A_y & A_z\\B_x& B_y&B_z\\ C_x&C_y&C_z} $$
A constraint of $SU(3)$ is that $\det{M}=1$. Therefore:
$$ A_x(B_yC_z-B_zC_x)+A_y(B_zC_x-B_xC_z)+A_z(B_xC_y-B_yC_x)=1 $$
Interestingly, this constraint is equivalent to the cross-product with the inner product being equal to 1:
$$ C\cdot(A\times B)=1 $$
A second constraint is inherited from the $U(3)$ group, $MM^*=I$. Consequently;
$$ M=\pmatrix{A_x & A_y & A_z\\B_x& B_y&B_z\\ C_x&C_y&C_z} \pmatrix{A_x^* & B_x^* & C_x^*\\A_y^*& B_y^*&C_y^*\\ A_z^*&B_z^*&C_z^*}=\pmatrix{1&0&0\\0&1&0\\0&0&1} $$
which generates three equations:
$$ A_xA_x^*+A_yA_y^*+A_zA_z^*=1\\ B_xB_x^*+B_yB_y^*+B_zB_z^*=1\\ C_xC_x^*+C_yC_y^*+C_zC_z^*=1 $$
Expanded as $A_x=a_x+id_x$, etc., these three equations become 5-sphere constraint:
$$ a_x^2+d_x^2+a_y^2+d_y^2+ a_z^2 + d_z^2=1\\ b_x^2+e_x^2 + b_y^2+e_y^2+b_z^2+e_z^2 =1\\ c_x^2+f_x^2 + c_y^2+f_y^2 + c_z^2+f_z^2=1 $$
Is it the case that $U(3)$ describes three 5-spheres? If not, what constraint am I missing to lower the number?
However, I have read (on wikipedia) that $SU(3)$ is the invariance group of the 5-sphere. Consequently, am I correct to understand that the constraint $\det M=1$ reduces the three 5-sphere of $U(1)$ to a single 5-sphere. If so, can someone show how this happens?