Number of independent components of a unitary matrix

Question

By definition, a $n$ dimensional unitary matrix $U$ satisfies the condition $U^{\dagger}U=I$, and $UU^{\dagger}=I$. I'd like to ask if these two equations are independent. If so, there will be $n^2$ independent equations of constrain, which is equal to the number of independent components of a general $n$ dimensional matrix. This is obviously impossible. If not, how to prove it?

Answer 1

You can remember that follows: the $n^2$ complex relations $U^*U=I$ are not algebraically independent because both matrices $U^*U$ and $I$ are hermitian. If we consider the upper triangular parts of these matrices (included diagonals), then we obtain $n^2$ real equalities (write them for $n=2$). The previous equalities are algebraically independent; that implies that a unitary matrix depends on $n^2$ real parameters.

In the body of text, I change the notation $SU_n$ with the standard one $U_n$.

Let $U_n=\{U=[u_{i,j}]\in M_n(\mathbb{C})|UU^*=I_n\}$ (the other equality $U^*U=I_n$ is useless). $f:U\rightarrow UU^*-I$ is not an algebraic function in the $(u_{i,j})$; yet, it is an algebraic function in the $2n^2$ real variables $Re(u_{i,j}),Im(u_{i,j})$. Then $U_n$ is a real algebraic set defined by $f(U)=0$; note that $U_n$ is a group; then it suffices to study $U_n$ in a neighborhood of $I_n$. One has $Df_I:H\rightarrow H+H^*$; then the tangent space of $U_n$ in $I_n$ is $\{H|Df_i(H)=H+H^*=0\}$, that is the vector space $SH$ of skew-hermitian matrices. The number of real independent parameters defining $U_n$ is the dimension over $\mathbb{R}$ of $SH$, that is $2\dfrac{n(n-1)}{2}+n=n^2$.

Note that $f(U)=0$ is equivalent to $n^2$ real relations; that shows that $U_n$ depends at least on $2n^2-n^2=n^2$ parameters. That is above shows that theses $n^2$ relations are algebraically independent.