Dtyjlui

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?

@ Wen Chern , since you do not know that $U^*U=I$ implies that $UU^*=I$, I think that you are not an eagle in the mathematical field. Since I have no response to my answer, I think that you did not understand one word of it. 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). $f:Urightarrow UU^*-I$ 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 $Df_I:Hrightarrow H+H^*$; 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 $2dfrac{n(n-1)}{2}+n=n^2$.

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.

Let us assume that $U$ is an $ntimes n$ unitary matrix, i.e.,

begin{equation}
U^dagger U=I~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)
end{equation}

The total number of entries in a unitary matrix is $n^2$ and the total number of real parameters is $2n^2$.

Let us further assume that $z_{pq}=a_{pq}+ib_{pq}$ where $a_{pq},~b_{pq}inmathbb{R}$.

From the equation (1), one can write

begin{eqnarray}
sum_{k=1}^n z_{ik}^dagger z_{kj}&=&delta_{ij}\
sum_{k=1}^n bar z_{ki} z_{kj}&=&delta_{ij}~~~~~~~~~~~~~~~~(2)
end{eqnarray}

For $i=j$, the eq. (2) reduces to

begin{equation}
sum_{k=1}^n |z_{ki}|^2=1~~~~~~~~~~~~~~~~~~~~~(3)
end{equation}

Therefore, for $i=1,2,...,n$, eq.(3) represents $n$ independent real conditions.

For if $ine j$, $i leftrightarrow j$, the eq. (2) remains the same. This gives $binom{n}{2}$ independent equations, therefore, $2timesbinom{n}{2}$ real conditions:

begin{eqnarray}
sum_{k=1}^n (a_{ki}a_{kj}+b_{ki}b_{kj})&=&0\
sum_{k=1}^n (a_{ki}b_{kj}+a_{kj}b_{ki})&=&0.
end{eqnarray}

Total number of independent real conditions is $n+2timesbinom{n}{2}=n^2$.

Therefore, the total number of independent real parameters (components) $=$ Total number of real parameters $-$ total number of independent real conditions.

begin{equation}
N=2n^2-n^2=n^2.
end{equation}