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

Let us assume that $U$ is an $n\times n$ 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 $a_{pq},~b_{pq}\in\mathbb{R}$.

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 $i\ne j$, $i \leftrightarrow j$, the eq. (2) remains the same. This gives $\binom{n}{2}$ independent equations, therefore, $2\times\binom{n}{2}$ 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 $n+2\times\binom{n}{2}=n^2$.

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}

Answer 2

Lemma. A unitary $n \times n$ matrix has $n^2$ real independent parameters.

Proof. An arbitrary complex $n\times n$ matrix $U$ has $2n^2$ real independent parameters. The unitarity condition $U^\dagger = U^{-1}$ is equivalent with both $UU^\dagger = I$, and $U^\dagger U = I$ (using the definition of an inverse matrix). So any one of these conditions contains all constraints on $U$. We pick the last one. \begin{equation} U^\dagger U = I \end{equation} Taking the complex conjugate of the product on the left hand side shows that $(U^\dagger U)$ is an Hermitian matrix. \begin{equation} (U^\dagger U)^\dagger = U^\dagger (U^\dagger)^\dagger = U^\dagger U \end{equation} This may seem obvious since $I$ is clearly Hermitian, but that is not the point. Knowing that the combination $U^\dagger U$ is Hermitian, we want to know how much $U^\dagger U$ has to be constrained to not be just any Hermitian matrix, but the particular Hermitian matrix $I$.

Before such constraints are imposed, the product $U^\dagger U$ (with same matrix twice) has the same $2n^2$ parameter-space as $U$. Every subsequent constraint imposed on the parameters of $U^\dagger U$ must then correspond to a constraint on the parameters of $U$. The total number of independent parameter constraints to make an arbitrary Hermitian matrix $U^\dagger U$ into $I$ is simply all of its parameters.

From the condition for an $n \times n$ Hermitian matrix (i.e. that $H^\dagger = H$) we find the number of free parameters. On the diagonal we require that the entries are purely real, which sums to $n$ real parameters. The conjugate symmetry condition on $H$ fixes all lower triangle elements in terms of the upper triangle (or vise versa). The number of elements in the upper triangle is $n(n-1)/2$, and each entry has $2$ real parameters. Summing these terms gives us all independent parameters in a Hermitian matrix. \begin{equation} n + 2 \frac{n(n-1)}{2} = n^2 \end{equation} So we remove $n^2$ parameters from $2n^2$ and we are done. $\blacksquare$

References. Credit goes to Yan Gobeil whose proof I adapted to this.

Answer 3

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.