# Degree of freedom of Unitary and Skew-Hermitian matrix (solved)

Reading an article (A. Veneziani, T. Pereira, A note on the volume form in normal matrix space) I didn't understand this sentence:

"Given a unitary matrix $U$ we may use the representation $U=e^W$ where $W$ is a skew-hermitian matrix $W=-W^*$. For our case, W will be a skew-hermitian matrix which zero diagonal. This is to equalise the degrees of freedom of $U$."

I don't understand why we have to force the elements of the diagonals to become zero. The degrees of freedom of a $n*n$ unitary matrix are $n^2$ and the same is true for skew-hermitian matrix ($n(n-1)+n$, being the diagonal elements imaginary).

Oh, I understood! The author did not consider the unitary group, but the unitary group minus the n-torus. My bad sorry, thanks any way!

• Surely you can not select all element of a unitary matrix at random? You must have $U*U = I$. Regardless, it would be good to edit the question to include the title of the paper and a link. – Carl Christian May 22 '18 at 7:48
• @CarlChristian Well the degrees of freedom of a unitary matrix are $n^2$ because a general $n*n$ (complex) matrix has $2n^2$ independent elements (real and imaginary part). The constrain $UU^*=1$ gives $n^2$ elements in function of the other. I linked the article. – Alex May 22 '18 at 7:56
• Ah, of course. Thanks for clarifying this for me. – Carl Christian May 22 '18 at 7:59