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I understand that the dimension of the $SU(n)$ matrix group is $n^2$ because there are $2n^2$ real variables (for the $n^2$ complex matrix elements) in each matrix, and there are $n^2$ equations (arising from the unitary condition) that relate the $2n^2$ real variables, so that the number of independent real parameters drops down to $n^2$. This is the dimension of the $U(n)$ group. The dimension of the $SU(n)$ group is $n^2 -1$ because an additional constraint of unit determinant reduces the number of independent parameters down by one.

My question is this:

Say we have a matrix $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then, $R^\dagger = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$. Therefore, $R^\dagger R=1$ implies that $|a|^2 + |b|^2 = 1$, $|c|^2 + |d|^2 = 1$, $a^*b + c^*d = 0$, and $ab^* + cd^* = 0$.

I was wondering if the last two equations are the same. If so, then the number of independent parameters does not really go down from $2n^2$ to $n^2$, does it?

nightmarish
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1 Answers1

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They are equivalent complex equations, but they are also expressible as two separate real equations, when you examine the real part and the complex part. Thus, you still have $4 = n^2$ conditions to be satisfied.

Rolf Hoyer
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  • Thank you! It's clear to me now. I was also wondering what effect the determinant equation $ad-bc = 1$ has on the number of parameters. I've learnt that the number of parameters decreases to $n^{2}-1$, but should it not decrease to $n^{2}-2$, because there are two real equations for the one complex determinant equation? – nightmarish Apr 06 '15 at 16:32
  • The equations governing $U(2)$ force the the determinant to be norm 1, leaving only one degree of freedom to be curtailed by the two equations governing $ad-bc = 1$. – Rolf Hoyer Apr 06 '15 at 16:47
  • This is most easily derived by seeing that $1 = \det(I) = \det(R^\dagger R ) = \overline{\det(R)}\det(R) = |\det(R)|^2$. – Rolf Hoyer Apr 06 '15 at 17:00