number of parameters of $SO(3)$ group

Question

The $SO(3)$ rotation group is defined by:

$A\cdot A^T=\mathbb{1}$ and $\det{A}=1$. The group is supposed to have 3 free parameters, as suggested by Euler's angles. However, I am doing something wrong with counting of the parameters because I cannot arrive at that number.

$3\times3$ matrix has $9$ free parameters. The requirement $A\cdot A^T=\mathbb{1}$ produces 6 equations for those parameters: $$ \vec{a_1}\cdot\vec{a_1}=1\\ \vec{a_1}\cdot\vec{a_2}=0\\ \vec{a_1}\cdot\vec{a_3}=0\\ \vec{a_2}\cdot\vec{a_2}=1\\ \vec{a_2}\cdot\vec{a_3}=0\\ \vec{a_3}\cdot\vec{a_3}=1\\ $$

($\vec{a_i}$ being the columns of $A$) and reduces the number of free parameters to 3. But with this requirement $\det{A}=\pm 1$ and so $\det{A}=1$ presents another restricting euqation, and so i get 2 free parameters.

If I do a similar counting for the $SU(2)$ group, i have 8 free parameters of $2\times2$ complex matrix, the requirement $A\cdot A^\dagger=\mathbb{1}$ provides 4 independent equations and the determinant provides 2 more conditions (complex and imaginary parts) which again reduces to only 2 free parameters.

Where am i doing the mistake in counting the parameters and equations?

Answer 1

The problem is that the restriction $\det A=1$ does not lead to a reduction on the number of parameters. Consider, for instance, $\Bbb R\setminus\{0\}$. It is $1$-dimensional. And if consider the subset $\{x\in\Bbb R\setminus\{0\}\mid\operatorname{sgn}(x)=1\}$, then you still have a $1$-dimensional thing.

It is similar here: $O(3,\Bbb R)$ is $3$-dimensional and $SO(3,\Bbb R)$, which is one-half of it, is still $3$-dimensional.

Answer 2

You have a $9$-dimensional space with coordinates $\{(a_{11},a_{12},\dots,a_{33})|A=(a_{ij})\}$ which are restricted exactly to those 6 equations that you wrote. If a point $x$ satisfy those equations then a point $y=-x$ also satisfy them since

$$(-a_1)\cdot (-a_1)=1$$ $$(-a_1)\cdot (-a_2)=0$$ $$(-a_1)\cdot (-a_3)=0$$ $$(-a_2)\cdot (-a_2)=1$$ $$(-a_2)\cdot (-a_3)=0$$ $$(-a_3)\cdot (-a_3)=1$$ i.e. $x$ and $-x$ belong to your hypersurface.

But in terms of the determinant, you just change the sign of it since $$det(-A)=(-1)^3det(A)=-det(A)$$ since $A$ is $3\times 3$ matrix.

So, as was said before, you just take a half of your hypersurface by setting the restriction $det(A)=1$.

Answer 3

I think the problem lies in the fact that in $O(3)$, the determinant can assume only two values, $\pm1$. Therefore, fixing one of the two values ($SO(3)$) does not remove any free parameter, as it needs to remain continuous. On the other hand, in $U(2)$, the determinant is a free parameter (with magnitude 1, but being a complex number, is continuous). Consequently, $SU(2)$ does indeed fix a free parameter.