Same number of independent parameters for $SO(n)$ and $O(n)$ Why is the number of independent parameters for $SO(n)$ and $O(n)$ same, in spite of an additional constraint of unit determinant for $SO(n)$?
 A: TL;DR: An orthogonal matrix $O^TO={\bf 1}$ must automatically have determinant $\det O = \pm 1$, i.e. the determinant condition does not cut any continuous$^1$ degrees of freedom.
More details:

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*We learn that the set $O(n)$ of orthogonal matrices consists of (at least) two connected components, one with determinant $+1$ and one with determinant $-1$. In fact one may show that $O(n)$ contains precisely 2 connected components.

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*If we look at the component with determinant $+1$, we see that the extra condition $\det O =+1$ for $SO(n)$ is automatically satisfied, and hence do not restrict this component in any way.


*If we look at the component with determinant $-1$, we see that the extra condition $\det O =+1$ for $SO(n)$ is not satisfied at all, so that the entire component is removed.


*In conclusion, the component with determinant $+1$ is $SO(n)$.




*One may show that each of the two components is a real differential manifold. The dimension is given by the dimension of the corresponding Lie algebra
$$o(n)~=~\{ m\in {\rm Mat}_{n\times n}(\mathbb{R})\mid  m^t=-m\},$$ which has dimension $\frac{n(n-1)}{2}$.
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$^1$ It is interesting to compare with the situation for $U(n)$ and $SU(n)$. Here a unitary matrix $U^{\dagger}U={\bf 1}$ must automatically have determinant $\det U\in U(1)\cong \mathbb{S}^1$ of unit modulus, i.e. the complex determinant (which are now 2 real conditions) condition only cut 1 real degrees of freedom, cf. above comment by V. Moretti.
