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To preserve the inner product of vectors $x'\cdot x' = x\cdot x $

The linear transformation $ x' = U x $, needs to satisfy

$UU^T=I$.

How many free parameters are we left with?

And also, if we require that $\det(U) = +1$ shouldn't that reduce the number of free parameters by $1?$

Adrian Keister
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Benjamin Fjell
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    The dimension of $O(n)$ is $\frac12n(n-1)$. – Angina Seng Jun 27 '18 at 17:08
  • But how do I prove it? – Benjamin Fjell Jun 27 '18 at 17:09
  • And also if we require that det(U)=+1, shouldn't that reduce the number by 1? So that the number of free parameters are 1/2n(n-1)-1 ? – Benjamin Fjell Jun 27 '18 at 17:14
  • @BenjaminFjell Not quite. It just means that you are considering just one of the two connected components. For instance, a matrix in $O(2)$ is either in the form $\begin{pmatrix} \cos\theta&-\sin\theta\\sin\theta&\cos\theta\end{pmatrix}$ (if $\det =1$) or $\begin{pmatrix} \cos\theta&\sin\theta\\sin\theta&-\cos\theta\end{pmatrix}$ (if $\det=-1$). Now, you could make some trick to make all this into a two-parameter parametrization and say that $SO(2)$ reduces it to a single parameter, but it wouldn't quite be a natural (or meaningful) thing. –  Jun 27 '18 at 18:05
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    No. $O(n)$ has two connected components, one of which is $SO(n)$, the special orthogonal group with determinant $+1$. Taking one of two connected components doesn't change the dimension. $\det U=+1$ would only reduce number of free parameters if $\det U$ had been free to vary continously about $1$; but it was already restricted to $\pm1$. – joriki Jun 27 '18 at 18:05

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We can count degrees of freedom and constraints. Since $U \in \mathbb R^{n\times n}$ then it has $n^2$ values (or degrees of freedom). However because it is orthogonal, there are several constraints. First $UU^T$ produces ones on the diagonal, which places a constraint on the length of each row of $U$. The number of length constraints is $n$. Second, because $UU^T$ produces zeros on the off-diagonals this means that the rows are all orthogonal to each other. This introduces another ${n \choose 2} = \frac{n(n-1)}{2}$ constraints. So the degrees of freedom minus the constraint yields:

Net DOF = $n^2 - n - \frac{n(n-1)}{2} = \frac{n(n-1)}{2}$

Tpofofn
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