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It is very well known that the dimension of $SO(n)$ is $n(n-1)/2$, which is obtained by the number of independent constraint equations we have from the fact that the matrix is orthogonal.

However, it is a little puzzling to me why the determinant constraint does not affect the dimension, because the determinant constraint seems to be another independent constraint equation to the matrices.

M. Zeng
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    The determinant doesn't induce an extra constraint because solutions of $x^T x =I$ (which defines $O(n)$) already satisfy $\det(x)=\pm 1$. – lcv Nov 20 '15 at 09:05
  • yes, but we still got to choose between 1 and -1. – M. Zeng Nov 20 '15 at 09:41
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    Yes but that doesn't diminish dimensionality. If you want only "half" of the matrices have determinant plus one. – lcv Nov 20 '15 at 17:39
  • yes, that's intuitively true, but I'm seeking for a more rigorous justification. – M. Zeng Nov 20 '15 at 17:48

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You can vary $n(n-1)/2$ parameters continuously and keep the determinant equal to $1$, or equal to $-1$, but to get from one to the other you have to make a discontinuous jump.

The split of $O(n)$ into $SO(n)$ and its complement is a little bit like how the equation $x^2=1$ in the $(x,y)$-plane defines two lines $x=1$ and $x=-1$, both one-dimensional. (In this analogy, $x^2=1$ corresponds to $O(n)$, and the extra condition $x>0$ corresponds to choosing the positive sign for the determinant, so that the line $x=1$ corresponds to $SO(n)$. But this choice of sign doesn't reduce the dimension.)

Hans Lundmark
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  • thanks for your answer, but how do we define the dimension of the region obeying $x^2=1$, since it's not even a vector space. I know that this is just an overly simplified example, but I'm trying to figure out the underlying reason for that. – M. Zeng Nov 20 '15 at 09:52
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    You should perhaps ask that question to yourself, since you're talking about the dimension of $SO(n)$, which also not a vector space! ;-) – Hans Lundmark Nov 21 '15 at 10:36
  • Here, we're talking about the dimension in the sense of differentiable manifolds. – Hans Lundmark Nov 21 '15 at 10:36