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Is there a simple proof that the dimension of $SO_n(\mathbb{R})$, a.k.a the group of rotations in $n$-dimensional space is $(n-1)n/2$?

It would be great to see some proofs based only on the algebraic definition: $$R \mid \left\{ R^T=R^{-1} \land \det(R)=1 \right\}$$ or alternatively proofs invoking geometrical arguments (though I'd like to stay away from proofs using Lie Algebra methods).

Any takers?

Nathaniel Bubis
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2 Answers2

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The orthogonality condition is equivalent to the columns of $R$ being an orthonormal basis. You can pick

  1. first column $v_1$ from $S^{n-1}$, with $(n-1)$ dimensions to choose from.
  2. second column $v_2$ from $S^{n-1}\cap \{v_1\}^\perp = S^{n-2}$, with $(n-2)$ dimensions.
  3. third column $v_3$ from $S^{n-1}\cap \{v_1,v_2\}^\perp = S^{n-3}$, with $(n-3)$ dimensions.

and so on, until you have $S^0$ for the last column (two vectors, of which only one gives positive determinant). The total is $$0+1+2+\dots+(n-1) = \frac{n(n-1)}{2}$$

user127096
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I don't want to work that hard.

Take any skew symmetric square matrix, $S^T = -S.$ We get $e^S \in SO_n(\mathbb R),$ and a neighborhood of the identity is automatically covered, in bijection with a neighborhood of the $0$ matrix.

Oh, $S$ and $S^T$ commute, so $e^S (e^S)^T = e^S e^{S^T} = e^{S + S^T} = e^0 = I,$ and $\det e^S = e^{\operatorname{trace} S} = e^0 = 1.$

Will Jagy
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