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The generators of $SO(n)$ are pure imaginary antisymmetric $n \times n$ matrices.

How can this fact be used to show that the dimension of $SO(n)$ is $\frac{n(n-1)}{2}$?

I know that an antisymmetric matrix has $\frac{n(n-1)}{2}$ degrees of freedom, but I can't take this idea any further in the demonstration of the proof.

Thoughts?

sagsg dagdgah
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    If we regard $SO(n)$ as a smooth submanifold of the space $M(n, \Bbb R)$ of $n \times n$ matrices, which we can identify in the usual way with $\Bbb R^{n^2}$, then the tangent space to $SO(n)$ at the identity matrix $I$ is the space of antisymmetric matrices. In saying this, we have used the fact that the tangent space $T_I M(n, \Bbb R)$ of the vector space $M(n, \Bbb R)$ can be canonically identified with $M(n, \Bbb R)$ itself. – Travis Willse Nov 18 '15 at 11:59
  • I understand what you say, but I'm not really sure how that relates the number of generators to the nature of the generators (i.e. antisymmetic). – sagsg dagdgah Nov 18 '15 at 12:10
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    I'm not sure I entirely understand the question, but the dimension of $SO(n)$ as a smooth manifold is the dimension of the tangent space $T_I SO(n)$. We can identify the latter with the space of antisymmetric matrices, which, like you suggest, is a vector space of dimension $\frac{1}{2} n (n - 1)$. – Travis Willse Nov 18 '15 at 12:15
  • Thanks! I get it! A further question: Do the generators of $SO(n)$ always have to be $n \times n$ matrices, or are representations of other dimensions possible as well? – sagsg dagdgah Nov 18 '15 at 12:27
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    i.e. If we consider representations other then the defining representation, then do we lose the property of antisymmetry of the generator matrices for the case of $SO(n)$? – sagsg dagdgah Nov 18 '15 at 12:35
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    Yes, in general one doesn't get antisymmetry: In fact, this is even true for conjugate copies of $SO(n)$ in $GL(n)$, i.e., for the matrix representations of special orthogonal groups for inner products on $\bbR^n$ with matrix representation other than usual one. – Travis Willse Nov 18 '15 at 18:25

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The group SO(n) consists of orthogonal matrices with unit determinant. Hence for $A\in SO(n)$, $A^{T}A=AA^{T}=1$, $\det(A)=1$. These matrices perform rotations in an n-dimensional space. To find the number of independent generators of the group, consider the group's fundamental representation in a real, n dimensional, vector space. Any rotation can be parametrised from the magnitude of its angle and the axis of rotation, which in turn, is parametrised via a pair of axis, which form a 2 dimensional surface. The number of independent surfaces is $\begin{pmatrix} n\\2 \end{pmatrix}$, hence the number of independent generators.

Spyros Konitopoulos
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  • To complement this answer I would add the following. Note that any finite, proper rotation in $n$ dimensions can be expressed as the exponential of a linear combination of the generators of $\operatorname{SO}(n)$, and this by definition. Then each generator captures an "infinitessimal" basic rotation, which are in one-to-one correspondence with the number of perpendicular planes in $\mathbb{R}^n$. A plane is defined by two non colinear vectors, so if we have $n$ basic vectors, we can build $\binom{n}{2}$ mutually perpendicular planes. – Albert Apr 12 '23 at 18:15