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I have a rather basic mis-understanding about Lie groups and Lie algebras. Consider the Lie group $SO(N)$ for $N>3$ of rotations on $\mathbb{R}^N$. On the one hand this Lie group has dimension $N(N-1)/2$, since every $SO(N)$ element can be parametrized as $e^{X}$, where $X$ is an anti-symmetric matrix with $N(N-1)/2$ free parameters.

On the other hand, $\mathbb{R}^N$ has $N$ cardinal axes. Can I not express every rotation in $\mathbb{R}^N$ as products of rotations about the axes, implying that $SO(N)$ has dimension $N$? Where do the additional degrees of freedom come from? A follow up related question is: each Lie algebra generator $X(i,j)$, for $1 \leq i < j \leq N$ generates a one-parameter group of rotations $O(\theta) = e^{\theta X_{ij}}$. Is there a simple geometric explanation for which axis this rotation is about?

Solarflare0
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    Well, what would it even mean to rotate about an axis in (say) $\mathbb{R}^4$? What would be the matrix for that? You should think about rotations in two-dimensional subspaces instead. – Hans Lundmark Feb 25 '24 at 21:50
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    Maybe look at https://math.stackexchange.com/a/1402376/27978 It is worthwhile noting that $\binom{n}{2} = {n (n-1) \over 2}$. – copper.hat Feb 25 '24 at 21:53
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    (That is, you need to consider rotating in 2d planes rather than about axes.) – copper.hat Feb 25 '24 at 21:59
  • To be concrete, write down an element of $SO(4)$ that is not rotation about any axis. In particular, $1$ will not be an eigenvalue of this linear map. – Ted Shifrin Feb 25 '24 at 22:07
  • Rotations can be characterized by points on a sphere which is one dimension lower already and some of those will be redundant. You should expect some loss of dimensionality. – CyclotomicField Feb 25 '24 at 23:50
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    @CyclotomicField Define rotation? – Ted Shifrin Feb 26 '24 at 00:31
  • @TedShifrin I define a rotation as an element of the special orthogonal group. They're rotations so you're allowed to define them circularly. – CyclotomicField Feb 26 '24 at 01:16
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    Bit they aren’t necessarily rotations about an axis in dimension $>3$, so your comment seems misleading and wrong. – Ted Shifrin Feb 26 '24 at 01:32

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Firstly note that already in just 2 dimensions this doesn't make sense. $SO(2)$ represents rotations in the plane and is 1 dimensional. So the naive assumption should be that we have $\frac{N}{2}$ ($\frac{N-1}{2}$ if $N$ is odd) dimensions for $SO(N)$ by splitting into orthogonal 2-dimensional subspaces . The problem being that products of these simple rotations don't generate all possible elements of $SO(N)$. Indeed they generate exactly a Cartan subgroup.

The next naive guess is that we shouldn't have made the 2-dimensional subspaces orthogonal and worked with all subspaces or at least all subspaces we get from a fixed orthonormal basis. But the number of those is exactly $N$ choose $2$ or $\frac{N(N-1)}{2}$ which is the correct answer. Obviously I haven't shown that there is no redundancy there but this is the right idea.

Callum
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Nobody seems to have mentioned $\mathbb R=\mathbb R^1$ yet, and yet it would seem to be a good answer to the question: the group of rotations of $\mathbb R^1$ is $0$-dimensional. Namely, it consists of only two "rotations": the identity map and the antipodal map $x\mapsto -x$.

Mikhail Katz
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