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Because there are $\binom{n}{2}$ distinct planes spanned by the elements of any given orthonormal basis for $\mathbb{R}^n$, it seems to follow readily (because apparently planar rotations generate all rotations, cf. [1][2][3]) that $SO(n)$ has $\binom{n}{2}$ generators (as a group).

For example, given the plane spanned by $e_i$ and $e_j$, if $i < j$, then let the corresponding planar rotation be the one taking $e_i$ to $e_j$ (the other rotation taking $e_j$ to $e_i$ is just the group inverse).

This seems like a very simple, straightforward, intuitive, combinatorial, etc. argument for why the dimension of $SO(n)$ "should" be $\binom{n}{2}$. However all proofs of the fact I've seen are much more involved, which leads me to suspect a major flaw in the reasoning.

In particular, the argument seems to be implicitly making the following

Claim: The following numbers are the same:

  • The number of generators (as a group) of a Lie group
  • The dimension (as a manifold) of the Lie group
  • The dimension (as a vector space) of the associated Lie algebra

The equality of the last two is largely tautological, to the extent that the definition of the dimension of a manifold is the vector space dimension of its tangent spaces, and the Lie algebra is the tangent space at the identity.

So obviously it is the purported equality of the first with the last two that is questionable. Any counterexamples, or if it's actually true then references to proofs, would be appreciated.

Related question: cf. Dimension of $SO_n(\mathbb{R})$

hasManyStupidQuestions
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    Finite products of your chosen elements have finite orders and thus are not group generators in the common sense. For example the rotation that only rotates the plane $\operatorname{span} {e_i,e_j}$ by $\dfrac{2 \pi}{3} \operatorname{rad}$ can not be expressed by finite product. – Zerox Dec 26 '21 at 16:09
  • To clarify, you mean that they only include "multiples of 90 degree" rotations, and not rotations of arbitrary degree? So $SO(n)$ does not have a finite number of generators as a group? (I mean yeah, even in the simple $SO(2)$ case, you can't get infinitely many angles from finitely many generators.) I guess maybe then I meant generators as an $V$-algebra, but then that doesn't make sense either because "hocus pocus" is needed for why we can apply the argument to the $SO(n)$ subset even though it isn't a subalgebra (the full generated algebra is elements of $End(V)$ with positive determinant?) – hasManyStupidQuestions Dec 26 '21 at 16:17
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    What you are talking about is in fact the dimension of the infinitesimal generators (i.e. the left-invariant vector fields) of the Lie group, which corresponds to the Lie algebra definition. – Zerox Dec 26 '21 at 16:20
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    Yeah, I think you're right, using the vector space dimension of the algebra of infinitesimal generators does seem to be the only way to rigorously capture the intuition that the dimension "should" be the same as the number of distinct orthogonal planes. https://math.stackexchange.com/questions/1259427/high-dimensional-rotation-matrices-as-product-of-in-plane-rotations/1267140#1267140 Because your observation about the finite vs. infinite number of generators is at the heart of why that's the best possible correct+intuitive explanation, feel free to make your comment an answer -- I will accept. – hasManyStupidQuestions Dec 26 '21 at 16:24
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    "All the proofs [...] I've seen are much more involved": It's easy to show that the velocity at $I$ of a path in $SO(n)$ is a skew-symmetric matrix, and conversely every skew-symmetric matrix $A$ gives rise to a path $\exp(tA)$ in $SO(n)$ having velocity $A$ at the identity, so your argument may be viewed as counting the dimension of the space of skew-symmetric matrices...? – Andrew D. Hwang Dec 26 '21 at 17:32
  • @AndrewD.Hwang Yeah exactly; it's just somewhat of a leap to explain to a layperson why we should want to count the dimension of the space of skew-symmetric matrices. It sounds intuitively appealing to say "oh we just need some non-trivial rotation for each orthogonal plane" but again as you pointed out, no single finite rotation will generate all of them, you need "infinitesimal generators" and thus to consider tangent spaces/Lie algebras. Again, I agree that the Lie algebra approach explains rigorously/correctly why something similar to/inspired by the naive idea can be made to work – hasManyStupidQuestions Dec 26 '21 at 21:28
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    As far as I knows $SO_n$ over reals or complex numbers is uncountable, so it cannot have finite number of generators. – markvs Dec 27 '21 at 01:32

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The second and third numbers are the same as you point out, essentially due to the standard isomorphism $\mathfrak{g}\cong T_eG$ and the basic fact about manifolds $\dim(T_pM)=\dim(M)$.

The first, however, is never the same in dimension $>0$. in fact, all Lie groups of poisitive dimension are infinitely generated, if we define "generator" in the sense used for abstract groups. The confusion seems to arise from the term "generator", which is often used loosely in the context of Lie groups, to the extent that I would not recommend using the term at all without first providing a definition.

Often, (in physics especially,) "a set of (infinitesimal) generators of a Lie group $G$" just means "a basis of the group's Lie algebra $\mathfrak{g}$". Used in this sense, the number of generators is equal to the dimension, but these "generators" do not correspond to any kind of generating set in the group theoretic sense (since, again, positive-dimensional Lie groups are not finitely generated). Alternately "an (infinitesimal) generator of a Lie group $G$" sometimes simply means "an element of the Lie algebra $\mathfrak{g}$", in which case the "number of generators" isn't a particularly meaningful piece of information. I suspect the term "generator" is used because a basis of $\mathfrak{g}$, together with the structure constants associated to it, give a complete description of the local structure of the group $G$. This correspondence, however, is more complicated that the group-theoretic notion of a generating set.

Kajelad
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