# gjm comments on What was your biggest recent surprise? - Less Wrong Discussion

11 09 June 2012 11:57PM

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Comment author: 10 June 2012 10:18:43AM 4 points [-]

First of all, terminology. SO(n) is orientation-preserving orthogonal transformations on n-space, or equivalently the orientation-preserving symmetries of an (n-1)-sphere in n-space. So Joshua's statement is about SO(n) for n>3.

OK. So the obvious way to interpret "rotation about an axis" in many dimensions is: you choose a 2-dimensional subspace V, then represent an arbitrary vector as v+w with v in V and w in its orthogonal complement, and then you rotate v. The dimension of the set of these things is (n-1)+(n-2) from choosing V -- you can pick one unit vector to be in V, and then another unit vector orthogonal to it -- plus 1 from choosing how far to rotate. So, 2n-2.

And yes, the dimension of SO(n) is n(n-1)/2. One way to see this: you've got matrices with n^2 elements, and n(n+1)/2 constraints on those elements because all the pairwise inner products of the columns (including each column with itself) are specified.

These dimensions are all topological dimensions rather than vector-space dimensions, since the sets we're looking at aren't vector subspaces of R^(n^2), but there's nothing abusive about that :-).

Comment author: 10 June 2012 10:52:19AM *  2 points [-]

It can't be 2n-2 because it's 3 when n=3. I get 2n-3 because the first vector is chosen with n-1 degrees of freedom, then the second with n-2, then subtract one because of the equivalence class of rotations, then add one for choosing how far to rotate.

EDIT: More generally, I think that the dimension of k-dimensional subspaces of an n-dimensional spaces is k(n-k), so where k=2 you get 2n-4, then add one for choosing how far to rotate. I'd feel better if I knew what I meant by "dimension" here though; it's not a vector space.

Comment author: 10 June 2012 05:39:50PM *  3 points [-]

These are the best references I know:

As for topological dimension, roughly, if you consider a neighborhood of a point in the space, what does space look like from there? Locally it's Euclidean if you're "on" a manifold. The rigorous definition involves charts. See also Lebesgue covering dimension.

Comment author: 10 June 2012 07:59:43PM 2 points [-]

Meh, you're right: the dimension of the space of 2-dimensional subspaces of n-space is 2n-4, not 2n-3. The reason why my handwavy dimension-counting above was wrong is ("of course") that I failed to "subtract one because of the equivalence class of rotations". And yes, you're right that in general it's k(n-k).

"Dimension" here means: locally the set looks like a that-many-dimensional vector space. That is, e.g., any element of SO(n) has a neighbourhood that's topologically the same as a neighbourhood in R^(n(n-1)/2).

Comment author: 11 June 2012 06:22:05AM 1 point [-]

EDIT: More generally, I think that the dimension of k-dimensional subspaces of an n-dimensional spaces is k(n-k)

This is correct.

Comment author: 10 June 2012 07:52:47PM 1 point [-]

I'd feel better if I knew what I meant by "dimension" here though; it's not a vector space.

The number of parameters you need to label each element (provided the labelling is a continuous function, otherwise you can label points of R^2 with a single parameter e.g. (3.1415..., 2.7182...) -> 32.174118...)

Comment author: 10 June 2012 07:59:41PM *  1 point [-]

To make this precise, you need the idea of "charts" and "atlases" that witzvo references.