Help with a proof that SO(n) is path-connected.

Question

I've found lots of different proofs that SO(n) is path connected, but I'm trying to understand one I found on Stillwell's book "Naive Lie Theory". It's fairly informal and talks about paths in a very intuitive way, but I found a more expanded version of it in these notes: http://mysite.science.uottawa.ca/asavag2/mat4144/notes/MAT%204144-5158%20-%20Lie%20Groups.pdf , page 29, Proposition 11.8.

I have trouble in only one part: the part where the author claims that because SO(2) is path-connected, there is path from I to R. Here's my doubt: what is this R? It says it's a rotation in the plane containing e1 and Ae1, but what about the rest of the vectors not contained in the plane? What does R do with them? Does it fix them, rotate them? Also, isn't R a nxn matrix? Doesn't SO(2) being path-connected only gives me information about 2x2 matrices? Am I supposed to have a function that maps 2x2 rotation matrices to nxn rotation matrices? If so, how does that function work? Is it continuous?

Thanks :)

Answer 1

The unspoken idea is that we should design $R$ to do as little as possible. $R$ acts on the plane containing $A e_1$ and $e_1$ in such a way that $A e_1$ is rotated to $e_1$. On the orthogonal complement to this plane, $R$ does nothing.

In particular, we can construct the matrix for $R$ as follows:

First of all, verify that if $A e_1$ is a multiple of $e_1$, then $A$ has an eigenvalue corresponding to the top-left entry, which means that we have a $1$ there, which means we have a matrix of the form that we're trying to achieve. So, WLOG, assume that this is not the case.

Since $A e_1$ and $e_1$ span a plane, we may take $v_1,v_2$ to be an orthonormal basis for that plane. Let $R_1$ from the plane to itself denote the rotation taking $A e_1$ to $e_1$, with respect to the basis $v_1,v_2$.

Extend $\{v_1,v_2\}$ to an orthonormal basis $\{v_1,\dots,v_n\}$ of $\Bbb R^n$. Let $V$ denote the matrix whose columns are $v_i$. We may write $$ R = V \pmatrix{R_1&0\\0&I_{n-2}} V^T $$ Now, $R_1 \in \text{SO}(2)$, so that we have a path from $R_1$ to $I_2$. Verify that this yields a path from $R$ to $I_n$.

Answer 2

SO(n) elements are all rotations. They are orthogonal real matrices so their eigenvectors / eigenvalues form complex conjugate pairs. Therefore you can always find a coordinate frame in which each conjugate pair aligns in a plane defined by two basis vectors. In this frame, the matrix looks like a block diagonal matrix with $2\times 2$ rotational blocks (plus a single $1$ at the end if $n$ is odd). This means that you just found a series of commuting 2D rotations in orthogonal planes. For instance, in 3D you just have to observe in the frame that aligns with the rotational axis and see that you can rotate gradually from angle $0$ to your final position. For higher dimensions, it's similar idea, just harder to visualize.