The following question is from Brian C. Hall's Lie Groups, Lie Algebras, and Representations.

Show that $\mathrm{SO}(n) $ is connected, using the following outline.

For the case $n =1$, there is nothing to show, since a $1 ×1$ matrix with determinant one must be $[1] $. Assume, then, that $n\ge 2$. Let $e_1$ denote the unit vector with entries $1, 0,\ldots,0 $ in $\Bbb R^n$. For every unit vector $v \in \Bbb R^n$, show that there exists a continuous path $R(t) \in \mathrm{SO}(n)$ with $R(0)=I$ and $R(1)v =e_1$. (Thus, any unit vector can be “continuously rotated” to $e_1$.)

I'm aware of alternate proofs, but I'm looking for $R (t) $ mentioned above.

I know that $\mathrm{SO}(n)$ acts transitively on $S^{n-1} $. So there exists a matrix $A\in \mathrm{SO}(n)$ such that $Av=e_1$. Define $R:[0,1]\to \mathrm{SO}(n)$ by $R (t)=(1-t)I+tA $. There are two things I need to verify; $R (t)^{\rm T}R (t)=I $ and $\det R (t)=1.$

But $$R (t)^{\rm T}R (t)=((1-t)I+tA^{\rm T})((1-t)I+tA )\\=(1-t)^2 I+(1-t)tA+(1-t)tA^{\rm T}+t^2I.$$

Any hints on how to find the $R(t)$? Thank you.


Your $R(t)$ is not necessarily orthogonal.

Suppose $e_1$ and $v$ are elements of a two-dimensional subspace $U$ with orthonormal basis $(e_1,w)$. Then $v=\cos\theta\, e_1+\sin\theta\, w$ for some theta. Let $R(t)$ map $e_1$ to $\cos t\theta\, e_1+\sin t\theta\,w$ and $w$ to $-\sin t\theta\,e_1+\cos t\theta\, w$ and also restrict to the identity on $U^\perp$, the orthogonal complement of $U$.

  • $\begingroup$ $v=\cos\theta\, e_1+\sin\theta\, w$ because $v$ is a unit vector, am I right ? $\endgroup$ – Thomas Shelby May 19 at 16:12
  • $\begingroup$ That's right....@ThomasShelby $\endgroup$ – Lord Shark the Unknown May 19 at 16:17

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