= Solution
Let
$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
$$
The vertical space is spanned by the fundamental vector $(y,-x)$ corresponding to $J\in\mathfrak o(2)$. Define
$$
\mathcal A_{(x,y)}(u,v)=\langle u,y\rangle J=-\langle v,x\rangle J.
$$
It sends $(y,-x)$ to $J$, is $O(2)$-equivariant, and therefore is a <principal connection>. Its kernel consists exactly of those $(u,v)$ for which
$$
\langle u,x\rangle=\langle u,y\rangle
=\langle v,x\rangle=\langle v,y\rangle=0.
$$
These are precisely the velocities satisfying the stated horizontality condition. Every tangent vector has the unique decomposition
$$
(u,v)=\langle u,y\rangle(y,-x)
+\bigl((u,v)-\langle u,y\rangle(y,-x)\bigr),
$$
into vertical and horizontal parts, proving uniqueness. This is the <Canonical principal connection on the Stiefel bundle over a Grassmannian>.
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