Choose with . Conjugation by identifies
with
It follows that the unit-vector stabilizer in a special orthogonal group is an embedded submanifold diffeomorphic to , and
At , its tangent space is
For any , both and contain the identity. At the identity,
Identify with the exterior square . The orthogonal complement of is . If and are linearly independent, the nonzero bivector belongs to the orthogonal complements of both tangent spaces. Their sum is therefore a proper subspace of , so the intersection is not transverse. If , then , whose tangent space is proper, so they are again not transverse. Consequently
as summarized by distinct unit-vector stabilizers are not transverse.