Transverse section

A transverse section of a smooth rank-$r$ vector bundle $E\to M$ is a smooth section $s:M\to E$ transverse to the zero section. In a local bundle trivialization write $s(x)=(x,f(x))$. At every zero, the condition isAt a zero this condition is independent of the trivialization: changing the frame multiplies the vertical derivative by an invertible matrix.

The preimage theorem then makes $s^{-1}(0)$ a smooth submanifold of codimension $r$. If $r>\dim M$, a transverse section has no zeros, since the required surjection cannot exist.

For a trivial real line bundle over ${\mathbb{R}}^2$, the section with coefficient $f(x,y)=x^2+y^2-1$ is transverse: its gradient $(2x,2y)$ never vanishes on the zero circle. The coefficient $x^2+y^2$ instead has a zero at the origin with zero derivative, so it fails transversality. Merely having a zero set that happens to be a manifold is weaker than being a transverse section.