Smooth submersion - Knowlpedia

Let $M$ and $N$ be smooth manifolds of dimensions $m$ and $n$ , and let $f : M \to N$ be a smooth map.

The map $f$ is a smooth submersion if for every $p\in M$ , the differential (pushforward)

d f_p : T_p M \longrightarrow T_{f(p)}N

is surjective, where $T_pM$ is the tangent space at $p$ .

Equivalent characterizations

Equivalently, $\operatorname{rank}(d f_p)=n$ for all $p\in M$ . In particular, if a submersion $M^m \to N^n$ exists then $m \ge n$ .

Examples

Projection from a product. For smooth manifolds $M$ and $F$ , the projection $\pi_M : M\times F \to M,\qquad \pi_M(m,f)=m$ is a smooth submersion: its differential is surjective at every point. (This is the basic model for bundle projections, and its fibers are the slices $\{m\}\times F$ .)

Coordinate projection. The map $\mathbb{R}^3\to \mathbb{R}^2$ , $(x,y,z)\mapsto (x,y)$ , is a smooth submersion since its Jacobian matrix has rank $2$ everywhere.

Determinant on $GL(n,\mathbb{R})$ . The determinant map $\det:GL(n,\mathbb{R}) \to \mathbb{R}\setminus\{0\}$ is a smooth submersion. At $A\in GL(n,\mathbb{R})$ , one has $d(\det)_A(B)=\det(A)\,\mathrm{tr}(A^{-1}B),$ and since $\det(A)\neq 0$ we can choose $B$ so that $d(\det)_A(B)$ is any prescribed real number, proving surjectivity.

Local normal form (Submersion theorem)

If $f : M^m \to N^n$ is a smooth submersion and $p\in M$ , then there exist smooth charts $(U,\varphi)$ around $p$ and $(V,\psi)$ around $f(p)$ (see coordinate charts) such that, in these coordinates,

(\psi\circ f \circ \varphi^{-1})(x^1,\dots,x^m)=(x^1,\dots,x^n)\in \mathbb{R}^n.

So a submersion is locally just a projection map.

A useful viewpoint is that if $f$ is a smooth submersion, then every $y\in N$ is a regular value of $f$ . Consequently, each fiber $f^{-1}(y)$ is (locally) a smooth submanifold of codimension $n$ .