Ehresmann's theorem

In mathematics , Ehresmann's theorem , named after Charles Ehresmann , is a fundamental theorem of differential topology .

Formulation of the sentence

Let be differentiable manifolds and ${\displaystyle M,N}$

${\displaystyle f:M\rightarrow N}$

a differentiable mapping with the following properties:

1. is a submersion , i. H. for all the differential is surjective , ${\displaystyle f}$ ${\displaystyle x\in M}$ ${\displaystyle D_{x}f:T_{x}M\rightarrow T_{f(x)}N}$

2. is surjective, d. H. for everyone is not empty, ${\displaystyle f}$ ${\displaystyle y\in N}$ ${\displaystyle f^{-1}(\left\{y\right\})}$

3. is actually, d. H. for all compact quantities is compact . ${\displaystyle f}$ ${\displaystyle K\subset N}$ ${\displaystyle f^{-1}(K)}$

Then is a bundle of fibers . ${\displaystyle f:M\rightarrow N}$

Note that the third condition is automatically satisfied when is compact. ${\displaystyle M}$

example

Levels of ${\displaystyle f(x,y)=x^{2}+y^{2}}$

One function provides a breakdown of the pre-image space into sets of levels ${\displaystyle f:M\rightarrow N}$ ${\displaystyle M}$

${\displaystyle f^{-1}(c):c\in N}$.

The picture on the right shows the decomposition of the function into level sets . ${\displaystyle \mathbb {R} ^{2}}$ ${\displaystyle f(x,y)=x^{2}+y^{2}}$

One can then ask whether this decomposition is locally trivial, i.e. a fiber bundle over with the level sets as fibers. (From this it would follow in particular that all level sets are diffeomorphic to one another.) ${\displaystyle N}$

The example is not a fiber bundle as an illustration from to , because it is not diffeomorphic to for . Ultimately, the reason for this is that there is no submersion at the point : the differential disappears at this point. ${\displaystyle f(x,y)=x^{2}+y^{2}}$ ${\displaystyle \mathbb {R} ^{2}}$ ${\displaystyle \mathbb {R} _{\geq 0}}$ ${\displaystyle f^{-1}(0)}$ ${\displaystyle f^{-1}(c)}$ ${\displaystyle c>0}$ ${\displaystyle f}$ ${\displaystyle (0,0)}$

In contrast, the restriction of to fulfills the requirements of Ehresmann's theorem, that is, the level sets of are the fibers of a fiber bundle . In this example, it is even a (globally) trivial fiber bundle, the mapping provides a diffeomorphism . ${\displaystyle f}$ ${\displaystyle \mathbb {R} ^{2}\setminus \left\{(0,0)\right\}}$ ${\displaystyle x^{2}+y^{2}}$ ${\displaystyle p:\mathbb {R} ^{2}\setminus \left\{(0,0)\right\}\rightarrow \mathbb {R} _{>0}}$ ${\displaystyle (r,\theta )\rightarrow (r\cos \theta ,r\sin \theta )}$ ${\displaystyle \mathbb {R} _{>0}\times S^{1}\rightarrow \mathbb {R} ^{2}\setminus \left\{(0,0)\right\}}$

Counterexample

Examples that fulfill the conditions 1. and 2., but neither condition 3. nor the conclusion, are obtained as follows: Let and be compact differentiable manifolds, an arbitrary point, and the through ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle (a_{0},b_{0})\in A\times B}$ ${\displaystyle M:=A\times B\setminus \left\{(a_{0},b_{0})\right\}}$ ${\displaystyle f:M\rightarrow B}$

${\displaystyle f(a,b)=b}$

defined figure. is a surjective submersion, but not a bundle of fibers, because it is not diffeomorphic to for . (Because is compact while is not compact.) ${\displaystyle f}$ ${\displaystyle f^{-1}(b_{0})}$ ${\displaystyle f^{-1}(b)}$ ${\displaystyle b\not =b_{0}}$ ${\displaystyle f^{-1}(b)}$ ${\displaystyle f^{-1}(b_{0})}$

literature

• Dundas: Differential Topology (PDF; 2.3 MB) with a proof of the theorem in Section 9.5.